Differential Geometry and Its Visualization 1032436662, 9781032436661

Table of contents :
Cover
Half Title
Title Page
Copyright Page
Contents
List of Figures
Preface
Symbol Description
Authors’ Biographies
CHAPTER 1: Curves in Three-Dimensional Euclidean Space
1.1. POINTS AND VECTORS
1.2. VECTOR-VALUED FUNCTIONS OF A REAL VARIABLE
1.3. THE GENERAL CONCEPT OF CURVES
1.4. SOME EXAMPLES OF PLANAR CURVES
1.5. THE ARC LENGTH OF A CURVE
1.6. THE VECTORS OF THE TRIHEDRON OF A CURVE
1.7. FRENET’S FORMULAE
1.8. THE GEOMETRIC SIGNIFICANCE OF CURVATURE AND TORSION
1.9. OSCULATING CIRCLES AND SPHERES
1.10. INVOLUTES AND EVOLUTES
1.11. THE FUNDAMENTAL THEOREM OF CURVES
1.12. LINES OF CONSTANT SLOPE
1.13. SPHERICAL IMAGES OF A CURVE
CHAPTER 2: Surfaces in Three-Dimensional Euclidean Space
2.1. SURFACES AND CURVES ON SURFACES
2.2. THE TANGENT PLANES AND NORMAL VECTORS OF A SURFACE
2.3. THE ARC LENGTH, ANGLES AND GAUSS’S FIRST FUNDAMENTAL COEFFICIENTS
2.4. THE CURVATURE OF CURVES ON SURFACES, GEODESIC AND NORMAL CURVATURE
2.5. THE NORMAL, PRINCIPAL, GAUSSIAN AND MEAN CURVATURE
2.6. THE SHAPE OF A SURFACE IN THE NEIGHBOURHOOD OF A POINT
2.7. DUPIN’S INDICATRIX
2.8. LINES OF CURVATURE AND ASYMPTOTIC LINES
2.9. TRIPLE ORTHOGONAL SYSTEMS
2.10. THE WEINGARTEN EQUATIONS
CHAPTER 3: The Intrinsic Geometry of Surfaces
3.1. THE CHRISTOFFEL SYMBOLS
3.2. GEODESIC LINES
3.3. GEODESIC LINES ON SURFACES WITH ORTHOGONAL PARAMETERS
3.4. GEODESIC LINES ON SURFACES OF REVOLUTION
3.5. THE MINIMUM PROPERTY OF GEODESIC LINES
3.6. ORTHOGONAL AND GEODESIC PARAMETERS
3.7. LEVI-CIVITÀ PARALLELISM
3.8. THEOREMA EGREGIUM
3.9. MAPS BETWEEN SURFACES
3.10. THE GAUSS-BONNET THEOREM
3.11. MINIMAL SURFACES
CHAPTER 4: Tensor Algebra and Riemannian Geometry
4.1. DIFFERENTIABLE MANIFOLDS
4.2. TRANSFORMATION OF BASES
4.3. LINEAR FUNCTIONALS AND DUAL SPACES
4.4. TENSORS OF SECOND ORDER
4.5. SYMMETRIC BILINEAR FORMS AND INNER PRODUCTS
4.6. TENSORS OF ARBITRARY ORDER
4.7. SYMMETRIC AND ANTI-SYMMETRIC TENSORS
4.8. RIEMANN SPACES
4.9. THE CHRISTOFFEL SYMBOLS
CHAPTER 5: Tensor Analysis
5.1. COVARIANT DIFFERENTIATION
5.2. THE COVARIANT DERIVATIVE OF AN (R, S)–TENSOR
5.3. THE INTERCHANGE OF ORDER FOR COVARIANT DIFFERENTIATION AND RICCI’S IDENTITY
5.4. BIANCHI’S IDENTITIES FOR THE COVARIANT DERIVATIVE OF THE TENSORS OF CURVATURE
5.5. BELTRAMI’S DIFFERENTIATORS
5.6. A GEOMETRIC MEANING OF THE COVARIANT DIFFERENTIATION, THE LEVI–CIVIT`A PARALLELISM
5.7. THE FUNDAMENTAL THEOREM FOR SURFACES
5.8. A GEOMETRIC MEANING OF THE RIEMANN TENSOR OF CURVATURE
5.9. SPACES WITH VANISHING TENSOR OF CURVATURE
5.10. AN EXTENSION OF FRENET’S FORMULAE
5.11. RIEMANN NORMAL COORDINATES AND THE CURVATURE OF SPACES
Bibliography
Index

Citation preview

Differential Geometry and Its Visualization Differential Geometry and Its Visualization is suitable for graduate level courses in differential geometry, serving both students and teachers. It can also be used as a supplementary reference for research in mathematics and the natural and engineering sciences. Differential geometry is the study of geometric objects and their properties using the methods of mathematical analysis. The classical theory of curves and surfaces in three-dimensional Euclidean space is presented in the first three chapters. The abstract and modern topics of tensor algebra, Riemannian spaces and tensor analysis are studied in the last two chapters. A great number of illustrating examples, visualizations and genuine figures created by the authors’ own software are included to support the understanding of the presented concepts and results, and to develop an adequate perception of the shapes of geometric objects, their properties and the relations between them. Features • Extensive, full colour visualisations • Numerous exercises • Self-contained and comprehensive treatment of the topic

Taylor & Francis Taylor & Francis Group

http://taylorandfrancis.com

Differential Geometry and Its Visualization

Eberhard Malkowsky

State University of Novi Pazar, Serbia

Ćemal Dolićanin

State University of Novi Pazar, Serbia

Vesna Veličković

University of Niš, Serbia

First edition published 2024 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 4 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN CRC Press is an imprint of Taylor & Francis Group, LLC © 2024 Eberhard Malkowsky, Ćemal Dolićanin, and Vesna Veličković Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-7508400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. ISBN: 978-1-032-43666-1 (hbk) ISBN: 978-1-032-44119-1 (pbk) ISBN: 978-1-00337056-7 (ebk) DOI: 10.1201/9781003370567 Typeset in Latin Modern font by KnowledgeWorks Global Ltd.

Publisher’s note: This book has been prepared from camera-ready copy provided by the authors

Contents List of Figures

ix

Preface

xvii

Symbol Description

xxi

Authors’ Biographies

xxv

Chapter

1 ■ Curves in Three-Dimensional Euclidean Space

1

1.1

POINTS AND VECTORS

2

1.2

VECTOR-VALUED FUNCTIONS OF A REAL VARIABLE

10

1.3

THE GENERAL CONCEPT OF CURVES

13

1.4

SOME EXAMPLES OF PLANAR CURVES

19

1.5

THE ARC LENGTH OF A CURVE

33

1.6

THE VECTORS OF THE TRIHEDRON OF A CURVE

44

1.7

FRENET’S FORMULAE

53

1.8

THE GEOMETRIC SIGNIFICANCE OF CURVATURE AND TORSION

58

1.9

OSCULATING CIRCLES AND SPHERES

63

1.10 INVOLUTES AND EVOLUTES

69

1.11 THE FUNDAMENTAL THEOREM OF CURVES

79

1.12 LINES OF CONSTANT SLOPE

84

1.13 SPHERICAL IMAGES OF A CURVE

99

Chapter

2 ■ Surfaces in Three-Dimensional Euclidean Space

105

2.1

SURFACES AND CURVES ON SURFACES

106

2.2

THE TANGENT PLANES AND NORMAL VECTORS OF A SURFACE

113

2.3

THE ARC LENGTH, ANGLES AND GAUSS’S FIRST FUNDAMENTAL COEFFICIENTS

116

THE CURVATURE OF CURVES ON SURFACES, GEODESIC AND NORMAL CURVATURE

130

2.4

v

vi ■ Contents

2.5

THE NORMAL, PRINCIPAL, GAUSSIAN AND MEAN CURVATURE

140

2.6

THE SHAPE OF A SURFACE IN THE NEIGHBOURHOOD OF A POINT

186

2.7

DUPIN’S INDICATRIX

195

2.8

LINES OF CURVATURE AND ASYMPTOTIC LINES

197

2.9

TRIPLE ORTHOGONAL SYSTEMS

220

2.10 THE WEINGARTEN EQUATIONS

Chapter

3 ■ The Intrinsic Geometry of Surfaces

228

245

3.1

THE CHRISTOFFEL SYMBOLS

246

3.2

GEODESIC LINES

263

3.3

GEODESIC LINES ON SURFACES WITH ORTHOGONAL PARAMETERS

281

3.4

GEODESIC LINES ON SURFACES OF REVOLUTION

285

3.5

THE MINIMUM PROPERTY OF GEODESIC LINES

293

3.6

298

3.7

ORTHOGONAL AND GEODESIC PARAMETERS LEVI-CIVITA` PARALLELISM

312

3.8

THEOREMA EGREGIUM

320

3.9

MAPS BETWEEN SURFACES

324

3.10 THE GAUSS-BONNET THEOREM

349

3.11 MINIMAL SURFACES

353

Chapter

4 ■ Tensor Algebra and Riemannian Geometry

373

4.1

DIFFERENTIABLE MANIFOLDS

374

4.2

TRANSFORMATION OF BASES

377

4.3

LINEAR FUNCTIONALS AND DUAL SPACES

379

4.4

TENSORS OF SECOND ORDER

381

4.5

SYMMETRIC BILINEAR FORMS AND INNER PRODUCTS

384

4.6

TENSORS OF ARBITRARY ORDER

384

4.7

SYMMETRIC AND ANTI-SYMMETRIC TENSORS

388

4.8

RIEMANN SPACES

389

4.9

THE CHRISTOFFEL SYMBOLS

393

Chapter 5.1

5 ■ Tensor Analysis COVARIANT DIFFERENTIATION

395 396

Contents ■ vii

5.2

THE COVARIANT DERIVATIVE OF AN (R, S)–TENSOR

400

5.3

THE INTERCHANGE OF ORDER FOR COVARIANT DIFFERENTIATION AND RICCI’S IDENTITY

405

BIANCHI’S IDENTITIES FOR THE COVARIANT DERIVATIVE OF THE TENSORS OF CURVATURE

413

5.5

BELTRAMI’S DIFFERENTIATORS

415

5.6

A GEOMETRIC MEANING OF THE COVARIANT ` PARALLELISM DIFFERENTIATION, THE LEVI–CIVITA

419

5.7

THE FUNDAMENTAL THEOREM FOR SURFACES

422

5.8

A GEOMETRIC MEANING OF THE RIEMANN TENSOR OF CURVATURE

432

SPACES WITH VANISHING TENSOR OF CURVATURE

436

5.4

5.9

5.10 AN EXTENSION OF FRENET’S FORMULAE

439

5.11 RIEMANN NORMAL COORDINATES AND THE CURVATURE OF SPACES

441

Bibliography

447

Index

459

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List of Figures 1.1

The Euclidean distance of two points and the length of a vector

2

1.2

Vector and position vector

4

1.3

The diagonals in a parallelogram

5

1.4

The heights in a triangle

5

1.5

The angle between two vectors ⃗x and ⃗y and the projection of ⃗y on ⃗x

6

1.6

Distance of a point from a plane

9

1.7

The surface area of a parallelogram

10

1.8

The parametric representation of a curve

13

1.9

A helix

14

1.10 Neil’s parabola

15

1.11 A suspended chain and a parametric representation for a catenary

17

1.12 Left: catenaries for varying c. Right: a catenary γ and a quadratic parabola γ∗

18

1.13 Left: the construction of a serpentine. Right: families of serpentines

19

1.14 Left: The construction of a versiera. Right: families of versieras

20

1.15 The construction of: Left: Diocles’s cissoid. Right: a strophoid

21

1.16 Families of Diocles’s cissoids and strophoids

22

1.17 Left: The construction of MacLaurin’s trisectrix. Right: A family of MacLaurin’s trisectrices

23

1.18 The construction (left) and families of Nikomedes’s conchoids (right)

25

1.19 The construction (left) and families of families of Pascal’s snails (right) 26 1.20 Left: The construction a κ–curve. Right: A family of κ–curves

26

1.21 The construction of a Cassini curve

28

1.22 Families of Cassini curves

28

1.23 Left: the construction of a rosette. Right: a family of rosettes

29

1.24 Left: the construction of a double egg line. Right: a family of double egg lines

30

1.25 Left: the construction of an astroid. Right: a family of astroids

31

1.26 A family of astroids and their orthogonal trajectories

33

1.27 A catenary with parameters t (left) and s (right)

34 ix

x ■ List of Figures

1.28 Suspended chains of constant length

37

1.29 The construction of an ordinary cycloid

39

1.30 Families of ordinary cycloids

39

1.31 Left: the construction of an epicycloid. Right: a family of epicycloids

39

1.32 Families of hypocycloids

40

1.33 Special hypocycloids

42

1.34 The construction of the curve of Visualization 1.5.8

43

1.35 Left: the construction of a tractrix. Right: a family of tractrices

46

1.36 Tangent vectors and vectors of curvature of catenaries (left) and tractrices (right)

49

1.37 The vectors of a trihedron of a helix

51

1.38 The vectors of a trihedron of the curve of Visualization 1.6.8

51

1.39 An epicycloid and its vectors of curvature

58

1.40 The torsion as a measure of the deviation of a curve from a plane

60

1.41 Osculating, normal and rectifying planes of a helix

61

1.42 The approximations of a helix in its osculating, normal and rectifying planes

62

1.43 An ellipse with osculating circles and centres of curvature

63

1.44 Osculating circles as second-order approximations of a curve

64

1.45 Osculating circle at a point of the curve of Visualization 1.9.3

65

1.46 A curve, the curve of its centres of curvature, the osculating plane, circle and sphere at a point

66

1.47 A helix, an osculating plane, circle and sphere, and the curve of its osculating spheres

68

1.48 The osculating sphere at a point of the curve γ of Visualization 1.9.7

69

1.49 Left: Tangent surface. Right: The definition of an involute

70

1.50 An involute in the tangent surface of a curve

71

1.51 Involutes of a curve on a sphere (left) catenoid (right)

71

1.52 The intersections of the tangents to a helix with planes orthogonal to its axis

73

1.53 The definition of a plane involute

74

1.54 Plane involutes of a curve on a catenoid

74

1.55 Plane involutes of helices

76

1.56 The evolutes of a curve on a catenoid

77

1.57 An evolute of a catenary

78

1.58 An evolute of a helix

79

1.59 Spherical curves

83

List of Figures ■ xi

1.60 Lines of constant slope of a given curvature

85

1.61 Lines of constant slope on a cone

89

1.62 The constant angle between a line of constant slope and a direction

91

1.63 A curve with κ(s) = s, hence ϕ(s) = s2 /2

93

1.64 Clothoids (left) and logarithmic spirals (right)

94

1.65 The orthogonal projection of a line of constant slope on a paraboloid

96

1.66 The orthogonal projection of a line of constant slope on a sphere

98

1.67 The spherical images of the tangent, principal normal and binormal vectors of γ

100

1.68 The spherical images of a helix

100

1.69 The spherical images of the tangent vectors of lines of constant slope on a catenoid

102

2.1

Curves on a surface (right) and in its parameter plane (left)

107

2.2

Parameter lines on a surface

107

2.3

Polar coordinates in the plane (left); the x1 x2 –plane with polar coordinates as parameters (right)

109

2.4

Curve on a plane with different parametric representations

110

2.5

The hemispheres of Visualization 2.1.7 (a) and (b)

111

2.6

Sphere with spherical parameters (left) and the parametrization of Visualization 2.1.7 (c)

112

2.7

Tangent planes and normal vectors of a surface

114

2.8

Generating a surface of revolution

120

2.9

Generating a plane

121

2.10 Generating a circular cylinder

121

2.11 Generating a circular cone

122

2.12 Generating a sphere

123

2.13 Generating a torus

123

2.14 Generating a catenoid

124

2.15 Loxodromes on a torus

125

2.16 A logarithmic spiral as a loxodrome of the plane in Example 2.3.5 (a) 127 2.17 A loxodrome on a sphere

127

2.18 Lines of constant slope

129

2.19 Lines of constant slope on a sphere (left) and catenoid (right)

130

2.20 Lines of constant slope on a torus with their bounding curve

130

2.21 The components of a vector of curvature

131

2.22 Representation of the normal curvature of a curve on a torus

132

xii ■ List of Figures

2.23 Intersections of a hyperboloid of one sheet with planes

137

2.24 Two families of straight lines on a hyperboloid of one sheet

137

2.25 Curvature, normal and geodesic curvature along a helix on a cylinder 138 2.26 Curvature, geodesic and normal curvature at a point on the parallel of a sphere

139

2.27 Normal and geodesic curvature κn = 0 and κg = κ at a point of a curve on a catenoid

140

2.28 Principal normal (left) and binormal (right) surfaces of a loxodrome on a sphere

141

2.29 Tangent, principal normal and binormal surfaces of a helix

142

2.30 A conoid and a helicoid

143

2.31 General screw surface (left) and screw surface (right)

143

2.32 Representation of the normal curvature of a curve on a torus

144

2.33 Curves on a cylinder with the same tangents and normal curvature

145

2.34 Projection of the vector of curvature of a curve on a cone to the surface normal vector of the cone

146

2.35 Normal section of a loxodrome on a torus

147

2.36 Normal curvature at a point of a curve on cylinder and curvature of the corresponding normal section

148

2.37 Illustration of Remark 2.5.11 2.38 Principal directions d⃗ 1 and d⃗ 2 at a point of a surface

149 150

2.39 A catenoid (left) and its Gaussian curvature represented as a surface of revolution (right)

155

2.40 A pseudo-sphere (left) and its mean curvature represented as a surface of revolution (right)

156

2.41 Hyperbolic (left) and elliptic (right) spherical surfaces

158

2.42 Parabolic (left), hyperbolic (middle) and elliptic (right) pseudospherical surfaces

158

2.43 Hyperbolic spherical and pseudo-spherical surfaces, and minimal and maximal radii of u2 –lines

159

2.44 Surfaces of revolution with Gaussian curvature K(u) = k/u2 for k = −0.5 (left) and k = −1 (right)

160

2

2.45 Surfaces of revolution with Gaussian curvature K(u) = k/u for k = 8 (left) and k = 2.5 (right)

161

2.46 Surfaces of revolution with given Gaussian curvature K(u), K(u) = exp(u) (left) and K(u) = sin u

162

2.47 The explicit surface S of Visualization 2.5.23 (a) for a = 3/2 (left), lines of constant Gaussian (middle) and constant mean curvature (right)

163

List of Figures ■ xiii

2.48 The Gaussian curvature of the explicit surface S of Visualization 2.5.23 (a) represented as an explicit surface (left) and its lines of constant Gaussian curvature (right)

164

2.49 The mean curvature of the explicit surface S of Visualization 2.5.23 (a) represented as an explicit surface (left) and its lines of constant mean curvature (right)

164

2.50 The explicit surface S of Visualization 2.5.23 (b) (left), lines of constant Gaussian (middle) and constant mean curvature (right)

165

2.51 The explicit surface of Visualization 2.5.23 (b)

165

2.52 Top: the Gaussian (left) and mean curvature (right) of the surface in Visualization 2.5.23 (a) as explicit surfaces. Bottom: the lines of constant Gaussian (left) and constant mean mean curvature (right) in Part (a), and the lines K(ui ) = const and H(ui ) = const on the surfaces on the top.

166

2.53 Ruled surfaces generated by the surface normal vectors along a curve on the surface of Visualization 2.5.23 (b)

167

2.54 The umbilical points of a hyperboloid of two sheets for distinct a, b and c

171

2.55 Representation of the first and second principal curvature along a curve on a hyperbolic paraboloid

176

2.56 Representation of the first and second principal curvature on the corresponding ruled surfaces

176

2.57 Representation of the Gaussian and mean curvature of a curve on a hyperbolic paraboloid

177

2.58 The umbilical points of an ellipsoid

182

2.59 Euler’s theorem

183

2.60 Hyperbolic, parabolic and elliptic points on an explicit surface

187

2.61 A torus with elliptic (outside), parabolic (dashed) and hyperbolic (inside) points

188

2.62 Tangent planes of an ellipsoid

191

2.63 Tangent planes of a hyperbolid of one sheet

191

2.64 Tangent planes of a torus at elliptic, parabolic and hyperbolic points 192 2.65 A paraboloid of revolution and the surface of Visualization 2.6.10

192

2.66 A flat point on the surface of Visualization 2.6.10

193

2.67 The monkey saddle

194

2.68 The monkey saddle and its intersection with the x1 x2 –plane 1

2

194

2.69 The monkey saddle and a normal section at (u , u ) = (0, 0)

195

2.70 Dupin’s indicatrix at an elliptic point

196

xiv ■ List of Figures

2.71 Dupin’s indicatrix at a parabolic point (left) and at a hyperbolic point with asymptotic directions (right)

197

2.72 Visualization of Remark 2.7.2

198

2.73 Lines of curvature on a helicoid

199

2.74 A general cone generated by an astroid

200

2.75 The second family of lines of curvature on the general cone of Figure 2.74 1

201

1

2.76 Asymptotic lines on the conoid with φ(u ) = log (1 + u )

203

2.77 Lines of curvature on a Moebius strip

204

2.78 Osculating and tangent planes at a point of an asymptotic line

206

2.79 Lines of curvature on a hyperbolic paraboloid

207

2.80 Lines of curvature on an elliptic cone

208

2.81 Asymptotic lines on a catenoid

209 1

2.82 Asymptotic lines on the surface of rotation with r(u ) = 1 and h(u1 ) = log u1

210

1

2.83 Asymptotic lines on the part of u ∈ (π/2, 3π/2) of a torus

210

2.84 Families of asymptotic lines on the pseudo-sphere of Visualization 2.8.12 1 2

2 3

2.85 Asymptotic lines on the explicit surface with f (u ) = (u ) + (u ) i

2

211 212

2.86 A developable surface and tangent planes along a u –line

212

2.87 A developable surface and tangent planes along a u2 –line

213

2.88 Developable surfaces: Top left tangent surface, right cone. Bottom cylinder

214

2.89 A circular cone (left) and a circular cylinder (right) generated by a line of curvature on a surface of rotation

217

2.90 A circular cylinder and cone generated by a line of curvature on a torus

218

2.91 A tangent surface generated by a line of curvature on an elliptic cone 219 2.92 A triple orthogonal system of spheres, circular cones and planes

222

2.93 The triple system of Visualization 2.9.3

223

2.94 A line of curvature on an elliptic cone which is the intersection of the elliptic cone and a sphere

224

2.95 The triple orthogonal system of ellipsoids and hyperboloids of one and two sheets (left). Lines of curvature on an ellipsoid (right)

226

2.96 Lines of curvature on hyperboloids of one (left) and two sheets (right)

228

2.97 The torsion (dashed) along an asymptotic line on a catenoid (left) and a pseudo-sphere (right)

232

List of Figures ■ xv

2.98 The ± torsion (dashed) along an asymptotic line on an explicit surface

232

2.99 A catenoid and a parallel surface

235

2.100 A hyperbolic spherical surface (left) and its parallel surface with constant mean curvature H = −1/2 (right)

237

2.101 An elliptic spherical surface (left) and its parallel surface with constant mean curvature H = −1/2 (right)

238

2.102 A pseudo-sphere (left) and its parallel surface (right) for a = 1

238

2.103 A curve on a monkey saddle and its image under the spherical Gauss map

239

2.104 A curve on a conoid and its image under the spherical Gauss map

239

2.105 The spherical images of the same domain of two different paraboloids

240

2.106 Curves and their orientations in the neighbourhoods of points with K > 0, K = 0 and K < 0 and their images under the Gauss map

241

2.107 Neighbourhoods of an elliptic and hyperbolic point on a torus

241

2.108 Circular neighbourhoods of zero on a hyperbolic paraboloid

242

3.1

The relation between the curvature and geodesic curvature of a curve on a surface (Theorem 3.1.1)

247

3.2

Representation of −0.9κg along a loxodrome (dashed) on the pseudo-sphere

254

3.3

Representation of −3κg along a loxodrome (dashed) on the unit sphere

255

3.4

The ruled surface generated by a loxodrome on the unit sphere and the vectors ⃗t(s)

256

3.5

The ruled surface generated by a loxodrome on a pseudo-sphere and the vectors ⃗t(s)

257

3.6

Geodesic curvature along the parameter lines of a hyperbolic paraboloid

260

3.7

Geodesic line on a cone and surface normal vector in the osculating plane at a point

264

3.8

Principal circles (left) and geodesic lines (right) on a sphere

270

3.9

Geodesic lines on a circular cone

271

3.10 Geodesic lines on a cone generated by a loxodrome on a sphere

275

3.11 Geodesic lines on a pseudo-sphere

279

3.12 Clairaut’s theorem

282

3.13 Geodesic lines of the first characteristic shape

287

xvi ■ List of Figures

3.14 Geodesic lines of the second characteristic shape with tangent vectors

289

3.15 Geodesic lines of the second characteristic shape

289

3.16 Geodesic lines of the second characteristic shape and the bounding lines for u1 = uˆ1

290

3.17 Geodesic lines of the third characteristic shape and their asymptotes 291 3.18 Parameter lines with respect to geodesic parameters on a paraboloid of revolution

303

3.19 A family of geodesic lines (dashed) on a cone and their orthogonal trajectories

303

3.20 Geodesic parallel and polar coordinates on surfaces of revolution

305

3.21 Parallel curves of loxodromes on a circular cone

306

3.22 Geodesic parallel coordinates on a cylinder

308

3.23 Geodesic parallel coordinates on a circular cone

311

3.24 Geodesic polar coordinates on a cylinder

312

3.25 Geodesic polar coordinates on a circular cone

312

3.26 Parallel movement along a loxodrome on a sphere

317

3.27 Parallel movement along a geodesic line on a cone and a torus

318

3.28 The stereographic projection

328

3.29 The principle of the stereographic projection

329

3.30 Intermediate image of the stereographic projection

330

3.31 Isothermal parameters on a sphere and a pseudo-sphere

333

3.32 Isometric maps of Visualization 3.9.19

342

3.33 Projection of a sphere to a cylinder

348

3.34 Geodesic triangles on a sphere (left) and a pseudo-sphere (right)

353

3.35 Enneper’s surface with respect to the parametric representation (3.162)

363

3.36 Enneper’s surface with respect to the parametric representation (3.163)

364

3.37 Enneper’s surface and lines of self-intersection

364

3.38 Several branches of Scherk’s minimal surface

367

Preface This book is based on the authors’ work, courses and workshops on differential geometry, its visualizations and software development at various universities in Germany, Serbia and South Africa over the period of more than thirty years. It contains the fundamental topics from the local theory of curves and surfaces in three-dimensional Euclidean space, tensor algebra and Riemannian geometry, and tensor analysis. Since visualization strongly supports the understanding of concepts and results in differential geometry, a great number of graphical representations and enlightening examples were included in the book. All the graphics in the book were generated by the authors’ own software package. This book is intended for students and teachers of courses in differential geometry at graduate level as well as in doctoral studies. It is also useful for mathematicians, physicists and engineers who need or are interested in its topics. The material of the book is presented in a comprehensive and self-contained way and strongly supported by a great number of examples and visualizations. The prerequisites are a solid background from the undergraduate courses in analytical geometry and analysis. As a reminder, the most important results from these areas are listed in Section 1.2. The book contains five chapters with a total of 54 sections including 215 figures. Major parts of the first three chapters are translations from Serbian of the authors’ book Diferencijalna geometrija i njena vizuelizacija [17]. Chapter 1 contains the study of the basic local geometric properties of curves in three-dimensional Euclidean E3 . The topics of special interest include the parametric representation of curves, the arc length of a curve as its natural parameter, many examples of planar curves and the geometrical principles of their constructions, the frame or the vectors of the trihedra of curves, Frenet’s formulae, the curvature and torsion of curves and their geometric significance, characteristic shape of a curve in a neighbourhood of any of its points, osculating circles and spheres, involutes and evolutes of curves, the fundamental theorem of curves and the natural or intrinsic equations of curves, lines of constant slope and finally spherical images of curves. Chapter 2 deals with the local differential geometry of surfaces in threedimensional Euclidean space E3 . This means the study of the geometric shape of surfaces in the neighbourhood of an arbitrary one of their points. The most important concepts arising in this task are those of the normal, principal, Gaussian and mean curvature. Fundamental important topics are curves on surfaces, tangent planes and normal vectors of surfaces, first and second fundamental coefficients, Meusnier’s theorem, principal directions, Euler’s formula, the local shape of surfaces and Dupin’s

xvii

xviii ■ Preface

indicatrix, lines of curvature and asymptotic lines, triple orthogonal systems and finally the Gauss and Weingarten equations. Chapter 3 deals with the intrinsic geometry of surfaces, that is, with the geometric properties of surfaces that only depend on measurements on the surface itself, namely on its first fundamental coefficients and their derivatives. The most important topics and subjects are the Christoffel symbols of first and second kind and the geodesic curvature, Liouville’s theorem, geodesic lines on surfaces with orthogonal parameters and on surfaces of revolution, the minimum property of geodesic lines, orthogonal and geodesic parameters, geodesic parallel and geodesic polar coordinates, the LeviCivit`a parallelism, the derivation formulae by Gauss, and Mainardi and Codazzi, the Theorema egregium, conformal, isometric and area preserving maps of surfaces, the Gauss-Bonnet theorem and finally minimal surfaces and the Weierstrass formulae. Chapter 4 deals with the redevelopment of the intrinsic geometry of surfaces independently of their embedding in three-dimensional space, based only on the definition of measurements of lengths in a point set. This concept was originally introduced by Riemann in 1854. For this purpose, the concept of a manifold in n–dimensional space is needed. Surfaces and n–dimensional Euclidean spaces are obtained as special cases of the manifolds. The main topics and subjects of this chapter are the definition of n–dimensional manifolds of class C k , the translation formulae between the bases of a finite dimensional vector space, and between the components of vectors with respect to bases, spaces V ∗ of linear functionals on finite dimensional vector spaces V , the isometry of V and V ∗ , the transformation formulae of the elements of V ∗ and their components, the concepts of contravariant and covariant vectors. Additional topics are the introduction of tensors of the second order and the transformation formulae for their components with respect to different bases, symmetric bilinear forms and inner products, the introduction of tensors of arbitrary order, the transformation formulae for their components, the characterization of tensors by the transformation properties of their components, and the identification of tensors with their components, sums, outer products, contractions and inner products of tensors, symmetric and anti-symmetric tensors of arbitrary order and the independence of their properties from the choice of the coordinate system, Riemann spaces and the metric tensor, the Christoffel symbols in a Riemann space and their transformation formulae. Chapter 5 deals with the study of the fundamentals of tensor analysis. Tensors first appeared in the theory of surfaces. Typical examples were the first and second fundamental coefficients. A tensor of a given order was assigned to every point of a manifold. It also turned out that certain derivatives of tensors were needed in various problems of differential geometry. It seems natural to try and define derivatives of a tensor in a way such that the result again is a tensor. This guarantees the independence of a derivative of the choice of a coordinate system. The most important topics and subjects are the study of covariant derivatives of contravariant, covariant vectors and (1, 1)–tensors, covariant derivatives of (r, s)–tensors, their basic properties and the Ricci identity for the covariant derivatives of the metric tensor, the mixed Riemann tensor of curvature and the interchange of the order of its covariant differentiation in Ricci’s identity, the Bianchi identities for the derivatives of the mixed Riemann tensor of curvature and the covariant Riemann tensor of curvature, the

Preface ■ xix

Beltrami differentiator of first order in a Riemann space, the divergence of contravariant and covariant vectors, a geometric meaning of the covariant differentiation, the Levi-Civit`a parallelism, the fundamental theorem of the theory of surfaces, a geometric interpretation of the Riemann tensor of curvature, spaces with vanishing tensor of curvature, the existence and uniqueness of autoparallel curves in Riemann spaces, an extension of Frenet’s formulae for curves in a Riemann spaces, Riemann normal coordinates and the curvature of spaces and the Bertrand-Puiseux theorem. Novi Pazar, Serbia Novi Pazar, Serbia Niˇs, Serbia

Eberhard Malkowsky ´ Cemal Doli´canin Vesna Veliˇckovi´c January, 2023

Taylor & Francis Taylor & Francis Group

http://taylorandfrancis.com

Symbol Description N Z Q R C E3 Rn xk d d(X, Y ) Vn xk ⃗x + ⃗y λ⃗y ∥·∥ ∥⃗x∥ En −→

PQ −→

OX ⃗x • ⃗y

={1, 2, . . . }, the set of natural numbers ={−N ∪ {0} ∪ N, the set of integers ={p/q : p ∈ Z, q ∈ N}, the set of rational numbers the set of real numbers the set of complex numbers three-dimensional (real) Euclidean space n–dimensional (real) point space k th coordinate of a point in Rn Euclidean metric in Rn Euclidean distance of the points X and Y in Rn n–dimensional (real) vector space k th component of a vector in Vn sum of the vectors ⃗x, ⃗y ∈ Vn scalar multiple of λ ∈ R and ⃗x ∈ Vn Euclidean norm in Vn Euclidean norm (length) of a vector ⃗x in Vn n–dimensional (real) Euclidean space

⃗x ⊥ ⃗y ⃗y⃗x ⃗x × ⃗y ⃗ ⊥γ N ⃗ ⊥ Pl N f⃗ fk Df⃗ f⃗(D ⃗) f

Nε (X0 )

the vectors ⃗x and ⃗y are orthogonal projection of the vector ⃗x to ⃗y vector product (or outer product) of the vectors ⃗x, ⃗y ∈ V3 ⃗ is orthogonal to the N straight line γ in E2 ⃗ is orthogonal to the plane N P l in E3 vector={f 1 , . . . , f n }, valued function k th component function of the vector-valued function f⃗ domain of the function f⃗ ={f (t) : t ∈ Df⃗}, range of the function f⃗

={X ∈ Rn : d(X, X0 ) < ε}, the ε–open ball of the point X0 Nε (⃗x0 ) ={⃗x ∈ En : ∥⃗x − ⃗x0 ∥ < ε}, the ε–open ball of the vector ⃗x0 ={X ∈ Rn : 0 < d(X, X0 ) < N˙ ε (X0 ) ε}, the ε–deleted open ball of the point X0 ={⃗x ∈ En : 0 < ∥⃗x − ⃗x0 ∥ < N˙ ε (⃗x0 ) ε}, the punctured ε–deleted vector that maps the point P ball of the vector ⃗x0 ¯ to the point Q S closure of the set S position vector of the point limt→t0 f⃗(t) limit of the function f⃗ as t approaches t0 X inner product of the vectors f⃗ ′ derivative of the vector func⃗x, ⃗y ∈ Vn tion f⃗ xxi

xxii ■ Symbol Description

f⃗ ′ (t0 ) df⃗ (t0 ) dt f⃗ ′′ f⃗ ′′′ f⃗ (m) f⃗ (0) C 0 (I) C m (I)

C ∞ (I)

(⃗x(t), I) s ⃗x˙ (s)

PQ ⃗t(t0 ) ⃗xt0 ⃗v1 ⃗v2 ⃗v3 ⃗v1 (s)

derivative of the vector func- ⃗v2 (s) tion f⃗ at t0 ⃗v3 (s) =f⃗ ′ (t0 ) second order derivative of ⃗x¨ ⃗x¨(s) the function f⃗

third order derivative of the κ function f⃗ th m order derivative of the κ(s) function f⃗ ρ =f⃗ class of all continuous funcρ(s) tions on the interval I class of all functions that have continuous mth order δik partial derivatives on the interval I class of all functions that τ have continuous partial τ (s) r derivatives of any order on C (D) the interval I parametric representation of a curve in Vn arc length along a curve; nat- (f, D) ural parameter of a curve d⃗x = (s), derivative of the ⃗x(ui ) ds parametric representation of a curve with respect to s, the arc length of the curve straight line through the points P, Q ∈ Rn ∂(ui ) (unit) tangent vector to a ∂(u∗k ) curve at t0 parametric representation of ⃗x1 (ui ) the tangent to a curve at t0 tangent unit vector of a curve principal normal vector of a ⃗x2 (ui ) curve =⃗v1 × ⃗v2 , binormal vector of a curve ⃗ (ui ) =⃗x˙ (s) tangent unit vector at N s of a curve with a parametric representation ⃗x(s)

principal normal vector of a curve at s =⃗v1 (s)×⃗v2 (s), binormal vector of a curve at s vector of curvature of a curve vector of curvature of a curve at s =∥⃗x¨∥, curvature of a curve =∥⃗x¨(s)∥, curvature of a curve at s =1/κ, radius of curvature of a curve =1/κ(s), radius of curvature of a curve at s Kronecker symbol; δik = 1 for i = k and δik = 0 for i ̸= k torsion of a curve torsion of a curve at s class of all functions that have continuous rth order partial derivatives on the domain D parametric representation of a surface given by a function f : D ⊂ R2 → R3 parametric representation of a surface in terms of the position vectors of its points with respect to the parameters ui (i = 1, 2) Jacobian of the functions ui = ui (u∗k ) =∂⃗x(ui )/∂u1 , vector in the direction of the u1 –line of the surface with the parametric representation ⃗x(ui ) =∂⃗x(ui )/∂u2 , vector in the direction of the u2 –line of the surface with the parametric representation ⃗x(ui ) =(⃗x1 (ui ) ×⃗x1 (ui ))/∥⃗x1 (ui ) × ⃗x1 (ui )∥, surface normal vector of the surface with

Symbol Description ■ xxiii

gik

(ds)2

(gik ) g

(g ik ) κg κn Lik

L

κ1 , κ2 H K Lik dui duk Lik cik [ikl] r ik

the parametric representation ⃗x(ui ) =⃗x1 • ⃗x2 , first fundamental coefficients of the surface with the parametric representation ⃗x =gik (uj )dui duk , first fundamental form of the surface with the parametric representation ⃗x(ui ) matrix of the first fundamental coefficients of a surface =det(gik ), determinant of the matrix of the first fundamental coefficients of a surface =(gik )−1 , the inverse of the matrix (gik ) geodesic curvature of a curve on a surface normal curvature of a curve on a surface ⃗ •⃗xik = −N ⃗ k •⃗xi , the sec=N ond fundamental coefficients of a surface =det(Lik ), determinant of the matrix of the second fundamental coefficients of a surface principal curvature of a surface =(κ1 + κ2 )/2, the mean curvature of a surface =κ1 · κ2 , the Gaussian curvature of a surface second fundamental form =g ij Ljk ⃗i • N ⃗ k , the third fun= N damental coefficients of a surace =⃗xik •⃗xl , the first Christoffel symbols of a surface =g lr [ikl], the second Christoffel symbols of a surface

m Rikj

(X, T ) TS SC(P ) R

T (P )

⃗b1 , . . . , ⃗bn ⃗x = xk⃗bk

V∗

⃗b1 , . . . , ⃗bn l = lk⃗bk

⃗x ∈ V ⃗x ∈ V ∗ bik

Riemannian tensor of curvature topological space relative topology for a subset S of a topological space (X, T ) set of all smooth curves through a point P relation on SC(P ) defined by γRγ ∗ if and only if γ and γ ∗ have the same tangents at P set of all equivalent classes with respect to R of tangent vectors through P basis vectors of an n– dimensional vector space representation of the vector ⃗x with respect to the basis vectors ⃗bk (k = 1, . . . , n); xk (k = 1, . . . , n) the components of ⃗x with respect to this representation dual space of the vector space V , that is, the space of all continuous linear functionals on V basis vectors of the dual space V ∗ of an n– dimensional vector space V representation of the vector l ∈ V ∗ with respect to the basis vectors ⃗bk (k = 1, . . . , n); lk (k = 1, . . . , n) the components of l with respect to this representation contravariant vector with components xk with respect to a basis covariant vector with components xk with respect to a basis components of a covariant tensor B of second order

xxiv ■ Symbol Description

bik

components of a contravari- Rj ilk ant tensor B of second order

bki

components of a mixed tensor B of second order

...is Tki11 ...k r

components of an (r, s) – tensor T , that is, T is contravariant of order r and coRijkl variant of order s ▽(Φ, Ψ) sum of (r, s)–tensors

T +U T ∗U

outer product tensors

of

(r, s)–

i ...i ,im+1 ...is i ...i jim+1 ...is Tk11 ...km−1 = Tk11 ...km−1 , con- ▽Φ n−1 ,kn+1 ...kr n−1 jkn+1 ...kr div T k traction of the tensor T

G

metric tensor with the components gik

g ik

components of the conjugate tensor of G

[ijk]

= (1/2) · (gik,j − gij,k + gjk,i ) △F first Christoffel symbols

  k ij

T;ji

Ti;j

n Tm;r

...js Tij11···i r ;k

= g kl [ijl] second Christoffel symbols ∂T i  i  k = + jk T covariant ∂xj derivative of a contravariant vector with the components Ti   ∂Ti = − ijk Tk covariant j ∂x derivative of a covariant vector T with the components Ti n   n ∂Tm l l = −Tln mr +Tm rl r ∂x covariant derivative of a (1, 1)–tensor with the comn ponents Tm ...js ∂Tij11...i r k ∂x    ...js m − rl=1 Tij11...i il k l−1 mil+1 ...ir

=







jn + sn=1 Ti11...irn−1 n+1 s pk covariant derivative of an (r, s)–tensor T with the com...js ponents Tij11...i r j ...j

pj

...j

div Tk

δξ α δs

δξ α δs

∂  j  ∂  j  − il ik k l ∂x  m   j  ∂x m  j  + il − ik mk ml mixed Riemann tensor of curvature in a Riemann space with the metric tensor gik r = gir Rjkl = g ik Φ; i Ψ; k Beltrami differentiator of first order of two real–valued functions of class r ≥ 2 in a Riemann space = ▽(Φ, Φ) = T;kk divergence of a contravariant vector T with the components T k = g ki Tk; i divergence of a covariant vector T with the components Tk = div Fi , where F is a realvalued function and Fi = ∂F ; Laplacian operator ∂xi α   dξ α = + ξ β βγ u˙ γ = 0 ds (α = 1, 2); differential equations for the components of a surface vector under parallel movement along a geodesic line   dξ α α = ξ;ασ u˙ σ = + ξ δ δγ u˙ γ ds = 0 (α = 1, 2), parallel movement by Levi–Civit`a of a surface vector with the components ξ α along an arbitrary curve given by (u1 (s), u2 (s)) =

Authors’ Biographies Eberhard Malkowsky is a Full Professor of Mathematics in retirement at the State University of Novi Pazar in Serbia. He earned his Ph.D. degree and habilitation at the Department of Mathematics of the Justus-Liebig Universit¨at Giessen in Germany in 1982 and 1988, respectively. He was a professor of mathematics at universities in Germany, South Africa, Jordan, Turkey and Serbia, and a visiting professor in the USA, India, Hungary and France. Furthermore, he participated as an invited or keynote speaker with more than 100 lectures in international scientific conferences and congresses. He is a member of the editorial boards of twelve journals of international repute. His list of publications contains 175 research papers in international journals. He is the author or co-author of nine books, and the editor or co-editor of six proceedings of international conferences. He supervised 6 Ph.D. theses and a great number of B.Sc. and M.Sc. theses in mathematics. His research and work areas include functional analysis, differential geometry and software development for the visualization of mathematics. ´ Cemal Doli´ canin is a Professor Emeritus at the Department of Sciences and Mathematics at the State University of Novi Pazar in Serbia and a member of the Serbian Academy of Non-linear Sciences. He was dean of the Electro-Technical Faculty and the Faculty of Technical Sciences in Priˇstina, vice-rector of the University of Priˇstina and founder and rector of the State University of Novi Pazar, Serbia. He obtained his M.Sc. degree at the Faculty of Mathematics of the University of Belgrade in 1974, and his Ph.D. degree at the Faculty of Sciences and Mathematics of the University of Priˇstina in 1980. He published more than 20 books, 23 papers in national scientific journals, 41 papers in international scientific journals, and gave more than 50 lectures at international scientific conferences. He was a visiting professor in Germany, Belorussia and Russia. He supervised 14 PhD theses and a great number of B.Sc. and M.Sc. theses in mathematics. His research and work areas include Euclidean and non-Euclidean geometry, differential geometry and applied mathematics. He is very active in promoting mathematics, and has established the Center for the Advancement and Popularization of Mathematics at the State University of Novi Pazar. He participated in the implementation of several national scientific projects TEMPUS projects and was the coordinator of Master’s study programs with the World University Service Austria (WUS). Vesna Veliˇ ckovi´ c is a Professor at the Department of Computer Science at the Faculty of Sciences and Mathematics of the University of Niˇs, Serbia. She obtained her xxv

xxvi ■ Authors’ Biographies

magister degree at the Faculty of Mathematics of the University of Belgrade in 1996, and her Ph.D. degree at the Faculty of Sciences and Mathematics of the University of Niˇs in 2012. She published 3 books, 30 papers in international scientific journals, 12 papers in national scientific journals, and gave more than 50 lectures at international scientific conferences. She participated in 3 international scientific and 4 software projects, and 7 study visits in Serbia, Germany, Bulgaria, Romania and Turkey. Together with Professor Malkowsky, she is developing the software MV-Graphics for visualization of mathematics. With large-format graphics, they participated in three exhibitions of mathematical art. For ten years she worked with pupils talented in mathematics and programming. Her students won a number of medals at International Olympiads in Informatics. She is still very active in organizing contests in programming and promoting of mathematics. Her research areas are software development, computer graphics and visualization of mathematics.

CHAPTER

1

Curves in Three-Dimensional Euclidean Space

Elementary differential geometry is the study of geometric objects using the methods of analysis. In particular, the introductory theory studies curves and surfaces embedded in three-dimensional Euclidean space E3 . The first chapter deals with the basic geometric properties of curves in threedimensional Euclidean E3 . The studies related to topics of special interest include the • parametric representation of curves in Definition 1.3.1 of Section 1.3

• geometric principles and construction of important planar curves in Section 1.4

• arc length of a curve as its natural parameter in Section 1.5

• the vectors of the trihedra of curves in Definition 1.6.3 of Section 1.6

• Frenet’s formulae in Theorem 1.7.1 of Section 1.7

• the curvature and torsion of curves in Definition 1.6.3 of Section 1.6 and in Definition 1.7.2 of Section 1.7 and their geometric significance in Remark 1.6.4 of Section 1.6 and in Theorem 1.8.1 of Section 1.8 • the characteristic shape of a curve in a neighbourhood of any of its points in Section 1.8 • osculating circles and spheres in Section 1.9

• involutes and evolutes of curves in Section 1.10

• fundamental theorem of curves, Theorem 1.11.1 and the natural or intrinsic equations of curves in Definition 1.11.4 of Section 1.11 DOI: 10.1201/9781003370567-1

1

2 ■ Differential Geometry and Its Visualization

• lines of constant slope in Section 1.12

• spherical images of curves in Section 1.13.

Throughout, we use the standard notations N = {1, 2, . . . }, Z, Q, R and C for the sets of all nonnegative integers (natural numbers), integers, rational, real and complex numbers, respectively.

1.1

POINTS AND VECTORS

Points and vectors are the simplest geometric objects. Therefore, we recall the definitions of n–dimensional point and vector spaces. For any nonnegative integer n, the set 





Rn = X = x1 , x2 , . . . , xn : xk ∈ R (k = 1, 2, . . . , n)



of all n–tuples of real numbers, xk is called n–dimensional (real) point space; its elements X are called points, and the reals xk are referred to as the coordinates of X. The map d : Rn × Rn → [0, ∞) defined by   n  (xk − y k )2 for all X, Y ∈ Rn (Figure 1.1) d(X, Y ) =  k=1

is called the Euclidean metric and the real number d(X, Y ) is called the Euclidean distance of the points X and Y (Figure 1.1). The set 





Vn = ⃗x = x1 , x2 , . . . , xn : xk ∈ R (k = 1, 2, . . . , n)

Figure 1.1



The Euclidean distance of two points and the length of a vector

Curves in Three-Dimensional Euclidean Space ■ 3

of all n–tuples of real numbers xk together with the operations 

⃗x + ⃗y = x1 + y 1 , x2 + y 2 , . . . , xn + y n and



λ⃗x = λx1 , λx2 , . . . , λxn





for all ⃗x, ⃗y ∈ Vn

for all λ ∈ R and all ⃗x ∈ Vn

is called the n–dimensional (real) vector space; its elements ⃗x are called vectors; the real numbers xk are referred to as the components of ⃗x. In contrast to the familiar notation in analysis to write the indices of coordinates or components as lower indices or subscripts, it is usual in differential geometry to place indices of coordinates or components as upper indices or superscripts. This notation distinguishes indices of coordinates and components from the lower indices of partial derivatives. It is helpful in the use of Einstein’s convention of summation in the theory of surfaces in Chapter 2. The Euclidean norm or length ∥⃗x∥ of a vector ⃗x ∈ Vn is given by   n  (xk )2 ∥⃗x∥ = 

(Figure 1.1).

k=1

The space Vn together with this norm is called the n–dimensional Euclidean vector space and denoted by En . Every vector ⃗x = {x1 , x2 , . . . , xn } ∈ Vn may be interpreted as a map ⃗x : Rn → Rn that assigns to every point P = (p1 , p2 , . . . , pn ) the point 



Q = ⃗x(P ) = q 1 , q 2 , . . . , q n , where q k = pk + xk (k = 1, 2, . . . , n). −→

The vector that maps P to Q will be denoted by P Q (left in Figure 1.2). Thus the Euclidean distance of the points P and Q is equal to the Euclidean norm of the vector −→

−→

P Q, that is, d(P, Q) = ∥P Q∥. If the origin of Rn is denoted by O, then the position −→

vector ⃗x of a point X ∈ Rn is given by ⃗x = OX. If a point X has the coordinates xk (k = 1, 2, . . . , n), then its position vector ⃗x has the components xk (k = 1, 2, . . . , n); in this sense, we may – and indeed shall frequently do so – identify points with their position vectors (right in Figure 1.2). A vector ⃗x with Euclidean norm ∥⃗x∥ = 1 is called a unit vector. If ⃗x ̸= ⃗0, then the length of the vector ⃗u = ⃗x/∥⃗x∥ is    ⃗ x   = 1 ∥⃗ x∥ = 1, ∥⃗x∥  ∥⃗x∥

∥⃗u∥ =  

and ⃗u is a unit vector in the direction of ⃗x. \ n , then the real Let ⃗x = {x1 , x2 , . . . , xn } and ⃗y = {y 1 , y 2 , . . . , y n } be vectors in V number n ⃗x • ⃗y =



k=1

xk y k

is called scalar or inner or dot product of ⃗x and ⃗y . It is easy to see that • : Vn ×Vn → R is a symmetric bilinear map. Obviously ⃗x • ⃗x = ∥⃗x∥2 for all ⃗x ∈ En .

4 ■ Differential Geometry and Its Visualization

Figure 1.2

Vector and position vector

Proposition 1.1.1 (Cauchy–Schwarz inequality). We have |⃗x • ⃗y | ≤ ∥⃗x∥∥⃗y ∥ for all ⃗x, ⃗y ∈ En .

(1.1)

Proof. The inequality obviously holds for ⃗y = ⃗0. If ⃗y = ⃗0, then we put λ = (⃗x • ⃗y )∥⃗y ∥−2 , and obtain

0 ≤ (⃗x − λ⃗y ) • (⃗x − λ⃗y ) = ∥⃗x∥2 − 2λ(⃗x • ⃗y ) + λ2 ∥⃗y ∥2 = ∥⃗x∥2 − 2

(⃗x • ⃗y )2 (⃗x • ⃗y )2 (⃗x • ⃗y )2 2 + = ∥⃗ x ∥ − . ∥⃗y ∥2 ∥⃗y ∥2 ∥⃗y ∥2

This implies (⃗x • ⃗y )2 ≤ (∥⃗x∥ ∥⃗y ∥)2 , and the inequality in (1.1) is an immediate consequence. The angle α ∈ [0, π) between two vectors ⃗x, ⃗y ∈ Vn \ {⃗0} is defined by cos α =

⃗x • ⃗y . ∥⃗x∥∥⃗y ∥

The vectors ⃗x, ⃗y ∈ V n are called orthogonal if ⃗x • ⃗y = 0; we denote this by ⃗x ⊥ ⃗y . Orthogonal unit vectors are said to be orthonormal. We use the scalar product and orthogonality to show two well–known elementary results. Example 1.1.2. (a) A parallelogram is a rectangle if and only if its diagonals have equal lengths (Figure 1.3). (b) The heights in a triangle intersect in one point (Figure 1.4). Proof. (a) Let ⃗x and ⃗y be the vectors that span the parallelogram. Then its diagonals are given by the vectors ⃗x + ⃗y and ⃗y − ⃗x and we have ∥⃗x + ⃗y ∥2 − ∥⃗y − ⃗x∥2 = (⃗x + ⃗y ) • (⃗x + ⃗y ) − (⃗y − ⃗x) • (⃗y − ⃗x) = 4⃗x • ⃗y = 0

if and only if ⃗x ⊥ ⃗y . This shows Part (a).

Curves in Three-Dimensional Euclidean Space ■ 5

Figure 1.3

The diagonals in a parallelogram

Figure 1.4

The heights in a triangle

(b) Let A, B and C denote the vertices of a triangle and P be the point of inter−→ −→ −→ section of the heights on AB and AC. We put ⃗a = BC, ⃗b = AC, ⃗c = AB, −→ −→ −→ ⃗h⃗a = AP , ⃗h⃗ = BP and ⃗h = P C (Figure 1.4). Then we obtain ⃗c = ⃗b − ⃗a, b ⃗h = ⃗a − ⃗h⃗ = ⃗b − ⃗h⃗a , and ⃗h⃗a ⊥ ⃗a and ⃗h⃗ ⊥ ⃗b, hence b

b

⃗c • ⃗h = (⃗b − ⃗a) • ⃗h = ⃗b • (⃗a − ⃗h⃗b ) − ⃗a • (⃗b − ⃗h⃗a ) = ⃗b • ⃗a − ⃗b • ⃗h⃗ − ⃗a • ⃗b + ⃗a • ⃗h⃗a = ⃗a • ⃗b − ⃗b • ⃗a = 0. b

Thus we have ⃗h ⊥ ⃗c. This shows Part (b). \ n, ⃗ Let ⃗x, ⃗y ∈ V x ̸= ⃗0 and ⃗u be the unit vector in the direction of ⃗x. Then the vector ⃗y⃗x = (⃗y • ⃗u)⃗u is called the projection of ⃗y to ⃗x. If α denotes the angle between the vectors ⃗x, ⃗y ̸= ⃗0, then the projection ⃗y⃗x of ⃗y to ⃗x is given by

⃗y⃗x = (⃗y • ⃗x)

⃗u = ∥⃗y ∥ cos α ⃗u (Figure 1.5). ∥⃗x∥

For any two vectors ⃗x = {x1 , x2 , x3 }, ⃗y = {y 1 , y 2 , y 3 } ∈ V3 , the vector ⃗x × ⃗y = {x2 y 3 − x3 y 2 , x3 y 1 − x1 y 3 , x1 y 2 − x2 y 1 } is called the vector or outer product of ⃗x and ⃗y . The following results are well known from analytical geometry.

6 ■ Differential Geometry and Its Visualization

Figure 1.5

The angle between two vectors ⃗x and ⃗y and the projection of ⃗y on ⃗x

Proposition 1.1.3. We have ⃗x × ⃗y = ⃗0 if and only if ⃗x and ⃗y are linearly dependent. Furthermore, the following identities hold for all ⃗x, ⃗y , ⃗z ∈ Vn and for all λ ∈ R ⃗x × ⃗y = −(⃗y × ⃗x)

(anticommutative law),

(1.2)

⃗x × (⃗y + ⃗z) = ⃗x × ⃗y + ⃗x × ⃗z

(distributive law),

(1.3)

⃗x × (⃗y × ⃗z) = (⃗x • ⃗z)⃗y − (⃗x • ⃗y )⃗z

(Grassmann identity),

(1.5)



(1.6)

(λ⃗x) × ⃗y = λ(⃗x × ⃗y ),

(1.4)



and

x1 y 1 z 1  2 2 2 (⃗x × ⃗y ) • ⃗z = det x y z  x3 y 3 z 3 (⃗x × ⃗y ) • (⃗z × w) ⃗ = (⃗x • ⃗z)(⃗y • w) ⃗ − (⃗x • w)(⃗ ⃗ y • ⃗z).

(1.7)

Proof. (i) The identities in (1.2), (1.3) and (1.4) are immediate consequences of the definition of the vector product. (ii) To prove (1.5), we put w ⃗ = ⃗y × ⃗z and obtain ⃗x × (⃗y × ⃗z) = {x2 w3 − x3 w2 , x3 w1 − x1 w3 , x1 w2 − x2 w1 } =

= {x2 (y 1 z 2 − y 2 z 1 ) − x3 (y 3 z 1 − y 1 z 3 ), x3 (y 2 z 3 − y 3 z 2 ) − x1 (y 1 z 2 − y 2 z 1 ), x1 (y 3 z 1 − y 1 z 3 ) − x2 (y 2 z 3 − y 3 z 2 )} =

= {y 1 (x1 z 1 + x2 z 2 + x3 z 3 ) − z 1 (x1 y 1 + x2 y 2 + x3 z 2 ), y 2 (x1 z 1 + x2 z 2 + x3 z 3 ) − z 2 (x1 y 1 + x2 y 2 + x3 y 3 ),

y 3 (x1 z 1 + x2 z 2 + x3 z 3 ) − z 3 (x1 y 1 + x2 y 2 + x3 y 3 )} =

= ⃗y (⃗x • ⃗z) − ⃗z(⃗x • ⃗y ).

Curves in Three-Dimensional Euclidean Space ■ 7

(iii) The identity (1.6) follows from 



x1 y 1 z 1  2 2 2 det x y z  = x1 (y 2 z 3 − y 3 z 2 ) − x2 (y 1 z 3 − y 3 z 1 ) + x3 (y 1 z 2 − y 2 z 1 ) x3 y 3 z 3

= z 1 (x2 y 3 − x3 y 2 ) + z 2 (x3 y 1 − x1 y 3 ) + z 3 (x1 y 2 − x2 y 1 ) = ⃗z • (⃗x × ⃗y ).

(iv) Finally, it follows from (1.6) that (⃗x × ⃗y ) • ⃗v = −(⃗x × ⃗v ) • ⃗y for all ⃗x, ⃗y , ⃗v ∈ V3 . Applying this with ⃗v = ⃗z × w ⃗ and using (1.5), we obtain (⃗x × ⃗y ) • (⃗z × w) ⃗ = −(⃗x × (⃗z × w)) ⃗ • ⃗y

= −((⃗x • w)⃗ ⃗ z + (⃗x • ⃗z)w) ⃗ • ⃗y

= (⃗x • ⃗z)(⃗y • w) ⃗ − (⃗x • w)(⃗ ⃗ y • ⃗z).

We state the following useful and frequently used results which are immediate consequences of Proposition 1.1.3. Remark 1.1.4. Using (1.6) in Proposition 1.1.3 with ⃗z = ⃗x and with ⃗z = ⃗y , we see that (⃗x × ⃗y ) ⊥ ⃗x, ⃗y for all ⃗x, ⃗y ∈ V3 . Furthermore, it follows from (1.6) in Proposition 1.1.3 that (⃗x × ⃗y ) • ⃗z = (⃗z × ⃗x) • ⃗y = (⃗y × ⃗z) • ⃗x = −(⃗y × ⃗x) • ⃗z = −(⃗x × ⃗z) • ⃗y \ 3. = −(⃗z × ⃗y ) • ⃗x for all ⃗x, ⃗y , ⃗z ∈ V

Example 1.1.5. (a) Let γ be a straight line in E2 , given by an equation ax1 +bx2 = d, and P = (p1 , p2 ) and X = (x1 , x2 ) be a given and an arbitrary point of γ. We write ⃗n = {a, b}. Since P, X ∈ γ, it follows that  −→

−→

⃗n • OX − OP





= {a, b} • {x1 , x2 } − {p1 , p2 }



= ax1 + bx2 − (ap1 + bp2 ) = d − d = 0.

Thus the vector ⃗n is orthogonal to every vector of the straight line γ; we say that ⃗n ⃗ in the direction of ⃗n is orthogonal to γ and denote this by ⃗n ⊥ γ. The unit vector N is called normal vector of γ, and given by ⃗ = √{a, b} . N a2 + b 2 ⃗ 1 and N ⃗ 2 ; they have Any straight line in E2 has exactly two unit normal vectors N ⃗ ⃗ opposite directions, that is, N1 = −N2 . (b) The distance ρ of a point P = (p1 , p2 ) ∈ E2 from a straight line γ ∈ E2 is the −→ ⃗ of γ. If length of the projection of any vector P X (X ∈ γ) to a normal vector N

8 ■ Differential Geometry and Its Visualization

p⃗ and ⃗x denote the position vectors of the points P and X, and γ is defined by an equation as in Part (a), then ρ is given by −→

⃗ )N ⃗ ∥ = |P X • ⃗n| ρ =∥P X N⃗ ∥ = ∥(P X • N ∥⃗n∥ 1 2 |⃗p • ⃗n − ⃗x • ⃗n| |p a + p b − d| √ = = ; ∥⃗n∥ a2 + b 2 −→

−→

if we replace X ∈ γ by X ′ ∈ γ in our last identities, then we see that our definition of the distance of a point from a straight line is independent of the choice of the point X on γ. If we choose P = 0, then the distance of γ from the origin is ρ= √

|d| . a2 + b 2

Now we extend Example 1.1.5 to planes in E3 . Visualization 1.1.6. (a) Let P l be a plane in E3 , given by an equation ax1 + bx2 + cx3 + d = 0, and let P = (p1 , p2 , p3 ) and X = (x1 , x2 , x3 ) be a given and an arbitrary point of P l. We write ⃗n = {a, b, c}. Then it follows as in Example 1.1.5 (a) that the vector ⃗n = {a, b, c} is orthogonal to every vector of the plane P l; we say that ⃗n is orthogonal to P l and denote this by ⃗ in the direction of ⃗n is called normal vector of P l and ⃗n ⊥ P l. The unit vector N given by ⃗ = √ {a, b, c} . N a2 + b 2 + c 2 ⃗ 1 and N ⃗ 2 ; they have opposite Any plane in E3 has exactly two unit normal vectors N ⃗ 1 = −N ⃗ 2. directions, that is, N (b) The distance ρ of a plane P l ∈ E3 from a point P = (p1 , p2 , p3 ) ∈ E3 is the length −→ ⃗ of P l. As in of the projection of any vector P X (X ∈ P l) to a normal vector N Example 1.1.5 (b), we obtain ρ=

|p1 a + p2 b + p3 c − d| √ ; a2 + b2 + c2

and we also see that our definition of the distance of a point from a plane is independent of the choice of the point X in the plane. If we choose P = 0, then the distance of P l from the origin is ρ= √

a2

|d| (Figure 1.6). + b2 + c 2

(c) Let Pk = (p1k , p2k , p3k ) be non–collinear points with position vectors p⃗k (k = 1, 2, 3) that span a plane P l, and X = (x1 , x2 , x3 ) be an arbitrary point of P l. Then (⃗x − p⃗1 ) × (⃗x − p⃗2 ) is a vector orthogonal to P l, and so ((⃗x − p⃗1 ) × (⃗x − p⃗2 )) • (⃗x − p⃗3 ) = 0

Curves in Three-Dimensional Euclidean Space ■ 9

Figure 1.6

Distance of a point from a plane

is an equation of P l. We may rewrite this by identity (1.6) in Proposition 1.1.3 as 

x1 − p11 x1 − p12 x1 − p13

  2 2 det  x − p1 

2

x −

p12

2

x −

  

p23  

x3 − p31 x3 − p32 x3 − p33



= 0.

Visualization 1.1.7. Let P g be the parallelogram spanned by the vectors ⃗x = {x1 , x2 } and ⃗y = {y 1 , y 2 }. Then the vector ⃗x⊥ = {−x2 , x1 } is orthogonal to ⃗x, and the surface area of P g is    ⃗x⊥   A(P g) = ∥⃗x ∥ ⃗y • ⊥  = |⃗y • ⃗x⊥ |,  ∥⃗x ∥ 

since ∥⃗x ∥ = ∥⃗x⊥ ∥. If ϕ denotes the angle between ⃗x and ⃗y , then

    π  |⃗y • ⃗x | = ∥⃗x ∥ · ∥⃗y ∥ cos ( − ϕ) = ∥⃗x ∥ · ∥⃗y ∥ | sin ϕ|, 2 ⊥

hence

A(P g) = ∥⃗x ∥ · ∥⃗y ∥ | sin ϕ|.

On the other hand, we have by (1.7) in Proposition 1.1.3 

∥⃗x ∥ · ∥⃗y ∥ | sin ϕ| = ∥⃗x ∥ · ∥⃗y ∥ 1 − cos2 ϕ = =



(⃗x × ⃗y )2 = ∥⃗x × ⃗y ∥.



∥⃗x ∥2 ∥⃗y ∥2 − (⃗x • ⃗y )2

Thus the surface area of a parallelogram is the length of the vector product of the vectors along its edges, We may also write

A(P g) = ∥⃗x × ⃗y ∥ (Figure 1.7). 



⃗x • ⃗x ⃗x • ⃗y A (P g) = (⃗x • ⃗x )(⃗y • ⃗y ) − (⃗x • ⃗y ) = det . ⃗x • ⃗y ⃗y • ⃗y 2

2

10 ■ Differential Geometry and Its Visualization

Figure 1.7

1.2

The surface area of a parallelogram

VECTOR-VALUED FUNCTIONS OF A REAL VARIABLE

In this section, we recall the definition of vector–valued functions of a real variable and state their well–known basic properties that are relevant in differential geometry. Let S ⊂ R be a subset of the real line. A function f⃗ : S → Vn that assigns to every real t ∈ S a unique vector f⃗(t) = {f 1 (t), f 2 (t), . . . , f n (t)} ∈ Vn is called a vector–valued function of a real variable, and the real–valued functions f 1 , f 2 , . . . , f n are referred to as the component functions of f⃗; we write f⃗ = {f 1 , f 2 , . . . , f n }. The set S is called the domain of f⃗, also denoted by Df⃗; and the set f⃗(S) = {f (t) : t ∈ S} is called the image or range of f⃗. Let X0 be a point of the n–dimensional Euclidean space and ⃗x0 be a vector of the n–dimensional Euclidean vector space. Then, for every real ε > 0, the sets Nε (X0 ) = {X ∈ Rn : d(X, X0 ) < ε} and Nε (⃗x0 ) = {⃗x ∈ En : ∥⃗x − ⃗x0 ∥ < ε} are called the ε–open ball or ε–neighbourhood of the point X0 and the vector ⃗x0 , and we write N˙ ε (X0 ) = Nε (X0 ) \ {X0 } and N˙ ε (⃗x0 ) = Nε (⃗x0 ) \ {⃗x0 } for the ε–deleted neighbourhoods of X0 and ⃗x0 , respectively. ¯ Let ⃗a ∈ En , S ⊂ R, If S is a subset of R, then we denote the closure of S by S. t0 ∈ S¯ and f⃗ : S → En . Then the vector-valued function f⃗ is said to have a limit ⃗a as t approaches t0 , written as lim f⃗(t) = ⃗a or f⃗(t) → ⃗a (t → t0 ),

t→t0

Curves in Three-Dimensional Euclidean Space ■ 11

if for every ε > 0, there is a real δ = δ(ε) > 0 such that f⃗(t) ∈ Nε (⃗a) for all t ∈ N˙ δ (t0 ) ∩ S, that is, for all t ∈ S with 0 < |t − t0 | < δ, or equivalently, f⃗(N˙ δ (t0 ) ∩ S) ⊂ Nε (⃗a). Since √ max |xk | ≤ ∥⃗x∥ ≤ n max |xk | for all ⃗x = {x1 , x2 , . . . , xn } ∈ En , 1≤k≤n

1≤k≤n

a vector–valued function f⃗ = {f 1 , f 2 , . . . , f n } has a limit ⃗a = {a1 , a2 , . . . , an } as t approaches t0 if and only if f k (t) → ak (t → t0 ) for all k ∈ {1, 2, . . . , n}. The following results are well known from analysis. Proposition 1.2.1. Let λ(t) → l0 , f⃗(t) → ⃗a and ⃗g (t) → ⃗b (t → t0 ). Then we have lim (f⃗(t) ± ⃗g (t)) = ⃗a ± ⃗b, lim (λ(t)f⃗(t)) = l0⃗a

t→t0

t→t0

lim ∥f⃗(t)∥ = ∥⃗a∥, lim (f⃗(t) • ⃗g (t)) = ⃗a • ⃗b

t→t0

and

t→t0

lim (f⃗(t) × ⃗g (t)) = ⃗a × ⃗b in V 3 .

t→t0

Furthermore, if lims→s0 τ (s) = t0 then lims→s0 f⃗(τ (s)) = ⃗a. We recall that a vector–valued function f⃗ : S ⊂ R → En is said to be continuous at t0 ∈ S if for every ε > 0 there is a real δ = δ(ε) > 0 such that f⃗(t) ∈ Nε (f⃗(t0 )) for all t ∈ Nδ (t0 ) ∩ S, or equivalently, f⃗(Nδ (t0 ) ∩ S) ⊂ Nε (f⃗(t0 )). The function f⃗ is said to be continuous on a subset S0 of S if it is continuous at all t ∈ S0 . The function f⃗ is said to be continuous (everywhere) if it is continuous on its domain. Obviously f⃗ is continuous at t0 ∈ S if and only if lim f⃗(t) = f⃗(t0 ).

t→t0

Furthermore, f⃗ is continuous at t0 if and only if each of its component functions is continuous at t0 . Finally, the sum, difference, dot product, and scalar and vector products (in V3 ) of continuous functions are continuous, and the continuous function of a continuous real–valued function is a continuous function. From now on let I always denote an interval on the real line. A function f⃗ : I → En is said to be differentiable at t0 ∈ I if the limit f⃗(t) − f⃗(t0 ) f⃗ ′ (t0 ) = lim t→t0 t − t0 exists; the vector f⃗ ′ (t0 ) is called the derivative of f⃗ at t0 , and we also write df⃗ f⃗ ′ (t0 ) = (t0 ). dt

12 ■ Differential Geometry and Its Visualization

The function f⃗ is said to be differentiable on a subset S0 of I if it is differentiable at every t0 ∈ S0 ; it is said to be differentiable (everywhere) if it is differentiable on I. The vector–valued function f⃗ ′ : Df⃗ ′ = {t ∈ I : f⃗ ′ (t) exists} → En is called the derivative of f⃗. Obviously the function f⃗ is differentiable at t0 if and only if each component function is differentiable at t0 . The following results are well known from analysis. Proposition 1.2.2. (a) The function f⃗ : I → En is differentiable at t0 ∈ I if and only if there exist a vector ⃗ct0 and a function ⃗rt0 such that ⃗rt (t) ⃗ f⃗(t) = f⃗(t0 ) + ⃗ct0 (t − t0 ) + ⃗rt0 (t) and lim 0 = 0. t→t0 t − t0 (b) If f⃗ is differentiable at t0 then it is continuous at t0 . Proposition 1.2.3. Let the functions f⃗, ⃗g and λ be differentiable at t0 . Then the functions f⃗ ± ⃗g , λf⃗, f⃗ • ⃗g and, in E3 , f⃗ × ⃗g are differentiable at t0 and d(f⃗ ± ⃗g ) (t0 ) = dt d(λf⃗) (t0 ) = dt d(f⃗ • ⃗g ) (t0 ) = dt

df⃗ d⃗g (t0 ) ± (t0 ), dt dt dλ df⃗ (t0 )f⃗(t0 ) + λ(t0 ) (t0 ), dt dt ⃗ d⃗g df (t0 ) • ⃗g (t0 ) + f⃗(t0 ) • (t0 ) dt dt

(1.8)

and, in E3 ,

df⃗ d⃗g d(f⃗ × ⃗g ) (t0 ) = (t0 ) × ⃗g (t0 ) + f⃗(t0 ) × (t0 ). (1.9) dt dt dt Furthermore, if h is differentiable at s0 and f⃗ differentiable at t0 = h(s0 ), then ⃗g = f⃗ ◦ h is differentiable at s0 and df⃗ dh d⃗g (s0 ) = (t0 ) (s0 ) (chain rule). ds dt ds

(1.10)

If the vector–valued function f⃗ is differentiable on an interval I, then its derivative ′ ⃗ f also is a vector–valued function, which again may be differentiable on I. This yields the second-order derivative of f⃗ denoted by f⃗ ′′ . Higher-order derivatives are defined similarly. If m is a nonnegative integer, then C m (I) denotes the class of all vector–valued functions that are m times continuously differentiable on the interval I, C 0 (I) is the class of all continuous functions on I, and C ∞ (I) is the class of all functions which have continuous derivatives of all orders on I. The following results are well known from analysis.

Curves in Three-Dimensional Euclidean Space ■ 13

Proposition 1.2.4. (a) A vector–valued function f⃗ is of class C m (I) if and only if all its component functions are of class C m (I). (b) If f⃗, ⃗g , h ∈ C m (I) then hf⃗, f⃗ ± ⃗g , f⃗ • ⃗g ∈ C m (I) and, in E3 , f⃗ × ⃗g ∈ C m (I).

Proposition 1.2.5 (Taylor’s formula). Let f⃗ ∈ C m (I). Then for every t and t0 in I m ⃗  1 ⃗(k) ⃗ m,t0 (t), where lim Rm,t0 (t) = ⃗0. f (t0 )(t − t0 )k + R f⃗(t) = t→t0 (t − t0 )m k! k=0

1.3

THE GENERAL CONCEPT OF CURVES

Our studies of curves take place in the three-dimensional Euclidean space with a Cartesian coordinate system. We start, however, with the more general definition of a curve in the n–dimensional space Rn . Definition 1.3.1. Let f : R → Rn be a continuous map on the real interval I = [a, b]. The point set 





γ = X ∈ Rn : X = f (t) = f 1 (t), f 2 (t), . . . , f n (t)

(t ∈ I)

(1.11)

is called a curve in Rn ; (f, I), t and I are called a parametric representation, the parameter and the parameter interval of γ, respectively (Figure 1.8). Remark 1.3.2. (a) A curve γ may also be given in terms of the position vectors of its points 



⃗x(t) = x1 (t), x2 (t), . . . , xn (t)

(t ∈ I), where xk (t) = f k (t) for (k = 1, 2, . . . , n) (1.12) and (1.12) is also referred to as a parametric representation of γ; in fact we shall rather use (1.12) than (1.11). (b) A parametric representation induces an orientation on a curve γ; as the parameter t moves through the interval [a, b] from a to b, the points X = X(t) move along the curve from the point X(a) to the point X(b). The points X(a) and X(b) are called the initial and end points of the curve γ, respectively (Figure 1.8).

Figure 1.8

The parametric representation of a curve

14 ■ Differential Geometry and Its Visualization

Figure 1.9

A helix

Visualization 1.3.3. (a) A straight line through the point A with position vector ⃗a in the direction of the vector ⃗b ̸= ⃗0 has a parametric representation ⃗x(t) = ⃗a + t⃗b (t ∈ R). (b) A parametric representation of a circle line of radius r > 0 and centre M in a plane spanned by the orthonormal vectors ⃗e 1 = {1, 0, 0} and ⃗e 2 = {0, 1, 0} is given by ⃗x(t) = m ⃗ + r⃗e 1 cos t + r⃗e 2 sin t (t ∈ [0, 2π]),

where m ⃗ is the position vector of M . (c) A parametric representation for a helix with its axis along the x3 –axis is given by ⃗x(t) = {r cos t, r sin t, ht} (t ∈ R)

where r > 0 and h ∈ R are constants (Figure 1.9).

In differential geometry, we have to assume that the function ⃗x in the parametric representation of a curve has a certain number of derivatives. Furthermore, we want to exclude trivial or pathological cases. Definition 1.3.4. A parametric representation (1.12) is called admissible if the following conditions are satisfied ⃗x ∈ C r (I) where r ≥ 1 is appropriately chosen

(1.13)

Curves in Three-Dimensional Euclidean Space ■ 15

and ⃗x ′ (t) =





dx1 dx2 dxn (t), (t), . . . , (t) dt dt dt

̸= ⃗0 for all t ∈ I.

(1.14)

The condition in (1.14) of Definition 1.3.4 implies that one of the equations xk = x (t) in (1.12) can locally be solved for t: k

t = t(xk ) for some k ∈ {1, 2, . . . , n},

(1.15)

and the function in (1.15) is r times continuously differentiable on its domain. Since, in future, we shall only deal with curves that have an admissible parametric representation, we shall omit the term admissible. In general, a curve γ has various different parametric representations. We may obtain a new vector–valued function from (1.12) by introducing a new parameter t∗ by t = t(t∗ ). (1.16) We have to make sure that the range of the function in (1.16) contains the interval I for t, and that the conditions of admissibility also hold for the new representation. Definition 1.3.5. A parameter transformation (1.16) is called admissible if the following conditions are satisfied: I ⊂ t(I ∗ ) for the domain I ∗ of the function t in (1.16), t ∈ C (I ), where r ≥ 1 is appropriately chosen, dt ∗ (t ) ̸= 0 for all t∗ ∈ I ∗ . dt∗ r

∗

(1.17) (1.18) (1.19)

Again, we shall omit the term admissible in future. Visualization 1.3.6 (Neil’s parabola). The curve given by ⃗x(t) = {t2 , t3 } (t ∈ R) is called Neil’s parabola (Figure 1.10).

Figure 1.10

Neil’s parabola

16 ■ Differential Geometry and Its Visualization

Since x1 (t) = t2 and x2 (t) = t3 , it is obvious that ⃗x ∈ C ∞ (R). Furthermore, ⃗x ′ (t) = t{2, 3t} = ̸ ⃗0 for all t ̸= 0. Thus ⃗x(t) = {t2 , t3 } (t ∈ J) is an admissible parametric representation for every interval J which does not contain 0. If we consider the part of Neil’s parabola in the upper half plane, that is, the part of the curve given by ⃗x(t) = {t2 , t3 } (t ∈ I = (0, ∞)), then we may write √ t = ϕ(t∗ ) = t∗ (t∗ ∈ I ∗ = (0, ∞)). (1.20) Then I ⊂ ϕ(I), ϕ ∈ C r (I ∗ ) for any r ≥ 1 and dϕ ∗ 1 (t ) = √ ∗ ̸= 0 for all t∗ ∈ I ∗ , ∗ dt 2 t and therefore, (1.20) defines an admissible parameter transformation. We obtain ⃗x ∗ (t∗ ) = {t∗ , t∗3/2 } (t∗ ∈ (0, ∞)) for the part of Neil’s parabola in the upper half plane; this is an admissible parametric representation. Visualization 1.3.7 (The catenary). The curve in which a uniform heavy cord or chain hangs is called a catenary (Figure 1.11). It has a parametric representation of the form 

⃗x(t) = t, c cosh

 

t c

(t ∈ R), where c is a constant (Figure 1.12).

(1.21)

Proof. We derive the parametric representation of a catenary in (1.21) (right in Figure 1.11). Let a uniform heavy cord or chain be attached to two distinct points P0 and P2 at the same level, and P1 denote the lowest point of the cord. We introduce a Cartesian coordinate system with its x–axis parallel to P0 P2 , such that P1 = (0, c) for some c > 0. Let P be a point on the catenary and ϕ denote the angle between the positive x–axis and the tangent to the catenary at P . We consider the equilibrium of the portion P1 P of the cord from P1 to P . If the length of the arc from P1 to P equals s, and w is the weight of the cord per unit length, the weight of P1 P is equal to ws. The other forces acting on P1 P are the tensions T1 and T at P1 and P , respectively, and the tangent at P1 is horizontal. Resolving horizontally and vertically, we obtain T cos ϕ = T1

(1.22)

Curves in Three-Dimensional Euclidean Space ■ 17

Figure 1.11

A suspended chain and a parametric representation for a catenary

and T sin ϕ = ws.

(1.23)

If we put T1 = wc, then dividing (1.23) by (1.22), we get tan ϕ =

s or s = c tan ϕ; c

(1.24)

this is called the intrinsic equation of the catenary. It gives the relation between the length of arc of the curve from the lowest point to any other point of the curve and the inclination of the tangent at the latter point. We can find a parametric representation of the curve as follows (right in Figure 1.11). It follows from dy dy = tan ϕ and = sin ϕ dx ds and (1.24) that dy ds c dy = = sin ϕ 2 , dϕ ds dϕ cos ϕ and so y = y(ϕ) =

c + d, where d ∈ R is a constant of integration. cos ϕ

Since y(0) = c, it follows that d = 0, hence y = y(ϕ) =

c . cos ϕ

(1.25)

We get from (1.25) and (1.24) y2 =

c2 = c2 (1 + tan2 ϕ) = c2 + s2 , cos2 ϕ

(1.26)

18 ■ Differential Geometry and Its Visualization

Figure 1.12

Left: catenaries for varying c. Right: a catenary γ and a quadratic parabola

γ∗ and then



dy s y 2 − c2 = tan ϕ = = ± . dx c c Taking the positive root, we have     x dy dx −1 y = = + d, cosh =  2 2 c c c y −c

where d ∈ R is a constant of integration, that is,   x y = y(x) = c cosh +d . c It follows from y(0) = c that d = 0, hence   x . y = y(x) = c cosh c Writing t = x, we obtain the parametric representation in (1.21) for the catenary. A catenary with a parametric representation (1.21) looks very much like a quadratic parabola. The Taylor series expansion T (f ; x) of the function f with f (x) = c cosh (x/c) is ∞  (x/c)2k T (f ; x) = c · ; (2k)! k=0

it converges to f (x) on the whole real line. We consider the partial sums T2n (f ; x) = c ·

n  (x/c)2k

k=0

(2k)!

of T (f ; x). If n = 1, then T2 (f ; x) is a quadratic polynomial, and the quadratic parabola with a parametric representation ⃗x(t) = {t, T2 (f ; t)} =



t2 t, c + 2c



(t ∈ R)

is a second order approximation of the catenary for |x| small (right in Figure 1.12).

Curves in Three-Dimensional Euclidean Space ■ 19

1.4

SOME EXAMPLES OF PLANAR CURVES

In this section, we consider some examples of planar curves; most of them are defined by equations, such as algebraic curves. We also give the geometrical principles of their constructions. If P and Q are given points in Rn , then P Q denotes the straight line through P and Q. Throughout, we write x = x1 and y = x2 . Let n be a non–negative integer. Then the set Cn =

 

X = (x, y) ∈ R2 :





0≤k+j≤n

ajk xj y k = 0 (ajk ∈ R)

is an algebraic curve of order n. First, we consider algebraic curves of order 3.

  

Visualization 1.4.1 (The serpentine). Let Sr (M ) be the circle line of radius r > 0 with its centre in M , and P1 be a point on Sr (M ) such that the position vectors m ⃗ and p⃗1 of M and P1 are not parallel. Furthermore, let P2 be the intersection of the straight line OP1 and the straight line γ ∗ parallel to m ⃗ at a distance a from M . Finally, let P denote the orthogonal projection of the point P2 on the parallel of γ ∗ through P1 . Then the points P generate a serpentine, as the point P1 moves along the circle line Sr (M ) (Figure 1.13). A serpentine γa,r is given by the equation 



f (x, y; a, r) = y x2 + a2 − 2arx = 0, and has a parametric representation 

⃗xa,r (t) = t,

Figure 1.13

2art 2 t + a2



(t ∈ R).

(1.27)

(1.28)

Left: the construction of a serpentine. Right: families of serpentines

20 ■ Differential Geometry and Its Visualization

Proof. We write C = Sr (M ), for short, and choose a Cartesian coordinate system such that M = (r, 0) and γ ∗ is parallel to the x–axis. Let P = (x, y), P1 = (x1 , y1 ) and P2 = (x2 , y2 ), and observe that x = x2 , y2 = a and y = y1 by construction. It follows from elementary geometry that y2 y a xy y1 . = , hence = and x1 = x1 x2 x1 x a Since P1 ∈ C, we have (x1 − r)2 + y12 = r2 , that is, x21 − 2rx1 + y12 = 0. Substituting x1 = (xy)/a and y1 = y and multiplying by a2 , we obtain x2 y 2 − 2raxy + a2 y 2 = 0, and (1.27) follows. Writing x = t and solving (1.27) for y, we obtain (1.28). Visualization 1.4.2 (The versiera). Let C = Sr (M ) be the circle line of radius r > 0 with its centre in M , P1 and P2 be −→

distinct points of C, γ1∗ be the straight line through M orthogonal to the vector P1 M , and P3 be the point of intersection of γ1∗ with the straight line through P1 and P2 . −→

Furthermore, let P denote the intersection of the parallel γ2∗ of P1 M through P3 with the parallel γ3∗ of γ1∗ through P2 . Then the points P generate a versiera as the straight line through P1 and P2 is rotated about P1 (Figure 1.14). A versiera γr is given by the equation f (x, y; r) = y(x2 + r2 ) − 2r3 = 0, (1.29) and has a parametric representation ⃗xa,r (t) =

Figure 1.14



2r3 t, 2 t + r2



(t ∈ R).

Left: The construction of a versiera. Right: families of versieras

(1.30)

Curves in Three-Dimensional Euclidean Space ■ 21

Proof. We choose a Cartesian coordinate system with its origin in P1 and its x–axis parallel to γ1∗ . Let P2 = (x2 , y2 ), P3 = (x3 , y3 ) and P = (x, y). Then we have x = x3 , y = y2 and y3 = r by construction, and it follows from elementary geometry that y y3 r yx y2 . = = = , hence x2 = x2 x2 x3 x r Furthermore, P2 ∈ C implies

r2 = x22 + (y2 − r)2 = x22 + y 2 − 2ry + r2 ,

hence

x22 r2 + y 2 r2 − 2r3 y = y 2 x2 + y 2 r2 − 2r3 y = y 2 (x2 + r2 ) − 2r3 y = 0,

and (1.29) follows. Writing x = t and solving (1.29) for y, we obtain (1.30).

Visualization 1.4.3 (Cissoids). Let γ1 and γ2 be curves, Q be a point and γ be a straight line that intersects γ1 in P1 and γ2 in P2 . A cissoid is generated by the points P for which    −→  QP  =  

as γ is rotated about the point Q. We consider the following special cases.

   −→  P1 P2  ,  

(1.31)

(a) Diocles’s cissoid If Q is the origin, γ1 the circle of radius r > 0 and its centre in M = (r, 0) and γ2 is the straight line through the point (2r, 0) parallel to the y–axis, then we obtain Diocles’s cissoid (left in Figures 1.15 and 1.16); it is given by the equation f (x, y; r) = x3 + xy 2 − 2ry 2 = 0, (1.32)

Figure 1.15

The construction of: Left: Diocles’s cissoid. Right: a strophoid

22 ■ Differential Geometry and Its Visualization

Figure 1.16

Families of Diocles’s cissoids and strophoids

or in polar coordinates (ρ, ϕ) with x = ρ cos ϕ and y = ρ sin ϕ sin2 ϕ ρ(ϕ) = 2r cos ϕ





π π ϕ∈ − , 2 2



(1.33)

.

This yields a parametric representation 





⃗x(t) = 2r sin2 t, 2r sin2 t tan t which is not admissible at t = 0.



π π t∈ − , 2 2



(1.34)

,

(b) Strophoids Choosing the point Q and γ1 as in Part (a) and γ2 as the straight line through the point (r, 0) parallel to the y–axis, we obtain a strophoid (right in Figures 1.15 and 1.16); it is given by the equation f (x, y; r) = x3 + xy 2 − rx2 + ry 2 = 0,

(1.35)

or in polar coordinates (ρ, ϕ) ρ(ϕ) = r

cos (2ϕ) cos ϕ





π π ϕ∈ − , 2 2



(1.36)

.

This yields a parametric representation ⃗x(t) = {r cos (2t), r cos (2t) tan t} which is admissible everywhere.





π π t∈ − , 2 2



,

(1.37)

Curves in Three-Dimensional Euclidean Space ■ 23

Left: The construction of MacLaurin’s trisectrix. Right: A family of MacLaurin’s trisectrices

Figure 1.17

(c) MacLaurin’s trisectrix Choosing Q = (0, 0), γ1 as the circle of radius 2r and its centre in (2r, 0) and γ2 as the straight line through the point (r, 0) parallel to the y–axis, we obtain MacLaurin’s trisectrix (Figure 1.17); it is given by the equation f (x, y; r) = x3 + xy 2 + ry 2 − 3rx2 = 0,

(1.38)

or in polar coordinates (ρ, ϕ) ρ(ϕ) = r



sin2 ϕ − 3 cos2 ϕ cos ϕ



π π ϕ∈ − , 2 2



(1.39)

.

This yields a parametric representation 

2

2

2

2

 



π π t∈ − , 2 2

⃗x(t) = r(sin t − 3 cos t), r(sin t − 3 cos t) tan t



. (1.40)

Proof. −→

(a) We write OP = {x, y} and obtain for x > 0 −→

OP2 =

−→ −→ −→ 2r −→ −→ 2rx 2ry 2 OP , OP1 = OP and P P = OP , 1 2 x x(x2 + y 2 ) (d(0, P ))2

and (1.31) yields 1=

2ry 2 or (1.32). x(x2 + y 2 )

Substituting x = ρ cos ϕ and y = ρ sin ϕ in (1.32), we get 



ρ3 cos3 ϕ + ρ3 cos ϕ sin2 ϕ = ρ3 cos ϕ cos2 ϕ + sin2 ϕ = ρ3 cos ϕ = 2r2 ρ2 sin2 ϕ

24 ■ Differential Geometry and Its Visualization

and (1.33) follows, since ρ > 0. We put t = ϕ and (1.34) is an immediate consequence. Finally, it follows from (1.34) that 



⃗x (t) = 4r sin t cos t, 4r sin t cos t tan t + 2r sin2 tcos2 t ′





= 2r sin2 (2t), sin2 (2t) + tan2 (2t) ,

′ hence ⃗x (0) = ⃗0, and the parametric representation (1.34) is not admissible at t = 0.

(b) Similarly as in Part (a), we obtain for x > 0 −→

OP2 =

−→ −→ r −→ −→ 2rx r(x2 − y 2 ) −→ OP , OP1 = OP and P P = OP , 1 2 x x(x2 + y 2 ) (d(0, P ))2

and (1.31) yields 1=

r(x2 − y 2 ) or (1.35). x(x2 + y 2 )

Substituting x = ρ cos ϕ and y = ρ sin ϕ in (1.35), we get 



ρ3 cos ϕ = rρ2 cos2 ϕ − sin2 ϕ = rρ2 cos (2ϕ) (1.36) follows, since ρ > 0. We put t = ϕ and (1.37) is an immediate consequence. Finally, it follows from (1.37) that 

r cos (2t) ⃗x (t) = −2r sin (2t), −2r sin (2t) tan t + cos2 t ′







π π ̸ ⃗0 for all t ∈ − , = . 2 2

(c) Part (c) follows in the same way as Parts (a) and (b).

Now we consider algebraic curves of order 4. Visualization 1.4.4 (Conchoids). The conchoid of a given curve γ is obtained by extending the position vector of every point of γ by a constant length ±r. If the curve γ is given by the equation in polar coordinates fγ (ρ, ϕ) = ρ − g(ϕ) = 0 for some function g, then the equation for the conchoid γ±r of γ is f±r (ρ, ϕ) = ρ − (g(ϕ) ± r) = 0. We consider some special cases. (a) Nikomedes’s conchoid Nikomedes’s conchoid is the conchoid of the straight line through the point (a, 0),

Curves in Three-Dimensional Euclidean Space ■ 25

Figure 1.18

The construction (left) and families of Nikomedes’s conchoids (right)

orthogonal to the x–axis (Figure 1.18). Putting t = ϕ where ϕ is the polar angle, we obtain the following parametric representation for Nikomedes’s conchoid 

⃗x(t) = {a ± r cos t, a tan t ± r sin t}



π π t∈ − , 2 2



(1.41)

,

or the equation in polar coordinates f±r (ρ, ϕ; a) = ρ −





a ±r =0 cos ϕ





π π ϕ∈ − , 2 2



.

This yields the equation in Cartesian coordinates 



∗ (x, y; a) = (x − a)2 x2 + y 2 − r2 x2 = 0. f±r

(b) Pascal’s snail Pascal’s snail is the conchoid of the circle of radius a/2 > 0 with its centre in the point (a/2, 0) (Figure 1.19). Putting t = ϕ where ϕ is the polar angle, we obtain the following parametric representation for Pascal’s snail ⃗x(t) = {a cos2 (t) ± r cos t, a sin t cos t ± r sin t}





π π t∈ − , 2 2



or the equation in polar coordinates f±r (ρ, ϕ; a) = ρ − (a cos ϕ ± r) = 0





π π ϕ∈ − , 2 2



This yields the equation in Cartesian coordinates 

∗ (x, y; a) = x2 + y 2 − ax f±r

2





− r2 x2 + y 2 = 0.

.

(1.42)

26 ■ Differential Geometry and Its Visualization

Figure 1.19

The construction (left) and families of families of Pascal’s snails (right)

Visualization 1.4.5 (The κ–curve). Let γ ∗ be straight line, P0 be a point on γ ∗ and C be a family of circles lines of a fixed radius r > 0 with their centres on γ ∗ . A κ–curve is generated by the points where the tangents from P0 to the circle lines of C touch the circles (Figure 1.20). A κ–curve is given by the equation f (x, y; r) = y 4 + x2 y 2 − r2 x2 = 0,

(1.43)

and has a parametric representation for two branches 

⃗x(k) (ϕ) = r



| cos ϕ| | cos ϕ| cos ϕ, r sin ϕ (ϕ ∈ Ik ) | sin ϕ| | sin ϕ| for k = 1, 2 where I1 = (0, π) and I2 = (ϕ, 2π). (1.44)

Proof. We choose a Cartesian coordinate system with its origin in the point P0 and its x–axis along γ ∗ . Let Ca ∈ C be the circle with its centre in Ma = (a, 0) and

Figure 1.20

Left: The construction a κ–curve. Right: A family of κ–curves

Curves in Three-Dimensional Euclidean Space ■ 27

P = (x, y), P ∗ = (x, y ∗ ) be the points where the tangents from P0 touch the circle, b = a − x, and α denote the angle between the x–axis and the tangents. Then we have y ∗ = −y. It follows that sin α =

r r2 b = , that is, b = and b(x + b) = r2 , hence b2 = r2 − bx. r a a

Since y 2 = (y ∗ )2 = r2 − b2 , we obtain y 2 = (y ∗ )2 = r2 − (r2 − bx) = bx.

(1.45)

Furthermore, the quadratic equation b2 + bx − r2 = 0 has the solutions b= so (1.45) yields

 1 √ 2 ± x + 4r2 − x , 2 

x√ 2 x2 x + 4r2 and =± y + 2 2

x2 y + 2

2

and finally

2

2

−

 x2  2 x + 4r2 = 0 4

y 4 + x2 y 2 − r2 x2 = 0.

We introduce polar coordinates ρ and ϕ, that is, x = ρ cos ϕ and y = ρ sin ϕ, and obtain from (1.43) 

ρ4 sin4 ϕ + ρ4 sin2 ϕ cos2 ϕ − r2 ρ2 cos2 ϕ = ρ2 ρ2 sin2 ϕ(sin2 ϕ + cos2 ϕ) − r2 cos2 ϕ 



= ρ2 ρ2 sin2 ϕ − r2 cos2 ϕ = 0,

and, since ρ > 0, ρ=r



|cosϕ| for ϕ ̸= kπ (k ∈ Z), | sin ϕ|

and (1.44) is an immediate consequence.

Visualization 1.4.6 (Cassini curves and lemniscates). Let P1 and P2 be distinct points and c = (1/2)d(P1 , P2 ). (a) A Cassini curve is generated by the points P with a constant product a2 of their distances from P1 and P2 (Figures 1.21 and 1.22); it is given by the equation 

f (x, y; a, c) = x2 + y 2 or in polar coordinates by

2





− 2c2 x2 − y 2 + c4 − a4 = 0,

g(ρ, ϕ; a, c) = ρ4 − 2c2 ρ2 cos (2ϕ) + c4 − a4 = 0.

(1.46)

(1.47)

28 ■ Differential Geometry and Its Visualization

Figure 1.21

The construction of a Cassini curve

Figure 1.22

Families of Cassini curves

(b) A lemniscate is the special case a = c of a Cassini curve; it is given by the equation  2   f (x, y; c) = x2 + y 2 − 2c2 x2 − y 2 = 0, (1.48) or parametric representations in polar coordinates ⃗x(k) (ϕ) =

√



2c cos (2ϕ) cos ϕ,

 √  2c cos (2ϕ) sin ϕ for ϕ ∈ Ik (k = 1, 2, 3),

where I1 = (0, π/4), I2 = (3π/4, 5π/4) and I3 = (7π/4, 2π). (1.49) Proof. We choose a Cartesian coordinate system with its origin in the midpoint of the straight line segment P1 P2 and its x–axis along P1 P2 . Then P1 = (−c, 0), P2 = (c, 0) and the distances of P = (x, y) from P1 and P2 are given by d(P1 , P ) =



(x + c)2 + y 2 and d(P2 , P ) =



(x − c)2 + y 2 .

The points P of a Cassini curve must satisfy by definition 

a4 = (d(P1 , P )d(P2 , P ))2 = (x + c)2 + y 2 

= x2 − c2

2







(x − c)2 + y 2

+ (x + c)2 + (x − c)2 y 2 + y 4



Curves in Three-Dimensional Euclidean Space ■ 29

Figure 1.23

Left: the construction of a rosette. Right: a family of rosettes 



= x4 − 2c2 x2 + c4 + 2y 2 x2 + c2 + y 4 

= x2 + y 2

2





− 2c2 x2 − y 2 + c4 ,

which yields (1.46). Introducing polar coordinates ρ and ϕ, we obtain with x = ρ cos ϕ and y = ρ sin ϕ 



g(ρ, ϕ; a, c) = ρ4 − 2c2 ρ2 cos2 ϕ − sin2 ϕ + c4 − a4 = ρ4 − 2c2 ρ2 cos (2ϕ) + c4 − a4 = 0,

that is, (1.47). If a = c then (1.48) √ is an immediate consequence of (1.46), and solving (1.47) for ρ, we obtain ρ = 2c cos (2ϕ) for cos (2ϕ) ≥ 0, which yields (1.49) and the interval Ik (k = 1, 2, 3). Now we consider algebraic curves of order 5. Visualization 1.4.7 (Rosettes). Let γ = AB be the straight line segment of a fixed length a > 0 with its end points A and B on the x1 – and x2 –axes of a Cartesian coordinate system of R 2 . Furthermore, let P = γ ∩ γ ⊥ , where γ ⊥ is the straight line orthogonal to γ and through the origin O. A rosette is the curve generated by the points P , when the end points of γ move along axes of the coordinate system (Figure 1.23); it is given by the equation 

f (x, y; a) = x2 + y 2

3



− a2 x2 y 2

2

= 0,

and has a parametric representation in polar coordinates a ⃗x(ϕ) = {sin (2ϕ) cos ϕ, sin (2ϕ) sin ϕ} (ϕ ∈ (0, 2π)). 2

(1.50)

(1.51)

Proof. We obtain d(O, B) = a cos ϕ, ρ = d(O, P ) = d(O, B) sin ϕ = a sin ϕ cos ϕ = and (1.51) is an immediate consequence.

a sin 2(ϕ) 2

30 ■ Differential Geometry and Its Visualization

It also follows from x = (a/2) sin (2ϕ) cos ϕ and y = (a/2) sin (2ϕ) sin ϕ that



and a

2



2 2

x y

2

=a

x2 + y 2

2



3

=

a6 sin6 (2ϕ) 64

a2 sin2 (2ϕ) cos ϕ sin ϕ 4 

sin (2ϕ) a6 = sin4 (2ϕ) 16 2

2

2

=

a6 sin6 (2ϕ), 64

and (1.50) is an immediate consequence. Visualization 1.4.8 (Double egg lines). Let C be a circle line of radius a > 0 and its centre in the origin, and let A and B be −→

−→

distinct points of C. Furthermore, let F be the orthogonal projection of OB on OA −→

and P be the orthogonal projection of F on OB. A double egg line is generated by the points P , when B moves along C (Figure 1.24); it is given by the equation 

f (x, y; a) = x2 + y 2 and has a parametric representation

3

− a4 x2 = 0,

⃗x(ϕ) = a{cos3 ϕ, cos2 ϕ sin ϕ} (ϕ ∈ (0, 2π)).

Figure 1.24

lines

(1.52)

(1.53)

Left: the construction of a double egg line. Right: a family of double egg

Curves in Three-Dimensional Euclidean Space ■ 31

Figure 1.25

Left: the construction of an astroid. Right: a family of astroids

Proof. We choose polar coordinates (ρ, ϕ) such that A = (0, a), and B = (a cos ϕ, a sin ϕ). It follows that F = (a cos ϕ, 0) and P = (ρ cos ϕ, ρ sin ϕ), where ρ = d(F, 0) cos ϕ = a cos2 (ϕ) and (1.53) is an immediate consequence. If follows from x = a cos3 ϕ and y = a cos2 ϕ sin ϕ that 

x2 + y 2

3



− a4 x2 = a6 cos2 ϕ(cos2 ϕ + sin2 ϕ) = a6 cos6 ϕ − a6 cos6 ϕ = 0,

3

− a6 cos6 ϕ

and (1.52) is satisfied. Visualization 1.4.9 (Astroids). Let γ be the straight line segment of a fixed length a > 0 with its end points A and B on the x1 – and x2 –axes of a Cartesian coordinate system of R2 . Furthermore, let P be the orthogonal projection of the point E of the rectangle BOAE on γ. An astroid is generated by the points P , when the end points of γ move along the coordinate axes (Figure 1.25); it is given by the equation (x1 )2/3 + (x2 )2/3 − a2/3 = 0. Proof. We obtain d(E, B) = a cos ϕ, d(P, B) = d(E, B) cos ϕ = a cos2 ϕ, hence

x = d(P, B) cos ϕ = a cos3 ϕ and similarly y = a sin3 ϕ,

and (1.54) is an immediate consequence.

(1.54)

32 ■ Differential Geometry and Its Visualization

Remark 1.4.10. If (p) = (p1 , p2 , . . . , pn ) is an n–tuple of positive real numbers p1 , p2 , . . . , pn , and H = max{1, p1 , . . . , pn }, then d(p) (X, Y ) =



n 

k=1

k

k pk

|x − y |

1/H

(X = (x1 , . . . , xn ), Y = (y 1 , . . . , y n ) ∈ R n )

defines a metric d(p) on Rn . In the special case, where (p) = (p, p, . . . , p) for some p ≥ 1, then ∥⃗x∥p = dp (X, 0) = is a norm on R n . The sets



n 

k=1

k p

|x |

(1.55)

1/p

∂B(p) (r; X0 ) = {X ∈ IR n : d(p) (X, X0 ) = r} (r > 0; X0 ∈ Rn ) are the boundaries of neighbourhoods in the metric d(p) . Thus the astroids of Visualization 1.4.9 are a boundaries of neighbourhoods of the origin in the metric d2/3 of R2 . We close this section by introducing the concept of orthogonal trajectories. Let Γ be a family of planar curves γc in an open connected set D ⊂ R 2 such that there is one and only one curve of Γ through every point of D. Then the curves γ ⊥ that intersect every curve of Γ at right angles are called the orthogonal trajectories of Γ. If the curves γc of the family Γ are given by the equations f (x, y; c) = 0, then the orthogonal trajectories of Γ are obviously given by f1 (x, y; c)

dy − f2 (x, y; c) = 0, where dx ∂f (x, y; c) ∂f (x, y; c) f1 (x, y; c) = and f2 (x, y; c) = . ∂x ∂y

(1.56)

Visualization 1.4.11. Let α, β ̸= 2 be positive reals, and Γα,β be a family of curves given by the equations f (x, y; c) = xα + y β − c = 0. Then the orthogonal trajectories γ ⊥ of Γα,β are given by the equations g(x, y; d) =

1 1 x2−α − y 2−β − d = 0, α(2 − α) β(2 − β)

(1.57)

where d ∈ R is a constant (Figure 1.26). If α = β = 2, then the orthogonal trajectories of Γ2,2 are given by f (x, y; d) = y − dx = 0,

that is, the orthogonal trajectories of concentric circles with their centres in the origin are straight lines through the origin.

Curves in Three-Dimensional Euclidean Space ■ 33

Figure 1.26

A family of astroids and their orthogonal trajectories

Proof. The orthogonal trajectories of Γα,β must satisfy by (1.56) αxα−1 hence for α, β ̸= 2

1 1 x2−α = α(α − 2) α



dy − βy β = 0, dx

x1−α dx =

1 β



y 1−β dy =

for some constant of integration d. This yields (1.57). If α = β = 2, then we obtain from (1.56) x

1 y 2−β + d β(2 − β)

dy − y = 0, that is log |x| = log |y| + log d˜ for some constant d˜ > 0. dx

This yields the statement for α = β = 2.

1.5

THE ARC LENGTH OF A CURVE

The arc length of a curve plays an eminent role as a parameter of a curve due to its significance as a geometric invariant. Definition 1.5.1. Let γ be a curve with a parametric representation ⃗x(t) (t ∈ I). Then the arc length of γ between the points X(t0 ) and X(t) (t0 , t ∈ I) is given by s(t) =

t

t0

∥⃗x ′ (τ )∥ dτ.

(1.58)

We always denote the arc length of a curve by s, and use a dot to indicate differentiation with respect to s, that is, we write d⃗x (s). ⃗x˙ (s) = ds

34 ■ Differential Geometry and Its Visualization

Proposition 1.5.2. (a) Equation (1.58) defines an admissible parameter transformation. (b) The arc length of a curve is invariant under admissible parameter transformations. (c) Every curve can be parameterized with respect to its arc length. (d) If ⃗x(s) (s ∈ I) is a parametric representation of a curve γ with respect to its arc length s then ∥⃗x˙ (s)∥ = 1 for all s.

Conversely, if ⃗x(t) (t ∈ [a, b]) is a parametric representation of a curve γ and ∥⃗x ′ (t)∥ = 1 for all t ∈ [a, b], then the value t − a is equal to the arc length of γ between its initial point X(a) and the point X(t).

This means that if the parameter interval I is divided into subintervals by the equidistant points sk (k = 0, 1, . . . , n), then the arcs of the curve between X(sk−1 ) and X(sk ) (k = 1, 2, . . . , n) have equal lengths (Figure 1.27); or kinematically, a point moves along the curve with constant speed. Proof. (a) Let γ be a curve with an admissible parametric representation ⃗x(t) (t ∈ I) and t0 ∈ I be given. We write ϕ : I → R for the function with ϕ(t) = s(t) =

t

t0

∥⃗x ′ (τ )∥ dτ for all t ∈ I.

Since ∥⃗x ′ (t)∥ > 0 for all t ∈ I by (1.14) in Definition 1.3.4, the inverse function ϕ−1 : I ∗ = ϕ(I) → R exists, is differentiable on I ∗ , and we have for every s0 = ϕ(t0 ) ∈ I ∗ dϕ−1 (s0 ) = ds

Figure 1.27



−1

dϕ (t0 ) dt

= ∥⃗x ′ (t0 )∥−1 ̸= 0.

A catenary with parameters t (left) and s (right)

Curves in Three-Dimensional Euclidean Space ■ 35

It also follows from this that if ⃗x ∈ C r (I) for some r ≥ 1, then ϕ ∈ C r (I ∗ ).

(b) Now let γ be a curve with a parametric representation as in Part (a), and t = ψ(t∗ ) (t∗ ∈ I ∗ ) with ψ(I ∗ ) = I be an admissible parameter transformation. Then ⃗x ∗ (t∗ ) = ⃗x(ψ(t∗ )) (t ∈ I ∗ ) is a parametric representation of γ. Let t∗0 , t∗1 ∈ I ∗ be arbitrary parameters, and t0 = ψ(t∗0 ) and t1 = ψ(t∗1 ). Then the rule of substitution yields the arc length s∗ (t∗0 , t∗1 ) along γ from t∗0 to t∗1 in terms of the parameters t0 and t1 ∗

∗

 t1  ∗  t1     d⃗  d⃗ x x ∗ ∗ ∗ ∗ ∗ dψ ∗     s (t0 , t1 ) =  ∗  dt =  (ψ(t )) ∗ (t ) dt∗ dt dψ dt t∗0

t∗0

ψ(t∗1 )

=



ψ(t∗0 )

   d⃗   x (ψ) dψ = s(ψ(t∗ ), ψ(t∗ )) = s(t0 , t1 ). 0 1  dψ 

This shows that the arc length is invariant under admissible parameter transformations. (c) Part (c) is an immediate consequence of Part (a). (d) If γ is a curve with a parametric representation ⃗x(t) (t ∈ I) and t is the arc length along γ, then s = t and it follows from (1.58) in Definition 1.5.1 that ds (t) = 1 = ∥⃗x ′ (t)∥ = ∥⃗x˙ (s)∥ for all s ∈ I. dt The converse part is obvious.

Owing to its geometric significance the parameter s is referred to as the natural parameter. It is convenient to use s as a parameter for theoretical purposes, since the formulae with respect to s are less complicated. Geometric properties of a curve should not depend on some particular choice of its representation, they must be invariant under (i) transformations of the coordinate system, (ii) admissible parameter transformations. Then the invariance of the conditions in (i) and (ii) is satisfied by the use of vectors and the use of the arc length of a curve as parameter, respectively. Example 1.5.3. (a) The arc length along the straight line in Visualization 1.3.3 (a) measured from the point A is given by s(t) =

t 0

s (s ∈ R). ∥⃗b∥ dτ = ∥⃗b∥t (t ∈ IR), and t(s) = ∥⃗b∥

36 ■ Differential Geometry and Its Visualization

So a parametric representation of the straight line with respect to its arc length is given by ⃗b s (s ∈ R). ⃗x ∗ (s) = ⃗a + ∥⃗b∥ (b) The arc length along the circle line in Visualization 1.3.3 (b) is given by s(t) =

t 0

r dτ = rt (t ∈ [0, 2π]), and t(s) =

s (s ∈ [0, 2πr]). r

So a parametric representation of the circle line of Visualization 1.3.3 (b) with respect to the arc length is given by  

s ⃗x (s) = m ⃗ + ⃗e r cos r 1

∗

2

+ ⃗e r sin

 

s r

(s ∈ [0, 2πr]).

(c) The arc length along the helix of Visualization 1.3.3 (c) is given by s(t) =

t √

r2 + h2 dτ =

√

0

r2 + h2 t (t ∈ R)

and t(s) = √

r2

s (s ∈ R). + h2

√ We put ω = 1/ r2 + h2 . Then a parametric representation of the helix with respect to its arc length is given by ⃗x ∗ (s) = {r cos (ωs), r sin (ωs), hωs} (s ∈ R). (d) We consider the catenary of Visualization 1.3.7 with a parametric representation (1.21). Then it it follows from 

⃗x (t) = 1, sinh ′

 

t c

and 2

∥⃗x (t)∥ = 1 + sinh ′

that s(t) =

t 0

cosh

2

 

τ c

 

t c

= cosh

dτ = c sinh

and so t = t(s) = c sinh

−1

 

s . c

2

 

t c

 

t , c

Curves in Three-Dimensional Euclidean Space ■ 37

Figure 1.28

Suspended chains of constant length

Now

 

t c cosh c

=c



1 + sinh2

 

t c

=c



1+

s2 √ 2 = c + s2 , c2

and (1.21) yields the following parametric representation of the catenary with respect to its arc length s (Figure 1.28) 

⃗x (s) = c sinh ∗

−1

  √ s

c

c2

,

+

s2



(s ∈ R).

(1.59)

Example 1.5.4. We define the functions ϕ, h : (0, ∞) → R by 

ϕ(t) = log e + t



e2t



− 1 and h(t) =

Let γ be the curve with a parametric representation 

  

⃗x(t) = e−t cos ϕ(t), e−t sin ϕ(t), h(t)

1 − e−2t dt.

for t > 0.

(1.60)

Then the parametric representation of γ with respect to the arc length s along γ is ⃗x (s) = ∗

Proof. We write



1 1 cos s, sin s, s − tanh s cosh s cosh s





for s > 0.

(1.61)



⃗y (t) = −e−t cos ϕ(t), −e−t sin ϕ(t), h′ (t) and

⃗z(t) = e−t {− sin ϕ(t), cos ϕ(t), 0} ϕ′ (t), where 



1 e2t et ϕ (t) = t √ 2t et + √ 2t = t √ 2t e + e −1 e −1 e + e −1 t e 1 1 = √ 2t = ′ , =√ −2t h (t) e −1 1−e ′

√

e2t − 1 + et √ e2t − 1



38 ■ Differential Geometry and Its Visualization

hence, since obviously ⃗y (t) ⊥ ⃗z(t), ∥⃗x ′ (t)∥2 = ∥⃗y (t)∥2 + ∥⃗z(t)∥2 = e−2t + (h′ (t))2 + e−2t (ϕ′ (t))2 = 1 + = This implies

1 1 = ′ . 1 − e−2t (h (t))2

s(t) =



and so hence and then

∥⃗x ′ (t)∥ dt =



1 dt = ′ h (t)

es = exp ϕ(t) = et +





e−2t 1 − e−2t

ϕ′ (t) dt = ϕ(t),

e2t − 1,

e2s − 2es+t + e2t = e2t − 1, that is, e2s − 2es et = −1, et =

Finally, we obtain  

e2s + 1 = cosh s. 2es 

 

dt(s) dh dh(s) (t(s)) · ds = dt ds ds    1 1− ds = s − tanh s. = cosh2 s

h(t(s)) =

2

ds

Remark 1.5.5. Of course, the statement of Example 1.5.5 could have been verified by directly showing that ∥⃗x˙ (s)∥ = 1.

Visualization 1.5.6 (Cycloids, hypocycloids and epicycloids). Let C be a circle line of radius r > 0 and centre in M , and P ̸= M be a point attached to C. We write λ = d(M, P )/r, that is, λ < 1, λ = 1 or λ > 1 depending on whether P is inside of C, on C or outside of C. A cycloid is a curve that is generated by P as C rolls along a given curve γ without sliding. When γ is a straight line then the cycloid is said to be an ordinary cycloid (Figures 1.29 and 1.30). An epicycloid is the special case of a cycloid where C rolls along the outside of a given circle line C0 (Figure 1.31). A hypocycloid is the special case of a cycloid where C rolls along the inside of a given circle line C0 (Figure 1.32). (a) Let γ be the x1 –axis. Then the ordinary cycloid has a parametric ⃗x(t) = {r(t − λ sin t), r(1 − λ cos t)} (t ∈ R).

(1.62)

Curves in Three-Dimensional Euclidean Space ■ 39

Figure 1.29

The construction of an ordinary cycloid

Figure 1.30

Families of ordinary cycloids

Figure 1.31

Left: the construction of an epicycloid. Right: a family of epicycloids

40 ■ Differential Geometry and Its Visualization

Figure 1.32

Families of hypocycloids

(b) Let γ be the circle line C0 in the x1 x2 –plane with its centre in the origin and radius r0 . We write R = r0 + r for the epicycloid and R = r0 − r for the hypocycloid. Then the epicycloid and hypocycloid have parametric representations 







⃗x(t) = R cos t − λr cos and







(t ∈ R)

(1.63)







(t ∈ R).

(1.64)

R R t , R sin t − λr sin t r r

R R t , R sin t − λr sin t ⃗x(t) = R cos t + λr cos r r

(c) The arc length of the epicycloid in the special case λ = 1 is given by s(t) = 8





Rr r0 t sin2 r0 4r



for t ∈ 0,



2πr . r0

(1.65)

Proof. (a) Let P (t) denote the point attached to C and M (t) be the centre of C, both at t. Furthermore, let the initial position of M (0) and P (0) be given by M (0) = (0, r) and       3π 3π P (0) = (0, r(1 − λ)) = cos , r 1 + λ sin (Figure 1.29). 2 2 When the circle line C has rolled the distance rt along the positive x1 –axis, we obtain −→

−→

−→

OP (t) = OM (t) + M (t)P (t)      3π 3π = {rt, r} + λr cos − t , sin −t 2 2 = {rt, r} − λr {sin t, cos t} = {r(t − λ sin t), r(1 − λ cos t)}.

Curves in Three-Dimensional Euclidean Space ■ 41

(b) Let P (t) denote the point attached to C and M (t) be the centre of C, both at t. Furthermore let the initial position P (0) be given by P (0) = (r0 + (1 − r)λ, 0) (left in Figure 1.31). We have −→

−→

−→

OP (t) = OM (t) + M (t)P (t) and, with

ϕ=

obviously

r0 t, r

−→

OM (t) = (r0 + r){cos t, sin t} and

−→

M (t)P (t) = −λr{cos (t + ϕ), sin (t + ϕ)}.

−→

Putting ⃗x(t) = OP (t), we conclude 

⃗x(t) = R cos t − λr cos









R R t , R sin t − λr sin t r r

for all t ∈ R.

The parametric representation in (1.64) for a hypocycloid is obtained analogously. It follows from (1.63) that 









d⃗x R R (t) = R − sin(t) + λ sin t , cos t − λ cos t dt r r

(1.66)

and   2 2  2   d⃗  x R R 2  (t) = R − sin t + λ sin ( t) + − cos t + λ cos ( t)  dt  r r 



R R = R2 1 + λ2 − 2λ sin t sin ( t) − 2λ cos t cos ( t) r r 

= R2 1 + λ2 − 2λ cos (







R−r r0 t) = R2 1 + λ2 − 2λ cos ( t) . r r

If λ ̸= 1, then 1 + λ2 − 2λ cos ( rr0 t) ≥ (1 − λ)2 > 0, and the parametric representation is admissible for all t ∈ R. If λ = 1, then cos ( rr0 t) = 1 if and only if tk = 2kπr (k ∈ Z), and the r0 parametric representation is not admissible if and only if t = tk for k ∈ Z (Figure 1.32).

42 ■ Differential Geometry and Its Visualization

Figure 1.33

Special hypocycloids

(c) Now let λ = 1. Then we have

and so

    √   d⃗  x  (t) = R 2 1 − cos ( r0 t) = 2R sin ( r0 t) for t ∈ 0, 2πr ,  dt  r 2r r0

s(t) =

t 0

=4 From we obtain



Rr r0 r 0 t 2R sin ( τ ) dτ = −4 cos ( τ ) 2r r0 2r 0 



Rr r0 Rr r0 1 − cos ( t) = 8 sin2 ( t). r0 2r r0 4r 

(1.67)



r r0 2πr s(t) = 8R sin2 ( t) for t ∈ 0, , r0 4r r0 4r t(s) = sin−1 r0







r0 r , s for s ∈ 0, 8R 8rR r0

and this yields a parametric representation of the epicycloid with respect to its arc length.

Remark 1.5.7. Ellipses and astroids are special cases of hypocycloids for r0 = 2r and arbitrary λ, and for r0 = 4r and λ = 1, respectively (Figure 1.33). It is also interesting to combine the principles of generating ordinary cycloids and epicycloids.

Curves in Three-Dimensional Euclidean Space ■ 43

Figure 1.34

The construction of the curve of Visualization 1.5.8

Visualization 1.5.8. Let C1 be a circle line of radius r1 that rolls along the x1–axis, and C2 be a circle line of radius r2 that rolls along the outside of C1 at the same time (Figure 1.34). We put R = r1 + 2r2 . Then the curve γ of the point P attached to the circle line C2 by this movement has a parametric representation ⃗x(t) = {r1 (t − sin (2t)), r1 (1 − cos (2t))}−        R R − r2 sin (2t) + sin t , r2 cos (2t) + cos t (1.68) r2 r2

Since it is often the case that planar curves are given in polar coordinates, it is useful to have a formula for their arc lengths. Proposition 1.5.9. Let γ be a planar curve with a parametric representation ⃗x(ϕ) = {ρ(ϕ) cos ϕ, ρ(ϕ) sin ϕ} (ϕ ∈ I = (ϕ0 , ϕ1 ) ⊂ (0, 2π)) .

Then the arc length along γ is s(ϕ, ϕ0 ) =

ϕ 

ϕ0

ρ 2 (φ) + (ρ ′ (φ))2 dφ (ϕ ∈ I), where ρ ′ (ϕ) =

dρ(ϕ) . dϕ

(1.69)

Proof. We obtain ⃗x ′ (ϕ) = {ρ ′ (ϕ) cos ϕ − ρ(ϕ) sin ϕ, ρ ′ (ϕ) sin ϕ + ρ(ϕ) cos ϕ}

and 2

∥⃗x ′ (ϕ)∥ = (ρ ′ (ϕ))2 cos2 ϕ − 2ρ(ϕ)ρ ′ (ϕ) sin ϕ cos ϕ + ρ 2 (ϕ) sin2 (ϕ)

+ (ρ ′ (ϕ))2 sin2 ϕ + 2ρ(ϕ)ρ ′ (ϕ) sin ϕ cos ϕ + ρ 2 (ϕ) cos2 (ϕ)

= (ρ (ϕ))2 + ρ 2 (ϕ) ′

and (1.69) is an immediate consequence.

44 ■ Differential Geometry and Its Visualization

Example 1.5.10. (a) First we consider the rosette of Visualization 1.4.7 with a parametric representation (1.51). Then we have ρ(ϕ) = (a/2) sin (2ϕ), ρ ′ (ϕ) = a cos (2ϕ) and a2 4 a2 = 4

ρ 2 (ϕ) + (ρ (ϕ))2 = ′

 



sin2 (2ϕ) + 4 cos2 (2ϕ) 

1 + 3 cos2 (2ϕ) .

Thus we obtain for the arc length along the rosette by (1.69) a s(ϕ, 0) = 2

ϕ  0

1 + 3 cos2 (2φ) dφ (ϕ ∈ (0, 2π)).

(1.70)

(b) Now we consider the double egg line of Visualization 1.4.8 with a parametric ′ representation (1.53). Then we have ρ(ϕ) = a cos2 ϕ, ρ (ϕ) = −2a sin ϕ cos ϕ and 

ρ 2 (ϕ) + (ρ (ϕ))2 = a2 cos2 ϕ cos2 ϕ + 4 sin2 ϕ ′





= a2 cos2 ϕ 1 + 3 sin2 ϕ .



Thus we obtain for the arc length along the double egg line by (1.69) s(ϕ, 0) = a

ϕ0 0

1.6



| cos φ| 4 − 3 cos2 φ dφ (ϕ ∈ (0, 2π)).

(1.71)

THE VECTORS OF THE TRIHEDRON OF A CURVE

From now on we confine our studies of curves to three-dimensional E 3 . With every curve, we may associate the vectors of the trihedron or the frame at any of its points. We start with the tangent vectors of a curve. The tangent to a curve γ at a point X is a first order approximation of γ at X by a straight line, and gives the direction of γ at X. Let γ be a curve with a parametric representation ⃗x(t) (t ∈ [a, b]). We consider the difference vector for t, t + h ∈ I, 



⃗x(t + h) − ⃗x(t) = x1 (t + h) − x1 (t), x2 (t + h) − x2 (t), x3 (t + h) − x3 (t) , and divide it by h to obtain ⃗x(t + h) − ⃗x(t) = h



x1 (t + h) − x1 (t) x2 (t + h) − x2 (t) x3 (t + h) − x3 (t) , , h h h

If the component functions xk (k = 1, 2, 3) are differentiable, then this yields d⃗x ⃗x(t + h) − ⃗x(t) = ⃗x ′ (t) = (t) = lim h→0 h dt





dx2 dx3 dx1 (t), (t), (t) . dt dt dt



.

Curves in Three-Dimensional Euclidean Space ■ 45

Definition 1.6.1 (Tangent, tangent vector). Let γ be a curve in R3 with a parametric representation ⃗x(t) (t ∈ I) and t0 ∈ I. (a) If the vector   dx1 dx2 dx3 ′ ⃗x (t0 ) = (t0 ), (t0 ), (t0 ) dt dt dt ′ exists and ⃗x (t0 ) ̸= ⃗0, then the unit vector

′ ⃗t(t0 ) = ⃗x (t0 ) ∥⃗x ′ (t0 )∥

is called the tangent vector of γ at the point X(t0 ). (b) The straight line through the point X(t0 ) in the direction of the tangent vector ⃗t(t0 ) is called the tangent of γ at the point X(t0 ); it has a parametric representation ⃗xt0 (λ) = ⃗x(t0 ) + λ⃗t(t0 ) (λ ∈ R). Visualization 1.6.2 (The tractrix). Let γ and γ ∗ be a curve and a straight line in a plane that have no points of intersection. Given a point P on γ, let P ∗ denote the point of intersection of γ ∗ with the tangent to γ at P . If the distances between the points P and P ∗ have a constant value d, then the curve γ is called a tractrix (Figure 1.35). We introduce a Cartesian coordinate system in the plane such that the straight line γ ∗ is the y–axis. Then γ has a parametric representation 





⃗x(ϕ) = d sin ϕ, d log tan

 

ϕ 2

+ cos ϕ



(ϕ ∈ (0, π/2)).

(1.72)

Proof. Let ⃗x(t) = {t, y(t)} be a parametric representation of γ. Then the tangent to γ at t is given by ⃗x(t) + λ{1, y ′ (t)} (λ ∈ R), and so we obtain for P ∗ −→

OP ∗ = {0, p∗ } = {t, y(t)} + λ{1, y ′ (t)}, that is, λ = −t. We observe that γ ∩ γ ∗ = ∅ implies d > 0, hence we have



d = d(P, P ∗ ) = ∥⃗x(t) − (⃗x(t) − t{1, y ′ (t))}∥ = |t| 1 + (y ′ (t))2 , and so, since |t| > 0,

y (t) = ± ′

√

d 2 − t2 for |t| < d. |t|

We choose the upper sign and t ∈ (0, d), and obtain  √ 2 d − t2 y(t) = dt. t

(1.73)

(1.74)

46 ■ Differential Geometry and Its Visualization

Figure 1.35

Left: the construction of a tractrix. Right: a family of tractrices

Since t ∈ (0, d), we may put t = d · sin ϕ for ϕ ∈ (0, π/2) and this yields  √ 2   2 d − t2 d − d 2 sin2 ϕ dt = d · cos ϕ dϕ t d · sin ϕ  cos2 ϕ dϕ = d · (I1 + I2 ) , =d· sin ϕ

where

I1 = and I2 = −



sin ϕ dϕ = cos ϕ =





(1.75)

dϕ sin ϕ

2

1 − sin ϕ + c2 =

√

d 2 − t2 + c2 , d

(1.76)

where c2 is a constant of integration. To evaluate the integral I1 , we make the substitution z = tan (ϕ/2). Then z ∈ (0, 1) and we obtain dϕ d 2 = (2 tan−1 z) = , dz dz 1 + z2 sin ϕ = 2 sin (ϕ/2) cos (2ϕ/2) = 2 tan (ϕ/2) cos2 (ϕ/2) cos2 (ϕ/2) cos2 (ϕ/2) + sin2 (ϕ/2) 2 tan (ϕ/2) 2z = = 2 1 + tan (ϕ/2) 1 + z2

= 2 tan (ϕ/2)

and I1 =



1 + z2 2 dz = 2z 1 + z 2





 

dz ϕ = log z + c1 = log tan z 2

+ c1 ,

(1.77)

Curves in Three-Dimensional Euclidean Space ■ 47

where c1 is a constant of integration. We put c = d · (c1 + c2 ). Then it follows from (1.74), (1.75), (1.76) and (1.77) that 



 

ϕ y(ϕ) = d · log tan 2



+ cos ϕ + c.

If we choose c such that limϕ→π/2 y(ϕ) = 0, then we have c = 0, and (1.72) is an immediate consequence. We now assume that γ is a curve with a parametric representation ⃗x(s) (s ∈ I) of class C r (I) (r ≥ 2). Then the derivative ⃗x˙ (s) is a unit vector by Proposition 1.5.2 (d), which means that the movement along the curve takes place at a constant value of speed equal to one (Figure 1.27). Since ⃗x ∈ C r (I) (r ≥ 2), the derivative ⃗x¨ is a continuous function on the interval I, and ⃗x¨(s) stands for the speed of change in the direction of γ at s. Furthermore, if ⃗x¨(s) ̸= ⃗0, then we may consider the unit vectors ⃗x¨(s) ⃗x˙ (s) × ⃗x¨(s) and . ∥⃗x¨(s)∥ ∥⃗x¨(s)∥

Definition 1.6.3 (The vectors of the trihedron and curvature). Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I) of class C r (I) (r ≥ 2), and ⃗x¨(s) ̸= ⃗0. Then the vectors ⃗x¨(s) ⃗v1 (s) = ⃗x˙ (s), ⃗v2 (s) = and ⃗v3 (s) = ⃗v1 (s) × ⃗v2 (s) ∥⃗x¨(s)∥

are called the tangent, principal normal and binormal vectors of γ at s, respectively. Attached to the point X(s) of the curve, the vectors ⃗vk (s) (k = 1, 2, 3) form the trihedron or frame of the curve at s (Figure 1.37). The vector ⃗x¨(s) is called vector of curvature, its length κ(s) = ∥⃗x¨(s)∥ curvature and ρ(s) = 1/κ(s) for κ(s) ̸= 0 radius of curvature of the curve at s.

Remark 1.6.4. (a) In many textbooks, the notations ⃗t, ⃗n and ⃗b are used instead of ⃗v1 , ⃗v2 and ⃗v3 . Obviously, we have ⃗vi (s) • ⃗vk (s) = δik =



1 (i = k) 0 (i = ̸ k)

for all s.

(1.78)

(b) The curvature of a curve at a point is a measure of the deviation of the curve from a straight line at that point. Remark 1.6.5. The trihedron of a curve is only defined at points where ⃗x¨(s) ̸= ⃗0. We shall exclude curves or segments of curves with ⃗x¨ ≡ ⃗0 in which case ⃗x(s) = ⃗as +⃗b for some constant vectors ⃗a and ⃗b. The corresponding curves are straight lines.

48 ■ Differential Geometry and Its Visualization

Visualization 1.6.6. (a) We consider the catenary of Example 1.5.3 (d) with a parametric representation (1.59) in E3 

⃗x (s) = c sinh ∗

−1

  √ s

,

c

s2

+



c2 , 0

(s ∈ R)

with respect to the natural parameter s, where c is a positive constant. Then we obtain ⃗v1∗ (s)

= ⃗x˙ ∗ (s) =





s c √ ,√ ,0 2 2 2 s +c s + c2

for the tangent vectors at s (left in Figure 1.36), ¨∗

⃗x (s) =





1 s2 √ − , − ,0 s2 + c2 (s2 + c2 )3/2 (s2 + c2 )3/2 cs

=

c (s2

+ c2 )3/2

s2

c + c2

{−s, c, 0}

for the vector of curvature at s (left in Figure 1.36), κ ∗ (s) = ∥⃗x¨ ∗ (s)∥ =

c (s2 +

c2 )3/2

√

s2 + c2 =

for the curvature at s, ⃗v2∗ (s) =

1 ⃗x¨ ∗ (s) s2 + c2 c {−s, c, 0} {−s, c, 0} = √ = 3/2 ∗ 2 ¨ c ∥⃗x (s)∥ s + c2 (s2 + c2 )

for the principal normal vector at s, and ⃗v3∗ (s) = ⃗v1∗ (s) × ⃗v2∗ (s) = =

s2

1 {c, s, 0} × {−s, c, 0} + c2

 1  2 2 0, 0, s + c = ⃗e 3 s2 + c2

for the binormal vector at s. (b) We consider the tractrix of Visualization 1.6.2. Writing c = −1/d, we obtain from (1.73) √ 1 1 − c2 t2 ′ y (t) = for 0 < t < − , |c|t c hence

and so

∥⃗x ′ (t)∥2 = 1 + s(t) =

1 |c|

t

t0

1 − c2 t2 1 = 2 2, 2 2 c t c t

1 dt = (log t − log t0 ). t |c|

Here, we may choose t0 = 1 and then the arc length along the tractrix is given by 



1 1 1 s(t) = log t for s < log . |c| |c| |c|

Curves in Three-Dimensional Euclidean Space ■ 49

Tangent vectors and vectors of curvature of catenaries (left) and tractrices (right)

Figure 1.36

Since we have t(s) = exp (|c|s) and (dt/ds)(s) = |c| exp (|c|s), and writing y ∗ (s) = y(t(s)), we obtain dt y˙ (s) = y (t(s)) (s) = ds ∗

′



and consequently ⃗v1∗ (s)

 1 − c2 exp (2|c|s) |c| exp (|c|s) = 1 − c2 exp (2|c|s), |c| exp (|c|s)



= |c| exp (|c|s),



1−

c2



exp (2|c|s), 0

for the tangent vector at s in E3 (right in Figure 1.36), ⃗x¨ ∗ (s) =





c2 |c| exp (2|c|s) ,0 c exp (|c|s), −  1 − c2 exp (2|c|s) 2





−|c| exp (|c|s) ,0 = c2 exp (|c|s) 1,  1 − c2 exp (2|c|s)

for the vector of curvature at s (right in Figure 1.36), 

c2 exp (|c|s) 1 − c2 exp (2|c|s) + c2 exp (2|c|s)  κ (s) = 1 − c2 exp (2|c|s) c2 exp (|c|s) = 1 − c2 exp (2|c|s) ∗

50 ■ Differential Geometry and Its Visualization

for the curvature at s, and   ⃗x¨(s) ∗ 2 1 − c exp (2|c|s), −|c| exp (|c|s), 0 ⃗v2 (s) = = ∥⃗x¨(s)∥

for the principal normal vector, and ⃗v3∗ (s) = ⃗e 3 (s) for the binormal vector at s. Visualization 1.6.7. (a) For the circle line in Example 1.5.3 (b) with a parametric representation s s ⃗x(s) = m ⃗ 0 + ⃗e 1 r cos + ⃗e 2 r sin (s ∈ [0, 2πr]), r r we have s s ⃗v1 (s) = −⃗e 1 sin + ⃗e 2 cos for the tangent vector, r r s s for the principal normal vector, ⃗v2 (s) = −⃗e 1 cos − ⃗e 2 sin r r ⃗v3 (s) = ⃗e 3 = {0, 0, 1} for the binormal vector, s 1 s 1 for the vector of curvature ⃗x¨(s) = − ⃗e 1 cos − ⃗e 2 sin r r r r and 1 for the curvature. r (b) For the helix in Example 1.5.3 (c) with a parametric representation κ(s) =

⃗x(s) = {r cos ωs, r sin ωs, hωs} (s ∈ R),

we have

and

⃗v1 (s) = ω{−r sin ωs, r cos ωs, h}

for the tangent vector,

⃗v2 (s) = −{cos ωs, sin ωs, 0}

for the principal normal vector,

⃗v3 (s) = ω{h sin ωs, −h cos ωs, r} for the binormal vector, ⃗x¨(s) = −ω 2 r{cos ωs, sin ωs, 0} for the vector of curvature κ(s) = rω 2 =

r r2 + h2

for the curvature

(Figure 1.37). Visualization 1.6.8. The vectors of the trihedra and the curvature of the curve γ of Example 1.5.4 with the parametric representation (1.61) are given by (Figure 1.38) ⃗v1 (s) = −

sinh s 1 {cos s, sin s, − sinh s + {− sin s, cos s, 0}, 2 cosh s cosh (s)

(1.79)

⃗v2 (s) = −

sinh s 1 {cos s, sin s, − sinh s} − {− sin s, cos s, 0}, 2 cosh s cosh (s)

(1.80)

⃗v3 (s) =

1 {sinh s cos s, sinh s sin s, 1} cosh s

(1.81)

Curves in Three-Dimensional Euclidean Space ■ 51

Figure 1.37

The vectors of a trihedron of a helix

Figure 1.38

The vectors of a trihedron of the curve of Visualization 1.6.8

and κ(s) =

2 . cosh s

Proof. We write ϕ(s) = 1/ cosh s, omit the argument in ϕ and obtain from ⃗x(s) = {ϕ cos s, ϕ sin s, s − tanh s} for s > 0 that ⃗v1 (s) = ⃗x˙ (s) = {ϕ˙ cos s, ϕ˙ sin s, 1 − ϕ2 } + {−ϕ sin s, ϕ cos s, 0} 1 sinh s {− sin s, cos s, 0} {cos s, sin s, − sinh s} + =− 2 cosh s cosh (s) which is (1.79).

(1.82)

52 ■ Differential Geometry and Its Visualization

This yields ˙ + 2ϕ{− ˙ ⃗x¨ ∗ (s) = {(ϕ¨ − ϕ) cos s, (ϕ¨ − ϕ) sin s, −2ϕϕ} sin s, cos s, 0}. It follows from

sinh s ϕ˙ = − = −ϕ2 sinh s 2 cosh (s)

that ˙ sinh s = −ϕ + 2ϕ3 sinh2 s ϕ¨ = −ϕ2 cosh s − 2ϕϕ = −ϕ + 2ϕ − 2ϕ3 = ϕ − 2ϕ3

and

ϕ¨ − ϕ = −2ϕ3 .

Therefore, we have ⃗x¨ ∗ (s) = −

sinh s 2 {cos s, sin s, − sinh s} − 2 {− sin s, cos s, 0}. 3 cosh (s) cosh2 (s)

Now we obtain for the curvature κ(s) along γ 

1 + sinh2 (s) sinh2 (s) + κ2 (s) = ∥⃗x¨ ∗ (s)∥2 = 4 cosh6 (s) cosh4 (s) 1 4 (1 + sinh2 (s)) = , =4 cosh4 (s) cosh2 (s) hence

κ(s) =



2 cosh s

which is (1.82). Now the principal normal vectors ⃗v2 (s) along γ are given by ⃗v2 (s) =

sinh s 1 1 ¨∗ {cos s, sin s, − sinh s} − ⃗x (s) = − {− sin s, cos s, 0}, 2 κ(s) cosh s cosh (s)

which is (1.80). Furthermore, since ⃗b(s) = {cos s, sin s, − sinh s} × {− sin s, cos s, 0} = {sinh s cos s, sinh s sin s, 1}, the binormal vectors ⃗v3 (s) along γ are given by ⃗v3 (s) = ⃗v1 (s) × ⃗v2 (s) = = which is (1.81).



1 sinh2 (s) + 3 cosh (s) cosh3 (s)

1 {sinh s cos s, sinh s sin s, 1} cosh s



⃗b

Curves in Three-Dimensional Euclidean Space ■ 53

1.7

FRENET’S FORMULAE

It follows from the identities in (1.78) that if γ is a curve with a parametric representation ⃗x(s) (s ∈ I) such that ⃗x¨(s) ̸= ⃗0, then the vectors ⃗vk (s) (k = 1, 2, 3) of the trihedron of γ are a basis of E3 for every s ∈ I. Consequently every vector of the space E3 can be represented as a unique linear combination of the vectors ⃗vk (s), in particular, this holds true for the derivatives ⃗v˙ k (s) of the vectors ⃗vk (s). The three formulae that give the vectors ⃗v˙ j (s) for j = 1, 2, 3 as linear combinations of the vectors ⃗vk (s) (k = 1, 2, 3) are called Frenet’s formulae. They are of the utmost importance in the theory of curves. Theorem 1.7.1 (Frenet’s formulae). Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I) such that ⃗x¨(s) ̸= 0 for all s ∈ I. We put τ (s) = ⃗v˙ 2 (s) • ⃗v3 (s) (s ∈ I),

and obtain

 ⃗v˙ 1 (s) =   

⃗v˙ (s) = −κ(s)⃗v1 (s)

2    ˙ ⃗v3 (s) =

κ(s)⃗v2 (s) +

τ (s)⃗v3 (s)

−τ (s)⃗v2 (s)

The formulae in (1.84) are called Frenet’s formulae.

(1.83)       

.

(1.84)

 Proof. We write ⃗v˙ k (s) = 3j=1 akj (s)⃗vj (s) (k = 1, 2, 3) with the functions akj (k, j = 1, 2, 3) to be determined, and omit the parameter s. First ⃗vk • ⃗vk = 1 for k = 1, 2, 3 implies d (⃗vk • ⃗vk ) = 2⃗v˙ k • ⃗vk = 0 (k = 1, 2, 3) ds and since ⃗vj • ⃗vk = 0 for j ̸= k, we conclude

0 = ⃗v˙ k • ⃗vk =

3  j=1

akj ⃗vj • ⃗vk = akk for k = 1, 2, 3.

Furthermore, we have for l ̸= k 0= =

d (⃗vk • ⃗vl ) = ⃗v˙ k • ⃗vl + ⃗vk • ⃗v˙ l ds

3  j=1

akj ⃗vj • ⃗vl +

that is akl = −alk . Finally, we obtain

3  j=1

alj ⃗vj • ⃗vk = akl + alk ,

a12 = ⃗v˙ 1 • ⃗v2 = ⃗x¨ • ⃗v2 = κ⃗v2 • ⃗v2 = κ, and a23 = ⃗v˙ 2 • ⃗v3 = τ by definition.

54 ■ Differential Geometry and Its Visualization

Frenet’s formulae may formally be written as  



⃗v˙ 1 0 κ ˙   −κ 0 = ⃗v2   ˙⃗v3 0 −τ

 

0 ⃗v1   τ  ⃗v2  . 0 ⃗v3

Now we define the torsion of a curve, and give some elementary results. Definition 1.7.2 (Torsion of a curve). Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I) such that ⃗x¨(s) ̸= ⃗0 for all s ∈ I. Then the value τ (s) = ⃗v˙ 2 (s) • ⃗v3 (s)

defined by the formula in (1.83) is called the torsion of γ at the point X(s). Example 1.7.3. We consider the helix of Visualization 1.6.7 (b). Then we have ⃗v2 (s) = −{cos ωs, sin ωs, 0},

⃗v3 (s) = ω{h sin ωs, −h cos ωs, r}, ⃗v˙ 2 (s) = ω{sin ωs, − cos ωs, 0}

and τ (s) = ⃗v˙ 2 (s) • ⃗v3 (s) = hω 2 =

h (s ∈ R). r2 + h2

for the torsion at s. Example 1.7.4. The torsion of the curve of Visualization 1.6.8 is given by τ (s) = 1.

(1.85)

Proof. We obtain from the definition of τ in Definition 1.7.2, and (1.80) and (1.81) τ (s) = −⃗v2 (s) • ⃗v˙ 3 (s) = =

1 sinh2 s 2 (1 + sinh (s)) + cosh4 (s) cosh2 s

1 2 2 (1 + sinh s) = 1. cosh s

Proposition 1.7.5. Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I) such that ⃗x¨(s) ̸= ⃗0 for all s ∈ I. Then the torsion τ of γ is given by ... ⃗x˙ (s) • (⃗x¨(s) × ⃗x (s)) τ (s) = (s ∈ I). (1.86) κ2 (s)

Curves in Three-Dimensional Euclidean Space ■ 55

Proof. We have by definition and the second identity in (1.8) of Proposition 1.2.3     ... ⃗x ⃗x˙ × ⃗x¨ d ⃗x¨ 1 ˙⃗v2 = d ¨ = , ⃗v3 = ⃗v1 × ⃗v2 = + ⃗x ds ∥⃗x¨∥ ds ∥⃗x¨∥ ∥⃗x¨∥ ∥⃗x¨∥ and

... ... ˙ × ⃗x¨) ⃗ x • ( ⃗ x ⃗x • (⃗x˙ × ⃗x¨) τ = ⃗v˙ 2 • ⃗v3 = = , κ2 ∥⃗x¨∥2

since

d ⃗x¨ ds



1 ∥⃗x¨∥



⊥ (⃗x˙ × ⃗x¨).

In many cases, it is useful to have formulae for the vectors of the trihedron, and the curvature and torsion of a curve given by a parametric representation with respect to an arbitrary parameter. Proposition 1.7.6. Let γ be a curve with a parametric representation ⃗x(t) (t ∈ I) and non–vanishing vector of curvature where t is an arbitrary parameter. Then the vectors of the trihedron, and the curvature and torsion of γ are given by ⃗x ′ (t) , ∥⃗x ′ (t)∥ ⃗x ′ (t) × (⃗x ′′ (t) × ⃗x ′ (t)) ⃗v2 (t) = ∥⃗x ′ (t) × (⃗x ′′ (t) × ⃗x ′ (t))∥ ⃗v1 (t) =

= ⃗v3 (t) = and κ(t) = for all t ∈ I.

⃗x ′′ (t)∥⃗x ′ (t)∥2 − ⃗x ′ (t)(⃗x ′ (t) • ⃗x ′′ (t)) , ∥⃗x ′ (t)∥ ∥⃗x ′ (t) × ⃗x ′′ (t)∥ ⃗x ′ (t) × ⃗x ′′ (t) , ∥⃗x ′ (t) × ⃗x ′′ (t)∥

⃗x ′ (t) • (⃗x ′′ (t) × ⃗x ′′′ (t)) ∥⃗x ′ (t) × ⃗x ′′ (t)∥ and τ (t) = ∥⃗x ′ (t)∥3 ∥⃗x ′ (t) × ⃗x ′′ (t)∥2

(1.87)

(1.88)

Proof. We write ⃗x ∗ (s) = ⃗x(t(s)) and ⃗vk (t) = ⃗vk∗ (s) for k = 1, 2, 3. Applying the chain rule in (1.10) of Proposition 1.2.3 and using the fact that (dt/ds)(s) = ∥⃗x ′ (t)∥ as in the proof of Proposition 1.5.2, we first obtain ⃗v1 (t) = ⃗v1∗ (s) = ⃗x˙ ∗ (s) =

d⃗x dt ⃗x ′ (t) d⃗x(t(s)) = (t(s)) (s) = , ds dt ds ∥⃗x ′ (t)∥

which is the identity for ⃗v1 (t) in (1.87).

56 ■ Differential Geometry and Its Visualization

We apply the chain rule again and get, using the second identity in (1.8) of Proposition 1.2.3 



d dt d⃗v1∗ (s) = ⃗x¨ ∗ (s) = ⃗x ′ (t(s)) (s) ds ds ds  2 2 d ⃗x d2 s ds (s) = 2 (t(s)) + ⃗x ′ (t(s)) 2 (s). dt dt dt

(1.89)

Now we apply the chain rule and the third identity in (1.8) of Proposition 1.2.3 to obtain d d2 s (s) = dt2 dt









1  ′ ⃗x (t(s)) • ⃗x ′ (t(s)) d (⃗x ′ (t(s)) • ⃗x ′ (t(s))) =−  3 ′ ′ ds 2 ⃗x (t(s)) • ⃗x (t(s))

=−

1 ∥⃗x ′ (t(s))∥ 1

d = dt

d⃗ x′ ds x ′ (t(s)) dt (t(s)) dt (s) • ⃗ ∥⃗x ′ (t(s))∥3

=−

⃗x ′′ (t(s)) • ⃗x ′ (t(s)) . ∥⃗x ′ (t(s))∥4

It follows from (1.89) and identity (1.5) in Proposition 1.1.3 that

⃗x ′′ (t(s)) ⃗x ′ (t(s))(⃗x ′ (t(s)) • ⃗x ′′ (t(s))) d⃗v1∗ (s) = − ds ∥⃗x ′ (t(s))∥2 ∥⃗x ′ (t(s))∥4 ⃗x ′′ (t(s))∥⃗x ′ (t(s))∥2 − ⃗x ′ (t(s))(⃗x ′ (t(s)) • ⃗x ′′ (t(s))) = ∥⃗x ′ (t(s))∥4 ′ ′′ ⃗x (t(s)) × (⃗x (t(s)) × ⃗x ′ (t(s))) = . ∥⃗x ′ (t(s))∥4

(1.90)

Since ⃗v2∗ (s) is a unit vector by definition, the last identity yields ⃗v2∗ (s) = ⃗v2 (t) =

⃗x ′ (t) × (⃗x ′′ (t) × ⃗x ′ (t)) , ∥⃗x ′ (t) × (⃗x ′′ (t) × ⃗x ′ (t))∥

which is the first identity for ⃗v2 (t) in (1.87). Furthermore, identity (1.7) in Proposition 1.1.3 yields 2

∥⃗x ′ (t) × (⃗x ′′ (t) × ⃗x ′ (t))∥2 = ∥⃗x ′ (t)∥2 ∥⃗x ′′ (t) × ⃗x ′ (t)∥2 − (⃗x ′ (t) • (⃗x ′′ (t) × ⃗x ′ (t))) = ∥⃗x ′ (t)∥2 ∥⃗x ′ (t) × ⃗x ′′ (t)∥2 ,

(1.91)

and we obtain from the third term in (1.90) ⃗v2 (t) =

⃗x ′′ (t)∥⃗x ′ (t)∥2 − ⃗x ′ (t)(⃗x ′ (t) • ⃗x ′′ (t)) , ∥⃗x ′ (t)∥ ∥⃗x ′ (t) × ⃗x ′′ (t)∥

which is the second term in identity (1.87) for ⃗v2 (t). We have ⃗v3 (t) = ⃗v1∗ (s) × ⃗v2∗ (s) by definition, and since ⃗x ′ (t) × ⃗x ′ (t) = ⃗0, the identity for ⃗v3 (t) in (1.87) immediately follows from the identity for ⃗v1 (t) and the second identity for ⃗v2 (t) in (1.87).

Curves in Three-Dimensional Euclidean Space ■ 57

It follows from the definition of the curvature, the last identity in (1.90), and from (1.90) and (1.7) in Proposition 1.1.3 that d⃗v ∗ ∥⃗x ′ (t)∥2 ∥⃗x ′ (t) × ⃗x ′′ (t)∥2 d⃗v1∗ (s) • 1 (s) = ds ds ∥⃗x ′ (t)∥8 ∥⃗x ′ (t)∥2 ∥⃗x ′′ (t)∥2 − (⃗x ′ (t) • ⃗x ′′ (t))2 = , ∥⃗x ′ (t)∥6

κ2 (t) =

which yields the identity for κ(t) in (1.88). Finally, we apply the chain rule in (1.89) and obtain  3 ... ∗ d3⃗x ds ⃗x (s) = 3 (t(s)) (s) + α(t(s))⃗x ′′ (t(s)) + β(t(s))⃗x ′ (t(s)). dt dt

Since the identity (1.86) for the torsion in Proposition 1.7.5 contains the expres... ∗ sion ⃗x¨ ∗ (s) × ⃗x (s), and ⃗x¨ ∗ (s) is a linear combination of ⃗x ′ (t(s)) and ⃗x ′′ (t(s)) by (1.89), we do not have to work out the expressions for α(t(s)) and β(t(s)), and obtain 

d3⃗x d2⃗x (t(s)) × (t(s)) = dt2 dt3



2

d2⃗x dt (s) (t(s)) 2 dt ds



d2 t d⃗x d3⃗x (t(s)) 2 (s) × 3 (s). + dt ds dt

Thus it follows from (1.86) and the identity for κ(t) that τ (t) = τ ∗ (s) =

⃗x ′ (t) • (⃗x ′′ (t) × ⃗x ′′′ (t)) ⃗x ′ (t) • (⃗x ′′ (t) × ⃗x ′′′ (t)) = . κ(t)2 ∥⃗x (t)∥6 ∥⃗x ′ (t) × ⃗x ′′ (t)∥2

This completes the proof.

Visualization 1.7.7. The curvature of the epicycloid in Visualization 1.5.6 (b) for all λ > 0 is given by (Figure 1.39)

κ(t) =

   1 + λ2 R  r − λ(1 +    2

R(1 + λ   2r + r0    4r(r + r0 )

−

 



R r0 r ) cos r t 2λ cos ( rr0 t))3/2

1

for all t ∈ R

2kπr ·   r0  for t ̸= (k ∈ Z) sin 2r t r0

if λ ̸= 1

(1.92)

if λ = 1.

Proof. We consider the epicycloid as a curve in the x1 x2 –plane of E3 . Then we obtain from (1.66) 

⃗x ′′ (t) = R − cos t +











λR R R λR cos t , − sin t + sin t ,0 r r r r

and then ⃗x ′ (t) × ⃗x ′′ (t) = ⃗e 3 R2





 





λR R sin t − r r       λR R R t cos t − cos t + − cos t − λ cos r r r

− sin t + λ sin

R t r

− sin t +

58 ■ Differential Geometry and Its Visualization











R R R − λ sin t sin t − λ sin t sin t − = ⃗e R 1 + λ r r r r     R R R − λ cos t cos t − λ cos t cos t r r r 3

2

2R



= ⃗e 3 R2 1 + λ2









R R r0 −λ 1+ t cos r r r

.

Now we have by the first identity in (1.88) of Proposition 1.7.6   1 + λ2 R ∥⃗x (t) × ⃗x (t)∥ r − λ(1 + κ(t) = =  ′ 3 ′

′′

∥⃗x (t)∥

R 1 + λ2 −

and, in the special case λ = 1, κ(t) = =

1.8

   1 − cos r0 t  2r + r0 r √   r0 3/2 =

r+R r

R · 2 2 1 − cos

r

t



 

R r0 r ) cos r t  3/2 2λ cos rr0 t

,

1 √  r(r + r0 ) 2 2 1 − cos  r0 t r

2r + r0 2πkr 1   r0  for t ̸= (k ∈ Z).   4r(r + r0 ) sin 2r t r0

(1.93)

THE GEOMETRIC SIGNIFICANCE OF CURVATURE AND TORSION

We already know from Remark 1.6.4 (b) in Section 1.6 that the curvature of a curve is a measure of its deviation from a straight line (Figure 1.39). The longer the vector

Figure 1.39

An epicycloid and its vectors of curvature

Curves in Three-Dimensional Euclidean Space ■ 59

of curvature, that is, the larger the curvature, the faster the curve moves away from its tangent. A similar geometric interpretation of the torsion can be obtained from the following theorem. Theorem 1.8.1. Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I), and ⃗x¨(s) ̸= ⃗0 for all s ∈ I. Then γ is a planar curve if and only if τ (s) = 0 for all s ∈ I. Proof.

(i) First we assume τ ≡ 0. Then ⃗v˙ 3 (s) = ⃗0 for all s by Frenet’s third formula in (1.83), and so ⃗v3 = ⃗c for some constant vector ⃗c. Now ⃗v1 • ⃗v3 = 0 implies d (⃗x • ⃗c) = ⃗v1 • ⃗c = ⃗v1 • ⃗v3 = 0, that is, ⃗x • ⃗c = const. ds Therefore, it follows that (⃗x(s) − ⃗x0 ) • ⃗c = 0 for all s with a suitable constant vector ⃗x0 . This shows that ⃗x(s) is in the plane through the point X0 and orthogonal to ⃗c. (ii) Conversely, we assume that ⃗x(s) is a planar curve. Then we can write ⃗x(s) = ⃗x0 + α(s)⃗a + β(s)⃗b, where the vectors ⃗a and ⃗b are independent of s. Differentiating this identity, we see that the vectors ⃗v1 , ⃗v2 , ⃗v˙ 1 , ⃗v˙ 2 are in the plane spanned by the vectors ⃗a and ⃗b, and since ⃗v3 = ⃗v1 × ⃗v2 by definition, it follows that τ (s) = ⃗v˙ 2 • ⃗v3 = 0 for all s.

Thus the torsion of a curve is a measure for its deviation from a plane (Figure 1.40). The longer the vector τ⃗v3 , that is, the larger the torsion, the faster the curve moves away from the plane spanned by the vectors ⃗v1 and ⃗v2 . We shall soon restate this more precisely. Now we apply Taylor’s formula (Proposition 1.2.5) to study the shape of a curve γ in the neighbourhood of an arbitrary given point X(s0 ) of γ, and obtain   1¨ 1 ...  2 3 ˙   x(s0 + ∆s) = ⃗x(s0 ) + ⃗x(s0 )∆s + ⃗x(s0 )(∆s) + ⃗x (s0 )(∆s) + ⃗r4 (∆s),   ⃗  2 6 ⃗r4 (∆s) ⃗     where lim = 0.   ∆s→0 (∆s)3 (1.94) ¨ By Definition 1.6.3, we have ⃗x(s) = κ(s)⃗v2 (s). Differentiating this identity and applying Frenet’s second formula in (1.84) of Theorem 1.7.1, we obtain ... ⃗x (s) = κ(s)⃗ ˙ v2 (s) + κ(s)⃗v˙ 2 (s) = κ(s)⃗ ˙ v2 (s) + κ(s) (−κ(s))⃗v1 (s) + τ (s)⃗v3 (s))

= −κ2 (s)⃗v1 (s) + κ(s)⃗ ˙ v2 (s) + κ(s)τ (s)⃗v3 (s).

60 ■ Differential Geometry and Its Visualization

Figure 1.40

The torsion as a measure of the deviation of a curve from a plane

We substitute these expressions for s = s0 , and ⃗v1 (s0 ) = ⃗x˙ (s0 ) in (1.94); this yields 

κ2 (s0 ) (∆s)2 ⃗x(s0 + ∆s) − ⃗x(s0 ) = ⃗v1 (s0 )∆s 1 − 6 





˙ 0) κ(s0 ) κ(s + ∆s 2 6 κ(s0 )τ (s0 ) + ⃗r4 (∆s). + ⃗v3 (s0 )(∆s)3 · 6 + ⃗v2 (s0 )(∆s)2

(1.95)

Thus we have shown: Remark 1.8.2. Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I). Then the tangent at s is a first-order approximation of γ at s. The second-order approximation of γ at s lies in the plane spanned by the vectors ⃗v1 (s) and ⃗v2 (s); the curvature κ(s) of γ appears in the coefficient of the second-order term and measures the deviation of γ from its tangent at s. Finally the torsion τ (s) appears in the thirdorder term and measures the deviation of γ from the plane spanned by the vectors ⃗v1 (s) and ⃗v2 (s). The so–called osculating, normal and rectifying planes can be defined at every point of a curve. Definition 1.8.3. Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I) and X(s0 ) be a point on γ. Then the planes through X(s0 ) orthogonal to the binormal vector ⃗v3 (s0 ), the tangent vector ⃗v1 (s0 ), and the principal normal vector ⃗v2 (s0 ) are called the osculating, normal and rectifying planes of γ at X(s0 ), respectively (Figure 1.41).

Curves in Three-Dimensional Euclidean Space ■ 61

Figure 1.41

Osculating, normal and rectifying planes of a helix

We continue our studies of a curve γ with a parametric representation ⃗x(s) in a neighbourhood of an arbitrary given point X(s0 ) of γ, and consider certain approximations of the orthogonal projections of ⃗x(s) to the osculating, normal and rectifying planes at X(s0 ). We choose a coordinate system for R3 with its origin in the point X(s0 ), and the directions of the x1 –, x2 – and x3 –axes given by the vectors ⃗v1 (s0 ), ⃗v2 (s0 ) and ⃗v3 (s0 ), respectively. Omitting the argument s0 , we obtain from (1.95) 



κ2 x = 1 − (∆s)2 ∆s, x2 = 6 1





κ κ˙ κτ + ∆s (∆s)2 and x3 = (∆s)3 . 2 6 6

Trying to eliminate the parameter ∆s in these equations, we get κ 1 2 (x ) + (∆s)3 (· · · ) in the osculating plane, 2 2τ 2 2 3 (x3 )2 = (x ) + (∆s)7 (· · · ) in the normal plane 9κ x2 =

and x3 =

κτ 1 3 (x ) + (∆s)3 (· · · ) in the rectifying plane. 6

Thus we obtain the following quantitative characterization of curves in a neighbourhood of any of their points. For κ, τ ̸= 0 and ∆s = 0, the last three equations yield that the approximations of the projections at the point X(s0 ) are a quadratic parabola in the osculating plane, a Neil’s parabola in the normal plane and a cubic parabola in the rectifying plane (Figure 1.42). They are referred to as the local canonical form of a curve.

62 ■ Differential Geometry and Its Visualization

Figure 1.42

planes

The approximations of a helix in its osculating, normal and rectifying

We close this section with an example. Visualization 1.8.4. We consider the helix with a parametric representation ⃗x(s) = {r cos ωs, r sin ωs, hωs} (s ∈ R). We obtain by Visualization 1.6.7 (b), κ(s) = rω 2 =

r r2 + h2

(s ∈ IR).

(1.96)

Furthermore it follows from ⃗v1 (s) = ω{−r sin ωs, r cos ωs, h}, ⃗v2 (s) = −{cos ωs, sin ωs, 0},

⃗v3 (s) = ω{h sin ωs, −h cos ωs, r} and ⃗v˙ 2 (s) = ω{sin ωs, − cos ωs, 0} that

h (s ∈ R). (1.97) + h2 Thus the approximations of the helix in an arbitrary point X(s) are given by τ (s) = ⃗v˙ 2 (s) • ⃗v3 (s) = hω 2 =

x2 =

κ(s) 1 2 1 2 1 2 (x ) = rω (x ) 2 2

r2

in the osculating plane,

Curves in Three-Dimensional Euclidean Space ■ 63

(x3 )2 =

2 τ 2 (s) 2 3 2 h2 ω 2 2 3 (x ) = (x ) 9 κ(s) 9 r

in the normal plane

and 1 1 x3 = κ(s)τ (s)(x1 )3 = rhω 4 (x1 )3 6 6

in the rectifying plane

(Figure 1.42).

1.9

OSCULATING CIRCLES AND SPHERES

Osculating circles and spheres are used for certain approximation purposes of curves. Definition 1.9.1. Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I) and ⃗x¨(s) ̸= ⃗0 for all s ∈ I, and let s0 ∈ I be fixed and X(s0 ) be a point of γ. In the osculating plane of γ at X(s0 ), we consider the circle line C(s0 ) with radius ρ(s0 ) = 1/κ(s0 ) and its centre at the point M (s0 ) with the position vector m(s ⃗ 0 ) = ⃗x(s0 ) + ρ(s0 )⃗v2 (s0 ). Then C(s0 ) is called the osculating circle of γ at s0 . The point M (s0 ) is called the centre of curvature of γ at s0 (Figure 1.43).

Figure 1.43

An ellipse with osculating circles and centres of curvature

64 ■ Differential Geometry and Its Visualization

Figure 1.44

Osculating circles as second-order approximations of a curve

Remark 1.9.2. The osculating circle C(s0 ) of a curve γ at s0 is a second-order approximation of γ at a point of intersection X(s0 ) ∈ γ ∩ C(s0 ), that is, the tangents at X(s0 ) of γ and C(s0 ) coincide, and γ and C(s0 ) have the same curvature at X(s0 ) (Figure 1.44). Proof. The osculating circle of γ at X(s0 ) has a parametric representation ⃗ 0 ) + ρ(s0 ) (⃗v1 (s0 ) sin t + ⃗v2 (s0 ) cos t) for t ∈ [0, 2π]. ⃗ys0 (t) = m(s This yields ⃗y ′s0 (t) = ρ(s0 )(⃗v1 (s0 ) cos t − ⃗v2 (s0 ) sin t) for t ∈ [0, 2π]. Choosing t = π, we obtain ⃗ 0 ) − ρ(s0 )⃗v2 (s0 ) = ⃗x(s0 ), ⃗ys0 (π) = m(s

⃗y ′s0 (π) = −ρ(s0 )⃗v1 (s0 ) and ⃗y ′s0 (π) = −⃗v1 (s0 ). ∥⃗ys′0 (π)∥

The last identity shows that the tangents at X(s0 ) of γ and C(s0 ) coincide. Furthermore, the circle C(s0 ) has the curvature κ(s0 ) by Visualization 1.6.7 (a), and κ(s0 ) is the curvature of γ at X(s0 ). Visualization 1.9.3. The centre of curvature at s along the curve γ with a parametric representation (1.61) is given by ⃗xm (s) =





1 cos s sin s 1 , , s − tanh s − sinh s{− sin s, cos s, 0}. 2 cosh s 2 cosh s 2 2

(1.98)

Curves in Three-Dimensional Euclidean Space ■ 65

Figure 1.45

Osculating circle at a point of the curve of Visualization 1.9.3

Hence, the osculating circle of γ at s has a parametric representation (Figure 1.45) ⃗ym,s (t) = ⃗xm (s) +

cosh s (cos t⃗v1 (s) + sin t⃗v2 (s)) for t ∈ (0, 2π). s

(1.99)

Proof. The radius of curvature of γ at s is by (1.82) ρ(s) =

cosh s 1 = . κ(s) 2

Therefore, we obtain from (1.61) and (1.80) for the centre of curvature of γ at s ⃗xm (s) = ⃗x ∗ (s) + ρ(s)⃗v2 (s) =     cos s sin s cos s sinh s sin s sinh s − − , ,s − , ,− cosh s cosh s cosh s 2 cosh s 2 cosh s 2 cosh s sinh s {− sin s, cos s, 0} = 2 



sin s 1 sinh s cos s , , s − tanh s − {− sin s, cos s, 0} 2 cosh s 2 cosh s 2 2

which is (1.98). Now (1.99) follows immediately from the definition of the osculating circle of a curve at s. Now we consider higher order approximations of curves. Definition 1.9.4. Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I) with non–vanishing curvature and torsion, and let s0 ∈ I be fixed. A sphere that is an approximation of highest possible order of γ at s0 is called an osculating sphere of γ at s0 (Figure 1.46).

66 ■ Differential Geometry and Its Visualization

A curve, the curve of its centres of curvature, the osculating plane, circle and sphere at a point Figure 1.46

Theorem 1.9.5. Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I) and non–vanishing curvature and torsion, and let s0 ∈ I be fixed. Then the osculating sphere of γ at s0 is uniquely determined; the position vector of its centre M (s0 ) and its radius r(s0 ) are given by m(s ⃗ 0 ) = ⃗x(s0 ) +

1 κ(s ˙ 0) ⃗v2 (s0 ) − ⃗v3 (s0 ) κ(s0 ) τ (s0 )κ2 (s0 )

(1.100)

and 1 r(s0 ) = κ(s0 )



1+

κ˙ 2 (s0 ) . 2 κ (s0 )τ 2 (s0 )

(1.101)

Proof. The sphere Cr (M ) with radius r and centre at M is given by the equation (⃗x − m) ⃗ 2 − r2 = 0. We define the function f : I → R by f (s) = (⃗x(s) − m) ⃗ 2 − r2 . If Cr (M ) is to be an approximation of ⃗x(s) of highest possible order at s0 , then the function f and its derivatives up to the third order should vanish at s0 . Since ⃗x˙ (s) = ⃗v1 (s), ⃗v1 (s) • ⃗v1 (s) = 1 and ⃗v˙ 1 (s) = κ(s)⃗v2 (s) it follows that f (s) = [⃗x(s) − m] ⃗ 2 − r2 = 0,

(1.102)

f˙(s) = 2[⃗x(s) − m] ⃗ • ⃗x˙ (s) = 2[⃗x(s) − m] ⃗ • ⃗v1 (s), 



f¨(s) = 2 ⃗x˙ (s) • ⃗v1 (s) + [⃗x(s) − m] ⃗ • ⃗v˙ 1 (s) = 2 (1 + κ(s)[⃗x(s) − m] ⃗ • ⃗v2 (s))

(1.103)

Curves in Three-Dimensional Euclidean Space ■ 67

and, since ⃗v1 (s) • ⃗v2 (s) = 0 for all s ∈ I, we obtain, applying Frenet’s second formula in (1.84) of Theorem 1.7.1   ... f (s) = κ(s)[⃗ ˙ x(s) − m] ⃗ • ⃗v2 (s) + κ(s)[⃗x(s) − m] ⃗ • ⃗v˙ 2 (s) 



= 2 [⃗x(s) − m] ⃗ • [κ(s)⃗ ˙ v2 (s) − κ2 (s)⃗v1 (s) + κ(s)τ (s)⃗v3 (s)] .

To establish the identity in (1.100), we write

(1.104)

⃗x(s) − m ⃗ = a1 (s)⃗v1 (s) + a2 (s)⃗v2 (s) + a3 (s)⃗v3 (s) for all s ∈ I, and substitute this in (1.102), (1.103) and (1.104) to obtain 0 = f˙(s0 ) = 2(a1 (s0 )⃗v1 (s0 ) + a2 (s0 )⃗v2 (s0 ) + a3 (s0 )⃗v3 (s0 )) • ⃗v1 (s0 ) = 2a1 (s0 ) which implies a1 (s0 ) = 0, 0 = f¨(s0 ) = 2 + 2κ(s0 )a2 (s0 ) which implies a2 (s0 ) = − and

1 κ(s0 )

  ... κ(s ˙ 0) 0 = f (s0 ) = 2 − + κ(s0 )τ (s0 )a3 (s0 ) κ(s0 )

which implies

a3 (s0 ) =

κ(s ˙ 0) . 0 )τ (s0 )

κ2 (s

This shows (1.100). Finally, (1.100) yields r2 (s0 ) = (⃗x(s0 ) − m) ⃗ 2= 



2

1 κ(s ˙ 0) ⃗v3 (s0 ) ⃗v2 (s0 ) − 2 κ (s0 ) κ(s0 )τ (s0 ) 

1 κ˙ 2 (s0 ) = 2 1+ 2 , κ (s0 ) κ (s0 )τ 2 (s0 ) and (1.101) follows.

Visualization 1.9.6. (a) Let γ be the catenary of Visualization 1.6.6 (a) in the xy–plane, given by the parametric representation 

⃗x(s) = c sinh It follows from ⃗v2 (s) = √

−1

  √ s

c

,

s2

+

c2



(s ∈ R).

1 c {c, −s} and κ(s) = 2 c + s2 c 2 + s2

(1.105)

68 ■ Differential Geometry and Its Visualization

Figure 1.47 A helix, an osculating plane, circle and sphere, and the curve of its osculating spheres

that the centres of curvature of γ are given by m(s) ⃗ = ⃗x(s) +



 

⃗v2 (s) s = c sinh−1 κ(s) c

−



√ s√ 2 c + s2 , 2 c 2 + s2 , } c

(s ∈ R).

(b) Let γ be the helix of Visualization 1.8.4 with a parametric representation ⃗x(s) = {r cos ωs, r sin ωs, hωs} (s ∈ R) where ω =

1 . r2 + h2

˙ = 0 and τ (s) = hω 2 for all s. Since κ(s) ˙ = 0 for Then we have κ(s) = rω 2 , κ(s) all s, the centres M (s) of the osculating circles and spheres of the helix coincide and their radii r(s) have identical values, given by m(s) ⃗ = {r cos ωs, r sin ωs, hωs} + =



−

1 {− cos ωs, − sin ωs, 0} rω 2

h2 h2 cos ωs, − sin ωs, hωs r r

and r(s) = 1/(rω 2 ) = r + h2 /r (s ∈ R). Thus the centres of the osculating circles and spheres of a helix again are on a helix (Figure 1.47). Visualization 1.9.7. The centre and radius of the osculating sphere at s along the curve γ with a parametric representation (1.61) are given by (Figure 1.48) m(s) ⃗ = and





1 1 1 cosh s cos s, cosh s sinh s, s − sinh s{− sin s, cos s, 0} 2 2 2 rm (s) =

1 2



cosh2 s + sinh2 s.

(1.106) (1.107)

Curves in Three-Dimensional Euclidean Space ■ 69

Figure 1.48

The osculating sphere at a point of the curve γ of Visualization 1.9.7

Proof. Since ρ(s) ˙ = sinh(s)/2 and τ (s) = 1 by (1.85), we obtain the centre of the osculating sphere of γ from (1.81), and (1.100) in Theorem 1.9.5 ρ(s) ˙ sinh s ⃗v3 (s) = ⃗xm (s) + ⃗v3 (s) τ (s) 2   sin s 1 1 cos s , , s − tanh s − sinh s{− sin s, cos s} = 2 cosh s 2 cosh s 2 2   2 2 sinh s cos s sinh s sin s tanh s + , , 2 cosh s 2 cosh s 2s

m(s) ⃗ = ⃗x ∗ (s) + ρ(s)⃗v2 (s) +

=





1 1 1 cosh s cos s, cosh s sinh s, s − sinh s{− sin s, cos s, 0} 2 2 2

which is (1.106). Finally, we obtain for the radius of the osculating sphere 2 rm (s)

2

= ρ (s) +



ρ(s) ˙ τ (s)

2

and (1.107) is an immediate consequence.

1.10

=

 1 cosh2 s + sinh2 s , 4

INVOLUTES AND EVOLUTES

With every curve we may associate two new curves that are closely related to one another, namely involutes and evolutes. We start with a geometric definition of involutes.

70 ■ Differential Geometry and Its Visualization

Figure 1.49

Left: Tangent surface. Right: The definition of an involute

Definition 1.10.1. Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I) and ⃗v1 (s) its tangent vector at s. Then ⃗y (s, t) = ⃗x(s) + t⃗v1 (s) ((s, t) ∈ I × R) is a parametric representation for the so–called tangent surface of γ (left in Figure 1.49). Curves on the tangent surface of γ which intersect the tangents to γ at right angles are called involutes of γ (in Figures 1.49 (right) and 1.50). The following result gives a parametric representation for involutes. Theorem 1.10.2. The involutes of a curve γ with a parametric representation ⃗x(s) and tangent vectors ⃗v1 (s) (s ∈ I) have a parametric representation ⃗x ∗ (s) = ⃗x(s) + (s0 − s)⃗v1 (s) (s ∈ I), where s0 ∈ R is an arbitrary constant. (1.108) There is one and only one involute that corresponds to a given value of the constant s0 ; thus every curve has a one–parametric collection of involutes (Figure 1.51). Proof. A curve γ ∗ in the tangent surface of γ is given by ⃗x ∗ (s) = ⃗x(s) + t(s)⃗v1 (s). By definition, γ ∗ is an involute of γ if d⃗x ∗ (s) • ⃗v1 (s) = 0, ds

Curves in Three-Dimensional Euclidean Space ■ 71

Figure 1.50

An involute in the tangent surface of a curve

Figure 1.51

Involutes of a curve on a sphere (left) catenoid (right)

that is,





dt dt ⃗v1 (s) + (s)⃗v1 (s) + t(s)⃗v˙ 1 (s) • ⃗v1 (s) = 1 + (s) = 0. ds ds

This implies t(s) = s0 − s where s0 ∈ R is a constant.

There is a simple relation between the curvature of an involute of a curve γ and the curvature and torsion of γ. Theorem 1.10.3. Let γ be curve with a parametric representation ⃗x(s) (s ∈ I), non–vanishing curvature κ, and torsion τ . Furthermore, let s0 ∈ I be given. Then the curvature κ∗ of the involute γ ∗ with a parametric representation (1.108) is given by κ (s) = ∗



κ2 (s) + τ 2 (s) for all s ∈ I \ {s0 }; |s0 − s|κ(s)

(1.109)

in particular, if γ is a planar curve, then κ∗ (s) =

1 for all s ∈ I \ {s0 }. |s0 − s|

(1.110)

72 ■ Differential Geometry and Its Visualization

Proof. It follows from (1.108) and Frenet’s first and second formulae in (1.84) of Theorem 1.7.1 that d⃗x ∗ (s) = ⃗v1 (s) − ⃗v1 (s) + (s0 − s)⃗v˙ 1 (s) = (s0 − s)κ(s)⃗v2 (s) ds

(1.111)

and d((s0 − s)κ(s)) d 2⃗x ∗ (s) + (s0 − s)κ(s)⃗v˙ 2 (s) = ⃗v2 (s) 2 ds ds d((s0 − s)κ(s)) = ⃗v2 (s) + (s0 − s)κ(s)(−κ(s)⃗v1 (s) + τ (s)⃗v3 (s)). ds Since ⃗vj (s) • ⃗vk (s) = δjk for j, k = 1, 2, 3, ⃗v2 (s) × ⃗v2 (s) = ⃗0, ⃗v2 (s) × ⃗v1 (s) = −⃗v3 (s) and ⃗v2 (s) × ⃗v3 (s) = ⃗v1 (s), we obtain  ∗ 3  d⃗   x (s)  = |s0 − s|3 κ3 (s),  ds 

d⃗x ∗ (s) d 2⃗x ∗ (s) × = (s0 − s)2 κ2 (s) (κ(s)⃗v3 (s) + τ (s)⃗v1 (s)) , ds ds2 and so by (1.88) in Proposition 1.7.6    d⃗ ∗ d 2⃗x ∗ (s)   x (s)  ×     2  ds  ds (s0 − s)2 κ2 (s) κ2 (s) + τ 2 (s) κ2 (s) + τ 2 (s) ∗ κ (s) = = , =  ∗ 3  d⃗  |s0 − s|3 κ3 (s) |s0 − s|κ(s)  x (s)   ds 

that is, (1.109) holds. If γ is a planar curve, then τ (s) = 0 for all s ∈ I, and (1.110) is an immediate consequence of (1.109). Visualization 1.10.4. The involutes of a helix with a parametric representation ⃗x(s) = {r cos ωs, r sin ωs, hωs} (s ∈ R), where ω =

1 , r2 + h2

are the lines of intersection of the tangents to the helix and the plane x3 = hωs0 , where s0 is the constant in the parametric representation (1.108) in Theorem 1.10.2 (Figure 1.52). If s∗ denotes the arc length along the lines of intersection, then their curvatures are given by κ∗ (s∗ ) = 

1 (s∗ ∈ R \ {0}). 2r|s∗ |

(1.112)

Proof. The tangent to the helix at s has a parametric representation ⃗y (s) = ⃗x(s) + λ⃗v1 (s) for λ ∈ R, where ⃗v1 (s) = {−ωr sin ωs, ωr cos ωs, hω} (1.113)

Curves in Three-Dimensional Euclidean Space ■ 73

Figure 1.52

axis

The intersections of the tangents to a helix with planes orthogonal to its

by Visualization 1.8.4, hence its intersection with the plane x3 = hωs0 is given by hωs + λ0 hω = hωs0 , that is, λ0 = s0 − s. Substituting this in the parametric representation (1.113) of the tangent to the helix, we obtain ⃗y (s; λ0 ) = ⃗x(s) + (s0 − s)⃗v1 (s) = x∗ (s) (s ∈ R)

for the line γ ∗ of intersection of the tangents to the helix and the plane x2 = hωs0 . Its curvature κ∗ (s) is given by (1.109) in Theorem 1.10.3, and by Visualization 1.8.4 κ (s) = ∗



κ2 (s)

τ 2 (s)

+ |s − s0 |κ(s)

=



1+

h2 r2

|s − s0 |

=

√

r2 + h2 1 =√ √ . r|s − s0 | r κ|s − s0 |

(1.114)

It follows from (1.111) in the proof of Theorem 1.10.3 that the arc length s∗ along γ ∗ is given by s∗ = that is,



|s0 − s|κ(s) ds =

  (s − s0 )2  κ

2 (s − s0 )2   −κ 2

for s ≥ s0 for s < s0 ,

 √ κ|s − s0 | = 2|s∗ |.

Substituting this in (1.114), we obtain κ∗ (s∗ ) = 

1 (s∗ ∈ R \ {0}) 2r|s∗ |

for the curvature of γ ∗ with respect to its arc length, that is, (1.112) is satisfied. Now we consider a different type of involutes.

74 ■ Differential Geometry and Its Visualization

Figure 1.53

The definition of a plane involute

Figure 1.54

Plane involutes of a curve on a catenoid

Definition 1.10.5. Let γ be a curve. A curve that intersects the osculating planes of γ orthogonally is called a plane involute (Figures 1.53 and 1.54). The next result gives a parametric representation for a plane involute. Theorem 1.10.6. Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I) and non–vanishing curvature κ and torsion τ . We put ψ(s) =

s 0

κ(σ) dσ for all s ∈ I.

Then the plane involutes of γ are given by a parametric representation 

⃗z(s) = ⃗x(s) − cos ψ(s) 

+ sin ψ(s)







cos ψ(s) ds + k1 + sin ψ(s) 

cos ψ(s) ds + k1 − cos ψ(s)

where k1 , k2 ∈ R are constants of integration.

 

sin ψ(s) ds + k2 sin ψ(s) ds + k2

 

⃗v1 (s) ⃗v2 (s),

(1.115)

Curves in Three-Dimensional Euclidean Space ■ 75

Proof. The vector ⃗z(s) − ⃗x(s) is in the osculating plane of the curve γ by Definition 1.10.5, that is, there are functions c1 , c2 : I → R such that ⃗z(s) − ⃗x(s) = c1 (s)⃗v1 (s) + c2 (s)⃗v2 (s) for all s ∈ I.

(1.116)

Furthermore, since the plane involute intersects the osculating plane of γ orthogonally, it follows that d⃗z(s) d⃗z(s) • ⃗v1 (s) = • ⃗v2 (s) = 0 for all s ∈ I. (1.117) ds ds First, differentiating (1.116) and applying Frenet’s first and second formulae in (1.84) of Theorem 1.7.1, we obtain, omitting the parameter s d⃗z = (1 + c˙1 ) ⃗v1 + c1⃗v˙ 1 + c˙2⃗v2 + c2⃗v˙ 2 ds = (1 + c˙1 ) ⃗v1 + c1 κ⃗v2 + c˙2⃗v2 − c2 κ⃗v1 + c2 τ⃗v3 = (1 + c˙1 − c2 κ) ⃗v1 + (c1 κ + c˙2 ) ⃗v2 + c2 τ⃗v3 .

(1.118)

We obtain the following system of differential equations from (1.117) c˙1 − c2 κ = −1 and c˙2 + c1 κ = 0.

(1.119)

Introducing the complex function w = c1 + ic2 , we can write the system in (1.119) as dw + iκw + 1 = 0. ds This differential equation has the general solution w(s) = − exp (−iψ(s))



exp (iψ(s)) ds + k,

(1.120)

where k = k1 + ik2 is a constant of integration. Since ⃗z(s) is a parametric representation for the plane involute, we must have ⃗z˙ ̸= ⃗0, that is, τ (s) ̸= 0 for all s ∈ I, by (1.118). It follows from (1.120) that w = − (cos ψ − i sin ψ) = − cos ψ





(cos ψ + i sin ψ) ds + k1 + ik2

cos ψ ds − k1 − sin ψ 

− i cos ψ

and this yields 

c1 = − cos ψ







sin ψ ds − k2

sin ψ ds − k1 − sin ψ

cos ψ ds + k1 + sin ψ







cos ψ ds + k1 − cos ψ





cos ψ ds + k2 ,

sin ψ ds + k2

and c2 = sin ψ



sin ψ ds − k2 .

This establishes the parametric representation given in (1.115).



76 ■ Differential Geometry and Its Visualization

Figure 1.55

Plane involutes of helices

Visualization 1.10.7. We consider the helix of Visualization 1.6.7 (b) where ⃗v1 (s) = {−ωr sin ωs, ωr cos ωs, hω},

⃗v2 (s) = {− cos ωs, − sin ωs, 0} and κ(s) = ω 2 r for all s ∈ R, hence ψ(s) = ω 2 rs, and (Figure 1.55) 





1 sin (ω 2 rs) + k1 ω2r    1 2 2 + sin (ω rs) − 2 cos (ω rs) + k2 ⃗v1 (s) ω r    1 2 2 sin (ω rs) + k1 + sin (ω rs) ω2r    1 2 2 − cos (ω rs) − 2 cos (ω rs) + k2 ⃗v2 (s) ω r

⃗z(s) = ⃗x(s) − cos (ω 2 rs)





= ⃗x(s) − k1 cos (ω 2 rs) + k2 sin (ω 2 rs) ⃗v1 (s) 



1 + k1 sin (ω 2 rs) − k2 sin (ω 2 rs) ⃗v2 (s) for all s ∈ R. + ω2r

Now we define the evolutes of curves. Definition 1.10.8. A curve γ ∗ is called evolute of a curve γ if γ is an involute of γ∗. The following result gives a parametric representation for evolutes.

Curves in Three-Dimensional Euclidean Space ■ 77

Figure 1.56

The evolutes of a curve on a catenoid

Theorem 1.10.9. Let γ be a curve with a parametric representation ⃗x(s), non– vanishing curvature κ, and principal normal and binormal vectors ⃗v2 (s) and ⃗v3 (s) (s ∈ I). Then the evolutes of γ have parametric representations 1 (⃗v2 (s) + ⃗v3 (s) cot α(s)) with α(s) = ⃗y (s) = ⃗x(s) + κ(s) ∗

s

τ (σ) dσ + c,

0

where c is a constant of integration (Figure 1.56). Proof. Let ⃗y ∗ be a parametric representation of an evolute γ ∗ of the curve γ with a parametric representation ⃗x(s). Since γ is an involute of γ ∗ , the tangents to γ ∗ intersect γ at right angles. Let ⃗a(s) be a unit vector in the normal plane of γ and orthogonal to γ; then ⃗y ∗ (s) = ⃗x(s) + t(s)⃗a(s). Here |t(s)| is the distance between the point X(s) on γ and the corresponding point Y ∗ (s) on γ ∗ . By the definition of an involute, ⃗a(s) has to be tangential to γ ∗ , hence dt d⃗a d⃗y ∗ (s) = λ⃗a(s) = ⃗x˙ (s) + (s)⃗a(s) + t(s) (s). ds ds ds

(1.121)

This implies λ = λ⃗a(s) • ⃗a(s) = ⃗v1 (s) • ⃗a(s) +

d⃗a dt dt (s) + t(s) (s) • ⃗a(s) = (s), ds ds ds

a since ⃗a(s) ⊥ ⃗v1 (s), and ∥⃗a(s)∥ = 1 implies ⃗a(s) • d⃗ ds (s) = 0. Therefore, λ = we obtain from (1.121) d⃗a ⃗v1 (s) + t(s) (s) = ⃗0. ds

dt ds (s)

and

(1.122)

78 ■ Differential Geometry and Its Visualization

Since ⃗a(s) is a unit vector in the normal plane, we may write ⃗a(s) = ⃗v2 (s) sin α(s) + ⃗v3 (s) cos α(s), where α(s) is the angle between ⃗a(s) and ⃗v3 (s). Omitting the argument s, we obtain d⃗a = sin α⃗v˙ 2 + α˙ cos α⃗v2 + cos α⃗v˙ 3 − α˙ sin α⃗v3 ds = −κ sin α⃗v1 + (α˙ cos α − τ cos α)⃗v2 + (τ sin α − α˙ sin α)⃗v3 by Frenet’s formulae, and substituting in (1.122), we conclude (1 − κt sin α)⃗v1 + t cos α(α˙ − τ )⃗v2 + t sin α(τ − α)⃗ ˙ v3 = ⃗0. Since the vectors ⃗v1 , ⃗v2 and ⃗v3 are linearly independent and cos and sin have no common zeros, this implies 1 − κt sin α = 0, α˙ − τ = 0, that is, ρ = 1/κ = t sin α and α˙ = τ, 

hence α(s) = 0s τ (σ) dσ+c with a constant c ∈ R. Now the parametric representation for the evolute γ ∗ is an immediate consequence. Remark 1.10.10. If γ is a planar curve with a parametric representation ⃗x(s) (s ∈ I) and non–vanishing curvature, then τ (s) = 0 for all s and the parametric representation for the evolutes given in Theorem 1.10.9 reduces, for the choice α(s) = π/2, to 1 ⃗y ∗ (s) = ⃗x(s) + ⃗v2 (s) (s ∈ I). κ(s) Consequently, this evolute of a planar curve coincides with the centres of curvature of the curve (Figure 1.57).

Figure 1.57

An evolute of a catenary

Curves in Three-Dimensional Euclidean Space ■ 79

Figure 1.58

An evolute of a helix

Visualization 1.10.11. Let γ be the helix of Visualization 1.10.7. Then, by Theorem 1.10.2, the involutes of the helix are given by ⃗x ∗ (s) = {r cos ωs, r sin ωs, hωs} + (s0 − s)ω{−r sin ωs, r cos ωs, h}

= {r(cos ωs − ω(s0 − s) sin ωs), r(sin ωs + ω(s0 − s) cos ωs), hωs0 } .

Furthermore, since

α(s) =

s

τ (σ) dσ + c = hω 2 s + c,

0

the evolutes of the helix are given by ⃗y ∗ (s) = {r cos ωs, r sin ωs, hωs} + +

1 {− cos ωs, − sin ωs, 0}+ rω 2

cot (hω 2 s + c) {h sin ωs, −h cos ωs, r} rω

by Theorem 1.10.9 (Figure 1.58).

1.11





s∈ −

c π−c , hω 2 hω 2



,

THE FUNDAMENTAL THEOREM OF CURVES

In this section, we will show that curves are completely determined by their curvature and torsion – up to linear transformations. This result is known as the fundamental theorem of the theory of curves. Theorem 1.11.1 (The fundamental theorem of curves). Let I be an open interval containing the origin, κ ¯ ∈ C 1 (I), τ¯ ∈ C(I) two functions and κ ¯ (s) > 0 for all s ∈ I. Furthermore, let ⃗c0 be a constant vector and ⃗ck (k = 1, 2, 3) be three orthonormal vectors with ⃗c1 •(⃗c2 ×⃗c3 ) = 1. Then there is one and only one curve γ with a parametric representation ⃗x(s) (s ∈ I) that has the following properties:

80 ■ Differential Geometry and Its Visualization

(i) s is the arc length along γ (ii) ⃗x(0) = ⃗c0 and ⃗vk (0) = ⃗ck (k = 1, 2, 3) for the vectors of the trihedron of γ at 0 (iii) κ(s) = κ ¯ (s) and τ (s) = τ¯(s) for the curvature κ and torsion τ of γ. Proof. First we find the solutions ⃗vk (k = 1, 2, 3) on I of Frenet’s formulae, and then we show that s ⃗x(s) =



⃗v1 (σ) dσ + ⃗c0

0

is a parametric representation of a curve with the desired properties. By the existence theorem for systems of first-order linear differential equations, the system   0 κ ¯ 0 3    ⃗v¯˙ j = κ 0 τ¯ ajk⃗v¯k with (ajk ) = −¯ (1.123) k=1 0 −¯ τ 0 has unique solutions ⃗v¯j with ⃗v¯j (0) = ⃗cj (j = 1, 2, 3). If there is a curve γ with a parametric representation ⃗x(s) (s ∈ I) that has the desired properties, then it is given by ⃗x(s) = ⃗c0 +

s

⃗v¯1 (σ) dσ.

(1.124)

0

This shows the uniqueness. Now we shall prove that ⃗x(s) in (1.124) has the desired properties. (i) First we show ⃗x ∈ C 3 (I). Since ⃗x˙ = ⃗v¯1 by (1.124), it follows from (1.123) that ⃗v¯˙ 1 = κ ¯⃗v¯2 . By hypothesis ˙ κ ¯ ∈ C 1 (I) and by (1.123), we obtain ⃗v¯2 ∈ C 1 (I) and ⃗v¯1 ∈ C 1 (I), hence ⃗x ∈ C 3 (I). (ii) Now we show ⃗v¯j • ⃗v¯k = δjk (j, k = 1, 2, 3) and ⃗v¯1 • (⃗v¯2 × ⃗v¯3 ) = 1 on I. It follows from (1.123) that 3 3   d ⃗ ⃗  ⃗˙ ⃗ v¯j • v¯k = v¯j • v¯k + ⃗v¯j • ⃗v¯˙ k = ajl⃗v¯l • ⃗v¯k + akl⃗v¯l • ⃗v¯j . ds l=1 l=1

Putting bjk = ⃗v¯j •⃗v¯k (j, k = 1, 2, 3), we obtain the following system of differential equations b˙ jk =

3 

(ajl blk + akl blj )

l=1

with blk = bkl and the initial conditions bjk (0) = δjk (j, k = 1, 2, 3)

Curves in Three-Dimensional Euclidean Space ■ 81

which has one and only one solution. Since 3 

(ajl δlk + akl δlj ) = ajk + akj = 0 = δ˙jk (j, k = 1, 2, 3),

l=1

the functions bjk = δjk (j, k = 1, 2, 3) are solutions. Consequently, we have bjk (s) = ⃗v¯j (s) • ⃗v¯k (s) = δjk (s) (j, k = 1, 2, 3) on I. Furthermore this implies ⃗v¯1 (s) • (⃗v¯2 (s) × ⃗v¯3 (s)) = ±1 for all s ∈ I. From the initial conditions, we obtain ⃗v¯1 (0) • (⃗v¯2 (0) × ⃗v¯3 (0)) = ⃗c1 • (⃗c2 × ⃗c3 ) = 1, and the continuity of the functions ⃗v¯k (k = 1, 2, 3) implies ⃗v¯1 (s) • (⃗v¯2 (s) × ⃗v¯3 (s)) = 1 for all s ∈ I. (iii) Now we show that s is the arc length of ⃗x. By (1.124) and Part (ii), we obtain (⃗x˙ (s))2 = (⃗v¯1 (s))2 = 1, and so s is the arc length of ⃗x by Proposition 1.5.2 (d). (iv) Now we show that the vectors ⃗v¯k (k = 1, 2, 3) are the vectors of the trihedra of ⃗x. Since ⃗x ∈ C 3 (I) by Part (i) and ¯ (s)⃗v¯2 (s) ̸= ⃗0 for all s ∈ I by Part (ii), ⃗x¨(s) = ⃗v¯˙ 1 (s) = κ

we conclude and

⃗v1 (s) = ⃗x˙ (s) = ⃗v¯1 (s) for all s ∈ I by Part (i), ⃗v2 (s) =

Finally and

⃗v¯˙ 1 (s) ⃗x¨(s) = ⃗v¯2 (s) for all s ∈ I. = κ ¯ (s) ∥⃗x¨(s)∥

⃗vj (s) • ⃗vk (s) = ⃗v¯j (s) • ⃗v¯k (s) = δjk (s) (j, k = 1, 2, 3) 



⃗v1 (s) • (⃗v2 (s) × ⃗v3 (s)) = ⃗v¯1 (s) • ⃗v¯2 (s) × ⃗v¯3 (s) = 1 for all s ∈ I by Part (ii) together imply ⃗v3 (s) = ⃗v¯3 (s) for all s ∈ I.

82 ■ Differential Geometry and Its Visualization

(v) Finally, we show that κ ¯ and τ¯ are the curvature and torsion of ⃗x. If we denote the curvature and torsion of ⃗x by κ and τ , then Frenet’s formulae and Part (iv) together imply κ(s) = ⃗v˙ 1 (s) • ⃗v2 (s) = ⃗v¯˙ 1 (s) • ⃗v¯2 (s) = κ ¯ (s) and τ (s) = −⃗v˙ 3 (s) • ⃗v2 (s) = −⃗v¯˙ 3 (s) • ⃗v¯2 (s) = τ¯(s) for all s ∈ I. We may restate Theorem 1.11.1 as follows: Theorem 1.11.2. Given any two functions κ and τ that satisfy the conditions in Theorem 1.11.1 there is – up to possible translations and rotations – one and only one curve that has κ, τ and s as its curvature, torsion and arc length, respectively. Remark 1.11.3. The significance of Theorem 1.11.2 is that it states that curvature and torsion are a complete system of invariants for a curve. By this we mean that, if I(s) is any invariant, that is, a functional independent of motion I(s) = J(⃗x(s)), defined for all curves, then, by Theorem 1.11.2, the curve is completely determined by its initial point X(0), the initial vectors of the trihedron and its curvature and torsion in terms of its arc length. Since by some linear transformation, the point X(0) can be moved into the origin and the vectors of the trihedron can be made to coincide with the coordinate axes, the function J(⃗x(s)) in fact depends on κ, τ and s only. In view of the fundamental theorem it is natural to define the concept of the natural or intrinsic equations of a curve. Definition 1.11.4. The equations κ = κ(s) and τ = τ (s), which express the curvature and torsion of a curve depending on its arc length, are called the natural or intrinsic equations of the curve. We close this section with two examples. Example 1.11.5. (a) The catenary of Example 1.5.3 (d) with a parametric representation (1.59) has the curvature κ = κ(s) =

a2

a for all s ∈ R + s2

by (1.105) in Visualization 1.9.6 (a), and the torsion τ = τ (s) = 0 for all s ∈ R

Curves in Three-Dimensional Euclidean Space ■ 83

Figure 1.59

Spherical curves

by Theorem 1.8.1. Therefore, its natural equations are κ(s) =

a2

a and τ (s) = 0 for all s ∈ R. + s2

(b) The natural equations for the helix of Visualization 1.8.4 with a parametric representation ⃗x(s) = {r cos ωs, r sin ωs, hωs} (s ∈ R), where ω = √

r2

1 + h2

are by (1.96) and (1.97) κ = κ(s) = rω 2 and τ = τ (s) = hω 2 for all s ∈ R. Visualization 1.11.6. A curve on a sphere is called a spherical curve (Figure 1.59). A curve with non–vanishing torsion τ is a spherical curve if and only if its natural equations satisfy the condition d ρ(s)τ (s) + ds

 

ρ˙ (s) = 0 for all s, τ

(1.125)

where ρ(s) = 1/κ(s) is the radius of curvature of the curve. Proof. (i) First we assume that γ is a curve with a parametric representation ⃗x(s) on a sphere S.

84 ■ Differential Geometry and Its Visualization

Then the osculating spheres of γ coincide with S, in particular their centres must be coincide. The position vector of centre of the osculating sphere at s is given by (1.100) in Theorem 1.9.5 ˙ 1 κ(s) ⃗v2 (s) − 2 ⃗v3 (s) κ(s) κ (s)τ 2 (s) ρ(s) ˙ = ⃗x(s) + ρ(s)⃗v2 (s) + ⃗v3 (s), τ (s)

m(s) ⃗ = ⃗x(s) +

(1.126)

and so by Frenet’s formulae (1.84) in Theorem 1.7.1 



ρ(s) ˙ ρ(s) ˙ ⃗v˙ 3 (s) ⃗v3 (s) + τ (s) τ (s)   ρ(s) ˙ = (1 − ρ(s)κ(s))⃗v1 (s) + ρ(s) ˙ − τ (s) ⃗v2 (s) τ (s)    d ρ(s) ˙ + ρ(s)τ (s) + ⃗v3 (s) ds τ (s)    d ρ(s) ˙ = ρ(s)τ (s) + ⃗v3 (s) = ⃗0, (1.127) ds τ (s)

dm ⃗ d ˙ v2 (s) + ρ(s)⃗v˙ 2 (s) + (s) = ⃗v1 (s) + ρ(s)⃗ ds ds

which implies (1.125).

(ii) Conversely, we assume that the condition in (1.125) is satisfied. We define the vector–valued function m ⃗ as in (1.126). Then it follows as in ⃗ (1.127) that dm(s)/ds ⃗ = 0 for all s, hence m(s) ⃗ =m ⃗ for some constant vector m. ⃗ Furthermore, we have r2 (s) = ∥⃗x(s) − m∥ ⃗ 2 = ρ2 (s) + and so



d dr2 ρ2 (s) + (s) = ds ds =2





ρ(s) ˙ τ (s)

2 

ρ(s) ˙ d ρ(s)τ (s) + τ (s) ds





ρ(s) ˙ τ (s)

= 2ρ(s)ρ(s) ˙ +2 ρ(s) ˙ τ (s)



2

,

ρ(s) ˙ d τ (s) ds



ρ(s) ˙ τ (s)



= 0.

Thus r(s) = r for some constant r, and so ∥⃗x(s) − m∥ ⃗ = r for all s, that is, γ is a spherical curve.

1.12

LINES OF CONSTANT SLOPE

There is a class of curves for which a parametric representation can easily be derived from the natural equations. Definition 1.12.1. A curve with a parametric representation ⃗x(s) (s ∈ I) is called a line of constant slope if there is a constant c such that τ (s) = cκ(s) for all s ∈ I (Figure 1.60).

Curves in Three-Dimensional Euclidean Space ■ 85

Figure 1.60

Lines of constant slope of a given curvature

Example 1.12.2. Helices are lines of constant slope, since it follows from Example 1.11.5 (b) that r τ (s) = κ(s) for all s ∈ R. h Now we give an explicit formula for the parametric representations of lines of constant slope. Theorem 1.12.3. All lines of constant slope that satisfy the natural equations τ (s) = cκ(s) (s ∈ I) are given by ⃗a ⃗x(s) = ω ˜

s

s0

s ⃗b  sin (˜ ω t(σ)) dσ − cos (˜ ω t(σ)) dσ + ⃗c1 s + ⃗c0 ω ˜

(1.128)

s0

for s0 ∈ I fixed (s ∈ I), where the constant vectors ⃗a, ⃗b, ⃗c1 and ⃗c0 satisfy ⃗a 2 = ⃗b 2 = 1, ⃗a • ⃗b = ⃗a • ⃗c1 = ⃗b • ⃗c1 = 0 and ⃗c12 = and t(s) =

s

c2 , 1 + c2

κ(σ) dσ and ω ˜ 2 = 1 + c2 .

(1.129)

(1.130)

s0

Proof. We omit the argument s in the proof. Since τ = c · κ, Frenet’s formulae (1.84) in Theorem 1.7.1 reduce to ⃗v˙ 1 = κ⃗v2 , ⃗v˙ 2 = −κ⃗v1 + cκ⃗v3 and ⃗v˙ 3 = −cκ⃗v2 .

We introduce a new parameter t by dt/ds = κ(s). Then Frenet’s formulae become ⃗v1 ′ = ⃗v2 , ⃗v2 ′ = −⃗v1 + c⃗v3 and ⃗v3 ′ = −c⃗v2 .

(1.131)

86 ■ Differential Geometry and Its Visualization

Differentiating the second equation, and substituting the first into the third equation yields ⃗v2 ′′ = −⃗v1 ′ + c⃗v3 ′ and

hence

⃗v1 ′′′ = ⃗v2 ′′ = −⃗v1 ′ + c⃗v3 ′ = −⃗v1 ′ − c2⃗v2 = −⃗v1 ′ (1 + c2 ), ⃗v1 ′′′ + ω ˜ 2⃗v1 ′ = ⃗0, where ω ˜ 2 = 1 + c2 .

(1.132)

The general solution of the system (1.132) of differential equations is ⃗v1 ′ = ⃗a cos ω ˜ t + ⃗b sin ω ˜ t,

(1.133)

where ⃗a and ⃗b are constant vectors. Observing that ⃗x¨ = ⃗v˙ 1 = ⃗x ′ κ(s), and integrating (1.132) twice, we obtain the representation in (1.128). Since we have ⃗b ⃗a ⃗x˙ (s) = ⃗v1 (s) = · sin ω ˜ t − · cos ω ˜ t + ⃗c1 ω ˜ ω ˜ and ⃗x¨(s) = κ⃗v2 = κ(⃗a cos ω ˜ t + ⃗b sin ω ˜ t), and ⃗v1 and ⃗v2 are orthonormal vectors, it follows, for t = 0, κ2⃗a 2 = κ2 , that is, π ⃗a 2 = 1, and for t = , κ2⃗b 2 = κ2 , that is, ⃗b 2 = 1. Furthermore, we have 2 ˜ t + 2⃗a • ⃗b cos ω ˜ t sin ω ˜ t + ⃗b 2 sin2 ω ˜t ⃗v22 = ⃗a 2 cos2 ω = 1 + 2⃗a • ⃗b cos ω ˜ t sin ω ˜ t = 1 for all t, that is, ⃗a • ⃗b = 0. The condition ⃗v1 • ⃗v2 = 0 implies (−⃗b/˜ ω + ⃗c1 ) • ⃗a = ⃗c1 • ⃗a = 0 for t = 0 and Finally ⃗v12 = 1 implies

(⃗a/˜ ω + ⃗c1 ) • ⃗b = ⃗c1 • ⃗b = 0 for t = π/2.

1 = (⃗a 2 /˜ ω 2 ) sin2 ω ˜ t + (⃗b 2 /˜ ω 2 ) cos2 ω ˜ t + ⃗c12 = 1/˜ ω 2 + ⃗c12 , hence ⃗c12 = 1 −

1 1 c2 = 1 − = . ω ˜2 1 + c2 1 + c2

Proposition 1.12.4. Every curve with constant curvature and torsion is a helix (including the special cases).

Curves in Three-Dimensional Euclidean Space ■ 87

Proof. We apply Theorem 1.12.3. Let s0 ∈ R be given, κ(s) = κ be constant, and τ (s) = c · κ for all s. Then we have t(s) = κ(s − s0 ) by (1.130) in Theorem 1.12.3.

(i) First we assume κ = 0. If κ = 0, then t(s) = 0 and it follows from (1.128) that the corresponding curve γ has a parametric representation ⃗x(s) = − We put ⃗b1 = −



⃗b (s − s0 ) + ⃗c1 s + ⃗c0 . ω ˜2

⃗b − ⃗c1 ω ˜



s0 and d⃗ = ⃗b + ⃗c0 ω ˜

and obtain from (1.129) and (1.130) ⃗x(s) = ⃗b1 s + d⃗ with ∥⃗b1 ∥ =



1 + ∥⃗c1 ∥2 = ω ˜2



1 c2 + = 1, 1 + c2 1 + c2

which is a parametric representation of a straight line through the point with position vector d⃗ and in the direction of the unit vector ⃗b1 . (ii) Now we assume κ ̸= 0. If κ ̸= 0, then it follows from (1.128) that the curve γ has a parametric representation ⃗a ⃗x(s) = ω ˜

s

s0

s ⃗b  sin (˜ ω κ(σ − s0 )) dσ − cos (˜ ω κ(σ − s0 )) dσ + ⃗c1 s + ⃗c0 ω ˜ s0

s s  ⃗b  ⃗a  ω κ(σ − s0 )) + 2 sin (˜ ω κ(σ − s0 )) + ⃗c1 s + ⃗c0 = − 2 cos (˜  ω ˜ κ ω ˜ κ s0 s0

⃗b ⃗a ⃗ ω κ(s − s0 )) + 2 sin (˜ ω κ(s − s0 )) + ⃗c1 s + d, = − 2 cos (˜ ω ˜ κ ω ˜ κ ⃗a where d⃗ = ⃗c0 + 2 . ω ˜ κ

(1.134)

(ii.1) First we assume c = 0. If c = 0, then τ = 0, and γ is a planar curve, and we have ⃗c1 = ⃗0 by (1.129), and it follows that  2   ⃗b ⃗ a 1   ⃗ = − ∥⃗x(s)−d∥ cos (˜ ω κ(s − s sin (˜ ω κ(s − s )) + )) ,  = 0 0  ω  ˜ 2κ ω ˜ 2κ (˜ ω 2 κ)2 2

that is, γ is a circle line with radius r = 1/(˜ ω 2 κ), its centre in the point ⃗ D with position vector d, and in the plane through D and spanned by the vectors ⃗a and ⃗b.

88 ■ Differential Geometry and Its Visualization

(ii.2) Now we assume c ̸= 0. If c ̸= 0, then we put 



⃗b1 = − ⃗a cos (˜ ω κs0 ) + ⃗b sin (˜ ω κs0 ) ,

⃗b2 = ⃗b cos (˜ ω κs0 ) − ⃗a sin (˜ ω κs0 ), ⃗b3 = ⃗c1 , r = 1 , h = cr and ω = √ 1 , ∥⃗c1 ∥ ω ˜ 2κ r2 + h2

and obtain from (1.129) ∥⃗b1 ∥ = ∥⃗b2 ∥ = ∥⃗b3 ∥ = 1, ⃗bi • ⃗bk = 0 for i ̸= k, ⃗c1 = ⃗b3 ∥⃗c1 ∥ = ⃗b3 and

ω ˜κ =



 

2

 h c2 ⃗b3  r2 2 = ⃗b3 |h|ω = 1 + c2 1 + hr2

1 1 1 = √ =√ = ω. 2 2 rω ˜ r 1+c r + h2

Thus (1.129) yields ⃗x(s) − |h|ωs⃗b3 − d⃗ =− 





⃗a (cos (˜ ω κs) cos (˜ ω κs0 ) + sin (˜ ω κs) sin (˜ ω κs0 )) ω ˜ 2κ 

⃗b ω κs) cos (˜ ω κs0 ) − cos (˜ + 2 (sin (˜ ω κs) sin (˜ ω κs0 )) ω ˜ κ 



= −r cos (˜ ω κs) ⃗a cos (˜ ω κs0 ) + ⃗b sin (˜ ω κs0 ) 



+ r sin (˜ ω κs) ⃗b cos (˜ ω κs0 ) ω κs0 ) − ⃗a sin (˜ = r cos (˜ ω κs)⃗b1 + r sin (˜ ω κs)⃗b2 = r cos (ωs)⃗b1 + r sin (ωs)⃗b2 .

This is a parametric representation of a helix with its axis through the ⃗ and in the direction of the vector ⃗b3 . point with position vector d,

Proposition 1.12.5. Let δ, c > 0, κ(s) = δ/s and τ (s) = c · κ(s) for all s > 0. Then all curves with curvature κ and torsion τ are given by ⃗x(s) =

s cos (˜ ω y(s)) ⃗ s sin (˜ ω y(s)) ⃗ |c|  b1 +  2 2 b2 + s⃗b3 + d⃗0 , 2 2 ω ˜ ω ˜ (˜ ω δ + 1) ω ˜ (˜ ω δ + 1)

(1.135)

where y(s) = δ log ( ss0 ) for s ∈ (0, ∞), and s0 is a constant, the vectors ⃗bk are constant ˜ 2 = c2 + 1. Every curve is on a orthonormal vectors, d⃗0 is a constant vector, and ω

Curves in Three-Dimensional Euclidean Space ■ 89

Figure 1.61

Lines of constant slope on a cone

⃗ its axis along the vector ⃗b3 cone with its vertex in the point with position vector d, and an angle of   1 −1  β = 2 tan |c| (˜ ω 2 δ 2 + 1) at its vertex (Figure 1.61).

Proof. Since κ(s) = δ/s for s > 0, we obtain from (1.130) in Theorem 1.12.3 s

t(s) =

κ(σ) dσ = δ log (

s0

s ) for some constant s0 > 0. s0

We put y(σ) = δ log (

s0 dσ σ = exp (y/δ). ), that is, σ = s0 exp (y/δ) and s0 dy δ

By (1.128) in Theorem 1.12.3, we have to evaluate the integrals I1 =

s

s0 sin (˜ ω t(σ)) dσ = δ

y(s) 

s

s0 cos (˜ ω t(σ)) dσ = δ

y(s) 

s0

and I2 =

s0

sin (˜ ω y) exp (y/δ) dy

0

cos (˜ ω y) exp (y/δ) dy.

0

Integration by parts yields δ exp (y/δ) I˜1 = I1 = s0 ω ˜ 2 + δ12



1 sin (˜ ω y) − ω ˜ cos (˜ ω y) δ

y(s)    0

90 ■ Differential Geometry and Its Visualization

= hence I1 =

  s ω ˜δ 2 1 , δ sin (˜ ω y(s)) − ω ˜ δ 2 cos (˜ ω y(s)) + 2 2 2 2 s0 ω ˜ δ +1 ω ˜ δ +1

ω ˜ δs0 , +1

(1.136)

s s0 (cos (˜ ω y(s)) + ω ˜ δ sin (˜ ω y(s))) − 2 2 . ω ˜ 2δ 2 + 1 ω ˜ δ +1

(1.137)

s

ω ˜ 2δ 2

+1

(sin (˜ ω y(s)) − ω ˜ δ cos (˜ ω y(s))) +

ω ˜ 2δ 2

and similarly I2 = We put

and

⃗ ˜ δ⃗a ⃗a − ω ˜ δ⃗b ⃗ ⃗c1 ⃗b1 = − √b + ω , ⃗b2 = √ , b3 = ∥⃗c1 ∥ ω ˜ 2δ 2 + 1 ω ˜ 2δ 2 + 1 d⃗0 = ⃗c0 +

s0 (˜ ω δ⃗a + ⃗b). ω ˜ (˜ ω 2 δ 2 + 1)

Then it follows from (1.129) in Theorem 1.12.3 that ∥⃗bk ∥ = 1 and ⃗bi • ⃗bk = 0 for i ̸= k,

and we obtain from (1.129) in Theorem 1.12.3, and from (1.136) and (1.137) I1 ⃗ I2 − b + ⃗c1 s + ⃗c0 ω ˜ ω ˜  s = ⃗a sin (˜ ω y(s)) − ⃗aω ˜ δ cos (˜ ω y(s)) ω ˜ (˜ ω 2 δ 2 + 1)   ω y(s)) + ⃗b˜ ω δ sin (˜ ω y(s)) − ⃗b cos (˜

⃗x(s) = ⃗a

+

s0 2 ω ˜ (˜ ω δ2

+ 1)

(˜ ω δ⃗a + ⃗b) + ⃗c1 s + ⃗c0

⃗b + ω ˜ δ⃗a s cos (˜ ω y(s))  =−  2 2 ω ˜ (˜ ω δ + 1) (˜ ω 2 δ 2 + 1) s sin (˜ ω y(s)) ⃗a + ω ˜ δ⃗b  +  2 2 + ⃗c1 s + d⃗0 ω ˜ (˜ ω δ + 1) (˜ ω 2 δ 2 + 1) =

s cos (˜ ω y(s)) ⃗ s sin (˜ ω y(s)) ⃗ |c|  b1 +  2 2 b2 + s⃗b3 + d⃗0 , 2 2 ω ˜ ω ˜ (˜ ω δ + 1) ω ˜ (˜ ω δ + 1)

where y(s) = y(σ) = δ log (σ/s0 )). This shows (1.135). Putting x1 = and tan β˜ =

s cos (˜ ω y(s)) s sin (˜ ω y(s)) s|c|  , x2 =  2 2 , x3 = 2 2 ω ˜ ω ˜ (˜ ω δ + 1) ω ˜ (˜ ω δ + 1) 1 , |c| (˜ ω 2 δ 2 + 1) 

Curves in Three-Dimensional Euclidean Space ■ 91

Figure 1.62

The constant angle between a line of constant slope and a direction

we see that

˜ 3 )2 = 0. (x1 )2 + (x2 )2 − tan2 β(x

Consequently the curve is on a cone with its vertex in the point with position vector d⃗0 , its axis along the vector ⃗b3 , and an angle of β = 2β˜ at its vertex. The following result explains the term line of constant slope. Theorem 1.12.6. A curve with a parametric representation ⃗x(s) (s ∈ I) and positive curvature κ(s) is a line of constant slope if and only if there is a unit vector ⃗u such that ⃗u • ⃗v1 (s) = constant for all s ∈ I (Figure 1.62). (1.138)

If ϕ denotes the constant angle between the vectors ⃗u and ⃗v1 , then we have τ (s) = cot ϕ · κ(s) for all s ∈ I. Proof. (i) First we show the sufficiency of the condition in (1.138). We assume that ⃗u • ⃗x˙ (s) = ⃗u • ⃗v1 (s) = cos ϕ for all s ∈ I,

where ϕ is the constant angle between ⃗u and the tangents of the curve. This implies ⃗u • ⃗x¨(s) = 0 for all s, hence ⃗u • ⃗v2 (s) = 0 for all s, since κ(s) ̸= 0.  Writing ⃗u = 3k=1 bk⃗vk (s), we obtain b1 = cos ϕ, b2 = 0 and b3 = sin ϕ, since ⃗u is a unit vector. Hence ⃗u = cos ϕ ⃗v1 (s) + sin ϕ ⃗v3 (s) for all s. The first and third formulae in (1.84) together imply ⃗0 = ⃗u˙ = cos ϕ ⃗v˙ 1 (s) + sin ϕ ⃗v˙ 3 (s) = κ(s) cos ϕ ⃗v2 (s) − τ (s) sin ϕ ⃗v2 (s),

that is,

0 = κ(s) cos ϕ − τ (s) sin ϕ, or equivalently τ (s) = κ(s) cot ϕ for all s ∈ I.

92 ■ Differential Geometry and Its Visualization

(ii) Now we show the necessity of the condition in (1.138). We assume that τ (s) = cκ(s) for all s ∈ I, where c is a constant. Putting c = cot ϕ, we obtain

and

⃗0 = (κ(s) cos ϕ − τ (s) sin ϕ)⃗v2 (s) = cos ϕ ⃗v˙ 1 (s) + sin ϕ ⃗v˙ 3 (s) ⃗u = ⃗v1 (s) cos ϕ + ⃗v3 (s) sin ϕ for all s,

by integration. This implies ⃗u • ⃗v1 (s) = const for all s ∈ I. The following interesting result holds. Theorem 1.12.7. Let γ be a line of constant slope with a parametric representation ⃗x(s) (s ∈ I) and curvature κ(s) ̸= 0 for all s. If φ denotes the constant angle between γ and the unit vector ⃗u, then the arc length s⊥ and the curvature κ⊥ of the orthogonal projection γ ⊥ of γ on to a plane orthogonal to the vector ⃗u are given by s⊥ = (s − s0 )| sin φ|

(1.139)

and κ⊥ (s) =

κ(s) sin2 φ

(φ ̸= 0, π).

(1.140)

Proof. Obviously the orthogonal projection γ ⊥ has a parametric representation First we observe that

⃗x⊥ (s) = ⃗x(s) − (⃗x(s) • ⃗u)⃗u.

d⃗x⊥ = ⃗v1 − (⃗v1 • ⃗u)⃗u = ⃗v1 − cos φ⃗u ds and    d⃗ ⊥ 2  x  u + cos2 φ = 1 − cos2 φ = sin2 φ   = 1 − 2 cos φ⃗v1 • ⃗  ds 

together imply (1.139) for φ ̸= 0, π. Furthermore it follows from

d2⃗x⊥ = ⃗v˙ 1 = κ⃗v2 , ds2 d⃗x⊥ d2⃗x⊥ × = κ(⃗v1 − cos φ⃗u) × ⃗v2 = κ(⃗v3 − cos φ(⃗u × ⃗v2 )) ds ds2

and  2  d⃗ ⊥ d2⃗x⊥   x  × u × ⃗v2 ) cos φ + cos2 φ)   = κ2 (1 − 2⃗v3 • (⃗  ds ds2 

= κ2 (1 + cos2 φ − 2⃗u • ⃗v1 cos φ) = κ2 sin2 φ

Curves in Three-Dimensional Euclidean Space ■ 93

Figure 1.63

A curve with κ(s) = s, hence ϕ(s) = s2 /2

that we have by (1.88) in Proposition 1.7.6  ⊥  2 ⊥  d⃗x  ds × dds⃗x2  κ(s) κ(s)| sin φ| = = κ⊥ (s) =   3  d⃗x⊥ 3 | sin φ| sin2 φ  ds 

(φ ̸= 0, π).

This shows (1.140).

There is a useful formula for the curvature of a planar curve. Theorem 1.12.8. Let γ be a planar curve with curvature κ(s) ̸= 0. If ϕ(s) denotes the oriented angle between the positive x1 –axis and the tangent vector of γ at s, then we have    dϕ(s)    for all s (Figure 1.63). κ(s) =  (1.141) ds 

Proof. The tangent vectors ⃗v1 of the planar curve γ satisfy ⃗v1 (ϕ) = {cos ϕ, sin ϕ} and hence ⃗v1 (ϕ) •

d⃗v1 (ϕ) = {− sin ϕ, cos ϕ}, dϕ

d⃗v1 (ϕ) = 0. dϕ

Thus the unit vector d⃗v1 /dϕ is orthogonal to the tangent vector at every s. Consequently we conclude d⃗v1 (ϕ) = ±⃗v2 (ϕ). (1.142) dϕ

94 ■ Differential Geometry and Its Visualization

Figure 1.64

Clothoids (left) and logarithmic spirals (right)

On the other hand, it follows from Frenet’s first formula (1.84) in Theorem 1.7.1 that dϕ d⃗v1 ⃗v˙ 1 (s) = (ϕ(s)) (s) = κ(s)⃗v2 (s). dϕ ds

(1.143)

Since κ(s) > 0 for all s, the identity in (1.141) follows from (1.142) and (1.143). Remark 1.12.9. We may choose the orientation of the curve in Theorem 1.12.8 so that dϕ(s) κ(s) = for all s. (1.144) ds Visualization 1.12.10. (a) Every planar curve with κ ≡ const > 0 is a circle line of radius r = 1/κ. (b) Every planar curve with κ(s) = c · s (s ∈ (0, ∞)), where c > 0 is a constant, is a so–called clothoid given by a parametric representation ⃗x(ϕ) =



1 2c



cos ϕ √ dϕ, ϕ





sin ϕ √ dϕ + ⃗c1 (ϕ ∈ R \ {0}) ϕ

where ⃗c1 is a constant vector (in Figure 1.63 and left in Figure 1.64). (c) Every planar curve with κ(s) = c/s (s ∈ (0, ∞)), where c > 0 is a constant, is a logarithmic spiral (right in Figure 1.64). Proof. (a) The identity κ(s) =

dϕ(s) = 1/r for all s ds

yields ⃗v1 (s) = {cos ϕ(s), sin ϕ(s)} =

dϕ 1 d⃗x d⃗x (ϕ(s)) (s) = (ϕ(s)), dϕ ds r dϕ

Curves in Three-Dimensional Euclidean Space ■ 95

hence

⃗x(ϕ) = r{sin ϕ, − cos ϕ} + ⃗c, where ⃗c is a constant vector.

This is a parametric representation of a circle line of radius r, with its centre in the point C with the position vector ⃗c. (b) Similarly as in Part (a), the identity dϕ c κ(s) = (s) = c · s, that is, ϕ(s) = s2 or s = ds 2



yields ⃗v1 (s) = {cos ϕ(s), sin ϕ(s)} = that is, ⃗v1 (ϕ) =



2 ϕ c

dϕ d⃗x d⃗x(ϕ)  d⃗x (ϕ(s)) (s) = (ϕ(s))cs = 2cϕ, dϕ ds dϕ dϕ

d⃗x 1 2cϕ , hence ⃗x(ϕ) = √ dϕ 2c



cos ϕ √ dϕ, ϕ





sin ϕ √ dϕ + ⃗c1 ϕ

for some constant vector ⃗c1 . (c) As in Part (b), we obtain κ(s) =

c 1 s dϕ (s) = , that is, ϕ(s) = log ds s c s0

for some constant s0 > 0, or s = s0 exp (ϕ/c), which yields ⃗v1 (s) = {cos ϕ(s), sin ϕ(s)} = =

dϕ d⃗x (ϕ(s)) (s) dϕ ds

c d⃗x(ϕ) d⃗x (ϕ(s)) = cs0 exp (−ϕ/c), dϕ s dϕ

that is, ⃗v1 (ϕ) = cs0 exp (−ϕ/c) hence

d⃗x , dϕ

exp (ϕ/c) exp (ϕ/c) dx2 dx1 = = cos ϕ and sin ϕ dϕ cs0 dϕ cs0

for the components of the vector ⃗x. Introducing the complex number z = x1 + ix2 , we obtain exp (ϕ(i + 1/c)) exp (ϕ(i + 1/c)) dz = and z = , dϕ cs0 (1 + ci)s0

96 ■ Differential Geometry and Its Visualization

Figure 1.65

The orthogonal projection of a line of constant slope on a paraboloid

or in polar coordinates r2 = |z|2 and ϕ, r2 =

exp (2ϕ/c) , s20 (1 + c2 )

and finally ⃗x(ϕ) =

s0

√

1 exp (ϕ/c){cos ϕ, sin ϕ} for ϕ ∈ R. 1 + c2

Remark 1.12.11. We could also have used Theorem 1.12.3 to establish the statements in Visualization 1.12.10. Now we consider two examples that are geometrically interesting as well as instructive in the sense that most results of the theory of curves are applied. Visualization 1.12.12. Let γ be a curve on a paraboloid of revolution such that γ has a constant angle with the axis of the paraboloid. Then the orthogonal projection γ ⊥ of γ onto a plane orthogonal to the axis of the paraboloid is the involute of a circle line (Figure 1.65). Proof. Let the x3 –axis be the axis of rotation of the paraboloid. Then the paraboloid may be given by the following equation (x1 )2 + (x2 )2 − 2ax3 = 0, where a ̸= 0 is a constant.

(1.145)

Curves in Three-Dimensional Euclidean Space ■ 97

If ⃗vk (k = 1, 2, 3) are the vectors of the trihedra of the curve γ, then ⃗v1 • ⃗e 3 = cos φ and d 0 = (⃗v1 • ⃗e 3 ) = ⃗v˙ 1 • ⃗e 3 = κ⃗v2 • ⃗e 3 for κ ̸= 0. ds This implies ⃗v2 • ⃗e 3 = 0 and consequently ⃗v3 • ⃗e 3 = sin φ. The osculating plane of γ −→

−→

at an arbitrary point P ∈ γ is given by the equation (0X − 0P ) • ⃗v3 = 0. Putting c(P ) = we obtain

1 −→ OP • ⃗v3 sin φ

(φ ̸= 0, π),

 1  1 1 x ⃗e • ⃗v3 + x2⃗e 2 • ⃗v3 + c(P ). (1.146) sin φ Let γ˜ denote the curve of intersection of the paraboloid and the osculating plane of γ at P . Then γ˜ is a second order approximation of γ. Obviously the orthogonal projection γ˜ ⊥ of γ˜ on to the x1 x2 –plane is also a second order approximation of γ ⊥ . Eliminating x3 from (1.145) and (1.146), we obtain an equation for γ˜⊥

x3 = −

(x1 )2 + (x2 )2 +

 2a  1 1 x ⃗e • ⃗v3 + x2⃗e 2 • ⃗v3 = 2ac(P ). sin φ

This is the equation of a circle line. The distance d of the centre of the circle line from the origin is given by d2 =

a2 2 l where l2 = (⃗e 1 • ⃗v3 )2 + (⃗e 2 • ⃗v3 )2 . sin2 φ

Thus |l| is the length of the projection of the vector ⃗v3 on to the x1 x2 –plane. Since ⃗e 3 •⃗v3 = sin φ, we have l2 = cos2 φ and consequently d2 = a2 cot2 φ. Thus the centres of curvature of γ ⊥ lie on a circle line, which is the evolute of γ ⊥ . Visualization 1.12.13. Let γ be a curve on a sphere of radius R > 0 such that γ has a constant angle with some constant unit vector ⃗u. Then the orthogonal projection γ ⊥ of γ on to a plane E orthogonal to the vector ⃗u is an epicycloid with λ = 1 (Figure 1.66). Proof. Using the notations of Visualizations 1.5.6 and 1.7.7 with R replaced by R0 and λ = 1, we obtain for a suitable choice s∗ along the epicycloid from (1.65) t(s∗ ) =





r0 ∗ 2r cos−1 s . r0 4rR0

Furthermore, since we have by (1.93)

(1.147)

2r + r0 2r + r0   = κ(t) = √ 2 2r(r0 + r) 1 − cos rr0 t 4r(r0 + r) 1 − cos2

r0 2r t

,

it follows from (1.147) that

1 ρ (s ) = = κ(t(s∗ )) ∗

∗



(4rR0 )2 − r02 (s∗ )2 R0 + r

.

(1.148)

98 ■ Differential Geometry and Its Visualization

Figure 1.66

The orthogonal projection of a line of constant slope on a sphere

If S denotes a sphere of radius R, then S is the osculating sphere of any curve on S. Let ρ and τ denote the radius of curvature and torsion of γ. Then it follows by Theorem 1.9.5 that ρ˙ 2 R 2 = ρ2 + 2 . (1.149) τ If φ denotes the constant angle between γ and ⃗u, then we have cos φ = ⃗u • ⃗v1 and 0 = ⃗u • ⃗v˙ 1 = κ⃗u • ⃗v2 , and κ ̸= 0 implies 0 = ⃗u • ⃗v˙ 2 . Since ⃗u is a unit vector orthogonal to ⃗v2 , we may write ⃗u = cos φ⃗v1 + sin φ⃗v3 . This implies and

⃗0 = ⃗u˙ = cos φ⃗v1 + sin φ⃗v˙ 3 = (κ cos φ − τ sin φ)⃗v2 κ cos φ = τ sin φ or τ =

1 cot φ (φ ̸= 0, π). ρ

Substituting this last result into (1.149), we obtain R2 − ρ2 = ρ2 ρ˙ 2 tan2 φ, and for φ ̸= 0, π

ρ˙ = ±

This yields ±







R2 − ρ2 . ρ| tan φ|

 1 ρ (s − s0 ). dρ = ∓ R2 − ρ2 = 2 2 | tan φ| R −ρ

We choose s0 = 0 to obtain ρ(s) =



R2 − s2 cot2 φ for |s| < R| tan φ|.

Curves in Three-Dimensional Euclidean Space ■ 99

We have for the arc length s⊥ and radius of curvature ρ⊥ of the orthogonal projection γ ⊥ of γ onto a plane orthogonal to the vector ⃗u, by (1.139) and (1.140) in Theorem 1.12.7, ρ⊥ (s⊥ ) = ρ⊥ (s(s⊥ )) = ρ(s(s⊥ )) sin2 φ = =



R2 −



(s⊥ )2 cot2 φ sin2 φ sin2 φ

R2 sin4 φ − (s⊥ )2 cos2 φ for |s⊥ | < R sin φ| tan φ|.

(1.150)

We consider the epicycloid with r0 = R cos φ, r = R(1 − cos φ)/2 and R0 = r + r0 . Then we obtain for the radius of curvature of this epicycloid from (1.148) ρ(s∗ ) =



R2 sin4 φ − cos2 φ(s∗ )2 .

Thus the radii of curvature ρ⊥ and ρ are the same, and the statement of the example follows from the fundamental theorem of curves.

1.13

SPHERICAL IMAGES OF A CURVE

A good illustration can be obtained of the change in direction of the vectors of a trihedron of a curve attached to the origin of the coordinate system of threedimensional space. Since these vectors are unit vectors this will yield three curves on the unit sphere. Definition 1.13.1. Let γ be a curve with a parametric representation ⃗x(s) (s ∈ I), and the vectors of the trihedra ⃗vk (s) (k = 1, 2, 3) for s ∈ I. The curves γ1 , γ2 and γ3 with parametric representations ⃗v1 (s), ⃗v2 (s) and ⃗v3 (s) (s ∈ I) are called the spherical images of the tangent, principal normal and binormal vectors (Figure 1.67). Visualization 1.13.2. We consider the helix of Visualization 1.6.7 (b) with a parametric representation ⃗x(s) = {r cos ωs, r sin ωs, hωs} (s ∈ R). Then we have ⃗v1 (s) = {−ωr sin ωs, ωr cos ωs, hω},

⃗v2 (s) = {− cos ωs, − sin ωs, 0} and

⃗v3 (s) = {ωh sin ωs, −ωh cos ωs, ωr} (s ∈ R) for its spherical images (Figure 1.68).

100 ■ Differential Geometry and Its Visualization

Figure 1.67

The spherical images of the tangent, principal normal and binormal vectors

Figure 1.68

The spherical images of a helix

of γ

There are some simple relations between the arc lengths of a curve and of its spherical images. Theorem 1.13.3. Let γ be a curve with non–vanishing curvature κ and with torsion τ , and s1 , s2 and s3 denote the arc lengths along the spherical images of its tangent, principal normal and binormal vectors, respectively. Then we have ds2 √ 2 ds3 ds1 = κ, = κ + τ 2 and = |τ |; ds ds ds

(1.151)

Curves in Three-Dimensional Euclidean Space ■ 101

in particular, we have

ds2 ds1 = ds ds

(1.152)

if and only if γ is a planar curve. Proof. Let ⃗x(s) (s ∈ I) be a parametric representation of γ, and ⃗vk (s) denote, as usual, the tangent, principal normal and binormal vectors of γ at s. The arc lengths of the spherical images γk are given by sk (s) = and so



s

s0

∥⃗v˙ k (σ)∥ dσ



dsk (s) = ∥⃗v˙ k (s)∥ = ⃗v˙ k (s) • ⃗v˙ k (s) (k = 1, 2, 3) for all s ∈ I. ds Applying Frenet’s formulae (1.84) in Theorem 1.7.1, we obtain for all s ∈ I  ds1 (s) = κ(s)⃗v2 (s) • κ(s)⃗v2 (s) = |κ(s)| = κ(s), ds

  ds2 (s) = (−κ(s)⃗v1 (s) + τ (s)⃗v3 (s)) • (−κ(s)⃗v1 (s) + τ (s)⃗v3 (s)) = κ2 (s) + τ 2 (s) ds and  ds3 (s) = −τ (s)⃗v3 (s) • (−τ (s))⃗v3 (s) = |τ (s)|. ds Thus we have proved the identities in (1.151). Since planar curves are characterized by the condition τ ≡ 0 by Theorem 1.8.1, the equality in (1.152) is an immediate consequence of the second identity in (1.151).

The next result characterizes lines of constant slope by the spherical images of their tangent vectors. Theorem 1.13.4. A curve is a line of constant slope if and only if the spherical image of its tangent vectors is a circle line (Figure 1.69). Proof. Let γ ∗ denote the spherical image of the tangent vectors of a curve γ, and ⃗x(s) be a parametric representation of γ. (i) First we assume that γ ∗ is a circle line. Then γ ∗ has a parametric representation 

s1 ⃗v1 (s1 ) = ⃗ar cos r





s1 + ⃗br sin r



+ ⃗c,

102 ■ Differential Geometry and Its Visualization

Figure 1.69

a catenoid

The spherical images of the tangent vectors of lines of constant slope on

where s1 and r are the arc length along γ ∗ and the radius of γ ∗ , and the constant vectors ⃗a, ⃗b and ⃗c satisfy the conditions ⃗a 2 = ⃗b 2 = 1, ⃗a • ⃗b = ⃗a • ⃗c = ⃗b · ⃗c = 0.

(1.153)

Now ∥⃗v1 (s1 )∥ = 1 implies ⃗c 2 = 1 − r2 . If κ denotes the curvature along γ, then we have by the first identity in (1.151) of Theorem 1.13.3 s1 (s) =

s

κ(σ) dσ,

s0

hence



s1 (s) ⃗v1 (s1 ) = ⃗ar cos r

and integration yields ⃗x(s) = ⃗ar

s

s0

cos









s1 (s) + ⃗br sin r

s1 (σ) dσ + ⃗br r

s

s0

sin





+ ⃗c,



s1 (σ) dσ + ⃗cs + ⃗c0 , r

where ⃗c0 is a constant vector. This is a parametric representation of a line of constant slope by Theorem 1.12.3. Hence γ is a line of constant slope. (ii) Now we assume that γ is a line of constant slope. Let c be the constant such that τ (s) = cκ(s). Then parametric representation ⃗x(s) of γ satisfies by Theorem 1.12.3 ⃗v1 (s) =

⃗b ⃗a sin (ωt(s)) − cos (ωt(s)) + ⃗c, ω ω

Curves in Three-Dimensional Euclidean Space ■ 103

where the constant vectors ⃗a, ⃗b and ⃗c satisfy the conditions in (1.153), and c2 ⃗c = ≤ 1, ω = 1 + c2 and t(s) = 1 + c2 2

s

κ(σ) dσ.

s0

It follows that ∥⃗v1 (s) − ⃗c∥ =

⃗b 2 ⃗a 2 1 1 2 sin (ωt(s)) + cos2 (ωt(s)) = 2 = 2 , 2 2 ω ω ω c

that is, ⃗v2 (s) is a parametric representation of a circle line.

Taylor & Francis Taylor & Francis Group

http://taylorandfrancis.com

CHAPTER

2

Surfaces in Three-Dimensional Euclidean Space

In this chapter, we deal with the local differential geometry of surfaces in threedimensional Euclidean space E3 . This means that we study the geometric shape of surfaces in the neighbourhood of an arbitrary one of their points. The most important concepts arising in this task are those of the normal, principal, Gaussian and mean curvature in Section 2.5. Additional topics are • curves on surfaces in Section 2.1

• tangent planes and normal vectors of surfaces in Section 2.2

• first and second fundamental coefficients in Sections 2.3 and 2.4 • Meusnier’s theorem, Theorem 2.5.7 in Section 2.5

• principal directions in Section 2.5

• Euler’s formula, (2.111) in Theorem 2.5.31 of Section 2.5

• the local shape of surfaces in Section 2.6 • Dupin’s indicatrix in Section 2.7

• lines of curvature and asymptotic lines in Section 2.8 • triple orthogonal systems in Section 2.9

• the Gauss and Weingarten equations, (2.42) of Section 2.4 and (2.141) in Theorem 2.10.1 of Section 2.10.

DOI: 10.1201/9781003370567-2

105

106 ■ Differential Geometry and Its Visualization

2.1

SURFACES AND CURVES ON SURFACES

In this section, we deal with the representation of surfaces in three-dimensional E3 and with curves on surfaces. Whereas a curve can be described by a parametric representation of the position vectors of its points with respect to one real parameter, two real parameters are needed in a parametric representation of the position vectors of the points of a surface. In general, it is not possible to globally relate all points of a surface to parameters in one region of the plane in a one–to–one way at the same time. For instance, spherical coordinates cannot be applied in the poles of a sphere. In local differential geometry, we confine our studies to sufficiently small sections of a surface. We start with the definition of a surface in three-dimensional R3 and recall two notations: An open connected subset of R2 is called a domain. By C r (D), we denote the class of all functions f = (f 1 , f 2 , f 3 ) : D → R3 for which the coordinate functions f k (k = 1, 2, 3) have continuous partial derivatives of order r on D. Definition 2.1.1. Let D ⊂ R2 be a domain and f = (f 1 , f 2 , f 3 ) : D → R3 be a function with f ∈ C 3 (D). The point set 

S = X ∈ R3 : X = f (u1 , u2 ) = (f 1 (u1 , u2 ), f 2 (u1 , u2 ), f 3 (u1 , u2 ))



is called a surface in R3 , if in addition the following condition holds 

∂f 1 ∂f 2 ∂f 3 , , ∂u1 ∂u1 ∂u1



×



∂f 1 ∂f 2 ∂f 3 , , ∂u2 ∂u2 ∂u2





(u1 , u2 ) ∈ D (2.1)

̸= ⃗0 on D;

(2.2)

(f, D) is called a parametric representation of S, and the values u1 and u2 are called parameters of S. A surface S may also be given in terms of the position vectors of its points 

 

⃗x = ⃗x(ui ) = ⃗x(u1 , u2 ) = x1 (u1 , u2 ), x2 (u1 , u2 ), x3 (u1 , u2 )



(u1 , u2 ) ∈ D , (2.3)

∂⃗x where xk (ui ) = f k (ui ) (k = 1, 2, 3), and the vectors ⃗xj = (j = 1, 2) satisfy the ∂uj condition (2.4) ⃗x1 × ⃗x2 ̸= ⃗0.

As in the case of curves, (2.3) will also be referred to as a parametric representation for S; in fact we shall rather use (2.3) than (2.1). For further studies in the theory of surfaces, the vector functions in (2.3) will often have to have higher-order continuous partial derivatives. In general, the existence of continuous third-order partial derivatives will be sufficient. Curves on surfaces are of special interest.

Surfaces in Three-Dimensional Euclidean Space ■ 107

Figure 2.1

Curves on a surface (right) and in its parameter plane (left)

Definition 2.1.2. (a) A curve γ is on a surface S with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D) if it is given by a parametric representation ⃗x(t) = ⃗x(ui (t)) (t ∈ I), where u1 (I) × u2 (I) ⊂ D,

(2.5)

and the functions uk (k = 1, 2) have continuous derivatives of order (r ≥ 1), and satisfy      1 ′   2 ′  (u ) (t) + (u ) (t) ̸= 0 for all t ∈ I.

The curve γ is called curve on the surface S (Figure 2.1). (b) Curves for which one parameter of the surface is constant are called parameter lines. The curves with u1 (t) = t and u2 (t) = const are called u1 –lines, those with u1 (t) = const and u2 (t) = t are called u2 –lines (Figure 2.2).

In general, a surface S may have various parametric representations. From (2.1), we can obtain a new vector function by introducing two new parameters u∗k (k = 1, 2) such that ui = ui (u∗k ) (i = 1, 2). (2.6)

Figure 2.2

Parameter lines on a surface

108 ■ Differential Geometry and Its Visualization

We have to take care that the range of the functions defined in (2.6) contains the domain D of the parameters (u1 , u2 ) and that the conditions for parametric representations of surfaces are satisfied. Definition 2.1.3. A parameter transformation (2.6) is called admissible if the following three conditions hold (i) The functions in (2.6) are defined on a domain D∗ such that D ⊂ u1 (D∗ ) × u2 (D∗ ). (ii) ui ∈ C r (D) (i = 1, 2) for some r ≥ 1.

(iii) For the Jacobian of the functions in (2.6), we must have   ∂u1   ∂u∗1 ∂(ui )  =  ∂(u∗k )  ∂u2 

∂u∗1



∂u1  ∂u∗2 

 ̸= 0 ∂u2  

∂u∗2





(u∗1 , u∗2 ) ∈ D∗ .

(2.7)

The same convention as in the case of curves is used concerning the term admissible. From now on we shall always make use of Einstein’s convention of summation: • If the same index appears in an expression twice, once as a subscript and once as a superscript, then summation takes place with respect to this index. For instance, we have ⃗xi • ⃗xk dui duk =

2 

i,k=1

⃗xi • ⃗xk dui duk .

Remark 2.1.4. If we use (2.6) to introduce new parameters u∗r for a surface with a parametric representation (2.3), then we have ⃗x(uk ) = ⃗x(uk (u∗r )) = ⃗x∗ (u∗j ). This implies ⃗x∗r = and

∂⃗x∗ ∂⃗x ∂uk ∂uk = = ⃗ x k ∂u∗r ∂uk ∂u∗r ∂u∗r

(r = 1, 2)

∂uk ∂uj × ⃗ x j ∂u∗1 ∂u∗2 1 ∂u ∂u2 ∂u1 ∂u2 = ⃗x1 ∗1 × ⃗x2 ∗2 − ⃗x1 ∗2 × ⃗x2 ∗1 ∂u ∂u ∂u ∂u

⃗x∗1 × ⃗x∗2 = ⃗xk

(2.8)

Surfaces in Three-Dimensional Euclidean Space ■ 109

Polar coordinates in the plane (left); the x1 x2 –plane with polar coordinates as parameters (right) Figure 2.3



∂u1 ∂u2 ∂u1 ∂u2 − = (⃗x1 × ⃗x2 ) ∂u∗1 ∂u∗2 ∂u∗2 ∂u∗1 = (⃗x1 × ⃗x2 ) ·



∂(uk ) , ∂(u∗r )

that is, ⃗x∗1 × ⃗x∗2 = (⃗x1 × ⃗x2 ) ·

∂(uk ) . ∂(u∗r )

(2.9)

Visualization 2.1.5. The x1 x2 –plane S has a parametric representation ⃗x(ui ) = {u1 , u2 , 0}



(u1 , u2 ) ∈ D = R2

with respect to the Cartesian coordinates u1 and u2 . Introducing polar coordinates u∗1 and u∗2 , we obtain ⃗x(u∗i ) = {u∗1 cos u∗2 , u∗1 sin u∗2 , 0}







u∗1 , u∗2 ) ∈ D∗ = (0, ∞) × (−π, π)

for S ∗ , which is the x1 x2 –plane with the negative x1 –axis and the origin removed (Figure 2.3). It follows for S that ⃗x1 = {1, 0, 0}, ⃗x2 = {0, 1, 0} and ⃗x1 × ⃗x2 = {0, 0, 1} = ̸ ⃗0 for all (u1 , u2 ) ∈ D, and for S ∗ that ⃗x1 = {cos u∗2 , sin u∗2 , 0}, ⃗x2 = {−u∗1 sin u∗2 , u∗1 cos u∗2 , 0}, and

⃗x1 × ⃗x2 = {0, 0, u∗1 } ̸= ⃗0 if and only if u∗1 ̸= 0.

We consider the two curves γ and γ ∗ in the parameter planes given by u1 (t) = u∗1 (t) = a exp (bt), and u2 (t) = u∗2 (t) = t (t ∈ R),

110 ■ Differential Geometry and Its Visualization

Figure 2.4

Curve on a plane with different parametric representations

where a, b > 0 are constants. The curve on S is given by ⃗x(ui (t)) = {a exp (bt), t, 0} (t ∈ R), and on S ∗ by ⃗x(u∗i (t)) = {a exp (bt) cos t, a exp (bt) sin t, 0} (t ∈ R) (Figure 2.4). Example 2.1.6. Let c1 ̸= c2 . We put 1

u (u ) = 2 tan ∗k

−1





u∗1 − u∗2 exp c1 − c2



−

π 2

and u2 (u∗k ) =

 c1 u∗1 − c2 u∗2  ∗1 ∗2 (u , u ) ∈ R2 . c1 − c2

Then we have, writing ϕ(u∗k ) = (u∗1 − u∗2 )/(c1 − c2 ),

  2 1 ∂u1 ∗k = · exp ϕ(u ) · 2 ∗1 ∗k ∂u c1 − c2 1 + (exp (ϕ(u )))

=

1 , (c1 − c2 ) · cosh (ϕ(u∗k ))

1 ∂u1 , = ∗2 ∂u (c1 − c2 ) · cosh (ϕ(u∗k )) c1 ∂u2 c2 ∂u2 = , =− ∗1 ∗2 ∂u c1 − c2 ∂u c1 − c2

Surfaces in Three-Dimensional Euclidean Space ■ 111

Figure 2.5

The hemispheres of Visualization 2.1.7 (a) and (b)

and c2 c1 ∂(ui ) =− + ∂(u∗k ) (c1 − c2 )2 · cosh (ϕ(u∗k )) (c1 − c2 ) · cosh (ϕ(u∗k )) =

1 ̸= 0. (c1 − c2 ) · cosh (ϕ(u∗k ))

Therefore, the parameter transformation is admissible. Surfaces in three-dimensional space may also be given be equations. Visualization 2.1.7. (a) The sphere of radius r > 0 centred at the origin has an equation 3 

(xk )2 = r2 ,

k=1

and we obtain a parametric representation for the upper semi–sphere with equator and north pole removed 

⃗x(ui ) = u1 , u2 ,



r2 − (u1 )2 − (u2 )2

 

0 < (u1 )2 + (u2 )2 < r2

(right in Figure 2.5). Furthermore, it follows that ⃗x1 = ⃗x2 = and ⃗x1 × ⃗x2 =





−u1 1, 0,  2 r − (u1 )2 − (u2 )2



−u2 0, 1,  2 r − (u1 )2 − (u2 )2





,

 

u2 u1   , ,1 r2 − (u1 )2 − (u2 )2 r2 − (u1 )2 − (u2 )2

̸= ⃗0.

112 ■ Differential Geometry and Its Visualization

Figure 2.6 Sphere with spherical parameters (left) and the parametrization of Visualization 2.1.7 (c)

We fix c with |c| < r. Then the u1 –line corresponding to u2 = c has a parametric representation 



r2 − t2 − c2

 











r2 − t2 − c2

 







⃗x(t) = t, c,

t ∈ (− r2 − c2 ,

r 2 − c2 ) ;

√ it is a semi–circle line in the plane x2 = c with radius r2 − c2 and centre in (0, c, 0). Similarly, the u2 –line corresponding to u1 = c has a parametric representation ⃗x(t) = c, t,

t ∈ (− r2 − c2 ,

r 2 − c2 ) ;

√ it is a semi–circle line in the plane x1 = c with radius r2 − c2 and centre in (c, 0, 0) (right in Figure 2.5). (b) The sphere of radius r > 0, centred at the origin, with north and south poles and a meridian removed has a parametric representation with respect to spherical coordinates 

1

2

1

2

1

⃗x(u ) = r cos u cos u , cos u sin u , sin u i

(Figures 2.5 and 2.6). We have

 



π π (u , u ) ∈ D = − , 2 2 1

2







× (0, 2π)



⃗x1 = r − sin u1 cos u2 , − sin u1 sin u2 , cos u1 , 



⃗x2 = r − cos u1 sin u2 , cos u1 cos u2 , 0 and



⃗x1 × ⃗x2 = −r2 cos u1 cos u1 cos u2 , cos u1 sin u2 , sin u1 = −r cos u1⃗x ̸= ⃗0





(u1 , u2 ) ∈ D .



The u1 –lines are the so–called meridians, and the u2 –lines are the so–called parallels (left in Figure 2.5).

Surfaces in Three-Dimensional Euclidean Space ■ 113

(c) Let c1 , c2 ∈ R with c1 ̸= c2 . We put ϕ = ϕ(ui ) = and ⃗x = ⃗x(ui ) =

u1 − u2 c 1 u1 − c 2 u2 , ψ = ψ(ui ) = c1 − c2 c1 − c2

  r i i i 1 2 2 {cos ψ(u ), sin ψ(u ), sinh ϕ(u )} (u , u ) ∈ R cosh ϕ(ui )

(2.10)

(right in Figure 2.6). This is one more parametric representation of a sphere of radius r and centre in the origin. Putting ⃗u(ψ) = {cos ψ, sin ψ, 0}, we obtain ⃗x =

r (⃗u(ψ) + ⃗e 3 · sinh ϕ), cosh ϕ 

and



r 1 −⃗u(ψ) · tanh ϕ + ⃗u ′ (ψ) · c1 + ⃗e 3 · , ⃗x1 = + (c1 − c2 ) cosh ϕ cosh ϕ   r 1 −⃗u(ψ) · tanh ϕ + ⃗u ′ (ψ) · c2 + ⃗e 3 · ⃗x2 = − (c1 − c2 ) cosh ϕ cosh ϕ 

r2 1 ⃗x1 × ⃗x2 = − e⃗3 · tanh ϕ + ⃗u(ψ) · 2 cosh ϕ (c1 − c2 ) cosh ϕ



̸= 0.

The u1 –line corresponding to u2 = u20 , and the u2 –line corresponding to u1 = u10 are given by ⃗x(t, u20 )

r = t − u20 cosh c1 − c2

and ⃗x(t, u10 )

2.2

r = u10 − t cosh c1 − c2



c1 t − c2 u20 c1 t − c2 u20 t − u20 cos , sin , sinh c1 − c2 c1 − c2 c1 − c2



(t ∈ R) (2.11)



c1 u10 − c2 t c1 u10 − c2 t u1 − t cos , sin , sinh 0 c1 − c2 c1 − c2 c1 − c2



(t ∈ R). (2.12)

THE TANGENT PLANES AND NORMAL VECTORS OF A SURFACE

In this section, we define the tangent plane and the normal vector of a surface at a given point of the surface. The direction of a curve γ on a surface S with a parametric representation (2.3) is given by ⃗x ′ (ui (t)) = ⃗xk (ui (t))(uk )′ (t), or ⃗x ′ = ⃗xk (uk )′ for short. The identity in (2.13) reduces to ⃗x ′ = ⃗x1 for the u1 –lines and ⃗x ′ = ⃗x2 for the u2 –lines.

(2.13)

114 ■ Differential Geometry and Its Visualization

Figure 2.7

Tangent planes and normal vectors of a surface

Thus the vector ⃗xk is a vector in the direction of the tangent to a uk –line. Condition (2.4) means that the vectors ⃗x1 and ⃗x2 are linearly independent. They span a plane at every point of the surface S, the so–called tangent plane. Definition 2.2.1. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D) and P = X(uk ) be a point of S. 0

(a) The plane with parametric representation

⃗tP = ⃗x(uk ) + λj ⃗xj (uk ) (λk ∈ R (k = 1, 2)) 0

0

is called the tangent plane of S at P (Figure 2.7). (b) The vector ⃗x1 (uk ) × ⃗x2 (uk ) 0 0 k ⃗ ⃗ NP = N (u ) = 0 ∥⃗x1 (uk ) × ⃗x2 (uk )∥ 0

0

is called the surface normal vector of S at P (Figure 2.7).

Remark 2.2.2. (a) Obviously the tangent plane of a surface S at a point P may be given by the equation ⃗ P = 0. (⃗tP − ⃗x(uk )) • N 0

(b) Under parameter transformations, the normal vector of a surface will at most reverse its direction; if ⃗x ∗ (u∗k ) = ⃗x(uj (u∗k )),

then we obtain from (2.9) ⃗∗ N P

=

∂(uj ) ∂(u ⃗ P  ∗k )  . N  ∂(uj )   ∂(u∗k ) 

Surfaces in Three-Dimensional Euclidean Space ■ 115

⃗ is as important (c) The trihedron of a surface spanned by the vectors ⃗x1 , ⃗x2 and N in the theory of surfaces as the trihedron of a curve is in the theory of curves. The vectors of the trihedron of a surface, however, are not orthonormal. We only have ⃗ 2 = 1 and N ⃗ • ⃗xk = 0 (k = 1, 2). N

Example 2.2.3. We consider the two parametric representations of the x1 x2 –plane given in Visualization 2.1.5 ⃗x(ui ) = {u1 , u2 , 0}

and



⃗x∗ (u∗i ) = {u∗1 cos u∗2 , u∗1 sin u∗2 , 0}



(u1 , u2 ) ∈ R2

u1 = u∗1 cos u∗2 and u2 = u∗1 sin u∗2 ∂(ui ) ∂(u∗k )

⃗ ∗ (u∗i ) = N



(u∗1 , u∗2 ) ∈ D = (0, ∞) × (−π, π) .

The formulae for the parameter transformation are

⃗ (ui ) = {0, 0, 1} and Now N







(u∗1 , u∗2 ) ∈ D .

= u∗1 together imply ∂(ui ) ∂(u ⃗ (ui (u∗k ))  ∗k )  N  ∂(ui )   ∂(u∗k ) 

= {0, 0, 1}.

Example 2.2.4. We consider the sphere of Visualization 2.1.7 (b) with a parametric representation ⃗x(ui ) = r{cos u1 cos u2 , cos u1 sin u2 , sin u1 }

It follows for the normal vectors that





(u1 , u2 ) ∈ D = (−π/2, π/2) × (0, 2π) .

⃗ (ui ) = −{cos u1 cos u2 , cos u1 sin u2 , sin u1 } = − 1 ⃗x(ui ) (u1 , u2 ) ∈ D , N r and consequently the tangent plane to the sphere at the point P = X(0, π/2) = (0, r, 0) is given by the equation 

Now let







{t1 , t2 , t3 } − {0, r, 0} • (−{0, 1, 0}) = −t2 + r = 0. 

⃗x(ui ) = u1 , u2 ,



r2 − (u1 )2 − (u2 )2



be a parametric representation of the upper half of the sphere in Visualization 2.1.7 (a). Then we have r2 − (u2 )2 , r2 − (u1 )2 − (u2 )2  2 r2 − (u1 )2 ⃗x2 (ui ) = 2 , r − (u1 )2 − (u2 )2 −u1 u2 ⃗x1 (ui ) • ⃗x2 (ui ) = 2 r − (u1 )2 − (u2 )2 

2

⃗x1 (ui )

and



=

 ⃗ (ui ) = 1 u1 , u2 , r2 − (u1 )2 − (u2 )2 N r



1 = ⃗x(ui ). r

116 ■ Differential Geometry and Its Visualization

2.3

THE ARC LENGTH, ANGLES AND GAUSS’S FIRST FUNDAMENTAL COEFFICIENTS

We recall from the fundamental theorem of curves, Theorem 1.11.1 that a curve is uniquely determined by two local, invariant quantities, namely its curvature and torsion as functions of its arc length. Similarly, the fundamental theorem of surfaces, which we will not prove, states that a surface is uniquely determined by certain local, invariant quantities, the so–called first and second fundamental forms. Lengths, arcs and surface areas can be measured when the first fundamental coefficients of a surface are known; this is why they are also referred to as metric coefficients. Let S be a surface with a parametric representation ⃗x(uj ) and γ be a curve on S with a parametric representation ⃗x(t) = ⃗x(uj (t)). Then the arc length along γ is given by s(t) =

t

t0

∥⃗x ′ (τ )∥ dτ.

(2.14)

It follows from dui (t) duk (t) • ⃗xk (uj (t)) dt dt i duk (t) du (t) = (⃗xi (uj (t)) • ⃗xk (uj (t))) · dt dt

⃗x ′ (t) • ⃗x ′ (t) = ⃗xi (uj (t))

that we are able to compute the arc length along γ provided we know the values gik (uj ) = ⃗xi (uj ) • ⃗xk (uj ) (i, k = 1, 2). Definition 2.3.1. Let S be a surface with a parametric representation ⃗x(uj ) ((u1 , u2 ) ∈ D ⊂ R2 ). (a) The functions gik : D → R with gik = ⃗xi • ⃗xk (i, k = 1, 2) are called the (Gauss’s) first fundamental coefficients of the surface S. (b) The function (ds)2 = gik (uj )dui duk is called the first fundamental form of the surface S. (c) The parameters of S are said to be orthogonal, if g12 (uj ) = 0.

Remark 2.3.2. (a) Obviously we have gik = gki (i, k = 1, 2).

(2.15)

Surfaces in Three-Dimensional Euclidean Space ■ 117

(b) In many text books, the notations E, F and G are used instead of g11 , g12 and g22 respectively. (c) We have gii = ⃗xi • ⃗xi > 0 (i = 1, 2) and, if we put g = det(gik ), then 2 = ⃗x12 · ⃗x22 − (⃗x1 • ⃗x2 )2 = (⃗x1 × ⃗x2 )2 > 0. g = g11 g22 − g12

Thus the first fundamental form (2.15) is positive definite, the inverse matrix (g ik ) of the matrix (gik ) exists, and g 11 = g22 , g 12 = −g12 and g 22 = g11 . (d) The first fundamental coefficients play an important role in calculating the arc length along a curve on a surface, the angle between two curves on a surface and the surface area. Let S be a surface with a parametric representation ⃗x(uj ) and γ be a curve on S with a parametric representation ⃗x(t) = ⃗x(uj (t)). Then the arc length along γ is given by s(t) =

t

t0



gik

dui duk dτ. dτ dτ

(2.16)

If γ ∗ is another curve on S with a parametric representation ⃗x(u∗j (t∗ )) and the curves γ and γ ∗ intersect at the point X(uj ) where t = t0 and t∗ = t∗0 , then the angle α 0 between the curves γ and γ ∗ is defined as the angle between the tangents to the curves at the point X(uj ). Therefore, 0

cos α = 





d⃗x(u∗j (t∗ ))  d⃗x(uj (t))  •    ∗ ∗ dt dt∗ t=t t =t

  d⃗  j  x(u (t))      dt

0

t=t0

     ∗j ∗  d⃗ x (u (t ))    ·  ∗   ∗ dt  t

0

     ∗ =t  0

dui du∗k (t0 ) ∗ (t∗0 ) 0 dt dt = . i k ∗l ∗m du du du du (t0 ) (t0 ) ∗ (t∗0 ) ∗ (t∗0 ) gik (uj )glm (uj ) 0 0 dt dt dt dt gik (uj )

Finally the surface area of a region of S defined by (u1 , u2 ) ∈ G is given by  

g(uj ) du1 du2 .

(2.17)

G

Now we establish the transformation formulae for the first fundamental coefficients and the invariance of the arc length, surface area and angles under parameter transformations.

118 ■ Differential Geometry and Its Visualization

Remark 2.3.3. Let S be given by the parametric representations ⃗x(uj ) = ⃗x(uj (u∗k )) = ⃗x∗ (u∗k ) ∗ and the fundamental coefficients gik and gik (k = 1, 2). (a) Then the first fundamental coefficients gik satisfy the transformation formulae ∗ = grs gik

∂ur ∂us for i, k = 1, 2. ∂u∗i ∂u∗k

(2.18)

(b) The coefficients g ik and g ∗ik satisfy the transformation formulae g ∗ik = g jm

∂u∗i ∂u∗k for i, k = 1, 2. ∂uj ∂um

(2.19)

(c) The arc length of a curve on S, angles, and the surface area of a part of S are invariant under parameter transformations. Proof. (a) By (2.8), the first fundamental coefficients transform as follows ∗ = ⃗x∗i • ⃗x∗k = ⃗xr • ⃗xs gik

∂ur ∂us ∂ur ∂us = g for i, k = 1, 2, rs ∂u∗i ∂u∗k ∂u∗i ∂u∗k

that is, the formulae in (2.18) are satisfied. (b) Interchanging u and u∗ in (2.18), we obtain ∗ gmr = gnt

∂u∗n ∂u∗t for m, r = 1, 2, ∂um ∂ur

hence by (2.8) and the interchange of the indices of summation j and m in the last step ∂u∗k ∂uj ∂u∗k ∂ur j = g ∗is δ j ∗s ∂u ∂u ∂uj ∂u∗s r ∗k ∂u∗k ∂ur ∂ur ∗ ∂u∗n ∂u∗t mj mj ∗is ∂u = g ∗is g g = g g g mr ∂uj ∂u∗s ∂uj ∂u∗s nt ∂um ∂ur ∗k ∗t ∗n ∂ur ∂u∗k ∂u∗n mj ∗ ∂u ∗is ∗ ∗t ∂u ∂u = g ∗is gnt g = g g δ g mj nt s ∂ur ∂u∗s ∂uj ∂um ∂uj ∂um ∗k ∗n ∗k ∗n ∗ mj ∂u ∂u ∗i mj ∂u ∂u = g ∗is gns g = δ g n ∂uj ∂um ∂uj ∂um ∗k ∗i ∗i ∗k ∂u ∂u ∂u ∂u = g mj = g jm j for i, k = 1, 2. ∂uj ∂um ∂u ∂um

g ∗ik = g ∗is δsk = g ∗is

Thus we have shown the formulae in (2.19).

Surfaces in Three-Dimensional Euclidean Space ■ 119

(c) Furthermore, the arc length of a curve on S is invariant under transformations of the parameters of S, since 

d⃗x∗ dt

2

du∗i du∗k ∂ur ∂us ∂u∗i ∂u∗k duj dun = grs ∗i ∗k dt dt ∂u ∂u ∂uj ∂un dt dt r ∗i s j n ∂u ∂u ∂u ∂u∗k duj dun r s du du = grs ∗i = g δ δ rs j n ∂u ∂uj ∂u∗k ∂un dt dt dt dt  2 j n du du d⃗x = . = gjn dt dt dt ∗ = gik

Similarly one may show the invariance of angles and surfaces areas under parameter transformations.

Example 2.3.4 (The first fundamental coefficients of explicit surfaces). Let f : D ⊂ R 2 → R be a function of class C r (D) for r ≥ 1. Then the surface with a parametric representation ⃗x(ui ) = {u1 , u2 , f (u1 , u2 )}



(u1 , u2 ) ∈ D



(2.20)

is called an explicit surface; the surface S is a representation of the function f , that is, the graph of f . Then we have ⃗x1 (ui ) = {1, 0, f1 (u1 , u2 )} and ⃗x2 (ui ) = {0, 1, f2 (u1 , u2 )},

and the first fundamental coefficients of S are given by 

2

, g12 (ui ) = f1 (ui )f2 (ui ), g22 (ui ) = 1 + f2 (ui )



2

+ f2 (ui )

g11 (ui ) = 1 + f1 (ui ) and

g(ui ) = 1 + f1 (ui )





2

2

(2.21)

.

Example 2.3.5. (The first fundamental coefficients of surfaces of revolution) Let γ be a curve in a plane through a point P and spanned by the unit vectors ⃗u1 and ⃗u3 in three-dimensional space. We assume that γ is given by a parametric representation

where

⃗x(t) = r(t)⃗u1 + h(t)⃗u3 (t ∈ I), r(t) > 0 and |r′ (t)| + |h′ (t)| > 0 for all t ∈ I.

Then a surface of revolution SR is generated by rotating γ about the axis defined by u3 (Figure 2.8). Putting ⃗u2 = ⃗u3 × ⃗u1 and u1 = t, and writing u2 for the angle of revolution, we obtain the following parametric representation for SR ⃗x(ui ) = r(u1 ) cos u2⃗u1 + r(u1 ) sin u2⃗u2 + h(u1 )⃗u3





(u1 , u2 ) ∈ D = I × (0, 2π) . (2.22)

120 ■ Differential Geometry and Its Visualization

Figure 2.8

Generating a surface of revolution

We may assume that the point P is the origin of E 3 , the generating curve γ is in the x1 x3 –plane, and the axis of revolution is the x3 –axis. Then the parametric representation (2.22) becomes ⃗x(ui ) = {r(u1 ) cos u2 , r(u1 ) sin u2 , h(u1 )}





(u1 , u2 ) ∈ D = I × (0, 2π) .

(2.23)

The u1 – and u2 –lines of a surface of revolution are called meridians and parallels. We obtain from (2.23) ⃗x1 (ui ) = {r′ (u1 ) cos u2 , r′ (u1 ) sin u2 , h′ (u1 )}, ⃗x2 (ui ) = {−r(u1 ) sin u2 , r(u1 ) cos u2 , 0}, hence 

2

g11 (ui ) = g11 (u1 ) = r′ (u1 )

g22 (ui ) = g22 (u1 ) = (r(u1 ))2



2

+ h′ (u1 )

, g12 (ui ) = g12 (u1 ) = 0, (2.24)

for the first fundamental coefficients of a surface of revolution, g(ui ) = g(u1 ) = (r(u1 ))2 · and



2

r′ (u1 )



2 

+ h′ (u1 )

⃗ (ui ) = {−h′ (u1 ) cos u2 , −h′ (u1 ) sin u2 , r′ (u1 )} ·  N

,

1 (r′ (u1 ))2 + (h′ (u1 ))2

(2.25)

for the surface normal vector.

Now we consider a few special cases of surfaces of revolution. Visualization 2.3.6 (Planes, cylinders and cones). Let the generating curve γ be a straight line segment in the positive x1 x3 –half–plane.

Surfaces in Three-Dimensional Euclidean Space ■ 121

Figure 2.9

Generating a plane

(a) Planes If γ is parallel to the x1 –axis and through a point with x3 = c, then r(u1 ) = u1 and h(u1 ) = c for u1 ∈ (0, ∞), and the surface of revolution is the plane through the point P = (0, 0, c) and parallel to the x1 x2 –plane, without the point P and the points of the straight line segment parallel to the positive x2 –axis (Figure 2.9). (b) Circular cylinders If γ is parallel to the x3 –axis and through a point with x1 = r > 0, then r(u1 ) = r > 0 and h(u1 ) = u1 for u1 ∈ R, and the surface of revolution is a circular cylinder of radius r (Figure 2.10). (c) Circular Cones If γ has an angle β ∈ (0, π) \ {π/2} with the positive x3 –axis, then r(u1 ) = u1 sin β and h(u1 ) = u1 cos β for u1 ∈ (0, ∞) and the surface of revolution is a circular cone with its vertex in the origin and an angle of 2β at its vertex (Figure 2.11).

Figure 2.10

Generating a circular cylinder

122 ■ Differential Geometry and Its Visualization

Figure 2.11

Generating a circular cone

(d) The first fundamental coefficients of planes, circular cylinders and cones The first fundamental coefficients are given by

for the plane,

g11 (u1 ) = 1, g12 (u1 ) = 0 and g22 = (u1 )2

(2.26)

g11 (u1 ) = 1, g12 (u1 ) = 0 and g22 (u1 ) = r2

(2.27)

for the circular cylinder, and g11 (u1 ) = 1, g12 (u1 ) = 0 and g22 = (u1 )2 sin2 β

(2.28)

for the circular cone. Visualization 2.3.7 (Spheres). If the generating curve γ is the semi–circle line in the positive x1 x3 –half–plane, of radius r > 0 and its centre in the origin, then r(u1 ) = r cos u1 and h(u1 ) = r sin(u1 ) for u1 ∈ (−π/2, π/2), and the surface of revolution is a sphere with radius r, its centre in the origin, without the circle line in the x1 x3 –plane, of radius r and the centre in the origin (Figure 2.12). The fundamental coefficients are g11 (u1 ) = r2 , g12 (u1 ) = 0 and g22 (u1 ) = r2 cos2 u1 .

(2.29)

Visualization 2.3.8 (Tori). If the generating curve γ is a circle line of radius r1 > 0 and its centre in the point (r0 , 0) in the x1 x3 –plane, where r0 > r1 , then r(u1 ) = r0 + r1 cos u1 and h(u1 ) = r1 sin u1 for u1 ∈ (0, 2π), and the surface of revolution is a torus without the circle lines C1 and C2 , where C1 is in the x1 x2 – plane and has the radius r0 + r1 and centre in the origin, and C2 is in the x1 x3 –plane and has the radius r1 and centre in (r0 , 0, 0) (Figure 2.13). The fundamental coefficients are 

g11 (u1 ) = r12 , g12 (u1 ) = 0 and g22 (u1 ) = r0 + r1 cos u1

2

.

(2.30)

Surfaces in Three-Dimensional Euclidean Space ■ 123

Figure 2.12

Generating a sphere

Figure 2.13

Generating a torus

Visualization 2.3.9 (Catenoids). If the generating curve γ is a catenary with a parametric representation x1 (t) = a cosh (t/a), x3 (t) = t (t ∈ R), where a > 0 is a constant, then the surface of revolution is a catenoid without γ (Figure 2.14). The first fundamental coefficients are 1

g11 (u ) = cosh

2





u1 , g12 (u1 ) = 0 and g22 (u1 ) = a2 cosh2 a





u1 . a

(2.31)

Example 2.3.10 (Some surface areas). Now we apply (2.17) in Remark 2.3.2 to compute the surface areas A(G) of some regions G of a circular cylinder, a circular cone, a sphere, torus and catenoid. (a) We obtain from (2.17) and (2.27) for the circular cylinder and the region G = (h0 , h1 ) × (0, 2π) with h1 > h0 A(G) =

2π h1 0 h0

r du1 du2 = 2πr(h1 − h0 ).

124 ■ Differential Geometry and Its Visualization

Figure 2.14

Generating a catenoid

(b) We obtain from (2.17) and (2.28) for the circular cone and the region G = (0, h) × (0, 2π) A(G) =

2π h 0

0

|u1 sin β| du1 du2 = πh2 sin β.

(c) We obtain from (2.17) and (2.29) for the sphere and the region G = (−π/2, π/2) × (0, 2π) A(G) =

2π π/2 0 −π/2

π/2 

r2 | cos u1 | du1 du2 = 2πr2 sin u1 

= 4πr2 .

−π/2

(d) We obtain from (2.17) and (2.30) for the torus and the region G = (0, 2π) × (0, 2π) A(G) =

2π 2π 0

0 2

1

1

2

r1 |r0 + r1 cos u | du du = 2πr1

= 4π r0 r1 .

2π

(r0 + r1 cos u1 ) du1

0

(e) We obtain from (2.17) and (2.31) for the catenoid and the region G = (h0 , h1 )× (0, 2π) with h1 > h0 A(G) =

2π h1

a cosh

0 h0

= 2πa

h1

h0



2



u1 a 



du1 du2

2u1 1 exp 4 a





2u1 + exp − a





+2

du1

Surfaces in Three-Dimensional Euclidean Space ■ 125

Figure 2.15

Loxodromes on a torus

= aπ

h1 

cosh

h0





2u1 a



a 2h1 sinh = aπ 2 a







+1

du1 

a 2h0 − sinh 2 a





+ h1 − h0 .

Visualization 2.3.11 (Loxodromes). A curve γ on a surface of revolution SR that intersects the meridians of SR at a constant angle β ∈ (−π/2, π/2] is called a loxodrome (Figure 2.15). Let ⃗x ∗ (ui (s)) (s ∈ I) be the parametric representation of a curve γ on SR, and s be the arc length along γ. If β ̸= π/2, then the loxodrome through the point P with the parameters (u10 , u20 ) is given by u2 (u1 ) + u20 = tan β

u1 

u10

= tan β

g11 (u) du g22 (u)

u1 

u10

(r′ (u))2 + (h′ (u))2 du. r(u)

(2.32)

If β = π/2, then the loxodrome through the point P with the parameters (u10 , u20 ) is the parallel corresponding to u1 = u10 . Proof. Since ∥⃗x˙ ∗ (ui (s)∥ = 1 and g12 = ⃗x1 • ⃗x2 = 0, g11 (ui ) = g11 (u1 ) and g22 (ui ) = g22 (u1 )

126 ■ Differential Geometry and Its Visualization

for surfaces of revolution by (2.24) in Example 2.3.5, the angle β(s) between the curve γ and the meridian at s is given by cos β(s) = This and

 ⃗x˙ ∗ (ui (s)) • ⃗x1 (ui (s)) ⃗x1 (ui (s)) • ⃗x1 (ui (s)) 1 = u ˙ (s) = g11 (u1 (s)) u˙ 1 (s). ∥⃗x1 (ui (s))∥ ∥⃗x1 (ui (s))∥ (2.33)  2 ˙ i  x (u (s)) = gik (u1 (s))u˙ i (s)u˙ k (s) ⃗  2

= g11 (u1 (s)) u˙ 1 (s)



= cos2 β(s) + sin2 β(s),

together yield

2

+ g22 (u1 (s)) u˙ 2 (s)

 

=1

 

|sin β(s)| =  g22 (u1 (s)) u˙ 2  ,

and, since g22 (u1 (s)) > 0 for all s, we obtain u˙ 2 (s) = 

sin β(s) g22 (u1 (s))

(2.34)

for a suitable choice of the orientation of the curve. For a loxodrome, the angle β(s) = β is constant, and we obtain for β ̸= π/2 from (2.33), (2.34) and (2.24) in Example 2.3.5 du2 = tan β du1



g11 (u1 ) , g22 (u1 )

and consequently, we obtain (2.32) for the loxodrome through (u10 , u20 ). If β = π/2, then it follows from (2.33) that u1 = u10 , and we obtain the parallel corresponding to u10 . Visualization 2.3.12. (a) Loxodromes in planes First we determine the loxodromes in the plane of Visualization 2.3.6 (a). We obtain from (2.32) and (2.26) that the loxodromes for β ∈ (0, π/2) are given by 2

u −

u20

= tan β

u1

u10



du u = tan β · log u u10



for u1 > 0 and (u10 , u20 ) ∈ (0, ∞) × (0, 2π). Putting b = 1/ tan β and a = exp (−u20 /b), we conclude u1 = u10 · a exp (bu2 ), that is, the loxodromes of the plane are logarithmic spirals; they intersect the straight lines through the origin, and hence the concentric circles with their centres in the origin, at a constant angle (Figure 2.16).

Surfaces in Three-Dimensional Euclidean Space ■ 127

Figure 2.16

A logarithmic spiral as a loxodrome of the plane in Example 2.3.5 (a)

(b) Loxodromes on spheres Now we determine the loxodromes on the sphere with radius r > 0 of Example 2.3.5 (b). For the sphere of radius r > 0 centred at the origin, we conclude from g11 (u1 ) = 1,

g12 = 0 and g22 (u1 ) = r2 cos2 u1

and (2.32) that the loxodromes for β ∈ (−π/2, π/2) \ {0} are given by 

u2 = tan β · log tan where c=

Figure 2.17

u20







u1 π + 2 4

u10 π + − tan β · log tan 2 4

A loxodrome on a sphere





+c

(Figure 2.17).

(2.35)

128 ■ Differential Geometry and Its Visualization

Proof of Part (b). We observe that the integral in (2.32) reduces to I(u) = The substitution t = tan u2 yields



u du. cos u

2 2t 1 − t2 du = , cos u = , 1 + t2 1 + t2 dt 1 + t2

sin u = and

 





1+t 1 dt 1 = dt = log + I(u) = 2 2 1−t 1−t 1+t 1−t      u 1 + tan 2 u π = log + + c˜ = log tan + c˜, 1 − tan u2 2 4 where c˜ is a constant of integration. Remark 2.3.13. The parameter lines of the parametric representation (2.10) of the sphere in Visualization 2.1.7 (c) are loxodromes. Proof. We write d = c1 − c2 , and put t = 2 · tan ∗

−1





t−c exp d



−

π 2

to obtain r · cos t∗ =



cosh

and

 t−c  ,





c1 − c2 t∗ π = c1 log tan + d 2 4

r

d



+c



sin t∗ t−c = sinh . ∗ cos t d Thus, if we write 



t∗ π u (t ) = t and u (t ) = c1 log tan − 2 4 ∗1

∗

∗

∗2

∗



+ c,

then identity (2.11) for the u1 –line corresponding to u2 = c yields ⃗x(t ∗ ) = r{cos t∗ cos u∗1 (t∗ ), sin t∗ cos u∗1 (t∗ ), sin t∗ }, that is, a loxodrome by (2.35) in Visualization 2.3.11 (b) on the sphere with respect to its usual parametric representation given in Visualization 2.1.7 (b). Visualization 2.3.14 (Lines of constant slope on surfaces of revolution). Now we determine all lines γβ of constant slope on surfaces of revolution that have a constant angle β ∈ [0, π) with the axis of revolution along ⃗e 3 .

Surfaces in Three-Dimensional Euclidean Space ■ 129

Figure 2.18

Lines of constant slope

If ⃗x∗ (ui (s)) is the parametric representation of a curve γ on the surface SR of revolution given by a paramerec representation (2.23), and s is the arc length along γ, then we must have for γβ ⃗x˙ ∗ • ⃗e 3 = h′ (u1 )u˙ 1 = cos β.

(2.36)

Let β ̸= π/2, and I1 denote the interval for the parameter u1 of the parametric representation of SR. Then solutions of (2.36) only exist in subintervals I ⊂ I1 for which h′ (u1 ) ̸= 0. From ∥⃗x˙ ∗ ∥ = 1 and we obtain 

du2 du1

2

1 (h′ (u1 ))2 = , (u˙ 1 )2 cos2 β

(h′ (u1 ))2 − g11 (u1 ) cos2 β , = g22 (u1 )

     du2  ′ 1 2 1 2 (h′ (u1 ))2 tan2 β − (r′ (u1 ))2 1 (h (u )) − g11 (u ) cos β   =  1 =  du  | cos β| g22 (u1 ) r(u1 )

and choose

du2 = du1



(h′ (u1 ))2 tan2 β − (r′ (u1 ))2 r(u1 )

(Figures 2.18, 2.19 and 2.20). Let I be a subinterval of I1 for which (r′ (u1 ))2 < tan2 β(h′ (u1 ))2 , and (u10 , u20 ) ∈ I × (0, 2π). Then γβ is given by cos β u1 (t) = t and u2 − u20 = | cos β|

u1

u10



(h′ (u))2 tan2 β − (r′ (u))2 r(u)

du.

(2.37)

130 ■ Differential Geometry and Its Visualization

Figure 2.19

Lines of constant slope on a sphere (left) and catenoid (right)

Figure 2.20

Lines of constant slope on a torus with their bounding curve

Now let β = π/2. In intervals with h′ (u1 ) = 0, we may choose u1 (s) = c, where c is a constant, and obtain from ∥x∥2 = g22 (u˙ 2 )2 = 1 that 

du2 =± ds

1 g22 (c)

and

2

u (s) =



1 · s + c˜ g22 (c)

where c˜ is a constant. In intervals for which h′ (u1 ) ̸= 0, we conclude from (2.36), that u˙ 1 = 0, thus we obtain u2 –lines.

2.4

THE CURVATURE OF CURVES ON SURFACES, GEODESIC AND NORMAL CURVATURE

In this section, we introduce the concepts of the geodesic and normal curvature in the study of the curvature of curves on surfaces. We will see that the geodesic curvature depends on the first fundamental coefficients, whereas the normal curvature depends on the so–called second fundamental coefficients.

Surfaces in Three-Dimensional Euclidean Space ■ 131

Figure 2.21

The components of a vector of curvature

Let S be a surface with a parametric representation ⃗x(uj ) and γ be a curve on S with a parametric representation ⃗x(s) = ⃗x(uj (s)). Then we have ⃗x˙ = ⃗xj u˙ j and ⃗x¨ = ⃗xj u¨j + ⃗xjk u˙ j u˙ k , where ⃗xjk = Furthermore, let

∂ 2⃗x (j, k = 1, 2). ∂uj ∂uk

(2.38)

⃗ × ⃗x˙ , ⃗t = N

(2.39)

⃗ (s). ⃗x¨(s) = κg (s)⃗t(s) + κn (s)N

(2.40)

⃗ is the surface normal vector of S. Then ⃗t is a unit vector in the tangent where N plane of S, and the vector of curvature can be split into two components, one in the tangent plane and one along the surface normal vector (Figure 2.21),

Definition 2.4.1. Let S be a surface with a parametric representation ⃗x(uj ) and γ be a curve on S with a parametric representation ⃗x(s) = ⃗x(uj (s)). Then the functions κg and κn defined in (2.40) are called the geodesic and normal curvature of the curve γ. Figure 2.22 shows a way to represent the normal curvature of a curve on a surface. Obviously we have ⃗ and κg = ⃗x¨ • ⃗t κ2 (s) = κ2g (s) + κ2n (s), and κn = ⃗x¨ • N

(2.41)

for the curvature κ, and the geodesic and normal curvature κg and κn of the curve γ. In order to determine κn and κg , we put ⃗ (i, k = 1, 2) ⃗xik = Γrik ⃗xr + Lik N

(2.42)

with the coefficients Γrik and Lik yet to be determined. The identities in (2.42) are referred to as the Gauss equations (for the derivatives). From (2.42), we obtain ⃗ Lik = ⃗xik • N

(i, k = 1, 2).

132 ■ Differential Geometry and Its Visualization

Figure 2.22

Representation of the normal curvature of a curve on a torus

⃗ • ⃗xi = 0 with respect to uk , we obtain Differentiating N ⃗ ⃗ • ⃗xi ) ∂(N ⃗ k • ⃗xi + N ⃗ • ⃗xik = 0, where N ⃗ k = ∂N , = N ∂uk ∂uk hence

⃗ • ⃗xik = −N ⃗ k • ⃗xi and Lik = Lki (i, k = 1, 2). Lik = N

(2.43)

Definition 2.4.2. Let S be a surface with a parametric representation ⃗x(uj ) ((u1 , u2 ) ∈ D ⊂ R2 ). The functions Lik : D → R (i, k = 1, 2) defined in (2.43) are called second fundamental coefficients of the surface S; we also write L = det(Lik ) = L11 L22 − L212 .

Remark 2.4.3. (a) In many textbooks the notations L, M and N are used instead of L11 , L12 and L22 . (b) Let S be given by the parametric representations ⃗x(uk ) = ⃗x(uk (u∗r )) = ⃗x∗ (u∗j ).

Then the transformation formulae for the surface normal vector and the second fundamental coefficients are k ⃗∗ = N ⃗ · signD, where D = ∂(u ) N ∂(u∗r )

(2.44)

and

∂um ∂un · sign(D) (i, k = 1, 2). ∂u∗i ∂u∗k (c) The normal and geodesic curvature are given by L∗ik = Lmn





κn = Lik u˙ i u˙ k and κg = u¨r + Γrik u˙ i u˙ k ⃗xr • ⃗t.

(2.45)

(2.46)

Surfaces in Three-Dimensional Euclidean Space ■ 133

With respect to an arbitrary parameter t for the curve, we obtain dui duk κn = Lik dt dt and κg =





i k d 2 ui i du du + Γ jk dt2 dt dt

dt ds

2



i

=

dt ds

duk dt i duk gik du dt dt

Lik du dt

2

(2.47) 

dui d 2 t + ⃗xi • ⃗t. dt ds2

(2.48)

Proof. (b) We conclude from for the surface normal vector (2.9)

Furthermore,

∗ ∗ ⃗ · signD. ⃗ ∗ = ⃗x1 × ⃗x2 = ⃗x1 × ⃗x2 · signD = N N ∥⃗x∗1 × ⃗x∗2 ∥ ∥⃗x1 × ⃗x2 ∥

⃗x∗i = ⃗xm

∂um ∂um ∂un ∂ 2 um ∗ and ⃗ x = ⃗ x + ⃗ x mn m ik ∂u∗i ∂u∗i ∂u∗k ∂u∗i ∂u∗k

together imply L∗ik



m n 2 m ⃗ ∗ • ⃗x∗ = N ⃗ • ⃗xmn ∂u ∂u + ⃗xm ∂ u =N ik ∂u∗i ∂u∗k ∂u∗i ∂u∗k ∂um ∂un = Lmn ∗i ∗k · signD (i, k = 1, 2). ∂u ∂u



signD

Thus we have shown (2.45). (c) We obtain from identities (2.41) and (2.42) 



⃗ u˙ i u˙ k ⃗x¨ = ⃗xi u¨i + ⃗xik u˙ i u˙ k = ⃗xi u¨i + Γrik ⃗xr + Lik N 



⃗, = u¨r + Γrik u˙ i u˙ k ⃗xr + (Lik u˙ i u˙ k )N

and the identities in (2.48) are an immediate consequence. Finally, if t is an arbitrary parameter of a curve ⃗x(ui (t)) on a surface, then writing ui (s) = ui (t(s)) for i = 1, 2, we obtain dui dt d 2 ui and u¨i = u˙ = dt ds dt2 i



dt ds

2

+

dui d 2 t for i = 1, 2, dt ds2

and the formulae for the normal and geodesic curvature in (2.47) and (2.48) follow directly from (2.46) and (2.16). Remark 2.4.4. The coefficients Lik and Γrik are introduced in a formal way. We want to give them a geometric meaning. The normal and geodesic curvature play as important a role in the theory of surfaces as the curvature and torsion do in the theory of curves.

134 ■ Differential Geometry and Its Visualization

Example 2.4.5 (The second fundamental coefficients for explicit surfaces). Let S be an explicit surface with a parametric representation ⃗x(ui ) = {u1 , u2 , f (u1 , u2 )} ((u1 , u2 ) ∈ D). Then we have Lik (uj ) = 

and

L(uj ) =

fik (uj ) for i, k = 1, 2 1 + (f1 (uj ))2 + (f2 (uj ))2

f11 (uj )f22 (uj ) − (f12 (uj ))2 . 1 + (f1 (uj ))2 + (f2 (uj ))2

(2.49)

(2.50)

In particular, if S is a plane, then f (uj ) = const for all (u1 , u2 ) ∈ D and so Lik (uj ) = 0 for i, k = 1, 2.

(2.51)

Proof. Since ⃗x1 = {1, 0, f1 (uj )}, ⃗x2 = {0, 1, f1 (uj )} and ⃗xik (uj ) = {0, 0, fik (uj )} for i, k = 1, 2, it follows from (2.43) and (1.6) the second fundamental coefficients of S are given by ⃗ j ) • ⃗xik (u ) = Lik (u ) = N (u j

j



⃗x1 (uj ) × ⃗x2 (uj ) ∥⃗x1 (uj ) × ⃗x2 (uj )∥

   0 1 0   1   0 1  j  0   fik (uj ) f1 (uj ) f2 (uj ) g(u )

and

=

L(uj ) =



• ⃗xik (uj )

fik (uj ) for i, k = 1, 2 1 + (f1 (uj ))2 + (f2 (uj ))2

f11 (uj )f22 (uj ) − (f12 (uj ))2 . 1 + (f1 (uj ))2 + (f2 (uj ))2

Thus we have shown (2.49) and (2.50). In particular, if S is a plane, then f (uj ) = const for all (u1 , u2 ) ∈ D, and so (2.51) is an immediate consequence of (2.49). Example 2.4.6. (The second fundamental coefficients for surfaces of revolution) Let SR be a surface of rotation with a parametric representation ⃗x(ui ) = {r(u1 ) cos u2 , r(u1 ) sin u2 , h(u1 )}. Then we have

Surfaces in Three-Dimensional Euclidean Space ■ 135

L11 (ui ) = L11 (u1 ) =

h′′ (u1 )r′ (u1 ) − h′ (u1 )r′′ (u1 )  , (r′ (u1 ))2 + (h′ (u1 ))2

L12 (ui ) = L12 (u1 ) = 0,

L22 (ui ) = L22 (u1 ) = 

and

L(ui ) = L(u1 ) =



h′ (u1 )r(u1 ) (r′ (u1 ))2 + (h′ (u1 ))2 

h′′ (u1 )r′ (u1 ) − h′ (u1 )r′′ (u1 ) h′ (u1 )r(u1 ) . (r′ (u1 ))2 + (h′ (u1 ))2

(2.52) (2.53) (2.54)

(2.55)

Proof. We have ⃗x1 (ui ) = {r′ (u1 ) cos u2 , r′ (u1 ) sin u2 , h′ (u1 )},

⃗x2 (ui ) = {−r(u1 ) sin u2 , r(u1 ) cos u2 , 0},

⃗x11 (ui ) = {r′′ (u1 ) cos u2 , r′′ (u1 ) sin u2 , h′′ (u1 )},

⃗x12 (ui ) = {−r′ (u1 ) sin u2 , r(u1 ) cos u2 , 0},

⃗x22 (ui ) = {−r(u1 ) cos u2 , −r(u1 ) sin u2 , 0},

and so L11 (ui ) = L11 (u1 )

 r ′′ (u1 ) cos u2   =  r′′ (u1 ) sin u2   h′′ (u1 )

= L12 (ui ) = L12 (u1 )

h′′ (u1 )r′ (u1 ) − h′ (u1 )r′′ (u1 )  , (r′ (u1 ))2 + (h′ (u1 ))2



 −r′ (u1 ) sin u2   =  r′ (u1 ) cos u2   0

= 0,

r′ (u1 ) cos u2 − sin u2  1  r′ (u1 ) sin u2 cos u2  ·  ′ 1 2  (r (u )) + (h′ (u1 ))2 h′ (u1 ) 0 

 −r(u1 ) cos u2   =  −r(u1 ) sin u2   0

r′ (u1 ) cos u2 − sin u2  1  r′ (u1 ) sin u2 cos u2  ·  ′ 1 2  (r (u )) + (h′ (u1 ))2 h′ (u1 ) 0 

1

L22 (u ) = L22 (u ) i



r′ (u1 ) cos u2 − sin u2  1  r′ (u1 ) sin u2 cos u2  ·  ′ 1 2  (r (u )) + (h′ (u1 ))2 h′ (u1 ) 0 

h′ (u1 )r(u1 ) = ′ 1 2 (r (u )) + (h′ (u1 ))2



136 ■ Differential Geometry and Its Visualization

and L(ui ) = L(u1 ) = L11 (u1 )L22 (u1 ) 



h′′ (u1 )r′ (u1 ) − h′ (u1 )r′′ (u1 ) h′ (u1 )r(u1 ) = . (r′ (u1 ))2 + (h′ (u1 ))2 Visualization 2.4.7. (a) (Lines of vanishing normal and geodesic curvature on a hyperboloid of one sheet) Let a, b, c ∈ R\{0} and H be the hyperboloid of one sheet given by the equation (x1 )2 (x2 )2 (x3 )2 + 2 − 2 = 1. a2 b c

Its intersection with the x1 x2 –plane is an ellipse with a parametric representation ⃗x(t) = {a cos t, b sin t, 0} (t ∈ (0, 2π)).

Given t0 ∈ (0, 2π), we consider the straight lines γ+ and γ− with the parametric representations ⃗y± (t) = ⃗x(t0 ) + t(±⃗x ′ (t0 ) + c⃗e 3 ) (t ∈ R) (Figure 2.23). It follows from 1 2 2 2 3 2 (y± ) ) ) (y± (y± + − = (cos t0 − t sin t0 )2 + (sin t0 + t cos t0 )2 − t2 2 2 2 a b c = 1 + t2 − t2 = 1

that γ+ and γ− are on H. Since t0 was arbitrary, the hyperboloid of one sheet contains two families of straight lines, which have vanishing curvature (Figure 2.24). Therefore, their geodesic and normal curvature vanish identically by (2.41). (b) The geodesic curvature vanishes along helices on circular cylinders √ 2 Let H, R > 0, ω = 1/ R + H 2 and γ be the helix with a parametric representation ⃗x(s) = {R cos (ωs), R sin (ωs), Hωs} (s ∈ R); its vector of curvature and curvature were given in Visualization 1.6.7 (b) by ⃗x¨(s) = −ω 2 R{cos ωs, sin ωs, 0} and κ(s) = Rω 2 . The helix is a curve on the circular cylinder of Example 2.3.5 (a) with radius R > 0 and axis along the x3 –axis; it is given by u1 (s) = Hωs and u2 (s) = ωs (s ∈ R). By (2.25), the surface normal vectors of the cylinder along the helix are ⃗ (ui (s)) = {−h′ (u1 (s)) cos u2 (s), −h′ (u1 (s)) sin u2 (s), r′ (u1 (s))}· N

Surfaces in Three-Dimensional Euclidean Space ■ 137

Figure 2.23

Intersections of a hyperboloid of one sheet with planes

Figure 2.24

Two families of straight lines on a hyperboloid of one sheet 1 (r′ (u1 (s)))2 + (h′ (u1 (s)))2 1 = {−Hω cos ωs, −Hω sin ωs, 0} √ H 2ω2 1 = −{cos ωs, sin ωs, 0} = 2 ⃗x¨(s). ω R ·

Therefore, it follows that kn (s) = κ(s) and κg (s) = 0 for all s ∈ R (Figure 2.25). (c) The normal and geodesic curvature along parallels on spheres Let S be the sphere with a parametric representation 



⃗x(ui ) = r{cos u1 cos u2 , cos u1 sin u2 , sin u1 } (u1 , u2 ) ∈ (−π/2, π/2) × (0, 2π) .

138 ■ Differential Geometry and Its Visualization

Curvature, normal and geodesic curvature along a helix on a cylinder

Figure 2.25

The u2 –line corresponding to the constant value u1 = α ∈ (−π/2, π/2) has a parametric representation ⃗x(t) = ⃗x(t, α) = r{cos α cos t, cos α sin t, sin α}

(t ∈ (0, 2π)).

Obviously the curvature of this u2 –line is given by κ=

1 . r cos α

⃗ = − 1 ⃗x, we obtain (Figure 2.26) Since N r

1 κn = κ cos α = . r

From κ2g = κ2 − κ2n =

1 r2

we obtain



|κg | =



tan2 α 1 − 1 = , cos2 α r2 | tan α| . r

(d) A curve with identically vanishing normal curvature on a catenoid Now let a > 0 be given, C be the catenoid of Example 2.3.5 (b) with a parametric representation ⃗x(u ) = i





u1 a cosh a



2

cos u , a cosh



u1 a



2

1

sin u , u





(u1 , u2 ) ∈ R × (0, 2π)



Surfaces in Three-Dimensional Euclidean Space ■ 139

Figure 2.26

sphere

Curvature, geodesic and normal curvature at a point on the parallel of a

and γ± be the curves on C given by u1 (t) = t and u2 (t) = ±t/a (t ∈ IR). Then we have 



u1 r(u1 ) = a cosh , r′ (u1 ) = sinh a







h(u1 ) = u1 , h′ (u1 ) = 1, h′′ (u1 ) = 0, and by Example 2.4.6 cosh



u1 a



cosh



u1 a



1  1  = − , = −  L11 (u1 ) = −   1  u a a cosh a  a sinh2 ua + 1

since cosh(u) > 0 for all u, L12 (u1 ) = 0 and L22 (u1 ) =

a cosh cosh





u1 a

u1 a



 = a.

Therefore, the normal curvature along γ± is by (2.47) and (2.24) 



du1 (t) κn (t) = L11 (u1 (t)) dt ·



2 1



du2 (t) + L22 (u1 (t)) dt

du (t) g11 (u1 (t)) dt

2

1



1 u1 u1 , r′′ (u1 ) = cosh , a a a



2  ·

du2 (t) + g22 (u1 (t)) dt

2 =

140 ■ Differential Geometry and Its Visualization

Figure 2.27 Normal and geodesic curvature κn = 0 and κg = κ at a point of a curve on a catenoid



1 1 − +a 2 a a



·



du1 (t) g11 (u1 (t)) dt

2

1



du2 (t) + g22 (u1 (t)) dt

2 = 0

(Figure 2.27).

2.5

THE NORMAL, PRINCIPAL, GAUSSIAN AND MEAN CURVATURE

In the previous section, we split the vector of curvature into two components the lengths of which were the geodesic and normal curvature κg and κn . First we are going to study the normal curvature of a curve on a surface. We will see in Theorem 2.5.6 that the normal curvature of curves at a point P on a surface only depends on the direction of the curves at P . The principal curvatures at P are the extreme values of κn at P which are the real solutions of a quadratic equation (Lemma 2.5.13 and Theorem 2.5.14); their geometric and arithmetic means are called the Gaussian and mean curvature (Definition 2.5.17). We introduce ruled surfaces, which among other things are useful for the visualization of the normal curvature along a given curve on a surface as in (2.58) and Figure 2.32. Definition 2.5.1. Let γ be a curve with a parametric representation ⃗y (t) (t ∈ I1 ) and ⃗z(t) be a unit vector for every t ∈ I1 . We write u1 = t. Then the surface generated by the curve γ and the family {⃗z(u1 ) : u1 ∈ I1 } with a parametric representation ⃗x(ui ) = ⃗y (u1 ) + u2⃗z(u1 ) is called a ruled surface.



(u1 , u2 ) ∈ D = I1 × I2 ⊂ R2



Surfaces in Three-Dimensional Euclidean Space ■ 141

Figure 2.28

sphere

Principal normal (left) and binormal (right) surfaces of a loxodrome on a

Remark 2.5.2. (a) It is obvious that any curve on a ruled surface that intersects any vector ⃗z(u1 ) exactly once with an angle unequal to zero can be chosen as a generating curve of the ruled surface. (b) The unit vectors ⃗z(u1 ), when attached to the origin of the coordinate system, define a curve on the unit sphere, the so–called spherical indicatrix. Visualization 2.5.3 (Some ruled surfaces). (a) Circular cylinders The circular cylinder with a parametric representation ⃗x(ui ) = {r cos u1 , r sin u1 , 0} + u2 {0, 0, 1}





(u1 , u2 ) ∈ D ⊂ (0, 2π) × R ,

where r > 0 is a constant, is a ruled surface; its generating curve is a circle line with radius r and centre in the origin in the x1 x2 –plane, here the vectors ⃗z(u1 ) = ⃗e 3 are constant. (b) Tangent, principal normal and binormal surfaces Let γ be a curve with a parametric representation ⃗y (s) (s ∈ I1 ) of class C 3 , where s denotes the arc length along γ. We write u1 = s. If γ has non–vanishing curvature κ on I, and ⃗vk denote the vectors of the trihedra of γ, we may consider the ruled surfaces given by the parametric representations (Figures 2.28 and 2.29) ⃗x(ui ) = ⃗y (u1 ) + u2⃗v1 (u2 ) ⃗x(ui ) = ⃗y (u1 ) + u2⃗v2 (u2 ) and ⃗x(ui ) = ⃗y (u1 ) + u2⃗v3 (u2 )



(u1 , u2 ) ∈ I1 × I2 , the tangent surface of γ,



(u1 , u2 ) ∈ I1 × I2 ) , the binormal surface of γ.







(u1 , u2 ) ∈ I1 × I2 , the principal normal surface of γ 

142 ■ Differential Geometry and Its Visualization

Figure 2.29

Tangent, principal normal and binormal surfaces of a helix

(c) Conoids and helicoids Let h : (0, 2π) → R be a function. Then the conoid is a ruled surface with a parametric representation ⃗x(ui ) = {0, 0, h(u1 )} + u2 {cos u1 , sin u1 , 0}



(u1 , u2 ) ∈ (0, 2π) × I2



(left in Figure 2.30). A helicoid is a special case of a conoid when h(u1 ) = cu1 for some constant c ̸= 0; it is constructed by joining every point of the helix with the point on the x3 –axis with the same third coordinate by a straight line (right in Figure 2.30). We introduced conoids and helicoids as special cases of ruled surfaces. They may also be considered as special cases of the class of general screw surfaces. Definition 2.5.4. Let f be a real–valued function on a domain D = I1 × I2 ⊂ R2 . Then a general screw surface is given by a parametric representation (left in Figure 2.31)   ⃗x(ui ) = {u1 cos u2 , u1 sin u2 , f (u1 , u2 )} (u1 , u2 ) ∈ D ; (2.56)

in the special case of f (ui , u2 ) = cu2 + g(u1 ), where c is a constant and g is a real valued function on I1 , we obtain the so–called screw surface with a parametric representation (right in Figure 2.31) ⃗x(ui ) = {u1 cos u2 , u1 sin u2 , cu2 + g(u1 )}





(u1 , u2 ) ∈ D .

(2.57)

Remark 2.5.5. (a) If c = 0 in (2.57) then the screw surface is a surface of revolution. (b) Similarly as an explicit surface may be used as a representation of a real valued function f of two real parameters over the x1 x2 –plane, where the parameters u1 and u2 are interpreted as the Cartesian coordinates of the plane, a general screw surface

Surfaces in Three-Dimensional Euclidean Space ■ 143

Figure 2.30

A conoid and a helicoid

Figure 2.31

General screw surface (left) and screw surface (right)

may be used as the representation of f over the x1 x2 –plane, where the parameters u1 and u2 are interpreted as the polar coordinates of the plane. (c) If we interchange u1 and u2 in (2.57) and put h = g and c = 0, then we obtain a helicoid as in Visualization 2.5.3 (c). (d) A screw surface S is generated by the simultaneous rotation of a curve about a fixed axis A and a translation along A such that the speed of translation is proportional to the speed of rotation. We choose A to be the x3 –axis and assume that the curve γ in the x1 x3 –plane is given by a parametric representation ⃗x(u1 ) = {u1 , 0, g(u1 )} with ⃗x(0) = ⃗0, where u1 denotes the distance between the x3 –axis and the points of γ. Let u2 denote the angle of rotation. Since the translation of γ is parallel to the x3 –axis and

144 ■ Differential Geometry and Its Visualization

Figure 2.32

Representation of the normal curvature of a curve on a torus

proportional to u2 , we obtain the parametric representation (2.57) for S. The u1 –lines of S are referred to as the meridians of S, and the u2 –lines are helices. The normal curvature along a curve on a surface can be visualized in a natural way as follows. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D ⊂ R2 ) and γ be a curve on S with a parametric representation ⃗x(t) = ⃗x(ui (t)) (t ∈ I) and normal curvature κn (t) = κn (ui (t)). Then we may represent κn (t) by the curve γn with a parametric representation ⃗ (t) (t ∈ I). ⃗x ∗ (t) = ⃗x(t) + κn (t)N

Writing u ∗1 = t, we see that γn is a curve on the ruled surface RS that has a parametric representation ⃗x ∗ (u ∗i ) = ⃗y (u ∗1 ) + u ∗2⃗z(u ∗1 )





(u∗ 1 , u ∗2 ) ∈ I × R ,

⃗ (ui (u ∗1 )), (2.58) where ⃗y (u ∗1 ) = ⃗x(ui (u ∗1 )) and ⃗z(u ∗1 ) = N and γn considered as a curve on RS is given by putting u ∗2 = κn (ui (u ∗1 )) in (2.58) (Figure 2.32). It turns out that the normal curvature of a curve at a point only depends on the direction of the curve. Theorem 2.5.6. All curves on a surface that intersect at a point P and have the same tangent at P , also have the same normal curvature at P (Figure 2.33). Proof. Since the direction of a curve with a parametric representation ⃗x(ui (s)) on a surface with a parametric representation ⃗x(ui ) is given by u˙ 1 (s) and u˙ 2 (s), the theorem is an immediate consequence of the formula for the normal curvature in (2.46).

Surfaces in Three-Dimensional Euclidean Space ■ 145

Figure 2.33

Curves on a cylinder with the same tangents and normal curvature

The following result gives a geometric interpretation of the normal curvature. Theorem 2.5.7 (Meusnier). The normal curvature of a curve on surface is equal to its curvature multiplied by the cosine of the angle between the principal normal vector of the curve and the normal vector of the surface (Figure 2.34). Proof. Let S be a surface with a parametric representation ⃗x(ui ) and γ be a curve on S, with a parametric representation ⃗x(ui (s)). If α(s) denotes the angle between ⃗ (s) of S the principal normal vector ⃗v2 (s) of γ at s and the surface normal vector N at s, then it follows from (2.40) that the normal curvature of γ at s is 



⃗ (s) = κ(s)⃗v2 (s) • N ⃗ (s) = κ(s) cos α(s), κn (s) = ⃗x¨(s) − κg (s)⃗t(s) • N

(2.59)

⃗ (s)∥ = 1. ⃗ (s) and ∥⃗v2 (s)∥ = ∥N since ⃗t(s) ⊥ N

The normal section of a surface is closely related to the normal curvature.

Definition 2.5.8. Let S be a surface, γ be a curve on S and P be a point on the curve γ. Furthermore, let E be the plane through the point P , and spanned by the tangent vector ⃗v1 of γ at P and the normal vector of the surface at P . Then the curve of intersection of the surface S and the plane E is called the normal section of the surface S at the point P along the direction of ⃗v1 (Figure 2.35). We will also say, the normal section of S at P along γ. The next result gives a relation between the normal curvature and the curvature of the normal section.

146 ■ Differential Geometry and Its Visualization

Projection of the vector of curvature of a curve on a cone to the surface normal vector of the cone

Figure 2.34

Theorem 2.5.9. The absolute value of the normal curvature of a curve on a surface at a point P is equal to the curvature of the normal section of the surface at P along the curve (Figure 2.36). Proof. Let P be a point on a surface S and κn be the normal curvature corresponding to a given direction d⃗ at P (Theorem 2.5.6). Then the curvature κ of all curves on S through P that have a tangent in the direction of d⃗ is given by (2.59) as κn = κ cos α where α is the angle between the principal normal vector ⃗v2 of the curves at P and ⃗ of S at P . If α = 0, then κ = κn , and if α = π then the surface normal vector N κ = −κn . Thus |κn | is the curvature of the intersection of the surface S with the plane through P that contains the direction d⃗ and the surface normal vector at P . Visualization 2.5.10. (Normal section of a cylinder along a helix and its curvature) Let r, h > 0 and Cyl be the circular cylinder with a parametric representation   ⃗x(ui ) = {r cos u2 , r sin u2 , u1 } (u1 , u2 ) ∈ R × (0, 2π) and γ be the helix on Cyl given by

u1 (s) = hωs and u2 (s) = ωs (s ∈ R) where ω = √

1 . r2 + h2

Then the normal section of Cyl along γ at s has a parametric representation ⃗ (ui (s)) + 1 sin t⃗v1 (ui (s)) (t ∈ (0, 2π)) ⃗ys (t) = ⃗x(ui (s)) + r(1 − cos t)N ω

(2.60)

Surfaces in Three-Dimensional Euclidean Space ■ 147

Figure 2.35

Normal section of a loxodrome on a torus

and its curvature at s is given by κs (t) =

r ω (r2

+

h2

cos2 t)3/2

(t ∈ (0, 2π)).

(2.61)

Proof. The tangent vector of the helix and the surface normal vector of the cylinder at any s ∈ R are by Visualizations 1.6.7 (b) and 2.4.7 (b) and

⃗v1 (s) = ⃗x˙ (ui (s)) = {−rω sin (ωs), rω cos (ωs), hω} ⃗ (s) = N ⃗ (ui (s)) = {− cos (ωs), − sin (ωs), 0}. N

A parametric representation for the plane of the normal section of Cyl along γ at s is ⃗ (s) for all λ, µ ∈ R. {x1 , x2 , x3 } = ⃗x(ui (s)) + λ⃗v1 (s) + µN The normal section of Cyl along γ at s has to satisfy the equation r2 = (x1 )2 + (x2 )2 = ((r − µ) cos (ωs) − λrω sin (ωs))2 + ((r − µ) sin (ωs) + λrω cos (ωs))2 or

= (r − µ)2 + λr2 ω 2 ,

λ2 (µ − r)2 + = 1. r2 1/ω 2

⃗ (s) and ⃗v1 (s). This is the equation of an ellipse in the plane spanned by the vectors N Hence the normal section of Cyl along γ at s has a parametric representation ⃗ (ui (s)) + ⃗ys (t) = ⃗x(ui (s)) + r(1 − cos t)N

1 sin t⃗v1 (ui (s)) (t ∈ (0, 2π)). ω

Thus we have shown (2.60). Furthermore, it follows that d⃗ys (t) ⃗ (ui (s)) + 1 cos t⃗v1 (ui (s)), = r sin tN dt ω

148 ■ Differential Geometry and Its Visualization

Normal curvature at a point of a curve on cylinder and curvature of the corresponding normal section

Figure 2.36

d 2 ⃗ys (t) ⃗ (ui (s)) − 1 sin t⃗v1 (ui (s)), = r cos tN 2 dt ω  2  d⃗   ys (t)  = r 2 sin2 t + 1 cos2 t  dt  ω2

and

   d⃗ d 2 ⃗ys (t)  r 2 r  ys (t)  × sin t + cos2 t = .  = 2  dt dt  ω ω

Hence, the curvature of the normal section of Cyl along γ at s is by (1.88)    d⃗ys (t) d 2⃗ y (t)   dt × dts2  r κs (t) = =    3/2  d⃗ys (t) 3 2 2 ω r sin t + ω12 cos2 t  dt 

=

r

ω (r2 + h2 cos2 t)3/2

(t ∈ (0, 2π)).

Thus we have shown (2.61). In particular, we have by Visualization 2.4.7 (b) κs (0) =

ω(r2

r = rω 2 = κn (s) for all s ∈ R (Figure 2.25). + h2 )3/2

Remark 2.5.11. Meusnier’s theorem may be restated as follows. The centres of curvature of all curves on a surface S that are tangent at a point P lie on a circle

Surfaces in Three-Dimensional Euclidean Space ■ 149

Figure 2.37

Illustration of Remark 2.5.11

line in their joint normal plane; this circle line has the diameter 1/|κn | and is tangent to the surface S at P (Figure 2.37). Proof. Let ⃗x0 be the position vector of a point P on a surface S. The centres of curvature of all curves with parametric representations ⃗x(s) through P are given by the position vectors 1 m ⃗ = ⃗x0 + ⃗v2 , κ where κ and ⃗v2 are the curvature and the principal normal vectors of the curves at P . All curves that are tangent to one another at P have the same tangent vector ⃗v1 at P which is orthogonal to ⃗v2 . This implies (m ⃗ − ⃗x0 ) • ⃗v1 = 0, hence the vector m ⃗ is in the joint normal plane of all these curves, since their normal plane is orthogonal to the vector ⃗v1 . Furthermore, all these curves have the same normal curvature by ⃗ , where N ⃗ is the surface normal vector of S at Theorem 2.5.6, namely κn = κ⃗v2 • N P . Therefore, we have 



1 ⃗ m ⃗ − ⃗x0 − N 2κn

2

=



1 ⃗ 1 ⃗v2 − N κ 2κn

2

=

1 1 ⃗ + 1 = 1 , − ⃗v2 • N κ2 κκn 4κ2n 4κ2n

that is, m ⃗ is on a circle line of radius 1/2|κn | with its centre given by ⃗x0 − 1/(2κn ). Obviously ⃗x0 is on this circle line, and the tangent to this circle line at P is orthogonal ⃗ . Hence this circle line is tangent to the surface S at P . to N We saw in Theorem 2.5.6 that the normal curvature of a curve on a surface at a point depends on the direction of the curve at that point only. So it is quite natural to find the directions for which the normal curvature has extreme values.

150 ■ Differential Geometry and Its Visualization

Figure 2.38

Principal directions d⃗ 1 and d⃗ 2 at a point of a surface

Definition 2.5.12. Let S be a surface and P a given point on S. The directions corresponding to the extreme values of the normal curvature at P are called the principal directions at the point P (Figure 2.38) and the extreme values of the normal curvature at P are called the principal curvature at the point P . We will see that there are two orthogonal principal directions at every point of a surface. The following lemma is needed for he proof of this statement. Lemma 2.5.13. Let S be s surface with first and second fundamental coefficients gik and Lik . Then the solutions λ of the quadratic equation det(Lik − λgik ) = 0

are always real.

2 > 0 implies Proof. First we observe that g = g11 g22 − g12





2 2g12 (g11 L12 − g12 L12 ) g11 2 g11 g22 − g12 +4· (g11 L12 − g12 L11 )2 2 g11 g12 = (g11 L22 − g22 L11 )2 − 4 · (g11 L22 − g22 L11 )(g11 L12 − g12 L11 ) g11 g22 +4· (g11 L12 − g12 L11 )2 g11

0 ≤ g11 L22 − g22 L11 −

4 · (g11 L12 − g12 L11 ) [g22 (g11 L12 − g12 L11 ) g11 −g12 (g11 L22 − g22 L11 )]

= (g11 L22 − g22 L11 )2 + = (g11 L22 − g22 L11 )2 +

4 · (g11 L12 − g12 L11 )(g11 g22 L12 − g11 g12 L22 ) g11

= (g11 L22 − g22 L11 )2 + 4 · (g11 L12 − g12 L11 )(g22 L12 − g12 L22 )

(2.62)

Surfaces in Three-Dimensional Euclidean Space ■ 151

= (g11 L22 + g22 L11 − 2g12 L12 )2 − 4 · g11 g22 L11 L22 + 4 · g12 L12 (g11 L22 + g22 L11 ) 2 2 − 4 · g12 L12 + 4 · (g11 L12 − g12 L11 )(g22 L12 − g12 L22 )

= (g11 L22 + g22 L11 − 2 · g12 L12 )2 − 4 · g11 g22 L11 L22 + 4 · g11 g12 L12 L22

2 2 + 4 · g12 g22 L11 L12 − 4 · g12 L12 + 4 · g11 g22 L212 − 4 · g11 g12 L12 L22 2 − 4 · g12 g22 L11 L12 + 4 · g12 L11 L22

= (g11 L22 + g22 L11 − 2 · g12 L12 )2 



2 − 4 · g11 g22 (L11 L22 − L212 ) − g12 (L11 L22 − L212 )

2 = (g11 L22 + g22 L11 − 2 · g12 L12 )2 − 4 · (g11 g22 − g12 )(L11 L22 − L212 )

= (g11 L22 + g22 L11 − 2 · g12 L12 )2 − 4 · gL.

The solutions of the equation



L11 − λg11 L12 − λg12 0 = det(Lik − λgik ) = det L12 − λg12 L22 − λg22



= (L11 − λg11 )(L22 − λg22 ) − (L12 − λg12 )2

2 = λ2 (g11 g22 − g12 ) − λ(g11 L22 + g11 L11 − 2g12 L12 ) + L11 L22 − L212

are λ1,2 and

= λ2 g − λ(g11 L22 + g11 L11 − 2 · g12 L12 ) + L 



 1 = g11 L22 + g22 L11 − 2g12 L12 ± (g11 L22 + g22 L11 − 2g12 L12 )2 − 4gL , 2g

(g11 L22 + g22 L11 − 2 · g12 L12 )2 − 4 · gL ≥ 0,

by what we have just shown. Thus the solutions of the equation in (2.62) are always real.

Theorem 2.5.14. There exist two orthogonal principal directions d⃗µ = ξµk ⃗xk (µ = 1, 2) at every point of a surface; they are given by the solutions of the equations (Lik − κµ gik ) ξµk = 0 (i = 1, 2) and gik ξµi ξµk = 1 for µ = 1, 2.

(2.63)

gjk ξ j ξ k = 1.

(2.64)

Proof. By Theorem 2.5.6, we may assume that the directions at a point P0 of the surface are given by ξ 1 and ξ 2 where ξ k ⃗xk = u˙ k ⃗xk . It follows that

To determine the extreme values of κn in the point P0 , we have to find the extreme values of the function f (ξ i ) = Ljk ξ j ξ k (2.65) under the side condition (2.64). Using the Lagrange multiplier rule, we have to solve ∂ (Ljk ξ j ξ k − λgjk ξ j ξ k ) = 0 (i = 1, 2), ∂ξ i

(2.66)

152 ■ Differential Geometry and Its Visualization

or

(Lik − λgik )ξ k = 0 (i = 1, 2).

(2.67)

det(Lik − λgik ) = 0.

(2.68)

The homogeneous system (2.67) has a solution if and only if

The quadratic equation (2.68) for λ has two real solutions κ1 and κ2 by Lemma 2.5.13; we denote by ξ1i and ξ2i (i = 1, 2) the corresponding solutions of (2.67) that also satisfy the side condition (2.64). Since we have by the first identity in (2.46) in Remark 2.4.3 (c) κµ = Lik ξµi ξµk with gik ξµi ξµk = 1 for µ = 1, 2, the values κµ are the extreme values of the normal curvature in the directions given by ξµi . The principal directions d⃗µ = ξ i⃗xi (µ = 1, 2) satisfy by (2.67) 

(Lik − κ1 gik )ξ1k = 0 (i = 1, 2), gik ξ1i ξ1k = 1 and

(Lik − κ2 gik )ξ2k = 0 (i = 1, 2), gik ξ2i ξ2k = 1.



(2.69)

We multiply the first equation in (2.69) by ξ2i and sum with respect to i to obtain (Lik − κ1 gik )ξ1k ξ2i = 0,

(2.70)

and multiply the third equation in (2.69) by ξ1i and sum with respect to i to obtain (Lik − κ2 gik )ξ1i ξ2k = 0.

(2.71)

We subtract equation (2.71) from (2.70), observing the symmetries gik = gki and Lik = Lki and interchanging the order of summation in (2.71) 0 = (Lik − κ1 gik )ξ1k ξ2i − (Lik − κ2 gik )ξ1k ξ2i = (κ2 − κ1 )gik ξ1k ξ2i = (κ1 − κ2 )d⃗1 • d⃗2 . If κ1 ̸= κ2 , then this implies d⃗1 •d⃗2 = 0, that is, the principal directions are orthogonal. If κ1 = κ2 , then Lik ξ i ξ k = k1 gik ξ i ξ k for all directions {ξ 1 , ξ√2 }, in particular, we obtain for {ξ 1 , ξ 2 } = {1, 0}, {ξ 1 , ξ 2 } = {0, 1} and {ξ 1 , ξ 2 } = 1/ 2{1, 1} that L11 = κ1 g11 , L22 = κ1 g22 and L12 = κ1 g12 . Conversely if Lik = c · gik (i, k = 1, 2) for some constant c then i duk Lik du dt dt κn = i duk = c. gik du dt dt Thus we have shown that κ1 = κ2 if and only if Lik = c · gik (i, k = 1, 2) for some constant c ∈ R. In this case, the equations in (2.67) are satisfied for arbitrary directions d⃗1 and d⃗2 , in particular, we may choose d⃗1 and d⃗2 to be orthogonal.

Surfaces in Three-Dimensional Euclidean Space ■ 153

Definition 2.5.15. A point on a surface with equal principal curvatures is called an umbilical point. Remark 2.5.16. (a) By Vieta’s formula, the principal curvatures κ1 and κ2 satisfy the conditions κ1 + κ2 1 1 L = (g22 L11 − 2g12 L12 + g11 L22 ) = g ik Lik and κ1 κ2 = , 2 2g 2 g

(2.72)

where g ik (i, k = 1, 2) are the entries of the inverse of the matrix (gik ) of the first fundamental coefficients (Remark 2.3.2 (b)). (b) We may choose any two orthogonal directions as principal directions at an umbilical point. The proof of Theorem 2.5.14 shows that points with κ1 = κ2 need a special treatment. The condition κ1 = κ2 is equivalent to Lik = c · gik (i, k = 1, 2) for some constant c ∈ R.

(2.73)

Definition 2.5.17. Let gik , Lik , κ1 and κ2 denote the first and second fundamental coefficients and the principal curvatures of a surface. Then the functions H=

κ1 + κ2 1 L = g ik Lik and K = κ1 · κ2 = 2 2 g

(2.74)

are called the mean and Gaussian curvature of the surface. Remark 2.5.18. It is clear from the definitions of the Gaussian curvature and umbilical points that a surface with negative Gaussian curvature cannot have any umbilical points. Visualization 2.5.19. (Gaussian and mean curvature of some surfaces of revolution) We consider a surface of revolution with a parametric representation ⃗x(ui ) = {r(u1 ) cos u2 , r(u1 ) sin u2 , h(u1 )} with the usual assumptions





(u1 , u2 ) ∈ D = I1 × I2 ⊂ R × (0, 2π) (2.75)

r(u1 ) > 0 and |r′ (u1 )| + h′ (u1 )| > 0 on I1 . Then it follows for the Gaussian and mean curvature K and H from (2.72) in Remark 2.5.16, (2.5.17) in Definition 2.5.17 and the formulae for the first and second fundamental coefficients of surfaces of revolution (2.24) in Example 2.3.5 and Example 2.4.6 that K(ui ) = K(u1 ), H(ui ) = H(u1 ), K= and H=

1 (g11 L22 + g22 L11 ) 2g

(h′′ r′ − h′ r′′ )h′ L = g r ((r′ )2 + (h′ )2 )2

(2.76)

154 ■ Differential Geometry and Its Visualization

1

=







h′ r (r′ )2 + (h′ )2 + (h′′ r′ − h′ r′′ ) (r)2

2(r)2 ((r′ )2 + (h′ )2 )3/2     1 ′ ′ 2 ′ 2 ′′ ′ ′ ′′ = h (r ) + (h ) + (h r − h r ) r . 2r ((r′ )2 + (h′ )2 )3/2



(2.77)

Since the Gaussian and mean curvature K and H of a surface of revolution only depend on the parameter u1 , we may represent K and H as surfaces with parametric representations ⃗x(K) (ui ) = {u1 cos u2 , u1 sin u1 , K(u1 )}

and

1

⃗x(H) (ui ) = {u1 cos u2 , u1 sinu , H(u1 )}.

In particular, if u1 is the arc length along the curve that generates the surface of revolution, that is, if (r′ (u1 ))2 +(h′ (u1 ))2 = 1 on I1 , then it follows that r′ r′′ +h′ h′′ = 0 r′ r′′ and, for h′ ̸= 0, h′′ = − ′ and so h 

r′′ (r′ )2 h r −hr =− + h′ r′′ h′ ′′ ′

′ ′′



=−

 r′′  ′ 2 r′′ ′ 2 (r ) + (h ) = − . h′ h′

(2.78)

In this case, the formulae for the Gaussian and mean curvature reduce to K=−

 1  ′ 2 r′′ ′′ and H = (h ) − r r . r 2rh′

(2.79)

(a) Spheres. First we consider the sphere with radius R > 0. Then we have r(u1 ) = R cos u1 , h(u1 ) = R sin u1 , (r′ (u1 ))2 + (h′ (u1 ))2 = R2 , r′′ (u1 ) = −r(u1 ) and h′ (u1 ) = r(u1 ), and we may apply (2.76), (2.77) and the second identity in (2.78) to obtain 1 r′′ (u1 ) K(u1 ) = − = 1 r(u )R R and 

′′ 1 1 1 2 ′ 1 2 r (u )r(u ) H(u ) = R h (u ) − R 2r(u1 )R3 h′ (u1 ) 1 1 (r(u1 ) + r(u1 )) = = 1 2Rr(u ) R 1



for the Gaussian and mean curvature of the sphere. (b) Catenoids. Now we consider the catenoid with r(u1 ) = a cosh (u1 /a), h(u1 ) = u1 where a > 0 is a constant. Then we have r′ (u1 ) = sinh (u1 /a), r′′ (u1 ) = (1/a) cosh (u1 /a), h′ (u1 ) = 1, h′′ (u1 ) = 0,

Surfaces in Three-Dimensional Euclidean Space ■ 155

Figure 2.39 A catenoid (left) and its Gaussian curvature represented as a surface of revolution (right)

(r′ (u1 ))2 + (h′ (u1 ))2 = sinh2 (u1 /a) + 1 = cosh2 (u1 /a) = r′′ (u1 )r(u1 ) and it follows from (2.76) and (2.77) that 1

K(u ) = −

1 a2 cosh4



u1 a

 cosh

2



u1 a



=−

1 a2 cosh2



u1 a



and H(u1 ) = 0 for the Gaussian and mean curvature of the catenoid (Figure 2.39). (c) Pseudo–spheres. Finally, we consider the surface of revolution with 1

1

1

r(u ) = exp (−u ) and h(u ) = We have

 

1 − exp (−2u1 ) du1 for u1 > 0.

r′ (u1 ) = −r(u1 ) = −r′′ (u1 ) = exp (−u1 ), h′ (u1 ) = and



1 − exp (−2u1 )

(r′ (u1 ))2 + (h1 (u1 ))2 = 1

and it follows from (2.79) K(u1 ) = −1 and H(u1 ) =

1 − 2 exp (−2u1 )  2 exp (−u1 ) 1 − exp (−2u1 )

(2.80)

for the Gaussian and mean curvature of the surface of revolution. A surface of revolution with constant negative Gaussian curvature is called a pseudo-sphere (Figure 2.40). Now we determine surfaces of revolution for given Gaussian curvature K. The cases of given constant Gaussian curvature K are of special interest; they yield spherical and pseudo-spherical surfaces or K > 0 and K < 0, respectively. Let S

156 ■ Differential Geometry and Its Visualization

Figure 2.40 A pseudo-sphere (left) and its mean curvature represented as a surface of revolution (right)

be a surface of revolution with a parametric representation (2.75), where we may assume that (r′ )2 + (h′ )2 = 1. The Gaussian curvature of S is given by (2.79) as r′′ (u) + K(u)r(u) = 0,

(2.81)

where we write u instead of u1 . Thus the surface with a given Gaussian curvature K(u) is given by the solutions r(u) of (2.81), and h(u) = ±

 

1 − (r′ (u))2 du,

(2.82)

where we may choose the upper sign in (2.82) without loss of generality. Surfaces of identically vanishing mean curvature will be studied in Section 3.11. Visualization 2.5.20. (Surfaces of revolution with constant K > 0: Spherical surfaces) If K(u) is constant and positive, we put K = 1/c2 with some constant c > 0. Then (2.81) reduces to 1 r′′ (u) + 2 r(u) = 0 (2.83) c with the general solution r(u) = λ cos

 

u c

with λ > 0

and we obtain h(u) =





 

u λ2 1 − 2 sin2 du. c c

(2.84)

There are three different types of spherical surfaces corresponding to the cases λ = c, λ > c or λ < c.

Surfaces in Three-Dimensional Euclidean Space ■ 157

(i) Case 1. λ = c Then the surface has a parametric representation 



u1 ⃗x(ui ) = c cos c





u1 cos u2 , c cos c



sin u2 , c sin 



u1 c





(u1 , u2 ) ∈ (−π/2, π/2) × (0, 2π) .

This is a sphere with radius c and centre in the origin.

(ii) Case 2. λ > c The corresponding surfaces are called hyperbolic spherical surfaces. Now the integral for h in (2.84) only exists for values of u with

that is,



     sin u  ≤ c ,  c  λ

u ∈ Ik = −c sin−1

 

c λ

+ kπ, c sin−1

 

c λ

+ kπ



for k = 0, ±1, ±2, . . . (left in Figure 2.41). Every interval Ik defines a region of the surface. The radii of the circle lines of√the u2 –lines are minimal at the end points of the intervals Ik and equal to r = λ2 − c2 , whereas the maximum radius R = λ is attained in the middle of each region (left in Figure 2.43). (iii) Case 3. λ < c The corresponding surfaces are called elliptic spherical surfaces (right in Figure 2.41). Now the integral for h in (2.84) exists for all u and the radii r of the circle lines of the u2 –lines attain all values r ≤ λ. Visualization 2.5.21. (Surfaces of revolution with constant K < 0: Pseudo–spherical surfaces) If K(u) is constant and negative, we put K = −1/c2 with some constant c > 0. The general solution of the differential equation in (2.83) is r(u) = C1 cosh with constants C1 and C2 . Again there are three cases.

 

u c

+ C2 sinh

 

u c

(i) Case 1. C1 = −C2 = λ ̸= 0 The corresponding surfaces are called parabolic pseudo-spherical surfaces (left in Figure 2.42). They have a parametric representation with 

u1 r(u ) = λ exp − c 1



158 ■ Differential Geometry and Its Visualization

Figure 2.41

Hyperbolic (left) and elliptic (right) spherical surfaces

and 1

h(u ) =









λ2 2u1 1 − 2 exp − du1 for u1 > c log (|λ|/c). c c

(ii) Case 2. C2 = 0 and C1 = λ ̸= 0 The corresponding surfaces are called hyperbolic pseudo-spherical surfaces (middle in Figure 2.42). They have a parametric representation with 

u1 r(u ) = λ cosh c 1

Figure 2.42

surfaces



Parabolic (left), hyperbolic (middle) and elliptic (right) pseudo-spherical

Surfaces in Three-Dimensional Euclidean Space ■ 159

Hyperbolic spherical and pseudo-spherical surfaces, and minimal and maximal radii of u2 –lines

Figure 2.43

and 1

h(u ) = for



|u1 | ≤ c · sinh−1





1−

c |λ|







λ2 u1 sinh2 2 du1 2 c c 

c + = c log  |λ|





c2 + 1. λ2

The radii r of the circle lines of the u2 –lines satisfy (right in Figure 2.43) √ |λ| ≤ r ≤ λ2 + c2 . (iii) Case 3. C1 = 0 and C2 = λ ̸= 0

The corresponding surfaces are called elliptic pseudo-spherical surfaces. They have a parametric representation with r(u1 ) = λ sinh 1

h(u ) = for all u1 with







u1 c







λ2 u1 1 − 2 cosh2 du1 c c

cosh



u1 c



≤

c ; |λ|

(2.85)

(since cosh u1 ≥ 1 for all u1 , we must have |λ| ≤ c) (right in Figure 2.42). The integral for h in (2.85) is elliptic. The radii r of the circle lines of the u2 –lines satisfy √ 0 ≤ r ≤ c 2 − λ2 .

160 ■ Differential Geometry and Its Visualization

Surfaces of revolution with Gaussian curvature K(u) = k/u2 for k = −0.5 (left) and k = −1 (right) Figure 2.44

Visualization 2.5.22 (Surfaces with Gaussian curvature K(u1 ) = k/(u1 )2 ). We determine the surfaces of revolution of Gaussian curvature K(u1 ) = k/(u1 )2 , where k ̸= 0 is a constant. We write u = u1 . Then the differential equation (2.81) reduces to k r′′ (u) + 2 · r(u) = 0; (2.86) u which is a differential equation of Euler type. We also use (2.82). (i) First we consider the case k < 1/4 (Figure 2.44). √ We put γ = 1 − 4k and obtain two linearly independent solutions 1

1

r1 (u) = u 2 (1+γ) and r2 (u) = u 2 (1−γ) . If r(u) = cr1 (u) where c ̸= 0 is a constant, then (2.82) yields h(u) =

1 2

 

u




(1 + γ)2 c2 4

1/(γ−1)

1/(1−γ)

if γ > 1, that is, k < 0, if 0 < γ < 1/4, that is, k > 0.

If r(u) = cr2 (u) where c ̸= 0 is a constant, then h(u) =

1 2

 

4 − (1 − γ)2 c2 u−(1+γ) du

Surfaces in Three-Dimensional Euclidean Space ■ 161

Surfaces of revolution with Gaussian curvature K(u) = k/u2 for k = 8 (left) and k = 2.5 (right)

Figure 2.45

for u>



(1 − γ)2 c2 4

1/(1+γ)

.

(ii) Now we consider the case k > 1/4 (Figure 2.45). √ We put δ = 4k − 1 and obtain two linearly independent solutions r1 (u) =

√

u cos



δ log u 2



and r2 (u) =

√

If r(u) = cr1 (u) where c ̸= 0 is a constant, then 





u sin





δ log u . 2



δ δ c cos log u − δ sin log u r (u) = √ 2 u 2 2 ′

and h(u) = for u>

 

1 − (r′ (u))2 du









c2 δ δ log u − δ sin log u cos 4 2 2

If r(u) = cr2 (u) where c ̸= 0 is a constant, then 







2

c δ δ log u + δ cos log u sin r (u) = √ 2 u 2 2 ′



.



162 ■ Differential Geometry and Its Visualization

Surfaces of revolution with given Gaussian curvature K(u), K(u) = exp(u) (left) and K(u) = sin u

Figure 2.46

and h(u) =

 

1 − (r′ (u))2 du

for u>









δ δ c2 sin log u + δ cos log u 4 2 2

2

.

Figure 2.46 shows surfaces of revolution with Gaussian curvature K(u) = exp(u) (left) and K(u) = sin u (right). Visualization 2.5.23 (Gaussian and mean curvature of explicit surfaces). (a) Let S be the explicit surface with a parametric representation ⃗x(ui ) = {u1 , u2 , (u2 )3 − a(u1 )2 u2 } where a ∈ R is a constant. Then



g(ui ) = 1 + 4a2 u1 u2

2





(u1 , u2 ) ∈ R2 ,



+ 3(u2 )2 − a(u1 )2

and the Gaussian and mean curvature of S are 

2



L(ui ) 12a u2 + 4a2 u1 = − K(u ) = g(ui ) (g(ui ))2 i

2

2

(2.87)

Surfaces in Three-Dimensional Euclidean Space ■ 163

The explicit surface S of Visualization 2.5.23 (a) for a = 3/2 (left), lines of constant Gaussian (middle) and constant mean curvature (right)

Figure 2.47

and 



 2 1 2 2 2 1 2 −2au 1 + 3(u ) − a(u ) H(u ) = (g(ui ))3/2 i



−8a2 u1

2











u2 3(u2 )2 − a(u1 )2 − 6u2 1 + 4a2 u1 u2

2 

, (2.88)

where a ∈ R is a constant. Figure 2.47 shows the surface S for a = 3/2, and lines of constant Gaussian and mean curvature on S. Figure 2.48 represents the Gaussian curvature of S as an explicit surface KS with a parametric representation ⃗x(ui ) = {u1 , u2 , K(u1 , u2 )} and the lines K(ui ) = const on KS. Figure 2.49 represents the mean curvature of S as an explicit surface HS with a parametric representation ⃗x(ui ) = {u1 , u2 , H(u1 , u2 )} and the lines H(ui ) = const on HS. (b) Let S be the explicit surface with a parametric representation 



⃗x(ui ) = u1 , u2 , (u1 )3 u2 − u1 (u2 )3 Then



 

g(ui ) = 1 + (u1 )2 + (u2 )2



(u1 , u2 ) ∈ R2 .

3

164 ■ Differential Geometry and Its Visualization

Figure 2.48 The Gaussian curvature of the explicit surface S of Visualization 2.5.23 (a) represented as an explicit surface (left) and its lines of constant Gaussian curvature (right)

The mean curvature of the explicit surface S of Visualization 2.5.23 (a) represented as an explicit surface (left) and its lines of constant mean curvature (right) Figure 2.49

and the Gaussian and mean curvature of S are K(u ) = −  i

and

H(u ) = −  i



9 (u1 )2 + (u2 )2

2

1 + ((u1 )2 + (u2 )2 )3 

6u1 u2 (u1 )4 − (u2 )4

1+

((u1 )2

+

(2.89)

2



3/2 (u2 )2 )3

.

(2.90)

Figure 2.50 shows the surface S, and lines of constant Gaussian and mean curvature on S. Figure 2.51 shows the surface S with its level lines, and its level lines in the parameter plane. Figure 2.52 represents the Gaussian and mean curvature of S as explicit surfaces as in Part (a), and the lines K(ui ) = const and H(ui ) = const.

Surfaces in Three-Dimensional Euclidean Space ■ 165

The explicit surface S of Visualization 2.5.23 (b) (left), lines of constant Gaussian (middle) and constant mean curvature (right) Figure 2.50

Figure 2.51

The explicit surface of Visualization 2.5.23 (b)

Figure 2.53 shows the ruled surface generated by the surface normal vectors along a curve on the surface of Part (b). Proof. We write x = u1 , y = u2 . (a) We put f (x, y) = (y 2 )3 − ax2 y. Then we have f1 (x, y) = −2axy, f2 (x, y) = 3y 2 − ax2 ,

f11 (x, y) = −2ay, f12 (x, y) = −2ax and f22 (x, y) = 6y,

and obtain for the first and second fundamental coefficients by (2.21) in Example 2.3.4 and (2.49) in Example 2.4.5 



g11 (x, y) = 1 + 4a2 (xy)2 , g12 (x, y) = −2axy 3y 2 − ax2 ,

166 ■ Differential Geometry and Its Visualization

Top: the Gaussian (left) and mean curvature (right) of the surface in Visualization 2.5.23 (a) as explicit surfaces. Bottom: the lines of constant Gaussian (left) and constant mean mean curvature (right) in Part (a), and the lines K(ui ) = const and H(ui ) = const on the surfaces on the top. Figure 2.52



g22 (x, y) = 1 + 3y 2 − ax2

and

L11 (x, y) = −  L(ui ) = −

2



, g(x, y) = 1 + 4a2 (xy)2 + 3y 2 − ax2

2

,

2ay 2ax 6y , L12 (x, y) = −  , L22 (x, y) =  g(x, y) g(x, y) g(x, y)

12ay 2 + 4a2 x2 . g(x, y)

This yields K(x, y) =

12ay 2 + 4a2 x2 L(x, y) , =− g(x, y) (g(x, y))2

that is, (2.87), and H(x, y) =

1 (L11 (x, y)g22 (x, y) − 2L12 (x, y)g12 (x, y) + L22 (x, y)g11 (x, y)) 2g(x, y)

Surfaces in Three-Dimensional Euclidean Space ■ 167

Ruled surfaces generated by the surface normal vectors along a curve on the surface of Visualization 2.5.23 (b) Figure 2.53





 2 1 2 2 −2ay 1 + 3y − ax = (g(x, y))3/2 







−8a2 x2 y 3y 2 − ax2 − 6y 1 + 4a2 (xy)2

that is, (2.88).



,

(b) We put f (x, y) = x3 y − xy 3 and obtain

f1 (x, y) = 3x2 y − y 3 , f2 (x, y) = x3 − 3xy 2 ,

f11 (x, y) = 6xy, f12 (x, y) = 3x2 − 3y 2 and f22 (x, y) = −6xy,

and as in Part (a)



2

,

2

,

g11 (x, y) = 1 + y(3x2 − y 2 ) 

g12 (x, y) = xy 3x2 − y 2 



g22 (x, y) = 1 + x(x2 − 3y 2 ) 

g(x, y) = 1 + y(3x2 − y 2

2





x2 − 3y 2 ,

+ x(x2 − 3y 2

2

= 1 + y 2 (9x4 − 6x2 y 2 + y 4 ) + x2 (9y 4 − 6x2 y 2 + x4 )

= 1 + 9x4 y 2 − 6x2 y 4 + y 6 + 9x2 y 4 − 6x4 y 2 + x6

= 1 + x6 + 3x4 y 2 + 3x2 y 4 + y 6 = 1 + (x2 + y 2 )3 ,

and

L11 (x, y) = 

6xy 3(x2 − y 2 ) = −L22 (x, y), L12 (x, y) =  g(x, y) g(x, y)

L(x, y) = −9

4x2 y 2 + (x2 − y 2 )2 (x2 + y 2 )2 =− . g(x, y) 1 + (x2 + y 2 )3

168 ■ Differential Geometry and Its Visualization

This implies K(x, y) =

(x2 + y 2 )2 L(x, y) = −9 , g(x, y) (1 + (x2 + y 2 )3 )2

that is, (2.89), and H(x, y) = =

1 (L11 (x, y)g22 (x, y) − 2L12 (x, y)g12 (x, y) + L22 (x, y)g11 (x, y)) 2g(x, y)    1 2 2 2 2 6xy 1 + x (x − 3y ) (g(x, y))3/2



−6xy(x2 − y 2 )(3x2 − y 2 )(x2 − 3y 2 ) 

− 6xy 1 + y 2 (3x2 − y 2 )2

=



 3xy x2 (x2 − 3y 2 )2 − x2 (3x2 − y 2 )(x2 − 3y 2 ) (g(x, y))3/2

+y 2 (3x2 − y 2 )(x2 − 3y 2 ) − y 2 (3x2 − y 2 )2

=

   3xy x2 (x2 − 3y 2 ) x2 − 3y 2 − (3x2 − y 2 ) 3/2 (g(x, y)) 





+y 2 (3x2 − y 2 ) x2 − 3y 2 − (3x2 − y 2 )

=− =−

 6xy(x2 + y 2 )  2 2 2 2 2 2 x (x − 3y ) + y (3x − y ) (g(x, y))3/2

6xy(x2 + y 2 )(x4 − y 4 ) 6xy(x2 + y 2 )(x4 − y 4 ) = − , (g(x, y))3/2 (1 + (x2 + y 2 )2 )3/2

that is, (2.90).

Remark 2.5.24. (a) The mean curvature at most changes its sign under parameter transformations. (b) The Gaussian curvature is invariant under parameter transformations. Proof. Let a parameter transformation be given by ui = ui (u∗1 , u∗2 ) for i = 1, 2. Then the formulae of transformations for g ik and Lik are by (2.19) in Remark 2.3.3 (b) and (2.45) in Remark 2.4.3 (b) g ∗ik = g jn where

∂u∗i ∂u∗k ∂ul ∂um ∗ and L = αL for i, k = 1, 2, lm ik ∂uj ∂un ∂u∗i ∂u∗k 



∂(ui ) . α = sign ∂(u∗k )

Surfaces in Three-Dimensional Euclidean Space ■ 169

(a) We obtain for the mean curvature ∂u∗i ∂u∗k ∂ul ∂um ∂uj ∂un ∂u∗i ∂u∗k l ∗i ∂u ∂u ∂um ∂u∗k = αg jn Llm ∗i ∂u ∂uj ∂u∗k ∂un jn = αg Llm δjl δnm = αg jn Ljn δjl = α2H.

2H ∗ = g ∗ik Lik = αg jn Llm

(b) Let Aik and A∗ik satisfy the formulae of transformation A∗ik = Alm Writing

∂ul ∂um for i, k = 1, 2. ∂u∗i ∂u∗k 

∂u1  ∗1 ∂u D=  ∂u2 ∂u∗1

and DT for the transpose of D, we obtain (AD)lk =

2 

Alm Dmk = Alm

m=1



∂u1   ∂u∗2 2  ∂u ∂u∗2 ∂um for l, k = 1, 2 ∂u∗k

and (DT (AD))ik =

2 

DilT (AD)lk =

l=1

= Alm hence

∂ul ∂um A lm ∂u∗i ∂u∗k

∂ul ∂um = A∗ik for i, k = 1, 2, ∂u∗i ∂u∗k

detA∗ = det(DT AD) = (detD)2 detA.

Therefore, we have detg ∗ = (detD)2 detg, detL∗ = (detD)2 detL and K∗ =

det L∗ L = =K ∗ det g g

for the Gaussian curvature.

Example 2.5.25 (The umbilical points on spheres). All points of a sphere are umbilical points.

170 ■ Differential Geometry and Its Visualization

Proof. Let S be a sphere with a parametric representation ⃗x(ui ) = r{cos u1 cos u2 , cos u1 sin u2 , sin u1 } Then we have





(u1 , u2 ) ∈ D = (−π/2, π/2) × (0, 2π) .

g11 (uk ) = r2 , g12 (uk ) = 0, g22 (uk ) = r2 cos2 u1 , L11 (uk ) = r, L12 (uk ) = 0 and L22 (uk ) = r cos2 u1 ,

hence

gik (uk ) = r · Lik (uk ) (i, k = 1, 2).

This is the characterization of umbilical points, by the identities in (2.73) of Remark 2.5.16 (b). Visualization 2.5.26. (The umbilical points on a hyperboloid of two sheets) We determine the umbilical points on a hyperboloid of two sheets given by a parametric representation ⃗x(ui ) = {a sinh u1 cos u2 , b sinh u1 sin u2 , ±c cosh u1 } where a, b, c ∈ R \ {0} are constants. (a) If a = b = c, then there are no umbilical points. (b) If a = b > c, then every point on the circle lines ⃗x(t) =





√ a2 a2 cos t, sin t, ± c2 + a2 c c

is an umbilical point. (c) If a > b ̸= c, then there are four umbilical points at 

(Figure 2.54).



(u1 , u2 ) ∈ D ⊂ R × (0, 2π) ,

X = 0, ±b





a2 − b 2 , ±c b 2 + c2





c2 + a2  b 2 + c2

Proof. We obtain ⃗x1 (ui ) = {a cosh u1 cos u2 , b cosh u1 sin u2 , ±c sinh u1 },

⃗x2 (ui ) = sinh u1 {−a sin u2 , b cos u2 , 0},

⃗x11 (ui ) = ⃗x(ui ),

⃗x12 (ui ) = cosh u1 {−a sin u2 , b cos u2 , 0},

⃗x22 (ui ) = − sinh u1 {a cos u2 , b cos u2 , 0}, 



g11 (ui ) = cosh2 u1 a2 cos2 u2 + b2 sin2 u2 + c2 sinh2 u1 ,

g11 (ui ) = (b2 − a2 ) sinh u1 cosh u1 sin u2 cos u2 ,

Surfaces in Three-Dimensional Euclidean Space ■ 171

Figure 2.54

The umbilical points of a hyperboloid of two sheets for distinct a, b and c 



g22 (ui ) = sinh2 u1 a2 sin2 u2 + b2 cos2 u2 , abc sinh u1 abc sinh3 u1 L11 (ui ) = ±  i , L12 (ui ) = 0 and L22 (ui ) = ±  i . g(u ) g(u )

(a) Let a = b = c. Then we have





g11 (ui ) = a2 cosh2 u1 + sinh2 u2 , g12 (ui ) = 0, g22 (ui ) = a2 sinh2 u1 , a3 sinh3 u1 a3 sinh u1 L11 (ui ) = ±  i , L12 (ui ) = 0 and L22 (ui ) = ±  i . g(u ) g(u )

The conditions in (2.73) for umbilical points now are

  a3 sinh u1 a3 sinh3 u1 = ka2 sinh2 u1 ±  i = ka2 cosh2 u1 + sinh2 u2 and ±  i g(u ) g(u )

for all (ui , u2 ) ∈ D and some constant k ∈ R. This implies

(cosh2 u1 + sinh2 u1 ) sinh2 u1 = sinh2 u1 ,

hence u1 = 0, since cosh2 u1 + sinh2 u1 > 0 for u1 ̸= 0; but ⃗x2 (0, u2 ) = ⃗0, and so the parametric representation is not admissible for u1 = 0. Thus there are no umbilical points. (b) Now let a = b > c. Then we have g11 (ui ) = a2 cosh2 u1 + c2 sinh2 u2 , g12 (ui ) = 0, g22 (ui ) = a2 sinh2 u1 , a2 c sinh u1 a2 c sinh3 u1 L11 (ui ) = ±  i , L12 (ui ) = 0 and L22 (ui ) = ±  i , g(u ) g(u )

and the conditions for umbilical points yields 



a2 cosh2 u1 + c2 sinh2 u2 sinh2 u1 = a2 sinh2 u1 ,

hence u1 = 0 or a2 − c2 sinh2 u1 = 0, which implies sinh u1 = ±a/c.

172 ■ Differential Geometry and Its Visualization

Thus every point on the circle lines ⃗x(t) =



√ a2 a2 cos t, sin t, ± c2 + a2 c c



(t ∈ (0, 2π))

is an umbilical point. (c) Finally, let a > b ̸= c. Then we have 







g11 (ui ) = cosh2 u1 a2 cos2 u2 + b2 sin2 u2 + c2 sinh2 u1 , g12 (ui ) = (b2 − a2 ) sinh u1 cosh u1 sin u2 cos u2 ,

g22 (ui ) = sinh2 u1 a2 sin2 u2 + b2 cos2 u2 ,

abc sinh u1 abc sinh3 u1 L11 (ui ) = ±  i , L12 (ui ) = 0 and L22 (ui ) = ±  i . g(u ) g(u )

First L12 (ui ) = kg12 (ui ) implies k = 0 or u2 = 0, π/2, π, 3π/2, since a2 ̸= b2 and u1 ̸= 0. If k = 0, then L11 (u1 ) = 0 implies u1 = 0, but u1 ̸= 0, since the parametric representation of the hyperboloid is not admissible at u1 = 0. Thus k has to be unequal to zero. Furthermore, since gij (u1 , 0) = gij (u1 , π), Lij (u1 , 0) = Lij (u1 , π),

gij (u1 , π/2) = gij (u1 , 3π/2) and Lij (u1 , π/2) = Lij (u1 , 3π/2) for i, j = 1, 2, we only have to study the cases u2 = 0 and u2 = π/2. (c.i) Let u2 = 0. Then we have g11 (u1 , 0) = a2 cosh2 u1 + c2 sinh2 u1 and g22 (u1 , 0) = b2 sinh2 u1 , and abc sinh u1 L11 (u1 , 0) = ±  1 = kg11 (u1 , 0) = k(a2 cosh2 u1 + c2 sinh2 u1 ) g(u , 0)

and

abc sinh3 u1 = kg22 (u1 , 0) = kb2 sinh2 u1 L22 (u1 , 0) = ±  1 g(u , 0)

imply a2 cosh2 u1 + c2 sinh2 u1 = b2 , since u1 ̸= 0. But a2 cosh2 u1 − b2 > 0 for all u1 , and so there is no umbilical point with u2 = 0.

Surfaces in Three-Dimensional Euclidean Space ■ 173

(c.ii) Finally, let u2 = π/2. Then we have g11 (u1 , π/2) = b2 cosh2 u1 + c2 sinh2 u1 and g22 (u1 , π/2) = a2 sinh2 u1 , and abc sinh u1 = kg11 (u1 , π/2) L11 (u1 , π/2) = ±  1 g(u , π/2) 2 = k(b cosh2 u1 + c2 sinh2 u1 ) and abc sinh3 u1 = kg22 (u1 , π/2) = ka2 sinh2 u1 L22 (u1 , π/2) = ±  1 g(u , 0)

imply b2 cosh2 u1 + c2 sinh2 u1 = a2 , that is, (b2 + c2 ) sinh2 u1 = a2 − b2 > 0 or   2 2 a − b c2 + a2 1 sinh u1 = ± 2 and cosh u = ± . b + c2 b 2 + c2 Therefore, there are four umbilical points on the hyperboloid of two sheets, at 

X = 0, ±b





a2 − b 2 , ±c b 2 + c2



c2 + a2  . b 2 + c2

Visualization 2.5.27 (Principal curvature on a hyperbolic paraboloid). We consider the hyperbolic paraboloid given by a parametric representation ⃗x(ui ) = {u1 , u2 , u1 u2 } Then it is easy to see that





(u1 , u2 ) ∈ R2 .

⃗x1 (uk ) = {1, 0, u2 }, ⃗x2 (uk ) = {0, 1, u1 }, g11 (uk ) = 1 + (u2 )2 , g12 (uk ) = u1 u2 ,

g22 (uk ) = 1 + (u1 )2 , g(uk ) = 1 + (u1 )2 + (u2 )2 , 1 1 . L11 (uk ) = L22 (uk ) = 0, L12 (uk ) =  and L(uk ) = − g(uk ) g(uk )

The principal curvature is given by κ1,2 (uk ) =

−u1 u2 ±



(1 + (u1 )2 ) (1 + (u2 )2 )

(1 + (u1 )2 + (u2 )2 )3/2

=

−g12 (uk ) ±



g11 (uk )g22 (uk )

(g(uk ))3/2

.

(2.91)

174 ■ Differential Geometry and Its Visualization

There are no umbilical points, since κ1 (uk ) ̸= κ2 (uk ) for all (u1 , u2 ) ∈ R2 . The mean and Gaussian curvature are given by H(uk ) = −

u1 u 2 (1 + (u1 )2 +

(u2 )2 )3/2

and K(uk ) = −

1 (1 +

(u1 )2

+ (u2 )2 )2

.

(a) First we find the principal curvature along a curve on the hyperbolic paraboloid. We omit the parameters u1 and u2 and obtain from (2.91) √ −g12 ± g11 g22 κ1,2 = . g 3/2 By (2.63), the principal direction corresponding to κ1 is given by         

κ1 g11 ξ11 = (L12 − κ1 g12 )ξ12 (L12 − κ1 g12 )ξ11 = κ1 g22 ξ12 and g11 (ξ11 )2 + 2g12 ξ11 ξ12 + g22 (ξ12 )2 = 1.

        

(2.92)

It follows from the first equation on the left hand side of (2.92) that √ 2 − g12 g11 g22 g12 1 √ + g g 3/2 L12 − κ1 g12 2 1 √ ξ1 = ξ12 ξ1 = −g12 + g11 g22 κ1 g11 g11 g 3/2 √ √ 2 − g12 g11 g22 2 g11 g22 − g12 g11 g22 2 g + g12  ξ = √  ξ = √ g11 −g12 + g11 g22 1 g11 g11 g22 − g12 1 √  √  g11 g22 g11 g22 − g12 2 g22 2 √  ξ1 = = ξ . g11 g11 g22 − g12 g11 1

Substituting this in the equation on the right hand side of (2.92), we obtain g11 (ξ11 )2 + 2g12 ξ11 ξ12 + g22 (ξ12 )2 = g22 (ξ12 )2 + 2g12 

= 2 g22 + g12





g22 2 2 (ξ ) + g22 (ξ12 )2 g11 1 

g22 (ξ12 )2 = 1, g11

and we may choose ξ12 =

√



1

2 g22 + g12



g22 g11

=

√



2

1  g22 √ g11 g22 + g12 g11

Surfaces in Three-Dimensional Euclidean Space ■ 175

and 

1 g22 1   =   √ g11 √ g22 g11 2 g22 + g12 2 g11 + g12 g11 g22 1  = √  g11 √ 2 g11 g22 + g12 g22

ξ11 =

as the principal direction corresponding to κ1 . Similarly we may choose ξ22 =

√



1

2 g22 − g12



g22 g11

=

√



2

1 g22 g11

√

g11 g22 − g12



and 

1 g22 1  =−    √ g11 √ g22 g11 2 g22 − g12 2 g11 − g12 g11 g22 1 = −  √  g11 √ 2 g11 g22 − g12 g22

ξ21 = −

as the principal direction corresponding to κ2 ; we observe that since g11 g22 > 2 2 g12 , g11 , g22 > 0 and g12 ≥ 0 for any surface, the expressions above are well defined and real. (b) We may represent the principal curvature along a given curve γ on the hyperbolic paraboloid as a curve γ ∗ on the ruled surface generated by γ and the principal direction corresponding to the principal curvature as follows. If γ is given by (u1 (t), u2 (t)), and κµ (t) = κ(ui (t)) and d⃗µ (t) = d⃗µ (ui (t)) (µ = 1, 2) denote the principal curvature and corresponding principal direction, then we write u ∗1 = t and consider the ruled surfaces RS µ (µ = 1, 2) given by the parametric representations ⃗x ∗(µ) (u ∗i ) = ⃗x((u1 (u ∗1 ), u2 (u ∗1 ))) + u ∗2 d⃗µ (u1 (u ∗1 ), u2 (u ∗1 )) for µ = 1, 2, (2.93) and represent κµ (ui (t)) as a curve γµ∗ on RS µ by substituting u ∗2(µ) = u ∗2 (t) = κµ (ui (t)) in (2.93). Thus each of the curves γµ∗ for µ = 1, 2 is given by a parametric representation ⃗x ∗(µ) (t) = ⃗x ∗(µ) (u ∗i (t)) = ⃗x(u1 (u1 (u ∗1 ), u2 (u ∗2 ))) + κµ (u1 (t), u2 (t))d⃗µ (u1 (u ∗1 ), u2 (u ∗1 )). (2.94)

176 ■ Differential Geometry and Its Visualization

Representation of the first and second principal curvature along a curve on a hyperbolic paraboloid

Figure 2.55

Representation of the first and second principal curvature on the corresponding ruled surfaces

Figure 2.56

Figures 2.55 and 2.56 show the representations of the first and second principal curvature along the curve γ given by u1 (t) = t and u2 (t) = t. (c) Figure 2.57 shows the representations of the Gaussian and mean curvature along γ as curves on the ruled surface generated by γ and the surface normal vectors of the hyperbolic paraboloid along γ. The next result is useful to find the normal, principal, Gaussian and mean curvature of a surfaces that are given by an equation. Theorem 2.5.28. Let S be a surface which is given by an equation F (x1 , x2 , x3 ) = 0. (a) Then the values of the principal curvature are κi = −

λi for i = 1, 2, F12 + F22 + F32

(2.95)

Surfaces in Three-Dimensional Euclidean Space ■ 177

Representation of the Gaussian and mean curvature of a curve on a hyperbolic paraboloid

Figure 2.57

where the values λi are the solutions of the quadratic equation for λ 



F11 − λ F12 F13 F1  F F22 − λ F23 F2    21 det   = 0.  F31 F32 F33 − λ F3  F1 F2 F3 0

(2.96)

(b) A direction given by the unit vector d⃗ = {ξ 1 , ξ 2 , ξ 3 } is a principal direction if and only if   F1 ξ k F1k ξ 1   det F2 ξ k F2k ξ 2  = 0 and ξ k Fk = 0. (2.97) k 3 F3 ξ F3k ξ

(c) Let A, B and C denote the coefficients of λ2 , λ and the constant term in the quadratic equation (2.96). Then the Gaussian and mean curvature are given by K=

C −B 1 1 and H = . · · A F12 + F22 + F32 2A 2 F1 + F22 + F32

(2.98)

Proof. The normal curvature of a curve γ with a parametric representation ⃗x(s) on ⃗ (s), and ⃗x˙ (s) • N ⃗ (s) = 0 implies ⃗x¨(s) • N ⃗ (s) + a surface is given by κn (s) = ⃗x¨(s) • N ˙ ⃗ (s) = 0, hence κn (s) = −⃗x˙ (s) • N ⃗ (s). If the surface S is given by an equation ⃗x˙ (s) • N 1 2 3 F (x , x , x ) = 0, then the gradient of F and the surface normal vector of S at s are linearly dependent, that is, gradF (s) ⃗ (s). = ±N ∥gradF (s)∥

178 ■ Differential Geometry and Its Visualization

This implies 

and



gradF (s) ⃗˙ (s) = d ±N ds ∥gradF (s)∥   1 ∂gradF (s) k 1 d = x˙ (s) , + gradF (s) ∂xk ∥gradF (s)∥ ds ∥gradF (s)∥ ⃗˙ (s) = ± ⃗x˙ (s) • N

∂gradF (s) 1 . ⃗x˙ (s) • x˙ k (s) ∥gradF (s)∥ ∂xk

(2.99)

(a) First we show (2.95). ⃗ = 1 and gradF • Let d⃗ = {ξ 1 , ξ 2 , ξ 3 } be the direction of a tangent, that is, ∥d∥ k ⃗ d = ξ Fk = 0. To find the principal directions, by (2.99), we have to find the extreme values of the function G(ξ 1 , ξ 2 , ξ 3 ) = ξ k ξ i Fik on the set 



M = {ξ 1 , ξ 2 , ξ 3 } : ξ k Fk = 0 and ∥{ξ 1 , ξ 2 , ξ 3 }∥ = 1 . Using Lagrange’s multipliers, we obtain ∂(ξ ξ Fik ) ∂G(ξ ) = =λ ∂ξ m ∂ξ m j

that is,

i k

∂

 3

i 2 i=1 (ξ )

∂ξ m



+µ



µ Fm for m = 1, 2, 3. 2 and sum to obtain

ξ k Fmk = λξ m +

We multiply (2.100) by ξ m

ξ m ξ k Fmk = λ

3 

m=1

(ξ m )2 +



∂ ξ i Fi , ∂ξ m (2.100)

µ m ξ Fm = λ. 2

Putting t = µ/2, we can rewrite the last equation as 

 

F11 − λ F12 F13 F1 ξ1  F  2 F22 − λ F23 F2    ξ  21    3  = 0,  F31 F32 F33 − λ F3  ξ  F1 F2 F3 0 t

(2.101)

and {ξ 1 , ξ 2 , ξ 3 , t} is a solution of (2.101) if and only if (2.96) holds; (2.96) is a quadratic equation for λ and, by (2.99), the values of the principal curvature are as stated in (2.95). (b) Now we show (2.97). We eliminate λ and t in (2.101). First, we multiply the first equation in (2.101) by F2 and the second equation in (2.101) by F1 and obtain ξ k F1k F2 − λξ 1 F2 + tF1 F2 = 0 and ξ k F2k F1 − λξ 2 F1 + tF1 F2 = 0;

Surfaces in Three-Dimensional Euclidean Space ■ 179

subtraction of the second from the first equation now yields ξ k (F1k F2 − F2k F1 ) − λ(ξ 1 F2 − ξ 2 F1 ) = 0.

(2.102)

Similarly, multiplication of the first and second equations in (2.101) by ξ 2 and ξ 1 , respectively, followed by subtraction, yields ξ 1 ξ k F2k − ξ 2 ξ k F1k + t(ξ 1 F2 − ξ 2 F1 ) = 0.

(2.103)

Now we multiply the third equation in (2.101) by ξ 1 F2 − ξ 2 F1 and substitute (2.102) in (2.103) to obtain (ξ 1 F2 − ξ 2 F1 )ξ k F3k + ξ k (F2k F1 − F1k F2 ) + (ξ 2 ξ k F1k − ξ 1 ξ k F2k ) = ξ 1 (ξ k F3k F2 − ξ k F2k F3 ) − ξ 2 (ξ k F3k F1 − ξ k F1k F3 ) + ξ 3 (ξ k F2k F1 − ξ k F12 F2 ) 



ξ 1 ξ k F1k F1  2  = − det ξ ξ k F2k F2  = 0. 3 k ξ ξ F3k F3

This and the last equation in (2.101) are equivalent to (2.97). (c) Finally, we prove the formulae for K and H. Writing (2.96) as Aλ2 + Bλ + C = 0, we obtain from Vieta’s rule λ1 λ2 =

B C and λ1 + λ2 = − , A A

hence K = κ1 κ2 =

1 1 C C · = · 2 2 A ∥{F1 , F2 , F3 }∥ A F1 + F22 + F32

and H=

κ1 + κ2 B 1 =− · . 2 2A F12 + F22 + F32

Visualization 2.5.29 (The umbilical points on ellipsoids). The umbilical points of an ellipsoid defined by the equation (x1 )2 (x2 )2 (x3 )2 + 2 + 2 = 1, a2 b c where a > b > c > 0 are constants, are given by X=

 



±a



a2 − b 2 , 0, ±c a2 − c 2





b2 − c2  . a2 − c 2 

(2.104)

180 ■ Differential Geometry and Its Visualization

Proof. We apply Theorem 2.5.28 to find the umbilical points of the ellipsoid. We have 2x1 2x2 2x3 F1 (xi ) = 2 , F2 (xi ) = 2 , F3 (xi ) = 2 , a b c 2 2 2 F11 (xi ) = 2 , F22 (xi ) = 2 , F33 (xi ) = 2 a b c and Fjk (xi ) = 0 for j ̸= k.

A direction d⃗ = {ξ 1 , ξ 2 , ξ 3 } is a principal direction by (2.97) in Theorem 2.5.28 if and only if  1  2x 2ξ 1 ξ1  2  a2  a22  2x  ξ 1 x1 ξ 2 x2 ξ 3 x3 2ξ  D = det  = 0 for + 2 + 3 = 0. ξ2  2 a2 b b b23  b3   2x  2ξ ξ3 c2 c2 We have 



x1 ξ 1 ξ1  2  a2 a22  x  ξ  D = 4 det  ξ2  2 2  b 3 b3  x  ξ ξ3 2 2 c  c       1 2  3  1 x 1 x 1 x 1 1 1 =4 ξ2ξ3 2 − 2 − 2 ξ1ξ3 − + 2 ξ1ξ2 − a2 b c b a2 c 2 c a2 b 2 =0

if and only if x1 ξ 2 ξ 3 (b2 − c2 ) + x2 ξ 1 ξ 3 (c2 − a2 ) + x3 ξ 1 ξ 2 (a2 − b2 ) = 0.

(2.105)

We saw in Remark 2.5.16 (b) and at the end of the proof of Theorem 2.5.14 that any direction can be chosen as a principal direction at an umbilical point. Therefore, the equation in (2.105) has to be satisfied at an umbilical point for all directions d⃗ = {ξ 1 , ξ 2 , ξ 3 } with ξ 1 x1 ξ 2 x2 ξ 3 x3 + 2 + 3 = 0. (2.106) a2 b b (i) First we assume x2 ̸= 0.

(i.1) If x1 = 0, then the equations in (2.105) and (2.106) reduce to (c2 − a2 )x2 ξ 1 ξ 3 + (a2 − b2 )x3 ξ 1 ξ 2 = 0 for all d⃗ = {ξ 1 , ξ 2 , ξ 3 } with

x 2 ξ 2 x3 ξ 3 + 2 = 0. (2.107) b2 c

Surfaces in Three-Dimensional Euclidean Space ■ 181

If we choose

x 3 ξ 3 b2 x2 c2 then the first equation in (2.107) is not satisfied, since ξ 1 , ξ 3 ̸= 0 and ξ 2 = −

(c2 − a2 )(x2 )2 ξ 2 − (a2 − b2 )(x3 )2

b2 3 ξ ̸= 0. c2

Thus we cannot have x1 = 0. (i.2) It can be shown as in Part (i.1) that we cannot have x3 = 0. (i.3) Thus we assume x1 , x3 ̸= 0. We choose ξ 2 = 0. Then the equation in (2.105) reduces to (c2 − a2 )x2 ξ 1 ξ 3 = 0,

which implies ξ 1 = 0 or ξ 2 = 0. If ξ 3 = 0, then the equation in (2.106) reduces to x3 ξ 3 = 0. But (ξ 1 )2 + (ξ 2 )2 + (ξ 3 )2 ) = 1 implies ξ 3 = ±1, hence x3 ξ 3 = 0 would yield x3 = 0, which is a contradiction to the assumption x3 ̸= 0. Similarly it can be shown that ξ 3 = 0 cannot hold. Thus there cannot be umbilical points with x2 ̸= 0.

(ii) Now let x2 = 0. Then we must have

(b2 − c2 )x1 ξ 2 ξ 3 + (a2 − b2 )x3 ξ 1 ξ 2 = 0

(2.108)

x 1 ξ 1 x3 ξ 3 + 2 = 0. a2 c

(2.109)

for all directions d⃗ = {ξ 2 , ξ 2 , ξ 3 } with

Since the equation in (2.108) has to hold for all directions, it follows that x1 ̸= 0, for otherwise x3 would also be equal to zero. But the point (0, 0, 0) is not on the ellipsoid. If ξ 1 , ξ 2 , ξ 3 ̸= 0, then the equation in (2.108) implies b2 − c2 x3 ξ 1 = 1 · 3, 2 2 b −a x ξ

and the equation in (2.109) implies b 2 − c2 = a2 − b 2



x3 x1

2

a2 or (b2 − c2 )c2 (x1 )2 = (a2 − b2 )a2 (x3 )2 . c2

Since a2 (x3 )2 + c2 (x1 )2 = a2 c2 , the equation in (2.110) yields (b2 − c2 )(x1 )2 c2 = (a2 − b2 )c2 (a2 − (x1 )2 )

(2.110)

182 ■ Differential Geometry and Its Visualization

Figure 2.58

The umbilical points of an ellipsoid

and

(x1 )2 (b2 − c2 + a2 − b2 ) = (a2 − b2 )a2 ,

hence

x1 = ±a



Now the first identity in (2.110) yields c x =± a 3



Finally, since

the points

b2 − c 2 a a2 − b 2



a2 − b 2 . a2 − c 2

a2 − b 2 == ±c a2 − c 2



b2 − c2 . a2 − c 2

(x1 )2 (x3 )2 a2 − b2 b2 − c 2 a2 − c2 + = + = = 1, a2 c2 a2 − c 2 a2 − c 2 a2 − c2 X=

 



±a



a2 − b 2 , 0, ±c a2 − c 2





b2 − c2  a2 − c 2 

are the umbilical points of the ellipsoid (Figure 2.58).

Remark 2.5.30. The second fundamental coefficients of a hyperboloid of one sheet with a parametric representation 



⃗x(ui ) = {a cosh u1 cos u2 , b cosh u1 sin u2 , c sinh u1 } (u1 , u2 ) ∈ D ⊂ R × (0, 2π) , where a, b, c ∈ R \ {0}, are

abc cosh u1 abc cosh3 u1 L11 (ui ) = −  i , L12 (ui ) = 0 and L22 (ui ) =  i , g(u ) g(u )

hence its Gaussian curvature is negative at every point. Consequently there are no umbilical points on hyperboloids of one sheet by Remark 2.5.18.

Surfaces in Three-Dimensional Euclidean Space ■ 183

Figure 2.59

Euler’s theorem

The next result gives the normal curvature in terms of the principal curvature. Theorem 2.5.31 (Euler). Let κ1 and κ2 denote the principal curvatures. Then the normal curvature κn of a curve on a surface and the angle ϕ between the principal direction of κ1 and the curve satisfy the relation (Figure 2.59) κn = κ1 cos2 φ + κ2 sin2 φ.

(2.111)

Proof. Let S be a surface with a parametric representation ⃗x(ui ), and γ be a curve on S with a parametric representation ⃗x(s) = ⃗x(ui (s)). The normal curvature of γ at s is given by ⃗ (s) and ⃗v1 (s) is the corresponding direction. κn (s) = ⃗v1 (s) • N

(2.112)

If d⃗µ (µ = 1, 2) are the principal directions at the point X(s), then we have ⃗v1 (s) = d⃗1 cos φ(s) + d⃗2 sin φ(s).

(2.113)

If γµ (µ = 1, 2) are curves on S, given by the parametric representations ⃗yµ (s) = ⃗yµ (ui (s)) such that d⃗yµ (s) = d⃗µ (s) cos φ(s) for µ = 1, 2, ds and if sµ denotes the arc length along γµ , then we obtain ds2 ds1 = cos φ(s) and = sin φ(s) ds ds

(2.114)

for suitable orientations of s1 and s2 . Writing d⃗µ∗ (sµ ) = d⃗µ∗ (sµ (s)) = d⃗µ (s) for µ = 1, 2,

(2.115)

184 ■ Differential Geometry and Its Visualization

we obtain κµ (sµ ) =



dd⃗µ∗ (sµ ) dsµ



⃗ (sµ ), •N

(2.116)

and it follows from (2.112), (2.113), (2.114), (2.115) and (2.116) that κn (s) =









dd⃗1 (s) dφ(s) ⃗ cos φ(s) − sin φ(s) d1 (s) ds ds +

= cos φ(s)











dφ(s) ⃗ dd⃗2 (s) ⃗ (s) sin φ(s) + cos φ(s) d2 (s) • N ds ds

ds1 dd⃗1∗ (s1 ) ds2 dd⃗2∗ (s2 ) • N⃗(s) + sin φ(s) • N⃗(s) ds ds1 ds ds2

= cos2 φ(s)κ1 (s) + sin2 φ(s)κ2 (s), ⃗ (s) = κν (s) for µ = 1, 2. since d⃗µ (s) • N

Theorem 2.5.32. Let P be a point on a surface S. (a) If ϕ denotes the angle between the principal direction corresponding to κ1 and the direction corresponding to κn , then the mean curvature H at P is given by H=

2π

κn dϕ.

0

(b) Let κn and κ⊥ n be the values of the normal curvatures at P corresponding to two orthogonal directions. Then the mean curvature H at P is given by 1 H = (κn + κ⊥ n ). 2 Proof. (a) Applying (2.111) in Euler’s Theorem 2.5.31, we obtain 1 2π

2π 0

1 κn dϕ = 2π κ1 = 2π

2π  0

2π 0



κ1 cos2 ϕ + κ2 sin2 ϕ dϕ

κ2 1 (1 + cos 2ϕ) dϕ + 2 2π

κ1 + κ2 = = H. 2

2π 0

1 (1 − cos 2ϕ) dϕ 2

(b) Let ϕ and ϕ⊥ denote the angles between the principal direction corresponding to κ1 and the directions d⃗ and d⃗ ⊥ corresponding to κn and κ⊥ n , respectively. Then it follows from (2.111) in Euler’s theorem, Theorem 2.5.31, that 2 ⊥ 2 ⊥ κn = κ1 cos2 ϕ + κ2 sin2 ϕ and κ⊥ n = κ1 cos ϕ + κ2 sin ϕ .

(2.117)

Surfaces in Three-Dimensional Euclidean Space ■ 185

Since the directions d⃗ and d⃗ ⊥ are orthogonal, we have cos2 ϕ⊥ = sin2 ϕ and sin2 ϕ⊥ = cos2 ϕ, and obtain from (2.117) κn + κ⊥ n = κ1 + κ2 = 2H.

Example 2.5.33 (An application of Euler’s theorem). We consider the cone given by a parametric representation ⃗x(ui ) =



1 1 √ u1 cos u2 , √ u1 sin u2 , u1 a a

 



(u1 , u2 ) ∈ D = (0, ∞) × (0, 2π) ,

where a ̸= 0 is a positive constant. Omitting the arguments (uk ), we obtain a (u1 )2 , g12 = 0, g22 = , a+1 a

g11 =

L11 = L12 = 0 and L22 = √

u1 . a+1

The two values of the principal curvature are given by κ1 =

L22 a = 1√ and κ2 = 0. g22 u a+1

We obtain the principal direction corresponding to κ2 = 0 from 

L11 ξ21 + L12 ξ22 = 0 · ξ21 + 0 · ξ22 = 0 L12 ξ21 + L22 ξ22 = 0 · ξ21 + L22 ξ22 = 0



, that is, ξ22 = 0.

Consequently, we may choose the principal directions corresponding to κ2 and κ1 as ⃗x1 ⃗x2 and d⃗1 = . d⃗2 = ∥⃗x1 ∥ ∥⃗x2 ∥

Now let c > 0 be given. Putting β=



a + 1√ √ √ 1 + c2 = g11 1 + c2 , a

we consider the curve γ on the cone given by u1 (s) =

√ s s and u2 (s) = c a + 1 log β β

Since gik u˙ i u˙ k = g11 (u˙ 1 )2 + g22 (u˙ 2 )2 = and

(s > 0).

(a + 1)(1 + c2 ) =1 aβ 2

√ c2 a + 1 , Lik u˙ u˙ = L22 (u˙ ) = βs i k

2 2

186 ■ Differential Geometry and Its Visualization

the normal curvature is given by √ c2 a + 1 . κn = βs We have for the angle φ between the curve γ and the principal direction given by d⃗1 cos φ = ⃗xk u˙ k • and consequently Euler’s formula yields

⃗x2 √ = u˙ 2 g22 , ∥⃗x2 ∥

√ (u˙ 2 )2 g22 L22 (a + 1)c2 s c2 a + 1 κn = cos φκ1 + sin φκ2 = = 2 √ . = g22 βs s β a+1 2

2.6

2

THE SHAPE OF A SURFACE IN THE NEIGHBOURHOOD OF A POINT

We know from Theorem 2.5.9 that the absolute value of the normal curvature of a curve at a point is equal to the curvature of the corresponding normal section. The study of how the change of direction of the tangent to the normal section affects the normal curvature leads to three different cases for the shape of the surface in the neighbourhood of a point with non–vanishing second fundamental form Lik dui duk depending on the sign of L = det(Lik ). Definition 2.6.1. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D ⊂ R2 ) and second fundamental coefficients Lik . Then a point P = X(u10 , u20 ) on S is called an elliptic point if L(u10 , u20 ) > 0, a hyperbolic point if L(u10 , u20 ) < 0, and a parabolic point if L(u10 , u20 ) = 0. Remark 2.6.2. (a) Since the first fundamental form is positive definite, the sign of κn depends on the second fundamental form only. We will see in Theorem 2.6.6: If P is an elliptic point, that is, L > 0 at P , then κn has the same sign at P for all directions of the normal sections, and consequently the centres of curvature of all normal sections are on one side of the surface. If P is a parabolic point, that is, L = 0 at P , then κn does not change sign, there is, however, one and only one direction for which κn = 0. If P is a hyperbolic point, that is, L < 0 at P , then κn changes sign. Part of the centres of curvature of the normal sections are on one side of the surface and part on the other. There are two directions for which κn = 0, and the normal curvature has the same sign in a sector bounded by these two directions. The normal curvature changes sign when the direction of the tangent moves across the direction corresponding to κn = 0. Summarizing we have L>0 L=0 L 0 are elliptic, with u2 = 0 parabolic, and with u2 < 0 hyperbolic (Figure 2.60). Proof. We have f1 (uj ) = 2u1 , f2 = 3(u2 )2 , f11 (uj ) = 2, f12 (uj ) = 0 and f22 (uj ) = 6u2 , hence by (2.49) in Example 2.4.5 L11 (uj ) =  L22 (uj ) = 

2 1+

4(u1 )2 2

+

9(u2 )4

, L12 (uj ) = 0,

6u 12u2 j and L(u ) = . 1 + 4(u1 )2 + 9(u2 )4 1 + 4(u1 )2 + 9(u2 )4

188 ■ Differential Geometry and Its Visualization

Figure 2.61

points

A torus with elliptic (outside), parabolic (dashed) and hyperbolic (inside)

Visualization 2.6.4. (a) Every point of an ellipsoid with a parametric representation ⃗x(ui ) = {a cos u1 cos u2 , b cos u1 sin u2 , c sin u1 }





(u1 , u2 ) ∈ (−π/2, π/2) × (0, 2π) ,

where a, b and c are distinct positive constants, is an elliptic point. (b) Every point of an elliptic cone is a parabolic point. (c) Every point of the hyperbolic paraboloid of Visualization 2.5.27 is a hyperbolic point. (d) We consider the torus with a parametric representation ⃗x(ui ) = {(r0 +r1 cos u1 ) cos u2 , (r0 +r1 cos u1 ) sin u2 , r1 sin u1 }



where r0 and r1 are constants with 0 < r1 < r0 . Then all points with u1 ∈ (0, π/2) ∪ (3π/2, 2π) are elliptic points, all points with u1 = π/2 or u1 = 3π/2 are parabolic points, and all points with u1 ∈ (π/2, 3π/2) are hyperbolic points (Figure 2.61). Proof. (a) It follows that ⃗x1 (ui ) = {−a sin u1 cos u2 , −b sin u1 sin u2 , c cos u1 }, ⃗x2 (ui ) = {−a cos u1 sin u2 , b cos u1 cos u2 , 0},

⃗x11 (ui ) = {−a cos u1 cos u2 , −b cos u1 sin u2 , −c sin u1 }, ⃗x12 (ui ) = {a sin u1 sin u2 , −b sin u1 cos u2 , 0},

⃗x22 (ui ) = {−a cos u1 cos u2 , −b cos u1 sin u2 , 0},  −a cos u1 cos u2 1  i L11 (u ) = √  −b cos u1 sin u2 g  −c sin u1



(u1 , u2 ) ∈ D = (0, 2π)2 ,



−a sin u1 cos u2 −a sin u2   −b sin u1 sin u2 b cos u2  cos u1   c cos u1 0

Surfaces in Three-Dimensional Euclidean Space ■ 189

abc abc = √ cos u1 (− sin u1 (− sin u1 ) − cos u1 (− cos u1 )) = √ cos u1 , g g

  sin u2  abc sin u cos u  L12 (ui ) = − cos u2 √  g  0 1

=

abc sin u1 cos2 u1 (sin u2 cos u2 − sin u2 cos u2 ) = 0, √ g

 − cos u2 abc cos u  i L22 (u ) =  − sin u1 √  g  0 2

=



− sin u1 cos u2 − sin u2   − sin u1 sin u2 cos u2   cos u1 0 

1

1



− sin u1 cos u2 − sin u2   − sin u1 sin u2 cos u2   1 cos u 0 

abc cos3 u1 abc cos3 u1 (cos2 u2 + sin2 u2 ) = √ √ g g

and so L(ui ) = L11 (ui )L22 (ui )−L212 (ui ) =

a2 b2 c2 cos4 u1 > 0 for all u1 ∈ (−π/2, π/2). g

(b) Let a1 , a2 > 0 and the elliptic cone be given by a parametric representation ⃗x(u ) = i



u1 u1 √ cos u2 , √ sin u2 , u1 a1 a2

Then we have ⃗x1 (u ) = i

⃗x2 (u ) = i

 







(u1 , u2 ) ∈ D ⊂ (0, ∞) × (0, 2π) .



1 1 √ cos u2 , √ sin u2 , 1 , a1 a2



u1 u1 − √ sin u2 , √ cos u2 , 0 , a1 a2

cos2 u2 sin2 u2 + + 1, g12 (ui ) = g11 (u ) = a1 a2 (u1 )2 (u1 )2 g22 (ui ) = sin2 u2 + cos2 u2 , a1 a2 i





1 1 − u1 sin u2 cos u2 , a2 a1

 √  √ 1 2 2 ⃗ (ui ) =  − a cos u , − a sin u , 1 , N 1 2 a1 cos2 u2 + a2 sin2 u2 + 1

⃗x11 (u ) = ⃗0, ⃗x12 (u ) = i

⃗x22 (u ) = i

i







1 1 − √ sin u2 , √ cos u2 , 0 , a1 a2 

u1 u1 − √ cos u2 , − √ sin u2 , 0 , a1 a2

u1 L11 (ui ) = L12 (ui ) = 0 and L22 (ui ) =  , a1 cos2 u2 + a2 sin2 u2 + 1

hence L(ui ) = 0 on D.

190 ■ Differential Geometry and Its Visualization

(c) We have by Visualization 2.5.27, L(ui ) = −1/g(ui ) < 0 for all u1 and u2 .

(d) Since r′ (u1 ) = −r1 sin u1 , h′ (u1 ) = r1 cos u1 , r′ (u1 (u1 ))2 + (h′ (u1 ))2 = r12 and r′′ (u1 ) = −r1 cos u1 , it follows from (2.76) in Visualization 2.5.19 as in the proof of (2.78) that cos u1 r′′ (u1 )r12 K(u1 ) = − , (2.118) = 4 r(u1 )r1 r1 (r0 + r1 cos u1 ) and the conclusion is an immediate consequence of (2.118). The following result gives a geometric interpretation of the second fundamental form. Theorem 2.6.5. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D) and X0 be a point on S with position vector ⃗x(uk0 ). If the second fundamental form of S does not vanish identically at (u10 , u20 ), then – up to terms in ((du1 )2 + (du2 )2 )1/2 of order higher than two – it is equal to twice the distance (including the sign) of a point Q ∈ S with position vector ⃗x(uk0 + duk ) from the tangent plane of S at X0 . Proof. We obtain for the point Q by Taylor’s formula

  1 ⃗x(ui0 + dui ) = ⃗x(ui0 ) + duk ⃗xk (ui0 ) + duk duj ⃗xkj (ui0 ) + ⃗o (du1 )2 + (du2 )2 2

and the distance d of Q from the tangent plane to S at the point X0 is given by 



⃗ (ui ) d = ⃗x(ui0 + dui ) − ⃗x(ui0 ) • N 0

  1 ⃗ (ui ) + o (du1 )2 + (du2 )2 = duk duj ⃗xkj (ui0 ) • N 0 2   1 = Lkj (ui0 )duk duj + o (du1 )2 + (du2 )2 . 2

The sign of the distance depends on which side of the tangent plane the point Q is, hence depends on the orientation. The next result is an immediate consequence of Theorem 2.6.5. Theorem 2.6.6. Let P be a point of a surface S and T (P ) denote the tangent plane of S at P . If P is an elliptic or parabolic point, then a sufficiently small neighbourhood of P in T (P ) lies completely on one side of S (Figure 2.62). If P is a hyperbolic point, then every neighbourhood of P in T (P ) contains points on either side of S (Figure 2.63). Tangent planes at elliptic, parabolic and hyperbolic points of a torus are shown in Figure 2.64. Since g is always positive and the Gaussian curvature is given by K = L/g, the sign of K is equal to that of L. Thus the following result holds.

Surfaces in Three-Dimensional Euclidean Space ■ 191

Figure 2.62

Tangent planes of an ellipsoid

Figure 2.63

Tangent planes of a hyperbolid of one sheet

Theorem 2.6.7. The Gaussian curvature at a point P of a surface is positive, if P is an elliptic point, negative, if P is a hyperbolic point and it vanishes, if P is a parabolic point. Remark 2.6.8. We already know from (2.73) in Remark 2.5.16 (b) that L = c2 g in an umbilical point, for some constant c. Since g > 0 and c2 ≥ 0, it follows that there are only elliptic umbilical points for c ̸= 0, and parabolic umbilical points for c = 0. A parabolic umbilical point is called planar or flat point. Example 2.6.9. (a) Every point of a sphere is an elliptic umbilical point. (b) Every point in a plane is a planar point. Proof. (a) Every point of a sphere of radius r > 0 is an umbilical point by Example 2.5.25. Since L11 (ui ) = r, L12 (ui ) = 0 and L22 (ui ) = r cos2 u2 , we have L(ui ) = r2 cos2 u2 > 0 for all u2 ∈ (−π/2, π/2).

192 ■ Differential Geometry and Its Visualization

Figure 2.64

Tangent planes of a torus at elliptic, parabolic and hyperbolic points

Figure 2.65

A paraboloid of revolution and the surface of Visualization 2.6.10

(b) Let the plane be given by a parametric representation ⃗x(ui ) = {u1 , u2 , 0} ((u , u2 ) ∈ D ⊂ R 2 ). Then it is easy to see that g11 (ui ) = g22 (ui ) = 1, g12 (ui ) = 0, L11 (ui ) = L12 (ui ) = L22 (ui ) = 0 for all (u1 , u2 ) ∈ D, and consequently every point of the plane is a flat point. 1

Visualization 2.6.10. (A flat point on a biquadratic parabola of revolution) Let RS be the surface of revolution generated by the rotation about the x3 –axis of the curve with the equation x3 = (x1 )4 in the x1 x3 –plane (Figure 2.65). Then RS is given by the parametric representation 



⃗x(ui ) = u1 , u2 , (u1 )2 + (u2 )2

2  



(u1 , u2 ) ∈ R2 .

Surfaces in Three-Dimensional Euclidean Space ■ 193

Figure 2.66

A flat point on the surface of Visualization 2.6.10

Then 

g11 (ui ) = 1 + 16(u1 )2 (u1 )2 + (u2 )2 

g22 (ui ) = 1 + 16(u2 )2 (u1 )2 + (u2 )2 8u1 u2 L12 (ui ) = √ , g

2 2



, g12 (ui ) = 16u1 u2 (u1 )2 + (u2 )2 ; L11 (ui ) =

12(u1 )2 + 4(u2 )2 , √ g

L22 (ui ) =

12(u2 )2 + 4(u1 )2 , √ g

2

,

and so 

1 2

2 2

g(u ) = 1 + 16 (u ) + (u ) i

3



2

48 (u1 )2 + (u2 )2 . and L(u ) = g i

Thus Lik (u1 , u2 ) = 0 for all i, k = 1, 2 if and only if (u1 , u2 ) = (0, 0). Therefore, the surface has a plane point in the origin (Figure 2.66), and all other points are elliptic points. Visualization 2.6.11 (The monkey saddle). Let S be the so–called monkey saddle with a parametric representation (Figure 2.67) 

⃗x(ui ) = u1 , u2 , u2 (u2 − Then

√

3u1 )(u2 +

√

 

3u1 )



(u1 , u2 ) ∈ R2 .





g11 (u1 , u2 ) = 1 + 36(u1 u2 )2 , g12 (u1 , u2 ) = 18u1 u2 (u2 )2 − (u1 )2 , 

g22 (u1 , u2 ) = 1 + 9 (u1 )2 − (u2 )2

2



, g(u1 , u2 ) = 1 + 9 (u1 )2 + (u2 )2

6u2 6u1 L11 (u1 , u2 ) = − √ , L12 (u1 , u2 ) = − √ , g g

2

,

194 ■ Differential Geometry and Its Visualization

Figure 2.67

The monkey saddle

Figure 2.68

The monkey saddle and its intersection with the x1 x2 –plane 



6u2 36 (u1 )2 + (u2 )2 L22 (u , u ) = √ and L(u1 , u2 ) = − . g g 1

2

The second fundamental form vanishes identically at (0, 0). Thus the monkey saddle has a planar point in the origin. Since ⃗x1 (0, 0) = ⃗e 1 and ⃗x2 (0, 0) = ⃗e 2 , the x1 x2 – plane is the tangent plane of the monkey saddle at the origin. It intersects the monkey saddle in three straight lines that are given by the following equations in the x1 x2 – plane (Figure 2.68) √ √ x2 = 0, x2 = 3x1 and x3 = − 3x1 . (2.119) Finally we consider a normal section at the origin, different from the three lines of intersection given in (2.119). If the tangent to the curve has direction ⃗v = {ξ 1 , ξ 2 }, that is, if ′ ⃗x = ξ 1⃗x1 (0, 0) + ξ 2⃗x2 (0, 0) = ξ 1⃗e 1 + ξ 2⃗e 2 ,

Surfaces in Three-Dimensional Euclidean Space ■ 195

Figure 2.69

The monkey saddle and a normal section at (u1 , u2 ) = (0, 0)

then a parametric representation of the plane of the corresponding normal section is given by −→

′ ⃗ (0, 0) = λξ 1⃗e 1 + λξ 2⃗e 2 + µ⃗e 3 (λ, µ ∈ R). OX = λ⃗x (0, 0) + µN

The normal section is given by the following equation in the of the normal √ plane 3 2 2 2 1 2 2 2 1 section µ = λ (ξ ((ξ ) − 3(ξ ) ))). If ξ = ̸ 0 and ξ = ̸ ± 3ξ , then putting a =  2 2  2 1 2 3 ξ (ξ ) − 3(ξ ) ) ̸= 0, we obtain the cubic parabola µ = aλ (Figure 2.69).

2.7

DUPIN’S INDICATRIX

In this section, we explain the geometric meaning of elliptic, parabolic and hyperbolic points, of the principal directions and the directions for which κn = 0, the so–called asymptotic directions. This leads to Dupin’s indicatrix. Meusnier and Euler’s theorems (Theorems 2.5.7 and 2.5.31) give complete information on the curvature of any curve on a surface through a given point. Definition 2.7.1. Let P be a point on a surface S, and we assume that P is not an umbilical point. We introduce a Cartesian coordinate system in the tangent plane T (P ) of S at P with its origin at P and its x1 – and x2 –axes along the principal directions corresponding to the principal curvatures κ1 and κ2 . If φ denotes the angle between the positive x1 –axis and direction that corresponds to the normal curvature κn , then the point set 

DI = X = (x1 , x2 ) ∈ T (P ) : x1 =



1 cos φ, x2 = |κn |





1 sin φ (φ ∈ [0, 2π]) |κn |

is called Dupin’s indicatrix of the surface S at the point P .

196 ■ Differential Geometry and Its Visualization

Figure 2.70

Dupin’s indicatrix at an elliptic point

Remark 2.7.2. (a) If we multiply equation (2.111) in Euler’s theorem, Theorem 2.5.31, by 1/|κn |, then we obtain, since 1

x = that



1 cos φ and x2 = |κn |

κ1 (x1 )2 + κ2 (x2 )2 =



1 sin φ, |κn |

κ1 κ2 cos2 φ + sin2 φ = ±1. |κn | |κn |

If P is an elliptic point, then the principal curvatures κ1 and κ2 have the same sign, and consequently Dupin’s indicatrix is an ellipse (Figure 2.70). If P is a hyperbolic point, then κ1 and κ2 have different signs, and consequently Dupin’s indicatrix consists of two hyperbolas with a joint pair of asymptotes. The so– called asymptotic directions on the surface, that is, the directions for which κn = 0, coincide with the directions of the asymptotes of the hyperbolas. This explains the terms elliptic and hyperbolic points and asymptotic directions (right in Figure 2.71). If P is a parabolic point, then κ1 = 0, say, and the equation for Dupin’s indicatrix reduces to 1 (x2 )2 = ± , κ2 which is the equation for a pair of straight lines parallel to the x1 –axis, which intersect the x2 –axis at the points ± √1 (left in Figure 2.71). |κ2 |

(b) The principal directions bisect the angle between the asymptotic directions. The directions of equal normal curvatures are symmetric with respect to the principal directions. (c) Dupin’s indicatrix at a point P of a surface S is closely related to the lines of intersection of S with planes parallel to the tangent plane T (P ) of S at P . If we choose two planes parallel to T (P ) at a distance ε on either side of S, then an equation of the

Surfaces in Three-Dimensional Euclidean Space ■ 197

Dupin’s indicatrix at a parabolic point (left) and at a hyperbolic point with asymptotic directions (right)

Figure 2.71

line of intersection – up to terms of higher than second order in ((du1 )2 + (du2 )2 )1/2 – is given by Theorem 2.6.5 1 Lik dui duk = ±ε. 2 Choosing the parameters such that for each parameter line its tangents at each point coincide with one principal direction, we obtain L11 (du1 )2 + L22 (du2 )2 = ±2ε,

as we shall see in Theorem 2.8.6. Finally κ1 =

L11 L22 and κ2 = g11 g22

together imply κ1 g11 (du1 )2 + κ2 g22 (du2 )2 = ±2ε (Figure 2.72).

2.8

LINES OF CURVATURE AND ASYMPTOTIC LINES

In view of the results of the previous sections, it is interesting to find those curves on surfaces the directions of which coincide with the principal or asymptotic directions. Definition 2.8.1. (a) A curve on a surface such that the direction of its tangent at any point coincides with one principal direction is called a line of curvature. (b) Asymptotic directions are directions for which the normal curvature vanishes. A curve on a surface such that the direction of its tangent at any point coincides with an asymptotic direction is called asymptotic line. The next result gives the differential equations for lines of curvature and asymptotic lines.

198 ■ Differential Geometry and Its Visualization

Figure 2.72

Visualization of Remark 2.7.2

Theorem 2.8.2. (a) There are two lines of curvature through every point of a surface that is not an umbilical point (Theorem 2.5.14). They are the solutions of the differential equations   L1k duk g1k duk det = 0. (2.120) L2k duk g2k duk (b) Since

′

′

Lik ui uk κn = gik ui ′ uk ′

for the normal curvature, a direction given by {ξ 1 , ξ 2 } is an asymptotic direction, if and only if Lik ξ i ξ k = 0. Thus the differential equation for an asymptotic line is Lik ui uk = 0. ′

′

(2.121)

The existence of a (real) asymptotic line depends on the shape of the surface. If a point P on a surface S is an elliptic point, then L > 0 in some neighbourhood of P and the second fundamental Lik dui duk is positive definite. Consequently there exists no (real) asymptotic line through P . If P is a hyperbolic point, then L < 0 in some neighbourhood of P and there are two asymptotic directions at P . Consequently there are two asymptotic lines through P in a neighbourhood of P . If P is a parabolic point, then L = 0 at P and there is one and only one asymptotic direction at P .

Surfaces in Three-Dimensional Euclidean Space ■ 199

Figure 2.73

Lines of curvature on a helicoid

Visualization 2.8.3 (Lines of curvature on a helicoid). The lines of curvature on the helicoid of Visualization 2.5.3 with a parametric representation   ⃗x(ui ) = {u2 cos u1 , u2 sin u1 , cu1 } (u1 , u2 ) ∈ D ,

where c ̸= 0 is a constant, are given by (Figure 2.73) ±(u1 + d) = sinh−1

u2 , or u2 (u1 ) = ±c sinh (u1 + d), c

(2.122)

where d is a constant. Proof. We obtain ⃗x1 (ui ) = {−u2 sin u1 , u2 cos u1 , c}, ⃗x2 (ui ) = {cos u1 , sin u1 , 0}, ⃗x11 (ui ) = {−u2 cos u1 , −u2 sin u1 , 0}, ⃗x12 (ui ) = {− sin u1 , cos u1 , 0}, ⃗x22 (ui ) = ⃗0, g11 (ui ) = (u2 )2 + c2 , g12 (ui ) = 0, g22 (ui ) = 1, g(ui ) = (u2 )2 + c2 ,   cos u1 −u2 sin u1 cos u1      g(ui )L11 (ui ) = −u2  sin u1 u2 cos u1 sin u1  = 0,    0 c 0     − sin u1 −u2 sin u1 cos u1        g(ui )L12 (ui ) = − cos u1 u2 cos u1 sin u1  = g(ui )c    0 c 0  

200 ■ Differential Geometry and Its Visualization

Figure 2.74

A general cone generated by an astroid

and L22 (ui ) = 0. Thus the differential equation (2.120) for the lines of curvature on the helicoid reduce to  2   2 c du2 − c (u2 )2 + c2 du1 = 0

and since c ̸= 0, to

with the solutions in (2.122).



du2 = ±du1 ((u2 )2 + c2 )

Visualization 2.8.4 (Lines of curvature on a general cone). Let γ be a curve of non–vanishing curvature in the x1 x2 –plane, and ⃗y (s) (s ∈ I) be a parametric representation of γ where s denotes the arc length along γ. A general cone GC is generated by the straight lines through the point (0, 0, 1) and the points of γ (Figure 2.74). Writing u1 = s, we obtain the following parametric representation for GC ⃗x(ui ) = u2 ⃗y (u1 ) + (1 − u2 )⃗e 3 The lines of curvature on GC are given by





(u1 , u2 ) ∈ I × (0, ∞) .

c2 u2 (u1 ) =  (⃗y (u1 ))2 + 1

(2.123)

Surfaces in Three-Dimensional Euclidean Space ■ 201

for some constant c ̸= 0 (Figure 2.75).

Figure 2.75

The second family of lines of curvature on the general cone of Figure 2.74

Proof. We obtain ⃗x1 (ui ) = u2 ⃗y ′ (u1 ), ⃗x2 (ui ) = ⃗y (u1 ) − ⃗e 3 ,

⃗x11 (ui ) = u2 ⃗y ′′ (u1 ), ⃗x12 (ui ) = ⃗y ′ (u1 ), ⃗x22 (ui ) = ⃗0, ⃗x1 (ui ) × ⃗x2 (ui ) = u2 ⃗y ′ (u1 ) × (⃗y (u1 ) − ⃗e 3 ) ̸= ⃗0,

g11 (ui ) = (u2 )2 , ∥⃗y ′ (u1 )∥ = 1, since u1 is the arc length along γ, 

2

g12 (ui ) = u2 ⃗y ′ (u1 ) • ⃗y (u1 ), g22 (ui ) = ⃗y (u1 ) 

+ 1, 

∥⃗x1 (ui ) × ⃗x2 (ui )∥L11 (ui ) = (u2 )2 ⃗y ′′ (u1 ) • ⃗y ′ (u1 ) × (⃗y (u1 ) − ⃗e 3 ) , L12 (ui ) = L22 (ui ) = 0.

Since ∥⃗y ′ ∥ = 1 implies ⃗y ′ ⊥ ⃗y ′′ in the x1 x2 –plane, we conclude that ⃗y ′ × ⃗y ′′ is in the direction of ⃗e 3 and ⃗y ′ − ⃗e 3 is not orthogonal to ⃗e 3 . Thus we have L11 (ui ) ̸= 0 for u2 > 0. The differential equation for the lines of curvature (2.120) on GC reduces to 



L11 (ui )du1 g12 (ui )du1 + g22 (u2 )du2 = 0. It follows from L11 (ui ) ̸= 0 that the u2 –lines are lines of curvature and the second family of lines of curvature is given by 



u2 ⃗y ′ (u1 ) • ⃗y (u1 )du1 = − (⃗y (u1 ))2 + 1 du2 , that is, log u2 = −



  1 ⃗y ′ (u1 ) • ⃗y (u1 ) 1 du = − log (⃗y (u1 ))2 + 1 + log k 1 2 (⃗y (u )) + 1 2

for some constant k > 0, hence (2.123) follows.

202 ■ Differential Geometry and Its Visualization

Visualization 2.8.5 (Asymptotic lines on ruled surfaces). We consider the ruled surface RulS with a parametric representation ⃗x(ui ) = ⃗y (u1 ) + u2⃗z(u1 )



(u1 , u2 ) ∈ D



with ∥⃗z(u1 )∥ = 1 and L11 (ui ) ̸= 0. Omitting the parameter u1 , we obtain ⃗x1 = ⃗y ′ + u2⃗z ′ , ⃗x2 = ⃗z, ⃗x11 = ⃗y ′′ + u2⃗z ′′ , ⃗x12 = ⃗z ′ , ⃗x22 = ⃗0,   1 ⃗x1 × ⃗x2 L11 (ui ) = ⃗x11 •  i =  i (⃗y ′′ + u2⃗z ′′ ) • (⃗y ′ + u2⃗z ′ ) × ⃗z , g(u ) g(u ) 1 ⃗x1 × ⃗x2 L12 (ui ) = ⃗x12 •  i =  i ⃗z ′ • (⃗y ′ × ⃗z) and L22 (ui ) = 0. g(u ) g(u )

Thus the differential equation (2.121) for the asymptotic lines on RulS reduces to 



du1 L11 (ui )du1 + 2L12 (ui )du2 = 0, and the u2 –lines are one family of asymptotic lines. Further solutions are obtained from   du2 2⃗z ′ • (⃗y ′ × ⃗z) 1 + (⃗y ′′ + u2⃗z ′′ ) • (⃗y ′ + u2⃗z ′ ) × ⃗z = 0. (2.124) du If ⃗z ′ • (⃗y ′ × ⃗z) = 0, then there are no more solutions. We assume ⃗z ′ • (⃗y ′ × ⃗z) ̸= 0 and put ⃗y ′′ • (⃗y ′ × ⃗z) , a(u1 ) ⃗y ′′ (⃗z ′ × ⃗z) + ⃗z ′′ • (⃗y ′ × ⃗z) g(u1 ) = − a(u1 )

a(u1 ) = 2⃗z ′ • (⃗y ′ × ⃗z), f (u1 ) = −

and h(u1 ) = −

⃗z ′′ • (⃗z ′ × ⃗z) . a(u1 )

Then we see that (2.124) is a differential equation of Riccati type du2 = f (u1 ) + u2 g(u1 ) + (u2 )2 h(u1 ), du1

(2.125)

and we obtain one more family of asymptotic lines from the solutions of (2.125). (a) First we consider the conoid of Visualization 2.5.3 (c) with ⃗y (u1 ) = {0, 0, φ(u1 )}, φ′ (u1 ) ̸= 0 and z(u1 ) = {cos u1 , sin u1 , 0}.

Surfaces in Three-Dimensional Euclidean Space ■ 203

Figure 2.76

Asymptotic lines on the conoid with φ(u1 ) = log (1 + u1 )

Then we obtain ⃗y ′ (u1 ) = {0, 0, φ′ (u1 )}, ⃗y ′′ (u1 ) = {0, 0, φ′′ (uť1 )},

⃗z ′ (u1 ) = {− sin u1 , cos u1 , 0} and ⃗z ′′ (u1 ) = −⃗z(u1 ).

Therefore, we have 







⃗z ′′ (u1 ) • ⃗y ′ (u1 ) × ⃗z(u1 ) = ⃗z ′′ (u1 ) • (⃗z ′ (u1 ) × ⃗z(u1 ) 



= ⃗y ′′ (u1 ) • ⃗y ′ (u1 ) × ⃗z(u1 ) = 0,

a(u1 ) = −2φ(u1 ), f (u1 ) = 0,

g(u1 ) = −

φ′′ (u1 ) ⃗y (u1 ) ′′ • (⃗z ′ (u1 ) × ⃗z(u1 )) = and h(u1 ) = 0. a(u1 ) 2φ′ (u1 )

Hence the differential equation (2.125) reduces to u2 φ′′ (u1 ) du2 = du1 2φ′ (u1 ) with the solutions

1 log |u | = 2 2

for u2 ̸= 0, that is,

 

1 u = c exp 2 2

φ′′ (u1 ) 1 du φ′ (u1 )





φ′′ (u1 ) 1 du φ′ (u1 ) 



 1 log |φ′ (u1 )| = c |φ′ (u1 )| = c exp 2

for some constant c ̸= 0 (Figure 2.76).

(b) Now we consider the ruled surface with ⃗y (u1 ) = r{cos u1 , sin u1 , 0} and ⃗z(u1 ) = {cos u1 , 0, sin u1 }

204 ■ Differential Geometry and Its Visualization

Figure 2.77

Lines of curvature on a Moebius strip

for some constant r > 0, the so–called Moebius strip or Moebius band (Figure 2.77). Then we have ⃗y ′ (u1 ) = r{− sin u1 , cos u1 , 0}, ⃗y ′′ (u1 ) = −⃗y (u1 ), ⃗z ′ (u1 ) = r{− sin u1 , 0, cos u1 }, ⃗z ′′ (u1 ) = −⃗z(u1 ),

a(u1 ) = 2r(− sin2 u1 cos u1 − cos3 u1 ) = −2r cos u1 ̸= 0 for u1 ̸=

π 3π , , 2 2

⃗y ′′ (u1 ) • (⃗y ′ (u1 ) × ⃗z(u1 )) = −r2 sin u1 , ⃗y ′′ (u1 ) • (⃗z ′ (u1 ) × ⃗z(u1 )) = −r sin u1 ,

⃗z ′′ (u1 ) • (⃗y ′ (u1 ) × ⃗z(u1 )) = ⃗z ′′ (u1 ) • (⃗z ′ (u1 ) × ⃗z(u1 )) = 0, r 1 π 3π f (u1 ) = − tan u1 , g(u1 ) = − tan u1 and h(u1 ) = 0 for u1 ̸= , . 2 2 2 2 Now (2.125) reduces to 1 du2 = − tan u1 (r + u2 ) du1 2 with the solutions for some constant c > 0.



u2 (u1 ) = c | cos u1 | − r

The differential equations for lines of curvature and asymptotic lines reduce considerably for suitable choices of the parameters of the surface.

Surfaces in Three-Dimensional Euclidean Space ■ 205

Theorem 2.8.6. (a) The parameter lines of a surface are lines of curvature if and only if g12 = L12 = 0. (b) The parameter lines of a surface are asymptotic lines if and only if L11 = L22 = 0 and L12 ̸= 0. Proof. (a) First we prove Part (a).

(a.i) We prove the necessity of the condition g12 = L12 = 0. We assume that the parameter lines of the surface S are lines of curvature. Then it follows that g12 = 0, since the principal directions are orthogonal by Theorem 2.5.14. Furthermore, since du1 = 0 for the u2 –lines, the differential equations (2.120) for the lines of curvature reduce to 

L12 du2 0 det 2 L22 du g22 du2



= L12 g22 (du2 )2 = 0.

This implies L12 = 0, since g22 ̸= 0.

(a.ii) Now we show the sufficiency of the condition L12 = g12 = 0. We assume that L12 = g12 = 0. Then the differential equations (2.120) for the lines of curvature on S reduce to 

L11 du1 g11 du1 det L22 du2 g22 du2



= (L11 g22 − L22 g11 )du1 du2 = 0

and the parameter lines of S are lines of curvature. (b) Now we prove Part (b). (b.1) First, we show the necessity of the condition L11 = L22 = 0 and L12 ̸= 0. We assume that the parameter lines of the surface S are asymptotic lines. Since du2 = 0 for the u1 –lines, the differential equations (2.121) for the asymptotic lines reduce to L11 (du1 )2 = 0; this implies L11 = 0. Similarly the differential equations (2.121) reduce to L22 (du2 )2 = 0 for the u2 –lines, which implies L22 = 0. (b.2) Finally, we show the sufficiency of the conditions L11 = L22 = 0 and L12 ̸= 0. If L11 = L22 = 0, then the differential equations (2.121) for the asymptotic lines on S reduce to L12 du1 du2 = 0 and the parameter lines of S are asymptotic lines, since L12 ̸= 0. Theorem 2.8.7. At every point of an asymptotic line with non–vanishing curvature κ, its binormal vector and the surface normal vector coincide, or equivalently, the osculating planes of an asymptotic line coincide with the tangent planes of the surface (Figure 2.78).

206 ■ Differential Geometry and Its Visualization

Figure 2.78

Osculating and tangent planes at a point of an asymptotic line

Proof. The normal curvature of a curve is given by (2.40). Thus we have for an asymptotic line ⃗ 0 = κn = κ⃗v2 • N

⃗ . Since ⃗v3 ⊥ ⃗v1 , ⃗v2 and N ⃗ ⊥ ⃗v1 it follows and κ ̸= 0 implies that ⃗v2 is orthogonal to N ⃗ that ⃗v3 = ±N .

Visualization 2.8.8 (Lines of curvature on a hyperbolic paraboloid). We consider the hyperbolic paraboloid of Visualization 2.5.27 with a parametric representation   ⃗x(ui ) = {u1 , u2 , u1 u2 } (u1 , u2 ) ∈ R2 .

The lines of curvature on the hyperbolic paraboloid are the u2 –lines and the curves given by u1 = sinh (± sinh−1 u2 ), (2.126) where c ∈ R is a constant (Figure 2.79). Proof. The differential equations (2.120) for the lines of curvature reduce to 

L12 du2 g11 du1 + g12 du2 det L12 du1 g12 du1 + g22 du2 or equivalently



g22 du2 hence 

2



− g11 du1

2



= 0,

= 0,

du2 du1  = ± . 1 + (u2 )2 1 + (u1 )2

Surfaces in Three-Dimensional Euclidean Space ■ 207

Lines of curvature on a hyperbolic paraboloid

Figure 2.79

This implies

sinh−1 u1 = ± sinh−1 u2 + c,

where c ∈ R is a constant, or (2.126).

Remark 2.8.9. Alternatively, since L11 = L22 = 0 and L12 ̸= 0 for the hyperbolic paraboloid of Visualization 2.8.8, the asymptotic lines are the parameter lines by Theorem 2.8.6 (b). Visualization 2.8.10 (Lines of curvature on an elliptic cone). Let a1 , a2 be two positive constants. We consider the elliptic cone of Visualization 2.6.4 with a parametric representation ⃗x(u ) = i



1 1 √ u1 cos u2 , √ u1 sin u2 , u1 a1 a2







(u1 , u2 ) ∈ (0, ∞) × (0, 2π) .

Then the lines of curvature are the u1 –lines and the curves given by (Figure 2.80) k , u1 (u2 ) =  a2 cos2 u2 + a1 sin2 u2 + a1 a2

(2.127)

where k > 0 is a constant.

Proof. We have L12 = g12 = 0 if and only if a1 = a2 and consequently the parameter lines are lines of curvature by Theorem 2.8.6 (a) if and only if the cone is circular. Now let a1 ̸= a2 . Then we have L11 (uk ) = L12 (uk ) = 0 by Visualization 2.6.4 (b) and the differential equations (2.120) for the lines of curvature reduce to 



L22 (uk )du2 g11 (uk )du1 + g12 (uk )du2 = 0.

208 ■ Differential Geometry and Its Visualization

Figure 2.80

Lines of curvature on an elliptic cone

The u1 –lines are one family of lines of curvature. Furthermore, L22 (uk ) ̸= 0 implies u1 g12 (uk ) du1 dg11 (u1 ) = − = − du2 g11 (u1 ) 2g11 (u1 ) du2

and consequently

1 log u1 = − log g11 (u1 ) + c, 2

where c ∈ R is a constant. Since, by the proof of Visualization 2.6.4 (b), g11 (u1 ) =

 1  a2 cos2 u2 + a1 sin u2 + a1 a2 , a1 a2

the second family of lines of curvature is given by (2.127).

Visualization 2.8.11 (Asymptotic lines on surfaces of rotation). Let RS be a surface of rotation with a parametric representation ⃗x(ui ) = {r(u1 ) cos u2 , r(u1 ) sin u2 , h(u1 )}

with r(u1 ) > 0 and |r′ (u1 | + |h′ (u1 )| > 0 for all u1 . Then we have by Example 2.4.6 L11 (ui ) = L11 (u1 ) = and



(u1 , u2 ) ∈ D



h′′ (u1 )r′ (u1 ) − h′ (u1 )r′′ (u1 )  , L12 (ui ) = 0 (r′ (u1 ))2 + (h′ (u1 ))2

L22 (ui ) = L22 (u1 ) = 

h′ (u1 )r(u1 ) . (r′ (u1 ))2 + (h′ (u1 ))2

The differential equations for asymptotic lines (2.121) reduce to 



du2 L11 (u1 ) = ± − =± du1 L22 (u1 )

h′ (u1 )r′′ (u1 ) − h′′ (u1 )r′ (u1 ) , h′ (u1 )r(u1 )

which have real solutions for those u1 for which h′ (u1 )r′′ (u1 ) − h′′ (u1 )r′ (u1 ) ≥ 0. h′ (u1 )r(u1 )

(2.128)

Surfaces in Three-Dimensional Euclidean Space ■ 209

Figure 2.81

Asymptotic lines on a catenoid

(a) Asymptotic lines on the catenoid. Let a > 0 be a constant. We consider the catenoid with u1 and h(u1 ) = u1 for u1 ∈ R. a Since h′′ (u1 ) = 0, (2.128) reduces to r(u1 ) = a cosh

du2 =± du1



1 r′′ (u1 ) = ± for all u1 ∈ R r(u1 ) a

with the solutions u2 (u1 ) = ±

where c ∈ R is a constant (Figure 2.81).

u1 + c, a

(b) We consider the surface of rotation with

r(u1 ) = u1 and h(u1 ) = log u1 for u1 ∈ (0, ∞).

Since r′′ (u1 ) = 0, (2.128) reduces to 

du2 h′′ (u1 ) 1 = ± − = ± 1 for all u1 ∈ (0, ∞) 1 ′ 1 1 du h (u )r(u ) u with the solutions u2 (u1 ) = ± log u1 + log c, where c > 0 is a constant (Figure 2.82). (c) Now we consider the torus with r(u1 ) = r0 + r1 cos u1 and h(u1 ) = r1 sin u1 for u1 ∈ (0, 2π),

where r0 > r1 . Then (2.128) reduces to (Figure 2.83) 

r1 du2 for u1 ∈ =± − 1 1 du cos u (r0 + r1 cos u1 )





π 3π , . 2 2

210 ■ Differential Geometry and Its Visualization

Figure 2.82

Asymptotic lines on the surface of rotation with r(u1 ) = 1 and h(u1 ) =

Figure 2.83

Asymptotic lines on the part of u1 ∈ (π/2, 3π/2) of a torus

log u1

Visualization 2.8.12 (Asymptotic lines on a pseudo-sphere). The asymptotic lines on the pseudo-sphere of Visualization 2.5.19 (c) with its parametric representation defined by 1

1

1

r(u ) = exp (−u ) and h(u ) = are given by (Figure 2.84) 

u1 (t) = t and u2 (t) = ± log et + where c ∈ R is a constant of intgration.

  

1 − exp (−2u1 ) du1 for u1 > 0 

e2t − 1 + c for all t > 0,

Proof. We write u = u1 , and observe that r(u) = e−u , h′ (u) =



1 − e−2u and r′′ (u) = r(u).

(2.129)

Surfaces in Three-Dimensional Euclidean Space ■ 211

Figure 2.84

Families of asymptotic lines on the pseudo-sphere of Visualization 2.8.12

Hence the differential equations (2.121) for the asymptotic lines reduce to du2 1 1 =± ′ =± ′ , 1 du |h (u)| h (u)

since h′ (u) > 0 for all u > 0. Therefore, we have to solve the integral I=



du = h′ (u)



The substitution t = eu > 1 yields I=



√

du √ = 1 − e−2u



√

eu e2u − 1

du.

      dt 2 − 1 = log eu + 2u − 1 . t + t e = log t2 − 1

We choose the upper sign and c = 0 in (2.129), that is, we consider the asymptotic line given by 

u1 (t) = t and u2 (t) = log et +





e2t − 1 for all t > 0.

(2.130)

Visualization 2.8.13. (Asymptotic lines on the explicit surface of Visualization 2.6.3) The differential equations for the asymptotic lines on the explicit surface of Visualization 2.6.3 reduce to 2(du1 )2 + 6u2 (du2 )2 = 0 with the real solutions (Figure 2.85) 2  2 3 −u + u10 u1 (u2 ) = ± √ 3

for u2 < 0, where u10 is a constant.

212 ■ Differential Geometry and Its Visualization

Figure 2.85

Asymptotic lines on the explicit surface with f (ui ) = (u1 )2 + (u2 )3

Figure 2.86

A developable surface and tangent planes along a u2 –line

Definition 2.8.14. A ruled surface with a parametric representation ⃗x(ui ) = ⃗y (u1 ) + u2⃗z(u1 ) ((u1 , u2 ) ∈ D ⊂ R2 )

(2.131)

is said to be a torse, or developable, if all the tangent planes along a u2 –line coincide (Figures 2.86 and 2.87). First we give a characterization of developable surfaces. Theorem 2.8.15. A ruled surface is a developable if and only if ⃗y ′ (u1 ) • (⃗z(u1 ) × ⃗z ′ (u1 )) = 0 for all u1 .

(2.132)

Surfaces in Three-Dimensional Euclidean Space ■ 213

Figure 2.87

A developable surface and tangent planes along a u2 –line

Proof. A developable surface is characterized by the fact that the normal vectors ⃗ 2 = ⃗0. Now N ⃗ 2 = 1 implies N ⃗ •N ⃗ 2 = 0, hence along a u2 –line are constant, that is, N ′ 2 ⃗ ⃗ ⃗ N2 = ⃗0 and N • N2 are equivalent conditions. We have ⃗x1 = ⃗y + u ⃗z ′ and ⃗x2 = ⃗z, hence ⃗ = ⃗x1 × ⃗x2 = √1 (⃗y ′ + u2⃗z ′ ) × ⃗z, N ∥⃗x1 × ⃗x2 ∥ g

and

⃗ √g ∂ ⃗ 2 = √1 (⃗z ′ × ⃗z) + N N g ∂u2



1 √ g



,

hence by the Grassmann identity (1.5) and the identities in Remark 1.1.4 ⃗ ×N ⃗ 2 = 1 (⃗y ′ × ⃗z) × (⃗z ′ × ⃗z) N g 1 = (((⃗y ′ × ⃗z) • ⃗z)⃗z ′ − ((⃗y ′ × ⃗z) • ⃗z ′ )⃗z) g 1 1 = − ((⃗y ′ × ⃗z) • ⃗z ′ )⃗z = − (⃗y ′ • (⃗z × ⃗z ′ ))⃗z. g g ⃗ ×N ⃗ 2 = ⃗0 if and only if the condition in (2.132) holds. Thus N Developable surfaces can be characterized explicitly.

Theorem 2.8.16. A surface is developable if and only if it is a cylinder, cone or tangent surface (Figure 2.88).

214 ■ Differential Geometry and Its Visualization

Figure 2.88

Developable surfaces: Top left tangent surface, right cone. Bottom cylinder

Proof. (i) First we show that cylinders, cones and tangent surfaces are developable surfaces. Let γ be a curve with a parametric representation ⃗x ∗ (t) (t ∈ I1 ). We put u1 = t and ⃗y (u1 ) = ⃗x ∗ (u1 ). (i.1) Then the general cylinder generated by γ has a parametric representation ⃗x(ui ) = ⃗y (u1 ) + u2⃗z





(u1 , u2 ) ∈ D ,

where ⃗z is a constant vector. Since ⃗z ′ = ⃗0, the condition in (2.132) for a developable surface is satisfied. (i.2) The tangent surface generated by γ has a parametric representation 



⃗x(ui ) = ⃗y (u1 ) + u2 ⃗y ′ (u1 ) (u1 , u2 ) ∈ D . Since now ⃗z = ⃗y ′ , the condition in (2.132) for a developable surface is satisfied. (i.3) We may assume that the general cone generated by γ has its vertex in the origin; it has a parametric representation 

⃗x(ui ) = ⃗0 + u2 ⃗y (u1 ) (u1 , u2 ) ∈ D



and the condition in (2.132) for a developable surface is satisfied.

Surfaces in Three-Dimensional Euclidean Space ■ 215

(ii) Conversely, we assume that a ruled surface RulS with a parametric representation (2.131) is a developable surface. Then the condition in (2.132) holds by Theorem 2.8.15. We may assume that u1 is the arc length along ⃗y . Furthermore, we may assume by Remark 2.5.2 (a) that ⃗y ′ ⊥ ⃗z. Let γ be a curve on RulS which is given by u2 = φ(u1 ), that is, γ has a parametric representation ⃗x ∗ (u1 ) = ⃗x(u1 , φ(u1 )). Denoting the derivatives with respect to u1 by a dot, we obtain ⃗x˙ ∗ = ⃗y˙ + φ⃗z˙ + φ⃗ ˙ z. ⃗ are orthogonal to one another, we may express Since the unit vectors ⃗y˙ , ⃗z and N ⃗z˙ as a linear combination of them

and

⃗ , where a = ⃗y˙ • ⃗z˙ , ⃗z˙ = a⃗y˙ + b⃗z + cN 1 d  2 ⃗z = 0 b = ⃗z • ⃗z˙ = 2 du1 ⃗ • ⃗z˙ = √1 ⃗z˙ • (⃗y˙ × ⃗z) = 0, by (2.132). c=N g

Therefore, it follows that ⃗x˙ ∗ = ⃗y˙ (1 + φ(⃗y˙ • ⃗z˙ )) + φ⃗ ˙ z. The tangent of the curve γ is in the direction of ⃗z, if 1 + φ(⃗y˙ • ⃗z˙ ) = 0.

(2.133)

If ⃗y˙ • ⃗z˙ = 0 for all u1 , then we have ⃗z˙ = ⃗0, hence ⃗z is a constant vector and the developable surface is a cylinder. If ⃗y˙ • ⃗z˙ ̸= 0, then (2.133) implies u2 = φ(u1 ) = −

1 . ˙⃗y • ⃗z

Consequently, there is at most one curve on RulS with its tangents in the direction of ⃗z, and one and only one curve for those parts of RulS for which ⃗0 ̸= vx ˙ ∗ = ⃗zφ. ˙ Parts of the surface with φ = const have ⃗x ∗ = ⃗c, where ⃗c is a constant vector. Then the straight lines in the directions of the vectors ⃗z have a common point; such parts of the surface are cones.

The next result shows that developable surfaces can be generated by lines of curvature on surfaces.

216 ■ Differential Geometry and Its Visualization

Theorem 2.8.17. A curve γ on a surface S is a line of curvature if and only if the ruled surface generated by γ and the surface normal vectors along γ is a developable surface. Proof. Let S have a parametric representation ⃗y (ui ) ((u1 , u2 ) ∈ D) and we may assume that the parameters are orthogonal. Furthermore, let γ be given by ⃗y ∗ (s) = ⃗y (ui (s)) where s is the arc length along γ. (i) First we assume that γ is a line of curvature on S. We consider the ruled surface RulS with a parametric representation ⃗ ∗ (s), ⃗x(s, t) = ⃗y ∗ (s) + tN ⃗ ∗ (s) = N ⃗ (ui (s)). It follows that where N ⃗x1 =

⃗∗ d⃗y ∗ dN ∂⃗x ⃗∗=N ⃗ i u˙ i and ⃗x2 = ∂⃗x = N ⃗. = +t = ⃗yi u˙ i + tN ∂s ds ds ∂t

⃗˙ ∗ = dN ⃗ ∗ /ds, we obtain by the Grassmann identity Writing ⃗y˙ ∗ = d⃗y ∗ /ds and N (1.5), the fact that the parameters on S are orthogonal, and the definition of the first and second fundamental coefficients ⃗ ∗×N ⃗˙ ∗ ) = ⃗yi • (N ⃗ ×N ⃗ k )u˙ i u˙ k ⃗y˙ ∗ • (N   √ ⃗ k u˙ i u˙ k = g⃗yi • (⃗y1 × ⃗y2 ) × N   √ ⃗ k × (⃗y1 × ⃗y2 ) u˙ i u˙ k = − g⃗yi • N   √ ⃗ k • ⃗y2 ) − ⃗y2 (N ⃗ k • ⃗y1 ) u˙ i u˙ k = − g⃗yi • ⃗y1 (N √ √ = − gg1i (−L2k )u˙ i u˙ k + gg2i (−L1k )u˙ i u˙ k  √  = g g1i L2k u˙ i u˙ k − g2i L1k u˙ i u˙ k .

The principal curvature λ satisfies

L1k u˙ k = λg1i u˙ i and L2k u˙ k = λg2i u˙ i , hence Thus we have

g1i L2k u˙ i u˙ k = λg2i g1l u˙ i u˙ l = L1k g2i u˙ i u˙ k . ⃗ ∗×N ⃗ ∗) = 0 ⃗y˙ ∗ • (N

(2.134)

and RulS is a developable surface by condition (2.132) in Theorem 2.8.15. (b) Conversely, we assume that RulS is a developable surface. Then (2.134) is satisfied and we have g1i L2k u˙ i u˙ k = L1k g2i u˙ i u˙ k and consequently the principal directions are given by ui .

Surfaces in Three-Dimensional Euclidean Space ■ 217

A circular cone (left) and a circular cylinder (right) generated by a line of curvature on a surface of rotation

Figure 2.89

Visualization 2.8.18. Let RS be a surface of revolution, γ be a u2 –line on RS that corresponds to u10 . Then γ and the surface normal vectors along γ generate a circular cone, if h′ (u10 ) ̸= 0, and a circular cylinder, if h′ (u10 ) = 0 (Figure 2.89). Proof. Let RS have a parametric representation ⃗x(ui ) = {r(u1 ) cos u2 , r(u1 ) sin u2 , h(u1 )}





(u1 , u2 ) ∈ D .

Then we have g12 = L12 , and consequently the parameter lines are lines of curvature by Theorem 2.8.6 (a). The surface normal vectors along the u2 –line corresponding to u10 are given by ⃗ (u1 , u2 ) =  ⃗ ∗ (u2 ) = N N 0

1 (r′ (u10 ))2

+

(h′ (u10 ))2

{−h′ (u10 ) cos u2 , −h′ (u10 ) sin u2 , r′ (u10 )}.

If h′ (u10 ) ̸= 0, then the ruled surface with a parametric representation

⃗ (u10 , u2 ) ((u1 , u2 ) ∈ R × (0, 2π)) ⃗x ∗ (ui ) = ⃗x(u10 , u2 ) + u1 N

is a circular cone with its vertex in the point P = since, for u11 = we have



0, 0, h(u10 )



r′ (u10 )r(u10 ) + , h′ (u10 )

r(u10 )  ′ 1 2 · (r (u0 )) + (h′ (u10 ))2 , h′ (u10 ) −→

⃗ (u10 , u2 ) = OP for all u2 ∈ (0, 2π). ⃗x ∗ (u11 , u2 ) = ⃗x(u10 , u2 ) + u11 N

If h′ (u10 ) = 0, then we obtain

⃗x ∗ (u1 , u2 ) = ⃗x(u10 , u2 ) + u1⃗e 3 , that is, a circular cylinder with its axis along the x3 –axis.

218 ■ Differential Geometry and Its Visualization

A circular cylinder and cone generated by a line of curvature on a torus

Figure 2.90

Visualization 2.8.19. Let r0 > r1 > 0. We consider the torus with a parametric representation ⃗x(ui ) = {(r0 + r1 cos u1 ) cos u2 , (r0 + r1 cos u1 ) sin u2 , r1 sin u1 } Then we have





(u1 , u2 ) ∈ (0, 2π)2 .

r(u1 ) = r0 + r1 cos u1 , h(u1 ) = r1 sin u1 ,

r′ (u1 ) = −r1 sin u1 and h′ (u1 ) = r1 cos u1 . If u10 ̸= π/2, 3π/2, then h′ (u10 ) ̸= 0, and the u2 –line that corresponds to u10 and the surface normal vectors of the torus along that u2 –line generate a cone with vertex in the point 

−r1 sin u10 (r0 + r1 cos u10 ) P = 0, 0, + r1 sin u10 r1 cos u10



= (0, 0, −r0 tan u10 ),

by Visualization 2.8.18 (right in Figure 2.90). If u10 = π/2, then h′ (u10 ) = 0 and we obtain the cylinder with a parametric representation (left in Figure 2.90) ⃗x ∗ (ui ) = {r0 cos u2 , r0 sin u2 , r1 } − u1⃗e 3





(u1 , u2 ) ∈ R × 2π .

Visualization 2.8.20. Let a1 and a2 be distinct positive constants. We consider the elliptic EC cone with a parametric representation ⃗x(u ) = i



1 1 √ u1 cos u2 , √ u1 sin u2 , u1 a1 a2







(u1 , u2 ) ∈ (0, ∞) × (0, 2π) ,

Surfaces in Three-Dimensional Euclidean Space ■ 219

Figure 2.91

A tangent surface generated by a line of curvature on an elliptic cone

and the curve γ on EC given by u1 (u2 ) = 

k 1 1 cos2 u2 + sin2 u2 + 1 a1 a2

,

where k > 0 is a constant. Then the ruled surface generated by γ and the surface normal vectors of EC along γ is a tangent surface (Figure 2.91). Proof. We know from Visualization 2.8.10 that γ is a line of curvature on EC, hence the ruled surface RulS generated by γ and the surface normal vectors of EC along γ is a developable surface by Theorem 2.8.17, hence a cone, a cylinder or a tangent surface by Theorem 2.8.16. (i) We are going to show that RulS is neither a cone nor a cylinder. The surface normal vectors along γ are given by ⃗ =N ⃗ (u2 ) = √ N

a1

cos2

u2

√ √ 1 {− a1 cos u2 , − a1 sin u2 , 1}. 2 2 + a1 cos u + 1

We put u¯1 = u2 and ⃗y (¯ u1 ) = ⃗x(u1 (u2 ), u2 ) =

k 1 1 cos2 u¯1 + sin2 u¯1 + 1 a1 a2





1 1 √ cos u¯1 , √ sin u¯1 , 1 a1 a2

220 ■ Differential Geometry and Its Visualization

and

⃗ (¯ ⃗z(¯ u1 ) = N u1 ).

Then RulS has a parametric representation ⃗x(¯ ui ) = ⃗y (¯ u1 ) + u¯2⃗z(¯ u1 ). (i.1) Since ⃗z(¯ u1 ) • ⃗e 3 = 

1 a1 cos2 u¯1 + a2 sin2 u¯1 + 1

,

the vectors ⃗z do not have a constant direction, hence RulS cannot be a cylinder. (ii.1) If RulS were a cone, the straight lines with the parametric representations ⃗x(t) = ⃗y (0) + t⃗z(0) and ⃗x ∗ (t ∗ ) = ⃗y (π/2) + t ∗⃗z(π/2) for t, t ∗ ∈ R would have a point of intersection P . Comparing the first two components of ⃗x(t) and ⃗x ∗ (t ∗ ), we obtain √ √ a1 a2 k k ∗ √ − t√ = 0 and √ −t √ = 0, 1 + a1 1 + a1 1 + a2 1 + a2 hence

k k t = √ and t ∗ = √ . a1 a2

Thus we have √ OS = ⃗x(k/ a1 ) =



√ OS = ⃗x ∗ (k/ a2 ) =



−→

and

−→

 √ k a1 + 1 {0, 0, √ a1  √ k a2 + 1 {0, 0, √ , a2

but a1 ̸= a2 implies

√ √ k a2 + 1 k a1 + 1 = ̸ , √ √ a1 a2 √ √ and consequently ⃗x(k/ a1 ) ̸= ⃗x ∗ (k/ a2 ). Thus RulS cannot be a cone either.

2.9

TRIPLE ORTHOGONAL SYSTEMS

Triple orthogonal systems can be used to determine lines of curvature on certain surfaces, as we will see in Theorem 2.9.4.

Surfaces in Three-Dimensional Euclidean Space ■ 221

Definition 2.9.1. We assume that the function ⃗y : D ⊂ R3 → R3 has continuous partial derivatives of order 2, that the vectors ⃗yi = ∂⃗y /∂ui (i = 1, 2, 3) are linearly independent and ⃗yi • ⃗yk = 0 (i ̸= k). Then the surfaces given by uj = const form three families of mutually orthogonal surfaces, a so–called triple orthogonal system of surfaces. Visualization 2.9.2 (Some triple orthogonal systems). (a) Let ⃗y (u1 , u2 , u3 ) = {u1 cos u2 , u1 sin u2 , u3 } for (u1 , u2 , u3 ) ∈ R × (0, 2π) × R. If u1 = r > 0 is constant, then we obtain a circular cylinder of radius r. If u2 = ϕ is constant, then we obtain a plane through the x3 –axis with an angle ϕ to the x1 –axis. If u3 = h is constant, then we obtain a plane through (0, 0, h), parallel to the x1 x2 – plane. Obviously ⃗y has continuous second-order partial derivatives, ⃗y1 = {cos u2 , sin u2 , 0}, ⃗y2 = {−u1 sin u2 0, u1 cos u2 , 0},

⃗y3 = {0, 0, 1} and ⃗yi • ⃗yk = 0 for i ̸= k.

Thus the surfaces form a triple orthogonal system. (b) Let ⃗y (u1 , u2 , u3 ) = {u1 cos u2 cos u3 , u1 cos u2 sin u3 , u1 sin u2 }

for (u1 , u2 , u3 ) ∈ (0, ∞) × (−π/2, π/2) × (0, 2π). If u1 = r > 0 is constant, then we obtain a sphere of radius r with its centre in the origin. If u2 = α ∈ (−π/2, π/2) is constant, then we obtain a circular cone with its vertex in the origin and meridians with an angle α to the x1 x2 –plane. If u3 = const is constant, then we obtain a plane orthogonal to the x1 x2 –plane. Obviously ⃗y has continuous second order partial derivatives ⃗y1 = {cos u2 cos u3 , cos u2 sin u3 , sin u2 },

⃗y2 = {−u1 sin u2 cos u3 , −u1 sin u2 sin u3 , u1 cos u2 },

⃗y3 = {−u1 cos u2 sin u3 , u1 cos u2 cos u3 , 0} and ⃗yi • ⃗yk = 0 for i ̸= k.

Thus the surfaces form a triple orthogonal system (Figure 2.92). Visualization 2.9.3 (A triple system which is not orthogonal). Let S˜ = {(r, ϕ, θ) : r ∈ (0, ∞), ϕ ∈ (0, 2π), θ ∈ (−π, π)}

and



⃗y (u1 , u2 , u3 ) = u3 (1 + cos u1 ) cos u2 , u3 (1 + cos u1 ) sin u2 , u3 sin u1



222 ■ Differential Geometry and Its Visualization

Figure 2.92

A triple orthogonal system of spheres, circular cones and planes

˜ for (u1 , u2 , u3 ) ∈ S. 3 If u = r > 0 is constant, then ⃗y (u1 , u2 , r) for (u1 , u2 ) ∈ (−π, π) × (0, 2π) is a parametric representation of a torus Tr . If u2 = ϕ ∈ (0, 2π) is constant, then 

⃗y (u1 , ϕ, u3 ) = u3 (1 + cos u1 ) cos ϕ, u3 (1 + cos u1 ) sin ϕ, u3 sin u1



for (u1 , u3 ) ∈ (−π, π) × (0, ∞) is a parametric representation for a half plane Hϕ . If u1 = θ ∈ (−π, π), then 

⃗y (θ, u1 , u2 ) = u3 (1 + cos θ) cos u2 , u3 (1 + cos θ) sin u2 , u3 sin θ



for (u2 , u3 ) ∈ (0, 2π) × (0, ∞) is a parametric representation of half a cone Cθ . ˜ and observe that for We consider the triple system S = {Tr , Hϕ , Cθ : (r, ϕ, θ) ∈ S} 1 u ̸= 0 ∂⃗y ∂⃗y ∂⃗y ∂⃗y ∂⃗y ∂⃗y • 2 = • 3 = 0 and • 3 = −u3 sin u1 ̸= 0. 1 2 1 ∂u ∂u ∂u ∂u ∂u ∂u Consequently the system S is not an orthogonal system (Figure 2.93).

The significance of triple orthogonal systems for lines of curvature lies in the next result. Theorem 2.9.4 (Dupin, 1813). The surfaces of a triple orthogonal system mutually intersect in lines of curvature. Proof. Partial differentiation of ⃗yi • ⃗yk = 0 (i ̸= k) yields ⃗y12 • ⃗y3 + ⃗y1 • ⃗y23 = ⃗y23 • ⃗y1 + ⃗y2 • ⃗y13 = ⃗y21 • ⃗y3 + ⃗y2 • ⃗y31 = 0,

Surfaces in Three-Dimensional Euclidean Space ■ 223

Figure 2.93

The triple system of Visualization 2.9.3

hence

⃗y12 • ⃗y3 = ⃗y31 • ⃗y2 = ⃗y23 • ⃗y1 = 0.

Since the vectors ⃗y1 , ⃗y2 and ⃗y12 are orthogonal to the vector ⃗y3 , they are linearly dependent, that is, ⃗y1 • (⃗y2 × ⃗y12 ) = 0. This means that we have L12 = 0 and g12 = ⃗y1 • ⃗y2 = 0 for the surfaces given by u3 = const. Consequently, the parameter lines of these surfaces are lines of curvature by Theorem 2.8.6 (a). The same holds true for the surfaces of the other two families. The next example shows that the converse of Dupin’s theorem, Theorem 2.9.4, does not hold, in general. Visualization 2.9.5. We consider the families of surfaces consisting of spheres with their centres in the origin, elliptic – non-circular – cones with their vertices in the origin, and planes through the x3 –axis. The lines of intersection of the planes and spheres are meridians on the spheres, hence lines of curvature on the spheres, by Dupin’s theorem, Theorem 2.9.4. The lines of intersection of the planes and cones are the u1 –lines of the cones, hence lines of curvature on the cones, by Visualization 2.8.10. The line of intersection of a cone, given by the parametric representation ⃗x(u ) = i



1 1 √ u1 cos u2 , √ u1 sin u2 , u1 a1 a2







(u1 , u2 ) ∈ R × (0, 2π) ,

where a1 ̸= a2 are constants, and a sphere of radius r > 0, given by the equation (x1 )2 + (x2 )2 + (x3 )2 − r2 = 0

224 ■ Differential Geometry and Its Visualization

Figure 2.94 A line of curvature on an elliptic cone which is the intersection of the elliptic cone and a sphere

is determined by the equation 1 1 1 2 (u ) cos2 u2 + (u1 )2 sin2 u2 + (u1 )2 = r2 , a1 a2 hence

u1 (u2 ) = 

r 1 1 cos2 u2 + sin2 u2 + 1 a1 a2

with respect to the parametric representation of the cone. Thus the line of intersection is a line of curvature on the cone by (2.127) in Visualization 2.8.10 (Figure 2.94). Obviously, all points of a plane are umbilical points, and all points of any sphere are umbilical points by Example 2.5.25. Consequently, all curves on planes and spheres are lines of curvature. Therefore, the surfaces of the families mutually intersect in their lines of curvature. But the surfaces do not form a triple orthogonal system, since the cones and planes do not intersect orthogonally, as we are going to see. The surface normal vectors of a cone along its u1 –line corresponding to u20 are given by 1 ⃗ = {−√a1 cos u2 , −√a2 sin u2 , 1}  , N 0 0 a1 cos2 u20 + a2 sin2 u20 + 1

and the surface normal vector of the corresponding plane of intersection is given by 1 ⃗ ∗ = {−√a1 sin u2 , √a2 cos u2 , 0}  N . 0 0 a1 sin2 u20 + a2 cos2 u20

Surfaces in Three-Dimensional Euclidean Space ■ 225

Therefore, we have ⃗ •N ⃗∗= N

(a1 − a2 ) sin u20 sin u20 

a1 cos2 u20 + a2 sin2 u20 + 1 a1 sin2 u20 + a2 cos2 u20

for u20 ̸= π/2, π, 3π/2.

̸= 0

The next result and Dupin’s theorem can be applied to determine the lines of curvature on ellipsoids and hyperboloids of one and two sheets. Theorem 2.9.6. Let (a, b, c) ∈ R3 be given with 0 < a2 < b2 < c2 and the functions gP : R \ {a2 , b2 , c2 } → R be defined for every P = (x1 , x2 , x3 ) ∈ R3 by gP (λ) =

(x1 )2 (x2 )2 (x3 )2 + + − 1. a2 − λ b2 − λ c2 − λ

Then the surfaces given by the equations

Eλ = {P ∈ R3 : gP (λ) = 0} for λ < a2 ,

Hλ1 = {P ∈ R3 : gP (λ) = 0} for a2 < λ < b2 , and Hλ2 = {P ∈ R3 : gP (λ) = 0} for b2 < λ < c2 form a triple orthogonal system of the families E = {Eλ : λ < a2 }, H1 = {Hλ1 : a2 < λ < b2 } and H2 = {Hλ2 : b2 < λ < c2 } (Figure 2.95). Proof. First, we observe that obviously Eλ is an ellipsoid for λ < a2 , Hλ1 is a hyperboloid of one sheet for a2 < λ < b2 , and Hλ2 is a hyperboloid of two sheets for b2 < λ < c2 . Furthermore, for every fixed P ∈ R3 \ {(0, 0, 0)}, the function gP is continuous on R \ {a2 , b2 , c2 }, and we have lim gp (λ) = lim2 gp (λ) = lim2 gp (λ) = ±∞

λ→a2 ∓

and

λ→b ∓

λ→a ∓

lim gP (λ) = −1.

λ→±∞

Therefore, there are at least one λ1 , one λ2 and one λ3 such that λ1 < a2 < λ2 < b2 < λ3 < c2 and gP (λk ) = 0 (k = 1, 2, 3). On the other hand gP has at most three zeros, since gP (λ) = 0 is equivalent to a cubic equation. Thus, there is one and only one surface of each family through any given point P ∈ R3 \ {(0, 0, 0)}.

226 ■ Differential Geometry and Its Visualization

The triple orthogonal system of ellipsoids and hyperboloids of one and two sheets (left). Lines of curvature on an ellipsoid (right)

Figure 2.95

Furthermore, the surface normal vector at a point P ∈ R3 \ {(0, 0, 0)} of a surface given by the equation gP (λk ) = 0 has the direction 1 d⃗k = 2





∂ ∂ ∂ (gP (λk )), 2 (gP (λk )), 3 (gP (λk )) 1 ∂x ∂x ∂x   1 2 3 x x x = , 2 , 2 , 2 a − λk b − λk c − λk

and we obtain for i ̸= k d⃗i • d⃗k =

(a2

(x2 )2 (x3 )2 (x1 )2 + 2 + 2 2 2 − λi )(a − λk ) (b − λi )(b − λk ) (c − λi )(c2 − λk )

1 = λk − λi =



(x1 )2 (x2 )2 (x2 )2 (x3 )2 (x3 )2 (x1 )2 − + − + − a2 − λi a2 − λk b2 − λi b2 − λk c2 − λi c2 − λk



1 (gP (λi ) − gP (λk )) = 0. λk − λi

Thus we have shown that the families E, H1 and H2 form a triple orthogonal system. Now we apply Dupin’s Theorem and Theorem 2.9.6 to determine the lines of curvature on ellipsoids, and hyperboloids of one and two sheets.

Surfaces in Three-Dimensional Euclidean Space ■ 227

Visualization 2.9.7. (The lines of curvature on ellipsoids and hyperboloids) We define the function for every point P = (x1 , x2 , x3 ) ∈ R3 \ {(0, 0, 0)} by ϕP (λ) = (a2 − λ)(b2 − λ)(c2 − λ)gP (λ),

(2.135)

where gP is the function defined in Theorem 2.9.6. We have ϕP (λ) = (x1 )2 (b2 − λ)(c2 − λ) + (x2 )2 (a2 − λ)(c2 − λ)

+ (x3 )2 (a2 − λ)(b2 − λ) − (a2 − λ)(b2 − λ)(c2 − λ) (2.136)

and ϕP (λ) = 0 is a cubic equation with zeros λk , where λ1 < a2 < λ2 < b2 < λ3 < c2 . Therefore, we may write ϕP (λ) = (λ − λ1 )(λ − λ2 )(λ − λ3 ).

(2.137)

Choosing λ = a2 , we obtain from (2.136) and (2.137) ϕP (a2 ) = (x1 )2 (b2 − a2 )(c2 − a2 ) = (a2 − λ1 )(a2 − λ2 )(a2 − λ3 ), hence (x1 )2 =

(a2 − λ1 )(a2 − λ2 )(a2 − λ3 ) . (a2 − b2 )(a2 − c2 )

(2.138)

Similarly, the choices λ = b2 and λ = c2 yield

(b2 − λ1 )(b2 − λ2 )(b2 − λ3 ) (c2 − λ1 )(c2 − λ2 )(c2 − λ3 ) 3 2 and (x . ) = (b2 − a2 )(b2 − c2 ) (c2 − a2 )(c2 − b2 ) (2.139) If we choose one λi = const, then the equations in (2.138) and (2.139) yield a parametrization of the surface given by gP (λ) = 0 with respect to the other two parameters λj and λk . We put (x2 )2 =

α = a2 − λi , β = b2 − λi , γ = c2 − λi , u1 = λj − λi and u2 = λk − λi . Then the surface given by the equation (x1 )2 (x2 )2 (x3 )2 + + =1 α β γ has a parametric representation       1 2 1 2 1 2  α(α − u )(α − u ) β(β − u )(β − u ) γ(γ − u )(γ − u ) ⃗x(ui ) = ± ,± , ± .  (α − β)(α − γ) (β − α)(β − γ) (γ − α)(γ − β) 

The lines of curvature of this surface are the parameter lines with respect to this parametric representation (right in Figure 2.95 and Figure 2.96).

228 ■ Differential Geometry and Its Visualization

Figure 2.96

2.10

Lines of curvature on hyperboloids of one (left) and two sheets (right)

THE WEINGARTEN EQUATIONS

In this section, we prove the Weingarten equations which give a relation between the partial derivatives of the normal vector of a surface with a parametric representation ⃗x(ui ) and the vectors ⃗xk (k = 1, 2). We also give a geometric interpretation of the Gaussian curvature. Theorem 2.10.1 (The Weingarten equations). Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D) and first and second fundamental coefficients gik and Lik (i, k = 1, 2). We put Lik = g ij Ljk for i, k = 1, 2.

(2.140)

⃗ k of the surface normal vector of S satisfy the following Then the partial derivatives N identities (2.141) N⃗k = −Lik ⃗xi for k = 1, 2, which are referred to as the Weingarten equations.

⃗ 2 = 1 implies N ⃗ •N ⃗ k = 0 for k = 1, 2, the vectors N ⃗ k are in the Proof. Since N tangent plane of the surface, hence we may express them as linear combinations of the vectors ⃗x1 and ⃗x2 ⃗ k = B m⃗xm for k = 1, 2. (2.142) N k It follows from (2.142) and the definition of the second fundamental coefficients i ⃗k Bkm = g ij gjm Bkm = g ij ⃗xj • ⃗xm Bkm = g ij ⃗xj • N Bki = δm

= −g ij Ljk = −Lik for i, k = 1, 2.

This implies (2.141).

We observed in Example 2.4.5 that the second fundamental coefficients of a plane vanish identically. Applying the Weingarten equations, we can show that the converse implication also holds true.

Surfaces in Three-Dimensional Euclidean Space ■ 229

Example 2.10.2. If the second fundamental coefficients of a surface S vanish identically, then S is a plane. Proof. Let Lik = 0 for i, k = 1, 2. Then we have Lki = Lij g jk = 0 for i, k = 1, 2 by (2.140), and so by (2.141) ⃗ k = −Li ⃗xi = ⃗0 for k = 1, 2, N k ⃗ = ⃗c, where ⃗c is a constant vector. This shows that S is a plane. hence N

The Weingarten equations (2.141) reduce when the parameter lines of a surface are its lines of curvature. Theorem 2.10.3 (The Rodrigues formulae). The Weingarten equations (2.141) reduce to the Rodrigues formulae ⃗ k = −κk ⃗xk for k = 1, 2, N

(2.143)

where κ1 and κ2 are the principal curvatures, if and only if the parameter lines of the surface are its lines of curvature. Proof. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D).

(i) First we show the necessity of the conditions in (2.143). We assume that the parameter lines are the lines of curvature of S. Then it follows from Theorem 2.8.6 (a) that g12 = L12 = 0, hence L11 L11 = g 1j Lj1 = g 11 L11 = , L12 = L21 = g 2j Lj1 = 0 g11 and L22 = g 2j Lj2 = g 22 L22 =

L22 . g22

Since the principal curvatures satisfy L11 L22 κ1 = and κ2 = , g11 g22 we obtain for the Weingarten equations (2.141) ⃗ 1 = −Lj ⃗xj = − L11 ⃗x1 = −κ1⃗x1 and N ⃗ 2 = −Lj ⃗xj = − L22 ⃗x2 = −κ2⃗x2 , N 1 2 g11 g22 (2.144) that is, the Rodrigues formulae (2.143) are satisfied. (ii) Now we show the sufficiency of the conditions in (2.143). We assume that the identities in (2.144) hold. Then it follows that L21 = L12 = 0, hence ⃗ 1 • ⃗x1 = −L1 = L11 , N ⃗ 1 • ⃗x2 = −L2 = L12 = 0 N ⃗ 2 • ⃗x1 = N

1 −L12

1

⃗ 2 • ⃗x2 = −L2 = L22 = L21 = 0 and N 2

230 ■ Differential Geometry and Its Visualization

and L12 = −g 1j Lj2 = −g 12 L22 = 0 and L21 = −g 2j Lj1 = −g 12 L11 = 0. If κ1 and κ2 are not both equal to zero, then we have g 12 = 0 and so g12 = 0. If κ1 = κ2 = 0, then we may choose g12 = 0. Thus we have in both cases g12 = L12 = 0 and consequently the parameter lines of the surface are its lines of curvature by Theorem 2.8.6 (a).

Now we establish the characterization of all surfaces that have only umbilical points. Theorem 2.10.4. If a surface S of class C r (D) for r ≥ 3 has only umbilical points, then S is a sphere or a plane. Proof. Umbilical points satisfy by (2.73) in Remark 2.5.16 (b) Lik (uj ) = λgik (uj ) for i, k = 1, 2. First λ = 0 implies Lik = 0 for i, k = 1, 2 and so S is a plane by Example 2.10.2. If λ ̸= 0, then every curve on S is a line of curvature, and we obtain by the Rodrigues formulae (2.143) in Theorem 2.10.3 ⃗ k = −κk ⃗xk for k = 1, 2, N where κk are the principal curvatures, and so ⃗ k = − Lkk ⃗xk = −λ⃗xk for k = 1, 2. N gkk

(2.145)

It follows from (2.145) that ⃗ jk = −λk ⃗xj − λ⃗xjk , where λj = ∂λ/∂uj (j = 1, 2), ⃗ kj = −λj ⃗xk − λ⃗xkj and N N hence

⃗ 12 − N ⃗ 21 = λ1⃗x2 − λ2⃗x1 . ⃗0 = N

Since ⃗x1 and ⃗x2 are linearly independent, we must have λ1 = λ2 = 0, that is, λ = const. Integrating (2.145), we obtain ⃗ = −λ⃗x + ⃗c for some constant vector ⃗c. N Writing λ = −1/r and ⃗c = −⃗x0 /r, we have ⃗ ) • (rN ⃗ ) = (⃗x − ⃗x0 )2 , r2 = (rN that is, S is a sphere of radius r and centre in X0 .

Surfaces in Three-Dimensional Euclidean Space ■ 231

The next result gives a relation between the torsion of an asymptotic line and the Gaussian curvature. Theorem 2.10.5 (Beltrami and Enneper, 1870). Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D). Then the torsion of an asymptotic line which is not a straight line is given by √ |τ | = −K. (2.146) Proof. First, we observe that K < 0 along the asymptotic line by Remark 2.8.2 (b). Let the asymptotic line γ be given by a parametric representation ⃗x(s) = ⃗x(ui (s)), where s is the arc length along γ, and ⃗vk (k = 1, 2, 3) denote the vectors of the trihedra along γ. Since γ has non–vanishing curvature, it follows from Theorem 2.8.7 ⃗ = ±⃗v3 , hence N ⃗˙ = ±⃗v˙ 3 = ∓τ⃗v2 by Frenet’s third formula (1.84). Using the that N Weingarten equations (2.141), we obtain

that is,

⃗˙ • (⃗v3 × ⃗v1 ) ⃗˙ • ⃗v2 = ∓N τ = ∓N ⃗˙ • (N ⃗ × ⃗x˙ ) = ∓N ⃗ k • (N ⃗ × ⃗x˙ )u˙ k = ∓N ⃗ × ⃗xj )u˙ k u˙ j = ±Li N ⃗ • (⃗xj × ⃗xi )u˙ k u˙ j = ±Lik ⃗xi • (N k    √ √  = ± g L2k u˙ k u˙ 1 − L1k u˙ k u˙ 2 = ± g g 2j Ljk u˙ k u˙ 1 − g 1j Ljk u˙ k u˙ 2 ,

    √  τ = ± g g 12 L1k u˙ k + g 22 L2k u˙ k u˙ 1 − g 11 L1k u˙ k + g 12 L2k u˙ k u˙ 2 .

(2.147)

Since the asymptotic line satisfies the differential equation (2.121) in Remark 2.8.2 (b) Lik u˙ i u˙ k = L11 (u˙ 1 )2 + 2L12 u˙ 1 u˙ 2 + L22 (u˙ 2 )2 = 0, we obtain 

L211 (u˙ 1 )2 + 2L11 L12 u˙ 1 u˙ 2 + L212 (u˙ 2 )2 = L11 u˙ 1 + L12 u˙ 2

2

= L212 (u˙ 2 )2 − L11 L22 (u˙ 2 )2 = −L(u˙ 2 )2 , hence

√ √ L1k u˙ k = ±u˙ 2 −L, and similarly L2k u˙ k = ∓u˙ 1 −L.

(2.148)

Substituting (2.148) in (2.147), we obtain

  g12 √ g11 √ g22 √ g12 √ √ 1 2 1 2 2 2 1 2 τ =± g − −Lu˙ u˙ − −L(u˙ ) − −L(u˙ ) − −Lu˙ u˙ g g g g  √ √ −L −L = ∓ −K. = ± √ (−gik u˙ i u˙ k ) = ∓ g g

This shows (2.146).

232 ■ Differential Geometry and Its Visualization

The torsion (dashed) along an asymptotic line on a catenoid (left) and a pseudo-sphere (right) Figure 2.97

Visualization 2.10.6. The Gaussian curvature of the catenoid Kc and the pseudosphere Kp are given by Visualization 2.5.19 as Kc = −

a2

1 1 and Kp = −1. cosh ua

Hence, the absolute values of the torsions τc and τp of the asymptotic lines on the catenoid and pseudo-sphere (Figure 2.97) are by (2.146) in the Beltrami–Enneper theorem, Theorem 2.10.5, |τc (u1 )| =

a2

1 1 and |τp | = 1. cosh ua

Figure 2.98 shows a representation of the ± torsion along an asymptotic line on an explicit surface.

Figure 2.98

The ± torsion (dashed) along an asymptotic line on an explicit surface

Surfaces in Three-Dimensional Euclidean Space ■ 233

Now we establish some useful relations between the derivatives of surface normal vectors and the Gaussian and mean curvature of surfaces. Theorem 2.10.7. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D). Then the surface normal vectors of S satisfy the relation ⃗ 2 = K √g N ⃗; ⃗1 × N (2.149) N furthermore, we have

Lki Lkj − Lkj Lki = 0 for i, j = 1, 2.

(2.150)

⃗˙ )2 = (2HLik − Kgik ) u˙ i u˙ k , (N

(2.151)

If γ is a curve on S given by a parametric representation ⃗x(ui (s)), where s denotes ⃗ along γ satisfy the relation the arc length along γ, then the surface normal vectors N

where K and H are the Gaussian and mean curvature along γ. Proof. (i) First we prove (2.149). We apply the Weingarten equations (2.141) and obtain   ⃗1 × N ⃗ 2 = Lℓ Lj (⃗xℓ × ⃗xj ) = L1 L2 − L2 L1 √g N ⃗ N 1 2 1 2 1 2  √ ⃗ = g 1m Lm1 g 2n Ln2 − g 2i Li1 g 1k Lk2 gN 

= g 11 L11 g 21 L12 + g 11 L11 g 22 L22 + g 12 L21 g 21 L12 + g 12 L21 g 22 L22 

− g 21 L11 g 11 L12 + g 21 L11 g 12 L22 + g 22 L12 g 11 L12  √ ⃗ + g 22 L12 g 12 L22 gN





= g 11 g 22 L11 L22 − g 12 





2

= L11 L22 g 11 g 22 − g 12 

11 22

= g g



− g

12

2  



L212 − g 12

2 

L11 L22 − g 11 g 22 L212





− L212 g 11 g 22 − g 12

L11 L22 − L212

Thus we have shown that (2.149) holds.

2

√

⃗ = gN

2  √



√ ⃗ gN

⃗ gN

L √ ⃗ ⃗ √g. · gN = K N g

(ii) Now we prove (2.150). We have Lki Lkj − Lkj Lki = g km Lmi Lkj − g kℓ Lℓj Lki = g km Lmi Lkj − g mℓ Lℓj Lmi = g km Lmi Lkj − g km Lkj Lmi = 0 for i, j = 1, 2.

Thus we have shown (2.150).

234 ■ Differential Geometry and Its Visualization

(iii) Finally we prove (2.151). Applying the Weingarten equations (2.141) again, we obtain ⃗ 1 = Lℓ Lj gℓj = g ℓm Lm1 g jn Ln1 gℓj = δ m Lm1 g jn Ln1 = Lj1 g jn Ln1 ⃗1 • N N 1 1 j

= L11 g 1n Ln1 + L21 g 2n Ln1 = L11 g in Lni + L21 g 2n Ln1 − L11 g 2n Ln2 = 2L11 H + g 2n (L12 Ln1 − L11 Ln2 )

= 2L11 H + g 21 (L12 L11 − L11 L12 ) + g 22 (L12 L21 − L11 L22 )   L = 2L11 H + g 22 L212 − L11 L22 = 2L11 H − g11 g = 2L11 H − g11 K,

and similarly ⃗1 • N ⃗ 2 = Lj1 g jn Ln2 = L12 g 2n Ln2 + L11 g 1n Ln2 N

= L12 g in Lni + g 1n (−L12 Ln1 + L11 Ln2 ) 



= 2L12 H + g 12 L11 L22 − L212 g12 L = 2L12 H − g12 K = 2L12 H − g and Thus we have shown

⃗ 2 = 2L22 H − g22 K. ⃗2 • N N

⃗i • N ⃗ k = 2HLik − Kgik for i, k = 1, 2. N

(2.152)

This implies ⃗˙ 2 = N ⃗i • N ⃗ k u˙ i u˙ k = (2HLik − Kgik ) u˙ i u˙ k which is (2.151). N

In view of (2.151), we define the third fundamental coefficients. Definition 2.10.8 (Third fundamental coefficients). Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D). Then the functions ⃗i • N ⃗ k = 2HLik − Kgik (i, k = 1, 2) cik = N

(2.153)

are called the third fundamental coefficients of S. Visualization 2.10.9 (Parallel surfaces). Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D) and a ∈ R be a constant. Then the surface S ∗ with a parametric representation ⃗ (ui ), ⃗x ∗ (ui ) = ⃗x(ui ) + aN

Surfaces in Three-Dimensional Euclidean Space ■ 235

Figure 2.99

A catenoid and a parallel surface

⃗ denotes the surface normal vector of S, is called a parallel surface (Figure where N 2.99). If K and H are the Gaussian and mean curvature of S and 1 − 2aH + a2 K ̸= 0, then the Gaussian and mean curvature K ∗ and H ∗ of S ∗ are given by K∗ =

K 1 − 2aH + a2 K

(2.154)

H∗ =

H − aK . 1 − 2aH + a2 K

(2.155)

and

⃗ k (k = 1, 2) by (2.149) in Theorem Proof. We note that it follows from ⃗xk∗ = ⃗xk + aN 2.10.7 and the Weingarten equations (2.141) that ⃗ 1 ) × (⃗x2 + aN ⃗ 2) ⃗x1∗ × ⃗x2∗ = (⃗x1 + aN ⃗2 + N ⃗ 1 × ⃗x2 ) + a2 (N ⃗1 × N ⃗ 2) = ⃗x1 × ⃗x2 + a(⃗x1 × N   √ ⃗ = ⃗x1 × ⃗x2 − a Lk2 ⃗x1 × ⃗xk + Lk1 ⃗xk × ⃗x2 + a2 gK N 



= ⃗x1 × ⃗x2 1 + a2 K − a L22 + L11 



= (⃗x1 × ⃗x2 ) 1 + a2 K − 2aH .





= (⃗x1 × ⃗x2 ) 1 + a2 K − ag ik Lik



(2.156)

236 ■ Differential Geometry and Its Visualization

⃗ ∗ of the parallel surface S ∗ Therefore, we have for the surface normal vector N ⃗ , if 1 − 2aH + a2 K ̸= 0. ⃗∗=N N

(2.157)

We observe that (2.157) means that if S ∗ is a parallel surface of S, then S is also a parallel surface of S ∗ . (i) First we prove (2.154). It follows from (2.156) that √ ∗ √ g = ∥⃗x1∗ × ⃗x2∗ ∥ = g|1 − 2aH + a2 K|. If |a| is sufficiently small, then 1 − 2aH + a2 K > 0, and we obtain for the Gaussian curvature K ∗ of the parallel surface S ∗ by (2.149) √ gK 1 ⃗ ∗  ⃗ ∗ ⃗ ∗ 1 ⃗ K ∗ ⃗ ⃗ K = √ ∗ N • N1 × N2 = √ ∗ N • (N1 × N2 ) = √ ∗ = , g g g 1 − 2aH + a2 K that is, (2.154) holds.

(ii) Now we prove (2.155). Since S is obviously a parallel surface of S ∗ , we obtain by interchanging the roles of K and H with K ∗ and H ∗ , and replacing a by −a K=

K∗ . 1 + 2aH ∗ + a2 K ∗

This implies 2aH ∗ K = K ∗ − a2 K ∗ K − K = K ∗ (1 − a2 K) − K 



1 − a2 K =K −1 1 − 2aH + a2 K K = (1 − a2 K − 1 + 2aH − a2 K), 1 − 2aH + a2 K

and we obtain for K ̸= 0

H∗ = that is, (2.155) holds.

H − aK , 1 − 2aH + a2 K

Remark 2.10.10. The identities in (2.154) and (2.155) may be used to find surfaces with given Gaussian or mean curvature.

Surfaces in Three-Dimensional Euclidean Space ■ 237

A hyperbolic spherical surface (left) and its parallel surface with constant mean curvature H = −1/2 (right)

Figure 2.100

Visualization 2.10.11 (Parallel surfaces of a surface of revolution). Now we consider the surface of revolution RS, namely the elliptic spherical surface of Visualization 2.5.20, Case 3, with a parametric representation r(u1 ) = λ cos



1

u c



     1  λ2 1 2 u  and h(u ) = 1 − 2 sin du1

c

c

for u1 ∈ I1 ⊂ (−π/2, π/2), where λ and c are constants with 0 < λ < c. Then RS has constant Gaussian curvature K = 1/c2 . If we choose a = c, then it follows from (2.155) for the mean curvature H ∗ of the parallel surface RS ∗ for a of RS 1 c = − 1 · 1 − cH = − 1 , H∗ = 2(1 − cH) 2c 1 − cH 2c H−

that is, RS ∗ has constant mean curvature H ∗ = −1/(2c) (Figure 2.100). The same argument holds true if we choose RS as above but with (Figure 2.101) 

λ > c and u1 ∈ −c sin−1



c c , c sin−1 . λ λ

Now we consider the pseudo-sphere which has constant Gaussian curvature K = −1 by (2.80) in Visualization 2.5.19. If we choose a = 1, then it follows from (2.154) and (2.80) for the Gaussian curvature K ∗ of the parallel surface RS ∗ for a of RS (Figure 2.102) 

1 exp (−u1 ) 1 − exp (−2u1 ) 1 K =− = = . 1 − 2H − 1 2H 1 − 2 exp (−2u1 ) ∗

Now we give a geometric interpretation of the Gaussian curvature. The following definition is important.

238 ■ Differential Geometry and Its Visualization

An elliptic spherical surface (left) and its parallel surface with constant mean curvature H = −1/2 (right) Figure 2.101

Figure 2.102

A pseudo-sphere (left) and its parallel surface (right) for a = 1

Definition 2.10.12. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D). The map that assigns to every point P of S the surface normal vector at P is called the spherical Gauss map (Figures 2.103 and 2.104); the set of all image points on the sphere under the spherical Gauss map is called spherical image of S. Remark 2.10.13. The third fundamental coefficients (Definition 2.10.8) of a surface are the first fundamental coefficients of its spherical Gauss map. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D), P ∈ S be a point given by the parameters u10 and u20 , and U be a neighbourhood of P , given by a subset DU of D. The image of U under the spherical Gauss map defines a part U ∗ of the unit sphere. If AU and AU ∗ denote the surface areas of U and U ∗ – including their signs –, then it follows that 





1 2 ⃗ i ⃗ i ⃗ i Du N (u ) • N1 (u ) × N2 (u ) du du

AU ∗ =  . ⃗ i AU x1 (ui ) × ⃗x2 (ui )) du1 du2 Du N (u ) • (⃗

(2.158)

The ratio in (2.158) is the bigger, the more the surface S is curved, since then the surface normal vectors cover a larger area of the unit sphere (Figure 2.105).

Surfaces in Three-Dimensional Euclidean Space ■ 239

Figure 2.103

A curve on a monkey saddle and its image under the spherical Gauss

Figure 2.104

A curve on a conoid and its image under the spherical Gauss map

map

If the neighbourhood U of P shrinks to the point P , that is, if we take the limit U → {P } in (2.158), then we obtain using (2.149) in Theorem 2.10.7 



 ⃗ 2 (ui ) ⃗ 1 (ui ) × N ⃗ (ui ) • N N ⃗ (ui ) g(ui )K(ui ) ⃗ (ui ) • N 0 0 0 AU ∗ N 0 0 0  lim =   = i i i ⃗ U →{P } AU i N (u0 ) • ⃗x1 (u0 ) × ⃗x2 (u0 ) g(u0 )

= K(ui0 ).

(2.159)

The value in (2.159) only depends on the point P on S and was originally introduced by Gauss as the measure of the curvature of a surface at a point. The sign of K which determines the elliptic, parabolic or hyperbolic property of P , has the following geometric significance: K is positive or negative depending on whether the Gauss map preserves or reverses the orientation (Figure 2.106).

240 ■ Differential Geometry and Its Visualization

Figure 2.105

The spherical images of the same domain of two different paraboloids

Visualization 2.10.14. (a) We verify (2.159) without the use of the above results in the case of a torus with a parametric representation (Visualization 2.3.8) ⃗x(ui ) = {r(u1 ) cos u2 , r(u1 ) sin u2 , h(u1 )} where





(u1 , u2 ) ∈ D = (0, 2π)2 ,

r(u1 ) = (r0 + r1 ) cos u1 and h(u1 ) = r1 sin u1 ,

and r0 and r1 are constants with 0 < r1 < r0 . Furthermore, we choose a point P0 on the torus given by the parameters (u10 , u20 ) ∈ D and a neighbourhood U of P0 defined by the rectangle R = [u10 − w1 , u10 + w1 ] × [u20 − w2 , u20 + w2 ], where w1 and w2 are positive reals and R ⊂ D (Figure 2.107). Then the surface area of U is, since g(ui ) = r12 (r0 + r1 cos u1 )2 by (2.30), AU =

  U

g(ui ) du1 du2 = 

u10 −w1

r1 (r0 + r1 cos u1 ) du1 du2

u10 −w1 u20 −w2

u10 +w1

= 2r1 w2

u10+w1 u20+w2

(r0 + r1 cos u1 ) du1 



= 4r0 r1 w1 w2 + 2r12 w2 sin (u10 + w1 ) − sin (u10 + w1 ) = 4r0 r1 w1 w2 + 4r12 w2 sin w1 cos u10 



= 4r1 w2 r0 w1 + r1 sin w1 cos u10 .

Surfaces in Three-Dimensional Euclidean Space ■ 241

Figure 2.106 Curves and their orientations in the neighbourhoods of points with K > 0, K = 0 and K < 0 and their images under the Gauss map

Figure 2.107

Neighbourhoods of an elliptic and hyperbolic point on a torus





Since g(u1 ) = r1 (r0 + r1 cos u1 ), we have K(u1 ) = cos u1 / g(u1 ) by (2.118) in Visualization 2.6.4 (d), and it follows for the surface area of the image U ∗ under the Gauss map by (2.149) in Theorem 2.10.7 AU ∗ =



⃗ 1 (u1 ) × N ⃗ 2 (u2 )) du1 du2 ⃗ (u1 ) • (N N

=





U∗

U∗

= 2w Thus it follows that

2

g(u1 )K(ui ) du1 du2 = 2w2

(sin (u10

1

+w )−

sin w1 cos u10 AU ∗ = = AU r1 (r0 w1 + r1 sin w1 cos u10 ) −→

sin (u10

1

u10+w1

cos u1 du1

u10 −w1

− w )) = 4w2 sin w1 cos u10 . cos u10

r1 (r0

w1 + r1 cos u10 ) sin w1

cos u10 = K(u10 ) for (w1 , w2 ) → (0, 0). r1 (r0 + r1 cos u10 )

242 ■ Differential Geometry and Its Visualization

Figure 2.108

Circular neighbourhoods of zero on a hyperbolic paraboloid

(b) Now we verify (2.159) without the use of the above results in the case of the hyperbolic paraboloid with a parametric representation ⃗x(ui ) = {u1 , u2 , u1 u2 }





(u1 , u2 ) ∈ D = R2 .

We choose the point P0 on the hyperbolic paraboloid given by the parameters (u10 , u20 ) = (0, 0) ∈ D and a neighbourhood Ur (r > 0) of P0 , given by the circle line in the parameter plane with radius r and centre in (0, 0) (Figure 2.108). It follows from ⃗x1 (ui ) = {1, 0, u2 }, ⃗x2 (ui ) = {0, 1, u1 }, ⃗x12 = {0, 0, 1}, ⃗x11 (ui ) = ⃗x22 (ui ) = ⃗0, ⃗n(ui ) = ⃗x1 (ui ) × ⃗x2 (ui ) = {−u2 , −u1 , 1}, g11 (ui ) = 1 + (u2 )2 , g12 (ui ) = u1 u2 , g22 (ui ) = 1 + (u1 )2 , ⃗ (ui ) = ⃗n(ui )  1 , g(ui ) = 1 + (u1 )2 + (u2 )2 , N g(ui )

⃗ (ui ) • ⃗xkk (ui ) = 0 (k = 1, 2) and L12 (ui ) = N ⃗ (ui ) • ⃗x12 (ui ) =  1 Lkk (ui ) = N g(ui )

that the Gaussian curvature of the hyperbolic paraboloid is given by K(ui ) =

L212 (ui ) 1 L(ui ) = − =− , i i g(u ) g(u ) (1 + (u1 )2 + (u2 )2 )2

hence K(0, 0) = −1. We put u1 = u1 (ρ, ϕ) = ρ cos ϕ and u2 = u2 (ρ, ϕ) = ρ sin ϕ for (ρ, ϕ) ∈ (0, r) × (0, 2π)

Surfaces in Three-Dimensional Euclidean Space ■ 243

and obtain for the surface area of Ur AUr =

  Ur

g(ui ) du1 du2



2π = 3 Furthermore, we have

=

2π r 

ρ 1 + ρ2 dρdϕ

0

3 r  1 + ρ2  

ρ=0

0

  3 2π √ 2 = 1+r −1 . 3

⃗ 1 (ui ) = {0, −1, 0}  1 + α1 (ui )⃗n(ui ) N g(ui )

and

⃗ 2 (ui ) = {−1, 0, 0}  1 + α2 (ui )⃗n(ui ), N g(ui )

where

∂ αk (u ) = ∂uk i



and it follows that 



1  g(ui )

⃗ (u ) • N ⃗ 2 (u ) = N ⃗ (u ) • ⃗ 1 (u ) × N N i

i

i

and A

Ur∗

=

 ur



i





for k = 1, 2,



ρ=0



1 × {−1, 0, 0}  i + α2 (ui )⃗n(ui ) g(u ) ⃗ (ui ) 1 • ({0, 1, 0} × {1, 0, 0}) = −  1  , =N  3 g(ui ) g(ui ) 

⃗ 2 (ui ) du1 du2 = − ⃗ (ui ) • N ⃗ 1 (ui ) × N N

r 2π  =   1 + ρ2 



1 {0, −1, 0}  i + α1 (ui )⃗n(ui ) g(u )

  √ 2π 1 − 1 + r2 √ = . 1 + r2

2π r 0

0



Therefore, we obtain √ √ AUr∗ 1 + r2 − 1 1 + r2 − 1   = −3 √ = −3 √  √ 3 AUr (1 + r2 )2 − 1 + r2 1 + r2 1 + r2 − 1

ρ 1 + ρ2

3 dρdϕ

244 ■ Differential Geometry and Its Visualization

r2 r2 + o(r3 ) − 1 + o(r3 ) 2 2 = −3 2 = −1 + o(1) (r → 0), = −3 r2 3r 2 3 3 3 + o(r ) + o(r ) 1 + 2r + o(r ) − 1 − 2 2 1+

hence lim

r→0

AUr∗ = −1 = K(0, 0). AUr

CHAPTER

3

The Intrinsic Geometry of Surfaces

In this chapter, we study the intrinsic geometry of surfaces. By this, we mean the geometric properties of surfaces that depend on measurements on the surface itself, that is, on the first fundamental coefficients and their derivatives only. The most important topics and subjects in this chapter are • the Christoffel symbols of first and second kind and the geodesic curvature in Section 3.1 • Liouville’s theorem, Theorem 3.1.12 in Section 3.1

• geodesic lines on surfaces with orthogonal parameters and on surfaces of revolution in Sections 3.3 and 3.4 • the minimum property of geodesic lines, Theorem 3.5.2 in Section 3.5 • orthogonal and geodesic parameters in Section 3.6

• geodesic parallel and geodesic polar coordinates in Section 3.6 • Levi–Civit`a parallelism in Section 3.7

• the derivation formulae by Gauss, and Mainardi and Codazzi, Theorem 3.8.1 in Section 3.8 • Theorema egregium, Theorem 3.8.3 in Section 3.8

• conformal, isometric and area preserving maps of surfaces in Section 3.9 • the Gauss–Bonnet theorem, Theorem 3.10.1 in Section 3.10

• minimal surfaces and the Weierstrass formulae, (3.158) in Section 3.11.

DOI: 10.1201/9781003370567-3

245

246 ■ Differential Geometry and Its Visualization

3.1

THE CHRISTOFFEL SYMBOLS

In Section 2.4, we introduced the geodesic and normal curvature κg and κn of a curve γ on a surface. This was done by splitting the vector ⃗x¨ of curvature of γ into two components ⃗ (s), ⃗x¨(s) = κg (s)⃗t(s) + κn (s)N (3.1) ⃗ (s) denotes the normal vector of the surface at s, and where N ⃗ (s) × ⃗x˙ (s) ⃗t(s) = N

(3.2)

is a vector in the tangent plane of the surface at s. In order to find κn and κg , we introduced the functions Γrik and the second fundamental coefficients Lik by (Section 2.4) ⃗ (i, k = 1, 2). ⃗xik = Γrik ⃗xr + Lik N (3.3) The remainder of Chapter 2 was mainly devoted to the study of the normal curvature and various concepts arising in this context. Now we are going to take a closer look at the geodesic curvature of curves on a surface. The first step will be to give some geometric meaning to the coefficient functions Γrik that have only been formally introduced. We start with a result for the geodesic curvature of a curve which is similar to Meusnier’s theorem, Theorem 2.5.7, for the normal curvature. Theorem 3.1.1. The geodesic curvature of a curve on a surface is equal to its curvature multiplied by the Cosine of the angle between the binormal vector of the curve and the normal vector of the surface, that is, the Cosine of the angle between the osculating plane of the curve and the tangent plane of the surface (Figure 3.1). Proof. Let γ be a curve on a surface given by a parametric representation ⃗x(s) = ⃗x(ui (s)) (s ∈ I), where s denotes the arc length along γ, and let ⃗vk be the vectors of the trihedra of γ. Then we have ⃗ × ⃗v1 ) = κN ⃗ • (⃗v1 × ⃗v2 ) = κN ⃗ • ⃗v3 = κ cos α. κg = ⃗x¨ • ⃗t = κ⃗v2 • (N Definition 3.1.2. Let S be a surface with parametric representation ⃗x(uj ) ((u1 , u2 ) ∈ D ⊂ R2 ) and first fundamental coefficients gjk . We recall that the inverse of the matrix (glr ) is denoted by (g lr ). (a) The functions [ikl] : D → R with [ikl] = ⃗xik • ⃗xl (i, k, l = 1, 2) are called the first Christoffel symbols of the surface S.

The Intrinsic Geometry of Surfaces ■ 247

Figure 3.1 The relation between the curvature and geodesic curvature of a curve on a surface (Theorem 3.1.1)

(b) The functions

r ik

: D → R with 

r ik



= g lr [ikl] (i, k, r = 1, 2)

are called the second Christoffel symbols of the surface S. Example 3.1.3 (The Christoffel symbols for explicit surfaces). Let f : D ⊂ R2 → R be a function of class C r (D) for r ≥ 2 and ES be the explicit surface with a parametric representation ⃗x(ui ) = {u1 , u2 , f (u1 , u2 )}





(u1 , u2 ) ∈ D .

Then the first and second Christoffel symbols of ES are given by [ijk] = [ijk](u1 , u2 ) = ⃗xij (u1 , u2 ) • ⃗xk (u1 , u2 ) = fij (ui , u2 )fk (u1 , u2 )

(3.4)

for i, j, k = 1, 2, and 

i jk



=

fi fjk for i, j, k = 1, 2. 1 + f12 + f22

(3.5)

248 ■ Differential Geometry and Its Visualization

Proof. (i) First we prove (3.4). We have for the first Christoffel symbols by Definition 3.1.2 (a) and Example 2.3.4 [ijk] = [ijk](u1 , u2 ) = ⃗xij (u1 , u2 ) • ⃗xk (u1 , u2 ) = fij (ui , u2 )fk (u1 , u2 ) for i, j, k = 1, 2, that is, (3.4) holds. (ii) Now we prove (3.5). It follows from Example 2.3.4 that g 11 =

1 + f22 f1 f2 g22 g12 = =− , g 12 = − g 1 + f12 + f22 g 1 + f12 + f22

and g 22 =

1 + f12 g11 = . g 1 + f12 + f22

Now we obtain by Definition 3.1.2 (b), omitting the arguments u1 and u2 , for the second Christoffel symbols 

1 jk



= g m1 [jkm] =

and similarly

1





(1 + f 2 )2 fjk f1 − f1 f2 fjk f2 =

1+

f12

+

f22





=

f2 fjk for j, k = 1, 2. 1 + f12 + f22

2 jk

f1 fjk , 1 + f12 + f22

Thus we have shown that (3.5) holds.

It is useful to establish some important formulae for the first and second Christoffel symbols. Lemma 3.1.4. The following formulae hold for the first and second Christoffel symbols. (a) (b) (c) (d)

[ikl] = [kil] (i, k, l = 1, 2)    l l = (i, k, l = 1, 2) ik ki   1 ∂gil ∂gik ∂gkl − + (i, k, l = 1, 2) [ikl] = 2 ∂uk ∂ul ∂ui ∂gik = [ilk] + [kli] (i, k, l = 1, 2) ∂ul



With respect to new parameters u∗k (k = 1, 2) given by uj = uj (u∗k ) = uj (u∗1 , u∗2 )

(j = 1, 2),

The Intrinsic Geometry of Surfaces ■ 249

we have

(f) (g)

∂u∗k ∂u∗l j m  ∂u ∂u  α ∂u ∂uβ ∂ 2 uα ∂uγ ∗ [ikl] = [αβγ] ∗i ∗k + gαγ ∗i ∗k ∂u ∂u ∂u ∂u ∂u∗l g ∗kl = g jm

(e)



r ik

∗

=



γ αβ



∂uα ∂uβ ∂ 2 uγ + ∂u∗i ∂u∗k ∂u∗i ∂u∗k



∂u∗r ∂uγ

(k, l = 1, 2) (i, k, l = 1, 2) (i, k, r = 1, 2).

Proof. (a,b) Parts (a) and (b) are immediate consequences of the definitions of the first and second Christoffel symbols in Definition 3.1.2 and the fact that ⃗xik = ⃗xki (i, k = 1, 2). (c) We obtain from gil = ⃗xi • ⃗xl ∂gil = ⃗xik • ⃗xl + ⃗xlk • ⃗xi = [ilk] + [lki] ∂uk and similarly −

∂gik = −⃗xil • ⃗xk − ⃗xkl • ⃗xi = −([ilk] + [kli]) ∂ul

and ∂gkl = ⃗xki • ⃗xl + ⃗xli • ⃗xk = [ilk] + [kli], ∂ui and so by Part (a).

∂gil ∂gik ∂gkl − + = 2[ikl] ∂uk ∂ul ∂ui

(d) Part (d) is the third identity in Part (c) of the proof. (e) We obtain from the formula (2.18) of transformation for the first fundamental coefficients by interchanging u and u∗ ∗ gmr = gnt

∂u∗n ∂u∗t (m, r = 1, 2) ∂um ∂ur

and ∗l r ∂u∗l ∂uj ∗ks ∂u ∂u = g δj ∂uj ∂u∗s ∂uj ∂u∗s r ∗l r ∗n ∂u∗l ∂ur ∂u∗t mj mj ∗ks ∂u ∂u ∗ ∂u = g ∗ks j g g = g g g mr ∂u ∂u∗s ∂uj ∂u∗s nt ∂um ∂ur ∗t ∗l ∗n ∂ur ∂u∗l ∂u∗n mj ∗ ∂u mj ∗ks ∗ ∗t ∂u ∂u = g ∗ks gnt g = g g g δ nt s ∂ur ∂u∗s ∂uj ∂um ∂uj ∂um

g ∗kl = g ∗ks δs∗l = g ∗ks

250 ■ Differential Geometry and Its Visualization ∗l ∗n ∂u∗l ∂u∗n mj ∗k ∂u ∂u = g δ n ∂uj ∂um ∂uj ∂um ∗l ∗k ∗k ∗l ∂u ∂u ∂u ∂u = g mj j = g jm (k, l = 1, 2). ∂u ∂um ∂uj ∂um ∗ = g mj g ∗ks gns

(f) From ⃗x∗i = ⃗xα and 

∂uα (i = 1, 2) ∂u∗i



∂ ∂ ∂uα ∂uα ∂ ⃗ x = (⃗ x ) + ⃗xα ∗k α α ∗k ∗i ∗k ∗i ∂u ∂u ∂u ∂u ∂u α β 2 α ∂u ∂u ∂ u = ⃗xαβ ∗i ∗k + ⃗xα ∗i ∗k (i, k = 1, 2), ∂u ∂u ∂u ∂u

⃗x∗ik =



∂uα ∂u∗i



we obtain [ikl] = ∗

⃗x∗ik 



∂uα ∂uβ ∂ 2 uα = ⃗xαβ ∗i ∗k + ⃗xα ∗i ∗k ∂u ∂u ∂u ∂u

• ⃗x∗l

∂ 2 uα ∂uα ∂uβ = [αβγ] ∗i ∗k + gαγ ∗i ∗k ∂u ∂u ∂u ∂u





• ⃗xγ

∂uγ ∂u∗l

∂uγ (i, k, l = 1, 2). ∂u∗l

(g) Finally, we obtain from Parts (e) and (f) 

r ik

∗

= g ∗rl [ikl]∗ =g

∂u∗l ∂uϕ ∂uϵ

ϕϵ ∂u

∗r





∂ 2 uα ∂uα ∂uβ [αβγ] ∗i ∗k + gαγ ∗i ∗k ∂u ∂u ∂u ∂u

∂uγ ∂u∗l ϕϵ ∂uα ∂uβ ∂ 2 uα = g [αβγ] + g αγ ∂u∗l ∂uϵ ∂u∗i ∂u∗k ∂u∗i ∂u∗k =

δϵγ g ϕϵ 



∂uα ∂uβ ∂ 2 uα [αβγ] ∗i ∗k + gαγ ∗i ∗k ∂u ∂u ∂u ∂u



∂uα ∂uβ ∂ 2 uα = g [αβγ] ∗i ∗k + g ϕγ gαγ ∗i ∗k ∂u ∂u ∂u ∂u = =

ϕγ

 

ϕ αβ γ αβ

 

∂uα ∂uβ ∂ 2 uϕ + ∂u∗i ∂u∗k ∂u∗i ∂u∗k ∂ 2 uγ ∂uα ∂uβ + ∗i ∗k ∗i ∗k ∂u ∂u ∂u ∂u

 





∂uγ ∂u∗l ∂u∗r ∂uϕ

∂u∗r ∂uϕ



∂u∗r ∂uϕ

∂u∗r ∂uϕ ∂u∗r (r, i, k = 1, 2), ∂uγ

where we changed the index of summation ϕ to γ in the last step.

We obtain as an immediate consequence of Lemma 3.1.4

The Intrinsic Geometry of Surfaces ■ 251

Corollary 3.1.5. We obtain for i ̸= k (no summation)

1 ∂gii 1 ∂gii , [iki] = [kii] = , 2 ∂ui 2 ∂uk

[iii] = 1 [iik] = 2  

i ii



∂gii ∂gik ∂gik − k + i ∂u ∂u ∂ui



=

∂g12 1 ∂gii − , ∂ui 2 ∂uk

= g ri [iir] = g ii [iii] + g ik [iik] 





1 ∂gii ∂g12 1 ∂gii 1 = gkk − g12 − i g 2 ∂u ∂ui 2 ∂uk    1 ∂g12 ∂gii ∂gii = gkk i + g12 −2 i , 2g ∂u ∂uk ∂u

 

k ii

= g lk [iil] = g kk [iik] + g ik [iii] 







1 ∂g12 1 ∂gii ∂gii = gii − − g12 i i k g ∂u 2 ∂u ∂u     1 ∂g12 ∂gii ∂gii = gii 2 i − k − g12 i 2g ∂u ∂u ∂u

and 

k ik



=



k ki 



= g rk [ikr] = g kk [ikk] + g ik [iki] 

1 1 ∂gkk 1 ∂gii gii − g12 = i g 2 ∂u 2 ∂uk   1 ∂gkk ∂gii = gii − g . 12 2g ∂ui ∂uk

For orthogonal parameters (Definition 2.3.1 (c)), where g12 = 0 and g = g11 g22 , we have for i ̸= k 1 ∂gii 1 ∂gii 1 ∂gii , [iik] = − , [iki] = [kii] = , i k 2 ∂u 2 ∂u  2 ∂uk  √     ∂ log gii 1 ∂gii 1 ∂gii i k = = , =− ii ii 2gii ∂ui ∂ui 2gkk ∂uk [iii] =

(3.6) (3.7)

and 

k ik



=



k ki



 √  ∂ log gkk 1 ∂gkk = = . 2gkk ∂ui ∂ui

(3.8)

If the parameters are orthogonal and the first fundamental coefficients satisfy gkk (ui ) = gkk (u1 ) for k = 1, 2,

(3.9)

252 ■ Differential Geometry and Its Visualization

then it easily follows from (3.7) and (3.8) that 











1 1 1 1 dg11 1 dg22 = , =− = 0, = 1 1 11 22 12 2g11 du 2g11 du         1 dg22 2 2 2 2 = = 0 and = = . 11 22 12 21 2g22 du1



1 21



= 0, (3.10)

Remark 3.1.6. By Definition 3.1.2 and Lemma 3.1.4, the Christoffel symbols only depend on the first fundamental coefficients and their derivatives. Quantities in the theory of surfaces that depend on the first fundamental coefficients only are called quantities of the internal or intrinsic geometry. Example 3.1.7 (The Christoffel symbols for surfaces of revolution). Let RS be a surface of revolution with a parametric representation ⃗x(ui ) = {r(u1 ) cos u2 , r(u1 ) sin u2 , h(u1 )}.

Then the parameters are orthogonal, hence g12 = 0. We also have gii (u1 , u2 ) = gii (u1 ) (i = 1, 2) and consequently ∂gii = 0. ∂u2 It follows from (3.6) that 1 ∂g11 1 ∂g11 = r′ (u1 )r′′ (u1 ) + h′ (u1 )h′′ (u1 ), [112] = − = 0, 1 2 ∂u 2 ∂u2 1 ∂g22 1 ∂g11 = 0, [222] = =0 [121] = [211] = 2 ∂u2 2 ∂u2 [111] =

and 1 ∂g22 [221] = − · = −r(u1 )r′ (u1 ). 2 ∂u1 Furthermore, we have from (3.7) 



1 ∂g11 r′′ (u1 )r′ (u1 ) + h′′ (u1 )h′ (u1 ) 1 = = , 11 2g11 ∂u1 (r′ (u1 ))2 + (h′ (u1 ))2     1 ∂g11 1 1 = = = 0, 12 21 2g11 ∂u2   1 1 ∂g22 r′ (u1 )r(u1 ) =− =− ′ 1 2 , 1 22 2g11 ∂u (r (u )) + (h′ (u1 ))2   1 ∂g11 2 =− = 0, 11 2g22 ∂u2     1 ∂g22 r′ (u1 ) 2 2 = = = 12 21 2g22 ∂u1 r(u1 ) and 

2 22



=

1 ∂g22 = 0. 2g22 ∂u2

The Intrinsic Geometry of Surfaces ■ 253 1

In particular, it follows for the pseudo-sphere, where g11 = 0 and r(u1 ) = e−u , that 















1 1 1 1 1 = = = 0, = e−2u , 11 12 21 22         2 2 2 2 = = 0 and = = −1. 11 22 12 21

(3.11) (3.12)

Theorem 3.1.8. Let S be a surface with a parametric representation ⃗x(ui ) and γ a curve on S given by a parametric representation ⃗x(ui (s)), where s is the arc length along γ. We put ⃗ • (⃗xm × ⃗xi ) for m, i = 1, 2, ϵmi = N that is,

ϵ11 = ϵ22 and ϵ12 = −ϵ21 = ∥⃗x1 × ⃗x2 ∥ = g.

Then the geodesic curvature of γ is given by κg = ϵmi u˙

m



u¨ + i







i u˙ j u˙ k . jk

(3.13)

Thus the geodesic curvature of a curve is a quantity of the internal geometry. Proof. We know from the second identity in (2.46) of Remark 2.4.3 (c) that 



κg = u¨r + Γrjk u˙ j u˙ k ⃗xr • ⃗t for r = 1, 2, ⃗ × ⃗x˙ . Now where ⃗t = N

(3.14)

⃗ × ⃗xl )u˙ l = N ⃗ • (⃗xl × ⃗xr )u˙ l = ϵlr u˙ l ⃗xr • ⃗t = ⃗xr • (N for r = 1, 2 implies





κg = ϵlr u˙ l u¨r + Γrjk u˙ j u˙ k .

Furthermore, it follows from (3.3) that

⃗ for j, k = 1, 2, ⃗xjk = Γijk ⃗xi + Ljk N so ⃗xjk • ⃗xl = [jkl] = Γijk ⃗xi • ⃗xl = Γijk gil for j, k, l = 1, 2,     r for r, j, k = 1, 2 Γrjk = δir Γijk = g mr gim Γijk = g mr [jkm] = jk

and consequently



κg = ϵmi u˙ m u¨i +







i u˙ j u˙ k . jk

(3.15)

254 ■ Differential Geometry and Its Visualization

Figure 3.2

Representation of −0.9κg along a loxodrome (dashed) on the pseudo-sphere

We use the formula in (3.13) of Theorem 3.1.8 to find the geodesic curvature of a curve on a surface of revolution. We may use the same principle as in the case of the normal curvature in (2.58) in Section 2.5 to represent the geodesic curvature along a curve. If S is a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D ⊂ R2 ) and γ is a curve on S with a parametric representation ⃗x(t) = ⃗x(ui (t)) (t ∈ I) and geodesic curvature κg (t) = κg (ui (t)), then we may represent κg (t) by the curve γg , given by a parametric representation ⃗x ∗ (t) = ⃗x(t) + κg (t)⃗t(t) (t ∈ I),

⃗ × ⃗v1 . Writing u ∗1 = t, we see that γg is a curve on the ruled surface RS where ⃗t = N that has a parametric representation ⃗x ∗ (u ∗i ) = ⃗y (u ∗1 ) + u ∗2⃗z(u ∗1 )





(u∗ 1 , u ∗2 ) ∈ I × R ,

(3.16)

where ⃗y (u ∗1 ) = ⃗x(ui (u ∗1 )) and ⃗z(u ∗1 ) = ⃗t(ui (u ∗1 )), and γg considered as a curve on RS is given by putting u ∗2 = κg (ui (u ∗1 )) in (3.16) (Figures 3.2–3.5 ). Visualization 3.1.9. (Geodesic curvature of certain curves on surfaces of revolution) Let RS be a surface of revolution with a parametric representation ⃗x(ui ) = {r(u1 ) cos u2 , r(u1 ) sin u2 , h(u1 )}





(u1 , u2 ) ∈ D = I1 × I2 ⊂ R × (0, 2π) ,

where we assume, as always, that r(u1 ) > 0 and |r′ (u1 )|+|h1 (u1 )| > 0 on I1 . Furthermore, let γ be a curve on RS given by a parametric representation ⃗x(ui (s)) (s ∈ I),

The Intrinsic Geometry of Surfaces ■ 255

Figure 3.3

Representation of −3κg along a loxodrome (dashed) on the unit sphere

where I is such that u1 (I) ⊂ I1 and s denotes the arc length along γ. Then the geodesic curvature κg along γ is given by (3.13) and the formulae for the second Christoffel symbols in Example 3.1.7 κg (s) = κg (u1 (s), u2 (s))









 i  1 = ϵmi u (s), u (s) u˙ (s) u¨ (s) + u (s), u2 (s) u˙ j (s)u˙ k (s) jk     ′ 1 r (u (s)) = g(u1 (s)) u˙ 1 (s) u¨2 (s) + 2 u˙ 1 (s)u˙ 2 (s) r(u1 (s)) 

1

2



m



2

−u˙ (s) u¨1 (s) +

i

r′′ (u1 (s))r′ (u1 (s)) + h′′ (u1 (s))h′ (u1 (s)) 1 (u˙ (s))2 (r′ (u1 (s)))2 + (h′ (u1 (s)))2

r′ (u1 (s))r(u1 (s)) − ′ 1 (u˙ 2 (s))2 (r (u (s)))2 + (h′ (u1 (s)))2



.

If we assume that u1 is the arc length along the curve that generates the surface of revolution RS, then we have (r′ (u1 (s))2 + (h′ (u1 (s))2 = 1 and consequently r′′ (u1 (s))r′ (u1 (s)) + h′′ (u1 (s))h′ (u1 (s)) = 0, hence 





r′ (u1 (s)) 1 u˙ (s)u˙ 2 (s) κg (s) = r(u (s)) u˙ (s) u¨ (s) + 2 r(u1 (s)) 1

1

2



− u˙ 2 (s) u¨1 (s) − r′ (u1 (s))r(u1 (s))(u˙ 2 (s))2



.

(3.17)

256 ■ Differential Geometry and Its Visualization

Figure 3.4 The ruled surface generated by a loxodrome on the unit sphere and the vectors ⃗t(s)

As a special case, we consider a loxodrome λβ on RS, where β ∈ (0, π/2) is the constant angle between the loxodrome λβ and the meridians of RS. Then λβ is given by (2.32) in Visualization 2.3.11 2

1

u (u ) = tan β

 

g11 (u1 ) 1 du = tan β g22 (u1 )



1 du1 . r(u1 )

Writing u1 (t) = t and c = tan β, we obtain a parametric representation ⃗x ∗ (t) = ⃗x(t, u2 (t)) for λβ , so  2  ∗ 2 2  d⃗  x (t) du (t) c2   = g11 (t) + g22 (t) = 1 + g (t) = 1 + c2 , 22  dt  dt g22 (t)

√ 2 hence the arc length √ along λβ is given by s(t) = 1 + c t. Writing c1 = 1/ 1 + c2 , we obtain a parametric representation of λβ with respect to its arc length s 



⃗x(s) = ⃗x ∗ c1 s, u2 (u1 ((c1 s))) = ⃗x ∗ (u∗1 (s), u∗2 (s)), and it follows that

du1 dt = c1 , u¨∗1 (s) = 0, dt ds du2 cc1 u∗2 (s) = u2 (u1 (t(s))), u˙ ∗2 (s) = 1 u˙ ∗1 (s) = du r(u∗1 (s))

u∗1 (s) = u1 (t(s)) = u1 (c1 s), u˙ ∗1 (s) =

The Intrinsic Geometry of Surfaces ■ 257

The ruled surface generated by a loxodrome on a pseudo-sphere and the ⃗ vectors t(s)

Figure 3.5

and u¨∗2 (s) =

′ ∗1 d2 u2  ∗1 2 du2 ∗1 2 r (u (s)) u ˙ + u ¨ = −cc . 1 d(u1 )2 du1 r2 (u∗1 (s))

Thus we obtain for the geodesic curvature κ∗g along λβ from (3.17) κ∗g (s)





= r(u (s)) u˙ (s) − ∗1

∗1

cr′ (u∗1 (s))  ∗1 2 u˙ (s) r2 (u∗1 (s))



r′ (u∗1 (s)) ∗1 c u˙ (s) u˙ ∗1 (s) +2 ∗ ∗1 r(u (s)) r(u (s))  2  c ∗1 ′ ∗1 ∗1 ∗1 + u ˙ (s)r (u (s))r(u (s)) u ˙ (s) r(u∗1 (s)) 



= c u˙ ∗1 (s)





′ ∗1   r′ (u∗1 (s)) ∗1 3 r (u (s)) 2 ∗1 + r(u (s)) = cc 1 + r (u (s)) . 1 r(u∗1 (s)) r(u∗1 (s))

In particular, we obtain for the unit sphere where r(u1 ) = cos u1 



κ∗g (s) = −cc31 tan (c1 s) 1 + cos2 (c1 s) , and for the pseudo-sphere with

r(u1 ) = exp(−u1 ) and h(u1 ) =

 

1 − exp (−2u1 ) du1

κ∗g (s) = −cc31 (1 + exp (−2c1 s)) .

258 ■ Differential Geometry and Its Visualization

Finally, we have for the geodesic curvature κg along the loxodrome λβ with respect to the parameter t (Figures 3.2 and 3.3) 



κg (t) = −cc31 tan t 1 + cos2 t

and κg (t) = −cc31 (1 + exp (−2t)) .

Now we establish formulae for the geodesic curvature of the parameter lines of surfaces. Lemma 3.1.10. The geodesic curvature κgm of a um –line on a surface S with a parametric representation ⃗x(ui ) is given by √   g k κgm = (−1)m+1 for k ̸= m. (3.18) (gmm )3/2 mm

In particular, if the parameters are orthogonal and sk denotes the arc length along the uk –lines then d √ (log gmm ) for k ̸= m. (3.19) κgm = (−1)m dsk Proof. Let s1 denote the arc length along a u1 -line of S. Since u2 = const along the u1 –lines, the formula in (3.13) for the geodesic curvature κg1 of the u1 –lines reduces to     2 2 √ 1 1 2 κg1 = ϵ12 u˙ (u˙ ) = g (u˙ 1 )3 . 11 11 As s1 is the arc length along the u1 –line, it follows that g11 (u˙ 1 )2 = 1, hence √   g 2 . κg1 = (g11 )3/2 11 Similarly, we obtain for the geodesic curvature κg2 of a u2 –line √   g 1 κg2 = − . 3/2 22 (g22 )

Thus we have proved (3.18). If the parameters are orthogonal then it follows from (3.7) and (3.8) that √   g22 1 ∂g11 1 1 ∂g11 κg1 = − =− √ 2 g11 2g22 ∂u 2 g22 g11 ∂u2

(3.20)

and κg2 = Finally, we obtain

and

√

g11 1 ∂g22 1 1 ∂g22 = √ . 1 g22 2g11 ∂u 2 g11 g22 ∂u1

 √  d log g11 1 ∂g11 du2 1 1 ∂g11 = = √ ds2 2g11 ∂u2 ds2 2 g22 g11 ∂u2  √  d log g22 1 ∂g22 du1 1 1 ∂g22 = = √ . ds1 2g22 ∂u1 ds1 2 g11 g22 ∂u1

Thus we have shown (3.19).

(3.21)

The Intrinsic Geometry of Surfaces ■ 259

Visualization 3.1.11. (Geodesic curvature of parameter lines on explicit surfaces) We apply (3.18) in Lemma 3.1.10 to compute the geodesic curvature along the parameter lines of an explicit surface with a parametric representation 

⃗x(ui ) = {u1 , u2 , f (u1 , u2 )}



(u1 , u2 ) ∈ D) .

First we consider the u1 –line corresponding to u20 . We obtain for the geodesic curvature κg1 along this u1 –line by (3.18) and (3.5) in Example 3.1.3 κg1 (t) = κg1 (t, u0 ) = = =





g(t, u20 )

(g11 (t, u20 ))3/2



2 11

1 + (f1 (t, u20 ))2 + (f2 (t, u20 ))2



·

3/2

f2 (t, u20 )f11 (t, u20 ) 1 + (f1 (t, u20 ))2 + (f2 (t, u20 ))2

(1 + (f1 (t, u20 ))2 ) f2 (t, u20 )f11 (t, u20 ) 3/2

(1 + (f1 (t, u20 ))2 )



1 + (f1 (t, u20 ))2 + (f2 (t, u20 ))2

.

Similarly, we obtain for the geodesic curvature κg2 along the u2 –line corresponding to u10 κg2 (t) = κg2 (u10 , t) = −

f1 (u10 , t)f22 (u10 , t)

3/2

(1 + (f2 (u10 , t))2 )



1 + (f1 (u10 , t))2 + (f2 (u20 , t))2

.

As a special case, we consider the hyperbolic paraboloid as an explicit surface given by f (u1 , u2 ) = (u1 )2 − (u2 )2 ((u1 , u2 ) ∈ R 2 ). Then we have f1 (ui ) = 2u1 , f2 (ui ) = −2u2 , f11 (ui ) = −f22 (ui ) = 2 and the geodesic curvature along the u1 –line corresponding to u20 and the u2 –line corresponding to u10 are given by (Figure 3.6) κg1 (t) = −

4u20

(1 + 4t2 )3/3

and κg2 (t) =



1 + 4(t2 + (u20 ))

4u10

(1 + 4t2 )3/3



1 + 4(t2 + (u21 ))

.

The next result expresses the geodesic curvature of a curve on a surface with orthogonal parameters in terms of the geodesic curvature of the parameter lines and the angle between the curve and the u1 –lines; it is the analogue for the geodesic curvature of Euler’s theorem, Theorem 2.5.31, for the normal curvature.

260 ■ Differential Geometry and Its Visualization

Figure 3.6

Geodesic curvature along the parameter lines of a hyperbolic paraboloid

Theorem 3.1.12 (Liouville). Let S be a surface with a parametric representation ⃗x(ui ) and orthogonal parameters, γ be a curve on S with a parametric representation ⃗x(ui (s)), where s is the arc length along γ, and Θ(s) be the angle between γ and the u1 –lines of S. Furthermore, let κgi denote the geodesic curvature of the ui –lines (i = 1, 2). Then we have dΘ κg = + κg1 cos Θ + κg2 sin Θ. ds Proof. Since s is the arc length along γ and g12 = 0, we have cos Θ = hence

⃗x˙ • ⃗x2 √ ⃗x˙ • ⃗x1 √ = g11 u˙ 1 and sin Θ = = g22 u˙ 2 , ∥⃗x1 ∥ ∥⃗x2 ∥ cos Θ sin Θ u˙ 1 = √ and u˙ 2 = √ . g11 g22

We put  √ √  A = g u˙ 1 u¨2 − u˙ 2 u¨1 = g

=

√



cos Θ d √ g11 ds



sin Θ √ g22





sin2 Θ dΘ d cos2 Θ dΘ +√ + u˙ 1 sin Θ g √ g11 g22 ds g11 g22 ds ds 

dΘ √ d + g u˙ 1 sin Θ = ds ds



1 √ g22



d − u˙ cos Θ ds 2

sin Θ d −√ g22 ds 

1 √ g22





1 √ g11





cos Θ √ g11



d − u˙ cos Θ ds 2

.



1 √ g11



The Intrinsic Geometry of Surfaces ■ 261

Now we obtain by (3.7) and (3.8) in Corollary 3.1.5 d ds



1 √ g22



=−

1

1 ∂g22 j u˙ = − √ j 3/2 ∂u g22 2(g22 )





1 2 u˙ 1 − √ 12 g22





2 u˙ 2 , 22

and similarly d ds



1 √ g11



1 = −√ g11





1 1 u˙ 1 − √ 11 g11





1 u˙ 2 . 21

Thus we have            2 2 1 1 dΘ √ − g u˙ 1 u˙ 1 u˙ 2 + (u˙ 2 )2 − u˙ 2 (u˙ 1 )2 + u˙ 1 u˙ 2 . A= 12 22 11 21 ds Finally, we obtain

κg = A + =

√



g u˙

1







2 u˙ i u˙ k − u˙ 2 ik 







1 u˙ i u˙ k ik







= 

dΘ √ 2 2 u˙ 1 u˙ 2 + (u˙ 1 )2 + g u˙ 1 12 11 ds        1 1 √ + g u˙ 2 − u˙ 1 u˙ 2 + (u˙ 2 )2 , 12 22

and by Corollary 3.1.5, (3.20) and (3.21) √

√ − g









2 2 1 (u˙ 1 )2 u˙ 2 = cos2 Θ sin Θ √ 12 g11 12 1 ∂g22 1 = cos2 Θ sin Θ √ = κg2 cos2 Θ sin Θ, g11 2g22 ∂u1 √     g 2 2 √ g (u˙ 1 )3 = cos3 Θ 3/2 = κg1 cos3 Θ, 11 11 g11

g









1 1 2 u˙ 1 )(u˙ 2 )2 = − sin2 Θ cos Θ √ 12 g11 12 1 ∂g11 1 = − sin2 Θ cos Θ √ = κg1 sin2 Θ cos Θ g22 2g11 ∂u2

and √ − g



√    g 1 1 (u˙ 2 )3 = sin3 Θ 3/2 = κg2 sin3 Θ, 22 22 g22

whence     dΘ + κg1 cos Θ cos2 Θ + sin2 Θ + κg2 sin Θ cos2 Θ + sin2 Θ ds dΘ + κg1 cos Θ + κg2 sin Θ. = ds

κg =

We close this section with a few examples.

262 ■ Differential Geometry and Its Visualization

Example 3.1.13 (Geodesic curvature of parallels on spheres). (a) For the sphere with a parametric representation ⃗x(ui ) = r{cos u1 cos u2 , cos u1 sin u2 , sin u1 } we obviously have





(u1 , u2 ) ∈ (−π/2, π/2) × (0, 2π) ,

⃗ (ui ) = −⃗x(ui ) · 1 . N r 2 1 The curvature of the u –line corresponding to u = α is κ=

1 , r cos α

and the binormal vector of this u2 –line is given by ⃗v3 = ⃗e 3 . Therefore, by Theorem 3.1.12, the geodesic curvature of the u2 –line is κg = −

1 1 · sin α = − · tan α. r cos α r

(b) A curve of constant geodesic curvature on a sphere is a circle line. Since ⃗ • ⃗x¨ = − 1 (⃗x • ⃗x¨) = 1 , κn = N r r κg = const implies κ = const. Furthermore, we have ⃗˙ = − 1 ⃗x˙ = − 1 ⃗v1 , N r r hence





⃗ • ⃗v2 ) = τ . ⃗ • d⃗v3 = −κτ (N 0 = κ˙ g = κ N ds r Thus τ = 0 and the curve is in a plane, that is, a circle. Example 3.1.14 (Geodesic curvature of helices on circular cylinders). Helices on circular cylinder with a parametric representation ⃗x(ui ) = {r cos u2 , r sin u2 , u1 }





(u1 , u2 ) ∈ R × (0, 2π) ,

where r > 0 is a constant, have identically vanishing geodesic curvature. Proof. Since the first fundamental coefficients are g11 = 1, g12 = 0, g22 = r2 , g = r2 , we obtain the first and second Christoffel symbols by Example 3.1.7 [ijk] = 0 and



i jk



= 0 (i, j, k = 1, 2).

Furthermore, it follows that ϵ11 = ϵ22 = 0 and ϵ12 = −ϵ21 =

√

g = r.

The Intrinsic Geometry of Surfaces ■ 263

We compute the geodesic curvature of a helix with a parametric representation ⃗x(ui (s)) = {r cos ωs, r sin ωs, hωs},

√ where ω = 1/ r2 + h2 , by applying identity (3.13). From

u1 (s) = hωs, u2 (s) = ωs, u˙ 1 (s) = hω, u˙ 2 (s) = ω, u¨1 (s) = 0 and u¨2 (s) = 0, we conclude 









2 u˙ j u˙ k + ϵ21 u˙ 2 u¨1 + jk = rhω(0 + 0) − rω(0 + 0) = 0.

κg = ϵ12 u˙ 1 u¨2 +

Since





1 u˙ j u˙ k jk



⃗ = −{cos u2 , sin u2 , 0}, ⃗v3 = ω{h sin ωs, −h cos ωs, r} and N

we obtain from Theorem 3.1.12 for the geodesic curvature of the helix ⃗ • ⃗v3 ) = 0. κg = κ(N

3.2

GEODESIC LINES

In the plane, curves of vanishing curvature, that is, straight lines, play an eminent role, because they are the curves of minimum length joining two points. It will turn out in Theorem 3.5.2 that, generally on surfaces, curves with vanishing geodesic curvature have the same minimum property. Definition 3.2.1. A curve on a surface with identically vanishing geodesic curvature is called a geodesic line. There is an analogy in the definitions of asymptotic and geodesic lines, namely that along asymptotic lines the normal curvature vanishes identically, whereas along geodesic lines the geodesic curvature vanishes identically. The next result will give characterizations of geodesic lines. Theorem 3.2.2. Geodesic lines on a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D) are characterized by either of the following conditions u¨ + i

or





i u˙ j uk = 0 (i = 1, 2) jk

(3.22)

⃗ × ⃗x˙ ) = 0. ⃗x¨ • (N

(3.23)

264 ■ Differential Geometry and Its Visualization

Figure 3.7 Geodesic line on a cone and surface normal vector in the osculating plane at a point

Proof. The condition in (3.23) is an immediate consequence of (3.1) and (3.3). By Remark 2.4.3, we have 

⃗x¨ = u¨r + Now (3.1) implies







r ⃗. u˙ i u˙ k ⃗xr + (Lik u˙ i u˙ k )N ik



κg⃗t = u¨r +







r u˙ i u˙ k ⃗xr . ik

Since the vectors ⃗x1 and ⃗x2 are linearly independent, it follows that κg = 0 if and only if the conditions in (3.22) hold. We obtain the following geometric characterization of geodesic lines as an immediate consequence of equation (3.23) in Theorem 3.2.2. Remark 3.2.3. If ⃗x¨ ̸= ⃗0 for a curve on a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D), then the geodesic lines are characterized by the fact that their osculating planes contain the normal vectors of the surface (Figure 3.7). Whereas asymptotic lines do not exist in the neighbourhood of elliptic points of a surface, that is, at points with L > 0, by Theorem 2.8.2 (b), geodesic lines exist at every point of any surface in some neighbourhood of the point, as the next result will show. Theorem 3.2.4. Let P be any point of a surface S, and d⃗ be a unit vector in the tangent plane of S at P . Then there is one and only one geodesic line through P in ⃗ and the geodesic line exists in some neighbourhood of the the direction given by d, point P .

The Intrinsic Geometry of Surfaces ■ 265

Proof. Let P have the parameters uk0 (k = 1, 2) and the direction d⃗ at P be given by d⃗ = ξ k ⃗xk where gik (uj0 )ξ i ξ k = 1. A geodesic line through P has to be a solution of the system (3.22) of differential equations with the initial conditions ui (0) = ui0

and u˙ i (0) = ξ i

(i = 1, 2).

(3.24)

It follows from the existence and uniqueness theorem for systems of ordinary differential equations that there is one and only one solution (u1 (s), u2 (s)) of (3.22) that satisfies the initial conditions in (3.24). To show that this solution is a geodesic line we have to prove that s is the arc length along the curve given by uk (s) (k = 1, 2). Then the assertion follows from (3.22). We consider the function f defined by f (s) = gik (uj (s))u˙ i u˙ k , and shall show f˙(s) = 0 for all s. Then we have f (0) = gik (uj0 )ξ i ξ k = 1 by the initial conditions, and this implies f (s) = 1 for all s, which means that s is the arc length along the curve. Now we obtain ∂gik j i k f˙(s) = u˙ u˙ u˙ + gik u¨i u˙ k + gik u˙ i u¨k . ∂uj Lemma 3.1.4 (d) yields ∂gik = [ijk] + [kji] = gkr ∂uj



r ij



+ gir



r kj



,

hence f˙(s) = gkr

 

r ij



u˙ u˙ u˙ + gir

= gkr u¨r +

j i k



r ij







r kj



u˙ j u˙ i u˙ k + gik u¨i u˙ k + gik u˙ i u¨k 

u˙ j u˙ i u˙ k + gir u¨r +



r kj





u˙ j u˙ k u˙ i .

Since the functions ui (s) (i = 1, 2) are solutions of the system (3.22), the terms in the parentheses vanish, and so f˙(s) = 0. It is useful to give the differential equations for geodesic lines with respect to an arbitrary parameter t. Theorem 3.2.5. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D). (a) If a geodesic line on S is given by ((u1 (t), u2 (t)), where t is an arbitrary parameter, then we have d 2 ui + dt2



i jk



dui d2 t duj duk =λ (i = 1, 2) where λ = − 2 dt dt dt ds



ds dt

2

.

(3.25)

(b) A curve on S given by (u1 (t), u2 (t)), where t is an arbitrary parameter, is a geodesic line if and only if d2 u1 du2 d2 u2 du1 − + dt2 dt dt2 dt



du2 dt



1 ik



du1 − dt



2 ik



dui duk = 0. dt dt

(3.26)

266 ■ Differential Geometry and Its Visualization

Proof. Let u¯i (s) = ui (t(s)) (i = 1, 2). Then we have dui dt d2 ui and u¨¯i (s) = 2 u¯˙ i (s) = dt ds dt



dt ds

2

+

dui d2 t for i = 1, 2. dt ds2

u1 (s), u¯2 (s)), then it follows from (3.22) in (a) If γg is a geodesic line given by (¯ Theorem 3.2.2 that 

hence



i ˙j ˙k u¯ u¯ jk     j   i du duk dt 2 d2 ui dt 2 dui d2 t = 2 + + for i = 1, 2, jk dt dt ds dt ds dt ds2

0 = u¨¯i +

d2 ui + dt2



i jk



dui d2 t 1 dui d2 t duj duk =− = −   2 dt dt dt ds2 dt dt ds2 ds i du for i = 1, 2, =λ dt



ds dt

2

that is, (3.25) holds. (b) Now we prove Part (b). (b.i) First we show the necessity of (3.26) for a curve to be a geodesic line. If γg is a geodesic line given by (u1 (t), u2 (t)), where t is an arbitrary parameter, then it follows from Part (a) that 

d2 u1 1 + 2 jk dt





du1 d2 u 2 duj duk 2 =λ and + 2 jk dt dt dt dt



du2 duj duk =λ . (3.27) dt dt dt

We multiply the first equation in (3.27) by du2 /dt and the second equation in (3.27) by du1 /dt, and obtain the difference d2 u1 du1 d1 u1 du2 − + dt2 dt dt2 dt



1 jk



du2 − dt



2 jk



du1 dt



duj duk = 0, dt dt

that is, (3.26) holds. (b.ii) Finally, we show the sufficiency of (3.26) for a curve to be a geodesic line. We assume that (3.26) holds for a curve γ given by (u1 (t), u2 (t)). Writing ui (t) = u¯i (s(t)) (i = 1, 2), we obtain 

dui ds d2 ui ds = u¯˙ i , = u¨¯i dt dt dt2 dt

2

+ u¯˙ i





d2 s for i = 1, 2, dt2

The first step is (3.26) and the second step is (3.13) for the geodesic curvature κg , 



ds 0 = u¨¯1 dt

2



d2 s ds ds + u¯˙ 1 2 u¯˙ 2 − u¨¯2 dt dt dt

2



d2 s ds + u¯˙ 2 2 u¯˙ 1 dt dt

The Intrinsic Geometry of Surfaces ■ 267

+ 





1 ˙2 u¯ − jk









2 ˙1 ˙j ˙k u¯ u¯ u¯ jk 





ds dt

3





ds 3 ˙ 1 ˙ 2 d2 s ds 1 ˙ 2 ˙ j ˙ k ds 3 u¯ u¯ u¯ + = u¨¯1 u¯˙ 2 + u¯ u¯ 2 jk dt dt dt dt  3    3 2 ds d s ds 2 ˙ 1 ˙ j ˙ k ds u¯ u¯ u¯ − − u¯¨2 u¯˙ 1 − u¯˙ 2 u¯˙ 1 2 jk dt dt dt dt =



ds dt

3 

=



ds dt

3 



u¯˙ 2 u¨¯1 +

−

κg g











1 ˙j ˙k u¯ u¯ − u¯˙ 1 u¨¯2 + jk





2 ˙j ˙k u¯ u¯ jk



.

Since ds/dt ̸= 0 this implies κg = 0, and so γ is a geodesic line. Now we give a few examples. First we determine the geodesic lines on a circular cylinder. We already know from Example 3.1.14 that helices are geodesic lines on circular cylinders. Now we show that if a curve on a circular cylinder is a geodesic line, then it is a helix or a parameter line. Example 3.2.6 (Geodesic lines on circular cylinders). We consider the circular cylinder with a parametric representation ⃗x(ui ) = {r cos u2 , r sin u2 , hu1 }





(u1 , u2 ) ∈ R × (0, 2π) ,

where r and h are positive constants. Then the geodesic line through the point with (u1 (0), u2 (0)) = (u10 , u20 ) at an angle Θ0 ∈ (−π/2, π/2) to the u2 –line through u1 (0) is given by u1 (s) = sin Θ0 · s + u10 and u2 (s) =

cos Θ0 · s + u20 for s ∈ R. r

(3.28)

Proof. Since r(u1 ) = r and h(u1 ) = u1 , it follows from (2.24) in Example 2.3.5 that g11 (ui ) = g11 (u1 ) = 1, g12 (ui ) = 0 and g22 (ui ) = g22 (u1 ) =

1 r

and we obtain from the formulae in Example (3.1.7) 

i jk



= 0 for all i, j, k = 1, 2.

So the differential equations (3.22) in Theorem 3.2.2 for the geodesic lines on the circular cylinder reduce to u¨i = 0 for i = 1, 2

268 ■ Differential Geometry and Its Visualization

with the solutions

ui = ci s + di (i = 1, 2) for all s ∈ R,

where ci and di (i = 1, 2) are constants. The initial conditions ui (0) = ui0 yield di = ui0 for i = 1, 2. We also have sin Θ0 = sin Θ0 = c1 u˙ 1 (0) =  g11 (ui (0))

and

u˙ 2 (0) = 

cos Θ0 cos Θ0 = = c2 . i r g22 (u (0))

Now we determine the geodesic lines on the hyperbolic plane or Poincar´e half– plane, which is a model of the hyperbolic non–Euclidean geometry. Definition 3.2.7. The hyperbolic plane or Poincar´e half–plane is the set H = {(x, y) ∈ R2 : y > 0}

with the first fundamental form

(ds)2 =

 1  2 2 · (dx) + (dy) . y2

Example 3.2.8 (Geodesic lines on the Poincar´ e half–plane). The geodesic lines on the Poincar´e half–plane are segments of Euclidean circle lines or straight lines that meet the real axis y = 0 at right angles. Proof. We put u1 = y and u2 = x. Since g11 (ui ) = g22 (ui ) =

1 and g12 (ui ) = 0, (u1 )2

we obtain from (3.10) in Corollary 3.1.5 





1 ∂g11 1 ∂g22 1 1 = · = · =− 1 1 11 22 2g ∂u 2g ∂u    11    11  1 1 2 2 = = = =0 12 21 11 22



=−

1 , u1

and 

2 12



=



2 21



=

∂g22 1 1 · = − 1. 2g22 ∂u1 u

Consequently the differential equations (3.22) in Theorem 3.2.2 for the geodesic lines on H reduce to  1  2u˙ 1 u˙ 2 u¨1 = 1 · (u˙ 1 )2 − (u˙ 2 )2 and u¨2 = . (3.29) u u1

The Intrinsic Geometry of Surfaces ■ 269

(i) If u˙ 2 = 0, then u2 = const and we obtain a straight line segment that meets the x–axis at a right angle. (ii) If u˙ 2 ̸= 0, then the second equation in (3.29) yields 

u˙ 2 (u1 )2

.

hence



1 2u˙ 1 u˙ 2 u¨2 (u1 )2 − 2u˙ 2 u˙ 1 u1 2 = · u ¨ − = (u1 )4 (u1 )2 u1



= 0,

u˙ 2 = c · u1 ̸= 0 for some constant c ̸= 0.

Since the arc length is the parameter of the geodesic lines, we have that gik ui uk =

 1  1 2 2 2 ( u ˙ ) + ( u ˙ ) = 1, that is, (u˙ 1 )2 = (u1 )2 − (u˙ 2 )2 (u1 )2

and it follows that



This implies

2

=

du1 du2

2

u˙ 1 u˙ 2



(u1 )2 − (u˙ 2 )2 1 = 2 1 2 − 1. 2 2 (u˙ ) c (u )

=

Since the improper integral 1/|c| 

t=0

1 − c2 (u1 )2 1 for u1 ≤ . c2 (u1 )2 |c|

√

t dt exists, 1 − c2 t2

we obtain for all u1 ≤ 1/|c| 2

u = ±|c|

u1

This yields and so

t

√

  u1 1  1  t dt 2 t2  2 (u1 )2 − 1 . 1 − c 1 − c = ∓ = ∓ t=0 |c| |c| 1 − c2 t2 

2

|c|u2 ∓ 1 

u2 ∓

= c2 u2 ∓ 2|c|u2 + 1 = 1 − c2 (u1 )2

1 |c|

2

+ (u1 ) =

1 1 for u1 ≤ . |c| |c|

This is the equation of a semi–circle line in H, with the radius r = 1/|c| and centre in the point (x, y) = (±1/|c|, 0); such a semi–circle line meets the x–axis at a right angle.

270 ■ Differential Geometry and Its Visualization

Figure 3.8

Principal circles (left) and geodesic lines (right) on a sphere

Visualization 3.2.9 (Geodesic lines on planes and spheres). (a) The geodesic lines in a plane are exactly the straight lines. (b) The geodesic lines on a sphere are exactly the principal circle lines, that is, the intersections of the sphere with planes through the centre of the sphere (left in Figure 3.8). Proof. ⃗ , hence ⃗a = ⃗x˙ × N ⃗ ̸= ⃗0. Furthermore, (a) If ⃗x(s) is a curve in a plane, then ⃗x˙ ⊥ N ⃗ , ⃗a ⊥ ⃗x˙ and ⃗x¨ ⊥ ⃗x˙ , N ⃗ , identity (3.23) implies ⃗x¨ = ⃗0. Thus ⃗x is a since ⃗a ⊥ N straight line. Conversely, any straight line satisfies ⃗x¨ = 0, hence identity (3.23). (b) Now we prove Part (b). (b.i) First we show that any geodesic line on a sphere is a principal circle line. If ⃗x(s) is a geodesic line on a sphere of radius r, then ⃗x(s) is a circle line by Example 3.1.13 (b). Furthermore, κg = 0 implies κ2n = κ2 . On the sphere, we have ⃗x • ⃗x = r, hence ⃗x • ⃗x˙ = 0 and ⃗x¨ • ⃗x + ⃗x˙ • ⃗x˙ = 0.

It follows that

⃗ • ⃗x¨ = − 1 (⃗x • ⃗x¨) = 1 , ⃗x¨ • ⃗x = −1, and consequently κn = N r r and the circle line is a principal circle line. (b.ii) Now we show that any principal circle line is a geodesic line on the sphere. If ⃗x(s) is a principal circle line, then we have κ = 1/r for its curvature and therefore 1 1 κ2g = κ2 − κ2n = 2 − 2 = 0. r r Thus ⃗x(s) is a geodesic line (right in Figure 3.8).

The Intrinsic Geometry of Surfaces ■ 271

Visualization 3.2.10 (Geodesic lines on a circular cone). We consider the circular cone with a parametric representation ⃗x(ui ) =



1 1 √ · u1 cos u2 , √ · u1 sin u2 , u1 a a

for given a >0. We put α = a/(a + 1).

 



(u1 , u2 ) ∈ D = (0, ∞) × (0, 2π)

(a) Then the geodesic lines are the meridians and the curves with   1 )2 + β 2  u (s) = α2 (s + s0  u2 (s) =



 

k α for s ∈ R, · tan−1 (s + s0 ) + d  αβ β

(3.30)

where k and d are real constants and β 2 = k 2 /a (Figure 3.9).

(b) Let (u10 , u20 ) ∈ D be given and Θ0 denote the angle between the parallel through u10 and the geodesic line through the point with position vector ⃗x0 = ⃗x(ui0 ). Then the constants k, s0 , β and d in (3.30) are given by k=

√ u1 sin Θ0 a · u10 cos Θ0 , s0 = 0 , β = u10 cos Θ0 α

and d = u20 −

Figure 3.9

√

˜ 0) a + 1 · tan−1 (tan Θ

Geodesic lines on a circular cone

272 ■ Differential Geometry and Its Visualization

with ˜0 = Θ

Proof.

   Θ0



Θ0 ∈











−π π ,   2 2 π 3π , Θ0 ∈ . 2 2

  Θ0 − π

(a) We obtain from Example 3.1.7 









u1 1 1 1 1 , = = = 0, =− 11 12 21 22 a+1         1 2 2 2 2 = = 0 and = = 1 11 22 12 21 u and the differential equations (3.22) in Theorem 3.2.2 reduce to

and

u¨1 −

u1 (u˙ 2 )2 = 0 a+1

(3.31)

u¨2 +

2 · u˙ 1 u˙ 2 = 0. u1

(3.32)

First it follows from (3.32) that u¨2 (u1 )2 + 2u˙ 1 u˙ 2 u1 = whence

u˙ 2 =

 d  2 u˙ · (u1 )2 = 0, ds

k

(u1 (s))2

,

(3.33)

where k ∈ R is a constant.

(a.i) First we assume that k = 0. Then u˙ 2 = 0, and since s is the arc length along the geodesic line, it follows that 1 a (u˙ 1 )2 = = = α2 g11 a+1 and so

d˜ u1 = αs + d˜ for all s > − , α

where d˜ ∈ R is a constant. Thus we have obtained themeridians, which are geodesic lines by (3.18) 2 in Lemma 3.1.10, since 11 = 0.

The Intrinsic Geometry of Surfaces ■ 273

(a.ii) Now we assume that k ̸= 0. Then we multiply (3.31) by u˙ 1 , substitute (3.33), and obtain 

k2 1 1 d k2 1 · 1 3 · u˙ 1 = · (u˙ 1 )2 + u¨ u˙ − a + 1 (u ) 2 ds a + 1 (u1 )2 1 1

hence



= 0,

1 k2 · 1 2, (u˙ 1 )2 = d˜ − a + 1 (u )

where d˜ > 0 is a constant. This implies u˙ 1 u1



2 ˜ 1 )2 − k d(u a+1

|k| . = ±1 for u1 >  ˜ + 1) d(a

Integration and solving for u1 yields 1

u (s) =



˜ + s0 )2 + d(s

k2 for all s, ˜ + 1) d(a

where s0 ∈ R is a constant. Since s is the arc length along the geodesic line, we obtain 

1 a+1 ˜ k2 1 = g11 (u˙ 1 )2 + g22 (u˙ 2 )2 = · 1 2 d− a a + 1 (u ) that is,

d˜ =



+

1 k2 · 1 2, a (u )

a = α2 . a+1

This yields the first identity in (3.30), where β 2 = k 2 /2. Substituting this in (3.33) and integrating, we finally obtain the second identity in (3.30). Thus we have proved Part (a). (b) It follows from

that Furthermore,

k=

√

cos Θ0 u˙ 2 (0) =  g22 (u10 ) a · u10 cos Θ0 and β 2 = (u10 )2 · cos2 Θ0 .

sin Θ0 u1 · sin Θ0 u˙ 1 (0) =  . implies s0 = 0 α g11 (u10 )

Since tan−1 is an odd function, we may choose β = u10 · cos Θ0 , and u20





k α · tan−1 · s0 + d = αβ β

274 ■ Differential Geometry and Its Visualization

finally yields

d = u20 −

with

˜0 = Θ

√

˜ 0) a + 1 · tan−1 (tan Θ

   Θ0

  Θ0 − π







−π π ,   2 2 π 3π , Θ0 ∈ . 2 2 Θ0 ∈

Visualization 3.2.11 (Geodesic lines on a general cone). Let S be the sphere with radius r > 0 and its centre in the origin, and γ be a curve on S with a parametric representation ⃗y (s∗ ) (s∗ ∈ I ⊂ R), where s∗ is the arc length along γ. By GC, we denote the general cone generated by the straight lines that join the origin and the points on γ. We put u2 = s∗ , and obtain the following parametric representation for the general cone GC ⃗x(ui ) = u1 ⃗y (u2 )





(u1 , u2 ) ∈ (0, ∞) × I .

(a) Then the u1 –lines are geodesic lines on GC. (b) The geodesic line through u1 (0) = u1 > 0, u2 (0) = u2 > 0 at an angle Θ0 ̸= 0

π/2, 3π/2 to u2 –line through u1 is given by (Figure 3.10)

0

0

 1   u1 (s) = (s + ru1 sin Θ0 )2 + (ru1 cos Θ0 )2   0 0  r   and    s + ru1 sin Θ0   2 0   + u2 − r tan−1 (tan Θ0 ) u (s) = r tan−1     0 ru1 cos Θ0 0

Proof. It follows that

              

for s > 0. (3.34)

⃗x1 = ⃗y , ⃗x2 = u1 ⃗y ′ , ⃗x11 = ⃗0, ⃗x12 = ⃗y ′ and ⃗x22 = u1 ⃗y ′′ . Since ⃗y 2 = r2 and (⃗y ′ )2 = 1, we have ⃗y ′ • ⃗y = ⃗y ′′ • ⃗y ′ = 0 and ⃗y ′′ • ⃗y = −1. Thus we obtain g11 = r2 , g12 = 0, g22 = (u1 )2 and g = (u1 )2 , and so the parameters are orthogonal and the first fundamental coefficients satisfy the condition in (3.9) in Corollary 3.1.5. Therefore, we have by (3.10) 

1 11



1 ∂g11 = = 0, 2g11 ∂u1



1 22



=−

1 ∂g22 u1 = − , 2g11 ∂u1 r2

The Intrinsic Geometry of Surfaces ■ 275

Figure 3.10

Geodesic lines on a cone generated by a loxodrome on a sphere 

1 12





2 12



=



1 21



= 0,

=



2 21



=



2 11



=



2 22



=0

and 1 ∂g22 1 = 1. 1 2g22 ∂u u

The system (3.22) of differential equations for the geodesic lines on the general cone reduces to u1 u¨1 − 2 (u˙ 2 )2 = 0 (3.35) r and 2 u¨2 + 1 u˙ 1 u˙ 2 = 0. (3.36) u We multiply equation (3.36) by (u1 )2 and obtain u¨2 (u1 )2 + 2u1 u˙ 1 u˙ 2 = hence

u˙ 2 (s) =

d  2 1 2 u˙ (u ) = 0, ds k

(u1 (s))2

,

where k ∈ R is a constant.

(a) First we assume k = 0. Since s is the arc length along the geodesic line, we have (u˙ 1 )2 =

1 1 = 2 g11 r

(3.37)

276 ■ Differential Geometry and Its Visualization

and the geodesic lines are given by 1 u1 (s) = (s + s0 ) (s > −s0 ) r for some constant s0 , and u2 (s) = c, where c ∈ I is a constant; they are the u1 –lines corresponding to c ∈ I.

(b) Now let k ̸= 0. We multiply equation (3.35) by u˙ 1 , substitute (3.37), and obtain 

k 2 u˙ 1 1 d k2 (u˙ 1 )2 + 2 1 2 u¨1 u˙ 1 − 2 1 3 = r (u ) 2 ds r (u ) This implies



and

u˙ 1 u1

c2 (u1 )2 −

= 0.

k2 = c2 , r2 (u1 )2

(u˙ 1 )2 + where c ̸= 0 is a constant,



k2 r2

= ±1







k u1 >   cr

k2 = c2 (s + s0 )2 , r 2 c2 where s0 ∈ R is a constant. Therefore, we have (u1 )2 −

u1 =



c2 (s + s0 )2 +

k2 for all s. r 2 c2

Now (3.37) implies u˙ 2 =

k c2 (s

+ s0

)2

2

k + 2 2 r c

=

k



r 2 c2 r 2 c4 (s+s k2

0)

2



+1

and 2

u (s) = r tan

−1



rc2 (s + s0 ) k



+ d,

where d ∈ R is a constant. Since s is the arc length along the geodesic line, we must have 1 = g11 (u˙ 1 )2 + g22 (u˙ 2 )2 = r2



k2 c2 − 1 2 (u )



+ (u1 )2

k2 = r 2 c2 , (u1 )4

The Intrinsic Geometry of Surfaces ■ 277

hence |c| = 1/r. Thus the geodesic lines are given by   1  1   (s + s0 )2 + k 2 r2   u (s) = 

and    2

r

u (s) = r tan

  −1 s+s0 kr

  

+d

where k ̸= 0, s0 and d are constants. For the initial conditions

(s ∈ R),

(3.38)

u1 (0) = u1 > 0, u2 (0) = u2 > 0 0

0

and the angle Θ0 between the geodesic line and the u2 –line through u1 , we 0 obtain cos Θ0 cos Θ0 u˙ 2 (0) =  = and k = u˙ 2 (0)(u1 (0))2 = u1 cos Θ0 , 0 u1 g22 (u2 ) 0 0

and then

sin Θ0 implies s0 = u1 r sin Θ0 . u˙ 1 (0) =  0 g11 (u1 ) 0

Finally, we have

2

d = u − r tan 0

=

−1

 u2 − rΘ0



s0 kr



= u2 − r tan−1 (tan Θ0 )

0

u2 − r(Θ0 + π) 0

0

(Θ0 ∈ (−π/2, π/2))

(Θ0 ∈ (π/2, 3π/2)).

Thus the geodesic lines for Θ0 ̸= π/2, 3π/2 are given by (3.34) (Figure 3.10). Remark 3.2.12. We can obtain the geodesic lines on the circular cone in Visualization 3.2.10 as a special case of the geodesic lines on a general cone as in Visualization 3.2.11. Putting   1 1 ⃗y (u2 ) = √ · cos u2 , √ · sin u2 , 1 , a a

we obtain ⃗x(ui ) = u1 ⃗y (u2 ). We also have 2

∥⃗y (u )∥ =



1 +1= a



a+1 =r a

and s(t) =

t 0

 t  t   1 1 1 t   ∥⃗y (τ )∥ dt =  − √ sin τ, √ cos τ, 0  dτ = √ dτ = √ , a a a a ′

0

0

278 ■ Differential Geometry and Its Visualization

that is, t =

√

as. We put u1 = u∗1 and u2 =

√

au∗2 , and obtain

⃗x∗ (u∗i ) = u∗1 ⃗y (u∗2 ) for the corresponding representation as a general cone. We obtain for the geodesic lines by Visualization 3.2.11 by the first identity in (3.34) u1 (s) = u∗1 (s) = =



1 r



2

2

∗1 (s + ru∗1 0 sin Θ0 ) + (u0 cos Θ0 )

2

2

(αs + u10 ) + (u∗1 0 cos Θ0 ) =



α2 (s + s0 )2 + β 2 ,

where αs0 = u10 sin Θ0 and β = u10 cos Θ0 . This is the first identity in (3.30). The second identity in (3.34) yields for the geodesic line u (s) =

√

=

√

2

au (s) = ∗2

√



a r tan

s 

a + 1 tan−1  r 

−1



s + ru∗1 0 sin Θ0 ru∗1 0 cos Θ0

+ u10 sin Θ0 u10

cos Θ0 

α k · tan−1 (s + s0 ) + d, = αβ β





+

u∗2 0

− r tan

−1



(tan Θ0 )

√  2 −1  + u0 − a + 1 tan (tan Θ0 )

where

s √ 1 + u10 sin Θ0 k au0 cos Θ0 √ α(s + s0 ) r = a + 1, = = 1 1 αβ αu0 cos Θ0 u0 cos Θ0 β √ and d = u20 − a + 1 tan−1 (tan Θ˜0 ). Thus we obtained the second identity in (3.30). Visualization 3.2.13 (Geodesic lines on a pseudo-sphere). Let P S be the pseudo-sphere of Visualization 2.5.19 (c), and (u10 , u20 ) ∈ (0, ∞)×(0, 2π) be given. Then the u1 –line corresponding to u20 is a geodesic line. Furthermore, if Θ0 ∈ [0, 2π) \ {π/2, 3π/2}, then the geodesic line through (u10 , u20 ) with an angle of Θ0 to the u2 –line through (u10 , u20 ) is given by (Figure 3.11)  1 u (s) = − log (|δ| cosh (s + s0 ))      and   

where

       

u2 (s) =

1 tanh(s + s0 ) + c1 δ   √ 1 + 1 − δ2 for s < log − s0 , |δ|

 1 1 δ = e−u0 cos Θ0 , c1 = u20 + eu0 tan Θ0     and    1 − sin Θ0   .  s0 = log

1 + sin Θ0

        

                

(3.39)

(3.40)

The Intrinsic Geometry of Surfaces ■ 279

Geodesic lines on a pseudo-sphere

Figure 3.11

Proof. By (3.11) and (3.12), the differential equations (3.22) for the geodesic lines on P S reduce to  2 1 u¨1 + e−2u u˙ 2 = 0 (3.41)

and

2

1

u¨2 − 2u˙ 1 u˙ 2 = 0.

(3.42) 2

If u˙ = 0, then we obtain the u –line corresponding to u = orientation of the arc length s,

u20 ,

that is, for a suitable

u1 (s) = s + u10 and u2 (s) = u20 for s > −u10 .

Now we assume u˙ 2 ̸= 0. Then (3.42) yields

u¨2 1 = 2u˙ 1 , that is, u˙ 2 = δe2u 2 u˙

(3.43)

for some constant δ ̸= 0. Substituting this in (3.41), we obtain 1

u¨1 + δ 2 e2u = 0, that is, hence 

u˙

1

2

2

2 2u1

=d −δ e

 d  1 2 1 (u˙ ) + δ 2 e2u = 0, ds

  d for some constant d = ̸ 0 and u ≤ log  . δ 1

(3.44)

Since s is the arc length along the geodesic line, we have by the second identity in (3.43) and (3.44) 

2

g11 (u1 (s)) u˙ 1 (s)



2

+g22 (u1 (s)) u˙ 2 (s)

= d 2 −δ 2 e2u

1 (s)

+e−2u

1 (s)

δ 2 e4u

1 (s)

= d 2 = 1,

280 ■ Differential Geometry and Its Visualization

that is, d = ±1. Now (3.44) yields u˙ 1 = ± that is, |δ| < 1, and



1 − δ 2 e2u1 for 0 < u1 ≤ − log |δ|,

I(u1 ) =



√

du1 1 − δ 2 e2u1

(3.45)

= ±(s + s0 ) 1

for some constant s0 . To evaluate the integral I(u1 ), we substitute z = 1/|δ|e−u , so du1 /dz = −1/z, observe that z > 1, since u1 < − log |δ|, and obtain 1

I(u ) = −



dz =− z 1 − z −2 √



√

   dz 2−1 . z + z = − log z2 − 1

Writing β = ∓(s + s0 ), we obtain z 2 − 1 = (eβ − z)2 , that is, 2zeβ = 1 + e2β , and so z = cosh β. Since cosh β = cosh (−β), this implies u1 (s) = − log (|δ| cosh (s + s0 )). We note that cosh (s + s0 ) ≥ 1 for all s implies u1 (s) ≤ − log |δ| for all s. Furthermore, u1 (s) > 0 implies cosh (s + s0 ) < 1/|δ|, that is,     √  1 1 1 + 1 − δ2 + − 1 − s0 = log − s0 . s < log |δ| δ2 |δ| Now it follows from the second identity in (3.43) that u˙ 2 = and so

1 , δ cosh (s + s0 ) 2

1 tanh (s + s0 ) + c1 δ for some constant c1 ∈ R. Therefore, we have shown that the general solution of the differential equations (3.41) and (3.42) is given by (3.39). Now we determine the constants δ, s0 and c1 such that ui (0) = ui0 for i = 1, 2 and Θ0 is the angle between the geodesic line and the u2 –line through (u10 , u20 ). Let ⃗y (s) be a parametric representation of the geodesic line. Then it follows from

that

u2 (s) =

1 ⃗y˙ (0) • ⃗x1 (u10 )  1 ) du (0) = − tanh s = sin Θ g (u = 11 0 0 0 ∥⃗x1 (u10 )∥ ds





1 1 − sin Θ0 s0 = tanh (− sin Θ0 ) = log , 2 1 + sin Θ0 −1

which is the third identity in (3.40). Furthermore,

du2 (0) ⃗y˙ (0) • ⃗x2 (u10 )  1 1 = g22 (u10 ) = e−u0 δe2u0 = cos Θ0 1 ∥⃗x2 (u0 )∥ ds

The Intrinsic Geometry of Surfaces ■ 281 1

implies δ = e−u0 cos Θ0 , which is the first identity in (3.40). Finally, u20 = yields c1 = u20 +

1 tanh s0 + c1 δ

1

eu0 sin Θ0 1 = u20 + eu0 tan Θ0 , cos Θ0

which is the second identity in (3.40) (Figure 3.11).

3.3

GEODESIC LINES ON SURFACES WITH ORTHOGONAL PARAMETERS

In general, it is difficult to find geodesic lines on surfaces by solving the system of differential equations (3.22). The solution can, however, explicitly be given by an integral, if the first fundamental coefficients gik of the surface satisfy the following conditions g12 (uk ) = 0 and g11 (uk ) = g11 (u1 ) and g22 (uk ) = g22 (u1 ).

(3.46)

We already know from (3.8) in Corollary 3.1.5 that then the second Christoffel symbols satisfy 















1 dg11 1 dg22 1 1 1 1 = , = = 0, , =− 1 11 12 21 22 2g11 du 2g11 du1         1 dg22 2 2 2 2 = 0, = = , = 0. 11 12 21 22 2g22 du1

(3.47)

For instance, the first fundamental coefficients of surfaces of revolution satisfy the conditions in (3.46). Theorem 3.3.1 (Clairaut). Let S be a surface with first fundamental coefficients that satisfy the conditions in (3.46), γ be a geodesic line on S, and Θ be the angle between γ and the u2 –lines of S. Then we have (Figure 3.12) √

g22 cos Θ = c = const.

(3.48)

Proof. Let the geodesic line γ be given by uk (s) (k = 1, 2). Then it follows from (3.22) and (3.47) that 2

u¨ +





1 dg22 1 2 2 u˙ i u˙ k = u¨2 + u˙ u˙ = 0, ik g22 du1

(3.49)

and so

 d  dg22 1 2 g22 u˙ 2 = g22 u¨2 + u˙ u˙ = 0 implies g22 u˙ 2 = c ds du1 for some constant c ∈ R.

(3.50)

282 ■ Differential Geometry and Its Visualization

Figure 3.12

Clairaut’s theorem

Furthermore, we have for the angle Θ between the geodesic line with a parametric representation ⃗xγ and the u2 –lines of S cos Θ =

⃗x˙ γ • ⃗x2 g22 u˙ 2 c =√ , = √ ∥⃗x2 ∥ g22 g22

(3.51)

since ⃗x1 • ⃗x2 = 0. Now (3.48) is an immediate consequence.

The following result gives the explicit solution of the differential equations (3.22) for surfaces with first fundamental coefficients satisfying (3.46). Theorem 3.3.2. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D) and first fundamental coefficients that satisfy the conditions in (3.46). Then we have: (a) The u1 –lines are geodesic lines. (b) The u2 –lines are geodesic lines if and only if dg22 = 0. du1

(3.52)

(c) If c ∈ R and u10 are chosen such that g22 (u10 ) > c2 , then a curve γ with a parametric representation ⃗xc (u1 ) = ⃗xc (u1 , u2 (u1 )) is a geodesic line in some neighbourhood of u1 0 if and only if 2

1

u (u ) −

u20

= ±c

u1

u10





g11 (u) du. g22 (u) g22 (u) − c2 

(3.53)

The Intrinsic Geometry of Surfaces ■ 283

Proof. By (3.22) and (3.47), the geodesic lines on S are the solutions of (3.50) and u¨1 +

1 dg11 1 2 1 dg22 2 2 (u˙ ) − (u˙ ) = 0. 1 2g11 du 2g11 du1

(3.54)

(a) For the u1 –lines, we have u˙ 2 = u¨2 = 0 and so (3.49) is satisfied. Furthermore, g11 (u˙ 1 )2 = 1 implies 0=





 dg11 1 3 1 dg11 1 2 d  g11 (u˙ 1 )2 = 2¨ u1 u˙ 1 g11 + (u˙ ) = 2u˙ 1 g11 u¨1 + (u˙ ) , 1 ds du g11 du1

and so (3.54) is also satisfied. Thus the u1 –lines are geodesic lines.

(b) For the u2 –lines, we have u˙ 1 = u¨1 = 0 and the differential equations (3.54) and (3.49) reduce to 1 dg22 2 2 (u˙ ) = 0 and u¨2 = 0. 2g11 du1 These differential equations are satisfied if and only if condition (3.52) holds. (c) Now we prove Part (c). (c.i) First we show the necessity of (3.53). We assume that a geodesic line γg is given by the functions uk (s) (k = 1, 2). Then these functions satisfy the differential equations (3.54) and (3.49). As in (3.50), (3.49) implies g22 u˙ 2 = c for some constant c ∈ R. Since s is the arc length along γg , we have 1 = gik u˙ i u˙ k = g11 (u˙ 1 )2 + g22 (u˙ 2 )2 = g11 (u˙ 1 )2 +

c2 , g22

and so      1 g22 − c2 c2 1  1− =± for g22 ≥ c2 . u˙ = ±

g22

g11

g11 g22

(3.55)

Let u10 > 0 be chosen such that g22 (u10 ) > c2 . Then we have by (3.55) and the fact that g22 u˙ 2 = c u˙ 2 c du2 = =± 1 1 du u˙ g22



c g11 g22 = ±√ 2 g22 − c g22



g11 . g22 − c2

This implies (3.53). (c.ii) Finally, we show the sufficiency of (3.53). We assume that the curve γg is given by ut = t and u2 (t) as in (3.53). Then we have 

du1 g11 (t) du2 d2 u1  = 0, , = 1, = ± c 2 dt dt dt g22 (t) g22 (t) − c2

284 ■ Differential Geometry and Its Visualization

and we obtain by (3.47) dg11 (t) c 1 d2 u2 1   · =±  2 2 dt 2 g22 (t) g22 (t) − c dt g11 (t) 



g11 (t) dg22 (t) c 2 dg22 (t) + g22 ∓  3 (g22 − c )  2 dt dt g22 (t) g22 (t) − c2 





g11 (t) 1 dg11 (t)  = ± c 2 g22 (t) g22 (t) − c 2g11 (t) dt   g22 (t) 1 2 dg22 (t) (g22 (t) − c ) + g22 (t) − 2g22 (t)(g22 (t) − c2 ) dt dt

=



1 11





du  1 11 dt



−2







du2 = dt 2

 

du2  1 = 11 dt

c2 1 dg22 (t) dg22 (t) − − 2 g22 (t) dt 2g22 (t)(g22 − c ) dt

+



2 12

2 11



−



1 dg22 (t) 2g11 (t) dt

du2 dt

2





 



2  du  2

dt

2  −2 , 12

hence (3.26) is satisfied, and so γ g is a geodesic line by Theorem 3.2.5. We close this section with an example. Example 3.3.3 (Geodesic lines on general cylinders and cones). (a) Geodesic lines on a general cylinder Let γ be a curve with a parametric representation ⃗y (u1 ) (u1 ∈ I1 ) in the x1 x2 –plane. Then a general cylinder GCyl is generated by the straight lines through the points of γ orthogonal to the x1 x2 –plane. A parametric representation of GCyl is given by ⃗x(ui ) = ⃗y (u1 ) + u2⃗e 3



It follows from ⃗x1 (ui ) = ⃗y ′ (u1 ) and ⃗x2 (ui ) = ⃗e 3 that 

2

g11 (ui ) = ⃗y ′ (u1 )



(u1 , u2 ) ∈ I1 × R .

, g12 (ui ) = 0 and g22 (ui ) = 1,

and the first fundamental coefficients of GCyl satisfy the conditions in (3.46). By Theorem 3.3.2 (a) and (b), the parameter lines are geodesic lines on GCyl. Furthermore, by Theorem 3.3.2 (c), the geodesic line through the point with parameters (u10 , u20 ) at an angle if Θ0 ̸= 0, π/2, π, 3/2π to the u2 –line through u10 is given by 2

1

u (u ) =

u20

cos Θ0 + | sin Θ0 |

u1

u10

∥⃗y ′ (u)∥ du.

The Intrinsic Geometry of Surfaces ■ 285

If we denote the arc length along the curve γ between u10 and u1 by s(u1 ), then the geodesic line is given by u2 (u1 ) = u20 +

cos Θ0 · s(u1 ). | sin Θ0 |

(b) Geodesic lines on a general cone The first fundamental coefficients of the general cone GC in Visualization 3.2.11 satisfy the conditions in (3.46). By Theorem 3.3.2, the geodesic line that starts at the point given by (u10 , u20 ) at an angle Θ0 ∈ (0, π/2) to the u2 –line through u10 is given by u2 (u1 ) = u20 + c

u1

u10

√

u

1 u2

− c2

(3.56)

du,

where c = u10 cos Θ0 . To evaluate the integral in (3.56), we substitute t = c/u, hence dt/du = −c/u2 , and obtain c

u1

u10

c

√

u

1 u2

−

c2

du = −

u1

c u1 0

√

= cos

−1

= cos−1

1 dt 1 − t2





c u1



− cos

−1

u10 cos Θ0 u1





c u10



− cos−1 cos Θ0 .

Therefore, the geodesic line on the general cone CG is given by 2

1

u (u ) =

3.4

u20

+ cos

−1



u10 cos Θ0 u1



− Θ0

(u1 > u10 cos Θ0 ).

GEODESIC LINES ON SURFACES OF REVOLUTION

We already observed that the first fundamental coefficients of surfaces of rotation satisfy the conditions in (3.46). Therefore, we can apply the results of Section 3.3 to find the geodesic lines on surfaces of revolution; it turns out that they have one of three characteristic shapes. Throughout this section, we assume that RS is a surface of rotation with a parametric representation ⃗x(ui ) = {r(u1 ) cos u1 , r(u1 ) sin u1 , h(u1 )}





(u1 , u2 ) ∈ I1 × I2 ⊂ R × (0, 2π)

and we make the usual assumptions r(u1 ) > 0 and |r(u1 )| + |h′ (u1 | > 0 on I1 . We recall that the first fundamental coefficients of RS are g11 (ui ) = (r′ (u1 ))2 + (h′ (u1 ))2 , g12 (ui ) = 0 and g22 (ui ) = (r(u1 ))2 .

(3.57)

286 ■ Differential Geometry and Its Visualization

We assume that a geodesic line γ with a parametric representation ⃗xc (u1 , u2 (u1 )) starts at a point P0 with parameters u10 and u20 at an angle Θ0 ∈ [0, π/2] to the parallel that corresponds to u1 = u10 . We obtain from Theorem 3.3.2: If Θ0 = π/2, then we obtain the meridian through P0 as a geodesic line. If Θ0 = 0, then the parallel through P0 is a geodesic line if and only if dg22 1 (u ) = 2r′ (u10 )r(u10 ) = 0. du1 0 Since r(u10 ) ̸= 0, this is the case if and only if r′ (u10 ) = 0. If Θ0 ∈ (0, π/2), then we have 0 u10 or has a removable or non–removable singularity. First characteristic shape If r(u1 ) > c for all u1 ∈ I1 with u1 > u0 , then the integral on the right hand side of (3.58) exits for all such values of u1 . Visualization 3.4.1 (Geodesic line on a circular cylinder and a catenoid). (a) We consider the circular cylinder with radius r > 0 and a parametric representation given by r(u1 ) = r and h(u1 ) = u1 . Let (u10 , u20 ) ∈ I1 × (0, 2π) be an arbitrary point. If Θ0 = π/2, then we obtain the meridian through (u10 , u20 ) as a geodesic line. If Θ0 = 0, then r′ (u10 ) = 0 implies that the parallel through (u10 , u20 ) is a geodesic line. If Θ0 ∈ (0, π/2), then we have c = r(u10 ) cos Θ0 < r(u10 ) = r = r(u1 ) for all u1 ∈ I1 , hence the geodesic line takes the first characteristic shape. Observing c = r cos Θ0 , we obtain from (3.58) for the geodesic line u2 (u1 ) = u20 + c

u1

u10

  c du1 2 1 1 √ = u + · u − u 0 0 r r 2 − c2 r r 2 − c2

√

   cos Θ0  1 r cos Θ0 1 1 2 1 = u20 + √ 2 u − u = u + u − u 0 0 0 r sin Θ0 r r − r2 cos2 Θ0   cot Θ0 = u20 + u1 − u10 r

for all u1 ∈ I1 .

The Intrinsic Geometry of Surfaces ■ 287

Figure 3.13

Geodesic lines of the first characteristic shape

(b) We consider the catenoid given by r(u1 ) = a cosh (u1 /a) and h(u1 ) = u1 . Let (u10 , u20 ) be an arbitrary point with u10 ≥ 0. Since r′ (u1 ) = sinh (u1 /a) = 0 if and only if u1 = 0, the parallel through (u10 , u20 ) is a geodesic line if and only if u0 = 0. If Θ0 ∈ (0, π/2) then c = r(u0 ) cos Θ0 = a cosh (u10 /a) < r(u1 ) = a cosh (u1 /a) for all u1 > u10 , since cosh is monotone increasing on (0, ∞). Hence the geodesic line takes the first characteristic shape. We obtain from (3.58) for the geodesic line (Figure 3.13) u2 (u1 ) = u20 + c

u1

u10

=

u20

c + 2 a

cosh  

u a cosh a

u1

u10

a2

 

u a

cosh

du



cosh

2

 

u a

c2 − 2 a

2

 

u a

du −

c2

.

The common properties of the second and third characteristic shapes Now let r(u1 ) = c for some u1 > u10 . We put uˆ1 = inf{u1 > u10 : r(u1 ) = c},

(3.59)

and observe that r(ˆ u1 ) = c, since r is continuous. We study the geodesic line as u1 → uˆ1 , that is, the improper integral I(ˆ u1 ; c) =

uˆ1 

u0

(r′ (u))2 + (h′ (u))2  . r(u) (r(u))2 − c2

(3.60)

288 ■ Differential Geometry and Its Visualization

We assume r ∈ C m+1 (I1 ) and k = min{m ∈ N : r(m) (ˆ u1 ) ̸= 0}. It follows from Taylor’s formula that r(u1 ) = c +

  r(k) (ˆ u1 ) 1 (u − uˆ1 )k + o (u1 − uˆ1 )k (u1 → uˆ1 ), k!

(r(u1 ))2 − c2 = 2c

  u1 ) 1 r(k) (ˆ (u − uˆ1 )k + o (u1 − uˆ1 )k (u1 → uˆ1 ). k!

hence

(3.61)

Thus, by (3.61), the improper integral converges if and only if k = 1. If Θ denotes the angle between the parallel through u1 and the tangent vector of the geodesic line, then we obviously have d⃗xc ⃗x1 ⃗x2 1  du  = sin Θ · + cos Θ ·  d⃗  x ∥⃗x1 ∥ ∥⃗x2 ∥  c  du1  =

and consequently



1−

c ⃗x2 ⃗x1 c2 + · , · 1 2 1 (r(u )) ∥⃗x1 ∥ r(u ) ∥⃗x2 ∥

d⃗xc (u1 , u2 (u1 )) ⃗x2 du1   → (u1 → uˆ1 ).  d⃗  1 2 1 ∥⃗ x ∥ x (u , u (u )) 2  c      du1

(3.62)

The second characteristic shape

We assume k = 1, that is, r′ (ˆ u1 ) = ̸ 0. Then the improper integral in (3.60) converges, that is, the limit uˆ2 = 1lim 1 u2 (u1 ) u →ˆ u

exists and the geodesic line is tangent to the parallel that corresponds to uˆ1 , which is not a geodesic line, since r′ (u1 ) ̸= 0 (Figures 3.14 and 3.15). If a geodesic line takes the second characteristic shape for some initial values of 1 u0 , u20 and Θ0 , then it will also take the second characteristic shape in some open interval IΘ that contains Θ0 , when u10 and u20 remain fixed. Then the value of uˆ1 depends on Θ ∈ IΘ , say uˆ1 (Θ), where r(ˆ u1 (Θ)) = c(Θ) = r(u10 ) cos Θ. We assume that the function r has an inverse ϕ on an interval Ir ⊃ {ˆ u1 (Θ) : Θ ∈ IΘ }. Then we consider the bounding line for the geodesic lines for Θ varying in IΘ (Figure 3.16), given by uˆ1 (Θ) = ϕ(c(Θ)) and uˆ2 (Θ) = u20 + c(Θ)

lim

u1 →ˆ u1 (Θ)

 1  u  ′ 2 ′ 2 (r (u)) + (h (u))    du (Θ ∈ IΘ ).  2 2 u10

r(u) r (u) − c (Θ)

The Intrinsic Geometry of Surfaces ■ 289

Figure 3.14

Geodesic lines of the second characteristic shape with tangent vectors

Figure 3.15

Geodesic lines of the second characteristic shape

Visualization 3.4.2 (Geodesic lines on a circular cone). We consider the circular cone with a parametric representation 

1 1 ⃗x(ui ) = (1 − u1 ) √ cos u2 , (1 − u1 ) √ sin u2 , u1 a a



for (u1 , u2 ) ∈ (0, 1) × (0, 2π), where a > 0 is a constant (Figure 3.14). This Visualization is similar to Visualization 3.2.11. Let (u10 , u20 ) ∈ (0, 1) × (0, 2π) be given. If Θ0 = 0, then the meridian through u20 is a geodesic line. √ If Θ0 ̸= 0, then r′ (u10 ) = −1/ a ̸= 0 and the parallel through u10 is not a geodesic line.

290 ■ Differential Geometry and Its Visualization

Geodesic lines of the second characteristic shape and the bounding lines for u1 = uˆ1

Figure 3.16

If Θ0 ∈ (0, π/2), then we have 

c = cos Θ0 g22 (u10 ) = cos Θ0 and Putting α :=



(1 − u1 ) (1 − u10 ) √ < √ 0 , a a

u1 ) ̸= 0 for uˆ1 = 1 − cos Θ0 (1 − u10 ). r′ (ˆ

(a + 1)/a, we obtain u2 (u1 ) = u20 + c

u1

u10

=

u20

+ acα



1 +1  a du 1 − u (1 − u)2 √ − c2 a a u1

u10

= u20 − acα

du (1 − u) (1 − u)2 − ac2

1−u  1

1−u10



dt √ . 2 t t − ac2

To evaluate the integral, we make the substitution y = 1−u  1

1−u10

dt √ = 2 t t − ac2

1−u  1

1−u10

t2

√

ac/t and obtain

√ 1) c a/(1−u 

1 dy  =− √ c a√ 1 − y2 ac2 c a/(1−u10 ) 1− 2 t



dt

The Intrinsic Geometry of Surfaces ■ 291

Figure 3.17

Geodesic lines of the third characteristic shape and their asymptotes 

1 = √ cos−1 c a



 √  √  c a c a −1 − cos . 1 − u1 1 − u10

√ Since c a = (1 − u10 ) cos Θ0 and Θ0 ∈ (0, π/2), we finally obtain u2 (u1 ) = u20 −

√



a + 1 cos−1



1 − u10 cos Θ0 1 − u1



− Θ0



for the geodesic line. The third characteristic shape We assume k ≥ 2, that is, in particular r′ (ˆ u1 ) = 0. Then the improper integral in 2 1 1 1 (3.60) diverges and u (u ) → ∞ as u → uˆ . It follows from (3.62) that ⃗xc approaches the u2 –line through uˆ1 which now is a geodesic line since r′ (ˆ u1 ) = 0. The geodesic line cannot be tangent to this u2 –line because of the uniqueness of geodesic lines stated in Theorem 3.2.4; it asymptotically approaches the u2 –line (Figure 3.17). Visualization 3.4.3 (Geodesic lines on a hyperboloid of one sheet). We consider the hyperboloid of one sheet given by parametric representation ⃗x(u ) = i



1 1 √ cosh u1 cos u2 , √ cosh u1 sin u2 , a a



1 − sinh u1 b



for ((u1 , u2 ) ∈ D = R × (0, 2π)), where a > 0 and b < 0 are constants. Let (u10 , u20 ) ∈ D be given. If we choose Θ0 = cos

−1





1 , cosh u10

(3.63)

292 ■ Differential Geometry and Its Visualization

then it follows that

1 r(u1 ) = √ cosh u1 a

implies

1 1 c = r(u10 ) cos Θ0 = √ cosh u10 · a cosh u10

and r(u1 ) > 0 for all u1 ̸= 0, and we obtain for uˆ1 = 0

1 r(ˆ u1 ) = c and r′ (ˆ u1 ) = √ sinh uˆ1 = 0. a

Therefore, the geodesic line takes the third characteristic shape. √ Putting α = 1 − a/b > 1 (since b < 0) and observing c = 1/ a, we obtain for u1 < 0 2

1

u (u ) =

We substitute

u20

√  +c a



α cosh2 u − 1 

cosh u cosh2 u − c2 a   √ α cosh2 u − 1 = u20 − c a du. cosh u sinh u

x= cosh u =

 

α cosh2 u − 1,

du

dx α sinh u cosh u , = du α cosh2 u − 1 

x2 + 1 and sinh u = − α

x2 + 1 − α . α

Then the integral becomes  



α cosh2 u − 1 x2 du = α dx cosh u sinh u (x2 + 1)(x2 + 1 − α)   dx dx + = (α − 1) (x2 + 1 − α) x2 + 1 √    α−1 dx dx √ √ = + tan−1 x − 2 x− α−1 x+ α−1   √ √ α−1 x− α−1 √ + tan−1 x + k, log = 2 x+ α−1

I=

where k is a constant of integration. We put   √ √   2 1 α cosh u − 1 − α − 1 α − 1 0 2 1 1 −1   k(u0 ) = log  + tan α cosh u0 − 1 √ 2 α cosh2 u10 − 1 + α − 1

and finally obtain for the geodesic line (Figure 3.17) √   √ √ α−1 α cosh2 u1 − 1 − α − 1 2 1 2 log  u (u ) = u0 − c a √ 2 α cosh2 u1 − 1 + α − 1 + tan

−1



α cosh

2

u1



−1 −



k(u10 )

for u1 < 0.

The Intrinsic Geometry of Surfaces ■ 293

3.5

THE MINIMUM PROPERTY OF GEODESIC LINES

A curve of minimal arc length between two distinct points in the plane is a section of a straight line. We need the next lemma to be able to prove that geodesic lines on surfaces have the same minimal property. Lemma 3.5.1 (Fundamental lemma of calculus of variation). Let D be a domain in n–dimensional space En , f be continuous on D and 

D

f g = 0 for every g ∈ C ∞ (D) with g|∂D ≡ 0,

where ∂D denotes the boundary of the domain D. Then f ≡ 0 on D.

Proof. We assume f (x0 ) ̸= 0 for some ⃗x0 ∈ D. It follows from the continuity of f that there exist a point with the position vector ⃗x0 in the interior of D, and a neighbourhood Nρ (⃗x0 ) = {⃗x : ∥vx − ⃗x0 ∥ < ρ} of ⃗x0 in D such that f (⃗x) ̸= 0 for all ⃗x ∈ Nρ (⃗x0 ). We may assume f (vx) > 0 on Nρ (⃗x0 ) and define the function g : En → R by    1  exp − (x ∈ Nρ (⃗x0 )) ρ2 − ∥⃗x − ⃗x0 ∥2 g(⃗x) =  0 (x ̸∈ Nρ (⃗x0 )). Then we have g ∈ C ∞ (D), g|∂D ≡ 0 for the restriction of g to the boundary of D, and D f g > 0. Theorem 3.5.2. If γ is a curve of minimum length between two points of a surface, then γ is a section of a geodesic line.

Proof. Let S be a surface, and (ui (t) = (u1 (t), u2 (t))) (t ∈ I = [t0 , t1 ]) be a parametric representation of a curve γ on S of minimal arc length between the distinct points P0 and P1 of S with position vectors ⃗x(t0 ) and ⃗x(t1 ). Furthermore, let Cτ = {γε : |ε| < τ } for τ > 0 be a family of curves γε ∈ S with the parametric representations ui (t; ε), with ui (t; 0) = ui (t), ui (t0 ; ε) = ui (t0 ) and ui (t1 ; ε) = ui (t1 ) for i = 1, 2 that join the points P0 and P1 and contains the curve of minimal arc length for ε = 0. Then the lengths L(ε) of the curves γε ∈ C must satisfy L′ (0) = 0. We write ui = ui (t; ε) (i = 1, 2), for short, and w(t; ε) =



L(ε) =

t1

and obtain

t0

gik (uj )ui uk , ′

w(t; ε) dt.

′

(3.64)

294 ■ Differential Geometry and Its Visualization

Partial differentiation of the integrand in (3.64) with respect to ε yields 1 ∂w = · ∂ε 2w 1 = · 2w

 

∂ 2 ui k ′ ∂ 2 uk i′ ∂gik ∂uj i′ k′ u u u · u + g · + g · ik ik ∂uj ∂ε ∂t∂ε ∂t∂ε 



∂ 2 uj i′ ∂gik ∂uj i′ k′ · u + 2 · g · u u , ij ∂uj ∂ε ∂t∂ε

since gik = gki . If ui (t; ε) = ui (t) + εv i (t), where v i (t0 ) = v i (t1 ) = 0 for i = 1, 2, and t is the arc length along the curve γ0 , then we have 

This implies

w(t; 0) = 1 and

1 L (0) = 2 ′

t1 

t0

∂ui = v i for i = 1, 2. ∂ε

∂gik i k j · u˙ u˙ v + 2gij u˙ i v˙ j ∂uj



ds.

Partial integration of the second term of the integrand yields, since v i (t0 ) = v i (t1 ) = 0, 1 L (0) = 2 ′

t1 

t0

d(gij u˙ i ) j ∂gik i k j · u ˙ u ˙ v − 2 v˙ ∂uj ds



ds = 0.

Since L′ (0) = 0 for all functions v j , it follow from Lemma 3.5.1 that 1 ∂gik i k d(gij u˙ i ) · u˙ u˙ − · = 0 for j = 1, 2. 2 ∂uj ds Also

∂gij k i d(gij u˙ i ) = u˙ u˙ + grj u¨r ds ∂uk

and so (3.65) implies 1 ∂gik k i ∂gij k i · u˙ u˙ − u˙ u˙ − grj u¨r = 0 for j = 1, 2. 2 ∂uj ∂uk If we write

∂gij k i 1 u˙ u˙ = ∂uk 2





∂gij k i ∂gkj k i u˙ u˙ + u˙ u˙ , ∂uk ∂ui

then it follows from Lemma 3.1.4 (c) that [ikj]u˙ i u˙ k + grj u¨r = 0 for j = 1, 2,

(3.65)

The Intrinsic Geometry of Surfaces ■ 295

and so, by the definition of the second Christoffel symbols in Definition 3.1.2 (b), g

lj





grj u¨ + [ikj]u˙ u˙ r

i k

δrl u¨r

=

+

= u¨ +



l





l u˙ i u˙ k ik 

l u˙ i u˙ k = 0 for l = 1, 2. ik

Thus the curve γ satisfies the differential equations (3.22) in Theorem 3.2.2 for geodesic lines. Remark 3.5.3. Obviously the converse of Theorem 3.5.2 is not true in general. We shall, however, show later that if a section γ of a geodesic line is a member of a socalled geodesic field, then γ is a curve of minimum length between the initial and the final points of γ. To achieve this, we need to consider transformations of parameters. Let S be a given surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D). We draw the u1 – and u2 –lines in the ui –parameter plane with a Cartesian coordinate system, and introduce new parameters u¯i (i = 1, 2) as follows: Given two families of curves in the ui –plane, we choose new parameters u¯i (i = 1, 2) such that the curves of one family become the u¯1 –lines and the other ones of the other family become the u¯2 –lines. Based on the families of curves C1 (¯ u2 ) :





u1 = u11 (t; u¯2 ) u2 = u21 (t; u¯2 )

and

C2 (¯ u1 ) :



u1 = u12 (t; u¯1 ) u2 = u22 (t; u¯1 )



we search for a one–to–one correspondence (u1 , u2 ) ↔ (¯ u1 , u¯2 ). It is our aim to 1 1 1 2 2 2 1 2 determine u = u (¯ u , u¯ ) and u = u (¯ u , u¯ ). Example 3.5.4 (Straight lines in the ui –plane). We consider the families of curves C1 (¯ u2 ) :



u1 = t u2 = c1 t + u¯2



and

C2 (¯ u1 ) :



u1 = t u2 = c2 t + u¯1

where c1 and c2 are constants. We obtain 

u2 = c1 u1 + u¯2 u2 = c2 u1 + u¯1

If c1 ̸= c2 , then this yields u1 = u1 (¯ ui ) = with the Jacobian 

∂ui ∂ u¯k



 1  ∂u   1 =  ∂ u¯1  ∂u  2

∂ u¯



,

hence



u¯1 = −c2 u1 + u2 u¯2 = −c1 u1 + u2



.

u¯1 − u¯2 c1 u¯1 − c2 u¯2 and u2 = u2 (¯ ui ) = c1 − c2 c1 − c2 

∂u2  1 ∂ u¯1  = ∂u2  (c1 − c2 )2  ∂ u¯2

   1 1 c1   · . = −1 −c1  c1 − c2



,

296 ■ Differential Geometry and Its Visualization

Now we study the technique of computing parameter transformations. u1 , u¯2 ). (i) For given (u1 , u2 ), we determine (¯ (i.1) We search u¯2 and t1 such that 

u1 = u11 (t1 ; u¯2 ) u2 = u21 (t1 ; u¯2 )



.

We eliminate t1 and compute u¯2 = u¯2 (u1 , u2 ). (i.2) We search u¯1 and t2 such that 

u1 = u12 (t2 ; u¯1 ) u2 = u22 (t2 ; u¯1 )



.

We eliminate t2 and compute u¯1 = u¯1 (u1 , u2 ). (ii) For given (¯ u1 , u¯2 ), we determine (u1 , u2 ). We eliminate t1 and t2 from 

and obtain

u11 (t1 ; u¯2 ) = u12 (t2 ; u¯1 ) u21 (t1 ; u¯2 ) = u22 (t2 ; u¯1 )



t1 = t1 (¯ u1 , u¯2 ) and t1 = t1 (¯ u1 , u¯2 ).

There are two ways to compute (u1 , u2 ): (ii.1)



u1 = u11 (t1 (¯ u1 , u¯2 ); u¯2 ) = u1 (¯ u1 , u¯2 ) 2 2 1 2 2 2 1 2 u = u1 (t1 (¯ u , u¯ ); u¯ ) = u (¯ u , u¯ )





u1 = u12 (t2 (¯ u1 , u¯2 ); u¯1 ) = u1 (¯ u1 , u¯2 ) u2 = u22 (t2 (¯ u1 , u¯2 ); u¯1 ) = u2 (¯ u1 , u¯2 )



or (ii.2)

.

Example 3.5.5. (A surface of revolution with loxodromes as parameter lines) We consider a surface RS of revolution with a parametric representation ⃗x(ui ) = {r(u1 ) cos u2 , r(u1 ) sin u2 , h(u1 )},





(u1 , u2 ) ∈ D = I × (0, 2π) ,

The Intrinsic Geometry of Surfaces ■ 297

and introduce new parameters u¯i for i = 1, 2 such that the u¯i –lines are the loxodromes at a constant angle βi to the u2 –lines of RS. We assume that β2 ̸= β1 + kπ (k ∈ Z). First we put   g11 (t) ck = | cot βk | for k = 1, 2 and φ(t) =  dt, g22 (t) and consider the families of curves 

2

C1 (¯ u ): and 1

u ): C2 (¯



We obtain



u1 = u11 (t) = t u2 = u21 (t; u¯2 ) = c1 · φ(t) + u¯2 u1 = u12 (t) = t u2 = u22 (t; u¯1 ) = c2 · φ(t) + u¯1



.

u¯1 − u¯2 = (c1 − c2 ) · φ(u1 ).

Since φ′ (u1 ) ̸= 0 for all u1 , the inverse function φ−1 exists, hence 1

u =φ

−1



u¯1 − u¯2 c1 − c2



.

Furthermore, we have (c1 − c2 )u2 = c1 u¯1 − c2 u¯2 , and so, because of c2 ̸= c1 , u2 = Finally, we obtain for the Jacobian  1  ∂u   1 ∂(ui ) =  u¯ 1 k ∂(¯ u )  ∂u  2

c1 u¯1 − c2 u¯2 . c1 − c2







  1 c1 ∂u2      c −c  1 1 c − c u¯  = 1 2 1 2   · 2  1 c ′ 1 2 ∂u  φ (u ) −  −    2 c − c c − c 1 2 1 2 u¯ u¯ c1 − c2 1 1 1 = ′ 1 · = · ′ 1 . 2 φ (u ) (c1 − c2 ) c1 − c2 φ (u )

In particular, we obtain for the sphere with



π π r(u1 ) = r cos u1 and h(u1 ) = r sin u1 for u1 ∈ − , 2 2



where r > 0 is a constant, that 



u1 π φ(u ) = log tan + 2 4 1



,

whence 1

1

2

u , u¯ ) = 2 · tan u (¯

−1



u¯1 − u¯2 exp c1 − c2



−

π c1 u¯1 − c2 u¯2 and u2 (¯ u1 , u¯2 ) = . 2 c1 − c2

298 ■ Differential Geometry and Its Visualization

It follows from 1 − tan2 x 2 tan x 2 2 and cos 2x = cos x − sin x = 1 + tan2 x 1 + tan2 x

sin 2x = 2 · sin x cos x = that 

sin u1 = − cos 2 · tan−1



u¯1 − u¯2 exp c1 − c2



=−



1 − exp 2 · 

1 + exp 2 ·

and 

cos u1 = sin 2 · tan−1



u¯1 − u¯2 exp c1 − c2



=

2 · exp





u ¯1 −¯ u2 c1 −c2

1 + exp 2 ·

u ¯1 −¯ u2 c1 −c2

 = tanh 2

u ¯1 −¯ u c1 −c2



u ¯1 −¯ u2 c1 −c2



 =

u¯1 − u¯2 c1 − c2

1 u cosh u¯c1 −¯ −c2 1

2

and we obtain the following parametric representation of the sphere with respect to the new parameters u¯1 and u¯2 (right in Figure 2.6) ⃗x(¯ ui ) =

r u cosh u¯c1 −¯ −c2 1

2



c1 u¯1 − c2 u¯2 c1 u¯1 − c2 u¯2 u¯1 − u¯2 · cos , sin , sinh c1 − c2 c1 − c2 c1 − c2



.

This is the parametric representation of a sphere in Visualization 2.1.7 (c).

3.6

ORTHOGONAL AND GEODESIC PARAMETERS

It is often convenient to have orthogonal and geodesic parameters. Therefore, we study in this section how to construct them. First we deal with the construction of orthogonal parameters. If g12 ̸= 0 for the parameters ui of a surface, then we introduce new parameters u∗1 and u∗2 by u1 = u∗1 and u2 = u2 (u∗1 , u∗2 ) such that

∗ g12 =0

for the first fundamental coefficients of the surface with respect to the new parameters u∗k . Since ∂u1 ∂u1 = 1 and = 0, ∂u∗1 ∂u∗2 the transformation formula ∂ui ∂uk ∗ g12 = gik ∗1 ∗2 ∂u ∂u reduces to ∗ g12



∂u2 ∂u2 ∂u2 ∂u2 = g12 ∗1 + g22 ∗1 ∗2 = g12 + ∗1 g22 ∂u ∂u ∂u ∂u



∂u2 . ∂u∗2

(3.66)

The Intrinsic Geometry of Surfaces ■ 299

For admissible parameter transformations, we must have   ∂u1   ∗1 ∂(ui )  ∂u =  ∂(u∗k )  ∂u1  ∗2





∂u2    1 ∂u∗1   = ∂u2    0 ∂u∗2

∂u

Thus the differential equation (3.66) yields g12 (u∗1 , u2 (u∗1 , u∗2 )) +



∂u2   ∂u2 ∂u∗1  = ̸ 0. =  ∂u2  ∂u∗2  ∂u∗2

∂u2 g22 (u∗1 , u2 (u∗1 , u∗2 )) = 0. ∂u∗1

(3.67)

If we put x = u∗1 , y = u∗2 and z = u2 , then (3.67) becomes g12 (x, z(x, y)) + g22 (x, z(x, y))

∂z = 0. ∂x

Example 3.6.1. Let S be a screw surface with a parametric representation ⃗x(ui ) = {u1 cos u2 , u1 sin u2 , hu2 + ξ(u1 )}





(u1 , u2 ) ∈ I × (0, 2π) ,

where h ∈ R is a constant and ξ ∈ C 3 (I). We may assume h ̸= 0 for otherwise S is a surface of revolution with orthogonal parameters. Since ⃗x1 = {cos u2 , sin u2 , ξ ′ (u1 )}, ⃗x2 = {−u1 sin u2 , u1 cos u2 , h}, 

2

g11 = 1 + ξ ′ (u1 ) and

, g12 = hξ ′ (u1 ) ̸= 0 for h ̸= 0 and ξ ′ (u1 ) ̸= 0

g22 = (u1 )2 + h2 , we have to solve the differential equation 

hξ ′ (u∗1 ) + (u∗1 )2 + h2 From

 ∂u2

∂u∗1

= 0.

ξ ′ (u∗1 ) ∂u2 = −h , ∂u∗1 (u∗1 )2 + h2 we obtain u2 = −h



ξ ′ (u∗1 ) du∗1 + u∗2 (u∗1 )2 + h2

say, and with H(u∗1 ) = h



ξ ′ (u∗1 ) du∗1 (u∗1 )2 + h2

300 ■ Differential Geometry and Its Visualization

the following transformation formulae u1 = u∗1 and u2 = −H(u∗1 ) + u∗2 . Thus a parametric representation of the screw surface S with respect to the new orthogonal parameters u∗1 and u∗2 is given by 

⃗x∗ (u∗i ) = u∗1 cos (u∗2 − H(u∗1 )), u∗1 sin (u∗2 − H(u∗1 )),



h · (u∗2 − H(u∗1 )) + ξ(u∗1 ) .

Now we introduce the concept of geodesic parameters.

Definition 3.6.2. Orthogonal parameters u∗i (i = 1, 2) for a surface such that the u∗1 –lines are geodesic lines are called geodesic parameters. The following result holds. Theorem 3.6.3. The uk –lines of a surface are geodesic lines if and only if 

i kk



= 0 (i ̸= k).

(3.68)

Proof. (i) First we assume that a uk –line is a geodesic line. Then ui = const for (i ̸= k), hence u˙ i = u¨i = 0 and so 0 = u¨i + implies



i kk







i u˙ j u˙ k = jk





i (u˙ k )2 kk

= 0 for i ̸= k. 



i = 0 for k ̸= i. Then ui = const clearly satisfies the (ii) Conversely we assume kk differential equation for geodesic lines.

Theorem 3.6.4. (a) If g12 = 0, then a uk –line is a geodesic line if and only if gkk depends on uk only. (b) If a surface is given with respect to geodesic parameters, then we may assume g11 = 1 and g12 = 0.

(3.69)

(c) If a surface is given with respect to geodesic parameters, then the u2 –lines are geodesically parallel; this means that the distances between two u2 –lines measured along the u1 –lines are constant.

The Intrinsic Geometry of Surfaces ■ 301

Proof. (a) If g12 = 0, then we have by (3.10) in Corollary 3.1.5 

i kk



=−

1 ∂gkk · for (k ̸= i) 2gii ∂ui

and the statement follows by Theorem 3.6.3. (b) We assume that u¯i (i = 1, 2) are geodesic parameters. Then their orthogonality implies g¯12 = 0. (3.70) Since the u¯1 –lines are geodesic lines, we obtain by Part (a) g¯11 (¯ ui ) = g¯11 (¯ u1 ).

(3.71)

We introduce new parameters ui by putting u1 = Since

 

g¯11 (¯ u1 ) d¯ u1 and u2 = u¯2 .

∂u1 ∂u2 ∂u2 ∂u1  1 ), = g ¯ (¯ u = = 0 and = 1, 11 ∂ u¯1 ∂ u¯2 ∂ u¯1 ∂ u¯2

we get





 g¯ (¯   1 ∂(ui )  11 u ) 0 = u1 ) ̸= 0,   = g¯11 (¯ 0 1 ∂(¯ uk ) 

and consequently the parameter transformation is admissible. Furthermore, it follows from (3.70) and the transformation formulae for the first fundamental coefficients in (2.18) of Remark 2.3.3 (a), since ∂ u¯2 /∂u1 = 0 and ∂ u¯1 /∂u2 = 0, that g11

∂ u¯i ∂ u¯k = g¯ik 1 · 1 = g¯11 ∂u ∂u

g12 = g¯ik



∂ u¯1 ∂u1

2

+ g¯22



∂ u¯2 ∂u1

2

= g¯11 ·

1 = 1, g¯11

∂ u¯i ∂ u¯k ∂ u¯1 ∂ u¯1 ∂ u¯2 ∂ u¯2 · = g ¯ · + g ¯ · =0 11 22 ∂u1 ∂u2 ∂u1 ∂u2 ∂u1 ∂u2

and g22

∂ u¯i ∂ u¯k = g¯ik 2 · 2 = g¯11 ∂u ∂u



∂ u¯1 ∂u2

2

+ g¯22



∂ u¯2 ∂u2

2

= g¯22 .

(c) We may assume by Part (b) that g11 = 1. We obtain for the arc length of the sections of the u1 –lines between the u2 –lines with u1 = c1 and u1 = c2 , since u˙ 1 = 1 and u˙ 2 = 0, c2 

c1

gik

u˙ i u˙ k

ds =

c2

c1

√

g11 ds =

c2

c1

ds = c2 − c1 .

302 ■ Differential Geometry and Its Visualization

Definition 3.6.5. A family F of geodesic lines on a surface S is called a geodesic field for a region S ′ ⊂ S if there is one and only one line of F through every point of S′. Now we are able to prove the converse result of Theorem 3.5.2. Theorem 3.6.6. If a part of a geodesic line is a member of a geodesic field F for a region G, then it is a curve of minimum length joining its endpoints which is possible inside G. Proof. We choose geodesic parameters for the surface. Then the first fundamental form ((2.15) in Definition 2.3.1 (b)) is given by (ds)2 = (du1 )2 + g22 (uj )(du2 )2 .

(3.72)

Let γ, given by u2 = u2 (u1 ) (u1 ∈ [a, b]), be a piecewise smooth curve that joins the endpoints of the given geodesic arc. Then we obtain for the arc length of γ  b   b           ′ 2 1 1 2   1 + g22 (u ) du  ≥  du  = |b − a|.      a

a

Now let γ be a given curve which is not represented in the form u2 = u2 (u1 ), which does not necessarily intersect a line u1 = const only once. Then we divide γ into subarcs which can uniquely be represented by functions u2 (u1 ). Now the statement follows as in the first part of the proof. Visualization 3.6.7. (a) The parameters of a surface of revolution with a parametric representation ⃗x(ui ) = {r(u1 ) cos u2 , r(u1 ) sin u2 , h(u1 )}

are geodesic parameters (Figure 3.18). (b) On a surface with first fundamental coefficients

g11 = g11 (u1 ), g12 = 0 and g22 = g22 (u1 ), we introduce geodesic parameters u∗i (i = 1, 2) as follows: Let the u∗1 –lines be the geodesic lines that start on the u2 –line through u10 at an angle Θ0 ∈ (0, π/2) with the u2 –line, and let the u∗2 –lines be the orthogonal trajectories of the geodesic lines (Figure 3.19). ∗2 This  means, the u –lines intersect the geodesic lines at right angles. We put c = g22 (u10 ) cos Θ0 and obtain by Theorem 3.3.2  1   u1 (t)=t  

u21 (t)=c

t

u10





g11 (τ ) dτ + u20 g22 (τ ) g22 (τ ) − c2 

The Intrinsic Geometry of Surfaces ■ 303

Figure 3.18

revolution

Parameter lines with respect to geodesic parameters on a paraboloid of

for the geodesic lines in a suitable interval I. Their orthogonal trajectories given by u12 (t) = t and u22 (t) must satisfy the condition g11 + g22

Figure 3.19

tories

du21 du22 = 0. dt dt

A family of geodesic lines (dashed) on a cone and their orthogonal trajec-

304 ■ Differential Geometry and Its Visualization

This yields

 1   u2 (t)=t  



1 t g11 (τ )   =− g22 (τ ) − c2 dτ + u20 . c u1 g22 (τ ) 0

u22

We obtain for the new parameters u∗i u

∗1

−u

∗2

1 = φ(u ) = c 1

u1  g11 (u)g22 (u)  du.

g22 (u) − c2

u10

Since φ′ (u1 ) ̸= 0, this can be solved for u1 and thus yields a transformation formula u1 = u1 (u∗1 , u∗2 ). Finally, from 2

1

u (u ) = c

u1

u10





g11 (u) du + u∗2 , g22 (u) g22 (u) − c2 

we obtain a transformation formula

u2 = u2 (u∗1 , u∗2 ). In particular, if we consider the circular cone with a parametric representation ⃗x(u ) = i





  u1 u1 √ cos u2 , √ sin u2 , u1 (u1 , u2 ) ∈ (−∞, 0) × (0, 2π) , a a

where a > 0 is a constant, and put  2   u∗1 − u∗2  ∗i √ ψ(u ) = − tan Θ0 + 1,

a+1

then the transformation formulae are √  1 ∗k u (u ) = −c aψ(u∗k )    and   √    u2 (u∗k ) = a + 1 cos−1 −

1 ψ(u∗k )



for

0 ≤ u∗1 − u∗2


u1 (s) · sin β α

since sin β < 0, an interval

t>− 

It ⊂ −

b1 · sin β, α 

b1 · sin β, ∞ , α

      

.

+ (u1 (s))2 (1 − sin2 β)

308 ■ Differential Geometry and Its Visualization

and obtain the parallel curve of the loxodrome for each fixed t ∈ It from   1 u (s) = (c1 s + 1)2 − 2tα sin β · (c1 s + 1) + α2 t2  t     

u2t (s) =    

√

αt − tan β (c1 s + 1) cos β √ +c2 · log (c1 s + 1) + a + 1 · β

a + 1 tan−1

        

for s ∈ Is . If we introduce new parameters u¯i such that the u¯1 –lines are the parallel curves of the loxodrome, and the u¯2 –lines are the geodesic lines, then we obtain the transformation formulae in (3.73). Visualization 3.6.9. (Geodesic parallel coordinates on a circular cylinder) We consider the circular cylinder with a parametric representation ⃗x(ui ) = {r cos u2 , r sin u2 , u1 }





(3.74)

cos Θ0 · s for all s ∈ R. r

(3.76)

(u1 , u2 ) ∈ R × (0, 2π) .

Let γ be the geodesic line through the point X(0, 0) at an angle Θ0 to the u2 –line through u10 = 0. Then the transformation formulae for the geodesic parallel coordinates are  1  u1 = u¯1 sin Θ0 + u¯2 cos Θ0 and u2 = · u¯1 cos Θ0 − u¯2 sin Θ0 (3.75) r (Figure 3.22). Proof. By (3.28) in Example 3.2.6, the geodesic line γ is given by u1 (s) = sin Θ0 · s and u2 (s) =

Figure 3.22

Geodesic parallel coordinates on a cylinder

The Intrinsic Geometry of Surfaces ■ 309

Since g22 (ui ) = r, it follows from (3.48) in Clairaut’s theorem, Theorem 3.3.1, that the angle between γ and the parallels of the cylinder is constant. Therefore, at each s, we have Θ = Θ0 + π/2 for the angle between the geodesic line γs∗ with (vsi (t)) and the parallel through u1 (s). Thus γs∗ is given by vs1 (t) = cos Θ0 · t + u1 (s) and vs2 (t) = −

sin Θ0 · t + u2 (s) for all t ∈ R. r

We introduce new parameters u¯i (i = 1, 2) such that the u¯1 –lines are the parallel curves of the geodesic line γ and the u¯2 –lines are the geodesic lines γs∗ . Then the transformation formulae are given by (3.75). Visualization 3.6.10 (Geodesic parallel coordinates on a circular cone). We consider the circular cone of Visualization 3.6.8. By Visualization 3.2.10 (b), the geodesic line γ with the initial conditions u10 > 0, u20 = 0 and Θ0 ∈ (0, π/2) is given by    1 2 s2 + 2u1 α sin Θ · s + (u1 )2   u (s) = α   0 0 0       , (3.77) √ αs      u2 (s) = a + 1 tan−1  + tan Θ − Θ 0 0 u10 cos Θ0 

where α = a/(a + 1). In a sufficiently small neighbourhood of (u10 , u20 ), the angle Θ(s) between γ and the parallel through u1 (s) satisfies 



π , Θ(s) ∈ 0, 2 and the angle Θv (s) of the geodesic line γs∗ given by (vsi (t)) satisfies Θv (s) = Θ(s) + π/2. Applying Visualization 3.2.10 (b) again, we find that γs∗ is given by     vs1 (t) = (α(t + t0 (s)))2 + β 2   k 2 −1 α  · tan · (t + t0 (s)) + d  vs (t) =

αβ

where

β

    

,

u1 (s) sin Θv (s) , β = u1 (s) cos Θv (s), α√ √ ˜ v (s) = u2 (s) − a + 1(Θv (s) − π) d = u2 (s) − a + 1 · Θ

t0 (s) =

and k=

√ a · u1 (s) cos Θv (s).

Since Θ(s) ∈ (0, π/2) it follows from (3.48) in Clairaut’s theorem, Theorem 3.3.1 that  g22 (u10 ) u10 sin Θv (s) = cos Θ(s) = cos Θ0 ·  = cos Θ · , 0 u1 (s) g22 (u1 (s))

310 ■ Differential Geometry and Its Visualization

and so

u1 · cos Θ0 u1 (s) · cos Θ(s) = 0 α α

t0 (s) = and

vs1 (t) =

Furthermore, we obtain

and



α2 t2 + 2u10 α cos Θ0 · t + (u1 (s))2 .

√ a · u1 (s) cos Θv (s) √ k = = a + 1, αβ α · u1 (s) cos Θv (s)   1 α (t + t0 (s)) = 1 αt + u1 (s) · cos Θ(s) β u (s) · cos Θv (s)   1 αt + u10 · cos Θ0 =− 1 u (s) · sin Θ(s)   1 1 =− 1  αt + u · cos Θ 0 0 u (s) 1 − cos2 Θ(s)   1 = − αt + u10 · cos Θ0 (u1 (s))2 − (u10 )2 cos2 Θ0 1 ˜ v (s) = Θs − π = cos−1 u0 · cos Θ0 − π , Θ 2 u1 (s) 2

hence vs2 (t) = u2 (s) −

√





a + 1 tan−1  

+ cos−1



αt + u10 · cos Θ0

(u1

u0 · cos Θ0 u1 (s)

(s))2



−

−

π 2



(u10 )2

cos2

Θ0

 

(Figure 3.23). Now we construct geodesic polar coordinates as follows: Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D ⊂ R) and P ∈ S be a given point with position vector ⃗x(ui0 ). Furthermore, let γΘ with a parametric representation ⃗yΘ (s) denote the geodesic line through the point P at an angle Θ to the u2 –line through i P . We assume that s is the arc length along γΘ and ⃗yΘ (s) is given by vΘ (s) with i i vΘ (0) = u0 (i = 1, 2). For each fixed value of s, the vectors ⃗yΘ (s) (Θ ∈ [0, 2π]) generate a circle line γs∗ of constant distance s from the point P measured along the geodesic line γΘ . Choosing the new parameters u¯j (j = 1, 2) such that the u¯1 –lines are the geodesic lines γΘ (Θ ∈ [0, 2π]) and the u¯2 –lines are the circle lines γs∗ , we obtain a system of geodesic polar coordinates. The transformation formulae are ui = vui¯2 (¯ u1 ) (i = 1, 2).

(3.78)

The Intrinsic Geometry of Surfaces ■ 311

Figure 3.23

Geodesic parallel coordinates on a circular cone

Visualization 3.6.11 (Geodesic polar coordinates on a circular cylinder). Again we consider the circular cylinder of Visualization 3.6.9 with a parametric representation (3.74). It follows from the representation of the geodesic line γ through the point X(0, 0) at an angle Θ0 to the u2 –line through u10 = 0 and the transformation formulae (3.78) for the corresponding geodesic polar coordinates that u1 (¯ ui ) = vu1¯2 (¯ u1 ) = u¯1 sin u¯2 and u2 (¯ ui ) = vu2¯2 (¯ u1 ) =

u¯1 cos u¯2 . r

Hence the parametric representation of the circular cylinder with respect to the new parameters u¯i (i = 1, 2) in a suitable neighbourhood of the point X(0, 0) is given by ⃗x¯(¯ ui ) =







u¯1 · cos u¯2 , r sin r cos r





u¯1 · cos u¯2 , u¯1 · sin u¯2 r



(Figure 3.24). Visualization 3.6.12 (Geodesic polar coordinates on a circular cone). Again we consider the circular cone of Visualization 3.6.10. It follows from the representation (3.77) of the geodesic line γ through the point X(u10 , 0) at an angle Θ0 to the u2 –line through u10 > 0 and the transformation formulae (3.78) that the parameters u¯i (i = 1, 2) for the geodesic polar coordinates satisfy   1 i 1 i  u (¯ u ) = v (¯ u ) = α(¯ u1 )2 + 2u10 α¯ u1 · sin u¯2 + (¯ u10 )2  u ¯2        1

u2 (¯ ui ) = vu2¯2 (¯ ui ) =       

where α =



√

a + 1 arctan

a/(a + 1), as before (Figure 3.25).

α¯ u + tan u¯2 · cos u¯1   − arctan tan u ¯2

u10

               

,

312 ■ Differential Geometry and Its Visualization

Figure 3.24

Geodesic polar coordinates on a cylinder

Figure 3.25

Geodesic polar coordinates on a circular cone

3.7

LEVI-CIVITA` PARALLELISM

In Section 3.6, we studied the construction of geodesically parallel curves of a given curve. Now we consider the problem of a sensible definition of the parallel translation of special vectors, the so–called surface vectors, which are introduced as follows. Definition 3.7.1. Let S be a surface with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D), P be a point of S and T (P ) denote the tangent plane to S at P . Then a vector ⃗z at P in T (P ) is called a surface vector of S at P . We write ⃗z = (ξ 1 , ξ 2 ), where ξ i (i = 1, 2) are the components of the vectors ⃗xi at P .

The Intrinsic Geometry of Surfaces ■ 313

If the surface vectors ⃗z(s) of a plane are moved parallel along a straight line, that is, along a geodesic line, then the angle between the vector and the straight line, that is, the angle between ⃗z(s) and the tangent to the geodesic line, remains constant, that is, ⃗z˙ (s) = 0. The following generalization to arbitrary surfaces is due to Levi–Civit`a. Definition 3.7.2. Let S be a surface and γ be a curve on S with a parametric representation ⃗x(ui (s)). The surface vectors ⃗z(s) are said to be geodesically parallel along γ, if the components of ⃗z˙ (s) in the direction of the tangents to γ at s are equal to 0, that is, ⃗xi (ui (s)) • ⃗z˙ (s) = 0 for i = 1, 2. (3.79) Remark 3.7.3. We omit the parameters and write ⃗z = ⃗xk ξ k . Then the condition in (3.79) yields 0 = ⃗xi •

d (⃗xk ξ k ) = ⃗xi • ⃗xk ξ˙ik + ⃗xi • xkj u˙ j ξ k = gik ξ˙k + [kij]u˙ i ξ k . ds

Multiplying by g ir and summing with respect to i we obtain 0=g

ir

We write





gik ξ˙k + [kij]u˙ i ξ k = δrk ξ˙k + g ir [kij]u˙ i ξ k = ξ˙r +

dξ i δξ i = + δs ds



i kj





r kj

u˙ j ξ k for i = 1, 2.



u˙ j ξ k for r = 1, 2. (3.80) (3.81)

If t is an arbitrary parameter of the curve, the (3.80) becomes dξ i δξ i = + dt dt



i kj



dui k ξ for i = 1, 2. dt

(3.82)

z the curve Apart from the parameters ξ i (s) andui (s)  (i = 1, 2) of the vector field ⃗ i γ, only the second Christoffel symbols kj appear in the formula in (3.80); hence the geodesic parallelism is a quantity of the intrinsic geometry  of  a surface. i We consider the plane with Cartesian parameters. Then kj = 0 for all i, j and k, and (3.82) reduces to dξ i = 0 for i = 1, 2, dt that is, the components of the surface vectors are constant. The concept of geodesic parallelism is also independent of the parametric representation of the surface, since the vector ⃗z˙ , which is orthogonal to the surface, is defined independently of the parameters of the surface.

314 ■ Differential Geometry and Its Visualization

The next result is an immediate consequence of the existence and uniqueness theorem for the system (3.80) of differential equations. Theorem 3.7.4. If γ is a curve on a surface S and ⃗z0 is a surface vector at the point P0 ∈ S, then there exists a unique field of geodesically parallel surface vectors ⃗z(s) along γ such that ⃗z(0) = ⃗z0 . We refer to a curve on a surface given by ui (s) (i = 1, 2) as straightest curve, if its tangent vectors form a geodesically parallel vector field along itself. Then it follows from (3.80) that   r u¨r + u˙ j u˙ k = 0, jk and so the next result holds.

Theorem 3.7.5. The geodesic lines of a surface are also its straightest lines. Theorem 3.7.6. Let γ given by ui (s) (i = 1, 2) be a curve on a surface, and ⃗v (s) and w(s) ⃗ given by v i (s) and wi (s) (i = 1, 2) be two geodesically parallel vector fields along γ. Then the angle between the vectors ⃗v (s) and w(s) ⃗ along γ is constant and the lengths of all vectors ⃗v (s) are equal. Proof. We have  d  ∂gik j i k gik (uj (s))v i wk = u˙ v w + gik v˙ i wk + gik v i w˙ k ds ∂uj

and, since the vector fields are parallel, it follows from (3.80) that 







i k v m u˙ r and w˙ k = − wl u˙ r , v˙ = − mr lr i

hence

 ∂gik j i k d  gik (uj (s))v i wk = u˙ v w − gik ds ∂uj





i u˙ r v m wk + mr







k u˙ r v i wl . lr

Changing the indices of summation i, k and j to m, l and r in the first term on the right hand side of the last identity, k to l in the second term, and i to m in the last term we obtain  d  gik (uj (s))v i wk = ds



∂gml − gil ∂ur



i mr



− gmk



k lr



u˙ r v m wl .

Now it follows from the definition of the second Christoffel symbols, Definition 3.1.2, that gil



i mr



= gil g ji [mrj] = δlj [mrj] = [mrl]





j = gmk g jk [lrj] = δm [mrj] = [lrm],

and gmk

k lr

The Intrinsic Geometry of Surfaces ■ 315

hence by Lemma 3.1.4 (d)  d  gik (uj (s))v i wk = ds





∂gml − [mrl] − [lrm] u˙ r v m wl = 0. ∂ur

From this, we conclude that the lengths and angles are constant. Visualization 3.7.7. (Parallel movement along a curve on a surface of revolution) We consider the parallel movement along an arbitrary curve γ on any surface of revolution RS with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D). Let P0 be a point of RS with position vector ⃗x(ui0 ). We assume that the surface vector ⃗z(t) with ⃗z(0) = ⃗x(ui0 ) has an angle Θ0 ∈ [0, π/2] at P0 to the parallel with u1 = u10 . We obtain for the second Christoffel symbols of RS from Example 3.1.7 





1 1 dg11 = · , 11 2g11 du1     1 2 2 = = 12 21 2g22



1 1 dg22 =− · , 22 2g11 du1 dg22 · du1

and 

i jk



= 0 for all other values i, j and k.

Hence the differential equations in (3.82) are 1 1 dg11 1 du1 dg22 2 du2 dξ 1 + − =0 · ξ · ξ dt 2g11 du1 dt 2g11 du1 dt

(3.83)

1 1 dg22 1 du2 dg22 2 du1 dξ 2 + + = 0. · ξ · ξ dt 2g22 du1 dt 2g22 du1 dt

(3.84)

and

We may assume by Theorem 3.7.6 that the surface vector has constant length 1 when moved along γ, that is, g11 (ξ 1 )2 + g22 (ξ 2 )2 = 1, whence

ξ2 = ±

1  · 1 − g11 (ξ 1 )2 . g22

(3.85)

Since the surface vector ⃗z(0) at t = 0 satisfies ⃗z(0) = ξ 1 (0)⃗x1 (ui (0)) + ξ 2 (0)⃗x2 (ui (0)) =

sin Θ0 cos Θ0 ·⃗x1 (ui (0)) + ·⃗x2 (ui (0)), i g11 (u (0)) g22 (ui (0))

316 ■ Differential Geometry and Its Visualization

that is, ξ 1 (0) =

sin Θ0 cos Θ0 and ξ 2 (0) = , g11 (ui (0)) g22 (ui (0))

we choose the upper sign in (3.85), that is, ξ2 =

1  · 1 − g11 (ξ 1 )2 . g22

(3.86)

Substituting (3.86) in (3.83), we obtain an ordinary first order differential equation for ξ 1 the solution of which and ξ 2 from (3.86) describe the parallel movement of the surface vector along γ. In particular we consider the parallel movement along a loxodrome γ on RS given by   g11 (t) 1 2 u (t) = t and u (t) = | cot β| dt, g22 (t)

where β ∈ (−π/2, π/2) \ {0} is the constant angle between γ and the parallels of RS. Now we have to solve the differential equation dξ 1 1 dg22 dg11 1 1 =− · ξ + √ · · | cot β|ξ 2 1 dt 2g11 du 2 g du1

with 2

g = g11 g22 and ξ =



1 − g11 (ξ 1 )2 . √ g22

If RS is a sphere of radius r > 0, we obtain g11 = r2 and g22 = r2 cos2 u1 , and (3.87) yields

This implies 

dξ 1 | cot β| · sin t  =− · 1 − r2 (ξ 1 )2 . dt r cos t r  dξ 1 = sin−1 (rξ 1 ) = −| cot β| 1 − r2 (ξ 1 )2 = | cot β| · log | cos t| + c,



where c is a constant of integration, and so ξ 1 (t) = Since

1 · sin (| cot β| · log | cos t| + c). r

sin Θ0 sin Θ0 ξ 1 (0) = √ = , g11 r

sin t dt cos t

(3.87)

The Intrinsic Geometry of Surfaces ■ 317

Figure 3.26

Parallel movement along a loxodrome on a sphere

we obtain (Figure 3.26) ξ 1 (t) =

1 · sin (| cot β| · log | cos t| + Θ0 ) r    −Θ0 − π2 −1 for t < cos exp | cot β|

and ξ 2 (t) =

1 · cos (| cot β| · log | cos t| + Θ0 ). r cos t

Visualization 3.7.8. (Parallel movement along a geodesic line on a surface of revolution) Now we consider parallel movements on surfaces of revolution RS with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D) along geodesic lines γ which are parameterized with respect to their arc lengths, that is, ⃗xγ (ui (s)). Again we assume that the surface vector ⃗z(t) at t = 0 at the point P0 corresponding to the parameter point (u10 , u20 ) = (u1 (0), u2 (0)) has an angle Θ0 ∈ [0, π/2] with the parallel through u10 . The angle β between γ and the surface vector ⃗z(t) of constant length 1 is constant under the movement along γ. We obtain ξ 1 (s) = 

and

where c =

ξ 2 (s) = 

c · sin β + cos β · u˙ 1 (s) g11 (u1 (s))g22 (u1 (s)) 

(3.88) 

 1 c · cos β − sin β g22 (u1 (s)) − c2 , g22 (u1 (s))

g22 (u10 ) · cos Θ0 .

(3.89)

318 ■ Differential Geometry and Its Visualization

In particular, we obtain for the circular cone of Visualization 3.2.10 and the parallel movement along the geodesic line γ with the initial conditions u1 (0) = u10 > 0, u2 (0) = 0 and Θ0 ∈ (0, π/2) αs + u10 · sin (Θ0 + β) ξ 1 (s) = α ·  2 2 (αs + u10 · sin Θ0 ) + (u10 · cos Θ0 )

and

2

ξ (s) = where α =



√





a · −αs · sin β + u01 · cos (Θ0 + β) 2

2

(αs + u10 · sin Θ0 ) + (u10 · cos Θ0 )

for s ≥ 0,

(3.90)

(3.91)

a/(a + 1) (left in Figure 3.27).

Proof. We put δ = cos β and have to solve the equations gik ξ i

duk = g11 ξ 1 u˙ 1 + g22 ξ 2 u˙ 2 = δ for |δ| ≤ 1 ds

(3.92)

and gik ξ i ξ k = g11 (ξ 1 )2 + g22 (ξ 2 )2 = 1.

(3.93)

It follows from (3.48) in Clairaut’s theorem, Theorem 3.3.1 that g22 · u˙ 2 = g22 (u1 (s)) · u˙ 2 (s) = c = const. and so by (3.92) and (3.93) 

Figure 3.27

ξ2 · c − δ

2





= ξ 2 g22 u˙ 2 − ξ 1 g11 u˙ 1 + ξ 2 g22 u˙ 2 



= g11 1 − g22 (ξ 2 )2 (u˙ 1 )2 ,

2

2 = g11 (ξ 1 )2 (u˙ 1 )2

Parallel movement along a geodesic line on a cone and a torus

The Intrinsic Geometry of Surfaces ■ 319

hence, since s is the arc length along γ 



0 = (ξ 2 )2 c2 + g11 g22 (u˙ 1 )2 − 2ξ 2 cδ + δ 2 − g11 (u˙ 1 )2 



= g22 (ξ 2 )2 g22 (u˙ 2 )2 + g11 (u˙ 1 )2 − 2ξ 2 cδ + δ 2 − g11 (u˙ 1 )2 = g22 (ξ 2 )2 − 2ξ 2 cδ + δ 2 − g11 (u˙ 1 )2 .

This quadratic equation has the solutions 1 g22 1 = g22 1 = g22

2 ξ1,2 =

  

cδ ± cδ ±

 



c2 δ 2 − g22 (δ 2 − g11 (u˙ 1 )2 ) 

c2 δ 2 − g22 δ 2 + g22 − c2 = 



c · cos β ± | sin β| g22 − c2 .

If β ∈ [0, π], then it follows from c = 2 (0) =  ξ1,2

1 g22 (u10 )







 1 cδ ± (g22 − c2 )(1 − δ 2 ) g22

g22 (u10 ) · cos Θ0 that

(cos Θ0 · cos β ± sin Θ0 · sin β) =

Since on the other hand ξ2 =

cos (Θ0 + β) 

g22 (u10 )

cos (Θ0 ∓ β) 

g22 (u10 )

.

,

we obtain (3.89).  √ Furthermore, since g22 − c2 = g11 g22 · |u˙ 1 |, it follows from (3.89) by (3.92) δg22 − c2 cos β + c · sin β · δ − cξ 2 = g11 u˙ 1 g11 g22 u˙ 1 c · sin β = sgn(u˙ 1 ) · √ + cos β · u˙ 1 . g11 g22

ξ1 =

√ g11 g22 · |u˙ 1 |

Finally, the identity in (3.88) follows, since ξ 1 (0) =

sin (Θ0 + β) . √ g11 g22

We obtain for the geodesic line on the circular cone from Visualization 3.2.10 1

u (s) = and



2





α αs + u10 · sin Θ0 u˙ (s) = . u1 (s) 1

2

(αs + u10 · sin Θ0 ) + (u10 · cos Θ0 )

320 ■ Differential Geometry and Its Visualization

Since 

g22 −

c2

=

√

1

g11 g22 · |u˙ | =





a + 1 (u1 )2 α αs + u10 · sin Θ0 · · a a u1 (s)

 1  = √ αs + u10 · sin Θ0 , a



(3.89) yields (3.91) and (3.88) yields (3.90).

3.8

THEOREMA EGREGIUM

One of the most important results of the theory of surfaces states that the Gaussian curvature K of a surface is a quantity of its intrinsic geometry. This result is Gauss’s theorema egregium. To be able to prove this we have to express the Gaussian curvature K in terms of the first fundamental coefficients and their derivatives. Since K = L/g, we have to establish a relation between the first and second fundamental coefficients. We need the following result. Theorem 3.8.1. (Derivation formulae by Gauss, and Mainardi and Codazzi) Let S be a surface with a parametric representation ⃗x(uj ). We denote the partial differentiation with respect to the parameter uj by ;j , and put for i, k, j, m = 1, 2 m = Rikj



m ik



;j

−





m ij



;k

+



r ik

m m m Gm ikj = Rikj − Lik Lj − Lij Lk

and

Cikj = Lik;j − Lij;k +









r Lrj − ik

m rj





−



Lrk .

r ij



r ij



m rk



(3.94) (3.95)

(3.96)

Then we have Gm ikj = 0 (m, i, k, j = 1, 2) (Gauss)

(3.97)

Cikj = 0

(3.98)

and (i, k, j = 1, 2) (Mainardi and Codazzi).

Proof. We know from (3.15) in the proof of Theorem 3.1.8 that ⃗xik =





r ⃗ for i, k = 1, 2. ⃗xr + Lik N ik

(3.99)

The Intrinsic Geometry of Surfaces ■ 321

Those formulae are referred to as the Gauss derivation formulae. Now it follows from (3.99) and the Weingarten equations (2.141) in Theorem 2.10.1 that 











Lik Lm j





∂ r r ⃗ = r ⃗ + Lik N ⃗j ⃗xr + Lik N ⃗xr + ⃗xrj + Lik;j N j ik ik ;j ik ∂u        r r m ⃗ ⃗ − Lik Lrj ⃗xr = ⃗xr + ⃗xm + Lrj N + Lik;j N ik ;j ik rj

⃗xikj =

=



m ik



;j

+



r ik



m rj

−





⃗ Lik;j + ⃗xm + N







r Lrj . ik

Interchanging the roles of j and k, we obtain ⃗xijk =



m ij



;k

+



r ij



m rk



−

Lij Lm k





⃗ Lij;k + ⃗xm + N



r ij





Lrk ,

hence by (3.94), (3.95) and (3.96) ⃗0 = ⃗xikj − ⃗xijk =





m ik



;j



m ij





r ik













m r m m − + − − Lik Lm ⃗xm j + Lij Lk rj ij rk ;k       r r ⃗ + Lik;j − Lij;k + Lrj − Lrk N ik ij 

m m ⃗ − Lik Lm xm + Cikj N = Rijk j + Lij Lk ⃗

⃗ for i, k, j = 1, 2. = Gm xm − Cijk N ikj ⃗

⃗ are linearly independent, the identities in (3.97) and Since the vectors ⃗x1 , ⃗x2 and N (3.98) follow. m Remark 3.8.2. The values Rikj form the so–called Riemann tensor of curvature; they are quantities of the intrinsic geometry of a surface by (3.94). We define r Rijkl = gir Rjkl for i, j, k, l = 1, 2.

(3.100)

Theorem 3.8.3 (Theorema egregium; Gauss). The Gaussian curvature K is a quantity of the intrinsic geometry of a surface; it can be expressed by the first fundamental coefficients and their derivatives up to the second order. The following holds K=

r g2r R112 R1212 = . g g

(3.101)

Proof. It follows from the identities in (3.94) and (3.97) and the definition of Lrk in (2.140) that r Rikj = Grikj + (Lik Lrj − Lij Lrk ) = Lik Lrj − Lij Lrk = Lik g rm Lmj − Lij g rm Lmk

= g rm (Lik Lmj − Lij Lmk ) .

322 ■ Differential Geometry and Its Visualization

We obtain for i = k = 1 and j = m = 2 r g2r R112 = L11 L22 − L212 = L.

(3.102)

By the definition of the Riemann tensor of curvature in (3.94), the values on the lefthand side of (3.102) depend on the first fundamental coefficients and their derivatives up to the second order only, hence are quantities of the intrinsic geometry of the surface. Finally, (3.102) and (3.100) yield K=

r L g2r R112 R1212 = = . g g g

Example 3.8.4 (The Gaussian curvature of Poincar´ e’s half–plane). Poincar´e’s half–plane of Definition (3.2.7) has constant Gaussian curvature K = −1. Proof. We saw in the proof of Example 3.2.8 that 

1 11





=

2 12



=



2 21



=−



1 1 =− 1 22 u



and



i jk



= 0 otherwise.

Since g11 = g22 = 1/u2 and g12 = 0, we obtain by the definition of the Riemannian tensor of curvature in (3.94) 

2 11





−

2 12







+

r 11

2 r2









r 2 12 r1 ;2 ;1       1 1 2 2 2 =0− 1 2 + +0−0− 11 12 12 21 (u )         1 1 1 1 1 1 = − 1 2 + − 1 · − 1 − − 1 · − 1 = − 1 2, (u ) u u u u (u )

2 R112 =

−

and so by (3.101) K=

r 2 g22 R112 R2 1 g2r R112 = = 112 = − 1 2 · (u1 )2 = −1. g g g11 (u )

We obtain the following result as a consequence of (3.101). Lemma 3.8.5. The Gaussian curvature K of a surface with orthogonal parameters uk (k = 1, 2) is given by 1 K=− √ 2 g



∂ ∂u2



1 ∂g11 √ g ∂u2



∂ + ∂u1



If the surface has geodesic parameters then √ ∂ 2 ( g22 ) √ + K g22 = 0. ∂(u1 )2

1 ∂g22 √ g ∂u1



.

(3.103)

(3.104)

The Intrinsic Geometry of Surfaces ■ 323

Proof. Since g12 = 0, we have by (3.101) and (3.94) 

1 g22 2 R112 = K= g g11

2 11



;2

−



2 12



;1



+

r 11



2 r2



−



r 12



2 r1



.

(i) First we show the identity in (3.103). It follows from (3.7) and (3.8) in Corollary 3.1.5 that 







1 ∂g11 1 1 1 2 = = g11 ;1 , =− g11 ;2 , 1 11 11 2g11 ∂u 2g11 2g22       1 1 1 1 2 2 = g11 ;2 , = g22 ;1 and = g22 ;2 , 12 12 22 2g11 2g22 2g22 hence 



2 = 11 ;2    1 2 = 11 12    1 2 − = 12 11





g22 ;11 (g22 ;1 )2 2 g22 ;2 g11 ;2 g11 ;22 , − − = , 12 ;1 2(g22 )2 2g22 2g22 2(g22 )2    1 1 2 2 g11 ;1 g22 ;1 , =− g11 ;2 g22 ;2 , 11 22 4g11 g22 4(g22 )2    2 2 1 (g22 ;1 )2 2 (g11 ;2 ) , − =− , 12 21 4g11 g22 4(g22 )2

and so by the identity at the beginning of the proof 1 K= g11



g11 ;2 g22 ;2 g11 ;22 g22 ;11 (g22 ;1 )2 − − + + 2(g22 )2 2g22 2g22 2(g22 )2 g11 ;1 g22 ;1 g11 ;2 g22 ;2 (g11 ;2 )2 (g22 ;1 )2 − + − + 4g11 g22 4(g22 )2 4g11 g22 4(g22 )2



. (3.105)

On the other hand, we have  

g11 ;2 √ g g22 ;1 √ g



;2

g11 ;22 g11 ;2 =√ − (g11 ;2 g22 + g11 g22 ;2 ), g11 g22 2(g11 g22 )3/2

;1

g22 ;11 g22 ;1 =√ − (g11 ;1 g22 + g11 g22 ;1 ) g11 g22 2(g11 g22 )3/2



and 

1 g11 ;2 − √  √ 2 g g



;2

+



   1 g11 ;22 (g11 ;2 )2 g11 ;2 g22 ;2 g22 ;1  = − + + − √ 2

g

g11

;1

−

2g22

4g11 g22

g22 ;11 g22 ;1 g11 ;1 (g22 ;1 )2 + + 2g22 4g22 g11 4(g22 )2

Comparing (3.105) and (3.106), we obtain (3.103).

4(g22 )



. (3.106)

324 ■ Differential Geometry and Its Visualization

(ii) Now we show the identity in (3.104). For geodesic parameters, we have g12 = 0 and g11 = 1, and so g11 ;i = g11 ;ik = 0 for i, k = 1, 2. We obtain from (3.103) 1 − g22 K = 2 √





;1



g22 ;1 √ 2 g22

1 √ g22 ;1 g22

g22 ;11 (g22 ;1 )2 = √ − . 2 g22 4(g22 )3/2

On the other hand we have √

( g22 ) ;11 =



, ;1

√  √ hence K g22 + g22 ;11 = 0, which is (3.104).

3.9

MAPS BETWEEN SURFACES

In this section, we study certain maps between surfaces, namely conformal, isometric and area preserving maps. Some of the results presented here will be used later in the proof of the Gauss–Bonnet theorem, Theorem 3.10.1, and the study of minimal surfaces in Section 3.11. The most important application of maps between surfaces is in cartography. Definition 3.9.1. Let S and S ∗ be surfaces with admissible parametric representations ⃗x(ui ) and ⃗x¯(¯ ui ). Then a map from S onto S ∗ is said to be admissible if it satisfies the following conditions: (i) the map is one-to-one; (ii) it is given by

u¯k = hk (ui ) with hk ∈ C r (D) for k = 1, 2,

(3.107)

where r ≥ 1 is chosen according to need;

(iii) the corresponding Jacobian has no zeros. Remark 3.9.2. (a) Since we will always assume that maps between surfaces are admissible, we will omit the word admissible. (b) We may introduce new parameters u∗i (i = 1, 2) on the surface S ∗ by putting u¯k = hk (u∗i ) for k = 1, 2

(3.108)

with the functions hk from (3.107). Then the map from S to S ∗ has the simple form u∗k = uk for k = 1, 2. So the values of the parameters of each image point are equal to those of corresponding pre–image point. In such case, we say that S and S ∗ have the same parameter systems.

The Intrinsic Geometry of Surfaces ■ 325

Example 3.9.3 (Map of a circular cylinder onto a part of the plane). We consider the circular cylinder S = Cyl with a parametric representation ⃗x(ui ) = {r cos u2 , r sin u2 , u1 }





(u1 , u2 ) ∈ D = R × (0, 2π) ,

and the strip S ∗ of the plane with a parametric representation ⃗x¯(¯ ui ) = {¯ u1 , u¯2 , 0}





˜ = R × (0, 2π) . (¯ u1 , u¯2 ) ∈ D

Furthermore, let the functions hk be defined on D by u¯1 = h1 (ui ) = u1 and u¯2 = h2 (ui ) = ru2 .

The corresponding map from S to S ∗ obviously is one-to-one with hk ∈ C ∞ (D) for k = 1, 2 and Jacobian satisfying 



1 0 ∂(¯ ui )   =   = r ̸= 0 for all (u1 , u2 ) ∈ D. k   0 r ∂(u )

Thus the map is admissible. We introduce new parameters u∗i (i = 1, 2) on S ∗ by

u¯1 = u∗1 and u¯2 = ru∗2 . Then

⃗x∗ (u∗i ) = {u∗1 , ru∗2 , 0} for all (u∗i , u∗2 ) ∈ D∗ = D

is a new parametric representation for S ∗ . Finally, the part of the helix on S with

u1 (t) = ct and u2 (t) = t (t ∈ (0, α) ⊂ (0, 2π)) , (where c ∈ R is a given constant) between the points P0 = (1, 0, 0) and Pα = (r cos α, r sin α, cα} is mapped onto the straight line segment on S ∗ between the points P0∗ = (0, 0, 0) and Pα∗ = (cα, rα, 0). First we consider conformal or angle preserving maps. Definition 3.9.4. An admissible map of a surface S onto a surface S ∗ is said to be conformal or angle preserving, if for each pair of sections of curves on S that intersect at a point P ∈ S the angle of intersection of the image curves at P ∗ ∈ S ∗ is the same as the angle of intersection of the original curves at P . The next result is a simple characterization of conformal maps. Theorem 3.9.5. A map of a surface S onto a surface S ∗ is conformal if and only if, with respect to the same parameter systems on S and S ∗ , the first fundamental ∗ coefficients gik of S and gik of S ∗ are proportional at corresponding points, that is, if ∗ gik = η(uj )gik (η > 0; i, k = 1, 2).

(3.109)

326 ■ Differential Geometry and Its Visualization

Proof. (i) First we show the sufficiency of the condition in (3.109). We assume that the condition in (3.109) is satisfied. Let γ and γ¯ be curves on S given by ui (s) and u¯i (s) (i = 1, 2) and γ ∗ and γ¯ ∗ given by u∗i (s) and u¯∗i (s) (i = 1, 2) their images on S ∗ . Then the angles α and α∗ between the curves at their points of intersection on S and S ∗ are given by cos α = 

gik u˙ i u¯˙ k gik

u˙ i u˙ k

·



gik u¯˙ i u¯˙ k

and cos α∗ = 

∗ ∗i ˙ ∗k gik u˙ u¯ ∗ u gik ˙ ∗i u˙ ∗k ·

It follows from (3.109) that cos α = cos α∗ .



∗ u ¯˙ ∗i u¯˙ ∗k gik

.

(ii) Now we show the necessity of the condition in (3.109). We assume that a conformal map from S onto S ∗ is given and that both surfaces have the same parameter system, that is, the map is given by u∗k = uk (k = 1, 2). Let P be an arbitrary point on a segment γ1 of a curve on S; we assume that 1 2 {ξ(1) , ξ(1) } is the direction of the tangent to γ1 at P . Then the image point P ∗ of P on S ∗ has the same parameter values as P on S, the image γ1∗ on S ∗ of γ1 on S is given by the same parametric representation as γ1 on S, and, by the conformity of the map, the direction of the tangent of γ1∗ at P ∗ is also given by 1 2 {ξ(1) , ξ(1) }. Let γ2 be a segment of another curve on S, which intersects γ at a point P , 1 2 with {ξ(2) , ξ(2) } as the direction of its tangent at P . The angle α between γ1 and γ2 at P is given by cos α = 

i k ξ(2) gik ξ(1) i ξk · gik ξ(1) (1)



i ξk gik ξ(2) (2)

.

If we write ⃗t(1) and ⃗t(2) for the tangent vectors to γ1 and γ2 at P , then we obtain ⃗t(1) • ⃗t(2) cos α = ∥⃗t(1) ∥ · ∥⃗t(2) ∥

and

sin2 α = = hence



∥⃗t(1) ∥2 · ∥⃗t(1) ∥2 − ⃗t(1) • ⃗t(2) 

∥⃗t(1) ∥2 · ∥⃗t(1) ∥2

1 2 2 1 ξ(2) − ξ(1) ξ(2) g ξ(1)

2

i ξk · g ξi ξk gik ξ(1) ik (2) (2) (1)

2

=



⃗t(1) × ⃗t(2)

2

∥⃗t(1) ∥2 · ∥⃗t(1) ∥2

,

   1 2 2 1  ξ ξ − ξ ξ  √ (1) (2) (1) (2)   sin α = g ·  i ξk · gik ξ(1) (1)

i ξk gik ξ(2) (2)

.

(3.110)

The Intrinsic Geometry of Surfaces ■ 327

Analogously, we obtain for the angle α∗ between the image curves γ1∗ and γ2∗ at the image point P ∗ of P sin α∗ =

√

   1 2 2 1  ξ(2) ξ(1) ξ(2) − ξ(1)   g∗ ·  ∗ ξi ξk · gik (1) (1)

∗ ξi ξk gik (2) (2)

.

(3.111)

Since α = α∗ , (3.110) and (3.111) imply 



√ ∗ ξi ξk · ∗ ξi ξk gik gik g (1) (1) (2) (2)  √ ∗ = . i ξk · i ξk g gik ξ(1) gik ξ(2) (1) (2)

We choose γ2 as the u1 –line. Then γ2∗ is the u∗1 –line and we have 

or

 ∗ √ ∗ ξi ξk gik g g (1) (1) √ ∗ = · √ 11 i k g g11 gik ξ(1) ξ(1) 

√ √ ∗ ξi ξk  gik g g11 (1) (1) √ ∗· ∗ = = η(u1 , u2 ), i ξk g g11 gik ξ(1) (1)

(3.112)

where η only depends on P , but not on the considered direction of the tangents. 1 2 , ξ(1) }, (3.109) follows. Since (3.112) has to hold for all directions {ξ(1) Example 3.9.6. The map of Example 3.9.3 is conformal. Proof. It follows from Example 3.9.3 that ⃗x1 (ui ) = {0, 0, 1}, ⃗x2 (ui ) = r{− sin u2 , cos u2 , 0}, g11 = 1, g12 = 0, g22 = r2 , ∗ ∗ ∗ ⃗x∗1 (u∗i ) = {1, 0, 0}, ⃗x∗2 (u∗i ) = {0, r, 0}, g11 = 1, g12 = 0 and g22 = r2 .

Thus the identities in (3.109) are satisfied with η(ui ) = 1 > 0, and consequently the map is conformal by Theorem 3.9.5. Visualization 3.9.7 (The Riemann sphere and stereographic projection). The Riemann sphere and stereographic projection play an important role in complex analysis concerning the extended complex plane C. We embed the complex plane C in three-dimensional space R3 in which we introduce a Cartesian coordinate system with coordinates ξ, η and ζ in such a way that the ξ– and η–axes coincide with the x– and y–axes, respectively, of C. The sphere in R3

328 ■ Differential Geometry and Its Visualization

Figure 3.28

The stereographic projection

with radius 1/2 and its centre in the point (0, 0, 1/2) is referred to as the Riemann sphere; it is given by S = S1/2



1 0, 0, 2



=





1 (ξ, η, ζ) ∈ IR : ξ + η + ζ − 2 3

2

2

2

1 = 4



.

The point N = (0, 0, 1) ∈ S is referred to as the north pole. The stereographic projection from the complex plane to the Riemann sphere is defined as follows: Given a point z ∈ C, we determine the straight line through the points z and N which intersects the Riemann sphere in one and only one point (other than the north pole) which we denote by P (z) (Figure 3.28). In this way, we obtain a one–to–one correspondence between the points of the complex plane and the points of the Riemann sphere minus the north pole. We write ∞ for the element which is formally assigned to the north ˜ = C ∪ {∞}. pole and obtain the extended complex plane C (a) The stereographic projection is conformal. (b) Loxodromes on the Riemann sphere correspond to logarithmic spirals in the xy– plane. Proof. (a) Let S be the part of the Riemann sphere given by the parametric representation ⃗x(ui ) =

 1  · cos u1 cos u2 , cos u1 sin u2 , 1 + sin u1 2

for (u1 , u2 ) ∈ D = (−π/2, π/2) × (0, ∞). If (ξ, η, ζ) ∈ S is given and (x∗ , y ∗ ) is the corresponding point under the stereographic projection in complex plane, then ξ=

1 1 1 · cos u1 cos u2 , η = · cos u1 sin u2 and ζ = (1 + sin u1 ) 2 2 2

and we obtain (Figure 3.29)

The Intrinsic Geometry of Surfaces ■ 329

Figure 3.29

The principle of the stereographic projection

x∗ =

cos u1 cos u2 cos u1 sin u2 ξ η ∗ = = and y = . 1−ζ 1 − sin u1 1−ζ 1 − sin u1

Let S ∗ denote the image of S under the stereographic projection. Then S ∗ has a parametric representation ⃗x∗ (u∗i ) = We obtain ⃗x1 = ⃗x2 = g11 = ⃗x∗1 = = ⃗x∗2 = ∗ g11 =

  cos u∗1 ∗2 ∗2 · cos u , sin u , 0 for (u∗1 , u∗2 ) ∈ D. 1 − sin u∗1

 1  · − sin u1 cos u2 , − sin u1 sin u2 , cos u1 , 2  1  · − cos u1 sin u2 , cos u1 cos u2 , 0 , 2 1 1 , g12 = 0, g22 = · cos2 u1 , 4 4 ∗1 ∗1 − sin u (1 − sin u ) + cos2 u∗1 · {cos u∗2 , sin u∗2 , 0} 2 ∗1 (1 − sin u ) 1 · {cos u∗2 , sin u∗2 , 0}, 1 − sin u∗1 cos u∗1 · {− sin u∗2 , cos u∗2 , 0}, 1 − sin u∗1 1 cos2 u∗2 ∗ ∗ , g = 0 and g = . 12 22 (1 − sin u∗1 )2 (1 − sin u∗1 )2

We write u∗i = ui for i = 1, 2. Then identities in (3.109) are satisfied with η(ui ) =

4 > 0 on D (1 − sin u∗1 )2

and so the stereographic projection is conformal by Theorem 3.9.5.

330 ■ Differential Geometry and Its Visualization

(b) Any loxodrome γ on the Riemann sphere S intersects the meridians of S at a constant angle. Obviously the images γ ∗ of the meridians under the stereographic projection are the rays in the complex plane S ∗ that originate in the origin. Since the stereographic projection is conformal by Part (a), γ ∗ intersects those rays at a constant angle, and consequently is a logarithmic spiral by Visualization 2.3.12 (a). Figure 3.30 shows the Riemann sphere, a loxodrome on it and the surface generated by moving a constant distance along the rays of the stereographic projection of the points of the loxodrome. One of the most popular maps from the sphere into the plane in cartography is the Mercator projection. Example 3.9.8 (Mercator Projection). Let a sphere S be given by a parametric representation ⃗x(ui ) = {r cos u2 cos u1 , r cos u2 sin u1 , r sin u2 }



(u1 , u2 ) ∈ (0, 2π) ×



π π , 2 2



.

(3.113) (We note that u1 and u2 are interchanged in the usual parametric representation of a sphere.) We use the transformations x∗1

1

= u and

x∗2





u2 π + = log tan 2 4



for |u2 |
0 and f (t) = (r′ (t))2 + (h′ (t))2 > 0 for all t ∈ I. We put ϕ(t) =

t 

t0

f (τ ) dτ for t0 , t ∈ I. r(τ )

Then ϕ′ (t) > 0 on I, and so the inverse ψ of ϕ exists on ϕ(I) and ψ ′ (ϕ(t)) =

1

ϕ′ (t)

for all t ∈ I.

We introduce new parameters u∗i (i = 1, 2) by 1

1

u = u (u ) = ψ ∗i



u∗1 − u∗2 √ 2





(u1 , u2 ) ∈ D = I × (0, 2π)

and u2 =

u∗1 + u∗2 √ . 2

The Intrinsic Geometry of Surfaces ■ 333

Then we have 1 ∂u1 1 ∂u2 ∂u2 1 ∂u1 √ · = − = and = =√ . ∗1 ′ 1 ∗i ∗2 ∗1 ∗2 ∂u ∂u ∂u ∂u 2 ϕ (u (u )) 2 It is easy to see that the transformation is admissible and the transformation formulae (2.18) in Remark 2.3.3 (a) for the first fundamental coefficients ∗ gjk = glm

∂ul ∂um for j, k = 1, 2 ∂u∗j ∂u∗k

yield ∗ g11

= g11



∂u1 ∂u∗1

2

+ g22



∂u2 ∂u∗1

2

= g11



∂u1 ∂u∗2

2

+ g22



∂u2 ∂u∗2

2

f (u1 (u∗1 )) 1 1 · + · r2 (u1 (u∗1 )) 2 (ϕ′ (u1 (u∗1 )))2 2 f (u1 (u∗1 )) r2 (u1 (u∗1 )) 1 2 1 ∗1 = · + · r (u (u )) = r2 (u1 (u∗1 )) 2 f (u1 (u∗1 )) 2 ∗ = g22 =

and ∂u1 ∂u1 ∂u2 ∂u2 + g 22 ∂u∗1 ∂u∗2 ∂u∗1 ∂u∗2  2  2 ∂u1 ∂u2 = −g11 + g22 = 0. ∂u∗1 ∂u∗2

∗ = g11 g12

Thus u∗1 and u∗2 are isothermal parameters for S (Figure 3.31). The following important general result holds; it is also needed in the proof of the Gauss–Bonnet theorem, Theorem 3.10.1.

Figure 3.31

Isothermal parameters on a sphere and a pseudo-sphere

334 ■ Differential Geometry and Its Visualization

Theorem 3.9.12. Any simply connected part S of a surface of class r ≥ 3 can be mapped conformally into the plane. Proof. Let u∗k (k = 1, 2) denote the parameters of S. We show that we can introduce isothermal parameters uk (k = 1, 2) on S by proving the existence of an admissible parameter transformation uα = uα (u∗1 , u∗2 ) (α = 1, 2)

(3.116)

∗ such that the first fundamental coefficients gαβ and gik satisfy the relation in (3.109). Since the parameter transformation is admissible, the inverse transformation also exists. It follows that the transformation (3.116) must satisfy ∗ ηδµν = gαβ

µ δ µν ∂u∗α ∂u∗β ∂uν ∗αβ ∂u = g and ∂uµ ∂uν η ∂u∗α ∂u∗β

(η > 0; µ, ν = 1, 2).

If µ = 1 and ν = 2, then this yields g ∗αβ

∂u1 ∂u2 = 0, ∂u∗α ∂u∗β

(3.117)

and if µ = ν = 1 or µ = ν = 2, then we obtain ∂u1 ∂u1 ∂u2 ∂u2 1 1 = g ∗αβ ∗α ∗β or = g ∗αβ ∗α ∗β , η ∂u ∂u η ∂u ∂u whence g ∗αβ

2 ∂u1 ∂u1 ∂u2 ∗στ ∂u = g . ∂u∗α ∂u∗β ∂u∗σ ∂u∗τ

(3.118)

√ We put ϵµν = ϵαβ g µα g νβ (µ, ν = 1, 2) where ϵ11 = ϵ22 = 0 and ϵ12 = −ϵ21 = g. √ Then it is easy to see that ϵ11 = ϵ22 = 0, ϵ12 = −ϵ21 = 1/ g, ϵαγ ϵβγ = δβα and ϵαγ ϵγβ = −δβα (α, β = 1, 2). Now we show: (A) If the functions µα (α = 1, 2) in (3.116) satisfy

2 ∂u1 ∗ ∗κλ ∂u = g ϵ for α = 1, 2, ακ ∂u∗α ∂u∗λ

(3.119)

then they satisfy (3.117) and (3.118). First ϵµν = ϵαβ g µα g νβ for µ, ν = 1, 2 implies gµτ gνσ ϵµν = ϵαβ gµτ g µα gνσ g νβ = ϵαβ δτα δσβ = ϵτ σ for τ, σ = 1, 2.

(3.120)

Since ϵαγ ϵβγ = δβα for α, β = 1, 2, we have from (3.120) ϵαγ ϵβγ g βκ = δβα g βκ = g ακ = ϵαγ gµβ gνγ ϵµν g βκ = ϵαγ ϵµν gµβ g βκ gνγ = ϵαγ ϵµν δµκ gνγ = ϵ ϵ gνγ for α, κ = 1, 2, αγ κν

(3.121)

The Intrinsic Geometry of Surfaces ■ 335

and so by (3.119) g ∗αβ

2 2 ∂u1 ∂u2 ∂u2 ∂u2 ∗αβ ∗ ∗κλ ∂u ∗β ∗κλ ∂u = g g ϵ = δ ϵ ακ κ ∂u∗α ∂u∗β ∂u∗λ ∂u∗β ∂u∗λ ∂u∗β 2 2 2 2 ∂u ∂u ∂u ∂u ∂u2 ∂u2 = ϵ∗βλ ∗λ ∗β = ϵ∗12 ∗2 ∗1 − ϵ∗21 ∗1 ∗2 = 0. ∂u ∂u ∂u ∂u ∂u ∂u

Hence (3.117) holds. Furthermore, we have by (3.119) and (3.121) g ∗αβ

2 2 ∂u1 ∂u1 ∂u2 ∂u2 ∗αβ ∗ ∗κλ ∗ ∗δµ ∂u ∗β ∗κλ ∗ ∗δµ ∂u = g g ϵ g ϵ = δ ϵ g ϵ ακ βδ κ βδ ∂u∗α ∂u∗β ∂u∗λ ∂u∗µ ∂u∗λ ∂u∗µ 2 2 2 2 ∂u ∂u ∂u ∗ ∗δµ ∂u = ϵ∗βλ gβδ ϵ = g ∗λµ ∗λ ∗µ . ∗λ ∗µ ∂u ∂u ∂u ∂u

So (3.118) is also satisfied. This completes the proof of the statement in (A). The conditions of integrability for (3.119) are √ ∗ ∗αβ ∂ 2 u1 ∂ 2 u1 ∂ 2 u1 = or equivalently g ϵ = 0. ∂u∗1 ∂u∗2 ∂u∗2 ∂u∗1 ∂u∗α ∂u∗β √ √ Now (3.119), g ∗ ϵ∗αβ = ±1 or g ∗ ϵ∗αβ = 0 and (3.121) together imply √ ∗ ∗αβ g ϵ

√ ∗ ∗αβ ∂ ∂ 2 u1 = g ϵ ∂u∗α ∂u∗β ∂u∗β ∂ = ∂u∗β



√



∂u1 ∂u∗α



∗ ∗κλ g ∗ ϵ∗αβ gακ ϵ

=

√

g ∗ ϵ∗αβ

∂u2 ∂u∗λ



∂ ∂u∗β

∂ = ∂u∗β



∗ ∗κλ gακ ϵ



√

g ∗ g ∗βλ

∂u2 ∂u∗λ ∂u2 ∂u∗λ





= 0.

(3.122)

This is a linear partial differential equation with continuously differentiable coefficients. It has a non–constant solution in any sufficiently small part S1 of S, as can be shown by the method of successive approximations. A similar differential equation follows when we solve (3.119) for ∂u2 /∂u∗λ . From (3.120) and (3.119), we obtain 2 2 2 ∂u2 ∗β ∂u ∗βτ ∗ ∂u ∗βτ ∗ ∗ ∗µν ∂u = δ = ϵ ϵ = ϵ g g ϵ σ στ µσ ντ ∂u∗σ ∂u∗β ∂u∗β ∂u∗β 2 1 ∗ ∗µν ∗ ∗τ β ∂u ∗ ∗µν ∂u = −gµσ ϵ gντ ϵ = −gµσ ϵ , ∂u∗β ∂u∗ν

that is,

1 ∂u2 ∗ ∗βα ∂u = −g ϵ for σ = 1, 2. (3.123) σβ ∂u∗σ ∂u∗α The equations in (3.119) and (3.123) are generalizations of the well–known Cauchy– Riemann differential equations in the theory of complex functions. ∗ ∗ If u∗1 and u∗2 are isothermal parameters, then gβσ = ηδβσ for β, σ = 1, 2 and g ∗ = η 2 , hence by (3.119) 2 2 2 ∂u1 η ∂u2 ∂u2 ∗ ∗κλ ∂u ∗ ∗κλ ∂u ∗1λ ∂u √ = g ϵ = ηδ ϵ = ηε = = , 1κ 1κ ∂u∗1 ∂u∗λ ∂u∗λ ∂u∗λ g ∗ ∂u∗2 ∂u∗2

336 ■ Differential Geometry and Its Visualization

and by (3.123) 1 1 1 η ∂u1 ∂u1 ∂u2 ∗ ∗βα ∂u ∗ ∗βα ∂u ∗1α ∂u √ = −g ϵ = −ηδ ϵ = −ηϵ = − = − . 1β 1β ∂u∗1 ∂u∗α ∂u∗α ∂u∗α g ∗ ∂u∗2 ∂u∗2

If u2 (u∗1 , u∗2 ) = const is a family of curves satisfying (3.122), then we obtain the second family u1 = const of isothermal parameters u1 and u2 on S1 from (3.119). It follows from (3.109) that 2

η =g=g

∗



∂(u∗1 , u∗2 ) ∂(u1 , u2 )

2

.

We show that the Jacobian of the transformation (3.116) does not vanish. From (3.119), we obtain 2 2 2 2 ∂u1 ∂u2 ∂u1 ∂u2 ∂(u1 , u2 ) ∗ ∗κλ ∂u ∂u ∗ ∗κλ ∂u ∂u = − = g ϵ − g ϵ 1κ 2κ ∂(u∗1 , u∗2 ) ∂u∗1 ∂u∗2 ∂u∗2 ∂u∗1 ∂u∗λ ∂u∗2 ∂u∗κ ∂u∗1

=

∗ ∗12 g11 ϵ





∂u2 ∂u∗2

1 ∗ = √ ∗ g22 g

2

∗ ∗21 + g12 ϵ

− 

∂u2 ∂u∗1

∗ ∗21 g22 ϵ

2

−

2 2 ∂u2 ∂u2 ∗ ∗12 ∂u ∂u − g ϵ 21 ∂u∗1 ∂u∗2 ∂u∗2 ∂u∗1



∗ 2g12

∂u2 ∂u∗1

2

∂u2 ∂u2 ∗ + g11 ∗1 ∗2 ∂u ∂u



∂u2 ∂u∗2

2  .

This quadratic form is positive definite, it can only vanish when both partial derivatives ∂u2 /∂u∗α (α = 1, 2) vanish. This cannot happen for an admissible solution of (3.122). Therefore, we can find a conformal map of a part S1 of S into the plane. Similarly a local conformal map can be found for any other sufficiently small part of S. All these local maps have to be joined to a global conformal map of S into the plane. This can be achieved by a principle from complex analysis. Now we consider isometric or length preserving maps. Definition 3.9.13. An admissible map from a surface S to a surface S ∗ is said to be isometric or length preserving if the length of any curve γ on S is equal to the length of its image γ ∗ on S ∗ . The following result gives a characterization of isometric maps. Theorem 3.9.14. An admissible map from a surface S onto a surface S ∗ is isometric if and only if ∗ gik (ui , u2 ) = gik (u∗1 , u∗2 ) (i, k = 1, 2) (3.124) ∗ (i, k = 1, 2) are at all corresponding pairs (u1 , u2 ) and (u∗1 , u∗2 ), where gik and gik ∗ ∗ the first fundamental coefficients of S and S , and S and S have the same parameter systems.

The Intrinsic Geometry of Surfaces ■ 337

Proof. Let u1 and u2 be the parameters of S, and u∗1 and u∗2 be the parameters of S ∗ . Since we assume that S and S ∗ have the same parameter systems, the map from S to S ∗ is given by u∗α = uα for α = 1, 2. If the functions uα = ϕα (t) (t ∈ [0, t1 ]) for α = 1, 2

define a curve γ on S, then its image γ ∗ is given by

u∗α = ϕα (t) (t ∈ [0, t1 ]) for α = 1, 2

with the same functions ϕα , since S and S ∗ have the same parameter systems (Remark 3.9.2 (b)). A part γ˜ of γ with uα = ϕα (t) (t ∈ [0, t0 ]), where 0 ≤ t0 ≤ t1 has the arc length  s(t0 ) =

t0

gik (ϕ1 (t), ϕ2 (t))

0

dϕi dϕk · dt for 0 ≤ t0 ≤ t1 , dt dt

and the corresponding image γ˜ ∗ has the arc length s (t0 ) = ∗

t0 0



∗ (ϕ1 (t), ϕ2 (t)) gik

dϕi dϕk · dt for 0 ≤ t0 ≤ t1 . dt dt

∗ If gik (u1 , u2 ) = gik (u∗1 , u∗2 ) (i, k = 1, 2) for all pairs (u1 , u2 ) and corresponding pairs (u∗1 , u∗2 ), then it follows that s(t0 ) = s∗ (t0 ). Conversely if we assume that a curve γ and its image γ ∗ have the same lengths, and if this also holds for each part γ˜ of γ and its image γ˜ ∗ , then the integrands of both integrals above must be equal. If the length of any curve on S is to be the same as that of its image on S ∗ , then the integrands of both integrals above have to be equal for any arbitrary pair of functions ϕα and any point t, that is, for all pairs (u1 , u2 ) and corresponding pairs (u∗1 , u2∗ ), we must have ∗ gik (ui , u2 ) = gik (u∗1 , u∗2 ) (i, k = 1, 2).

Remark 3.9.15. (a) It follows from Theorems 3.9.5 and 3.9.14 that every isometric map is conformal. (b) The map of Example 3.9.3 is isometric. We are going to show that helicoids and catenoids are isometric. Example 3.9.16. Let a > 0 be a constant and D = R × (0, 2π). Then the helicoid given by   ⃗⃗x¯(¯ ui ) = {¯ u1 cos u¯2 , u¯1 sin u¯2 , a¯ u2 } (¯ u1 u¯2 ) ∈ D and the catenoid given by

⃗x(ui ) = {a cosh u1 cos u2 , a cosh u1 sin u2 , au1 }

are isometric.



(u1 , u2 ) ∈ D



338 ■ Differential Geometry and Its Visualization

Proof. The first fundamental coefficients g¯ik of the helicoid and gik of the catenoid are g¯11 = 1, g¯12 = 0, g¯22 = a2 + (¯ u1 )2 , and

g11 = g22 = a2 cosh2 u1 , and g12 = 0.

We choose suitable parameters u∗1 and u∗2 for the helicoid. By Theorem 3.9.14, we must have  2  2 1 2 ∂ u ¯ ∂ u ¯ ∗ = g¯11 = g¯11 + g¯22 . g11 ∂u∗1 ∂u∗1 Putting

u¯1 = a sinh u∗1 and u¯2 = u∗2 ,

we obtain

(3.125)

∗ g11 (u∗i ) = a2 cosh2 u∗1 = g¯11 (¯ ui ).

Obviously the transformation (3.125) is one-to-one. Since the Jacobian also satisfies 



∂(¯ u1 , u¯2 ) a cosh u∗1 0 = det = a cosh u∗1 > 0 for all (u∗1 , u+2 ) ∈ D, 0 1 ∂(u∗1 , u∗2 ) the transformation (3.125) is admissible. The parametric representation of the helicoid with respect to the parameters u∗1 and u∗2 is given by 

⃗x∗ (u∗i ) = a sinh u∗1 cos u∗2 , a sinh u∗1 sin u∗2 , au∗2



with the first fundamental coefficients satisfying in the same parameter systems u∗i and u∗2 ∗ ∗ g11 (u∗i ) = a2 cosh2 u∗1 = g11 (ui ), g12 (u∗i ) = 0 = g12 (ui ) and

∗ g22 (u∗i ) = a2 (1 + sinh2 u∗1 ) = a2 cosh2 u∗1 = g22 (u∗i ).

Thus the helicoid and catenoid are isometric by Theorem 3.9.14. Example 3.9.16 is a special case of a more general result which we are going to prove next, namely that every screw surface is isometric to some surface of revolution. The following result holds. Theorem 3.9.17 (Bour). Every screw surface S can be mapped isometrically onto a surface of revolution RS. Proof. We may assume that the surface of revolution RS and the screw surface S are given by the parametric representations ⃗x(ui ) = {u1 cos u2 , u1 sin u2 , h(u1 )}

(3.126)

The Intrinsic Geometry of Surfaces ■ 339

and

⃗x¯(¯ ui ) = {¯ u1 cos u¯2 , u¯1 sin u¯2 , ξ(¯ u1 ) + c¯ u2 }

with the first fundamental coefficients

g11 (ui ) = 1 + (h′ (u1 ))2 , g12 = 0, g22 (ui ) = (u1 )2 , ui ) = 1 + (ξ ′ (¯ u1 ))2 , g¯12 (¯ ui ) = cξ ′ (¯ u1 ) and g¯22 (¯ ui ) = (¯ u1 )2 + c2 , g¯11 (¯ and the first fundamental forms 



ds = 1 + (h′ (u1 ))2 (du1 )2 + (u1 )(du2 )2 and 



2  

d¯ s = 1 + ξ ′ (¯ u1 ) 



d¯ u1

2

2  

2  

u¯1 ξ ′ u¯1 = 1+ (¯ u1 )2 + c2 We put

du

d¯ u

1 = η

∗2



+ 2cξ ′ (¯ u1 )d¯ u1 d¯ u2 + 1



2

+



u¯

1

2

+c

2

u¯1



2

+ c2

(3.127) 

d¯ u2

2

2

cξ ′ (¯ u1 ) · d¯ u1 + d¯ u2 . 2 1 2 (¯ u ) +c (3.128) 

cξ ′ (¯ u1 ) · d¯ u1 + d¯ u2 , (¯ u1 )2 + c2

where η ̸= 0 is a constant, that is, we use the transformation u¯1 = u∗1 and u¯2 = −c ·



ξ ′ (u∗1 ) du∗1 + ηu∗2 . (u∗1 )2 + c2

Then u∗1 and u∗2 are orthogonal parameters for S ∗ (as in Example 3.6.1) such that the u∗1 –lines are helices and the u∗2 –lines are their orthogonal trajectories. Now it follows from (3.128) that the first fundamental form of the screw surface S ∗ with respect to the parameters u∗1 and u∗2 is given by 



2  

u∗1 ξ ′ u∗1 ds = 1 + (u∗1 )2 + c2 ∗

2  

du

∗1

2

+η

2



u

∗1

2

+c

2



du∗2

2

.

(3.129)

The surfaces RS and S are isometric by Theorem 3.9.14 if and only if their fundamental forms (3.127) and (3.129) are of the same form. This can be achieved by putting u2 = u∗2 , (3.130)

and

 

u

1

2 

=η

2

1



2 

1 + h (u ) ′

u

∗1

·

2

+c



du1 du∗1

2



2

for some constant η ̸= 0 

2  

u∗1 ξ ′ u∗1 =1+ (u∗1 )2 + c2

2

.

(3.131)

(3.132)

340 ■ Differential Geometry and Its Visualization

Remark 3.9.18. The relations in (3.130), (3.131) and (3.132) enable us to find a screw surface isometric to a given surface of revolution, and conversely, a surface of revolution isometric to a given screw surface. If a surface of revolution is given, then we use (3.131) to eliminate the parameter u1 in (3.132) and then solve (3.132) for ξ ′ (u∗1 ) to find the function ξ of the screw surface. Conversely, if a screw surface is given, then we use (3.131) to eliminate the parameter u∗1 in (3.132) and then solve (3.132) for h′ (u1 ) to find the function h of the surface of revolution. We apply this method to find screw surfaces which are isometric to a catenoid. Visualization 3.9.19. Let a > 0 and a catenoid S be given by the parametric representation 

⃗x(˜ ui ) = a cosh u˜1 cos u˜2 , a cosh u˜1 sin u˜2 , a˜ u1



˜ = (0, ∞) × (0, 2π). for (˜ u1 , u˜2 ) ∈ D

We apply the method described in Remark 3.9.18. Using the transformation 1

u˜ = cosh

−1



u1 a







= log u1 +

(u1 )2 − a2



for u1 > a and u˜2 = u2 ,

we obtain the following parametric representation for S in the new parameters ⃗x(u ) = i



1

2

1

2

u cos u , u sin u , a · cosh

−1



u1 a



for (u1 , u2 ) ∈ D = (a, ∞) × (0, 2π),

which is the parametric representation in (3.126) in the proof of Theorem 3.9.17 with h(u1 ) = a · cosh−1 (u1 /a) and 1

h (u ) =  ′



a

1

2

and 1 + h (u )

(u1 )2 − a2

′



2

u1 = . (u1 )2 − a2

First (3.131) yields u1 du1 = η 2 u∗1 du∗1 , that is, ∗1 du1 2 u = η · . du∗1 u1

Substituting this in (3.132), eliminating u1 and solving for ξ(u∗1 ), we obtain 



1

2 

1 + h (u ) ′



du1 du∗1





2



2



2

u1 u∗1 η 4 u∗1   =η · · = (u1 )2 − a2 (u1 )2 η 2 (u∗1 )2 + c2 − a2 4



2

η 2 u∗1 = (u∗1 )2 + c2 − =1+



u∗1

2 

′

a2 η2



ξ u∗1

(u∗1 )2

+ c2

2

for u

∗1

>



a a2 , − c2 and c < 2 η |η|

The Intrinsic Geometry of Surfaces ■ 341

and  

ξ ′ u1∗





2



2

u∗1 + c2  η 2 u∗1 = (u∗1 )2 (u∗1 )2 + c2 −

2





2

u∗1 + c2 = · 2 2 (u∗1 ) (u∗1 )2 + c2 − ηa2 1

a2 η2



u



− 1

∗1

2



a2 (η − 1) − c − 2 η 2

2



.

We choose η = 1 and put k 2 = a2 − c2 for a > c, and obtain  

ξ ′ u1∗

that is,

)=



u∗1

(u∗1 )2



(u∗1 )2 + c2 +



2



   ′ ∗ ξ (u ) = k ·  

This yields ξ(u∗1 ) = k · log

2



+ c2 k 2

(u∗1 )2 

−

u∗1 

k2 2

 for u∗1 > k > 0,

+ c2

(u∗1 )2 (u∗1 )2 − k 2

(u∗1 )2 − k 2



. 

    ∗1 2 2  u +c  ˜ k − c · tan−1  ·   + d, ∗1 2 2

(u ) − k

c

where d˜ is a constant. We observe that we may choose d˜ = 0, since a change in d˜ only results in a movement of the screw surface in the direction of the x3 –axis. For every k with 0 < k ≤ a, that is, for every c with 0 ≤ c < a, we obtain a screw surface Sk with ξ(u∗1 ) = k · log



(u∗1 )2 + c2 +



(u∗1 )2 − k 2





    ∗1 2 2 +c  k  u − c · tan−1  ·   2 ∗1 2

(u ) − k

c

for u∗1 > k, which is isometric to the catenoid. If k = a, that is, c = 0, then we obtain the original catenoid with ξ(u ) = a cosh ∗1

−1



u∗1 a





= log u∗1 +





(u∗1 )2 − a2 .

If k = 0, that is, c = a, then ξ = 0 and we obtain a helicoid (Figure 3.32). It turns out that the Gaussian and geodesic curvature are invariant under isometric maps. Theorem 3.9.20. The Gaussian curvature K of a surface and the geodesic curvature κg of a curve on a surface are invariant under isometric maps.

342 ■ Differential Geometry and Its Visualization

Figure 3.32

Isometric maps of Visualization 3.9.19

The Intrinsic Geometry of Surfaces ■ 343

Proof. The Gaussian curvature K of a surface depends only of the first fundamental coefficients by the theorema egregium, Theorem 3.8.3; also the geodesic curvature κg of a curve on a surface only depends on the first fundamental coefficients by Theorem 3.1.8. Remark 3.9.21. Isometric surfaces must necessarily have the same Gaussian curvature in corresponding points. Since the Gaussian curvature of a sphere of radius R is equal to 1/R and that of a plane is equal to 0, there is no isometric map between any part of a sphere and a plane; in particular, the stereographic projection is not isometric. In general, the equality of the Gaussian curvature is not sufficient for the isometry of surfaces except in the case of surfaces with constant Gaussian curvature. Theorem 3.9.22. Under an isometry, the image γ ∗ of any curve γ of minimal length between the points P1 and P2 is also a curve of minimal length between the image points P1 ∗ and P2∗ .

Proof. If γ ∗ were not the shortest connection between P1∗ and P2∗ , then there would be a shorter curve γ˜ ∗ between P1∗ and P2∗ . The inverse image γ˜ of γ˜ ∗ would then be shorter than γ by isometry, a contradiction to the assumed minimality of γ. The class of surfaces that are isometric to a part of the plane is relatively small, namely that of developable surfaces, as we will see in Theorem 3.9.25. The proof uses the fact that developable surfaces can be characterized as surfaces of identically vanishing Gaussian curvature; this is the result of Theorem 3.9.24. We need the next Lemma in the proof of Theorem 3.9.24. Lemma 3.9.23. A curve γ with a parametric representation ⃗x(t) (t ∈ I) for which the vectors ⃗x′′ (t) and ⃗x′ (t) are linearly dependent for all t is a straight line segment.

Proof. Let ⃗x(t) = {x1 (t), x2 (t), x3 (t)}. Since ⃗x′ (t) ̸= ⃗0, there exists a function c such that (xk )′′ (t) = c(t)(xk )′ (t) (k = 1, 2, 3) for all t ∈ I. Writing y k = (xk )′ for k = 1, 2, 3, we obtain (y k )′ (t) = c(t), y k (t) and by integration y k (t) = Ck exp





c(t) dt for k = 1, 2, 3,

where the Ck are constants of integration. One more integration yields x (t ) = Ck t + dk with t (t) = k

∗

∗

∗



exp





c(t) dt dt for k = 1, 2, 3,

where the dk are constants of integration. These are the component functions of a straight line segment with t∗ as their parameter.

344 ■ Differential Geometry and Its Visualization

Theorem 3.9.24. A surface S of class C r for r ≥ 2 is part of a developable surface if and only if its Gaussian curvature K vanishes identically. Proof. (i) First we assume that S is a developable surface. ⃗ are constant along each u1 –line by Definition Then its surface normal vectors N 2.8.14 and consequently its Gauss spherical image is a curve. This implies K = 0 by (2.159). (ii) Now we assume K = 0. Then K = L/g implies L = 0. If Lik = 0 for all i, k, then S is a plane by Example 2.10.2. So let L11 L22 = (L12 )2 , (3.133) where not all of the coefficients Lik vanish. The differential equations (2.121) for the asymptotic lines on S reduce to L11 (du1 )2 + 2L12 du1 du2 + L22 (du2 )2 =



|L11 |du1 +



|L22 |du2

2

= 0,

(3.134) and consequently only one family of asymptotic lines exists; this has to be the case by Remark 2.6.2 (a), when all points are parabolic points. We assume that the parameters u1 and u2 of S are chosen such that the asymptotic lines coincide with the u1 –lines. Then it follows from (3.134) that L11 du1 = 0, and then L12 = 0 by (3.133). Now we obtain from the Weingarten equations (2.141) and (2.140) 



⃗ 1 = −Lk ⃗xk = −g kj Lj1⃗xk = − g k1 L11 + g k2 L21 ⃗xk = ⃗0, N 1 ⃗ do not depend on u1 . Consequently they hence the surface normal vectors N are constant along each asymptotic line, which is given by u2 = const in our parameter system. If these asymptotic lines are straight lines, then the surface is part of a developable surface that is generated by these straight lines. This is indeed the case, since if L11 = L12 = 0 and L22 ̸= 0 the Mainardi–Codazzi equations (3.98) and (3.96) reduce for i = k = 1 and j = 2 to C112 = L11;2 − L12;2 −





r Lr2 − 11





r Lr1 = 12





2 L22 = 0 11

and so the Gauss derivation formulae (3.99) yield ⃗x11 =





k ⃗ = ⃗xk + L11 N 11





1 ⃗x1 11

Hence ⃗x′′11 and ⃗x′1 are linearly dependent and the asymptotic lines are straight line segments by Lemma 3.9.23.

The Intrinsic Geometry of Surfaces ■ 345

Theorem 3.9.25. Any sufficiently small part S of a surface of class C r for r ≥ 3 can be mapped isometrically into a plane if and only if S is part of a developable surface. Proof. (i) First we assume that S can be mapped isometrically into a plane. Then the Gaussian curvature K of S is equal to that of the plane by Theorem 3.9.20. Since the Gaussian curvature of a plane is identically equal to 0 it follows that K ≡ 0 and so S is a part of a developable surface by Theorem 3.9.24. (ii) Now we show that a sufficiently small part S of a developable surface can be mapped isometrically into a plane. We introduce geodesic parallel coordinates u1 and u2 on S as follows: We choose a geodesic line γg as the u2 –line corresponding to u1 = 0, the geodesic lines orthogonal to γg as the u1 –lines corresponding to u2 = const, and their orthogonal trajectories as the u2 –lines corresponding to u1 = const. The first fundamental form for a suitable choice of u1 is given by (3.72) as 

ds2 = du1

2



+ g22 du2

2

.

If we take the arc length along γg as the parameter u2 , then we have ds = du2 on γg , hence g22 (0, u2 ) = 1. (3.135) Since γg is a geodesic line it follows from (3.21) that 

∂g22  = 0. ∂u1 u1 =0

(3.136)

Since K = 0 by Theorem 3.9.24 and S has geodesic parameters, it follows from (3.104) that √ ∂ 2 g22 = 0. (∂u1 )2 Integrating twice we obtain √

g22 = c1 (u2 )u1 + c2 (u2 ).

It follows from (3.135) that c1 ≡ 1, and from (3.136) that c1 (u2 ) = 0. Thus we have g11 = g22 = 1 and g12 = 0. If u∗1 and u∗2 are the Cartesian parameters of the plane, then u∗i = ui for i = 1, 2 defines a map which is isometric by Theorem 3.9.14. Remark 3.9.26. If S is a developable surface then the orthogonal trajectories of geodesic lines are geodesic lines provided at least one of the trajectories is a geodesic line. This property only holds for developable surfaces, for otherwise parts of some other surfaces could be mapped isometrically into the plane in contradiction to Theorems 3.9.26 and 3.9.20.

346 ■ Differential Geometry and Its Visualization

Finally, we consider area preserving maps. Definition 3.9.27. An admissible map from a surface S onto a surface S ∗ is said to be area preserving if any part D of S is mapped onto a part D∗ of S ∗ which has the same surface area as D. We obtain a characterization for area preserving maps similar to the characterizations of isometric and conformal maps in Theorems 3.9.5 and 3.9.14, respectively. Theorem 3.9.28. An admissible map from a surface S onto a surface S ∗ is area preserving if and only if ∗ g = det(gik ) = g ∗ = det(gik )

(3.137)

∗ at every point on S and its image point on S ∗ , where gik and gik (i, k = 1, 2) are the ∗ ∗ first fundamental coefficients of S and S , and S and S have the same coordinate systems.

Proof. We consider the integral for the surface area of any part D of S A(D) =



√

g du1 du2 ,

D

and the corresponding integral for the surface area of the image D∗ on S ∗ A(D∗ ) =



√

g ∗ du ∗1 du ∗2 .

D∗

We conclude as in the proof of Theorem 3.9.5 that the map is area preserving if and only if (3.137) holds. The following result is a consequence of Theorems 3.9.5 and 3.9.14. Theorem 3.9.29. (a) Every isometric map is area preserving. (b) Every area preserving and conformal map is isometric. Proof. We assume that the surface S and its image S ∗ have the same parameter systems. (a) If a map is isometric then we have by (3.124) in Theorem 3.9.14 ∗ gik (u1 , u2 ) = gik (u∗1 , u∗2 ) (i, k = 1, 2),

and so the condition in (3.137) holds. This implies by Theorem 3.9.28 that the map is area preserving. Thus we have shown Part (a).

The Intrinsic Geometry of Surfaces ■ 347

(b) Let a map be conformal and area preserving. Since the map is conformal, we have by (3.109) in Theorem 3.9.5 ∗ gik (u∗1 , u∗2 ) = η(u1 , u2 )gik (u1 , u2 ) (i, k = 1, 2), for η > 0,

hence

g ∗ = η 2 g.

(3.138)

Since the map is also area preserving, it follows from (3.137) in Theorem 3.9.28 that g ∗ = g. (3.139) Now it follows from (3.138) and (3.139) that η 2 = 1, hence η = 1, since η > 0. ∗ Therefore, we have gik = gik (i, k = 1, 2), and so the map is isometric by (3.124) in Theorem 3.9.14. Thus we have shown (b). Combining Remark 3.9.15 (a) and Parts (a) and (b) of Theorem 3.9.29, we obtain Corollary 3.9.30. A map is isometric if and only if it is conformal and area preserving. Example 3.9.31 (Lambert Projection). Let the sphere be given by the parametric representation in (3.113) in Example 3.9.8. We put x∗1 = ru1 x∗2 = r sin u2

(−π < u1 < π)   π π 2 − π,

3 

βν < π,

ν=1

as for geodesic triangles on a sphere (left in Figure 3.34). If the Gaussian curvature of a geodesic triangle is negative throughout, then

ν=1

as for geodesic triangles on a pseudo-sphere (right in Figure 3.34). The difference between π and the sum of angles in a geodesic triangle is called the excess. By (3.147), the excess is equal to the total curvature of a triangle. We obtain from (3.147) for n = 0: Theorem 3.10.4. On a surface of class r ≥ 3 with nowhere positive Gaussian curvature, there are no closed geodesic lines such that one of them is a complete boundary of a part of the surface. We close this section with an example. Example 3.10.5. (a) The simplest example for Theorem 3.10.4 is a plane where the geodesic lines are the straight lines. (b) As another example we consider the circular cone. The u2 –lines are closed geodesic lines, but no such circle line is the complete boundary of a part of the surface.

The Intrinsic Geometry of Surfaces ■ 353

Figure 3.34

3.11

Geodesic triangles on a sphere (left) and a pseudo-sphere (right)

MINIMAL SURFACES

In this section, we study some minimal surfaces. The name minimal surface is connected to the following problem, known as Plateau’s problem. In 1866, the physicist Plateau posed the problem if, for every closed curve γ, there exists a surface with minimal surface area that has γ as its boundary curve. The answer to this problem is affirmative as we will see in Theorem 3.11.5. We start with the definition of the notion of minimal surfaces. The term minimal surface will become clear in Theorem 3.11.5. Definition 3.11.1. A minimal surface is a surface with identically vanishing mean curvature 1 H = Lik g ik ≡ 0. 2 Remark 3.11.2. The Gaussian curvature of a minimal surface cannot be positive. Proof. Let κ1 and κ2 denote the principal curvatures of the minimal surface. Then it follows from (2.74) in Definition 2.5.17 that 2H = κ1 + κ2 = 0, and so κ1 = −κ2 , hence K = κ1 κ2 ≤ 0. Example 3.11.3. (a) We already know from Visualization 2.5.19 (b) that the catenoid (Figure 2.39) is a minimal surface. (b) The helicoid of Visualization 2.5.3 (c) (right in Figure 2.30) is a minimal surface.

Proof. We only have to show Part (b). A helicoid has a parametric representation by Visualization 2.5.3 (c) ⃗x(ui ) = {u2 cos u1 , u2 sin u1 , cu1 }





(u1 , u2 ) ∈ D = (0, 2π) × I2 ,

354 ■ Differential Geometry and Its Visualization

where c ∈ R is a constant, and I2 ⊂ R is an interval. It follows that ⃗x1 = {−u2 sin u1 , u2 cos u1 , c}, ⃗x2 = {cos u1 , sin u1 , 0}, g12 = 0, ⃗ = 1 N · {−c sin u1 , c cos u1 , −u2 }, (u2 )2 + c2 ⃗x11 = −u2 {cos u1 , sin u1 , 0}, ⃗x22 = ⃗0, 2 ⃗ • ⃗x11 =  cu L11 = N · (sin u1 cos u1 − sin u1 cos u1 ) = 0, (u2 )2 + c2 ⃗ • ⃗x22 = 0, L22 = N and so by (2.72) and (2.74) H=

1 · (g22 L11 − 2g12 L12 + g11 L22 ) = 0. g

First we establish a characterization of minimal surfaces. Theorem 3.11.4. A surface S of class r ≥ 3 is a minimal surface or a sphere if and only if its spherical Gauss map is a conformal map. Proof. (i) First we show that if S is a minimal surface or a sphere then its spherical Gauss map is conformal. We assume that S is a minimal surface, that is, H = 0. Then we have by (2.153) in Definition 2.10.8 cik = −Kgik for i, k = 1, 2, (3.149)

for the third fundamental coefficients cik of S, which are the first fundamental coefficients of the image S ∗ of the spherical Gauss map of S. By Theorem 3.9.5 and Remark 3.11.2, the conditions in (3.149) imply that the spherical Gauss map is conformal. Trivially the spherical Gauss map of a sphere is conformal.

(ii) Now we show that if the spherical Gauss map of S is conformal, then S is a minimal surface. We assume that the spherical Gauss map of S is conformal. Then it follows by Theorem 3.9.5 that cik = η(ui , u2 )gik for i, k = 1, 2, and so by (2.153) in Definition 2.10.8 cik = 2HLk − Kgik = η(ui , u2 )gik , that is,

(K + η(ui , u2 ))gik = 2HLik for i, k = 1, 2,

The Intrinsic Geometry of Surfaces ■ 355

hence

(K + η)gik dui duk = 2HLik dui duk .

(3.150)

We may assume that the second fundamental form does not vanish identically. Otherwise the surface would be a plane by Example 2.10.2, in which case its spherical Gauss image would reduce to a single point and the study of the conformity of the spherical Gauss map would not make sense. If K = −η < 0, then H = 0 by (3.150) and S is a minimal surface. If we assume H ̸= 0, then a solution of (3.150) is given by K +η Lik dui duk = = κn , 2H gik dui du2

where κn denotes the normal curvature and the second equality holds by (2.47) in Remark 2.4.3 (c). Then the normal curvature at any point is the same in every direction and consequently every point of S is an umbilical point, and so S is a sphere by Theorem 2.10.4.

The following result explains the term minimal surface. Theorem 3.11.5. Let S be a surface bounded by a closed curve γ. If the surface area A(S) of S is less than or equal to the surface area of any sufficiently close surface bounded by γ, then H ≡ 0 for the mean curvature of S.

Proof. Let S be given by a parametric representation ⃗x(uj ) ((u1 , u2 ) ∈ D). We consider a family of parallel surfaces given by the parametric representations ⃗ (uj ) ((u1 , u2 ) ∈ D). ⃗x∗ (uj ) = ⃗x(uj ) + εv(uj )N

The Weingarten equations (2.141) yield ⃗x∗i = ⃗xi + ε

∂v ⃗ ⃗ i = ⃗xi − εvLki ⃗xk + ε ∂v N ⃗ for i = 1, 2 N + εv N ∂ui ∂ui

and 





∂v ⃗ ∂v ⃗ xm + ε j N N • ⃗xj − εvLm j ⃗ i ∂u ∂u     ∂v ∂v = ⃗xi − εvLki ⃗xk • ⃗xj − εvLm xm + ε2 i · j j ⃗ ∂u ∂u   ∂v ∂v k 2 = gij − εv Lm · for i, j = 1, 2. j gim + Li gkj + ε ∂ui ∂uj

∗ gij = ⃗x∗i • ⃗x∗j = ⃗xi − εvLki ⃗xk + ε

Since by the definition of Lm j in (2.140)

ml l Lm j gim = g Llj gim = δi Llj = Lij

and similarly Lki gkj = Lji = Lij , we obtain ∗ gij = gij − 2εvLij + ε2 (· · · ) for i, j = 1, 2,



356 ■ Differential Geometry and Its Visualization

and ∗ ∗ ∗ 2 g ∗ = g11 g22 − (g12 )





= g11 − 2εvL11 + ε2 (· · · )





2

g22 − 2εvL22 + ε2 (· · · ) − g12 − 2εvL12 + ε2 (· · · )

= g11 g22 − 2εv(g11 L22 + g22 L11 ) − (g12 12 − 4εvg123 L12 ) − ε2 (· · · )

2 = g11 g22 − g12 − 2εv (g11 L22 + g22 L11 − 2g12 L12 ) − ε2 (· · · ),

hence by (2.72) in Remark 2.5.16 (a) and (2.74) in Definition 2.5.17 g ∗ = g (1 − 4εvH) + ε2 (· · · ). This implies   √  g (1 − 4εvH) + ε2 (· · · )  ∂ ∂ ( g ∗ )   =   ∂ε ε=0 ∂ε    −4Hvg + 2ε(· · · )  =   2 g (1 − 4εvH) + ε2 (· · · ) 

√ = −2Hv g.

ε=0

ε=0

If the surface area A(S) of S is to be minimal for ε = 0, then we must have 

∂  ∂ε

 D

√

   1 2  ∗ g du du  

=− ε=0



√ 2Hv g du1 du2 = 0

D

for all admissible functions v(uj ). This implies H = 0 by the fundamental lemma of calculus of variations, Lemma 3.5.1. It is known from elementary geometry that the planar surface, which is bounded by a planar curve of length L and has maximum surface area A, is a disk; the following formula holds L2 A = πr2 = , (3.151) 4π where r is the radius of the disk. This result can be used to prove a similar relation between the surface area of a minimal surface and its bounding curve. Theorem 3.11.6. Let a part S of a minimal surface of class r ≥ 3 be bounded by a closed simple curve γ of length L. If the surface area A(S) of S corresponds to the minimum surface area of all surfaces of class r that are bounded by γ then the following estimate holds L2 A(S) ≤ . 4π Proof. Let S ∗ be the cone generated by joining the origin with the points of γ by straight lines. Then S ∗ is part of a developable surface by Theorem 2.8.16 and so

The Intrinsic Geometry of Surfaces ■ 357

can be mapped isometrically into a plane by Theorem 3.9.25. Furthermore, isometric maps are area preserving by Theorem 3.9.29 (a). Now (3.151) implies A(S ∗ ) ≤

L2 . 4π

This implies the statement, since A(S) is the minimum of the surface area of the part of the surface bounded by γ. We saw in Example 3.11.3 that catenoids and helicoids are minimal surfaces. It will turn out that they are the only minimal surfaces in the classes of surfaces of revolution and conoids, respectively. Theorem 3.11.7. (a) Catenoids are the only minimal surfaces of revolution. (b) Helicoids are the only minimal surfaces among the conoids. Proof. (a) Let S be a surface of revolution given by the parametric representation ⃗x(ui ) = {u1 cos u2 , u1 sin u2 , h(u1 )}





(u1 , u2 ) ∈ D .

Since first and second fundamental coefficients are by (2.24) in Example 2.3.5 and (2.52)–(2.54) in Example 2.4.6 

2

g11 = 1 + h′ (u1 ) and L11 = 

h′′ (u1 ) 1 + (h′ (u1 ))2



, g12 = 0, g22 = u1

2

, g = g11 g22

, L12 = 0 and L22 = 

u1 h′ (u1 ) 1 + (h′ (u1 ))2

,

the condition for S to be a minimal surface is H=

1 (L11 g22 + L22 g11 ) = 0, 2g

that is, 1 L11 L22 + = g11 g22 1 + (h′ (u1 ))2



h′ (u1 ) h′′ (u1 ) + u1 1 + (h′ (u1 ))2



= 0.

If h(u1 ) = const, then we obtain a plane parallel to the x1 x2 –plane. If h′ (u1 ) ̸= 0, then H = 0 is equivalent to h′′ (u1 )h′ (u1 ) 2

(h′ (u1 ))



2

1 + (h′ (u1 ))

 =−

1 . u1

We put z = (h′ )2 , hence 2h′ h′′ = z ′ and obtain z′ = z(1 + z)





1 2 1 − z′ = − 1 . z 1+z u

358 ■ Differential Geometry and Its Visualization

Integration yields

c2 z = log 1 2 , z+1 (u ) where c ̸= 0 is a constant of integration, whence log

c2 (h′ )2 = 1 + (h′ )2 (u1 )2

or

c h′ =  for |u1 | > |c|. 2 1 2 (u ) − c

Integrating once again we obtain for c > 0 and u1 > c h(u1 ) = c · cosh−1

u1 − d, c

where d is a constant of integration. We may choose d = 0, since a change in the value of d means a translation of S along the x3 –axis. Introducing a new parameters u∗i (i = 1, 2) by u1 = c · cosh we obtain

u∗1 and u2 = u∗2 , c





u∗1 u∗1 c · cosh cos u∗2 , c · cosh sin u∗2 , u∗1 , c c

⃗x(u ) = ∗i

that is, part of a catenoid. (b) Let S ∗ be a conoid given by a parametric representation ⃗x∗ (u∗i ) = {u∗1 cos u∗2 , u∗1 sin u∗2 , h(u∗2 )}

We obtain





(u1 , u2 ) ∈ D ⊂ R2 .

⃗x∗1 = {cos u∗2 , sin u∗2 , 0}, ⃗x∗2 = {−u∗1 sin u∗2 , u∗1 cos u∗2 , h′ (u∗2 )}  2  2 ∗ ∗ ∗ = 1, g12 = 0, g22 = u∗1 + h′ (u∗2 ) , g11 ⃗x∗11 = ⃗0, ⃗x∗12 = {− sin u∗2 , cos u∗2 , 0},  = −u∗1 cos u∗2 , −u∗1 sin u∗2 , h′′ (u∗2 ) ,   ⃗ ∗ = √1 · h′ (u∗2 ) sin u∗2 , −h′ (u∗2 ) cos u∗2 , u∗1 , N ∗ g ′ ∗2 (u ) u∗1 h(u∗2 ) ∗ ⃗ ∗ • ⃗x∗ = 0, L∗ = − h√ √ ∗ . and L = =N 22 11 12 g∗ g ⃗x∗22

L∗11

Now the condition for S ∗ to be a minimal surface is H=

 1  ∗ ∗2 u∗1 h′′ (u∗2 ) ∗ ∗ · L g + L g = = 0, 11 22 22 11 2g ∗ 2(g ∗ )3/2

that is, h′′ (u∗2 ) = 0. Hence h must be a linear function of u∗2 which means that S ∗ is a helicoid.

The Intrinsic Geometry of Surfaces ■ 359

One very important result of this section will be the Weierstrass formulae (3.157). They constitute a relation between minimal surfaces and the theory of complex functions. The following fundamental results, which are also interesting in themselves, are needed for the proof on the Weierstrass formulae. Theorem 3.11.8. Let S be a simply connected minimal surface of class r ≥ 3 with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D) with respect to isothermal parameters. Then the component functions of ⃗x are harmonic functions, that is, in vector notation ∆⃗x = ⃗x11 + ⃗x22 = ⃗0, where ∆ denotes the Laplace operator. Proof. The existence of isothermal parameters for S follows from Definition 3.9.9 and Theorem 3.9.12. Then the first fundamental form of S is given by (3.115) 

ds2 = η(u1 , u2 ) (du1 )2 + (du2 )2

2

(3.152)

,

where η > 0, and so g = g11 g22 = η 2 , hence g 11 g 22 = 1/η. The condition for S to be a minimal surface is now 1 1 1 ⃗ 1 ⃗ H = g ik Lik = (L11 + L22 ) = N • (⃗x11 + ⃗x22 ) = N • ∆⃗x = 0, 2 2η 2η 2η ⃗ of S. that is, ∆⃗x is orthogonal to the surface normal vector N It also follows from (3.152) that g11 = ⃗x1 • ⃗x1 = g22 = ⃗x2 • ⃗x2 and g12 = ⃗x1 • ⃗x2 = 0

(3.153)

and we obtain 1 ∂g11 1 ∂g22 ∂g12 = ⃗x1 • ⃗x11 = = ⃗x2 • ⃗x12 and = ⃗x1 • ⃗x22 + ⃗x2 • ⃗x12 = 0. 2 ∂u1 2 ∂u1 ∂u2 It follows that

⃗x1 • ⃗x11 = ⃗x2 • ⃗x12 = −⃗x1 • ⃗x22 ,

hence ⃗x1 • ∆⃗x = 0. Consequently ∆⃗x is orthogonal to ⃗x1 . If we interchange the partial differentiations in (3.153), then it follows similarly that ∆⃗x is orthogonal to ⃗x2 . ⃗ , ⃗x1 and ⃗x2 , which implies This we have shown that ∆⃗x is orthogonal to N ∆⃗x = ⃗0. We saw in Theorem 3.11.8 that each of the components xk of the parametric representation 

 

⃗x(ui ) = x1 (u1 , u2 ), x2 (u1 , u2 ), x3 (u1 , u2 )

(u1 , u2 ) ∈ D



360 ■ Differential Geometry and Its Visualization

of a minimal surface of class r ≥ 3 with respect to isothermal parameters u1 and u2 is a solution of the so–called potential equation ∆xk =

∂ 2 xk ∂ 2 xk + = 0, (∂u1 )2 (∂u2 )2

that is, the component functions xk are harmonic functions. By a known result from potential theory, they are the √real parts of complex functions z k of the complex variable u = u1 + iu2 with i = −1; we write ⃗x = Re(⃗z), where ⃗z(u) = ⃗x(ui ) + i⃗y (ui ).

(3.154)

The following result holds. Theorem 3.11.9. Let S be a simply connected surface of class r ≥ 3 with a parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D) with respect to isothermal parameters. Then S is a minimal surface if and only if there exists a complex analytic function ⃗z = ⃗z(u) of u = u1 + iu2 such that ′ 2

(⃗z ) =



d⃗z du

2

= 0 and ⃗x(ui ) = Re (⃗z(u)) .

Proof. (i) First we assume that S is a minimal surface with a parametric representation ⃗x(ui ) with respect to isothermal parameters. Then it follows by Theorem 3.11.8 that ∆⃗x = ⃗0. Writing ⃗z as in (3.154) the Cauchy–Riemann differential equations ⃗x1 = ⃗y2 and ⃗x2 = −⃗y1 imply d⃗z = ⃗x1 − i⃗x2 , du and so 

d⃗z du

2

= ⃗x1 • ⃗x1 − 2i⃗x1 • ⃗x2 − ⃗x1 • ⃗x2 = g11 − 2ig12 + g22 .

Since u1 and u2 are isothermal parameters, we have g11 = g22 and g12 = 0, and consequently  2 d⃗z = 0. (3.155) du (ii) Conversely we assume that ⃗x is harmonic and the real part of a complex function ⃗z that satisfies (3.155). This implies g11 = g22 = 0 and g12 = 0, that is, u1 and u2 are isothermal parameters for S, and ∆⃗x = ⃗0 implies as in the proof of Theorem 3.11.8 1 ⃗ H= N • ∆⃗x = 0. 2η

The Intrinsic Geometry of Surfaces ■ 361

Now we can derive the Weierstrass formulae which enable us to determine a minimal surface with a parametric representation ⃗x(ui ) for every given regular function F ̸≡ 0 of the complex variable u = (u1 + iu2 ). We use the notations of Theorem 3.11.9, and assume that f and g are arbitrary regular functions of u = u1 + iu2 on some domain. If we choose ⃗z(u) = {z 1 (u), z 2 (u), z 3 (u)} such that dz 1 dz 2 dz 3 = f 2 − g2, = i(f 2 + g 2 ) and = 2f g, du du du

(3.156)

then

 2  2 d⃗z d⃗z • = f 2 − g 2 − f 2 + g 2 + 4f 2 g 2 = 0, du du that is, (3.155) is satisfied. Integration of the equations in (3.156) yields the following general representation for the minimal surfaces of class r ≥ 3  1   x = Re(z 1 ) = Re (f 2 − g 2 ) du      x2 = Re(z 2 ) = Re i(f 2 + g 2 ) du     3 3

x = Re(z ) = Re ( 2f g du) .

   

(3.157)

  

Now we assume f ̸= 0 and (g/f )′ ̸= 0, that is, f g ′ − f ′ g ̸= 0, since if p ≡ 0 or (g/p)′ ≡ 0, then we obtain a part of a plane. We put t = g/f , hence du =

f2 · dt, g′f − f ′g

where ′ denotes differentiation with respect to u. We also put F (t) =

f4 . g′f − f ′g

Then f 2 − g 2 = f 2 (1 − g 2 /f 2 ) = (1 − t2 )f 2 and we obtain from the identity in the first line of (3.157) 1

x = Re = Re



2

2



(f − g ) du = Re

 

2

1−t









2

1−t





f4 dt · ′ g f − f ′g

F (t) dt .

Analogous arguments yield the second and third formulae of the following identities known as the Weierstrass equations.  1   x = Re (1 − t2 )F (t) dt      x2 = Re i(1 + t2 )F (t) dt     3

x = Re ( 2tF (t) dt)

      

.

(3.158)

362 ■ Differential Geometry and Its Visualization

Remark 3.11.10. The Weierstrass formulae (3.158) enable us to find a minimal surface for every regular function F of the complex variable t = t1 + it2 . Putting F = h′′′ then integration by parts yields  1   x = Re (1 − t2 )h′′ + 2th′ − 2h      x2 = Re i(1 + t2 )h′′ − 2ith′ + 2ih    3 ′′ ′

x = Re (2th − 2h ) dt) .

   

(3.159)

  

Example 3.11.11 (The helicoid). If F (t) = i/(2t2 ) (t ̸= 0) then the Weierstrass formulae (3.158) yield the minimal surface S with the parametric representation ⃗x(ui ) = {x1 (u1 , u2 ), x2 (u1 , u2 ), x3 (u1 , u2 )}, where tan x3 = x1 /x2 . Proof. We write t = t1 + it2 and obtain from the first formula in (3.158) 1

x = Re







i i 1 + t2 · (1 − t ) 2 dt = −Re 2t 2 t 2

Since 1/t = t¯/(|t|2 ) = (t1 − it2 )/(t21 + t22 ), we get 



.



t2 1 x = 1− 2 . 2 t1 + t22 1

Similarly the second formula in (3.158) yields x2 = − Hence we have





t1 1 1− 2 . 2 t1 + t22 t2 x1 =− . x2 t1

(3.160)

Furthermore, the third formula in (3.158) yields 



 



i t2 x = Re dt = Re (i log t) = Re i log |t| + i tan−1 t t1   t 2 = − tan−1 , t1 3

or by (3.160) tan (−x3 ) = − tan x3 =

t2 x1 = − 2. t1 x

We introduce the parameters u1 and u2 by x1 = ru1 sin u2 and x2 = ru1 cos u2 . Then (3.161) yields x3 = u2 , that is, the minimal surface is a helicoid.



(3.161)

The Intrinsic Geometry of Surfaces ■ 363

Figure 3.35

Enneper’s surface with respect to the parametric representation (3.162)

Visualization 3.11.12 (The Enneper surface). (a) The choice F (t) ≡ 1 in the Weierstrass formulae (3.158) yields Enneper’s minimal surface ES given by a parametric representation (Figure 3.35) ⃗x(u ) = i



(u1 )3 (u2 )3 + u1 (u2 )2 , u2 − + u2 (u1 )2 , (u1 )2 − (u2 )2 u − 3 3 1



(b) We introduce polar coordinates ρ and ϕ in R 2 by





(u1 , u2 ) ∈ R2 . (3.162)

u1 (ρ, ϕ) = ρ cos ϕ and u2 (ρ, ϕ) = ρ sin ϕ for (ρ, ϕ) ∈ (0, ∞) × (0, 2π).

Then Enneper’s surface ES has a parametric representation (Figure 3.36) ⃗x(ρ, ϕ) =



ρ3 ρ3 cos (3ϕ), ρ sin ϕ + sin (3ϕ), ρ2 cos (2ϕ) ρ cos ϕ − 3 3



((ρ, ϕ) ∈ (0, ∞) × (0, 2π)). (3.163)

(c) The components xk (ρ, ϕ) of the parametric representation (3.163) satisfy the relation 

2

x1 (ρ, ϕ)



2

+ x2 (ρ, ϕ)



2 4 3 ρ3 + x (ρ, ϕ) = ρ + 3 3

2

.

(3.164)

(d) The lines of self–intersection of Enneper’s surface with the parametric representation (3.163) are given by f1 (ρ, ϕ) = cos ϕ −

ρ2 cos (3ϕ) = 0 3

364 ■ Differential Geometry and Its Visualization

Figure 3.36

Enneper’s surface with respect to the parametric representation (3.163)

and f2 (ρ, ϕ) = sin ϕ +

ρ2 sin (3ϕ) = 0; 3

consequently they are in the planes x = 0 and y = 0, respectively (Figure 3.37). Proof. (a) The formulae in (3.158) yield for t = t1 + it2 x1 = Re



= t1 −

Figure 3.37





t3 (1 − t2 ) dt = Re t − 3



  1 t3 · Re (t1 )3 + 3it21 t2 − 3t1 t22 − it33 = t1 − 1 + t1 t22 , 3 3

Enneper’s surface and lines of self-intersection

The Intrinsic Geometry of Surfaces ■ 365

x2 = Re





 

t3 i(1 + t2 ) dt = Re i t + 3



   1 t3 = −t2 + Re · i (t1 )3 + 3it21 t2 − 3t1 t22 − it32 = −t2 − t21 t2 + 2 , 3 3

and 3

x = Re





 

2t dt = Re t2 = t21 − t22 .

Introducing new parameters u1 and u2 by t1 = u1 and t2 = −u2 , we obtain the parametric representation in (3.162). (b) We write xk (ui ) and xk (ρ, ϕ) (k = 1, 2, 3) for the components of the parametric representations of ES in (3.162) and (3.163). Then we obtain x3 (ρ, ϕ) = x3 (ui (ρ, ϕ)) = (u1 (ρ, ϕ))2 − (u2 (ρ, ϕ))2 



= ρ2 cos2 ϕ − sin2 ϕ = ρ2 cos (2ϕ).

Also using

cos (3ϕ) = Re(exp (3iϕ)) = Re (cos ϕ + i sin ϕ)3 = cos3 ϕ − 3 cos ϕ sin2 ϕ and sin (3ϕ) = Im(exp (3iϕ)) = Im (cos ϕ + i sin ϕ)3 = − sin3 ϕ + 3 cos2 ϕ cos ϕ, we conclude ρ3 cos3 ϕ + ρ3 cos ϕ sin2 ϕ 3  ρ3  3 ρ3 = ρ cos ϕ − cos 3ϕ cos ϕ − 3 cos ϕ sin2 ϕ = ρ cos ϕ − 3 3

x1 (ρ, ϕ) = x1 (ui (ρ, ϕ)) = ρ cos ϕ −

and

ρ3 sin3 ϕ + ρ3 sin ϕ cos2 ϕ 3  ρ3  3 ρ3 sin (3ϕ). = ρ sin ϕ − sin ϕ − 3 sin ϕ cos2 ϕ = ρ sin ϕ + 3 3

x2 (ρ, ϕ) = x2 (ui (ρ, ϕ)) = ρ sin ϕ −

(c) We have 

2

x1 (ρ, ϕ)

2 4 3 x (ρ, ϕ) = 3 ρ6 2ρ4 cos ϕ cos (3ϕ) + cos2 (3ϕ) ρ2 cos2 ϕ − 3 9



2

+ x2 (ρ, ϕ)

+

366 ■ Differential Geometry and Its Visualization

ρ6 2ρ4 sin ϕ sin (3ϕ) + sin2 (3ϕ) 3 9 4 + ρ4 cos2 (2ϕ) = 3 6 4 ρ 2ρ 4 ρ2 + − (cos ϕ cos (3ϕ) − sin ϕ sin (3ϕ)) + ρ4 cos2 (2ϕ) = 9 3 3 6 4 2ρ 4 ρ ρ2 + − cos (4ϕ) + ρ4 cos2 (2ϕ) = 9 3 3 6 4   4 2ρ ρ ρ2 + − cos2 (2ϕ) − sin2 (2ϕ) + ρ4 cos2 (2ϕ) = 9 3 3 6 4   4 2ρ ρ − ρ2 + 2 cos2 (2ϕ) − 1 + ρ4 cos2 (2ϕ) = 9 3 3  2 ρ6 2ρ4 ρ3 2 ρ + . + = ρ+ 9 3 3 + ρ2 sin2 ϕ +

(d) The points of self–intersection of Enneper’s surface given by a the parametric representation (3.163) must satisfy ⃗x(ρ1 , ϕ1 ) = ⃗x(ρ2 , ϕ2 ), that is,

xk (ρ1 , ϕ1 ) = xk (ρ2 , ϕ2 ) for k = 1, 2, 3,

and it follows from (3.164) that ρ1 +

ρ31 ρ3 = ρ2 + 2 . 3 3

Since the function f : R → R with f (t) = t + t3 /3 obviously is one–to-one, this implies ρ1 = ρ2 = ρ. Thus it follows from x3 (ρ, ϕ1 ) = x3 (ρ, ϕ2 ), that cos (2ϕ1 ) = cos (2ϕ2 ), hence ϕ2 = π − ϕ1 or ϕ2 = 2π − ϕ1 . If ϕ2 = π − ϕ1 , then x1 (ρ, ϕ1 ) = x1 (ρ, π − ϕ1 ) implies cos ϕ1 −

ρ2 ρ2 cos (3ϕ1 ) = cos (π − ϕ1 ) − cos (3(π − ϕ1 )) 3 3   ρ2 = − cos ϕ1 − cos (3ϕ1 ) 3

that is, x1 (ρ, ϕ1 ) = −x1 (ρ, ϕ1 ) = f1 (ρ, ϕ1 ) = 0. If ϕ2 = 2π − ϕ1 , then it can similarly be shown that x2 (ρ, ϕ1 ) = −x2 (ρ, ϕ1 ) = f2 (ρ, ϕ1 ) = 0. It is easy to see that if ϕ2 = π − ϕ1 or ϕ2 = 2π − ϕ1 , then ⃗x(ρ, ϕ1 ) = ⃗x(ρ, ϕ2 ).

The Intrinsic Geometry of Surfaces ■ 367

Figure 3.38

Several branches of Scherk’s minimal surface

Visualization 3.11.13 (The Scherk surface). The choice of F (t) = 2/(1 − t4 ) (t ̸= ±1, ±i) in the Weierstrass formulae (3.158) yields Scherk’s minimal surface with a parametric representation (Figure 3.38) ⃗x(u ) = i



1

2

u , u , log



cos u2 cos u1





(2k − 1)π (2k + 1)π , 2 2

(3.165)

for (u1 , u2 ) ∈ Dj,k =



(2j − 1)π (2j + 1)π , 2 2



×



(k, j ∈ Z).

Proof. Let (t ̸= ±1, ±i). The first formula in (3.158) yields 







 

1 1 − t2 1 1 x = Re + 2· dt = Re 2· dt = Re 1 − t4 1 + t2 1 − it 1 + it      1 + it 1 + it = Re i−1 · Log = arg + 2kπ (k ∈ Z), 1 − it 1 − it 1





dt

where Log and arg(z) denote the complex logarithm and the angle of the complex number z measured anti–clockwise from the positive real axis. We will only consider the principal branch of Log, that is, k = 0 in the formula for arg. Writing t = u + iv and t¯ = u − iv for the conjugate complex number of t, we obtain

hence

1 + it (1 + it)(1 + it¯) 1 − |t|2 + 2iRe(t) = = , 1 − it |1 − it|2 |1 − it|2 

1 + it arg 1 − it



= tan

−1





2u . 1 − (u2 + v 2 )

(3.166)

368 ■ Differential Geometry and Its Visualization

Similarly the second formula in (3.158) yields 





1 + t2 dt x = Re 2i · dt = Re 2i · 4 1−t 1 − t2       1+t 1+t = Re i ·Log = −arg . 1−t 1−t 2

We obtain



= Re

2

x = tan

−1



x = Re



1 1 + 1−t 1+t



−2v . 1 − (u2 + v 2 )

Finally the third formula in (3.158) yields 

i·





dt

(1 + t)(1 − t¯) 1 − |t|2 + 2iIm(t) 1+t = = , 1−t |1 − t2 | |1 − t2 |

hence

3





4t dt = Re 1 − t4

   1 + t2    = log  .  1 − t2 





1 1 2t + 2 1−t 1 + t2



(3.167) 





1 + t2 dt = Re Log 1 − t2



(3.168)

We have t2 = u2 − v 2 + 2iuv and

     1 + t2 2 (1 + t2 )(1 + t¯2 ) 1 + |t|4 + Re t2 + t¯2     =   =  1 − t2  (1 − t2 )(1 − t¯2 ) 1 + |t|4 − Re t2 + t¯2

=

1 + (u2 + v 2 )2 + 2(u2 − v 2 ) . 1 + (u2 + v 2 )2 − 2(u2 − v 2 )

Since cos2 φ = (1 + tan2 φ)−1 , we obtain from (3.166) and (3.167) 1

1

cos x = 2

1+

2

cos x =

1+ 

1



2u u2 +v 2 −1



2v u2 +v 2 −1

1



2



2

u2 + v 2 − 1 , 2 = 2 (u + v 2 − 1)2 + 4u2 u2 + v 2 − 1 , 2 = 2 (u + v 2 − 1)2 + 4v 2

2

cos x2 u2 + v 2 − 1 + 4u2 = cos x1 (u2 + v 2 − 1)2 + 4v 2 

2





u2 + v 2 − 2 u2 + v 2 + 1 + 4u2 = (u2 + v 2 )2 − 2 (u2 + v 2 ) + 1 + 4v 2 

2





1 + u2 + v 2 + 2 u2 − v 2 = 1 + (u2 + v 2 )2 + 2 (v 2 − u2 )

and by (3.168) and (3.169)





 1 + t2  1   x = · log   = log  1 − t2  2 3





cos x2 . cos x1

(3.169)

The Intrinsic Geometry of Surfaces ■ 369

Finally, if we write u1 = u and u2 = v, then we obtain the parametric representation (3.165) for (u1 , u2 ) ∈ D0,0 . Example 3.11.14 (The catenoid). The choice of F (t) = −1/(2t2 ) in the Weierstrass formulae (3.158) yields the catenoid with a parametric representation 

⃗x(ui ) = cosh u1 cos u2 , cosh u1 sin u2 , u1

 



(u1 , u2) ∈ R × (0, 2π) .

Proof. The Weierstrass formulae (3.158) yield 

  1  1 − t2 · x = Re − 2    i  2 x = Re − 1 + t2 · 2 1





(3.170)



1 1 1 dt = Re t + 2 t 2 t     1 1 1 −t dt = Re i · t2 2 t

and x3 = Re



 



1 −2t dt = −Re(Log(t)) = log  . 2 2t t

We write t = exp (u1 + iu2 ) and obtain

  1  1  u1 1 x1 = Re exp (u1 + iu2 ) + exp (−u1 − iu2 ) = e + e−u cos u2 2 2 = cosh u1 cos u2 ,   1  u1 1  1 e + e−u sin u2 x2 = Re i exp (−u1 − iu2 ) − exp (u1 + iu2 ) = 2 2 = cosh u1 sin u2

and

1

x3 = − log (|t|) = − log (eu ) = −u1 .

Finally, since cosh (−u1 ) = cosh u1 , we may replace u1 by −u1 , and obtain the parametric representation (3.170). Example 3.11.15 (Conjugate minimal surfaces). Let Sf and Sg be surfaces given by parametric representations ⃗x and ⃗y with the component functions f k and g k (k = 1, 2, 3). If the functions f k and g k satisfy the Cauchy–Riemann differential equations, that is, if f1k = g2k and f2k = −g1k for k = 1, 2, 3, (3.171) then the component functions are said to be conjugate harmonic. If ⃗x and ⃗y are the isothermal parametric representations of two minimal surfaces such that their component functions are pairwise conjugate harmonic, then the surfaces are said to be conjugate harmonic minimal surfaces. (a) The helicoid and catenoid are conjugate minimal surfaces.

370 ■ Differential Geometry and Its Visualization

(b) If two conjugate minimal surfaces with the parametric representations ⃗x and ⃗y are given, then the surface with the parametric representation ⃗z (λ) = (cos λ) · ⃗x + (sin λ) · ⃗y

(3.172)

is a minimal surface for each λ ∈ R. (λ) (c) The first fundamental coefficients gik of each surface with the parametric representation in (3.172) satisfy (λ)

(λ)

(λ)

g11 = ⃗x1 • ⃗x1 = ⃗y1 • ⃗y1 , g12 = 0 and g22 = ⃗x2 • ⃗x2 = ⃗y2 • ⃗y2 .

(3.173)

Proof. We note that (3.171) implies that the component functions are harmonic functions. (a) The catenoid is a minimal surface by Example 3.11.14; its parametric representation ⃗x(ui ) = {cosh u1 cos u2 , cosh u1 sin u2 , u1 } is isothermal, since



(u1 , u2 ) ∈ R × (0, 2π)



⃗x1 = {sinh u1 cos u2 , sinh u1 sin u2 , 1},

⃗x2 = {− cosh u1 sin u2 , cosh u1 cos u2 , 0} and g11 = 1 + sinh2 u1 = cosh2 u1 = g22 and g12 = 0. The helicoid is a minimal surface by Example 3.11.11; its parametric representation (for r = 1 in Example 3.11.11) ⃗y (u∗i ) = {u∗1 sin u∗2 , u∗1 cos u∗2 , u∗2 } however, is not isothermal, since 

∗ = 1 ̸= 1 + u∗1 g11





(u∗1 , u∗2 ) ∈ R × (0, 2π) ,

2

∗ = g22 .

We introduce new parameters u1 and u2 by u∗1 = sinh u1 and u∗2 = −u2 , and write ⃗y (ui ) = ⃗y (u∗i (uj )) = {− sinh u1 sin u2 , sinh u1 cos u2 , −u2 }. Then we obtain

⃗y1 = {− cosh u1 sin u2 , cosh u1 sin u2 , 0} = ⃗x2 ,

⃗y2 = {− sinh u1 cos u2 , − sinh u1 cos u2 , −1} = −⃗x1 , hence the component functions of ⃗x and ⃗y are harmonic and ⃗x(ui ) and ⃗y (ui ) are parametric representations with respect to isothermal parameters. The catenoid and helicoid are conjugate minimal surfaces.

The Intrinsic Geometry of Surfaces ■ 371

(b) We assume that ⃗x and ⃗y are the parametric representations of two conjugate minimal surfaces, and write gik and hik (i, k = 1, 2) for the first fundamental coefficients of ⃗x and ⃗y , respectively. Since the component functions are conjugate harmonic and the parameters are isothermal, it follows that ⃗x1 = ⃗y2 and ⃗x2 = −⃗y1 , and g11 = h22 , g22 = h11 , g12 = −h12 = 0 and g11 = g22 = h11 = h22 . We obtain (λ)

⃗z1

(λ)

= (cos λ)⃗x1 + (sin λ)⃗y1 and ⃗z2

= (cos λ)⃗x2 + (sin λ)⃗y2 ,

hence (λ)

(λ)

g11 = ⃗z1

(λ)

• ⃗z1









= cos2 λ g11 + 2 (cos λ sin λ) (⃗x1 • ⃗y1 ) + sin2 λ h11





= cos2 λ + sin2 λ g11 − 2 (cos λ sin λ) (⃗x1 • ⃗x2 ) = g11 = h11 , (λ) g12

(λ)

= ⃗z1

(λ)

• ⃗z2









= cos2 λ g12 + (cos λ sin λ) (⃗x1 • ⃗y2 + ⃗x2 • ⃗y1 ) + sin2 λ h12 = (cos λ sin λ) (⃗x1 • ⃗x1 − ⃗x2 • ⃗x2 ) = (cos λ sin λ) (g11 − g22 ) = 0 and (λ)

(λ)

g22 = ⃗z2

(λ)

• ⃗z2











= cos2 λ g22 + 2 (cos λ sin λ) (⃗x2 • ⃗y2 ) + sin2 λ h22 

= cos2 λ + sin2 λ g11 − 2 (cos λ sin λ) (⃗x1 • ⃗x2 ) = g22 = h22 . (c) Since ⃗x1 = ⃗y2 and ⃗x2 = −⃗y1 , we have (λ)

⃗z1

(λ)

× ⃗z2

= ((cos λ)⃗x1 + (sin λ)⃗y1 ) × ((cos λ)⃗x2 + (sin λ)⃗y2 ) = ((cos λ)⃗x1 − (sin λ)⃗x2 ) × ((cos λ)⃗x2 + (sin λ)⃗x1 )   √ ⃗ = (cos λ)2 + (sin λ)2 (⃗x1 × ⃗x2 ) = g N ,

⃗ is the surface normal vector of the surface given by ⃗x(ui ). So we have where N ⃗ (λ) = N ⃗ for the surface normal vector of the surface with ⃗z (λ) . Furthermore, N we have (λ)

⃗z11 = (cos(λ)) ⃗x11 + (sin(λ)) ⃗y11 = (cos(λ)) ⃗x11 − (sin(λ)) ⃗x12 and (λ)

⃗z22 = (cos(λ)) ⃗x22 + (sin(λ)) ⃗y22 = (cos(λ)) ⃗x22 + (sin(λ)) ⃗x12 .

372 ■ Differential Geometry and Its Visualization (λ)

Thus the second fundamental coefficients Lik of the surface given by ⃗z (λ) are given by (λ) ⃗ (λ) • ⃗z (λ) = N ⃗ • ((cos λ) ⃗x11 − (sin λ) ⃗x12 ) , L11 = N 11

(λ) ⃗ (λ) • ⃗z (λ) = N ⃗ • ((cos λ) ⃗x12 − (sin λ) ⃗x22 ) L12 = N 12

and (λ) ⃗ (λ) • ⃗z (λ) = N ⃗ • ((cos λ) ⃗x22 + (sin λ) ⃗x12 ) . L22 = N 22

Let H and H λ denote the mean curvature of the surfaces with ⃗x(ui ) and ⃗z (λ) (ui ). Then we have by Part (c) 

(λ) (λ)

(λ) (λ)

(λ) (λ)



(λ)

(λ)

2 g (λ) H (λ) = L11 g22 − 2L12 g12 + L22 g11 = L11 + L22



g11

= ((cos λ) L11 − (sin λ) L12 + (cos λ) L22 + (sin λ) L12 ) g11 √ = (cos λ) (L11 g22 + L22 g11 ) = 2(cos λ) gH = 0.

CHAPTER

4

Tensor Algebra and Riemannian Geometry

In this chapter, we redevelop the intrinsic geometry of surfaces independently of their embedding in three-dimensional space E3 , based only on the definition of measurements of lengths in a point set. This concept was originally introduced by Riemann in 1854. For our purpose, we need the concept of a manifold in n–dimensional space. We shall obtain surfaces and n–dimensional Euclidean spaces as special cases of the manifolds. There is one more reason for introducing manifolds. So far we only considered parts of surfaces and not surfaces as a whole. For instance, the whole sphere does not fall under the definition of surfaces given in Chapter 2, since there is no one–to–one correspondence between the whole sphere and a single part of the parameter plane. Furthermore, we deal with an introduction into tensor algebra. This will provide an extremely useful formalism in the solution of problems in Differential Geometry. In particular, we study • the definition of n–dimensional manifolds of class C k in Definition 4.1.1 of Section 4.1 • the transformation formulae between the bases of a finite dimensional vector space, and between the components of vectors with respect to bases in Proposition 4.2.1 • spaces V ∗ of linear functionals on finite-dimensional vector spaces V , the isometry of V and V ∗ in Proposition 4.3.1, the transformation formulae of the elements of V ∗ and their components in Proposition 4.3.3, and the concepts of contravariant and covariant vectors in Definition 4.3.4 • introduction of tensors of the second order in Definition 4.4.1 and the transformation formulae for their components with respect to different bases in 4.4.2 • symmetric bilinear forms and inner products in Section 4.5 DOI: 10.1201/9781003370567-4

373

374 ■ Differential Geometry and Its Visualization

• introduction of tensors of arbitrary order in Definition 4.6.1 the transformation formulae for their components in Theorem 4.6.2, the characterization of tensors by the transformation properties of their components, and the identification of tensors with their components in Remark 4.6.4; sums, outer products, contractions (Definition 4.6.5) and inner products of tensors • symmetric and anti–symmetric tensors of arbitrary order in Definition 4.7.1 and the independence of their properties from the choice of the coordinate system in Theorem 4.7.2 • Riemann spaces in Section 4.8 and the metric tensor in Definition 4.8.3

• the Christoffel symbols in a Riemann space in Definition 4.9 and their transformation formulae in Theorem 4.9.3.

4.1

DIFFERENTIABLE MANIFOLDS

In this section, we give the definition of an n–dimensional manifold of class C k . First we need to recall a few definitions and well–known facts from topology. Definition 4.1.1. (a) Let X be a nonempty set. A family T of subsets of X is called a topology for X, and the members of T are called open sets if T satisfies the following conditions (T.1) (T.2) (T.3)

X, ∅ ∈ T



O ∈ T for every subcollection O of T

O1 ∩ O2 ∈ T for all O1 , O2 ∈ T .

The pair (X, T ) is called a topological space; sometimes we write X for short. (b) Let S be a set in a topological space (X, T ). We say that a subset of S is S–open if it is the intersection of S with a set in T . The collection of S–open sets is called the relative topology of X for S, denoted by TS . Thus TS = {O ∩ S : O ∈ T }. (c) A topological space (X, T ) is said to be a Hausdorff space if for any two distinct elements x, x′ ∈ X there are open sets O and O′ such that x ∈ O, x′ ∈ O′ and O ∩ O′ = ∅. (d) Let (X, T ) be a topological space. A subset F of X is called closed if its complement X \ F is open. A point s ∈ X is called interior to a set S if there is an open set O with s ∈ O ⊂ S, in this case S is called a neighbourhood of s. (e) A local base at a point x for a topology T is a collection Σ of neighbourhoods of x such that given any neighbourhood N of x there is G ∈ Σ with G ⊂ N . A topological space (X, T ) is said to be first countable at x ∈ X if it has a countable local base at x; it is said to be first countable if it is first countable at each of its points. Example 4.1.2. A metric space (X, d) has a local base at any of its points x, for instance, the collection of all sets N1/n (x) = {x′ ∈ X : d(x′ , x) < 1/n} (n = 1, 2, . . .). Hence every metric space is first countable.

Tensor Algebra and Riemannian Geometry ■ 375

Definition 4.1.3. Let X and Y be topological spaces. A map f : X → Y is said to be continuous at x ∈ X if for every neighbourhood G of f (x) in Y the pre–image f −1 (G) of G is a neighbourhood of x in X; f is said to be continuous on a subset S of X if it is continuous at every point s in S; it is called continuous if it is continuous on X. The map f is called a homeomorphism if it is one–to–one, onto and continuous and if the inverse map f −1 : Y → X is also continuous. Now we give the definition of a manifold.

Definition 4.1.4. An n–dimensional manifold M of class C k is a first countable Hausdorff space with the following properties: There exist a family G = {Gα : α ∈ A} of open sets with M = ∪α∈A Gα , a so– called open covering of M , and a family F of homeomorphisms fα : Gα → Oα onto open subsets Oα of Rn , and for each pair (α, β) with Gα ∩ Gβ ̸= ∅ there is a map gα,β = fβ ◦fα−1 : fα (Gα ∩Gβ ) → fβ (Gα ∩Gβ ) of class C k with nonvanishing Jacobian.

Remark 4.1.5. For each α ∈ A, the map fα in Definition 4.1.4 is a continuous, one–to–one map from the open set Gα of M onto an open subset Oα of Rn which has a continuous inverse map fα−1 (Oα ) → Gα . If fα (P ) = (x1 , x2 , . . . , xn ) and fβ (P ) = α

α

α

(x1 , x2 , . . . , xn ) for a point P ∈ Gα ∩ Gβ then gα,β is given by β

β

β

gα,β (fα (P )) = gα,β (x1 , x2 , . . . , xn ) = (x1 , x2 , . . . , xn ). α

α

α

β

β

β

If we write xj = xj (xi ) (j = 1, 2, . . .) for the coordinate functions of gα,β , then the β

β

α

functions xj have to have continuous partial derivatives of order k and β



∂ x 

j

β

∂ xj α



 ̸= 0 for the Jacobian.

If the functions xj are analytic, that is, if they have power series expansions, then β

the manifold is said to be analytic or of class C ∞ . Example 4.1.6. (a) The Euclidean point set Rn is an n–dimensional manifold of class C ∞ . We may choose G = {Rn } and F = {id} where id : Rn → Rn is the identity map defined by id(x) = x for all x ∈ Rn . (b) Every surface S of class C r in R3 with parametric representation ⃗x(ui ) ((u1 , u2 ) ∈ D), where D ⊂ R2 is a domain, is a two-dimensional manifold of class C r . We have S = {X ∈ R3 : X = F (u1 , u2 ), ((u1 , u2 ∈ D)}, where the map F = (f 1 , f 2 , f 3 ) : D → R3 is given by f k (u1 , u2 ) = xk (u1 , u2 ) (k = 1, 2, 3) and satisfies the conditions of Definition 2.1.1. We choose G = {F (D)} and F = {f } with f = F −1 . (c) Let F : R3 → R be a differentiable map and c ∈ F (R3 ) with grad F (X) ̸= ⃗0. Then F −1 ({c}) is a manifold. If (∂F/∂x3 )(X0 ) ̸= 0 for some X0 = (x10 , x20 , x30 ) ∈ F −1 ({c}), then, by the implicit function theorem, there are a neighbourhood N = NX0 of (x10 , x20 ),

376 ■ Differential Geometry and Its Visualization

a neighbourhood V = VX0 of x30 and a differentiable map g = gX0 : N → V with F (x1 , x2 , g(x1 , x2 )) = c, and more precisely x ∈ (N × V ) ∩ F −1 ({c}) if and only if x3 = g(x1 , x2 ). Consequently the function fX0 on (N × V ) ∩ F −1 ({c}) is given by the projection PX0 (x1 , x2 , x3 ) = (x1 , x2 ) with the inverse map PX−10 (x1 , x2 , g(x1 , x2 )). Since F −1 ({c}) ⊂ R3 , F −1 ({c}) is a metric, hence a Hausdorff space, and is first countable.

Definition 4.1.7. Let M be an n–dimensional manifold of class C k , G = {Gα : α ∈ ˜ λ : λ ∈ A} ˜ be two open coverings of M , and fα : Gα → Oα and f˜λ : A} and G˜ = {G ˜ ˜ Gλ → Oλ be maps that satisfy the conditions of Definition 4.1.4. A transformation from the system G to the system G˜ is called an admissible coordinate transformation ˜ λ ) → f˜λ (Gα ∩ G ˜ λ ) is of class C k or C ∞ and has if each of the maps gα,λ : fα (Gα ∩ G nonvanishing Jacobian. Remark 4.1.8. (a) With the notations of Remark 4.1.5, a transformation from the system G to the system G˜ is an admissible coordinate transformation if for each ˜ λ , the maps x˜j = x˜j (xi ) are of class C k (or C ∞ ) with nonvanishing P ∈ Gα ∩ G λ

λ

α

Jacobian. (b) A differentiable manifold has only a topological structure, but no geometric one. The geometric properties will have to be independent of the choice of a coordinate system.

Definition 4.1.9. (a) The continuous image P (I) of an interval I = [a, b] is called a curve. A curve is said to be m times continuously differentiable (m ≤ k if the manifold is of class C k ), if the coordinate functions xi with respect to some coordinate system are m times continuously differentiable. This definition makes sense, since, by the chain rule, x¯j (t) = x¯j (xi (t)) is differentiable in any other admissible coordinate system. A smooth curve is a curve that is at least once continuously differentiable; a curve that consists of a finite number of smooth parts is called piecewise smooth. (b) Let P be a given point. Two curves through P with coordinates xi (t) and y i (τ ) are said to have the same tangent vector ⃗t at P if dy i dxi = at P . dt dτ Remark 4.1.10. (a) By the chain rule, the definition of tangent vectors is independent of the choice of the coordinate system. (b) Let P be a given point. We define a relation R on the set SC(P ) of all smooth curves through P by γRγ ∗ if and only if γ and γ ∗ have the same tangents at P . It is easy to see that R is an equivalence relation. Now we define the tangent vectors ⃗t as the equivalence classes [⃗t ] of R. Let T (P ) denote the set of all tangent vectors through P , that is, of all equivalence classes. We define a multiplication with scalars and an addition on T (P ). Let ⃗t ∈ T (P ) and c ∈ R be given. If γ is a representative of

Tensor Algebra and Riemannian Geometry ■ 377

the class ⃗t given by the coordinates xi (t), then we have for γ ∗ given by the coordinates z i (t) = xi (ct) dz i dxi =c at P , dt dt and we define c · ⃗t as the equivalence class of γ ∗ . Furthermore, let γ1 (⃗t) and γ2 (⃗t) be two representatives of the class ⃗t given by the coordinates xi1 (t) and xi2 (t), and γ1 (⃗u) and γ2 (⃗u) two representatives of the class ⃗u given by the coordinates y1i (t) and y2i (t). Then dxi dy i dy i dxi1 = 2 and 1 = 2 at P dt dt dt dt and so d i d (x + xi2 ) = (y1i + y2i ) at P . dt 1 dt Consequently we define ⃗t + ⃗u to be the equivalence class of the curve given by the coordinates xi1 (t) + y1i (t). In both cases the definition is independent from the choice of a coordinate system, as can be seen from the chain rule. It is easy to see that T (P ) is a vector space with the operations just defined. Since the curves given by xi (t) = tδki where δkk = 1 and δki = 0 for i ̸= k have the tangent vectors dxi = δki dt and any tangent vector can be represented as dy k dy i = ak δki where ak = , dt dt the vector space has dimension n.

4.2

TRANSFORMATION OF BASES

In Section 4.1, we saw that we can associate an n–dimensional vector space with every point of a differentiable manifold. Therefore, we take a closer look at vector spaces and first study transformations of bases. Let H = {⃗b1 , ⃗b2 , . . . , ⃗bn } be a basis of an n–dimensional vector space V . Then every vector ⃗x ∈ V has a unique representation ⃗x = xk⃗bk ,

(4.1)

and the real numbers xk (k = 1, 2, . . . , n) are called the components of ⃗x with respect ˜ = {⃗˜b1 , ⃗˜b2 , . . . , ⃗˜bn } is another basis, then, by (4.1), each vector ⃗˜bk to the basis H. If H has a unique representation in terms of the vectors ⃗bi and each vector ⃗bj has a unique representation in terms of the vectors ⃗˜bk ⃗˜b = ai ⃗b (k = 1, 2, . . . , n) and ⃗b = αk⃗˜b (j = 1, 2, . . . , n). k j k i j k The equations in (4.2) and describe a basis transformation.

(4.2)

378 ■ Differential Geometry and Its Visualization

˜ = {⃗˜b1 , ⃗˜b2 , . . . , ⃗˜bn } be bases of an Proposition 4.2.1. Let H = {⃗b1 , ⃗b2 , . . . , ⃗bn } and H n–dimensional vector space V and their transformation be given by (4.2). Then aik αil = αki ali = δkl

(k, l = 1, 2, . . . , n),

(4.3)

and the components xk and x˜k (k = 1, 2, . . . , n) of a given vector ⃗x ∈ V with respect ˜ transform as to H and H x˜k = αjk xj (k = 1, 2, . . . , n) and xj = ajk x˜k (j = 1, 2, . . . , n).

(4.4)

Proof. We have by (4.2) ⃗˜b = ai ⃗b = ai αl⃗˜b and ⃗b = αi ⃗˜b = αi al⃗b (k = 1, 2, . . . , n), k k k i k i l k i k i l ˜ are linearly independent. and this implies (4.3), since the sets H and H Furthermore, ⃗x = xj⃗bj = xj αjk⃗˜bk = x˜k⃗˜bk = xk ajk⃗bj implies (4.4). Example 4.2.2. In two-dimensional V2 , let a transformation of bases be defined by ⃗˜b = ⃗b + ⃗b and ⃗˜b = ⃗b . Then we have 1 1 2 2 2 

aik







1 1 = 0 1

and



αki







1 −1 = . 0 1

The components of a vector transform as follows: x˜k = αjk xj (k = 1, 2), hence x˜1 = x1 and x˜2 = −x1 + x2 . Example 4.2.3. Let En be the n–dimensional real Euclidean space with the dot product • defined as usual by ⃗x • ⃗y =

n 

k=1

xk y k for all ⃗x = {x1 , x2 , . . . , xn }, ⃗y = {y 1 , y 2 , . . . , y n } ∈ En .

A basis H = {⃗b1 , ⃗b2 , . . . , ⃗bn } is said to be orthonormal if ⃗bi • ⃗bk = δik =



1 (i = k) 0 (i = ̸ k)

(i, k = 1, 2, . . . , n).

Then the formulae in (4.2) transform the orthonormal basis H = {⃗b1 , ⃗b2 , . . . , ⃗bn } into  = {⃗˜b1 , ⃗˜b2 , . . . , ⃗˜bn } if and only if the orthonormal basis H aji = αji for all i, j = 1, 2, . . . , n.

(4.5)

Tensor Algebra and Riemannian Geometry ■ 379

Proof. (i) First we assume that the conditions in (4.5) are satisfied. Then we obtain from ⃗bi • ⃗bk = δik for i, k = 1, 2, . . . , n by (4.3) for all l, m = 1, 2, . . . , n ⃗˜b • ⃗˜b = ai⃗b • ak ⃗b = ai ak ⃗b • ⃗b = ai ak δ = αl ak δ = αl ak = δ l . l m k l i m k l m i l m ik i m ik k m m  are orthonormal and connected by the (ii) Now we assume that the bases H and H transformation formulae (4.2). Then we obtain for all i, j = 1, 2, . . . , n

αji = αjk δki = αjk⃗˜bk • ⃗˜bi = ⃗bj • ⃗˜bi = ⃗bj • ali⃗bl = ali δjl = aji ,

which is (4.5).

4.3

LINEAR FUNCTIONALS AND DUAL SPACES

In this section, we deal with linear functionals, dual spaces and transformations of bases of dual spaces. Let V be a real vector space. Then a map f : V → R is called a linear functional if f (a⃗x + b⃗y ) = af (⃗x) + bf (⃗y ) for all ⃗x, ⃗y ∈ V and all a, b ∈ R. (4.6)

The set of all linear functionals on V is denoted by V ∗ and called the dual space of V . We define f + g and αf for all f, g ∈ V and all α ∈ R by (f + g)(⃗x) = f (⃗x) + g(⃗x) and (αf )(⃗x) = αf (⃗x) (⃗x ∈ V ). Then it is easy to see that V ∗ is a real vector space. We recall that two vector spaces V and V˜ are said to be isomorphic if there exists a linear bijective map between them. The next result is well known from linear algebra. Proposition 4.3.1. The vector spaces V and V ∗ are isomorphic. Proof. Let H = {⃗b1 , ⃗b2 , . . . , ⃗bn } be a basis of V and ⃗x = xk⃗bk ∈ V be given. We define the maps fk : V → R (k = 1, 2, . . . , n) by fk (⃗y ) = y k for all ⃗y = y k⃗bk ∈ V,

(4.7)

and put f = xk fk . Now we define the map T on V by T (x) = f . Since obviously fk ∈ V ∗ for each k = 1, 2, . . . , n, it follows that f ∈ V ∗ , and so T : V → V ∗ . (i) First we show that T is linear. ˜⃗x˜ = y k⃗bk with Let α, α ˜ ∈ R and ⃗x = xk⃗bk , ⃗x˜ = x˜k⃗bk ∈ V . Then ⃗y = α⃗x + α k k k y = αx + α ˜ x˜ (k = 1, 2, . . . , n) and T (α⃗x + α ˜⃗x˜) = T (⃗y ) = y k fk = αxk fk + α ˜ x˜k fk = αT (⃗x) + α ˜ T (⃗x˜).

380 ■ Differential Geometry and Its Visualization

(ii) Now we show that T is one-to-one. Let T (⃗x) = T (⃗x˜), that is, xj = xk fk (⃗bj ) = (T (⃗x))(⃗bj ) = (T (⃗x˜)(⃗bj ) = x˜k fk (⃗bj ) = x˜j (j = 1, 2, . . . , n) by the definition of T and the linear functionals fk (k = 1, 2, . . . , n), and so ⃗x = xj⃗bj = x˜j⃗bj = ⃗x˜. (iii) Finally we show that T is onto. Let f ∈ V ∗ be given. We put xk = f (⃗bk ) (k = 1, 2, . . . , n). Then ⃗x = xk⃗bk ∈ V and, for all ⃗y = y j⃗bj , (T (⃗x))(⃗y ) = xk fk (⃗y ) =

n 

f (⃗bk )fk (⃗y ) = f (⃗bk )y k = f (⃗y ),

k=1

that is, T (x) = f .

Remark 4.3.2. Since an n–dimensional vector space V and its dual space V ∗ are isomorphic, V ∗ is an n–dimensional vector space. Let H = {⃗b1 , ⃗b2 , . . . , ⃗bn } be a basis of V . We define the vectors ⃗b i ∈ V ∗ (i = 1, 2, . . . , n) by ⃗b i (⃗bk ) = δki (i, k = 1, 2, . . . , n). Since ⃗b i ∈ V ∗ (i = 1, 2, . . . , n), we have ⃗b i (⃗x) = ⃗b i (xk⃗bk ) = xk⃗b i (⃗bk ) = xi (i = 1, 2, . . . , n) for all ⃗x = xk⃗bk ∈ V , and ⃗b i indeed is defined on all of V . Furthermore, λi⃗b i = ⃗0 implies λi⃗b i (⃗x) = ⃗0 for all ⃗x ∈ V , in particular, for ⃗x = ⃗bk (k = 1, 2, . . . , n), λi⃗b i (⃗bk ) = λi δki = λk = 0 (k = 1, 2, . . . , n). Thus the set H = {⃗b 1 , ⃗b 2 , . . . , ⃗b n } is linearly independent. Since V ∗ is an n–dimensional vector space, H is a basis of V ∗ , the so–called dual basis. Therefore, every vector l ∈ V ∗ has a unique representation l = li⃗b i , where li = l(⃗bi ) for i = 1, 2, . . . , n. ˜ = {⃗˜b1 , ⃗˜b2 , . . . , ⃗˜bn } be bases of V Proposition 4.3.3. Let H = {⃗b1 , ⃗b2 , . . . , ⃗bn } and H  = {⃗˜b 1 , ⃗˜b 2 , . . . , ⃗˜b n } be with the transformation (4.2), and H = {⃗b 1 , ⃗b 2 , . . . , ⃗b n } and H the corresponding dual bases, that is, ⃗b k (⃗b ) = δ k and ⃗˜b k (⃗˜b j ) = δ k (k, j = 1, 2, . . . , n). If l = li⃗b i = ˜li⃗˜b i , then

j

j

j

j

li = αik ˜lk (i = 1, 2, . . . , n) and ˜lk = aik li (k = 1, 2, . . . , n).

(4.8)

⃗b k = ak⃗˜b i (k = 1, 2, . . . , n) and ⃗˜b i = αi ⃗b k (i = 1, 2, . . . , n). i k

(4.9)

Furthermore,

Therefore, the components of the vectors of the dual space V ∗ transform like the vectors of the bases of V , and the vectors of the dual bases transform like the components of the vectors of V .

Tensor Algebra and Riemannian Geometry ■ 381

Proof. Since li = l(⃗bi ) for i = 1, 2, . . . , n and ˜lk = l(⃗˜bk ) for k = 1, 2, . . . , n, the linearity of l implies li = l(⃗bi ) = l(αik⃗˜bk ) = αik l(⃗˜bk ) = αik ˜lk (i = 1, 2, . . . , n) and ˜lk = l(⃗˜bk ) = l(ai ⃗bi ) = ai l(⃗bi ) = ai li (k = 1, 2, . . . , n), k k k which proves (4.8). Furthermore, we obtain ⃗˜b i = ⃗˜b i (⃗b )⃗b k = ⃗˜b i (αl ⃗˜b )⃗b k = αl ⃗˜b i (⃗˜b )⃗b k = αl δ i⃗b k = αi ⃗b k (i = 1, 2, . . . , n) k l k l k k l k and ⃗b k = ⃗b k (⃗˜bi )⃗˜b i = ⃗b k (al⃗bl )⃗˜b i = al⃗b k (⃗bl )⃗˜b i = al δ k⃗˜b i = ak⃗˜b i (k = 1, 2, . . . , n), i i i l i which proves (4.9). Definition 4.3.4. The vectors of V and V ∗ are called contravariant and covariant vectors, respectively. The components of contravariant vectors have a superscript, and those of covariant vectors have a subscript. Remark 4.3.5. For basis transformations given by ⃗˜b = ai ⃗b (k = 1, 2, . . . , n) and ⃗b = αk⃗˜b (j = 1, 2, . . . , n), k j k i j k the components xk of a contravariant vector and the components xk of a covariant vector transform as follows x˜k = αjk xj (k = 1, 2, . . . , n)

xj = ajk x˜k (j = 1, 2, . . . , n)

x˜k = ajk xj (k = 1, 2, . . . , n) and xj = αjk x˜k (j = 1, 2, . . . , n).

4.4

TENSORS OF SECOND ORDER

Now we generalize the concepts of contravariant and covariant vectors. Let X and Y be two vector spaces. A function B : X × Y → R is called a bilinear form if for every fixed vector ⃗y in Y , the map B(·, ⃗y ) : X → R and for every fixed vector ⃗x in X, the map B(⃗x, ·) : Y → R are linear functionals. Here, we are interested in those cases where X and Y are equal to an n–dimensional vector space V or its dual space. There are three different cases (a) X = Y = V , (b) X = Y = V ∗ , (c) X = V and Y = V ∗ . These three cases lead to the so–called tensors of second order.

382 ■ Differential Geometry and Its Visualization

Definition 4.4.1. (a) A bilinear form B on X = Y = V is called a covariant tensor of second order. (b) A bilinear form B on X = Y = V ∗ is called a contravariant tensor of second order. (c) A bilinear form B on X = V and Y = V ∗ is called a mixed tensor of second order.  = {⃗˜b1 , ⃗˜b2 , . . . , ⃗˜bn } be bases of V , Proposition 4.4.2. Let H = {⃗b1 , ⃗b2 , . . . , ⃗bn } and H  = {⃗˜b 1 , ⃗˜b 2 , . . . , ⃗˜b n } be bases of V ∗ with the transforand H = {⃗b 1 , ⃗b 2 , . . . , ⃗b n } and H mations (4.2) and (4.9).

(a) Every covariant tensor B of second order can be written as

B(⃗x, ⃗y ) = bik xi y k (⃗x = xi⃗bi , ⃗y = y i⃗bi ∈ V ),

(4.10)

˜bjl = ai ak bik and bjl = αi αk˜bik (j, l = 1, 2, . . . , n). j l j l

(4.11)

where bik = B(⃗bi , ⃗bk ) for all i, k = 1, 2, . . . , n, and conversely (4.10) defines a covariant tensor of second order. The components bik of B satisfy the transformation formulae

(b) Every contravariant tensor B of second order can be written as B(⃗x, ⃗y ) = bik xi yk (⃗x = xi⃗b i , ⃗y = yi⃗b i ∈ V ),

(4.12)

˜bjl = αj αl bik and bjl = aj al ˜bik (j, l = 1, 2, . . . , n). i k i k

(4.13)

where bik = B(⃗b i , ⃗b k ) for all i, k = 1, 2, . . . , n, and conversely (4.12) defines a covariant tensor of second order. The components bik of B satisfy the transformation formulae

(c) Every mixed tensor B of second order can be written as B(⃗x, ⃗y ) = bki xi yk (⃗x = xi⃗bi , ⃗y = yi⃗b i ∈ V ),

(4.14)

where bki = B(⃗bi , ⃗b k ) for all (i, k = 1, 2, . . . , n), and conversely (4.14) defines a mixed tensor of second order. The components bki of B satisfy the transformation formulae ˜bl = ai αl bk and bl = αi al ˜bk (j, l = 1, 2, . . . , n). (4.15) j j k i j j k i Proof. (a) Let B be a covariant tensor of second order and ⃗x = xi⃗bi , ⃗y = y i⃗bi ∈ V . Then B(⃗x, ⃗y ) = B(xi⃗bi , y k⃗bk ) = xi y k B(⃗bi , ⃗bk ) = xi y k bik . The converse part is trivial. Furthermore, we have by (4.2) ˜bjl = B(⃗˜bj , ⃗˜bl ) = B(ai⃗bi , ak⃗bk ) = ai ak B(⃗bi , ⃗bk ) = ai ak bik (j, l = 1, 2, . . . , n) j l j l j l and bjl = B(⃗bj , ⃗bl ) = B(αji⃗˜bi , αlk⃗˜bk ) = αji αlk B(⃗˜bi , ⃗˜bk ) = αji αlk˜bik (j, l = 1, 2, . . . , n).

Tensor Algebra and Riemannian Geometry ■ 383

(b) Let B be a contravariant tensor of second order and ⃗x = xi⃗b i , ⃗y = yi⃗b i ∈ V ∗ . Then B(⃗x, ⃗y ) = B(xi⃗b i , yk⃗b k ) = xi yk B(⃗b i , ⃗b k ) = xi yk bik . The converse part is trivial. Furthermore, we have by (4.9), ˜bjl = B(⃗˜b j , ⃗˜b l ) = B(αj⃗b i , αl ⃗b k ) = αj αl B(⃗b i , ⃗b k ) = αj αl bik (j, l = 1, 2, . . . , n) k i i k i k and bjl = B(⃗b j , ⃗b l ) = B(aji⃗˜b i , alk⃗˜b k ) = aji alk B(⃗˜b i , ⃗˜b k ) = aji αkl ˜bik (j, l = 1, 2, . . . , n). (c) Let B be a mixed tensor of second order and ⃗x = xi⃗bi ∈ V , ⃗y = yi⃗b i ∈ V ∗ . Then B(⃗x, ⃗y ) = B(xi⃗bi , yk⃗b k ) = xi yk B(⃗bi , ⃗b k ) = xi yk bki . The converse part is trivial. Furthermore, we have by (4.2) and (4.9), ˜bl = B(⃗˜bj , ⃗˜b l ) = B(ai⃗bi , αl ⃗b k ) = ai αl B(⃗bi , ⃗b k ) = ai αl bk (j, l = 1, 2, . . . , n) j j k j k j k i and blj = B(⃗bj , ⃗b l ) = B(αji⃗˜bi , alk⃗˜b k ) = αji alk B(⃗˜bi , ⃗˜b k ) = αji alk˜bki (j, l = 1, 2, . . . , n).

Example 4.4.3. (a) Products of the form B(⃗x, ⃗y ) = l(⃗x)m(⃗y ) of linear functionals l and m are covariant tensors of second order. We have for the components bik = B(⃗bi , ⃗bk ) = l(⃗bi )m(⃗bk ) = li mk for i, k = 1, 2, . . . , n, and all products of the components of the two covariant vectors appear in the formula. (b) Let ⃗x = {x1 , x2 , . . . , xn }, ⃗y = {y 1 , y 2 , . . . , y n } ∈ Vn be two contravariant vectors. We put bik = xi y k (i, k = 1, 2, . . . , n). Then a contravariant tensor B of second order is defined by B(l, m) = bik li mk (l, m ∈ (Vn )∗ ). The linearity is trivial. Furthermore, the bilinear form is independent of the basis, since ˜bjr ˜lj m ˜ r = bik αij αkr asj atr ls mt = bik δis δkt ls mt = bik li mk . (c) The outer product ⃗x ∧ ⃗y of two contravariant vectors is defined by sik = (⃗x ∧ ⃗y )ik = xi y k − xk y i (i, k = 1, 2, . . . , n).

(4.16)

Its tensor properties can be shown by the study of the transformation of the components s˜jr = x˜j y˜r − x˜r y˜j = αij αkr (xi y k − xk y i ) = αij αkr sik (j, r = 1, 2, . . . , n).

384 ■ Differential Geometry and Its Visualization

4.5

SYMMETRIC BILINEAR FORMS AND INNER PRODUCTS

Since symmetric covariant tensors of second order, that is, tensors that satisfy B(⃗x, ⃗y ) = B(⃗y , ⃗x) for all vectors ⃗x and ⃗y , are of special interest, we shall study them more closely. The corresponding quadratic form is positive definite, that is, B(⃗x, ⃗x) > 0 for all ⃗x ̸= ⃗0. We write ⃗x • ⃗y = B(⃗x, ⃗y ), and call ⃗x • ⃗y inner product and ⃗x • ⃗x the corresponding square of the length. If we choose a basis {⃗b1 , . . . , ⃗bn }, then gik = ⃗bi • ⃗bk

(i, k = 1, 2, . . . , n).

gij g jk = δik

(i, k = 1, 2, . . . , n).

It is a well–known fact from linear algebra that det(gik ) ̸= 0 and that the inverse matrix (g ik ) exists, that is, g ik = g ki (i, k = 1, 2, . . . , n) and (4.17)

Thus an inner product which is adjoint to the first one is defined by l · m = g ik li mk .

The map xi → gik xk (i = 1, 2, . . . , n) from a space into its dual space is invertible by (4.17), and the inverse map is given by mi → g ik mk (i = 1, 2, . . . , n). If ⃗z(1) , ⃗z(2) , . . . , ⃗z(n) are orthonormal vectors, that is, ⃗z(i) • ⃗z(k) = δ(ik) (i, k = 1, 2, . . . , n) then i k g ik = δ rs z(r) z(s) (i, k = 1, 2, . . . , n). (4.18) The set Z = {⃗z(1) , ⃗z(2) , . . . , ⃗z(n) } is a basis of the space, and so every vector ⃗x with the representation ⃗x = xi⃗bi can also be represented as k ⃗ ⃗x = ξ i⃗x(i) = ξ i z(i) bk .

We put

i k bij = δ rs z(r) z(s) gkj (i, j = 1, 2, . . . , n).

Then (4.18) will be proved, once we have shown bij = δji (i, j = 1, 2, . . . , n). But i k i i i l z(s) gkj ξ l z(l) = δ rs δsl z(r) ξ l = z(l) ξ = xi bij = δ rs z(r)

implies bij = δji .

4.6

TENSORS OF ARBITRARY ORDER

In this section, we generalize the concepts of the previous sections. Definition 4.6.1. A multilinear form T of r contravariant and s covariant vectors is called r–contravariant, s–covariant tensor or (r, s)–tensor. Once basis vectors ⃗bk and ⃗bi of the vector space and the dual space have been chosen, the components of the tensor are obtained from 



...is Tki11 ...k = T ⃗bi1 , . . . , ⃗bis ; ⃗bk1 , . . . , ⃗bkr . r

We put T = 0 if the value of T is equal to 0 for arbitrary arguments. Obviously this is the case if and only if all components of the tensor are equal to 0 with respect to some basis.

Tensor Algebra and Riemannian Geometry ■ 385

Theorem 4.6.2. Any superscript and any subscript transforms with a transformation of bases as the components of a contravariant and of a covariant vector, respectively. Conversely this rule of transformation is sufficient for the components of an invariant multilinear form. Proof. (i) We assume that the transformations of the bases are given by ⃗˜bi1 = αi1⃗bj1 , . . . , ⃗˜bis = αis⃗bjs and ⃗˜b = al1 ⃗b , . . . , ⃗˜b = alr ⃗b . k1 kr j1 js k 1 l1 k r lr Then we have 

...is Tki11 ...k = T ⃗˜bi1 , . . . , ⃗˜bis ; ⃗˜bk1 , . . . , ⃗˜bkr r





= αji11 · · · αjiss alk11 · · · alkrr T ⃗bj1 , . . . , ⃗bjs ; ⃗bl1 , . . . , ⃗blr ...js = αji11 · · · αjiss alk11 · · · alkrr Tlj11...l . r



This proves the stated rule of transformation. (1)

(s)

1 r (ii) Let xi1 , . . . , xis and xk(1) , . . . , xk(r) be the components of s covariant vectors (1) (s) ⃗x , . . . , ⃗x and r contravariant vectors ⃗x(1) , . . . , ⃗x(r) , respectively. We put

(1)

(s)

...is 1 r x · · · xis xk(1) · · · xk(r) . T (⃗x(1) , . . . , ⃗x(s) ; ⃗x(1) , . . . , ⃗x(r) ) = Tki11 ...k r i1

Then this obviously defines a multilinear form and ...ms (1) Tnm1 1...n x˜m1 · · · x˜(s) ˜n(1)1 · · · x˜n(r)r ms x r

(1)

(s)

...js i1 1 r = αjm1 1 · · · αjms s aln11 · · · alnrr Tlj11...l am1 · · · aims s xi1 · · · xis αtn11 . . . αtnrr xt(1) · · · xt(r) r (1)

(s)

...js i1 1 r = Tlj11···l δj1 · · · δjiss δtl11 · · · δtlrr xi1 · · · xis xt(1) · · · xt(r) r (1)

(s)

...is 1 r = Tti11...t xi1 · · · xis xt(1) · · · xt(r) . r

j1 ...js ...js ˜ j1 ...js Example 4.6.3. Let Tij11...i , Ti1 ...ir and T˜i1 ...ir that satisfy the transformation formur lae ...js mr l1 ...ls 1 = αlj11 · · · αljss am T˜ij11...i i1 · · · air Tm1 ...mr r

(4.19)

and j1 ...js

mr l1 ...ls 1 T˜i1 ...ir = βlj11 · · · βljss bm i1 · · · bir Tm1 ...mr .

(4.20)

386 ■ Differential Geometry and Its Visualization j1 ...js ...js Then we obtain the following transformation formulae between T˜ij11...i and T˜i1 ...ir as r follows: l1 ...ls n1 nr l1 s Tm = δm · · · δm δ · · · δtlss Tnt11...t ...nr 1 ...mr 1 r t1

p1 n1 pr nr q1 l1 s = αm a · · · αm a αt1 aq1 · · · αtqss alqss Tnt11...t ...nr 1 p1 r pr

p1 pr l1 s = αm · · · αm a · · · alqss αtq11 · · · αtqss anp11 · · · anprr Tnt11...t ...nr 1 r q1 p1 pr l1 ...qs = αm · · · αm aq · · · alqs T˜pq1...p , 1

r

1

s

1

r

hence j1 ...js

mr p1 pr l1 ls ˜ q1 ...qs 1 T˜i1 ...ir = βlj11 · · · βljss bm i1 · · · bir αm1 · · · αmr aq1 · · · aqs Tp1 ...pr .

Remark 4.6.4. (a) Covariant vectors are (1, 0)–tensors, contravariant vectors are (0, 1)–tensors, and scalars are (0, 0)–tensors. (b) Tensors are characterized by the transformation properties of their components. Therefore, we frequently identify tensors with their components. (c) The sum of two tensors T and U of the same structure with the components ...is Tki11 ...k r

is defined by

and

s Uki11...i ...kr

...is i1 ...is i1 ...is s Vki11...k = (T + U )ik11...i ...kr = Tk1 ...kr + Uk1 ...kr ; r

it is an (r, s)–tensor. (d) The outer product V = T ∗ U of an (r, s)–tensor T and an (r′ , s′ )–tensor U with the components ...is l1 ...ls′ Tki11 ...k and Um 1 ...mr ′ r

is defined by

1 ...is l1 ...ls′ s′ Vk11...ksr m = Tki11 ...k · Um ; 1 ...mr ′ 1 ...mr ′ r

i ...i l ...l

it is an (r + r′ , s + s′ )–tensor. Multiplication and addition are commutative T ∗U =U ∗T

and

T + U = U + T.

Furthermore, the distributive and associative laws hold, for instance, V ∗ (U + T ) = V ∗ U + V ∗ T and T ∗ (U ∗ V ) = (T ∗ U ) ∗ V. This follows from the corresponding laws for real numbers. The tensors of the same structure make up a vector space. Proof. We only need to show the statement concerning the outer product of two tensors. p ...p l1 ...ls Let T and U be (r, s)– and (r′ , s′ )–tensors with the components Tm and Uq11...qrs′′ 1 ...mr that satisfy the transformation formulae ...js mr l1 ...ls 1 Tij11...i = αlj11 · · · αljss am i1 · · · air Tm1 ...mr r

Tensor Algebra and Riemannian Geometry ■ 387

and  1 s′ = αn1 · · · αns′ aq1 · · · a r′ U p1 ...ps′ . U p1 ps′ k1 k1 ...kr′ kr′ q1 ...qr′ n ...n

q

Then we obtain for the transformation formulae of the outer product V = T ∗ U ...js  1 1 r′ Vi11...irsk1 ...k = Tij11...i Uk1 ...krs′ ′ r r′ j ...j n ...n

n ...n

mr l1 ...ls n1 ns′ q1 p1 ...ps′ 1 r′ = αlj11 · · · αljss am i1 · · · air Tm1 ...mr αp1 · · · αps′ ak1 · · · akr′ Uq1 ...qr′ q

mr q1 l1 ...ls p1 ...ps′ 1 r′ = αlj11 · · · αljss αpn11 · · · αpnss′′ am i1 · · · air ak1 · · · akr′ Vm1 ...mr q1 ...qr′ . q

...is Definition 4.6.5. Let T be a tensor with the components Tki11 ...k . The contraction r of T is defined by i ...i ,im+1 ...is i ...i jim+1 ...is Tk11 ...km−1 = Tk11 ...km−1 . n−1 ,kn+1 ...kr n−1 jkn+1 ...kr

Remark 4.6.6. The contraction of an (r, s)–tensor yields an (r − 1, s − 1)–tensor.

Proof. We obtain for the transformation formulae of the contraction of an (r, s)– tensor T j ...jq−1 jq+1 ...js j1 ...jq−1 ,jq+1 ...js j1 ...jq−1 tjq+1 ...js V˜i11...ip−1 ip+1 ...ir = Ti1 ...ip−1 ,ip+1 ...ir = Ti1 ...ip−1 tip+1 ...ir

q−1 t q+1 p−1 p+1 s = αlj11 · · · αlq−1 αlq αlq+1 · · · αljss ani11 · · · aip−1 at p aip+1 · · · anirr Tnl11...l ...nr

j

j

n

n

n

q−1 q+1 p−1 p+1 q−1 lq lq+1 ...ls = αlj11 · · · αlq−1 αlq+1 · · · αljss ani11 · · · aip−1 aip+1 · · · anirr αltq at p Tnl11...l ...np−1 np np+1 ...nr

j

j

n

n

n

q−1 q+1 p−1 p+1 q−1 lq lq+1 ...ls = αlj11 · · · αlq−1 αlq+1 · · · αljss ani11 · · · aip−1 aip+1 · · · anirr δlqp Tnl11...l ...np−1 np np+1 ...nr

j

j

n

n

n

q−1 q+1 p−1 p+1 q−1 np lq+1 ...ls = αlj11 · · · αlq−1 αlq+1 · · · αljss ani11 · · · aip−1 aip+1 · · · anirr Tnl11...l ...np−1 np np+1 ...nr

j

j

n

n

q−1 q+1 p−1 p+1 q−1 lq+1 ...ls = αlj11 · · · αlq−1 αlq+1 · · · αljss ani11 · · · aip−1 aip+1 · · · anirr Vnl11 ...l ...np−1 np+1 ...nr .

j

j

n

n

Example 4.6.7. (a) The contraction of a (1, 1)–tensor with the components Tki yields a scalar T,, = Tii = T11 + T22 + · · · + Tnn .

(b) Contracting a (3, 2)–tensor T with the components Tki11 ki22 k3 with respect to k2 and i1 we obtain the (2, 1)–tensor with the components Tk,i12,k3 = Tkji1 2jk3 .

Definition 4.6.8. The outer product of two tensors with contraction with respect to indices from different factors is called inner product of the two tensors. Remark 4.6.9. The inner product of an (r, s)–tensor and an (r′ , s′ )–tensor is an (r + r′ − 1, s + s′ − 1)–tensor.

388 ■ Differential Geometry and Its Visualization

Proof. The statement is an immediate consequence of Remarks 4.6.4 (d) and 4.6.6. Example 4.6.10. (a) We consider the first fundamental coefficients gik and g lm . The outer product is lm clm ik = gik g , and for k = l, we obtain the contraction km km c,m = δim . i, = cik = gik g

(b) We consider the values g ik and Llm . The values g ik transform like the components of a tensor, and the values Llm also transform like the components of a tensor, provided the transformation of the coordinates preserves the orientation of the surface normal vectors. We obtain the outer product ik cik lm = g Llm ,

and the contractions ik , k ik dkm = c,k ,m = g Lim and d, = dk = g Lik = 2H.

4.7

SYMMETRIC AND ANTI-SYMMETRIC TENSORS

In this section, we study symmetric and anti–symmetric tensors. Definition 4.7.1. A tensor is called symmetric in two of its indices, if it is invariant with respect to their interchange; it is called anti–symmetric if it changes sign. Theorem 4.7.2. Symmetry and anti–symmetry of tensors are independent of the choice of the coordinate system. Proof. We show that

We have

s s s s Tij11i...j = Tij21i...j implies Tij11i...j = Tij21i...j . 2 ...ir 1 ...ir 2 ...ir 1 ...ir

mr l1 l2 ...ls s 1 m2 Tij11i...j = αlj11 αlj22 · · · αljss am i1 ai2 · · · air Tm1 m2 ...mr 2 ...ir

mr l1 l2 ...ls 2 m1 j1 ...js = αlj11 αlj22 · · · αljss am i2 ai1 · · · air Tm2 m1 ...mr = Ti2 i1 ...ir .

Remark 4.7.3. Any (r, s)–tensor may be written as the sum of a symmetric and an anti–symmetric tensor, for instance, Tik =

1 1 (Tik + Tki ) + (Tik − Tki ) . 2 2

Tensor Algebra and Riemannian Geometry ■ 389

Example 4.7.4. The first fundamental coefficients gik are the components of a symmetric tensor. The values ϵik with √ ϵ11 = ϵ22 = 0 and ϵ12 = −ϵ21 = g are the components of an anti–symmetric tensor.

Definition 4.7.5. Two tensors with components Aik and B jl and non–vanishing determinants are called conjugated if Aik B kl = δil . Example 4.7.6. The values gik and g jl are the components of conjugated tensors.

4.8

RIEMANN SPACES

The concept of Riemann spaces is needed in the measurement of lengths of curves on a manifold. We shall see that we require a covariant symmetric tensor gik (x) on the manifold. Then the length of a segment of a curve given by a parametric representation ⃗x(t) (t ∈ [t1 , t2 ]) will be defined by s(t2 ) − s(t1 ) =

t2

t1



gik

dxi dk dt. dt dt

Let a coordinate system S with the coordinates (xi ) be given in some neighbourhood of a point P0 of an n–dimensional manifold and P0 = (x1 , x2 , . . . , xn ). Then a special 0 0 0 basis is defined for the tangent space T (P0 ) at P0 . The vectors ⃗bi of the basis of T (P0 ) are the tangents to the curves γ(i) with the parametric representations 

1

⃗x(i) (t) = x , . . . , x 0

0

i−1

, x + t, x i

0

i+1

0

,...,x 0

n



(i = 1, 2, . . . , n).

(4.21)

If we choose another coordinate system S˜ with coordinates (˜ xi ) then this yields another basis for the tangent space T (P0 ) at the point P0 , and we are going to find the corresponding basis transformation according to (4.2). Let γ be an arbitrary curve with the parametric representations ⃗x(t) = {x1 (t), . . . , xn (t)} and ⃗x˜(t) = {˜ x1 (t), . . . , x˜n (t)}

in the coordinate systems S and S˜ with the coordinates (xi ) and (˜ xi ). Since xi (t) = xi (˜ xk (t)) (i = 1, 2, . . . , n), the chain rule yields ∂xi d˜ xk dxi = dt ∂ x˜k dt

(i = 1, 2, . . . , n).

(4.22)

Since the tangent vector at P0 is independent of the bases {⃗b1 , . . . , ⃗bn } and {⃗˜b1 , . . . , ⃗˜bn }, we must have i xk ⃗bi dx = ⃗˜bk d˜ . dt dt

390 ■ Differential Geometry and Its Visualization

Substituting (4.21) and (4.2), we obtain i

k

k

k

x d˜ x d˜ x ⃗bi ∂x d˜ = ⃗˜bk = aik⃗bi . k ∂ x˜ dt dt dt Since this holds for arbitrary d˜ xk /dt, we must have aik =

∂xi (i, k = 1, 2, . . . , n). ∂ x˜k

(4.23)

Similarly, we obtain for the inverse αjk =

∂ x˜k (j, k = 1, 2, . . . , n). ∂xj

Example 4.8.1. Let S be a surface in R3 with a parametric representation ⃗x(ui ). Then the tangent vectors ⃗bi above are the partial derivatives ⃗xi , hence the tangent vectors of the ui –lines. We define gik = ⃗xi • ⃗xk (i, k = 1, 2). Then the following formulae of transformation hold g˜ik = glm

∂ul ∂um (i, k = 1, 2). ∂ u˜i ∂ u˜k j1 ...js

...js Example 4.8.2. We consider the tensors T˜ij11...i and T˜i1 ...ir of Example 4.6.3 that r satisfy the transformation formulae in (4.19) and (4.20) in the special case, where k

αik Then

 ∂x ∂xk ˜ ∂ x˜k k ∂xk k k = , a = , β = and b = i i i i for i, k = 1, 2, . . . , n. ∂xi ∂ x˜i ∂xi  ∂x ˜

∂ x˜j1 ∂ x˜js ∂xm1 ∂xmr l1 ...ls ...js T˜ij11...i = · · · · · · T r ∂xl1 ∂xlr ∂ x˜i1 ∂ x˜ir m1 ...mr and

imply

j1

js

j1

js

j1

js

  j1 ...js ˜ ˜ ∂xm1 ∂x ∂x ∂xmr l1 ...ls · · · · · · T˜i1 ...ir = i ir Tm1 ...mr 1 ∂xl1 ∂xls ∂ x   ˜ ∂x ˜ ˜ j1 ...js

T i1 ...ir

  ∂x ∂x ∂xmr ∂ x˜p1 ∂ x˜pr ∂xl1 ∂xls ˜q1 ...qs ˜ ˜ ∂xm1 = · · · · · · · · · · · · T i1 ir m ∂xl1 ∂xls ∂ x ∂xmr ∂ x˜q1 ∂ x˜qs p1 ...pr   ˜ ∂x ˜ ∂x 1 j1 js   ˜ ∂xl1 ˜ ∂xls ∂ x˜p1 ∂xm1 ∂x ∂x ∂ x˜pr ∂xmr ˜q1 ...qs = · · · · · · i1 ir Tp1 ...pr ∂xl1 ∂ x˜q1 ∂xls ∂ x˜qs ∂xm1 ∂ x ∂xmr ∂ x   ˜ ˜   ∂x ∂x ∂ x˜pr ˜q1 ...qs ˜ ˜ ∂ x˜p1 = · · · · · · i1 ir Tp1 ...pr . ∂ x˜q1 ∂ x˜qs ∂ x   ˜ ∂x ˜

Tensor Algebra and Riemannian Geometry ■ 391

Definition 4.8.3. Let G be a symmetric tensor in Rn with the components gik = ⃗bi • ⃗bk (i, k = 1, 2, . . . , n) in a coordinate system S with the coordinates (xi ), where the vectors ⃗bi (i = 1, 2, . . . , n) are the tangent vectors defined above. We assume that the functions gik are continuous. Then G is called metric tensor. Let γ be a curve in Rn with the parametric representations x1 (t), x˜2 (t), . . . , x˜n (t)} ⃗x(t) = {x1 (t), x2 (t), . . . , xn (t)} and ⃗x˜(t) = {˜

xi ). Then in the coordinate systems S and S˜ with the coordinates (xi ) and (˜ g˜ik =

d˜ xi d˜ ∂xl ∂xm ∂ x˜i dxj ∂ x˜k dxn xk = glm i dt dt ∂ x˜ ∂ x˜k ∂xj dt ∂xn dt dxj dxn dxj dxn dxi dxk = glm δjl δnm = gjn = gik . dt dt dt dt dt dt

We define the length s by s = s(t1 , t2 ) =

t2

t1



e · gik

dxi dxk dt, dt dt

where the indicator function e is defined such that 

dxi dxk dxi dxk e · gik ≥ 0, that is, e = sign gik dt dt dt dt



.

Thus a curve may be split into arcs with e > 0, e = 0 and e < 0. The length is independent of both the coordinate system and the parameters. Definition 4.8.4. Let S and S˜ be two coordinate systems in Rn , G be a metric tensor with the components gik (xj ) and g˜ik (˜ xj ) (i, k = 1, 2, . . . , n) and s be a length defined by  s=

t2

t1

e · gik

dxi dxk dt. dt dt

Then R with this length is called Riemann space. n

Example 4.8.5 (Euclidean En ). Let S be a Cartesian coordinate system and the components of the metric tensor be given by gik = δik (i, k = 1, 2, . . . , n). For a curve γ with a parametric representation ⃗x(t) = {x1 (t), x2 (t), . . . , xn (t)}, we have s=

t2

t1



dxi dxk dt = e · δik dt dt

and the transformation formulae

  2 t2  n  dxi  dt,

t1

i=1

dt

n  ∂xm ∂ x˜l ∂xm ∂xm g˜ik = δ˜ik = gml i = (i, k = 1, 2, . . . , n). ∂ x˜ ∂ x˜k m=1 ∂ x˜i ∂ x˜k

392 ■ Differential Geometry and Its Visualization

Example 4.8.6 (Surfaces in E3 ). Let a surface in R3 be given by some parametric representation ⃗x(uα ) = ⃗x(u1 , u2 ). A curve given by (u1 (t), u2 (t)) in the parameter plane generates a curve with a parametric representation ⃗x(t) = ⃗x(u1 (t), u2 (t)) on the surface. By Example 4.8.5, the arc length along this curve is given by     t2  t2  3 3 k 2   dx ∂xk duα ∂xk duβ s=  dt dt =  α β t1

=

t2

t1

k=1

dt



duα duβ dt = ⃗xα • ⃗xβ dt dt

∂u

dt ∂u

gαβ

duα duβ dt. dt dt

k=1

t1

t2

t1



dt

We have for parameter transformations g˜αβ =

∂uγ ∂uδ gγδ . ∂ u˜α ∂ u˜β

Let S˜ be some other coordinate system in E3 with x˜i = x˜i (xj ) and

x˜i (uα ) = x˜i (xj (uα )) (i = 1, 2, 3).

The length of a curve with a parametric representation 



⃗x˜(t) = x˜1 (xj (uα (t))), x˜2 (xj (uα (t))), x˜3 (xj (uα (t))) is given by s=

t2

t1



g˜ik

3  d˜ xi d˜ ∂xm ∂xm xk dt, where g˜ik = (i, k = 1, 2). dt dt ∂ x˜i ∂ x˜k m=1

This implies g˜ik

3  d˜ xi d˜ ∂xm ∂xm ∂ x˜i ∂xj duα ∂ x˜k ∂xl duβ xk = dt dt ∂ x˜i ∂ x˜k ∂xj ∂uα dt ∂xl ∂uβ dt m=1

= = =

3  ∂xm ∂ x˜i ∂xm ∂ x˜k ∂xj ∂xl duα duβ

m=1 3  m=1 3 

∂ x˜i ∂xj ∂ x˜k ∂xl ∂uα ∂uβ dt dt

δjm δlm

∂xj ∂xl duα duβ ∂uα ∂uβ dt dt

∂xm ∂xm duα duβ duα duβ = g , αβ ∂uα ∂uβ dt dt dt dt m=1

hence g˜ik

d˜ xi x˜k duα duβ = gαβ . dt dt dt dt

Tensor Algebra and Riemannian Geometry ■ 393

4.9

THE CHRISTOFFEL SYMBOLS

In this section, we study the Christoffel symbols which we introduce in a formal way. Definition 4.9.1. Let a metric tensor G with the components gik and its conjugated tensor be given in a Riemann space. We define the first and second Christoffel symbols by 1 1 [ijk] = (gik,j − gij,k + gjk,i ) = 2 2



∂gij ∂gjk ∂gik − + j k ∂x ∂x ∂xi



and Γkij

=



k ij



= g kl [ijl].

Remark 4.9.2. We note that the values Γkij are not the components of a tensor as we will see in the transformation formulae (4.28) below. Theorem 4.9.3. The following formulae hold [ikl] = glj





j , ik

[ijk] = [jik],



k ij



=

∂gik = [ijk] + [kji], ∂xj   ∂ 2 xl ∂xm ∂xl ∂ 2 xm ∂˜ gij = + i glm ∂ x˜k ∂ x˜i ∂ x˜k ∂ x˜j ∂ x˜ ∂ x˜j ∂ x˜k





k , ji

∂xl ∂xm ∂glm ∂xn , ∂ x˜i ∂ x˜j ∂xn ∂ x˜k   ∂xl ∂xn ∂xm ∂ 2 xl  [ijk] = [lnm] + g , lm ∂ x˜i ∂ x˜j ∂ x˜i ∂ x˜j ∂ x˜k

(4.24) (4.25) (4.26)

+

  

k ij

=



∂xp ∂xq ∂ x˜i ∂ x˜j



m pq



∂ 2 xm + i j ∂ x˜ ∂ x˜



∂ x˜k ∂xm

(4.27) (4.28)

and 



∂ 2 xp ∂xp  ∂xl ∂xm k = − ∂ x˜i ∂ x˜j ∂ x˜k ij ∂ x˜i ∂ x˜j





p . lm

(4.29)

Proof. The proof is analogous to that of Lemma 3.1.4 with the exception of the identities in (4.26) and (4.29). The transformation formula g˜ij =

∂xl ∂xm glm ∂ x˜i ∂ x˜j

and the product and chain rules together imply ∂˜ gij = ∂ x˜k



∂ 2 xl ∂xm ∂xl ∂ 2 xm + i j k ∂ x˜i ∂ x˜k ∂ x˜j ∂ x˜ ∂ x˜ ∂ x˜



glm +

∂xl ∂xm ∂glm ∂xn . ∂ x˜i ∂ x˜j ∂xn ∂ x˜k

394 ■ Differential Geometry and Its Visualization

It follows from (4.28) that 



∂xp  k = k ∂ x˜ ij



∂xl ∂xq ∂ x˜i ∂ x˜j

=



=

∂xl ∂xq ∂ x˜i ∂ x˜j

∂xl ∂xq ∂ x˜i ∂ x˜j 

 

m lq m lq

p lq



 

+

∂ 2 xm + i j ∂ x˜ ∂ x˜ ∂ 2 xm + i j ∂ x˜ ∂ x˜

 

∂ x˜k ∂xp ∂xm ∂ x˜k p δm

∂ 2 xp , ∂ x˜i ∂ x˜j

hence 



∂xp  ∂xl ∂xq k ∂ 2 xp = − ∂ x˜i ∂ x˜j ∂ x˜k ij ∂ x˜i ∂ x˜j





p . lq

The identity in (4.29) now follows if we rename the index of summation q by m.

CHAPTER

5

Tensor Analysis

Tensors first appeared in the theory of surfaces. Typical examples were the first and second fundamental coefficients. A tensor of a given order was assigned to every point of a manifold. We also saw that certain derivatives of tensors were needed in various problems of differential geometry. It seems natural to try and define derivatives of a tensor in a way such that the result again is a tensor. This guarantees the independence of a derivative of the choice of a coordinate system. In particular, we study • covariant derivatives of contravariant, covariant vectors and (1, 1)–tensors introduced in Definitions 5.1.2, 5.1.3 and 5.1.4 • covariant derivatives of (r, s)–tensors in Definition 5.2.1, their basic properties in Theorems 5.2.2 and 5.2.4, and the Ricci identity for the covariant derivatives of the metric tensor in Theorem 5.2.6 • the mixed Riemann tensor of curvature in a Riemann space with metric tensor gik in Definition 5.3.1 and the interchange of the order of its covariant differentiation in Ricci’s identity, Theorem 5.3.2 • the Bianchi identities for the derivatives of the mixed Riemann tensor of curvature and the covariant Riemann tensor of curvature in Theorem 5.4.1 • the Beltrami differentiator of first order in a Riemann space in Definition 5.5.1, the divergence of contravariant and covariant vectors in Definitions 5.5.3 and 5.5.6, and some properties and a few cases of special interest in Theorem 5.5.5, Remarks 5.5.2, 5.5.4, 5.5.7, and Example 5.5.8 • a geometric meaning of the covariant differentiation, the Levi–Civit`a parallelism in Definition 5.6.3 • the fundamental theorem of the theory of surfaces, Theorem 5.7 • a geometric interpretation of the Riemann tensor of curvature

• spaces with vanishing tensor of curvature, the existence and uniqueness of autoparallel curves in Riemann spaces in Theorem 5.9.3 DOI: 10.1201/9781003370567-5

395

396 ■ Differential Geometry and Its Visualization

• an extension of Frenet’s formulae for curves in Riemann spaces in Theorem 5.10.1 • Riemann normal coordinates and the curvature of spaces and the Bertrand– Puiseux theorem, Theorem 5.11.1.

5.1

COVARIANT DIFFERENTIATION

In this section, we deal with the following problem: differentiation should turn a tensor into a tensor and, in special cases, coincide with partial differentiation. Example 5.1.1. (a) Partial differentiation of a (0, 0)–tensor, that is, of a scalar, yields a covariant vector. Let F be a (0, 0)–tensor with the components F (x1 , x2 , . . . , xn ), that is, the values of a real–valued function on Rn . Furthermore let F ∗ (x∗1 , x∗2 , . . . , x∗n ) be its components in another coordinate system. Then F ∗ (x∗1 , . . . , x∗n ) = F (x1 (x∗1 , . . . , x∗n ), . . . , xn (x∗1 , . . . , x∗n )) and

∂F ∗ ∂F ∂xk ∂xk = = Fk for all i. ∂x∗i ∂xk ∂x∗i ∂x∗i Thus the components of the partial derivative of a (0, 0)–tensor transform in the same way as the components of a covariant vector. (b) Partial differentiation of a contravariant tensor T with the components T i and T ∗i in the coordinate systems S and S ∗ yields Fi∗ =

∂T ∗i ∂ = ∂x∗j ∂x∗j



∂x∗i m T ∂xm



=

∂xn ∂x∗i ∂T m ∂ 2 x∗i ∂xn + m n ∗j T m for all i and j. (5.1) ∗j m n ∂x ∂x ∂x ∂x ∂x ∂x

Only the first term of the result on the right hand side in (5.1) satisfies the law of transformation for the components of a tensor. The problem, however, is in the second term. We try to add certain terms to the partial derivative of a tensor such that the result will transform like a tensor. First we consider a contravariant vector T with the components T i and T ∗i in the coordinate systems S and S ∗ , and write for the new kind of derivative T,ji =

∂T i ∂T ∗i i k ∗i ∗k + Λ T and T = + Λ∗i for all i and j jk ,j jk T ∂xj ∂x∗j

(5.2)

with the values Λijk and Λ∗i jk yet to be determined. Since the terms T,ji are to transform like the components of a tensor, that is, satisfy T,j∗i =

∂x∗i ∂xn m T for all i and j, ∂xm ∂x∗j ,n

(5.3)

Tensor Analysis ■ 397

we have to choose the terms Λijk and Λ∗i jk such that identities in (5.3) hold. Substituting (5.2) into (5.3), we obtain ∂x∗i ∂xn m ∂x∗i ∂xn ∂T ∗i ∗i ∗k ∗i + Λ T = T = T = jk ,j ∂x∗j ∂xm ∂x∗j ,n ∂xm ∂x∗j



∂T m k + Λm nk T ∂xn



for all i and j.

Now this, (5.2) and (5.1) together yield ∗l ∂T ∗i ∂x∗i ∂xn ∂T m ∂ 2 x∗i ∂xn m ∗i ∗k ∗i ∂x + Λ T = + T + Λ Tk jk jl ∂x∗j ∂xm ∂x∗j ∂xn ∂xm ∂xn ∂x∗j ∂xk   ∂x∗i ∂xn ∂T m m k = + Λ T nk ∂xm ∂x∗j ∂xn

T,j∗i =

for all i and j. This implies

∂ 2 x∗i ∂xn m ∂x∗l k ∗i ∂x∗i ∂xn m k T + T Λjl = Λ T for all i and j. ∂xm ∂xn ∂x∗j ∂xk ∂xm ∂x∗j nk If we rename the index of summation m in the first sum on the left hand side to k, then we obtain 



∂x∗l ∗i ∂x∗i ∂xn m ∂ 2 x∗i ∂xn + Λ − Λ T k = 0 for all i and j. ∂xk ∂xn ∂x∗j ∂xk jl ∂xm ∂x∗j nk

The last identity has to hold true for all values of T k . Thus ∂x∗l ∗i ∂x∗i ∂xn m ∂ 2 x∗i ∂xn + Λ − Λ = 0 for all i, j and k. ∂xk ∂xn ∂x∗j ∂xk jl ∂xm ∂x∗j nk This implies ∂x∗j ∂xp



∂x∗l ∗i ∂x∗i ∂xn m ∂ 2 x∗i ∂xn + Λ − Λ ∂xk ∂xn ∂x∗j ∂xk jl ∂xm ∂x∗j nk

=



∂ 2 x∗i ∂x∗l ∂x∗j ∗i ∂x∗i m + Λ − Λ =0 ∂xk ∂xp ∂xk ∂xp jl ∂xm pk

for all i, k and p. We have by identity (4.29) −

∂x∗i ∂x∗l ∂x∗j ∗i ∂x∗i m Λ + Λ = ∂xk ∂xp jl ∂xm pk ∂xm

and we may choose Λkij

=



k ij





m pk



−

∂x∗l ∂x∗j ∂xk ∂xp



i jl

∗

for all i, k and p,

= Γkij for all i, j and k.

This gives rise to the following definition. Definition 5.1.2. The covariant derivative of a contravariant vector with the components T i is defined by T;ji

∂T i = + ∂xj





i T k for all i and j. jk

398 ■ Differential Geometry and Its Visualization

We introduce the covariant differentiation of covariant vectors by requiring that the product rule holds. Let T be a covariant vector with the components Ti and U be a contravariant vector with the components U i . Then we require for the covariant differentiation by Definition 5.1.2 (Ti U i );j = Ti;j U i + Ti U;ji = Ti;j U i + Ti



∂U i + ∂xj





i Uk jk



for all j.

Since Ti U i is a scalar, the result of its covariant differentiation is equal to its partial derivative by Example 5.1.1 (a). Thus we obtain ∂U i ∂U i ∂Ti i i i U + T = (T U ) = T U + T + Ti i i ,j i;j i ∂xj ∂xj ∂xj





i U k for all j, jk

and consequently ∂Ti i U = Ti;j U i + Ti ∂xj





∂Tk k i U k or U = Tk;j U k + Ti jk ∂xj





i U k for all j. jk

Since the last identity has to hold true for arbitrary contravariant tensors U , we must have   ∂Tk i Tk;j = − T for all k and j. i jk ∂xj This gives rise to the following definition.

Definition 5.1.3. The covariant derivative of a covariant vector T with the components Ti is defined by Ti;j =

∂Ti − ∂xj



k ij



Tk for all i and j.

Before we generalize the definition of the covariant differentiation to arbitrary tensors, we consider the covariant differentiation of a (1, 1)–tensor T with the components Tik . The following transformation formulae hold for all i nd k Ti∗k =

∂xm ∂x∗k n T . ∂x∗i ∂xn m

Partial differentiation yields n ∂xm ∂x∗k ∂Tm ∂xl ∂Ti∗k n = + Tm ∂x∗j ∂x∗i ∂xn ∂xl ∂x∗j



∂ 2 x∗k ∂xr ∂xm ∂ 2 xm ∂x∗k + ∂x∗j ∂x∗i ∂xn ∂xr ∂xn ∂x∗j ∂x∗i



for all i, j and k. Applying (4.29), we eliminate the second order partial derivatives ∂ 2 xm ∂x∗k ∂xm ∂ 2 x∗k ∂xr + ∗i r n ∗j ∂x∗j ∂x∗i ∂xn ∂x ∂x ∂x ∂x     ∂xm l ∗ ∂x∗k ∂xl ∂xq m ∂x∗k = − ∗i ∗j lq ∂xn ∂x∗l ji ∂xn ∂x ∂x

Tensor Analysis ■ 399

+

∂xm ∂x∗k ∂x∗i ∂xl



l rn



∂xr ∂x∗l ∂x∗q − ∂x∗j ∂xr ∂xn



k lq

∗

∂xm ∂xr ∂x∗i ∂x∗j

for all i, j, k, m and n. Thus 







m ∗k m ∗q ∗l r l ∗ k ∗ ∂Ti∗k n ∂x ∂x n ∂x ∂x ∂x ∂x − T + T m m ∂x∗j ∂x∗l ∂xn ji ∂x∗i ∂xn ∂xr ∂x∗j lq    ∗ m m ∗k ∗q ∂Ti∗k l ∗ k n ∂x ∂x n ∂x ∂x ∗l = − Tm ∗l + Tm ∗i δ ∂x∗j ∂x ∂xn ji ∂x ∂xn j lq  ∗  ∗ ∂Ti∗k l k ∗q ∗k = − T + T l i ∗j jq ∂x ij   n l m ∗k l q ∗k  r m ∗k  m l ∂Tm ∂x ∂x ∂x n ∂x ∂x ∂x n ∂x ∂x ∂x = − Tm ∗i ∗j + Tm ∗j ∗i ∂xl ∂x∗j ∂x∗i ∂xn ∂x ∂x ∂xn lq ∂x ∂x ∂xl rn

and interchanging n and l in the last term =





n l q ∂Tm ∂xl ∂xm n ∂x ∂x − T m ∂xl ∂x∗j ∂x∗i ∂x∗i ∂x∗j

m lq



+

l Tm

∂xr ∂xm ∂x∗j ∂x∗i







n rl



∂x∗k ∂xn

and interchanging m and l in the second term =



n q ∂Tm ∂xl n ∂x − T l ∂xl ∂x∗j ∂x∗j



l mq



r l ∂x + Tm ∂x∗j

n rl

∂xm ∂x∗k ∂x∗i ∂xn

and changing l to r in the first term and q to r in the second term =



n ∂Tm − Tln ∂xr



l mr



l + Tm



n rl



∂xr ∂xm ∂x∗k ∂x∗j ∂x∗i ∂xn

for all i, j and k. Consequently, we have 







∂Ti∗k l ∗ k ∗ ∗q ∗k − T + T l i ij jq ∂x∗j      r m ∗k n ∂x ∂x ∂x ∂Tm l n n l = − Tl + Tm ∗j ∗i n r mr rl ∂x ∂x ∂x ∂x for all i, j and k. Thus the terms n Tm;r

n ∂Tm = − Tln ∂xr



l mr



+

l Tm



n rl



for all m, n and r

satisfy the transformation formulae of a tensor. Definition 5.1.4. The covariant derivative of a (1, 1)–tensor with the components n is defined by Tm n Tm;r

n ∂Tm = − Tln ∂xr



l mr



+

l Tm



n rl



for all m, n and r.

400 ■ Differential Geometry and Its Visualization

Remark 5.1.5. We summarize. The covariant derivatives of a contravariant vector, a covariant vector and a (1, 1)–tensor with the components T i , Ti and Tji are defined by

Ti;k





∂T i i + Tl kl ∂xk   l ∂Ti = − Tl ik ∂xk

T;ki =

for all i and k, for all i and k,

and i Tj;k

5.2

∂Tji = − Tli ∂xk



l jk



+

Tjl



i kl



for all i, j and k.

THE COVARIANT DERIVATIVE OF AN (R, S)–TENSOR

In this section, we extend the concept of covariant derivatives to the general case of (r, s)–tensors. Furthermore we shall prove the Ricci identity. Definition 5.2.1. The covariant derivative of an (r, s)–tensor T with the components ...js Tij11...i r is defined by   ...js   Tij1···i 1 r ;k   

...js ∂Tij11...i r = ∂xk

− +

r 

l=1 s 

...js Tij11...i l−1 mil+1 ...ir

n=1



j ...j pj ...j Ti11...irn−1 n+1 s

m il k





jn pk



(5.4) .

Theorem 5.2.2. The covariant derivative of an (r, s)–tensor is an (r + 1, s)–tensor. Covariant differentiation increases the number of covariant indices by one. Proof. The proof is similar to that in the case of a (1, 1)–tensor. ...js 1 ...js Let T be an (r, s)–tensor with the components Tij11...i and Ti∗j in the coordinate r 1 ...ir systems S and S ∗ , respectively. We put A1 = A2 =

1 ...js ∂Ti∗j 1 ...ir , ∂x∗k

r 

1 ...js Ti∗j 1 ...il−1 mil+1 ...ir

l=1



m il k

∗

and A3 =

s 

n=1

∗j ...jn−1 pjn+1 ...js

1 Ti1 ...i r



jn pk

∗

Tensor Analysis ■ 401

so that

1 ...js Ti∗j = A1 − A2 + A3 1 ...ir ;k

(5.5)

by Definition 5.2.1. Using the transformation formula for the components of an (r, s)– tensor, we obtain ∂ A1 = ∂x∗k



∂xmr ∂x∗j1 ∂x∗js l1 ...ls ∂xm1 · · · · · · T ∂x∗i1 ∂x∗ir ∂xl1 ∂xls m1 ...mr



l1 ...ls ∂xm1 ∂xmr ∂xα ∂x∗j1 ∂x∗js ∂Tm 1 ...mr · · · · · · ∂x∗i1 ∂x∗ir ∂x∗k ∂xl1 ∂xls ∂xα r  ∂xmα−1 ∂xmα+1 ∂xmr ∂x∗j1 ∂x∗js ∂ 2 xmα ∂xm1 l1 ...ls + Tm · · · · · · · · · 1 ...mr ∂x∗iα ∂x∗k ∂x∗i1 ∂x∗iα−1 ∂x∗iα+1 ∂x∗ir ∂xl1 ∂xls α=1

=

+

s  ∂ 2 x∗jβ ∂xm1

∂xlβ ∂xγ ∂x∗i1 β=1

mr

···

We have by (4.29) l1 ...ls A11= Tm 1 ...mr

= − =

l1 ...ls Tm 1 ...mr l1 ...ls Tm 1 ...mr r 

∗j1

∗jβ−1

∗jβ+1

∗js

γ



∂x ∂x ∂x ∂x ∂x  ∂x ··· ··· . ∂x∗ir ∂xl1 ∂xlβ−1 ∂xlβ+1 ∂xls ∂x∗k

∂ 2 xmα ∂xm1 ∂xmα−1 ∂xmα+1 ∂xmr ∂x∗j1 ∂x∗js · · · · · · · · · ∂x∗iα ∂x∗k ∂x∗i1 ∂x∗iα−1 ∂x∗iα+1 ∂x∗ir ∂xl1 ∂xls α=1 r 

  r  ∂xmα β ∗ ∂xm1

α=1 r 

∂x∗β

∂xγ ∂xδ ∂x∗iα ∂x∗k α=1



l1 ...ls Tm 1 ...mα−1 mα mα+1 ...mr

α=1

∂x∗i1

iα k

mα γδ



···

∂xmα−1 ∂xmα+1 ∂xmr ∂x∗j1 ∂x∗js · · · · · · ∂x∗iα−1 ∂x∗iα+1 ∂x∗ir ∂xl1 ∂xls

∂xmα−1 ∂xmα+1 ∂xmr ∂x∗j1 ∂x∗js ∂xm1 · · · · · · · · · ∂x∗i1 ∂x∗iα−1 ∂x∗iα+1 ∂x∗ir ∂xl1 ∂xls

∂xm1 ∂xmα−1 ∂xmα ∂xmα+1 ∂xmr · · · · · · · ∂x∗i1 ∂x∗iα−1 ∂x∗β ∂x∗iα+1 ∂x∗ir 



∂x∗js ∂x∗j1 β ∗ · · · · ∂xl1 ∂xls iα k  r     ∂xmr ∂x∗j1 ∂x∗js ∂xδ m ∂xm1 l1 ...ls − Tm . . . . . . , 1 ...ml−1 mml+1 ...mr ml δ ∂x∗i1 ∂x∗ir ∂xl1 ∂xls ∂x∗k l=1 where we change the indices of summation mα and γ to m and ml , respectively, in the second sum. An application of the transformation formula in the first sum yields   r ∂xm1 m ∂xmr ∂xα ∂x∗j1 ∂x∗js  l1 ...ls A11 = A2 − ∗i1 · · · ∗ir ∗k ··· T . ∂x ∂x ∂x ∂xl1 ∂xls l=1 m1 ...ml−1 mml+1 ...mr ml α (5.6) Similarly it follows that l1 ...ls A22= Tm 1 ...mr

=

l1 ,...,ls Tm 1 ...mr

s  ∂ 2 x∗jβ ∂xm1

β=1 s 

∂xlβ ∂xγ ∂x∗i1

···

∂xmr ∂x∗j1 ∂x∗jβ−1 ∂x∗jβ+1 ∂x∗js ∂xγ · · · · · · ∂x∗ir ∂xl1 ∂xlβ−1 ∂xlβ+1 ∂xls ∂x∗k

∂xmr ∂x∗j1 ∂x∗jβ−1 ∂x∗jβ ∂x∗jβ+1 ∂x∗js ∂xγ ∂xm1 · · · · · · · · · l l ∂x∗i1 ∂x∗ir ∂xl1 ∂x β−1 ∂xδ ∂x β+1 ∂xls ∂x∗k β=1



δ lβ γ



402 ■ Differential Geometry and Its Visualization

−

l1 ...ls Tm 1 ...mr

β=1



= −

s  ∂xm1

s 

l1 ...lβ−1 plβ+1 ...ls Tm 1 ...mr

β=1

s  ∂xm1

β=1

∂x∗i1

···

∂x∗i1

∂xmr ∂x∗j1 ∂x∗jβ−1 ∂x∗t ∂x∗jβ+1 ∂x∗js · · · · · · · ∂x∗ir ∂xl1 ∂xlβ−1 ∂xlβ ∂xlβ+1 ∂xls



·







jβ tu

∗

∂x∗u ∂xγ ∂xγ ∂x∗k

∂xmr ∂x∗j1 ∂x∗js ∂xγ lβ  ∂xm1 · · · · · · pγ ∂x∗i1 ∂x∗ir ∂xl1 ∂xls ∂x∗k

∂xmr ∂x∗j1 ∂x∗jβ−1 ∂x∗t ∂x∗jβ+1 ∂x∗js l1 ...ls · · · ∗ir · · · · · · T ∂x ∂xl1 ∂xlβ−1 ∂xlβ ∂xlβ+1 ∂xls m1 ...mr



jβ tk

∗

.

Therefore, A22



s ∂xm1 ∂xmr ∂x∗j1 ∂x∗js ∂xα   = · · · · · · T l1 ...lβ−1 plβ+1 ...ls ∂x∗i1 ∂x∗ir ∂xl1 ∂xls ∂x∗k β=1 m1 ...mr

Finally, we obtain from (5.5), (5.6) and (5.7) 1 ...js Ti∗j = 1 ...ir ;k



 

lβ  − A3 . (5.7) pα

∂xm1 ∂xmr ∂xα ∂x∗j1 ∂x∗js · · · · · · ∂x∗i1  ∂x∗ir ∂x∗k ∂xl1 ∂xls   r l1 ...ls  ∂Tm m l1 ...ls 1 ...mr − T + m1 ...ml−1 mml+1 ...mr m ∂xα α l l=1 +

s 

l1 ...lβ−1 plβ+1 ...ls Tm 1 ...mr

β=1

=



 

lβ  pα

∂xm1 ∂xmr ∂xα ∂x∗j1 ∂x∗js l1 ...ls · · · · · · T , ∂x∗i1 ∂x∗ir ∂x∗k ∂xl1 ∂xls m1 ...mr ;k

and the components of the covariant derivative of an (r, s)–tensor satisfy the transformation formulae of an (r + 1, s)–tensor. Example 5.2.3. The covariant derivatives of (2, 0)–, (1, 1)– and (0, 2)–tensors T are given by 

Tij;k =

∂Tij − Tmj ∂xk

j = Ti;k

∂Tij j − Tm ∂xk



m ik

T;kij =

∂T ij + T mj ∂xk



m ik



− Tim



+ Tim

m ik









m , jk

j mk



(5.8) (5.9)

and + T im





j . mk

(5.10)

Tensor Analysis ■ 403 ...js ...js Theorem 5.2.4. (a) Let T and U be tensors with the components Tij11...i and Uij11...i . r r Then the covariant derivative of the sum and difference of T and U taken termwise, that is,   ...js ...js ...js ...js Tij11...i + Uij11...i = Tij11...i + Uij11...i . r r r ;k r ;k ;k

...js (b) Let T and U be tensors with the components Tij11...i and Um11 ...mpq . r Then the product rule holds for the covariant differentiation of the outer product of T and U , that is, n ...n



...js n1 ...np Tij11...i ∗ Um 1 ...mq r

Proof. (a) Let



;k

...js ...js n1 ...np n1 ...np ;k = Tij11...i ∗ Um + Tij11...i ∗ Um . 1 ...mq 1 ...mq r r ;k ...js ...js ...js Vij11...i = Tij11...i + Uij11...i . r r r

Then it follows by Definition 5.2.1 that ...js Vij11...i r ;k

...js ∂Vij11...i r = ∂xk

− +

=

...js ∂Tij11...i r ∂xk

− +

+

...js ∂Uij11...i r ∂xk

− +

l=1 s 

...js Vij11...i l−1 mil+1 ...ir j ...j

n=1 r  l=1 s 

pjn+1 ...js

...js Tij11...i l−1 mil+1 ...ir

n=1 r  l=1 s 

Vi11...irn−1

j ...j

Ti11...irn−1

 

m il k



pjn+1 ...js



...js Uij11...i l−1 mil+1 ...,ir



+



+

jn pk

m il k



j ...j pj ...j Ui11...irn−1 n+1 s n=1 ...js +Uij11...i . r; k

...js = Tij11...i r; k

(b) Let

r 

jn pk

m il k









jn pk



...js n1 ...nq Vi11...irsm11...mpq = Tij11...i · Um . 1 ...mq r j ...j n ...n

Then we have

j ,...,j n ...n

j ...j n ...n Vi11...irsm11...mpq ;k

= − − + +

1 p ∂Vi11...ir ms1 ...m q

r 

∂xk j ...j n ...n

s 1 p Vi11...il−1 mil+1 ...ir m1 ...mq

l=1

q 

j ...j n ...n Vi11...irsm11...mpl−1 mml+1 ...mq

l=1

s  l=1 p  l=1



j ...j

tj

l+1 Vi11...irl−1 m1 ...mq

j ...j n ...n

...js n1 ...np

Vi11...irsm11...ml−1 q



tnl+1 ...np

m il k 

jl tk





m ml k



nl tk





+

404 ■ Differential Geometry and Its Visualization

=



  ...js r  ∂Tij11...i m j1 ,...,js r − Ti1 ...il−1 mil+1 ...ir il k ∂xk l=1

+

j ...j tj ...j Ti11...irl−1 l+1 s

p 

n1 ...nl−1 tnl+1 ...np Um 1 ...mq

l=1





s 

jl tk



n1 ...np · Um 1 ...mq

  q n ...n  ∂Um11 ...mpq m n1 ...np + − U m1 ...ml−1 mml+1 ...mq ml k ∂xk l=1

+ =

l=1 ...js Tij11...i r ;k



nl tk



...js · Tij11...i r

...js n1 ...np · Um + Tij11...i · Um11 ...mpq ;k . 1 ...mq r n ...n

Example 5.2.5. We have by Theorem 5.2.4 (b) (Tij U mn );k = Tij;k U mn + Tij U;kmn . We put m = j and obtain the contraction 

Tij U jn



;k

= Tij;k U jn + Tij U;kjn .

Theorem 5.2.6 (Ricci). The covariant derivatives of the tensors with the components gik , g ik and δik are the zero tensor. Proof. We have by (5.8), (4.24) and (4.26) gij;k









∂gij ∂gij m m = − gmj − gim = − [ikj] − [jki] ik jk ∂xk ∂xk = [ikj] + [jki] − [ikj] − [jki] = 0.

Furthermore we obtain by (5.10) ij g;k

=

Finally gij g jm = δim implies

∂g ij + g mj ∂xk



i mk



+ g im





j . mk

∂g jm ∂gij jm g + g =0 ij ∂xk ∂xk and

∂gij jm li ∂g lm ∂g jm ∂gij jm li li g g + g g = g g + = 0. ij ∂xk ∂xk ∂xk ∂xk This yields by (4.25) and (4.24) ∂gij ∂g lm = − k g jm g li = −[ikj]g jm g li − [jki]g jm g li k ∂x ∂x     m li l =− g − g jm . ik jk

(5.11)

Tensor Analysis ■ 405

If we replace l and m by i and j and rename the indices of summation i and j by m then we conclude 



j ∂g ij =− g im − mk ∂xk





i g mj . mk

ij Now g;k = 0 is an immediate consequence of (5.11). By Example 5.2.5 and what we have just shown, we conclude



j = gim g mj δi;k



mj = gim;k g mj + gim g;k = 0.

;k

Example 5.2.7. The second order covariant derivative of a scalar F is given by F;ik = (F;i );k =



∂F ∂xi



;k

=

∂2F − ∂xi ∂xk



j ik



∂F . ∂xj

Thus Gauss’s formulae of differentiation may be written as xl;ik

∂ 2 xl = − ∂xi ∂xk = Lik N l .



j ik



∂xl = ∂xj



j ik





∂xl + Lik N l − ∂xj

j ik



∂xl ∂xj

Furthermore, the Mainardi–Codazzi identities 



r Lrj − ik



may be written as

5.3

r ij



Lrk +

∂Lik ∂Lij − =0 j ∂u ∂uk

Lik;j = Lij;k .

THE INTERCHANGE OF ORDER FOR COVARIANT DIFFERENTIATION AND RICCI’S IDENTITY

In general, the second order covariant derivative depends on the order of differentiation. A trivial exception is the case where the covariant derivative reduces to the normal partial derivative, because all Christoffel symbols vanish identically, as is the case with respect to Cartesian coordinates. Definition 5.3.1. In a Riemann space with the metric tensor gik , the mixed Riemann tensor of curvature is defined by j Rilk =

∂ ∂xk

 

j il

−

∂ ∂xl



j ik



+



Now we prove the important Ricci identity.

m il



j mk



−



m ik





j . ml

406 ■ Differential Geometry and Its Visualization

Theorem 5.3.2 (Ricci identity). ...is Let T be an (r, s)–tensor with components Tji11...j . Then r r 

...is ...is Tji11...j − Tji11...j = r ;lk r ;kl

ρ=1 s 

−

σ=1

...is Tji11...j Rjmρ kl ρ−1 mjρ+1 ...jr i ...i

Tj11...jrσ−1

miσ+1 ···is

(5.12)

iσ Rmkl .

The identity in (5.12) is called Ricci’s identity. Proof. (i) First we prove the theorem for a covariant vector T with the components Ti . We have   ∂Ti j Ti;l = − Tj , l il ∂x and









m m ∂ (Ti;l ) − Tm;l − Ti;m ik lk ∂xk         2 j ∂Tm n m ∂ Ti ∂ = − k Tj − − Tn k l l il ml ik ∂x ∂x ∂x ∂x      ∂Ti n m − − Tn m im lk ∂x

Ti;lk =

and similarly Ti;kl





j ik



∂ − k ∂x

 



∂Ti − m ∂x



∂ 2 Ti ∂ = − l Tj l k ∂x ∂x ∂x

j ik







 



∂Tm n m − − Tn k mk il ∂x      ∂Ti n m − − Tn . im kl ∂xm

Thus Ti;lk − Ti;kl = Tj



∂ ∂xl



∂Tm − ∂xl + = Tj



∂Tj ∂xl

∂ ∂xl







m ik

j ik



j ik



−

∂Tj ∂xk

j il

m lk









j m + ml ik     j m − mk il

∂Tm + ∂xk



m il



∂Ti + m ∂x

 

∂ − k ∂x

j il

 

j il







j m + ml ik     j m − mk il



m kl



Tensor Analysis ■ 407

∂Tj − ∂xl = Tj



∂ ∂xl





j ik

j ik



 

∂Tj + k ∂x



−

j il

 

∂ ∂xk



∂Tj + ∂xl

j il

j ik





∂Tj − k ∂x



 

j il



j m ml ik     j m − mk il

+

j = Tj Rikl .

(5.13)

(ii) Now we show the assertion for a (2, 0)–tensor T with the components Tri . We have Tri;l

− Trm



∂ (Tri;l ) − Tmi;l ∂xk



− Trm;l



∂ = (Tri;k ) − Tmi;k ∂xl



− Trm;k



∂Tri = − Tmi ∂xl

Tri;lk = Tri;kl



m rl



m rk m rl

 



m , il m il



m ik

− Tri;m







m , lk



m , kl





− Tri;m



hence Tri;lk − Tri;kl



m rl



∂ − Tmi k ∂x

∂Tmi ∂xl



m rk



+ Tmi

∂Tmi − ∂xl



m rk



∂Tmi ∂xk



m rl



∂Trm − ∂xl



m ik



∂Trm + ∂xk



m il



∂Tmi =− ∂xk

+

+

= Tmi



∂ ∂xl



m lr

∂ ∂xl



m rk

+ Tsi



s ml



m rk



− Tsi



s mk



m rl



+ Tsm



s rl



m ik



− Tsm



s rk



m rk



−



∂ ∂xk



∂Trm − ∂xk



+

m il





m il

∂ − Trm k ∂x







m il

∂Trm m ik ∂xl   ∂ m + Trm l ik ∂x

+ Tms

 

m rk

− Tms



s il



m rl



s ml



m ik



+ Trs − Trs







s ik



m rl



s mk



m il



408 ■ Differential Geometry and Its Visualization













s m s m + Tsi − ml rk mk rl      ∂ ∂ m m + Trm − k ik il ∂xl ∂x        s m s m + Trs − + A, ml ik mk il where   





















s m s m s m s m A = Tms − + Tsm − il rk ik rl rl ik rk il                 s m s m m s m s = Tms − + − = 0. il rk ik rl rl ik rk il

If we interchange the indices s and m in the terms of Tsi and Trs , then we obtain Tri;lk − Tri;kl =Tmi



∂ ∂xl

+ Trm





∂ ∂xl

m rk





m ik

∂ − k ∂x





∂ − k ∂x

m rl





m il







m s + sl rk     m s − sk rl









m s + sl ik     m s − sk il

m m = Tmi Rrkl − Trm Rikl .

(iii) Now we show the general formulae for (r, 0)–tensors Tm1 ...mr ;lk − Tm1 ...mr ;kl =

r 

(5.14)

s Tm1 ...mρ−1 smρ+1 ...mr Rm . ρ lk

ρ=1

We have by (5.4) Ti1 ...ir ;l

Ti1 ...ir ;lk

r ∂Ti1 ...ir  = − Ti1 ...iρ−1 miρ+1 ...ir ∂xl ρ=1

r  ∂ = (Ti1 ...ir ;l ) − Ti1 ...iρ−1 miρ+1 ...ir ;k ∂xk ρ=1





m iρ l





r ∂2 ∂  = (T ) − Ti ...i mi ...i i1 ...ir ∂xk ∂xl ∂xk ρ=1 1 ρ−1 ρ+1 r

−

r  ∂ 

ρ=1

∂xl

Ti1 ...iρ−1 miρ+1 ...ir





m iρ k



m iρ l



,

− Ti1 ...ir ;m 



m  iρ l



m lk



Tensor Analysis ■ 409

+

r ρ−1  

ρ=2 τ =1

+ +

Ti1 ...iτ −1 uiτ +1 ...iρ−1 miρ+1 ...ir

r r−1  

ρ=1 τ =ρ+1 r 



Ti1 ...iρ−1 miρ+1 ...iτ −1 uiτ +1 ...ir

Ti1 ...iρ−1 uiρ+1 ...ir

ρ=1





m iρ k



∂ m − m (Ti1 ...ir ;m ) lk ∂x

−



r 

u ml



m iρ k 



m iρ k

u iτ l



ρ=1





r   ∂2 ∂  = (Ti1 ...ir ) − Ti1 ...iρ−1 miρ+1 ...ir k l k ∂x ∂x ∂x ρ=1 r 

∂ − Ti1 ...iρ−1 miρ+1 ...ir k ∂x ρ=1 − + +

r  ∂ 

ρ=1

r ρ−1  

ρ=2 τ =1 r 

ρ=1

+

∂xl

r r−1  



m iρ k





−

We put 

m ∂ (Ti1 ...ir ) Akl = m lk ∂x



r 

u ml







Ti1 ...iρ−1 miρ+1 ...ir

and obtain





m ∂ Akl − Alk = (Ti ...i ) ∂xm 1 r lk 

∂ m − m (Ti1 ...ir ) kl ∂x

m iρ k





m iρ k

− 





− −





+

r 

ρ=1 r 

ρ=1



u iρ m 



m iρ l

m lk





r  ∂ 

ρ=1

∂xl



u iτ l

u iρ m

Ti1 ...iρ−1 uiρ+1 ...ir

ρ=1

m iρ l

u iτ l



Ti1 ...iρ−1 uiρ+1 ...ir

r 

m lk



ρ=1

and ∂xk

m iρ k

Ti1 ...iρ−1 miρ+1 ...iτ −1 uiτ +1 ...ir

m ∂ − m (Ti1 ...ir ) lk ∂x

ρ=1







Ti1 ...iτ −1 uiτ +1 ...iρ−1 miρ+1 ...ir

ρ=1 τ =ρ+1

Bkl =

m iρ l

Ti1 ...iρ−1 miρ+1 ...ir

Ti1 ...iρ−1 uiρ+1 ···ir

r  ∂ 





u iτ l

− Ti1 ...ir ;m

Ti1 ...iρ−1 uiρ+1 ...ir







− Ti1 ...ir ;m



u iρ m



Ti1 ...iρ−1 uiρ+1 ...ir Ti1 ...iρ−1 uiρ+1 ...ir





u iρ m u iρ m

m lk



m . lk

m lk

Ti1 ...iρ−1 miρ+1 ...ir













m lk m kl



m iρ k







=0

,

410 ■ Differential Geometry and Its Visualization

and Bkl − Blk =

r  ∂ 

ρ=1

∂xk

Ti1 ...iρ−1 miρ+1 ...ir +

r  ∂ 

ρ=1

−

r  ∂ 

ρ=1

∂xl

It follows that

r  ∂ 

ρ=1



m iρ l



Ti1 ...iρ−1 miρ+1 ...ir

Ti1 ...iρ−1 miρ+1 ...ir −

Ti1 ...ir ;kl − Ti1 ...ir ;kl

∂xl



∂xk





m iρ k



Ti1 ...iρ−1 miρ+1 ...ir

+ +

ρ=2 τ =1 r  ρ=1

+

Ti1 ...iτ −1 uiτ +1 ...iρ−1 miρ+1 ...ir

Ti1 ...iρ−1 uiρ+1 ···ir

r r−1  

ρ=1 τ =ρ+1 r 



m iρ k



u ml

− − − = +

ρ=2 τ =1 r  ρ=1

+

Ti1 ...iρ−1 uiρ+1 ···ir

ρ=1 τ =ρ+1 r 



Ti1 ...iρ−1 uiρ+1 ···ir

r ρ−1  

ρ=2 τ =1

m iρ l



u mk



∂ ∂xl



m iρ k



m iρ k





u ml

m iρ l











−





m iρ l



u iτ l

u iτ k



m iρ l

= 0.

u iτ l



∂ − k ∂x 





m iρ k

m iρ l





m iρ l

m iρ k



m iρ k



m iρ k

Ti1 ...iρ−1 miρ+1 ...iτ −1 uiτ +1 ...ir

Ti1 ...iρ−1 miρ+1 ...ir

ρ=1



Ti1 ...iτ −1 uiτ +1 ...iρ−1 miρ+1 ...ir

r r−1  

ρ=1 r 



Ti1 ...iρ−1 miρ+1 ...iτ −1 uiτ +1 ...ir

∂ + Blk + Ti1 ...iρ−1 miρ+1 ...ir l ∂x ρ=1 r ρ−1  









r 

∂ = −Bkl − Ti1 ...iρ−1 miρ+1 ...ir k ∂x ρ=1 r ρ−1  





− Akl



+ Alk



u iτ k

m iρ l





u mk



Ti1 ...iτ −1 uiτ +1 ...iρ−1 miρ+1 ...ir · 

m iρ k



u iτ l



−



m iρ l



u iτ k



Tensor Analysis ■ 411

+

r r−1  

ρ=1 τ =ρ+1

Ti1 ...iρ−1 miρ+1 ...iτ −1 uiτ +1 ...ir · 

m iρ k





−





m iρ k



∂ − k ∂x



m iρ l

u iτ l

m iρ l



u iτ k



.

We put C=

r 

ρ=1 r 

+

ρ=1

Ti1 ...iρ−1 miρ+1 ...ir Ti1 ...iρ−1 uiρ+1 ···ir



∂ ∂xl



m iρ k



u ml



−



m iρ l





u mk



and obtain interchanging the indices of summation u and m in the second sum C=

r 

Ti1 ...iρ−1 miρ+1 ...ir

ρ=1

=

r 



∂ ∂xl 



u iρ k

m iρ k





∂ − k ∂x

m ul



−





m iρ l

u iρ l





+ m uk



Ti1 ...iρ−1 miρ+1 ...ir Rimρ kl

ρ=1

by Definition 5.3.1. We also write D as follows and change the order of summation in the second sum D=

r ρ−1  

ρ=2 τ =1

+

Ti1 ...iρ−1 uiρ+1 ...iτ −1 miτ +1 ...ir · 

r r−1  

ρ=1 τ =ρ+1

=

r ρ−1  

ρ=2 τ =1

+

−1 r τ 

τ =2 ρ=1

m iρ k



u iτ l



−



m iρ l



u iτ k





m iρ l



u iτ k





m iρ l



u iτ k





m iρ l



u iτ k



Ti1 ...iτ −1 miτ +1 ...iρ−1 uiρ+1 ...ir · 

m iρ k



u iτ l



−

Ti1 ...iρ−1 uiρ+1 ...iτ −1 miτ +1 ...ir · 

m iρ k



u iτ l



−

Ti1 ...iτ −1 miτ +1 ...iρ−1 uiρ+1 ...ir · 

m iρ k



u iτ l



−

412 ■ Differential Geometry and Its Visualization

and change the indices of summation from τ to ρ and u to m in the second sum to obtain D=

r ρ−1  

ρ=2 τ =1

−

r ρ−1  

ρ=2 τ =1

Ti1 ...iρ−1 uiρ+1 ...iτ −1 miτ +1 ...ir · 

m iρ k



u iτ l



−



m iρ l



u iτ k





u iτ l



m iρ k



Ti1 ...iρ−1 uiρ+1 ...iτ −1 miτ +1 ...ir · 

u iτ k



m iρ l



−

= 0.

Finally we have Ti1 ...ir ;kl − Ti1 ...ir ;kl = C + D =

r 

Ti1 ...iρ−1 miρ+1 ...ir Rimρ kl .

ρ=1

(iv) The general formulae in (5.12) could be established in a similar way. Here we prove the formulae for the covariant derivative of a contravariant vector and a (0, 2)–tensor by applying suitable contractions. We have g;ijk = 0 by the Ricci Theorem, Theorem 5.2.6. Putting Tl = glm T m , i we conclude T i = δm T m = g il glm T m = g il Tl and 

T;ijk = g il Tl =



g;ilk Tl; j

; jk

=



g il Tl

  ;j

;k

+ g Tl; jk = g Tl; jk . il

il



= g;ilj Tl + g il Tl; j



;k

This implies together with (5.13) m Tm . T;ijk − T;ikj = g il (Tl; jk − Tl; kj ) = g il Rlkj m m . Then it follows that Rlkj = g mr Rrlkj and We put Rilkj = gim Rlkj

T;ijk − T;ikj = g il g mr Rrlkj Tm = g il g mr Tm Rrlkj = g il T r Rrlkj i = −g il T r Rlrkj = −T r g il Rlrkj = −T r Rrkj .

Similarly we obtain for a (0, 2)–tensor i l − T ir Rrkj T;iljk − T;ilkj = −T rl Rrkj

and now Ricci’s identity (5.12) for arbitrary tensors follows.

Theorem 5.3.3. The covariant derivative of a covariant vector in V3 is a symmetric tensor if and only if the vector itself is the gradient of a scalar of class r ≥ 2.

Tensor Analysis ■ 413

Proof. Let T be a covariant vector with the components Ti . Then for the   weobtain  j j covariant derivative Ti;k by Definition 5.1.3 and the fact that ik = ki for all i, j and k Ti; k − Tk; i =

∂Ti − Tj ∂xk



j ik



−



∂Tk − Tj ∂xi



j ki



=

∂Ti ∂Tk − k ∂x ∂xi

(5.15)

for all i and k. The last two terms in (5.15) are the components of the curl in V3 of a vector with the components Ti (i = 1, 2, 3). The curl of a vector in V3 vanishes if and only if the vector is the gradient of a scalar Φ, that is, Ti = Φi for i = 1, 2, 3. Thus the identities in (5.15) vanish if and only if Φ; ik − Φ; ki =

5.4

∂ ∂xk



∂Φ ∂xi



−

∂ ∂xi



∂Φ ∂xk



= 0 for all i and k.

BIANCHI’S IDENTITIES FOR THE COVARIANT DERIVATIVE OF THE TENSORS OF CURVATURE

In this section, we establish the Bianchi identities for the derivatives of the mixed Riemann tensor of curvature and the covariant Riemann tensor of curvature. Theorem 5.4.1 (Bianchi identities). The Riemann tensors of curvature satisfy the following identities for all i, j, k, l and m i i i Rjkl; (5.16) m + Rjmk; l + Rjlm; k = 0 and

Rijkl; m + Rijmk; l + Rijlm; k = 0.

(5.17)

The identities in (5.16) and (5.17) are called the Bianchi identities. Proof. (i) First we show: In any n–dimensional Riemann space R there exists a coordinate system such that the Christoffel symbols vanish in a given point. Let S be a coordinate system in R, P0 = (x1 , x2 , . . . , xn ) ∈ R be a point, and



i jk 0



0

0

0

be the Christoffel symbols at P0 with respect to S. We define a new

coordinate system S ∗ by 



1 i x =x +x − x∗j x∗k 0 2 jk i

∗i

i

0

(i = 1, 2, . . . , n).

Then P0 is the origin of S ∗ , and differentiation yields 







1 i 1 i ∂xi = δli − δlj x∗k − x∗j δlk ∗l ∂x 2 jk 2 jk 0

0

(5.18)

414 ■ Differential Geometry and Its Visualization

=









1 i 1 i − x∗k − x∗j = δli − 2 lk 2 jl

δli

0

0





i x∗k lk 0

for all i and l. Thus we have 

∂xi   = δli for all i and l, ∂x∗l P 0

and the transformation in (5.18) is admissible in a neighbourhood of P0 . Furthermore, we obtain 





∂ 2 xi i i k =− δm =− ∗l ∗m lk lm ∂x ∂x 0

0



(5.19)

for all i, l and m. The transformation formulae in (4.28) and (5.19) yield at the point P0 

i jk 0

∗

=



l jk



0

∗

∂xi = ∂x∗l



∂xp ∂xq ∂x∗j ∂x∗k 



m pq



∂ 2 xm + ∗j ∗k ∂x ∂x





∂x∗l ∂xi   ∂xm ∂x∗l P

0

        m  i i i  p q m = δj δk − − =0  δm =

pq

jk

0

jk

0

jk

0

0

for all i, j and k.

(ii) Now we show the identities in (5.16). If we introduce a coordinate system S such that the Christoffel symbols vanish in some fixed point P0 (as in Part (i) of the proof), then we have at P0 by the i definition of the terms Rjkl in Definition 5.3.1 i (P0 ) Rjkl

=



∂ ∂xl





∂ ∂xm

i jk



∂ − k ∂x

and i Rjkl; m (P0 )

= =





∂ ∂xl

∂2 ∂xl ∂xm





i jk

i jk





−



i jl

   

∂ − k ∂x



∂2 ∂xk ∂xm

P0

    P0   i   .  jl

i jl

P0

Replacing k, l and m by m, k and l, respectively, we obtain i Rjmk; l (P0 ) =



∂2 ∂xk ∂xl



i jm



∂2 − m l ∂x ∂x



i jk

   

, P0

Tensor Analysis ■ 415

and replacing m, k and l by l, m and k, respectively, we obtain i Rjlm; k (P0 )

=



∂2 ∂xm ∂xk



i jl



∂2 − l k ∂x ∂x



i jm

Therefore, we obtain for all i, k, j, l and m i i i Rjkl; m (P0 ) + Rjmk; l (P0 ) + Rjlm; k (P0 )



   

. P0

  =   P0        2 2 ∂ ∂ i i  + − m l  k l  jm jk ∂x ∂x ∂x ∂x P   0      2 2 ∂ i i ∂  + − l k  = 0. m k  jl jm ∂x ∂x ∂x ∂x

∂2 ∂xl ∂xm



i jk



∂2 − k m ∂x ∂x



i jl

P0

Thus we have proved the identities in (5.16) at P0 for our special choice of the coordinate system. Since the left hand side in (5.16) is the sum of the components of a tensor and therefore a component of a tensor itself and vanishes at P0 with respect to the coordinate system S, it must also vanish with respect to any coordinate system. Finally, since P0 was an arbitrary point, the identities in (5.16) must hold in any point. (iii) Finally, we show the identities in (5.17). We have by the Ricci identity, Theorem 5.2.6, 

i ir Rjkl; m = g Rrjkl



;m

= g ir Rrjkl; m ,

and the identities in (5.17) follow from (5.16).

5.5

BELTRAMI’S DIFFERENTIATORS

In this section, we define the Beltrami differentiator of first order in a Riemann space, and study a few special cases of interest. Definition 5.5.1. Let Φ and Ψ be two given real–valued functions of class r ≥ 2 in a Riemann space. We define the Beltrami differentiator of first order ▽ by ▽(Φ, Ψ) = g ik Φ; i Ψ; k , and we write ▽Φ = ▽(Φ, Φ), for short.

Remark 5.5.2. (a) Since Φ and Ψ are scalars, we have Φ; i =

∂Φ ∂Ψ for all i and Ψ; k = for all k, i ∂x ∂xk

416 ■ Differential Geometry and Its Visualization

hence ▽(Φ, Ψ) = g ik

∂Φ ∂Ψ ∂Φ ∂Φ and ▽ Φ = g ik i k . i k ∂x ∂x ∂x ∂x

Thus ▽Φ is the square of the length of the gradient of Φ. We have for Cartesian coordinates g 11 = g 22 = g = 1 and g 12 = 0, hence ▽Φ =

 n   ∂Φ 2 i=1

∂xi

.

(b) Curves given by Φ(xi ) = const are called level lines of the function Φ. If the values ξ i are the components of the tangent vectors to the level lines, then ∂Φ i ξ = 0. ∂xi This means that the gradient is orthogonal to the level lines; it is in the direction of maximum change of the function. (c) If the collections of curves given by the equations Φ(xi ) = const and Ψ(xi ) = const are members of a net, then the angle γ between the gradients of the functions is equal to the angle between the curves, since the gradient is always orthogonal to the corresponding curve, hence ▽(Φ, Ψ) cos γ =  . (▽Φ • ▽Ψ)

The collections of curves given by Φ(xi ) = const and Ψ(xi ) = const make up an orthogonal net if and only if ▽(Φ, Ψ) ≡ 0. Next we define the divergence of a contravariant vector.

Definition 5.5.3. The divergence of a contravariant vector T with the components T k is defined by div T k = T;kk for all k. Remark 5.5.4. Being the contraction of a (1, 1)–tensor the divergence of a contravariant vector is a scalar. We have by Definition 5.1.2 T;kk Theorem 5.5.5. We have

where g = det(gij ).

∂T k = + ∂xk





k T l for all k. kl

1 ∂ √ k  div T k = √ gT , g ∂xk

(5.20)

Tensor Analysis ■ 417

Proof. (i) First we show

We conclude from g = 





i ik

(k1 ,...,kn )



√ ∂(log g) = . ∂xk

(*)

sign(k1 , . . . , kn )g1k1 · · · gnkn that 

∂g ∂   = sign(k1 , . . . , kn )g1k1 · · · gnkn  k ∂x ∂xk (k ,...,k ) =

n 

1



j=1 (k1 ,...,kn )

=

n  n  ∂gjl j=1 l=1

∂xk

n

sign(k1 , . . . , kn )g1k1 · · · gj−1,kj−1

 ∂  gjkj gj+1,kj+1 · · · gnkn k ∂x

Gjl ,

where the terms Gjl denote the adjoints of the determinant g. We conclude from g jl = Gjl /g for all j and l that by (4.25) and (4.24) ∂g ∂gjl = g · g jl k = g · g jl ([jkl] + [lkj]) k ∂x ∂x 







= g g jl [jkl] + g jl [jkl] = g g il [ikl] + g il [ikl] = 2g Finally, we obtain

which is (*).





i . ik

√   ∂(log g) 1 ∂g i = = , ik ∂xk 2g ∂xk

(ii) Now we prove the identity in (5.20). We have   ∂T k ∂T k 1 ∂g j k j div T k = T;kk = + T = + T k k kj ∂x ∂x 2g ∂xj and

√ 1 k ∂( g) ∂T k 1 ∂g 1 ∂ √ k  ∂T k T gT = + = + Tk k. √ √ k k k k g ∂x ∂x g ∂x ∂x 2g ∂x

This shows (5.20).

Now we define the divergence of a covariant vector and the Laplacian of a real– valued function. Definition 5.5.6. (a) The divergence of a covariant vector T with the components Tk is defined by div Tk = g ki Tk; i .

418 ■ Differential Geometry and Its Visualization

(b) Let F be a real–valued function. Then the partial derivatives ∂F ∂xi are the components of a covariant vector. Then the Laplacian operator △ is defined by △F = div Fi . Fi =

Remark 5.5.7. (a) We have for conjugate vectors, that is, for vectors that satisfy Tk = gkj T j , 

div T k = T;kk = g kj Tj (b) We obtain from Theorem 5.5.5



= g kj Tj; k = g jk Tj; k = div Tk .

;k

1 ∂ √ i  1 ∂ √ ik  △F = div Fi = div F i = √ gF = gg Fk . √ g ∂xi g ∂xi

Example 5.5.8. Let S denote the Cartesian coordinate system of IR3 and S ∗ be the system of cylindrical coordinates defined by x1 = x∗1 cos x∗2 , x2 = x∗1 sin x∗2 and x3 = x∗3 . Since gik = δik , we have with respect to the system S △F =

3  ∂2F

k=1

(∂xk )2

.

We obtain for △∗ with respect to the system S ∗ from the transformation formulae ∗ gik =

Since 

i

∂x ∂x∗j



3  ∂xm ∂xn ∂xm ∂xm g = . mn ∂x∗i ∂x∗k ∂x∗i ∂x∗k m=1





cos x∗2 sin x∗2 0  ∗1  ∗2 ∗1 x cos x∗2 0 , = −x sin x 0 0 1

∗ ∗ ∗ ∗ = g33 = 1, g22 = (x∗1 )2 and gik = 0 otherwise g11

and



1



0 0  2     1   g ∗ik = 0 , 0 ∗1   x 0 0 1

it follows that     √ ∗ ∗ik ∂F √ ∗ ∗ii ∂F 1 ∂ 1 ∂ ∗ g g = √ ∗ ∗i g g △ F = √ ∗ ∗i g ∂x ∂x∗k g ∂x ∂x∗i        ∂ ∂ ∂ 1 ∂F 1 ∗1 ∂F ∗1 ∗1 ∂F = ∗1 x + x + x x ∂x∗1 ∂x∗1 ∂x∗2 (x∗1 )2 ∂x∗2 ∂x∗3 ∂x∗3 1 ∂F ∂2F 1 ∂2F ∂2F = ∗1 ∗1 + + + . x ∂x (∂x∗1 )2 (x∗1 )2 (∂x∗2 )2 (∂x∗3 )2

Tensor Analysis ■ 419

5.6

A GEOMETRIC MEANING OF THE COVARIANT DIFFERENTIATION, ` PARALLELISM THE LEVI–CIVITA

In this section, we give a geometric interpretation of the covariant differentiation. This will lead to the concept of Levi–Civit`a parallelism. We recall the concept of a surface vector (Definition 3.7.1). Definition 5.6.1. Let S be a surface in R3 with a parametric representation ⃗x(ui )  (u1 , u2 ) ∈ D , P be a point on S, and T (P ) be the tangent plane of S at P . Then a unit vector ⃗z in the tangent plane is called surface vector of S at P . We shall always  write ⃗z = ξ 1 , ξ 2 , where the values ξ i (i = 1, 2) are the components of the vector ⃗z along the vectors ⃗xi (i = 1, 2).



If we move the surface vectors of a plane in a parallel way along a straight line in the plane, that is, along a geodesic line in the plane, then the angles between the vectors and the tangents of the geodesic line are constant. Similarly, we require for the parallel movement of surface vectors of arbitrary surfaces along geodesic lines that the angles between the surface vectors and the geodesic line are constant. This leads to the following result for the components ξ i (i = 1, 2) of the surface vectors under parallel movement along a geodesic line. Theorem 5.6.2. The components ξ α (α = 1, 2) of a surface vector ⃗z = {ξ 1 , ξ2} under parallel movement along a geodesic line given by (u1 (s), u2 (s)) satisfy the differential equations   dξ α δξ α α β = +ξ u˙ γ = 0 (α = 1, 2). (5.21) βγ δs ds Proof. Let the surface vectors at the points of the geodesic line be given by ⃗z(s) = ξ α (s)⃗xα (uβ (s)), ⃗x(s) be a parametric representation of the geodesic line, s be the arc length along the geodesic line and ⃗x˙ (s) = u˙ β (s)⃗xβ (uγ (s)) ̸= ⃗z(s). We have gαβ ξ α ξ β = 1 for all s. The surface vectors are parallel, if their angles with the vectors ⃗x˙ (s) are constant. This means that  d  gαβ ξ α u˙ β = 0 ds

or

∂gαβ α β γ ξ u˙ u˙ + gαβ ξ˙α u˙ β + gαβ ξ α u¨β = 0. ∂uγ We rename β by τ in the last term above and apply (4.25) to obtain ([αγβ] + [βγα]) ξ α u˙ β u˙ γ + gαβ ξ˙α u˙ β + gατ ξ α u¨τ =





([αγβ] + [βγα]) ξ α u˙ γ + gαβ ξ˙α u˙ β + gατ ξ α u¨τ = 0.

420 ■ Differential Geometry and Its Visualization

Using the identities in (4.24), we obtain [βγα] = gατ



τ βγ



and ξ α [αγβ] = ξ δ [δγβ] = ξ δ gαβ



α δγ



,

and conclude 



0 = ([αγβ] + [βγα]) ξ α u˙ γ + gαβ ξ˙α u˙ β + gατ ξ α u¨τ 

= ξ δ gαβ 



α δγ



u˙ γ + ξ α gατ 







τ βγ





α u˙ γ u˙ β + gατ ξ α u¨τ + δγ     α = gαβ ξ˙α + ξ δ u˙ γ u˙ β , δγ = gαβ ξ˙α + ξ δ



u˙ γ + gαβ ξ˙α u˙ β + gατ ξ α u¨τ 

τ βγ



u˙ β u˙ γ



since (u1 (s), u2 (s)) is a geodesic line. Therefore, we have gαβ



ξ˙α + ξ δ



α δγ



u˙

γ



u˙ β = 0.

(5.22)

Now we show that the vector with the components δξ α = ξ˙α + ξ δ δs



α δγ



u˙ γ (α = 1, 2)

is the zero vector. It follows from gαβ ξ α ξ β = 1 that 

gαβ ξ α ξ β



;σ

= gαβ; σ ξ α ξ β + gαβ ξ;ασ ξ β + gαβ ξ α ξ;βσ = 2gαβ ξ;ασ ξ β = 0,

since gαβ; σ = 0 for all α and β by Theorem 5.2.6. Furthermore, we have δξ α ∂ξ α σ = u˙ + ξ δ δs ∂uσ





α u˙ σ = δσ



∂ξ α + ξδ ∂uσ

hence gαβ ξ;ασ ξ β u˙ σ = gαβ



α δσ



u˙ σ = ξ;ασ u˙ σ

δξ α β ξ = 0. δs

(α = 1, 2),

(5.23)

This means that the vector with the components δξδs is orthogonal both to the geodesic line by (5.22) and to the surface vector ⃗xα ξ α by (5.23). Since this is an arbitrary vector, we must have α

δξ α dξ α = ξ;ασ u˙ σ = + ξδ δs ds



α δγ



u˙ γ = 0 (α = 1, 2).

(5.24)

Now we extend the concept of parallel movement along geodesic lines to parallel movement along arbitrary curves.

Tensor Analysis ■ 421

Definition 5.6.3. The parallel movement of a surface vector with the components ξ α along an arbitrary curve given by (u1 (s), u2 (s)) is defined by dξ α δξ α = ξ;ασ u˙ σ = + ξδ δs ds



α δγ



u˙ γ = 0 (α = 1, 2);

(5.25)

this is called parallel movement by Levi–Civit`a. Remark 5.6.4. (a) Since δξδs = ξ;ασ u˙ σ , we may write (5.25) as ξ;ασ u˙ σ = 0 for α = 1, 2. Thus the contravariant surface vectors ⃗z(s) with the components ξ α (s) are parallel in the sense of Levi–Civit`a with respect to a curve on a surface given by (u1 (s), u2 (s)) if and only if the components of the contravariant vector vanish, which is the contraction of the covariant derivative of ⃗z(s) and the contravariant vector with the components u˙ α (s). (b) If t is an arbitrary parameter of the curve then (5.25) obviously becomes α

dξ α + ξδ dt



α δγ



duγ = 0 (α = 1, 2). dt

(5.26)

From ξ α = g αν ξν and g ∗ = det(g µν ) ̸= 0, we obtain the differential equations for a covariant surface vector dξν − ξδ dt



δ νγ



duγ = 0 (ν = 1, 2). dt

(c) The Christoffel symbols vanish for the Cartesian coordinates of the plane. Then (5.26) reduces to dξ α = 0 (α = 1, 2). dt Thus the components of the surface vectors are constant. This is the situation in the case described at the beginning of this section. (d) By (5.25), the parallel movement is an intrinsic property of the surface. (e) The parallel movement is independent of the choice of the parameters of the surface, since the left hand side of (5.25) is a tensor. (f) The tangents of a geodesic line γ are parallel to each other with respect to γ, since, for ξ α = u˙ α (α = 1, 2), we have δξ α = u¨α + δs



α δγ



u˙ δ u˙ γ = 0 (α = 1, 2).

This property is referred to as the autoparallelism of geodesic lines. Example 5.6.5. (a) We consider the circular cylinder with a parametric representation ⃗x(ui ) = {r cos u2 , r sin u2 , u1 }.

All Christoffel symbols vanish and the parallel movement by Levi–Civit`a is given by dξ α = 0, that is, ξ α (t) = ξ α (0) (α = 1, 2) dt

422 ■ Differential Geometry and Its Visualization

for all t ≥ 0. Thus the movement is independent of the choice of the curve along which it takes place. (b) We consider the circular cone with a parametric representation ⃗x(ui ) =



where a > 0 is a constant. Putting α = g11 =



1 1 √ u1 cos u2 , √ u1 sin u2 , u1 , a a 

a/(a + 1), we obtain

1 a+1 (u1 )2 = 2 , g12 = 0 and g22 = , a α a

and by (3.47) at the beginning of Section 3.3, the first differential equation in (5.26) reduces to dξ 1 1 dg22 2 du2 α 2 u1 2 u2 ξ = = ξ . dt g11 du1 dt a dt The choice of

√  a 1  1 − g11 (ξ 1 )2 = α2 − (ξ 1 )2 ξ =√ g22 αu1 2

yields

This implies

dξ 1 du2 1  2 α − (ξ 1 )2 =√ . dt dt a+1 



u2 (t) ξ1 dξ 1 √ = = arcsin + c, α a+1 α2 − (ξ 1 )2

where c ∈ R is a constant. If the curve for the movement is given by (u1 (t), u2 (t)) and we choose the initial conditions sin Θ0 cos Θ0 u2 (0) = 0, ξ 1 (0) =  and ξ 2 (0) =  , g11 (u1 ) g22 (u1 ) 0

0

2

where Θ0 is the angle between the u –line at t = 0 and the surface vector, then 1

ξ (t) = α sin



u2 (t) √ + Θ0 a+1



and 

5.7

   √  √   2  2  a u (t) a u (t)   ξ 2 (t) = 1 1 − sin2 √ + Θ0 = 1 cos √ + Θ0  .   u (t) u (t) a+1 a+1

THE FUNDAMENTAL THEOREM FOR SURFACES

In this section, we prove the fundamental theorem of the theory of surfaces, which is the analogue for surfaces of the fundamental theorem of curves, Theorem 1.11.1. It roughly states that a surface is uniquely defined by its first and second fundamental

Tensor Analysis ■ 423

coefficients. Thus the first and second fundamental coefficients gik and Lik play a similar role for surfaces as the curvature and torsion κ and τ in the case of curves. We need the following result from the theory of partial differential equations in the proof of the fundamental theorem for surfaces. Theorem 5.7.1. Let a system of partial differential equations ∂y i = fαi (xβ ; y k ) (i = 1, 2, . . . , n; α = 1, 2, . . . , m) ∂xα

(5.27)

be given. Furthermore, let D ⊂ Rm+n be a domain and f i ∈ C 2 (D) (i = 1, 2, . . . , n). The system is said to be totally integrable if given any point (xβ , y k ) ∈ D there exists 0

0

at least one solution y i (xα ) ∈ C 2 of (5.27) which satisfies the initial conditions y i (xα ) = y i 0

0

(i = 1, 2, . . . , n).

(5.28)

(a) Given initial conditions there is at most one solution of the system (5.27) (uniqueness). (b) The system (5.27) is totally integrable if and only if, for i i i i i Jαβ = fαβ − fβα + fαk fβk − fβk fαk

we have

(i = 1, 2, . . . , n; α, β = 1, 2, . . . , m), (5.29)

i Jαβ = 0 (i = 1, 2, . . . , n; α, β = 1, 2, . . . , m).

(5.30)

Proof. (i) First we show the necessity of condition in (5.30) for the total integrability of the system (5.27). Let y i (xα ) (i = 1, 2, . . . , n) be a solution of the system (5.27) which is two times continuously differentiable and satisfies the initial conditions in (5.28). Then we may interchange the order of differentiation, and so i i yαβ − yβα = 0 (i = 1, 2, . . . , n; α, β = 1, 2, . . . , m).

Now we have ∂  i β k  ∂  i β k  f (x ; y ) − f (x ; y ) α ∂xβ ∂xα β k k i i i ∂y i ∂y = fαβ − fβα + fαk − f βk ∂xβ ∂xα i i i k i k i = fαβ − fβα + fαk fβ − fβk fα = Jαβ =0

i i yαβ − yβα =

for i = 1, 2, . . . , n and α, β = 1, 2, . . . , m. This shows the necessity of the conditions in (5.30).

424 ■ Differential Geometry and Its Visualization

(ii) Now we show the sufficiency of the conditions in (5.30) for the total integrability of the system (5.27) and the uniqueness of the solution. We consider the system dy i dxα = fαi (xβ (t); y k ) dt dt

(i = 1, 2, . . . , n)

(5.31)

of ordinary differential equations. Since ∂y i dxα dy i = (i = 1, 2, . . . , n), dt ∂xα dt any solution y i (xβ ) (i = 1, 2, . . . , n) of (5.27) taken along the curve given by xβ (t) (β = 1, 2, . . . , m), that is, any solution with y i (t) = y i (xβ (t)) has to be a solution of the system (5.31). By the uniqueness theorem for systems of ordinary differential equations, there is one and only one solution y i (t) with y i (0) = y i (i = 1, 2, . . . , n) for xβ (0) = xβ (β = 1, 2, . . . , m). This contains the 0

0

uniqueness statement in Part (a) of the theorem, since any two solutions that satisfy the same initial conditions have to have the same values at arbitrary points that are joined with xβ by a curve. Thus they have to be equal. 0

To prove the existence of a solution of the system (5.27), we have to use (5.30). It suffices to prove that the integration of (5.31) along different curves given by xβ (t) and x∗β (t) with xβ (0) = x∗β (0) = xβ and xβ (1) = x∗β (1) = xβ (β = 1, 2, . . . , m) 0

1

always leads to the same value y i (xβ ), whenever y i (0) = y i (i = 1, 2, . . . , n) is 0

chosen as an initial condition. The functions y i (xβ ) (i = 1, 2, . . . , n) defined in some neighbourhood of xβ (β = 1, 2, . . . , m) then indeed satisfy the differential 0

equations in (5.27). We consider all curves given by xβ (t; ε) = (1 − ε)xβ (t) + εx∗β (t) (0 ≤ ε ≤ 1; β = 1, 2, . . . , m) of class C 2 (as are xβ and x∗β ), and which satisfy xβ = xβ (0, ε) and xβ = xβ (1; ε) (β = 1, 2, . . . , m). 0

1

Let y i (t; ε) (i = 1, 2, . . . , n) denote the corresponding solutions of (5.31) that satisfy y i (0; ε) = y i (i = 1, 2, . . . , n). 0

Since the right hand side in (5.31) is differentiable with respect to ε, each y i (t; ε) is also differentiable with respect to ε. Thus we may write z i (t; ε) =

dy i dxα − fαi (xβ (t; ε); y k (t; ε)) dε dε

(i = 1, 2, . . . , n).

(5.32)

Tensor Analysis ■ 425

Now (5.31) implies 

d dz i = dt dt d = dε



dxα dy i − fαi (xβ (t; ε); y k (t; ε)) dε dε



dy i dt

dxα d − fαi dε dt





−



i dx fαβ

β

dt

+

i dy fαk

k

dt

α





dxα dε



β α k α d dx i dx dx i dy dx − fαk fαi (xβ (t; ε); y k (t; ε)) − fαβ dε dt dt dε dt dε  α d dx − fαi dt dε   β α k α β α dx d dxα dx i i dy dx i dx dx + fαk + fαi = fαβ − fαβ dε dt dε dt dt dε dt dε  α k α d dx i dy dx − fαk (i = 1, 2, . . . , n), fαi dt dε dt dε

=

hence k α β α k α β α dz i i dy dx i dx dx i dy dx i dx dx = fαk + fαβ − fαk − fαβ dt dε dt dε dt dt dε dt dε

(5.33)

for i = 1, 2, . . . , n. We obtain by (5.32), (5.29) and (5.33) i zk fαk

α β dxα i dx dx + Jαβ dt dt dε k α α β α β dxβ dxα i dy dx i i dx dx i dx dx = fαk − fαk + fαβ − fβα fβk dε dt dε dt dt dε dt dε α β α β dx dx dx dx i i + fαk − fβk fβk fαk dt dε dt dε

by interchanging the indices of summation α and β in the last term in the second line i = fαk

β α β α dy k dxα dxα dxβ i dx dx i dx dx i + fαβ − fαβ − fβk fαk dε dt dε dt dt dε dt dε

by observing that fαk dxα /dt = dy k /dt and then interchanging the indices of summation β and α in the last term i = fαk

=

β α β α k α dy k dxα i dx dx i dx dx i dy dx + fαβ − fαβ − fαk dε dt dε dt dt dε dt dε

dz i dt

for i = 1, 2, . . . , n, that is, α β dz i dxα i i dx dx = fαk + Jαβ (i = 1, 2, . . . , n). zk dt dt dt dt

426 ■ Differential Geometry and Its Visualization i The assumption Jαβ = 0 for i = 1, 2, . . . , n now yields

dxα dz i i zk = fαk (i = 1, 2, . . . , n). dt dt

(5.34)

Since z i (0; ε) = 0 (i = 1, 2, . . . , n) by the initial conditions, z i (t; ε) = 0 (i = 1, 2, . . . , n) are the only solutions of (5.34) by the uniqueness theorem. This and (5.32) together imply dy i dxβ = 0 at t = 1, since = 0 at t = 1. dε dε Therefore, y i (xβ ) is independent of the choice of the curve. 1

Now we are able to prove the fundamental theorem of the theory of surfaces. Theorem 5.7.2 (Fundamental Theorem of the Theory of Surfaces). ∗ Let gik (u1 , u2 ) and L∗ik (u1 , u2 ) be given symmetric tensors of class C 2 and C 1 , respectively, that satisfy the conditions ∗ = L∗ik L∗hj − L∗ij L∗hk , Rhikj

L∗ik; j

=

(5.35) (5.36)

L∗ij; k

and ∗ i k gik ξ ξ > 0 for all ξ i ̸= 0..

(5.37)

Then there is – up to movements – one and only one surface with a parametric representation ⃗x(u1 , u2 ) of class C 3 the fundamental coefficients of which satisfy ∗ and Lik = L∗ik gik = gik

in suitable coordinates. Proof. (i) First we show that the system ⃗ ∗, ⃗xi;∗ k = L∗ik N ⃗ i∗ = −L∗r xr∗ N i ⃗

(5.38) (5.39)

of the Gauss and Weingarten equations for the derivatives ((3.99) with Defini⃗∗ tion 5.1.3, and (2.141)) has one and only one system of solutions ⃗xi∗ and N that satisfies the given initial conditions ⃗ ∗ (u1 , u2 ) = N ⃗, ⃗x ∗ (u1 , u2 ) = ⃗x, ⃗xi∗ (u1 , u2 ) = ⃗xi (i = 1, 2) and N 0

0

0

0

0

0

0

0

0

Tensor Analysis ■ 427

where ∗ ⃗ • ⃗x = 0 and N ⃗ 2 = 1. ⃗xi • ⃗xk = gik (u1 , u2 ) (i, k = 1, 2), N 0

0

0

0

0

0

0

To prove this, we have to show that the conditions of integrability in Theorem 5.7.1 are satisfied. These conditions follow from the fact that the order of the ⃗ ∗ and ⃗x ∗ may be interchanged. We have by the Ricci second derivatives of N i identity (5.13) ∗m ∗ ⃗xi;∗ kj − ⃗xi;∗ jk = −Rikj ⃗xm . (5.40) On the other hand (5.38), (5.39), (5.36) and (5.35) together imply

∗ ⃗∗ ⃗ ∗ + L∗ N ⃗∗ ⃗xi;∗ kj − ⃗xi;∗ jk = (L∗ik; j − L∗ij; k )N ik j − Lij Nk ⃗ ∗ − (L∗ L∗m − L∗ L∗m )⃗x ∗ = (L∗ − L∗ )N ik; j



ij; k

ik

j

ij



k

m

⃗ ∗ − L∗ g ∗ml L∗ − L∗ g ∗ml L∗ ⃗x ∗ =0·N ik lj ij lk m ∗ = −g ∗ml (L∗ik L∗lj − L∗ij L∗lk )⃗xm

∗ ∗ ∗m ∗ = −g ∗ml Rlikj ⃗xm = −Rikj ⃗xm .

Consequently the condition in (5.40) is satisfied. Furthermore, we have ⃗∗ ⃗ i;∗ k = ∂ Ni − N ∂uk and

m ik

=

m ki



 ⃗ ∗ m ∂N m ⃗∗ ∗ ∗ k ⃗ ⃗m − , Nm and Nk; i = N ik ki ∂ui

implies ⃗∗ = N ⃗∗ −N ⃗ ∗ = ⃗0 (i, k = 1, 2). ⃗ i;∗ k − N N k; i ik ki

(5.41)

On the other hand (5.39), (5.38), Theorem 5.2.6 and (5.36) together imply 





⃗ ∗ = − L∗r − L∗r ⃗x ∗ − L∗r ⃗x ∗ − L∗r ⃗x ∗ ⃗ i;∗k − N N k; i i; k k; i r i r; k k r; i =−



g ∗rj L∗ij 



;k



− g ∗rj L∗kj 

  ;i



∗ ∗r ∗ ⃗ ∗ ⃗xr∗ − (L∗r i Lrk − Lk Lri ) N

∗ ∗r ∗ ⃗ ∗ = −g ∗rj L∗ij; k − L∗kj; i ⃗xr∗ − (L∗r i Lrk − Lk Lri ) N





∗ ∗r ∗ ⃗ ∗ ∗lr ∗ ∗ ∗lr ∗ ∗ ⃗∗ = 0 · ⃗xr∗ − (L∗r i Lrk − Lk Lri ) N = g Lli Lrk − g Llk Lri N

(with the interchange of the indices l and r in the last term) ⃗ ∗ = ⃗0. = g ∗lr (L∗li L∗rk − L∗rk L∗li ) N Consequently the conditions in (5.41) are satisfied. (ii) Now we show that there exists a function ⃗x of class C 3 with ∂⃗x = ⃗xi∗ for i = 1, 2, and ⃗x(u1 , u2 ) = ⃗x. 0 0 0 ∂ui

428 ■ Differential Geometry and Its Visualization

We have by the symmetry of L∗ik ⃗ ∗ = ⃗0 (i, k = 1, 2), ⃗xi;∗ k − ⃗x∗k; i = (L∗ik − L∗ki )N

and so

∂⃗x ∗i ∂⃗x ∗k ⃗ − = 0 (i, k = 1, 2). ∂uk ∂ui The fact that the function is of class C 3 follows from (5.38) and (5.39). This step shows that there is at most one surface when an initial point and the ⃗ are given at this point. Furthermore, the surface has to be vectors ⃗xi and N 0

0

given by the function ⃗x. (Values without ∗ will refer to this surface.)

∗ ⃗ =N ⃗ ∗. (iii) Now we show gik = gik and N It follows from (5.38) and (5.39) that

 ⃗ ∗ • ⃗xj ) + L∗ (N ⃗ ∗ • ⃗xi ) (⃗xi • ⃗xj ); k = ⃗xi; k • ⃗xj + ⃗xi • ⃗xj; k = L∗ik (N  jk    ∗ ∗ ∗ ∗r ∗  (N ⃗ • ⃗xj ); k = N ⃗ • ⃗xj + N ⃗ • ⃗xj; k = −L (⃗xr • ⃗xj ) + L (N ⃗ ∗ )2

and       ⃗ ∗ )2  (N

We put

k

;k

k

jk

⃗ ∗ xr ). = −2L∗r k (N • ⃗

⃗ ∗ • ⃗xj and ϕ = (N ⃗ ∗ )2 . ϕij = ⃗xi • ⃗xj , ϕj = N

          

(5.42)

(5.43)

Then (5.42) is a system of partial differential equations for the functions ϕij , ϕj and ϕ. This system again has at most a unique solution that satisfies the given initial conditions. By construction, the functions ϕij , ϕj and ϕ in (5.42) are solutions for the initial conditions at (u1 , u2 ) = u = (u1 , u2 ), 0

0

0

ϕij (u) = gij (u), ϕj (u) = 0 and ϕ(u) = 1. 0

0

0

0

∗ , ϕj = 0 On the other hand, an application of Theorem 5.2.6 shows that ϕij = gij and ϕ = 1 is a system of solutions for these initial conditions. Using the terms ϕij , ϕj and ϕ, we obtain in (5.42) ∗ ∗r ϕij; k = L∗ik ϕj + L∗jk ϕi , ϕj; k = −L∗r k ϕrj + Ljk ϕ and ϕ; k = −2Lk ϕr .

∗ ∗ Since ϕij = gij , ϕj = 0 and ϕ = 1, and ϕij; k = gij; k = 0 by Theorem 5.2.6, we conclude

ϕij; k = 0 = L∗ik · 0 + L∗jk · 0

∗ ∗ ∗ ∗ ∗ ∗ ϕj; k = 0 = −L∗r k grj + Ljk = −Lkj + Ljk = −Ljk + Ljk

and ϕ; k = 0 = −2L∗r k ϕr = −2 · 0.

The two solutions consequently coincide. This proves (iii).

Tensor Analysis ■ 429

(iv) Now we show Lik = L∗ik . The Gauss equations for the derivatives imply that the surface satisfies ⃗xi; k = ⃗ and ⃗x ∗ = L∗ N ⃗ . Now it follows from N ⃗∗ = N ⃗ ∗ that Lik = L∗ . Lik N i; k ik ik (v) Finally, the fact that ⃗x is of class C 3 is an immediate consequence of the equations for the derivatives.

We give two applications of Theorem 5.7.2. Example 5.7.3. If the fundamental coefficients of a surface S satisfy g11 = g22 = 1, g12 = 0, L11 = −1 and L12 = L22 = 0, then S is a circular cylinder of radius r = 1. Proof. We obtain L11 = g 1j Lj1 = −1, L21 = g 2j Lj1 = 0, L22 = g 2j Lj2 = 0 



and, by (3.47) at the beginning of Section 3.3, jki = 0 for i, j, k = 1, 2. So the Gauss and Weingarten equations for the derivatives (3.99) and (2.141) reduce to ⃗, ⃗x11 = −N

(5.44)

⃗x12 = ⃗x22 = ⃗0, ⃗ 1 = ⃗x1 N

(5.45) (5.46)

⃗ 2 = ⃗0. N

(5.47)

and

⃗ 1 = −⃗x1 and integration with respect It follows from (5.44) and (5.46) that ⃗x111 = −N 1 to u yields ⃗x(ui ) = ⃗a(u2 ) cos u1 + ⃗b(u2 ) sin u1 + ⃗c(u2 ), (5.48) where ⃗a, ⃗b and ⃗c are vector–valued functions of u2 . Since ⃗x12 = ⃗0 in (5.45), we obtain ⃗x12 = −⃗a ′ (u2 ) sin u1 + ⃗b ′ (u2 ) cos u1 = ⃗0. This ⃗a ′ (u2 ) = ⃗b ′ (u2 ) = ⃗0, so ⃗a(u2 ) = ⃗a and ⃗b(u2 ) = ⃗b, where ⃗a and ⃗b are constant vectors. Finally ⃗x22 = ⃗0 in (5.45) implies ⃗x22 = ⃗c ′′ (u2 ) = ⃗0, hence ⃗c(u2 ) = ⃗c1 u2 + d⃗ ⃗ Thus we have with constant vectors ⃗c1 and d. ⃗ ⃗x(ui ) = ⃗a cos u1 + ⃗b sin u1 + ⃗c1 u2 + d. Since ⃗x1 = −⃗a sin u1 + ⃗b cos u1 and ⃗x2 = ⃗c1 , it follows from g11 = ⃗a 2 sin2 u1 + ⃗b 2 cos2 u1 − 2⃗a • ⃗b sin u1 cos u1 = 1,

(5.49)

430 ■ Differential Geometry and Its Visualization

g12 = −⃗a • ⃗c1 sin u1 + ⃗b • ⃗c1 cos u1 = 0 and g22 = ⃗c12 = 1 that ⃗c12 = 1 and

for u1 = 0: for u1 = π/2: for u1 = π/4:

and

⃗b • ⃗c1 = 0, ⃗b 2 = 1, ⃗a 2 = 1, ⃗a • ⃗c1 = 0 ⃗a • ⃗b = 0.

Consequently (5.49) is a parametric representation of a circular cylinder of radius 1. Example 5.7.4. If the fundamental coefficients of a surface S satisfy g11 = 1, g12 = 0, g22 = sin2 u1 ,

L11 = 1, L12 = 0 and L22 = sin2 u1 (u1 ∈ (0, π)),

then S is a sphere of radius 1. Proof. We obtain from (3.47) at the beginning of Section 3.3 















1 1 1 1 = = = 0, = − sin u1 cos u1 , 11 12 21 22         2 2 2 2 = = 0 and = = cot u1 . 11 22 12 21 Furthermore, we have 1 g11 g22 = 1, g 12 = 0, g 22 = = , g g sin2 u1 L11 = g 11 L11 = 1, L12 = g 11 L12 = 0 and L22 = g 22 L22 = 1.

g 11 =

Therefore, the Gauss and Weingarten equations for the derivatives (3.99) and (2.141) reduce to 



r ⃗ =N ⃗, ⃗x11 = ⃗xr + L11 N 11   r ⃗ = cot u1⃗x2 , ⃗xr + L12 N ⃗x12 = 12   r ⃗ = − sin u1 cos u1⃗x1 + sin2 u1 N ⃗, ⃗xr + L22 N ⃗x22 = 22 ⃗ 1 = −Li ⃗xi = −⃗x1 N 1

(5.50) (5.51) (5.52) (5.53)

and ⃗ 2 = −Li ⃗xi = −⃗x2 . N 2

(5.54)

Tensor Analysis ■ 431

First (5.50) and (5.53) together imply ⃗ 1 = −⃗x1 . ⃗x111 = N Integration with respect to u1 yields ⃗x(ui ) = ⃗a(u2 ) sin u1 + ⃗b(u2 ) cos u1 + ⃗c(u2 ),

(5.55)

where ⃗a, ⃗b and ⃗c are vector–valued functions of u2 . We obtain from this and (5.51) ⃗x12 = ⃗a ′ cos u1 − ⃗b ′ sin u1 = ⃗x2 cot u1 = ⃗a ′ cos u1 + ⃗b ′ cos u1 cot u1 + ⃗c ′ cot u1 and so ⃗b ′ (sin u1 + cos u1 cot u1 ) = −⃗c ′ cot u1 or equivalently ⃗b ′ = −⃗c ′ cos u1 . Since ⃗b ′ and ⃗c ′ are functions that depend on u2 only, we must have ⃗b ′ = ⃗c ′ = ⃗0, and so ⃗b and ⃗c are constant vectors. Furthermore (5.55), (5.52) and (5.50) together imply ⃗x22 = ⃗a ′′ sin u1 = − sin u1 cos u1⃗x1 + sin2 u1⃗x11 = −(sin u1 cos u1 )(⃗a cos u1 − ⃗b sin u1 ) + sin2 u1 (−⃗a sin u1 − ⃗b cos u1 )

= −⃗a sin u1 (cos2 u1 + sin2 u1 ) + ⃗b sin u1 (sin u1 cos u1 − sin u1 cos u1 ) = −⃗a sin u1 ,

that is, ⃗a ′′ = −⃗a, and so ⃗a = d⃗ cos u2 + ⃗e sin u2 with constant vectors d⃗ and ⃗e. Therefore, we have ⃗x(ui ) = (d⃗ cos u2 + ⃗e sin u2 ) sin u1 + ⃗b cos u1 + ⃗c and furthermore, 

⃗x2 • ⃗x2 = g22 = sin2 u2 = −d⃗ sin u2 + ⃗e cos u2 or equivalently,

2

sin2 u1

1 = d⃗ 2 sin2 u2 − 2(d⃗ • ⃗e) sin u2 cos u2 + ⃗e 2 cos2 u2 . For u2 = 0, π/2, π/4, we conclude ⃗e 2 = d⃗ 2 = 1 and d⃗ • ⃗e = 0. Now ⃗x1 • ⃗x2 = g12 = 0 = 







= (d⃗ cos u2 + ⃗e sin u2 ) cos u1 − ⃗b sin u1 • −d⃗ sin u2 + ⃗e cos u2 sin u1

= −d⃗ 2 sin u1 sin u2 cos u1 cos u2 + ⃗e 2 sin u1 sin u2 cos u1 cos u2 + ⃗b • d⃗ sin2 u1 sin u2 − ⃗b • ⃗e sin2 u1 cos u2 ⃗ sin2 u1 sin u2 − (⃗e • ⃗b) sin2 u1 cos u2 = (⃗b • d) implies ⃗b • d⃗ = ⃗e • ⃗b = 0. Finally ⃗x1 • ⃗x1 = g11 = 1

432 ■ Differential Geometry and Its Visualization



= (d⃗ cos u2 + ⃗e sin u2 ) cos u1 − ⃗b sin u1

2

= cos2 u1 cos2 u2 + cos2 u1 sin2 u2 + ⃗b 2 sin2 u1 = cos2 u1 + ⃗b 2 sin2 u1 ⃗ ⃗e and ⃗b are orthonormal. implies ⃗b 2 = 1. Therefore, the vectors d, Thus we have ∥⃗x − ⃗c∥2 = sin2 u1 cos2 u2 + sin2 u1 sin2 u2 + cos2 u1 = 1, and the surface is a sphere of radius 1, centred at the point C with the position vector ⃗c.

5.8

A GEOMETRIC MEANING OF THE RIEMANN TENSOR OF CURVATURE

In this section, we give a geometric interpretation of the Riemann tensor of curvature. Let γ be a curve with a parametric representation ⃗x(t) in a Riemann space and T0 be the tangent space to γ at ⃗x = ⃗x(t0 ). Then we may move the tangent space T0 0 along γ by applying the parallel movement by Levi–Civit`a to every surface vector in T0 . This means, we assign to every surface vector with the components ξ i in T0 the solutions ξ i (t) (i = 1, 2, . . . , n) of the system of differential equations dxj δξ i = ξ;ij = 0 with the initial conditions ξ i (t0 ) = ξ i (i = 1, 2, . . . , n) δt dt 0

(5.56)

(Remark 5.6.4 (b), (5.26)). Let t be a fixed parameter of the curve γ and Tt be the tangent space to γ at ⃗x(t). The parallel movement, which in general depends on the curve and the Christoffel symbols, defines a map A(γ; t) : T0 → Tt by ξ i (t) = Aik (γ; t)ξ k (i = 1, 2, . . . , n). The map A(γ; t) is linear, since the system (5.56) is linear in the components ξ i . Now let γ be a piecewise smooth curve. Then we obtain a linear map by joining the linear maps from the tangent planes at the initial points to the tangent planes at the end points of the smooth parts. If the end point of γ coincides with its initial point then we obtain a linear map A(γ) from T0 onto itself. Using vector notation, we may write ⃗z ∗ = A(γ)⃗z for all surface vectors ⃗z ∈ T0 .

Let γ1 and γ2 be piecewise smooth curves with initial and end points in a given point P . Then we define the product curve γ = γ2 γ1 to be the curve that results from first passing along γ1 and then along γ2 . Furthermore, let γ −1 be the curve that has the opposite orientation of γ. The corresponding linear maps from the tangent plane T (P ) at P onto itself clearly satisfy A(γ2 γ1 ) = A(γ2 ) ◦ A(γ1 ) and A(γ −1 ) = (A(γ))−1 .

Tensor Analysis ■ 433

Proposition 5.8.1. Let M be a differentiable manifold which is path connected, that is, any two points of M can be joined by a curve on M . Furthermore, let P ∈ M be a point and H(P ) denote the set of all maps A(γ) from T (P ) onto itself. Then H(P ) is a group, the so–called holonomy group of connectedness at the point P . Proof. Let A, B ∈ H(P ), that is, A = A(γ1 ) and B = A(γ2 ) for two curves γ1 and γ2 with initial and end points in P . Then BA = A(γ2 γ1 ) = A(γ2 ) ◦ A(γ1 ) ∈ H(P ). Obviously, we have A(γ3 (γ2 γ1 )) = A(γ3 ) ◦ A(γ2 γ1 ) = A(γ3 ) ◦ (A(γ2 ) ◦ A(γ1 )) = (A(γ3 ) ◦ A(γ2 )) ◦ A(γ1 ) = A(γ3 γ2 ) ◦ A(γ1 ) = A((γ3 γ2 )γ1 ).

Thus the law of associativity is satisfied. The identity map E defined by E(⃗z) = ⃗z for all surface vectors ⃗z ∈ T (P ) is an element of H(P ), since E = A(γ)◦A(γ −1 ) = A(γγ −1 ). Furthermore, we have E ◦A(γ) = A(γ) for all A(γ) ∈ H(P ). So E is the neutral element of H(P ). Finally, let A ∈ H(P ). Then A = A(γ) for some curve γ. We put A−1 = A(γ −1 ), and obtain A−1 ∈ H(P ) and AA−1 = A(γγ −1 ) = E. Now we consider special curves γε that are similar to parallelograms with one vertex at a point P . We assume that P is the origin  of the coordinate system and that the Christoffel symbols vanish at P , that is, jki (0) = 0 for all i, j and k. Such a choice of a coordinate system is possible by the proof of Theorem 5.4.1. Let η i and ζ i be the components of two linearly independent vectors ⃗y and ⃗z. We define the curve γε by   tε⃗y (t ∈ [0, 1])     ε⃗ y + (t − 1)ε⃗z (t ∈ [1, 2]) ⃗x(t) =  ε(⃗y + ⃗z) − (t − 2)ε⃗y (t ∈ [2, 3])     ε⃗ z (1 − (t − 3)) (t ∈ [3, 4]),

or using the components

xi (t) =

  tεη i     εη i + (t − 1)εζ i

 ε(η i + ζ i ) − (t − 2)εη i     εζ i (1 − (t − 3))

(t ∈ [0, 1]) (t ∈ [1, 2]) (t ∈ [2, 3]) (t ∈ [3, 4]).

The following result shows that the second order approximation of the change of a surface vector by parallel movement along γε depends on the Riemann tensor of curvature at P . Theorem 5.8.2. The Riemann tensor of curvature is a measure for the change of a surface vector under parallel movement along very small parallelograms γε . More precisely, the linear map A(γε ) is given by i (0)η j ζ l + ε3 (· · · ). Aik (γε ) = δik + ε2 Rkjl

(5.57)

This relation is independent of the choice of a coordinate system, since it is invariant.

434 ■ Differential Geometry and Its Visualization

Proof. Let ξ i be the components of a surface vector ⃗x at P that is to be moved along γε in a parallel way. The Taylor expansion yields ξ i (t + 1) = ξ i (t) + We write By the parallelism, we have



i jk



,l

dξ i (t) 1 d2 ξ i (t) + · + ··· . dt 2 dt2

∂ = ∂xl 



i jk



(5.58)

.



dxj (t) i dξ i (t) =− (t)ξ k (t) jk dt dt

(5.59)

and 

i d2 ξ i (t) =− 2 jk dt











dxj (t) dxs (t) i k . (t) (t)ξ r (t) jk rs dt dt (5.60) We neglect terms in ε of order higher than 2. The Taylor expansion yields 



i (t) = jk



dxj (t) dxl (t) + (t)ξ (t) dt dt ,l k



i (0) + jk



i jk



,l

(0)∆xl + · · · =



i jk



,l

(0)∆xl + · · · .

(5.61)

The change of the components ξ i under the parallel movement along γε is given by ξ i (4) − ξ i (0) =

4 

m=1

(ξ i (m) − ξ i (m − 1)).

(5.62)

We obtain from (5.59) and (5.60) 



dxj (0) i dξ i (0) =− =0 (0)ξ k (0) jk dt dt

(5.63)

and 

i d2 ξ i (0) =− jk dt



(0)ξ k (0)ε2 η j η l .

(5.64)

,l

Now we put f (1) (m) =

1 dxj (m) l 1 dxj (m) dxl (m) (2) x (m) and f (m) = ε2 dt ε2 dt dt

We have by (5.59) and (5.61) for m ≥ 1 





dxj (m) i i dξ i (m) =− =− (m)ξ k (m) jk jk dt dt



,l

(m = 1, 2, 3).

(0)ε2 f (1) (m)ξ k (m) + ε3 (· · · ),

Tensor Analysis ■ 435

and by (5.60) and (5.61) 



i d2 ξ i (m) =− jk dt2

,l

+ = −ε2



i jk



= −ε



i jk











dxj (m) dxs (m) i k (m) (m)ξ r (m) jk rs dt dt

(0)f (2) (m)ξ k (m)+ ,l

+ 2

dxj (m) dxl (m) dt dt

(m)ξ k (m)

,l



i jk



(0)x (m) + · · · l

,l



dxj (m) dxs (m) ξ (m) dt dt r





k (m) rs

(0)f (2) (m)ξ k (m) + ε3 (· · · ).

Applying (5.58), we obtain dξ i (m) = −ε2 dt



i jk





· · · = −ε2



dξ k (m − 1) 1 d2 ξ k (m − 1) + (0)f (1) (m) ξ k (m − 1) + + ··· dt 2 dt2 ,l

i jk



+ ε3 (· · · ) = · · ·

,l

(0)f (1) (m)ξ k (0) + ε3 (· · · ),

and similarly d2 ξ i (m) = −ε2 dt2



i jk



,l

(0)f (2) (m)ξ k (0) + ε3 (· · · ).

This yields dξ i (1) = −ε2 dt dξ i (2) = ε2 dt dξ i (3) = ε2 dt

 



i jk

i jk i jk

 



d2 ξ i (1) = −ε2 dt2

(0)ξ (0)η ζ k

l j

,l

(0)ξ (0)η (η + ζ ) k

j

l

l

,l

d2 ξ i (3) = −ε2 dt2

(0)ξ (0)ζ ζ k

d2 ξ i (2) = −ε2 dt2

j l

,l

  

i jk i jk i jk

  

(0)ξ k (0)ζ l ζ j ,l

(0)ξ k (0)η l η j ,l

(0)ξ k (0)ζ l ζ j . ,l

Now it follows from this, (5.62) and (5.63) that 3    1  i 1  i i i ξ (4) − ξ (0) = ξ (m + 1) − ξ (m) ε2 ε2 m=0



3 1  dξ i (m) 1 d2 ξ i (m) + · = 2 ε m=0 dt 2 dt2

=



i jk



,l





(0)ξ k (0) 0 − η l ζ j + η j (η l + ζ l ) + ζ i ζ l +



436 ■ Differential Geometry and Its Visualization

=

= = =



i jk



i jk



 

i jk

,l

,l





 1 + −η j η l − ζ l ζ i − η l η j − ζ l ζ j + ε(· · · ) 2

(0)ξ k (0) −η l ζ j + η j (η l + ζ l ) +







+ζ j ζ l − η j η l − ζ j ζ l + ε(· · · ) 

(0)ξ k (0) η j ζ l − η l ζ j + ε(· · · ) (0) −



,l i Rkjl (0)ξ k (0)η j ζ l

i lk



,j



(0) ξ k (0)η j ζ l + ε(· · · )

+ ε(· · · )





i , since jki (0) = 0. Thus by the definition of the Riemann tensor of curvature Rkjl the map A(γε ) is given by (5.57). The smaller we choose ε in (5.57) the better the difference ξ i (4) − ξ i (0) is approximated by the term that contains the tensor of curvature. The invariance of the relation in (5.57) follows from the fact that it is a relation between the components of tensors.

5.9

SPACES WITH VANISHING TENSOR OF CURVATURE

In this section, we study when the tensor of curvature in a Riemann space vanishes identically. In Euclidean En , the metric tensor in Cartesian coordinates is given by gik = δik .   i i = 0. Therefore, we have for a parallel This implies jk = 0 and consequently Rljk vector field along an arbitrary curve with parameter t dξ i δξ i = = 0 (i = 1, 2, . . . , n). δt dt Thus the parallel movement is independent of the choice of the curve. The next result shows that a Riemann space with a tensor of curvature that vanishes at a point is locally the same as Euclidean En in the sense that parallel movements locally are independent of the choice of a curve. i Theorem 5.9.1. Let P0 be a point in a Riemann space such that Rljk = 0 at P0 for the components of the tensor of curvature. Then there is  a coordinate system S ∗ with ∗ coordinates x∗i in some neighbourhood N of P0 such that jki = 0 on N . Therefore, parallel vectors with the components

ξ ∗i (x∗j ) and ξ ∗i (x∗j ) 1

2

at any two distinct points P1 = (x∗1 , x∗2 , . . . , x∗n ), P2 = (x∗1 , x∗2 , . . . , x∗n ) ∈ N 1

1

1

2

2

2

have the same components. The parallel movement is independent of the choice of a curve.

Tensor Analysis ■ 437 i Proof. First we show that if Rljk = 0 then there is a neighbourhood N1 of P0 such that for every vector with the components ξ0i there exists a vector field with the components ξ i (x) in N1 such that ξ i (x) = ξ0i and which consists of parallel vectors 0 independent of the choice of a curve. It is necessary and sufficient for this that the system   ∂ξ i i = − ξ l (i, k = 1, 2, . . . , n) k lk ∂x

of differential equations has a solution with ξ i (x) = ξ0i . But this is true, since the 0

i corresponding condition of integrability in Theorem 5.7.1 is Rljk = 0. This shows that the parallel movement is independent of the curve. Now we choose those n vector fields among the parallel fields for which i ξ(k) = δki (i, k = 1, 2, . . . , n).

(5.65)

i Since the functions ξ(k) are continuous, there is a neighbourhood N2 of P0 such that i det(ξ(k) ) ̸= 0 on N2 .

(5.66)

Now we consider the system

∂xi i = ξ(k) (x) ∂x∗k of differential equations. Since the Christoffel symbols are symmetric and

(5.67)

  i  ∂xl ∂ξ(k) ∂ 2 xi ∂  i i l ξ(k) (x) = ξ(j) = − = ξs ξl , ∗k ∗j l ∗j l sl (k) (j) ∂x ∂x ∂x ∂x ∂x

the system (5.67) satisfies the condition of integrability. Thus there exist solutions xi (x∗k ) of (5.67) in some neighbourhood N of P0 . It follows from (5.66) that the inverse functions x∗k (xi ) exist. They define an admissible transformation of coordinates. We have for the vector fields with respect to the new coordinates by the transformation formula for the components of contravariant vectors l ξ(k) =

∂xl ∗i ξ . ∂x∗i (k)

Using (5.67), we obtain l ∗i ξ(i) ξ(k) =

∂xl ∗i l ξ = ξ(k) , ∂x∗i (k)

∗i and so, by (5.66), ξ(k) = δki on N . Since the vector fields are parallel, we conclude

0=

∗i ∂ξ(k)

∂xj

+



i jr

∗

∗r ξ(k) =0+



i jr

∗

δkr =



i jk

∗

.

Thus the Christoffel symbols with respect to the new coordinates vanish on all of N .

438 ■ Differential Geometry and Its Visualization

We saw in Remark 5.6.4 (f) that the tangents of a geodesic line are parallel to each other. Therefore, we may define autoparallel curves. Definition 5.9.2. A curve in a Riemann space is called an autoparallel curve if its tangents are parallel in the sense of Levi–Civit`a. The next results establishes the local existence of an autoparallel curve through every point of a Riemman space in each direction. Theorem 5.9.3. There is an autoparallel curve through every point P of a Riemann space in each direction. The autoparallel curves are uniquely defined in a neighbourhood of P . Proof. For sufficiently small t, the system d2 xi + dt2





i jk

dxj dxk = 0 (i = 1, 2, . . . , n) dt dt

(5.68)

has a unique solution xi (t) that satisfies the conditions xi (t0 ) = xi and 0

dxi (t0 ) = ξ i (i = 1, 2, . . . , n). dt 0

It remains to be shown that the same solution satisfies the initial conditions xi (t1 ) = xi and 0

dxi (t1 ) = λ · ξ i (i = 1, 2, . . . , n). dt 0

Indeed, for each i, xi (t) = xi (λ(t − t1 ) + t0 ) is a solution for these initial conditions. This is the only solution by the uniqueness theorem for differential equations, since the new curve differs from the old one only by the choice of parameters. Now let t = t(τ ) be a parameter transformation for an autoparallel curve. Then x (t) = xi (τ (t)) implies i

d2 xi d2 x i = 2 dt dτ 2

dxi dτ dxi = , dt dτ dt



2

+

dxi d2 τ , dτ dt2

dτ dt

2

dxi d2 τ − dτ dt2

dxj dxk dxi d2 τ =− dτ dτ dτ dt2



1 . (dτ /dt)2

dτ dt

and so by (5.68) 1 d2 x i = 2 dτ (dτ /dt)2 1 = (dτ /dt)2

 

d2 xi dxi d2 τ − dt2 dτ dt2 

i − jk





dxj dxk dτ dτ





.

Thus (5.68) transforms to d2 xi + dτ 2



i jk





(5.69)

Tensor Analysis ■ 439

A parameter τ for which the right hand side in (5.69) vanishes is called a natural parameter of the autoparallel curves. Since dxi /dτ ̸= 0 for admissible parameter transformations, natural parameters τ are characterized by the condition d2 τ /dt2 = 0. This yields the following result. Theorem 5.9.4. If t is a natural parameter of an autoparallel curve, then all other natural parameters τ satisfy τ = c 0 t + c1 , where c0 ̸= 0 and c1 are constants.

We close this section with the differential equations for autoparallel curves.

Theorem 5.9.5. A solution of the system d2 xi + dτ 2



i jk



dxj dxk dxi = −µ(τ ) (i = 1, 2, . . . , n) dτ dτ dτ

is an autoprallel curve. Proof. We introduce a new parameter t by 

dτ d2 τ = µ(τ ) 2 dt dt

2

.

The assertion now follows from (5.69).

5.10

AN EXTENSION OF FRENET’S FORMULAE

In Section 1.7, we proved Frenet’s formulae in Euclidean E3 which gave the derivatives of the vectors of the trihedron of a curve in terms of the tangent, principal normal and binormal vectors. Now we study an extension of Frenet’s formulae for curves in a Riemann space. Theorem 5.10.1. Let γ be a curve with a parametric representation ⃗x(s) = {x1 (s), x2 (s), . . . , xn (s)} of class C n+1 in a Riemann space with the metric tensor 1 2 n gik , and s be the arc length along γ. We define the vectors ⃗x(k) = {ξ(k) , ξ(k) , . . . , ξ(k) } for k = 1, 2, . . . , n by

i ξ(k)

=

 i dx    

(k = 1)

ds

   ξ i

(k)

=

i δξ(k−1)

δs

=

i dξ(k−1)

ds

+

  i jl

dx l ξ(k−1)

j

ds

(i = 1, 2, . . . , n),

(2 ≤ k ≤ n)

and assume that the vectors ⃗x(k) are linearly independent. 1 2 n Let ⃗y(k) = {η(k) , η(k) , . . . , η(k) } (k = 1, 2, . . . , n) denote the vectors obtained from the vectors ⃗x(k) (k = 1, 2, . . . , n) by the Gram–Schmidt orthonormalization process. We put i δη(j) κ(j) = gik ηk (j = 1, 2, . . . , n − 1). δs (j+1)

440 ■ Differential Geometry and Its Visualization

Then we have

 i δη(1)     =   δs     i  δη(2)    =    δs       ···

i κ(1) η(2) i i −κ(1) η(1) + κ(2) η(3)

···

(5.70)

i   δη(j)  i i   + κ(j) η(j+1) = −κ(j−1) η(j−1)   δs      ··· ···        i  δη(n)    = −κ(n−1) η i (n−1)

δs

for i = 1, 2, . . . , n.

Proof. The vectors ⃗y(k) are linear combinations of the vectors ⃗x(1) , ⃗x(2) , . . . , ⃗x(k) by i i i i construction. Furthermore, δη(k) /δs is a linear combination of ξ(1) , ξ(2) , . . . , ξ(k+1) . i i Since each component ξ(j) can be given in terms of η(r) for r ≤ j, we may write i δη(k)

δs

=

k+1 

(5.71)

i akj η(j)

j=1

with the coefficients akj yet to be determined. The orthonormality of the vectors ⃗y(k) i k implies gik η(j) η(l) = δjl . From Ricci’s theorem, Theorem 5.2.6, we conclude i k i k i k η(l) ); m = gik η(j); 0 = δjl; m = (gik η(j) m η(l) + gik η(j) η(l); m .

Multiplying this by dxm /ds and summing with respect to m, we obtain from (5.71), since i δη(j) dxm i η(j); m = , ds δs 0 = gik =

i δη(j)

k+1 

δs

k i η(l) + gik η(j)

i k ajs gik η(s) η(l) +

s=1

= ajl + alj .

k δη(l)

δs

k+1  s=1

= gik

k+1 

i k ajs η(s) η(l) + gik

s=1

k i als gik η(s) η(j) =

k+1 

k i als η(s) η(j)

s=1

k+1 

(ajs δsl + als δsj )

s=1

Consequently, the matrix (ajl ) is skew symmetric. Since alj = 0 for j ≥ l + 2, at most the entries aj,j−1 and aj,j+1 are unequal to zero. We put κ(j) = aj,j+1 . Then we obtain i δη(j) κ(j) = gik ηk (j = 1, 2, . . . , n − 1). δs (j+1) This shows the identities in (5.70). The values κ(j) are invariants connected with the curve, since all terms in their definitions are components of vectors.

Tensor Analysis ■ 441

Remark 5.10.2. (a) In Euclidean E3 , we have ⃗x(1) =



i jk



= 0 for all i, j and k, and so

d⃗x(1) d⃗x = ⃗v1 = ⃗y(1) and ⃗x(2) = . ds ds

Since ⃗x(1) • ⃗x(2) = 0, we may choose ⃗y(2) =

⃗x(2) = ⃗v2 and ⃗y(3) = ⃗y(1) × ⃗y(2) = ⃗v1 × ⃗v2 = ⃗v3 . ∥⃗x(2) ∥

Furthermore, we have κ(1) = gik

i δη(1)

δs

k η(2) =

3 

k=1

v˙ 1k v2k = ⃗v˙ 1 • ⃗v2 ,

hence κ(1) = κ, the curvature, and κ(2) = gik

i δη(2)

δs

k η(3) = ⃗v˙ 2 • ⃗v3 ,

hence κ(2) = τ , the torsion. Thus the identities in (5.70) reduce to Frenet’s formulae of Section 1.7. (b) If S is a surface in Euclidean E3 , then there is only the first identity in (5.70) and κ(1) = κg , the geodesic curvature. We know from (3.14) in the proof of Theorem 3.1.8 that     i i j k κg⃗t = u¨ + u˙ u˙ ⃗xi , jk i . and so the assertion follows from the definition of the components ξ(2)

5.11

RIEMANN NORMAL COORDINATES AND THE CURVATURE OF SPACES

Cartesian coordinates play an important role in Euclidean space. Very useful coordinates can be introduced in Riemann spaces. They have many properties similar to those of Cartesian coordinates and are referred to as Riemann normal coordinates. Let P0 be a point in a Riemann space with the metric tensor gik and let the coordinates xi (i = 1, 2, . . . , n) be geodesic in P0 = (x) = (x1 , x2 , . . . , xn ), that is,



i jk



0

0

0

0

(x) = 0. There is one and only one geodesic line through P0 in the di0

rection of a unit vector with the components ξ i (i = 1, 2, . . . , n). Let ⃗x(s) = {x1 (s), x2 (s), . . . , xn (s)} be a parametric representation of the geodesic line γ with xi (0) = xi , x˙ i (0) = ξ i (i = 1, 2, . . . , n) and s be the arc length along γ. The Taylor 0 expansion yields xi (s) = xi + sx˙ i (0) + 0

s2 i s3 ...i x¨ (0) + x (0) + · · · (i = 1, 2, . . . , n). 2 3!

(5.72)

442 ■ Differential Geometry and Its Visualization

The differential equations for geodesic lines 



i x˙ j x˙ k (i = 1, 2, . . . , n) x¨ = − jk i

yield ...i x =−



...i x (0) = −









i i x˙ l x˙ j x˙ k − x¨j x˙ k − jk , l jk     i i l j k =− x˙ x˙ x˙ − 2 x¨j x˙ k , jk , l jk   i (x)ξ j ξ k = 0 x¨i (0) = − jk 0





i x˙ j x¨k jk

and i jk



(x)ξ l ξ j ξ k .

,l 0

for (i = 1, 2, . . . , n). Therefore (5.72) becomes 1 x (s) = x + sξ − s3 0 3! i

i

i



i jk



(x)ξ l ξ j ξ k + s4 (· · · ) (i = 1, 2, . . . , n)

,l 0

for sufficiently small s. We put x∗i = sξ i . Then a coordinate transformation is given in a neighbourhood of P0 by 1 x =x +x − 0 3! i

∗i

i



i jk



(x)x∗j x∗k x∗l + · · · (i = 1, 2, . . . , n).

,l 0

Furthermore, we have 







∂xi 1 1 i i i ∗j ∗k ∗l ∗k ∗l = δm − (x)δm x x − (x)x∗j δm x ∗m ∂x 3! jk , l 0 3! jk , l 0   1 i ∗l − (x)x∗j x∗k δm + · · · (i, m = 1, 2, . . .). 3! jk , l 0 Since we have x∗i (xi ) = 0 at xi = xi by the definition of x∗i , it follows that the 0

0

Jacobian of the transformation is equal to one at (x) = (x), hence unequal to zero in 0 some neighbourhood of P0 , and so the transformation is locally admissible. The differential equations for the geodesic lines with respect to the new coordinates are  ∗ i ∗i x¨ + x∗j x∗k = 0 (i = 1, 2, . . . , n). jk We put x∗i = sξ i (i = 1, 2, . . . , n), and obtain 

i jk

∗

ξ i ξ k = 0 (i = 1, 2, . . . , n),

(5.73)

Tensor Analysis ■ 443

and multiplication with s yields 

i jk

∗

(x∗ )x∗j x∗k = 0 (i = 1, 2, . . . , n).

The geodesic lines through P0 are given by the linear equations x∗i = ξ i s (i = 1, 2, . . . , n) with respect to the so–called Riemann normal coordinates x∗i with centre at P0 , in analogy of the case of straight lines in Euclidean space. The following result shows that the Gaussian curvature of a surface may easily be found by measuring lengths on the surface. Furthermore, its proof contains a very illustrating proof for the theorema egregium, Theorem 3.8.3. Theorem 5.11.1 (Bertrand–Puiseux (1848)). Let x∗1 and x∗2 be Riemann normal coordinates with centre at a point P for a twodimensional Riemann space. We introduce geodesic polar coordinates u1 and u2 by x∗1 = u1 cos u2 and x∗2 = u1 sin u2 , where u1 is the distance from P measured along the geodesic lines through P and u2 is the angle between a geodesic line and a given direction. If K0 denotes the Gaussian curvature at P and L(r) is the circumference of a geodesic circle of radius r, that is, the length of the u2 –line for u1 = r, then 3(2πr − L(r)) . r→0 πr3 Proof. We obtain from the transformation formulae K0 = lim

g22 =

This implies

∗i ∗ ∂x gik 2

∂x∗k ∗ = g11 ∂u ∂u2



−2

+

∂x∗2 ∂u2

2

∂x∗2 ∗ g22, 2 1



∗1 ∗ ∂x 2g12 2

∂x∗2 ∗ + g22 ∂u ∂u2







lim g22 = 0.

2 = 1 g22 + (u1 )2 u 

2

∗ ∗ ∗ = (u1 )2 g11 sin2 u2 − 2g12 sin u2 cos u2 + g22 cos2 u2 .

u1 →0

Furthermore, we have g22, 1

∂x∗1 ∂u2

(5.74)

∂x∗1 ∗ g12, 1 1 ∂u



+

∂x∗1 ∂x∗2 ∗ ∗ g11, + g 1 11, 2 ∂u1 ∂u1

∂x∗2 ∗ g12, 2 1 ∂u



2

2



sin u cos u +

sin2 u2 

∂x∗1 ∗ g22, 1 1 ∂u



+

∗

∂u

2

cos u



2

.

Since (5.73) holds for all directions at P , it follows that jki (P0 ) = 0 for all i, j ∗ and k, and so gij, (P ) = 0 for all i, j and k by (4.24) and (4.25). Now k 1 g22, 1 √ ( g22 ), 1 = √ 2 g22

444 ■ Differential Geometry and Its Visualization

implies

√ lim ( g22 ) ,1 = 0.

u1 →0

Since it holds in geodesic coordinates that √ ∂ 2 ( g22 ) √ = −K g22 1 2 (∂u ) for the Gaussian curvature K by (3.104) of Lemma 3.8.5, we also have √ ∂ 2 ( g22 ) → 0 (u1 → 0) (∂u1 )2 and

√ ∂ 3 ( g22 ) ∂K √ √ = − 1 g22 − K( g22 ), 1 → −K0 (u1 → 0), 1 3 (∂u ) ∂u

where K0 is the Gaussian curvature at the point P . The Taylor expansion at P yields √

1 g22 = u1 − K0 (u1 )3 + (u1 )4 (· · · ) 6

and

1 g22 = (u1 )2 − K0 (u1 )4 + (u1 )5 (· · · ). 3 Finally, we have for the circumference of the geodesic circle corresponding to u1 = r L(r) =

2π 0

√

1 g22 du2 = 2πr − K0 πr3 + r4 (· · · ), 3

hence (5.74) follows. Now we deal with the curvature of space. Let ⃗x and ⃗y be two vectors with the components ξ i and η i (i = 1, 2, . . . , n) in the tangent space at P0 . We introduce Riemann normal coordinates x∗i (i = 1, 2, . . . , n) with the centre at P0 such that ⃗x and ⃗y are the directions of the x∗1 – and x∗2 –lines, respectively. The geodesic lines with the tangent vectors α⃗x + β⃗y at P0 span a surface in a neighbourhood of P0 which is given by the equations x∗3 = · · · = x∗n = 0.

The Gaussian curvature of this surface which depends on x∗1 and x∗2 only is given by j g2j R112 K= (cf. (3.101) in Theorem 3.8.3). g r i i We write Rijkl = gir Rjkl and use the fact that Rhjk +Rhkj = 0 which is an immediate consequence of the definition of the tensor of curvature to conclude

K=−

R2121 R2121 ξ 1 η 2 ξ 1 η 2 =− . g11 g22 − g12 g21 (g11 g22 − g12 g21 )ξ 1 η 2 ξ 1 η 2

(5.75)

Tensor Analysis ■ 445

We have in arbitrary coordinates K=

Rijkl η i η k ξ j ξ l , (gil gjk − gik gjl )η i η k ξ j ξ l

(5.76)

since this invariant term reduces to (5.75) in the special coordinates. The term in (5.76) is the Gaussian curvature of the two-dimensional surface spanned by the geodesic lines through P0 in the directions that are linear combinations of the given vectors ⃗x and ⃗y in the tangent space at P0 ; it is called the curvature of the space at P0 in the two-dimensional direction spanned by ⃗x and ⃗y . The Ricci tensor Rjk is a contraction of the tensor of curvature, namely i Rjk = Rjki ,

and the scalar of curvature R is defined by R = g jl Rjl . The Ricci tensor and the scalar of curvature can be interpreted as certain means of the Gaussian curvature. Let ⃗x(1) , ⃗x(2) , . . . , ⃗x(n) be orthonormal vectors with the comi ponents ξ(k) (i = 1, 2, . . . , n) at a point P with the coordinates (x) = (x1 , x2 , . . . , xn ), and let K(r, s) be the curvature in (5.76) which corresponds to the direction of the plane spanned by the vectors ⃗x(r) and ⃗x(s) . Then we obtain j i l k ξ(s) ξ(r) ξ(r) . K(r, s) = Rijkl ξ(s)

We have by (4.18) g = il

n 

(5.77)

i l ξ(s) ξ(s) ,

s=1

and so, by the definition of the Ricci tensor, n 

K(1, s) =

s=1

n 

j j i l k k Rijkl ξ(s) ξ(s) ξ(1) ξ(1) = g il Rijkl ξ(1) ξ(1)

s=1 il

j j j r k r k l k = g gir Rjkl ξ(1) ξ(1) = δrl Rjkl ξ(1) ξ(1) = Rjkl ξ(1) ξ(1) j k = Rjk ξ(1) ξ(1) .

Since K(1, 1) = 0, the term

1 j k Rjk ξ(1) ξ(1) n−1 is the arithmetic mean of the curvatures corresponding to n − 1 orthogonal planes to ⃗x(1) . The mean is independent of the choice of the vectors ⃗x(2) , . . . , ⃗x(n) . Furthermore, we have n 

r,s=1

K(r, s) =

n  r=1

j k Rjk ξ(r) ξ(r) = g jk Rjk = R.

446 ■ Differential Geometry and Its Visualization

Consequently R/(n(n − 1)) is equal to the arithmetic mean of all curvatures that correspond to all possible directions of planes that can be constructed from n orthonormal vectors. In particular, for n = 2, there is only one direction of a plane, and R 1 K= = g ik g il Rijkl . 2 2

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Index (f, D), 106 (f, I), 13 (gik ), (g ik ), g, 117 C 0 (I), C m (I), C ∞ (I), 12 C r (D), 106 Cikj , 320 Df⃗, 10 Gm ikj , 320 H, 153 K, 153 Lik , 228 Lik , L, 132 Lik , 228 Nε (X0 ), 10 Nε (⃗x0 ), 10 m Rikj , 320 Rijkl , 321 j Rilk , 405 T;ji , 397 Ti;j , 398 ,...,js Tij11,...,i , 400 r ;k n Tm;r , 399 V ∗ , 379 [ikl],246  Γkij , ijk , 393 △F , 418 ▽(Φ, Ψ), ▽Φ, 415 ¯ 10 S, δik , 47 N˙ ε (X0 ), 10 N˙ ε (⃗x0 ), 10 ⃗x˙ (s), 33 df⃗ ⃗′ dt (t0 ), f (t0 ), 11 κ1 , κ2 , 152 κg , 131 κn , 131 λ⃗x, 3 r ik , 247

limt→t0 f⃗(t), 10 C, 2 E3 , 1 N, 2 Q, 2 R, 2 Rn , 2 Vn , 2 Z, 2 κ–curve, 26 div T k , 416 div Tk , 417 ∥ · ∥, ∥⃗x∥, 3 P Q, 19 τ (s), 54 ˜ 328 C, ⃗v1 , ⃗v2 , ⃗v3 , 47 ε–open ball of a point, 10 ε–open ball of a vector, 10 ⃗x + ⃗y , 3 ⃗ (ui ), 114 N f⃗, f⃗(t), 10 f⃗ ′′ , 12 f⃗ ′′′ , f⃗ (m) , 12 f⃗ ′ (t0 ), 11 ⃗t, ⃗n, ⃗b, 47 ⃗t(t0 ), 45 −→

P Q, 3 −→

0X, 3 ⃗x • ⃗y , 3 ⃗x ⊥ ⃗y , 4 ⃗x × ⃗y , 5 ⃗xk (ui ), 106 ⃗xt0 , 45 ⃗x(ui ), 106 ⃗y⃗x , 5 cik , 234 d, d(X, Y ), 2 459

460 ■ Index

gik , 116 gik , g ik , 391 xk , 2 [ijk], 393 ε–deleted neighbourhood, 10 n–dimensional (real) Euclidean vector space, 3 n–dimensional (real) point space, 2 n–dimensional (real) vector space, 3 u1 –lines, 107 u2 –lines, 107 dot product, 3 admissible map between surfaces, 324 angle preserving, 325 conformal, 325 admissible parameter transformation for a curve, 15 admissible parameter transformation for a surface, 108 admissible parametric representation of a curve, 14 algebraic curve, 19 κ–curve, 26 astroid, 31 Cassini curve, 27 cissoid, 21 conchoid, 24 Diocles’s cissoid, 21 double egg line, 30 lemniscate, 27 MacLaurin’s trisectrix, 23 Nikomedes’s conchoid, 24 Pascal’s snail, 25 rosette, 29 serpentine, 19 strophoid, 22 versiera, 20 analytic manifold, 375 angle between vectors in Vn , 4 angle preserving map, 325 anti–symmetric tensor, 388 arc length of a curve between two of its points, 33

area preserving map, 346 astroid, 31 asymptotic direction, 196 asymptotic line, 197 autoparallel curve, 438 autoparallelism of geodesic lines, 421 autoprallel curve natural parameter, 439 basis transformation, 377 Beltrami differentiator of first order, 415 Bianchi identities, 413 bilinear form, 381 binormal surface of a curve, 141 binormal vector of a curve, 47 Bour’s theorem, 338 Cassini curve, 27 catenary, 16 catenoid, 123 Cauchy–Schwarz inequality, 4 centre of curvature, 63 chain rule, 12 Christoffel symbols, 393 first, 393 second, 393 circular cone, 121 circular cylinder, 121 cissoid, 21 Diocles’s cissoid, 21 MacLaurin’s trisectrix, 23 strophoid, 22 Clairaut’s theorem, 281 closure of a set, 10 clothoid, 94 Collignon projection, 348 component function, 10 components of a vector, 3 conchoid, 24 Nikomedes’s conchoid, 24 Pascal’s snail, 25 conformal map, 325 conjugate harmonic functions, 369 conjugate harmonic minimal surfaces, 369

Index ■ 461

conjugated tensors, 389 conoid, 142 continuity of a function at a point, 11 continuity of a function on a set, 11 contiunity (everywhere), 11 contraction of a tensor, 387 contravariant tensor of second order, 382 contravariant vector, 381 coordinates of a point, 2 covariant derivative of a (1, 1)–tensor, 399 covariant derivative of a contravariant vector, 397 covariant derivative of a covariant vector, 398 covariant derivative of an (r, s)–tensor, 400 covariant tensor of second order, 382 covariant vector, 381 curvature of a curve, 47 curve, 376 m times continuously differentiable, 376 arc length, 33 binormal vector, 47 clothoid, 94 curvature, 47 frame, 47 intrinsic equations, 82 line of constant slope, 84 local canonical form, 61 logarithmic spiral, 94 loxodrome, 125 natural equations, 82 piecewise smooth curve, 376 principal normal vector, 47 radius of curvature, 47 same tangent vector, 376 smooth curve, 376 spherical curve, 83 straightest curve, 314 tangent vector, 45 tangent vector to a curve at a point, 45 torsion, 54

tractrix, 45 tridedron, 47 vector of curvature, 47 curve in Rn , 13 admissible parametric representation, 14 end point, 13 initial point, 13 orientation, 13 parameter, 13 parameter interval, 13 parametric representation, 13 curve on a surface, 107 geodesic curvature, 131 normal curvature, 131 cycloid, 38 epicycloid, 38 hypocycloid, 38 ordinary cycloid, 38 derivation formulae of Gauss, 320 derivation formulae of Mainardi and Codazzi, 320 derivative of f⃗ at t0 , 11 derivative of a function, 12 developable ruled surface, 212 differentiability (everywhere), 12 differentiability of a function on a set, 11 differentiable at a point, 11 distance of a point and a line, 7 distance of a point and a plane, 8 divergence of a contravariant vector, 416 divergence of a covariant vector, 417 domain of a function, 10 double egg line, 30 dual basis, 380 dual space of a vector space, 379 Dupin’s indicatrix of a surface at a point, 195 Dupin’s theorem, 222 Einstein’s convention of summation, 108 ellipsoid, 179 elliptic point of a surface, 186 elliptic pseudo-spherical surface, 159

462 ■ Index

elliptic spherical surface, 157 end point of a curve, 13 Enneper minimal surface, 363 epicycloid, 38 Euclidean distance of points, 2 Euclidean metric, 2 Euclidean norm, 3 Euler’s theorem, 183 evolute, 76 excess, 352 explicit surface, 119 extended complex plane, 328 first first first first

Christoffel symbols, 246, 393 countable at x, 374 countable topological space, 374 fundamental coefficents of a surface, 116 first fundamental form of a surface, 116 flat point of a surface, 191 frame of a curve, 47 Frenet’s formulae, 53 fundamental lemma of calculus of variation, 293 fundamental theorem of curves, 79 fundamental theorem of the theory of surfaces, 426 Gauss derivation formulae, 321 Gauss equations, 131 Gauss’s first fundamental coefficients, 116 Gauss–Bonnet theorem, 350 Gaussian curvature of a surface, 153 general cone, 200 general screw surface, 142 geodesic n–gon, 352 geodesic curvature, 131 geodesic field, 302 geodesic line, 263 geodesic parallel coordinates, 305 geodesic parameters, 300 geodesic polar coordinates, 310 geodesically parallel, 300, 313 Grassmann identity, 6

Hausdorff space, 374 helicoid, 142 helix, 14 higher order derivative, 12 holonomy group, 433 homeomorphism, 375 hyperbolic paraboloid, 173 hyperbolic plane, 268 hyperbolic point of a surface, 186 hyperbolic pseudo-spherical surface, 158 hyperboloid of one sheet, 136 hyperboloid of two sheets, 170 hypocycloid, 38 image of a function, 10 indicator function, 391 initial point of a curve, 13 inner product, 3, 384 inner product of tensors, 387 interior point, 374 intrinsic equation of the catenary, 17 intrinsic equations of a curve, 82 involute, 70 isometric map, 336 isomorphic vector spaces, 379 isothermal parameters, 332 Jacobian, 108 Lambert projection, 347 Laplacian operator, 418 lemniscate, 27 length of a vector, 3 length preserving map, 336 level lines, 416 limit of a vector function, 10 line of constant slope, 84 line of curvature, 197 linear functional, 379 local canonical form of a curve, 61 logarthmic spiral, 94 loxodrome, 125 MacLaurin’s trisectrix, 23 manifold, 375

Index ■ 463

admissible coordinate transformation, 376 analytic, 375 open covering, 375 manifold of class C ∞ , 375 mean curvature of a surface, 153 Mercator projection, 330 meridian of a screw surface, 144 metric coefficients, 116 metric tensor, 391 Meusnier’s theorem, 145 minimal surface, 353 conjugate harmonic minimal surfaces, 369 mixed Riemann tensor of curvature, 405 mixed tensor of second order, 382 Moebius strip, 204 monkey saddle, 193 natural equations of a curve, 82 natural parameter for autoparallel curves, 439 natural parameter of a curve, 35 Neil’s parabola, 15 Nikomedes’s conchoid, 24 normal curvature, 131 normal plane, 60 normal section of a surface, 145 normal vector of a plane, 8 normal vector of a straight line, 7 open covering of a manifold, 375 ordinary cycloid, 38 orientation on a curve, 13 orthogonal parameters, 116 orthogonal trajectories, 302 orthogonal trajectories of a curve, 32 orthogonal vectors, 4 orthonormal basis, 378 orthonormal vectors, 4, 384 osculating circle, 63 osculating plane, 60 osculating sphere, 65 outer product in V3 , 5 outer product of tensors, 386

parabolic point of a surface, 186 parabolic pseudo-spherical surface, 157 parallel curve, 305 parallel movement by Levi–Civit`a, 421 parallel surface, 234 parameter interval of a curve, 13 parameter lines, 107 parameter of a curve, 13 parameters of a surface, 106 parametric representation of a curve, 13 parametric represetation of a surface, 106 Pascal’s snail, 25 path connected, 433 planar point of a surface, 191 plane involute, 74 Plateau’s problem, 353 Poincar´e half–plane, 268 points, 2 coordinates, 2 Euclidean distance of points, 2 polar coordinates, 109 position vector of a point, 3 potential equation, 360 principal circle lines, 270 principal curvature at a point of a surface, 150 principal directions at a point of a surface, 150 principal normal surface of a curve, 141 principal normal vector of a curve, 47 projection vector, 5 pseudo-sphere, 155 pseudo-spherical surface, 157 elliptic pseudo-spherical surface, 159 hyperbolic pseudo-spherical surface, 158 parabolic pseudo-spherical surface, 157 quantities of the internal geometry, 252 quantities of the intrinsic quantity, 252 radius of curvature of a curve, 47 range of a function, 10

464 ■ Index

spherical surface, 157 elliptic spherical surface, 157 hyperbolic spherical surface, 157 square of the length, 384 stereographic projection, 328 straightest curve, 314 strophoid, 22 sum of tensors, 386 sum of vectors, 3 surface, 106 u1 –lines, 107 u2 –lines, 107 admissible parameter transformation, 108 asymptotic direction, 196 asymtotic lines, 197 same parameter systems, 324 curve on a surface, 107 scalar multiple of a vector, 3 developable ruled surface, 212 scalar product, 3 Dupin’s indicatrix, 195 Scherk minimal surface, 367 ellipsoid, 179 screw surface, 142 elliptic point, 186 meridian, 144 explicit surface, 119 second Christoffel symbols, 247, 393 first fundamental coefficients, 116 second fundamental coefficients of a first fundamental form, 116 surface, 132 flat point, 191 second fundamental form, 186 Gauss’s first fundamental second order derivative, 12 coefficients, 116 serpentine, 19 Gaussian curvature, 153 special curves, 15 general cone, 200 catenary, 16 general screw surface, 142 Neil’s parabola, 15 geodesic lines, 263 sphere, 111, 122 hyperbolic paraboloid, 173 meridians of a sphere, 112 hyperbolic point, 186 parallel of a sphere, 112 hyperboloid of one sheet, 136 spherical coordinates, 112 hyperboloid of two sheets, 170 spherical coordinates, 112 lines of curvature, 197 spherical curve, 83 mean curvature, 153 spherical Gauss map, 238 metric coefficients, 116 spherical image, 238 monkey saddle, 193 spherical image of the binormal vector, normal section, 145 99 normal vector, 114 spherical image of the principal normal parabolic point, 186 vector, 99 parallel surface, 235 spherical image of the tangent vector, 99 parameter lines, 107 spherical indicatrix, 141 parameters, 106 rectifying plane, 60 relative topology, 374 Ricci identity, 404, 406 Riemann space, 391 Riemann sphere, 328 Riemann tensor of curvature, 321 Rodrigues formulae, 229 rosette, 29 ruled surface, 140 binormal surface, 141 conoid, 142 helicoid, 142 principal normal surface, 141 spherical indicatrix, 141 tangent surface, 141

Index ■ 465

mixed Riemann tensor of curvature, 405 outer product V = T ∗ U of tensors, 386 sum of tensors, 386 symmetric, 388 tensor of second order, 382 conjugated, 389 contravariant, 382 covariant, 382 mixed, 382 symmetric covariant, 384 the Bertrand–Puiseux theorem, 443 Theorem of Beltrami and Enneper, 231 Theorem of Liouville, 260 theorema egregium (Gauss), 321 third fundamental coefficients, 234 topological space, 374 S–open, 374 continuity of a function, 375 continuity of a function at a point, 375 continuity of a function on a set, 375 first countable, 374 first countable at a point, 374 Hausdorff space, 374 homeomorphism, 375 topology, 374 closed set, 374 local base, 374 open set, 374 tangent plane of a surface at a point, 114 relative topology, 374 tangent surface, 70 torse, 212 tangent surface of a curve, 141 torsion of a curve, 54 tangent to a curve, 45 torus, 122 tangent vector, 45 total curvature of a surface, 351 Taylor’s formula, 13 tractrix, 45 tensor, 384 trihedron of a curve, 47 (r, s)–tensor, 384 triple orthogonal system of surfaces, 221 r–contravariant, 384 umbilical point, 153 s–covariant, 384 unit vector, 3 anti–symmetric, 388 contraction, 387 vector function, 10 inner product, 387 component function, 10 metric tensor, 391 parametric representation, 106 planar point, 191 principal curvature, 150 principal directions, 150 pseudo-sphere, 155 ruled surface, 140 screw surface, 142 second fundamental coefficients, 132 surface area, 117 surface of revolution, 119 tangent plane, 114 third fundamental coefficients, 234 torse, 212 umbilical point, 153 surface area, 117 surface normal vector at a point, 114 surface of revolution, 119 catenoid, 123 circular cone, 121 circular cylinder, 121 meridian, 120 parallel, 120 pseudo-sphere, 155 sphere, 122 torus, 122 surface vector, 419 surface vector at a point, 312 symmetric covariant tensor of second order, 384 symmetric tensor, 388

466 ■ Index

continuous (everywhere), 11 continuous at a point, 11 continuous on a set, 11 derivative, 12 derivative at a point, 11 differentiability on a set, 11 differentiable (everywhere), 12 differentiable at a point, 11 domain, 10 higher order derivative, 12 image of a function, 10 limit, 10 range, 10 second order derivative, 12 vector of curvature, 47 vector product in V3 , 5

vector–valued function of a real variable, 10 vectors, 3 components, 3 Euclidean norm, 3 length, 3 orthogonal, 4 orthonormal, 4 projection vector, 5 scalar multiple of a vector, 3 sum of vectors, 3 versiera, 20 Weierstrass equations, 361 Weingarten equations, 228