Finding Ellipses (Carus Mathematical Monographs) 147044383X, 9781470443832

Table of contents :
Cover
Title page
Copyright
Contents
Preface
Part 1
Chapter 1. The Surprising Ellipse
Chapter 2. The Ellipse Three Ways
Chapter 3. Blaschke Products
Chapter 4. Blaschke Products and Ellipses
Chapter 5. Poncelet’s Theorem for Triangles
Chapter 6. The Numerical Range
Chapter 7. The Connection Revealed
Intermezzo
Chapter 8. And Now for Something Completely Different… Benford’s Law
Part 2
Chapter 9. Compressions of the Shift Operator: The Basics
Chapter 10. Higher Dimensions: Not Your Poncelet Ellipse
Chapter 11. Interpolation with Blaschke Products
Chapter 12. Poncelet’s Theorem for ?-Gons
Chapter 13. Kippenhahn’s Curve and Blaschke’s Products
Chapter 14. Iteration, Ellipses, and Blaschke Products
On Surprising Connections
Part 3
Chapter 15. Fourteen Projects for Fourteen Chapters
15.1. Constructing Great Ellipses
15.2. What’s in the Envelope?
15.3. Sendov’s Conjecture
15.4. Generalizing Steiner Inellipses
15.5. Steiner’s Porism and Inversion
15.6. The Numerical Range and Radius
15.7. Pedal Curves and Foci
15.8. The Power of Positivity
15.9. Similarity and the Numerical Range
15.10. The Importance of Being Zero
15.11. Building a Better Interpolant
15.12. Foci of Algebraic Curves
15.13. Companion Matrices and Kippenhahn
15.14. Denjoy–Wolff Points and Blaschke Products
Bibliography
Index
Back Cover

Citation preview

AMS / MAA

THE CARUS MATHEMATICAL MONOGRAPHS

Finding Ellipses What Blaschke Products, Poncelet’s Theorem, and the Numerical Range Know about Each Other

Ulrich Daepp Pamela Gorkin Andrew Shaffer Karl Voss

VOL 34

Finding Ellipses What Blaschke Products, Poncelet’s Theorem, and the Numerical Range Know about Each Other

AMS/MAA

THE CARUS MATHEMATICAL MONOGRAPHS

VOL 34

Finding Ellipses What Blaschke Products, Poncelet’s Theorem, and the Numerical Range Know about Each Other

Ulrich Daepp Pamela Gorkin Andrew Shaffer Karl Voss

Committee on Books Jennifer J. Quinn, Chair 2017–2018 Editorial Committee Fernando Gouvea, Editor Francis Bonahon Alex Iosevich Kristin Estella Lauter Steven J. Miller

Gizem Karaali David P. Roberts

2010 Mathematics Subject Classification. Primary 47A05, 47A12, 30J10, 15-02, 15A60, 51-02, 51M04, 51N35. For additional information and updates on this book, visit www.ams.org/bookpages/car-34 Library of Congress Cataloging-in-Publication Data Names: Daepp, Ulrich, author. Title: Finding ellipses: What Blaschke products, Poncelet’s theorem, and the numerical range know about each other / Ulrich Daepp [and three others]. Description: Providence, Rhode Island: MAA Press, An imprint of the American Mathematical Society, [2018] | Series: The Carus mathematical monographs; volume 34 | Includes bibliographical references and index. Identifiers: LCCN 2018021655 | ISBN 9781470443832 (alk. paper) Subjects: LCSH: Ellipse. | Conic sections. | Geometry, Projective. | AMS: Operator theory – General theory of linear operators – General (adjoints, conjugates, products, inverses, domains, ranges, etc.). msc | Operator theory – General theory of linear operators – Numerical range, numerical radius. msc | Functions of a complex variable – Function theory on the disc – Blaschke products. msc | Linear and multilinear algebra; matrix theory – Research exposition (monographs, survey articles). msc | Linear and multilinear algebra; matrix theory – Basic linear algebra – Norms of matrices, numerical range, applications of functional analysis to matrix theory. msc | Geometry – Research exposition (monographs, survey articles). msc | Geometry – Real and complex geometry – Elementary problems in Euclidean geometries. msc | Geometry – Analytic and descriptive geometry – Questions of classical algebraic geometry. msc Classification: LCC QA559 .F56 2018 | DDC 516/.152–dc23 LC record available at https://lccn.loc.gov/2018021655 Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy select pages for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for permission to reuse portions of AMS publication content are handled by the Copyright Clearance Center. For more information, please visit www.ams.org/publications/pubpermissions. Send requests for translation rights and licensed reprints to [email protected].

© 2018 by the author. All rights reserved. Printed in the United States of America. ∞ The paper used in this book is acid-free and falls within the guidelines ⃝

established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

23 22 21 20 19 18

Contents Preface

vii

Part 1

1

Chapter 1 The Surprising Ellipse

3

Chapter 2 The Ellipse Three Ways

13

Chapter 3 Blaschke Products

23

Chapter 4 Blaschke Products and Ellipses

35

Chapter 5 Poncelet’s Theorem for Triangles

47

Chapter 6 The Numerical Range

61

Chapter 7 The Connection Revealed

75

Intermezzo

85

Chapter 8 And Now for Something Completely Different. . . Benford’s Law

87

Part 2

101

Chapter 9 Compressions of the Shift Operator: The Basics

103

Chapter 10 Higher Dimensions: Not Your Poncelet Ellipse

121

Chapter 11 Interpolation with Blaschke Products

133

Chapter 12 Poncelet’s Theorem for 𝑛-Gons

147

v

vi

Contents

Chapter 13 Kippenhahn’s Curve and Blaschke’s Products

159

Chapter 14 Iteration, Ellipses, and Blaschke Products

177

On Surprising Connections

195

Part 3

199

Chapter 15 Fourteen Projects for Fourteen Chapters 15.1 Constructing Great Ellipses 15.2 What’s in the Envelope? 15.3 Sendov’s Conjecture 15.4 Generalizing Steiner Inellipses 15.5 Steiner’s Porism and Inversion 15.6 The Numerical Range and Radius 15.7 Pedal Curves and Foci 15.8 The Power of Positivity 15.9 Similarity and the Numerical Range 15.10 The Importance of Being Zero 15.11 Building a Better Interpolant 15.12 Foci of Algebraic Curves 15.13 Companion Matrices and Kippenhahn 15.14 Denjoy–Wolff Points and Blaschke Products

201 201 201 206 210 213 222 224 228 231 234 237 241 245 251

Bibliography

255

Index

263

Preface [Mathematics] is security. Certainty. Truth. Beauty. Insight. Structure. Architecture. I see mathematics, the part of human knowledge that I call mathematics, as one thing—one great, glorious thing. Whether it is differential topology, or functional analysis, or homological algebra, it is all one thing. . . . They are intimately interconnected, and they are all facets of the same thing. That interconnection, that architecture, is secure truth and is beauty. That’s what mathematics is to me. –Paul Halmos, Celebrating 50 Years of Mathematics Discovering surprising connections between one area of mathematics and another area is not only exciting, it is useful. Laplace transforms allow us to switch between differential equations and algebraic equations, the fundamental theorem of algebra can be proved using complex analysis, and group theory plays a fundamental role in the study of public-key cryptography. If we can think of an object from different angles, these different perspectives aid our understanding. It is in that vein that we have written this book: We look at three seemingly different stories and a hidden connection between them. The first of our stories is complex—that is, it is about complex function theory. In complex analysis, linear fractional transformations are among the first functions that we study. We understand their geometric and function-theoretic properties, and we know that they play an important role in the study of analytic functions. In the class of linear fractional transformations, the automorphisms of the unit disk stand out for the geometric and analytic insight they provide. The functions that we focus on are a natural generalization of the disk automorphisms; they are

vii

viii

Preface

finite products of these automorphisms and are called Blaschke products, in honor of the German geometer Wilhelm Blaschke. While we know, for example, that automorphisms of the disk map circles to circles (if we agree that lines are circles), much less is known about the geometry of Blaschke products. The second story comes from projective geometry. Here we focus on Poncelet’s closure theorem, a beautiful geometric result about conics inscribed in polygons that are themselves inscribed in conics. We devote some attention to precursors of Poncelet’s theorem, including Chapple’s formula and Fuss’s theorem. In the final part of this book we will also consider variations on this theme, including Steiner’s porism. The final story is about the numerical range of a matrix. The numerical range is the range of a quadratic form associated with a matrix but restricted to the unit sphere of ℂ𝑛 . The numerical range always contains the eigenvalues of the matrix, and it often can tell you more about a matrix than the eigenvalues can. While questions about the numerical ranges of 2 × 2 matrices are (relatively) easily analyzed, consideration of 𝑛 × 𝑛 matrices for 𝑛 > 2 is deeper. Even 3 × 3 matrices lead to deep and interesting questions! What is the hidden connection between these three stories? Curiously, the connection is an object we have not discussed much yet: It is the ellipse. In fact, this book is really a story about the ellipse, an object studied by the Greek mathematicians Menaechmus, Euclid, Appolonius of Perga, and Pappus, among others. Later, Kepler developed a description of planetary motion, leading to his first law: The orbit of a planet is elliptical with one focus at the Sun. In spite of this long history, ellipses never cease to surprise us. It turns out that ellipses provide a remarkable amount of information about Blaschke products. Ellipses can tell us when a Blaschke product has a nontrivial factorization with respect to composition. And when we study the dynamics of Blaschke products, ellipses can tell us what to expect. The ellipse even establishes a connection between Blaschke products and a useful tool for detecting tax fraud, known as Benford’s law. But why does an ellipse know anything about these particular rational functions? That is the story we wish to tell, and it is a story that

Preface

ix

relies heavily on properties of the numerical range of a class of operators and that story, in turn, relies on projective geometry. The first part of this book (Chapter 1 through Chapter 7) focuses on Blaschke products that are products of three disk automorphisms, ellipses that are inscribed in triangles, and 2 × 2 matrices. In this case, ellipses suggest directions to study. Each Blaschke product of degree 3 is naturally associated with an ellipse, as is each 2 × 2 matrix. The relationship between Blaschke products, 2 × 2 matrices, and Poncelet’s theorem will be revealed in Chapter 7. This part of the book also introduces the basic ideas of two-dimensional projective geometry and provides a proof of Poncelet’s theorem in the event that the circumscribing polygons are triangles. After concluding Part 1, the reader will be treated to an intermezzo (Chapter 8), as we focus on the surprising connection between Poncelet’s theorem and Benford’s law. In the second part of the book (Chapter 9 through Chapter 14) we consider the connection between ellipses and Blaschke products that are products of more than three analytic disk automorphisms. We will see that though the boundaries of the numerical ranges of the matrices we study need not be elliptical, the boundaries always satisfy a Poncelet-like property. In addition, when the boundary is elliptical, it provides insight into the function-theoretic behavior of the corresponding Blaschke product. Because Poncelet’s ideas are intimately connected to our presentation, we provide a recent and beautiful proof of the general theorem. This part of the book is also a bit more challenging than the first part, mathematically speaking, but we strive for a self-contained treatment, providing appropriate proofs and references. Finally, the third part of the book provides a range of exercises and projects, from straightforward exercises for someone first entering the field to potential research problems. Each chapter has a corresponding project that is usually introduced with new material. To be prepared to work through a project, active reading is required. Projects include things we find particularly interesting; for example, we discuss Sendov’s conjecture, interpolation, inversion, and the spectral radius. In addition, we provide suggestions for the creation of several algorithms that will enhance one’s understanding. We hope that the reader will pick one

x

Preface

(or more) of these projects and use it (or them) to develop a deeper appreciation of the subject, to write an honors thesis, or to begin a research project. The necessary background and a starter bibliography for each project is provided, but we encourage readers to do a thorough search of the literature. Using this book. The first two parts of the book are meant to tell the story we have just described; a story that is meant to be read. However, this book can be used in various ways. It can serve as a reference for independent study, as the text for a capstone course in mathematics, or as a reference for a researcher. We have written this book for an active reader—paper and pencil in hand—and though there are a few exercises embedded in this reading, it is Part 3 of the text that includes a significant number of exercises as well as projects. These projects provide a range of both material and depth by including exercises well within the reach of every reader, suggested research papers, and research problems that have been open for many years. Applets. There are several interactive applets designed to go along with the book, and we encourage using the applets while reading. If we feel that the time is ripe for experimentation, we will direct the reader to the applet using the symbol ,.1 Our applets illustrate various results in the text, such as the main result on the connection of Blaschke products to ellipses. The proof of Poncelet’s theorem that we present depends on Pascal’s and Brianchon’s theorem, and we provide applets associated with each of these results. In addition, we have applets that illustrate the mapping behavior of Blaschke products (including locating fixed points in the closed unit disk), composition with Blaschke products, and these applets also produce the numerical range of particular matrices. It has been our experience that these tools are not only fun to work with, they also enhance one’s understanding and are great for testing conjectures. Acknowledgments. The authors wish to thank the MAA Editorial Board of the Carus Series for many suggestions that have improved the manuscript. We also thank the editors at the AMS for their invaluable help preparing this book. We are grateful to Bucknell University for its 1 http://pubapps.bucknell.edu/static/aeshaffer/v1/

Preface

xi

support and to the many students we have had who have played an important role in the creation of this book. In particular, we wish to thank Robert C. Rhoades, Benjamin Sokolowsky, Nathan Wagner, and Beth Skubak Wolf. We are also grateful to Kelly Bickel, David Farmer, Gunter Semmler, Elias Wegert, and Jonathan Partington for valuable discussions and to Bucknell L&IT for technical support. Pamela Gorkin’s work was supported by Simons Foundation Collaboration Grant 243653.

Part 1 A circle no doubt has a certain appealing simplicity at the first glance, but one look at an ellipse should have convinced even the most mystical of astronomers that the perfect simplicity of the circle is akin to the vacant smile of complete idiocy. Compared to what an ellipse can tell us, a circle has little to say. –Eric Temple Bell, Mathematics: Queen and Servant of Science, p. 277.

Chapter

1

The Surprising Ellipse The Ellipse, a 52-acre park located south of the White House in President’s Park, Washington D.C., is one of many geometric shapes located near the capital. According to a survey of the Ellipse (see [93]), the major axis is 1,058.26 feet and the minor axis is 902.85 feet. In case you are wondering, the area is 751,071.67 square feet and the perimeter is 3,086.87 feet. Is that enough for us to be certain that the creators intended to design an ellipse? The Colosseum in Rome, located east of the Roman Forum, also appears to be an ellipse, but is it? Some architects contend that the Colosseum is an ellipse, while others suggest that it is a curve made of circular arcs that are then connected in a smooth fashion. This is actually the subject of an important debate, because it would tell us what the Romans knew about ellipses.1 What is an ellipse, anyway? There are many ways to introduce the ellipse. If we take a cone and slice it in such a way that we have a bounded (nondegenerate) slice, we obtain an ellipse. We might, instead, connect two points (the foci of the ellipse) with a string, stretch the string tightly, and, using a pen to keep the length of the string constant, sketch out our ellipse. When introduced this way we say that the ellipse is the locus of points in the plane for which the sum of the distances to two fixed points (the foci) is constant. The line segments with endpoints on the ellipse that pass through the center are called diameters. The most important of these are the diameter through the foci or major axis and the shortest diameter or minor 1 See [73, p. 37] which says that it is likely “the parallel sequence of ovals for the Colosseum” was “laid out, one by one, as combinations of circular arcs”, while http:// www.wdl.org/en/item/4243 (accessed 12/15/2017) tells us in no uncertain terms that “It forms an ellipse, measuring approximately 190 meters long by 155 meters wide”.

3

4

Chapter 1. The Surprising Ellipse

Figure 1.1. The Colosseum in Rome—an ellipse?

axis. In Cartesian coordinates, ellipses are given by the equation 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0, where 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, and 𝐹 are real numbers and 𝐵2 − 4𝐴𝐶 < 0. If we use complex numbers and think of the foci as 𝑤1 and 𝑤2 , assuming our string has length 2𝑟, where 𝑟 ≥ 0, the points on the ellipse are those complex numbers 𝑧 for which |𝑧 − 𝑤1 | + |𝑧 − 𝑤2 | = 2𝑟. When the two foci are the same, we have a circle of radius 𝑟. If 𝑟 = 0, then we have a single point, which we consider a (degenerate) ellipse. One of the reasons for the widespread use of ellipses in the sciences is the reflective property of an ellipse: A ball that travels along a ray from one focus will hit the ellipse and bounce off passing through the other focus. Both light and sound are influenced by this property—perhaps you have seen this exhibited in museums or whisper chambers; if you stand at one focus and whisper something to a friend at the other focus, your friend will hear even the quietest whisper. Another interesting consequence of this property is that ellipses are used in medicine to treat kidney stones and gallstones using a process called lithotripsy. Ellipses also

Chapter 1. The Surprising Ellipse

5

appear in the laws of planetary motion, architecture, acoustics, and optics. Given the widespread use of ellipses, we should understand them well. But most of us know much more about circles than we do about ellipses. Consider the following natural questions. What is the area bounded by a circle? That is well known; it is 𝜋𝑟2 , where 𝑟 is the radius of the circle. What is the area bounded by an ellipse? You may not have memorized this, but you can figure it out: The area stays the same if we rotate and shift our ellipse, so let us assume that our ellipse has its center at the origin (the point (0, 0)), semi-major axis of length 𝑎, and semi-minor axis of length 𝑏. Then we can write the equation of our ellipse as 𝑥2 𝑦2 + = 1. 𝑎2 𝑏2 So we can find the area bounded by an ellipse if we can recall how to find the area bounded by 𝑦2 𝑎 √1 − 2 𝑏 and the 𝑦-axis. This gives us a chance to use calculus: Trigonometric substitution and a calculation that we are pretty sure you would like to do yourself tell us that the area bounded by our ellipse is 𝜋𝑎𝑏. Or, if ⃗ 𝑦) = you prefer, you can use Green’s theorem with the vector field 𝐹(𝑥, (1/2)[−𝑦 𝑥] to get the same result. If 𝑎 = 𝑏, then we have a circle of radius 𝑟 = 𝑎 = 𝑏 and we have the correct area, 𝜋𝑟2 ; that is comforting. What about the perimeter of a circle? That is well known too. It is 2𝜋𝑟. How about the perimeter, 𝑃, of an ellipse? We will give you a few minutes to think about it. You might try using the formula for the ellipse above, or you might try parametrizing the ellipse as 𝑥 = 𝑎 cos 𝜃, 𝑦 = 𝑏 sin 𝜃 for 0 ≤ 𝜃 < 2𝜋. No matter how you approach it though, you will find it to be a challenging problem. Using calculus once again you will discover that you need to compute 2𝜋 2

𝑃 = ∫ (𝑎2 sin 𝜃 + 𝑏2 cos2 𝜃)1/2 𝑑𝜃. 0

6

Chapter 1. The Surprising Ellipse

Though you have learned lots of integration techniques, you will probably soon come around to the thinking expressed in the website numericana.com. 2 “There is no simple exact formula: There are simple formulas but they are not exact, and there are exact formulas but they are not simple”.3

Figure 1.2. © by Sidney Harris, ScienceCartoonsPlus.com

The search for a formula (as well as estimates) for the perimeter of an ellipse has a fascinating history; a history that involves mathematicians like Gauss, Ramanujan, and—not too surprisingly—Kepler. An estimate for the perimeter of an ellipse was desirable because of its connection to the elliptical orbits of planets. In 1609 Kepler noted that 𝑃 ≈ 2𝜋√𝑎𝑏 and 𝑃 ≈ 𝜋(𝑎 + 𝑏). These were “the first legitimate approximations … although [Kepler’s] arguments were not very rigorous and 2𝜋√𝑎𝑏 was intended to be only a lower bound” for 𝑃 [4]. These approximations are best when 𝑎 is close to 𝑏. 2 http://numericana.com/answer/ellipse.htm#elliptic (accessed 12/15/2017) 3 Semjon Adlaj [1] disputes this statement in his article, An eloquent formula for the

perimeter of an ellipse.

Chapter 1. The Surprising Ellipse

7















 Figure 1.3. Four folds.

Returning to the Ellipse in Washington D.C. for a moment, you might check that the measurements are consistent with the claim that the Ellipse is actually an ellipse. In fact, Colin Maclaurin [105] was the first to publish an exact expression for the perimeter, now referred to as the Maclaurin expansion; the perimeter 𝑃 is given by 1 2 1 ⋅ 3 2 𝑘4 1 ⋅ 3 ⋅ 5 2 𝑘6 𝑃 = 2𝜋𝑎 [1 − ( ) 𝑘 2 − ( −( − ⋯] , ) ) 2 2⋅4 3 2⋅4⋅6 5 where 𝑘 2 = 1 − (𝑏2 /𝑎2 ). It is not likely that you guessed that! Perhaps Bell was referring to this rich history when he said, “Compared to what an ellipse can tell us, a circle has little to say” or he might have meant more than that. We will tell three stories about the ellipse here together for the first time: one from linear algebra, one from complex analysis, and one from geometry. But before we tell you what these

8

Chapter 1. The Surprising Ellipse

D

C

Figure 1.4. Many folds, using the applet.

stories are, we would like to present an interesting and, perhaps, unfamiliar way to construct an ellipse, a construction we first learned of in Martin Gardner’s book [53, pp. 173–183]. Here is how it works. Take out a piece of paper and choose (and label) two points, 𝑐 and 𝑑. Now using 𝑐 as your center, draw a circle of radius 𝑟 > |𝑐 − 𝑑|. Choose an arbitrary point, 𝑎1 , on your circle and fold your paper over until the point 𝑎1 lies on top of 𝑑. This will make a “crease” in your sheet, and you should make a fold there, creating a line segment. Choose another point, 𝑎2 , on your circle and repeat, always folding over to the point 𝑑. Figure 1.3 shows what happens when we do this four times. If you have a lot of time on your hands, you can do this with wax paper4 or you can obtain Figure 1.4 using our Ellipse Construction by Folding applet (we use this symbol ,5 to tell you to go to an applet). In this figure, it looks like all the folds made using the slash marks are 4 https://www.youtube.com/watch?v=psuTYtDfxPE

(accessed 12/15/2017)

5 http://pubapps.bucknell.edu/static/aeshaffer/v1/

Chapter 1. The Surprising Ellipse

9







 

 









Figure 1.5. Left: Point 𝑏 is on the ellipse. Right: Point 𝑥 is not on the ellipse.

tangent to the ellipse with the points 𝑐 and 𝑑 as foci. In the applet, you can drag one of the foci or enter new ones to see that this always happens. Why would that be? If we look at Figure 1.3 we see that the fold we made using points 𝑎1 and 𝑑 is the perpendicular bisector of the line segment 𝑑𝑎1 . That is the crucial observation, and we are ready to see why this yields an ellipse. Draw the fold obtained from folding point 𝑎 onto 𝑑 (see the picture on the left in Figure 1.5). Label the midpoint of the segment 𝑎𝑑 by 𝑚. Now draw the radius, 𝑎𝑐, and call the point of intersection of the fold and the radius 𝑏. We will show that 𝑏 lies on the ellipse with foci 𝑐 and 𝑑 and major axis of length 𝑟 by showing that |𝑏𝑑| + |𝑏𝑐| = 𝑟, where 𝑟 is the radius of the circle centered at 𝑐. Here is how to check that our lengths have the right property. The two triangles △𝑚𝑑𝑏 and △𝑚𝑎𝑏 are congruent because they share the side 𝑚𝑏, have right angles at 𝑚, and satisfy |𝑚𝑑| = |𝑚𝑎|; that is the “side-angle-side” argument you learned in geometry. In particular, 𝑏𝑑 has the same length as 𝑎𝑏. So if we look at the sum of the distances from 𝑏 to the two fixed points 𝑐 and 𝑑, we have |𝑏𝑐| + |𝑏𝑑| = |𝑏𝑐| + |𝑎𝑏| = 𝑟. In other words, the sum of the distances from the point 𝑏 to the two fixed points 𝑐 and 𝑑 is always the same constant, 𝑟.

10

Chapter 1. The Surprising Ellipse

If we now take any point on the fold different from 𝑏, call it 𝑥 (in the picture on the right in Figure 1.5), then the argument above shows that |𝑑𝑥|+|𝑐𝑥| = |𝑎𝑥|+|𝑐𝑥|. The points 𝑎, 𝑥, and 𝑐 are not collinear, so by the triangle inequality, we conclude that |𝑑𝑥| + |𝑐𝑥| > 𝑟. So the only point that is on our ellipse and the fold is the point 𝑏, and we see that the fold is a tangent line to the ellipse. This really does produce an ellipse! In fact, we can get every ellipse this way. (Think about how to do that.) There is another very important property hidden in this picture, and it will be useful later, so we isolate it as a proposition below. This proposition gives you what is often referred to as the optical or reflection property of an ellipse or what is sometimes more simply stated as “angle in equals angle out”. Proposition 1.1. Let 𝐸 be an ellipse with foci 𝑐 and 𝑑, and let 𝑏 be a point on 𝐸 that lies in the interior of the line segment 𝑧1 𝑧2 . Then the line containing 𝑧1 and 𝑧2 is tangent to 𝐸 if and only if ∠𝑐𝑏𝑧1 = ∠𝑧2 𝑏𝑑.

 




 








Figure 1.6. Tangents have equal angles with the lines to the foci.

Proof. We use the notation in Figure 1.6. Given the foci and a point on the ellipse 𝐸, the length of its major axis is determined; we call it 𝑟. Draw a circle 𝐶 with center 𝑐 and radius 𝑟.

Chapter 1. The Surprising Ellipse

11

The point of intersection of 𝐶 with the line through 𝑐 and 𝑏 is denoted by 𝑎. As shown above, the fold obtained by bringing 𝑎 on top of 𝑑 is a tangent line to this ellipse, 𝑚 is the midpoint of 𝑎𝑑, and the length of segment 𝑑𝑏 is equal to the length of 𝑎𝑏. Thus, the point 𝑏 is on the fold and the fold is the tangent line to 𝐸 through 𝑏. Now ∠𝑐𝑏𝑧1 and ∠𝑎𝑏𝑧2 are vertical angles, so they are equal. The triangles △𝑚𝑏𝑑 and △𝑚𝑏𝑎 are congruent since corresponding sides are of equal length. Thus, ∠𝑧2 𝑏𝑑 = ∠𝑚𝑏𝑑 = ∠𝑎𝑏𝑚 = ∠𝑎𝑏𝑧2 . Putting this together, we get ∠𝑧2 𝑏𝑑 = ∠𝑎𝑏𝑧2 = ∠𝑐𝑏𝑧1 , which is what we wanted to show. The converse follows from the fact that there is exactly one line through 𝑏 for which ∠𝑧2 𝑏𝑑 = ∠𝑐𝑏𝑧1 . Figure 1.4 suggests other questions, and we list a few here. Suppose that when we fold we think of one of the points of intersection of the fold and the circle as the initial point and the other as the starting point for the next fold. Try to fold so that the second fold starts where the first ends and the third starts where the second ends. Does something special happen? Here are some related questions: Scaling things, we may suppose the boundary of the disk is the unit circle. Then we can ask: When is an ellipse inscribed in a triangle that is inscribed in the unit circle? What about a quadrilateral? Or, more generally, what about a convex polygon? These are great questions, and they have great answers—answers that we present in the following chapters.

Chapter

2

The Ellipse Three Ways What can an ellipse tell us? We consider three answers to this question that, on the surface, appear to be quite different. Beneath that surface, however, lies a surprising connection between matrix theory, function theory, and projective geometry. Let us start with matrix theory and, in particular, with matrices that have complex entries. What does an ellipse know about a matrix? To tell this story, we need to review a bit of linear algebra and introduce the numerical range of a matrix. An introductory course in linear algebra covers eigenvalues but rarely studies the numerical range in depth—if at all. Yet the numerical range could easily be included as the definition relies only on some familiarity with inner products on ℂ𝑛 . So first recall that ℂ𝑛 consists of elements of the form 𝑥 = [𝑥1 𝑥2 … 𝑥𝑛 ]𝑇 with 𝑥𝑗 ∈ ℂ for all 𝑗. Now for 𝑥 and 𝑦 in ℂ𝑛 , we consider the standard inner product ⟨𝑥, 𝑦⟩, where 𝑛

⟨𝑥, 𝑦⟩ = ∑ 𝑥𝑗 𝑦𝑗 . 𝑗=1

In a general inner product space, the norm of 𝑥, or ‖𝑥‖, satisfies ‖𝑥‖2 = ⟨𝑥, 𝑥⟩. Thus, for 𝑥 ∈ ℂ𝑛 with the standard inner product, we get the Euclidean norm 1/2

𝑛 2

‖𝑥‖ = ( ∑ |𝑥𝑗 | )

.

𝑗=1

Eigenvalues and the numerical range are connected and each provides us with valuable information about a matrix. Recall that for an 13

14

Chapter 2. The Ellipse Three Ways

𝑛 × 𝑛 matrix 𝐴 = (𝑎𝑖𝑗 ) with complex entries, an eigenvalue of 𝐴 is a complex number 𝜆 for which there exists a nonzero vector 𝑥 such that 𝐴𝑥 = 𝜆𝑥. We call 𝑥 a corresponding eigenvector, but there are many of them; for example, 𝐴(𝑥/‖𝑥‖) = 𝜆(𝑥/‖𝑥‖) as well, so once we have an eigenvector we also have a corresponding unit eigenvector. (In fact, once we have one unit eigenvector we have many since all vectors 𝑒𝑖𝜃 𝑥/‖𝑥‖ with 𝜃 ∈ ℝ also work.) The vectors we use will usually have their entries in the complex plane, ℂ. So much for the familiar; let us now turn to the unfamiliar. The numerical range of an 𝑛 × 𝑛 matrix 𝐴 is defined by 𝑊(𝐴) = {⟨𝐴𝑥, 𝑥⟩ ∶ 𝑥 ∈ ℂ𝑛 , ‖𝑥‖ = 1}. The numerical range and eigenvalues are related: If 𝜆 is an eigenvalue of 𝐴 with corresponding unit eigenvector 𝑦, then ⟨𝐴𝑦, 𝑦⟩ = ⟨𝜆𝑦, 𝑦⟩ = 𝜆. So the eigenvalues are always in the numerical range. The numerical range of a matrix was a natural object to study in the early days of Hilbert space theory because there was great interest in quadratic forms, and the numerical range of 𝐴 is just the range of the quadratic form associated with 𝐴 restricted to the unit sphere. But sometimes the numerical range provides more information about a matrix than the eigenvalues do: If you fix a point 𝜆 ∈ ℂ, there are several matrices for which the set of eigenvalues consists of the single point 𝜆. But, as we will see in Theorem 6.2, there is only one matrix for which the numerical range is the set {𝜆}, and that matrix is a multiple of the identity matrix; that is, it is 𝜆𝐼. Example 2.1 presents a matrix for which the only eigenvalue is zero, but its numerical range is more than just a singleton. Example 2.1. Let 𝐴1 = [

0 0

1 ]. 0

What is the numerical range of 𝐴1 ? To find the answer to this question, let 𝑥 ∈ ℂ2 of norm 1 be an arbitrary vector. Then 𝑧 𝑥 = [ 1 ], 𝑧2 where ‖𝑥‖2 = |𝑧1 |2 + |𝑧2 |2 = 1. Thus, |𝑧1 | ≤ 1 and |𝑧2 | ≤ 1. Writing 𝑧1 = 𝑟 𝑒𝑖𝜃1 with 𝑟 = |𝑧1 | and 𝜃1 ∈ ℝ, we see that 0 ≤ 𝑟 ≤ 1 and

Chapter 2. The Ellipse Three Ways

15

𝑧2 = √1 − 𝑟2 𝑒𝑖𝜃2 for some appropriate choice of 𝜃2 ∈ ℝ. Thus, 𝑟 𝑒𝑖𝜃1 𝑥=[

]. √1 −

𝑟2 𝑒𝑖𝜃2

So letting 𝛾 = 𝑒𝑖(𝜃2 −𝜃1 ) , we have |𝛾| = 1 and ⟨𝐴1 𝑥, 𝑥⟩ = 𝛾𝑟√1 − 𝑟2 . Since 𝑥 is an arbitrary vector in ℂ2 , we may fix 𝑟 and let 𝛾 vary. Doing so, we get a circle of radius 𝑟√1 − 𝑟2 . Since 0 ≤ 𝑟√1 − 𝑟2 ≤ 1/2, we see that the numerical range of this matrix (and a very important matrix it is!) is the closed disk centered at 0 of radius 1/2 sketched in Figure 2.1.

Figure 2.1. The disk in Examples 2.1, 2.4, and 2.7.

Let us look at a more interesting example. Example 2.2. Let 0 √3/2 ]. 0 1/2 What is the numerical range of 𝐴2 ? 𝐴2 = [

16

Chapter 2. The Ellipse Three Ways

It is harder to figure out what happens here (which is why it is more interesting!). In fact, 𝑊(𝐴2 ) is the closed elliptical disk with foci at its eigenvalues, 0 and 1/2, minor axis of length √3/2, and major axis of length 1 sketched in Figure 2.2. Rather than show you the computations here, we use the following theorem, which we prove in Chapter 6: For a matrix 𝐴, let 𝐴⋆ denote the adjoint of 𝐴; that is, the matrix satisfying ⟨𝐴𝑥, 𝑦⟩ = ⟨𝑥, 𝐴⋆ 𝑦⟩ for all 𝑥, 𝑦 ∈ ℂ𝑛 .1 If 𝐴 = (𝑎𝑖𝑗 ), the trace of 𝐴 𝑛 is denoted 𝑡𝑟(𝐴) and defined by tr(𝐴) ∶= ∑𝑗=1 𝑎𝑗𝑗 . Then we have the following result [76, p. 109]. Theorem 2.3 (Elliptical range theorem). Let 𝐴 be a 2 × 2 matrix with eigenvalues 𝑎 and 𝑏. Then the numerical range of 𝐴 is an elliptical disk with foci at 𝑎 and 𝑏 and minor axis given by (tr(𝐴⋆ 𝐴) − |𝑎|2 − |𝑏|2 )1/2 . That is consistent with what we saw in Examples 2.1 and 2.2; we just need to think of the “two” foci as 𝑎 = 0 and 𝑏 = 0 and the ellipse as the circle of radius 1/2 centered at 0.

Figure 2.2. The elliptical disk in Examples 2.2, 2.5, and 2.8.

So the ellipse “knows” where the eigenvalues of the matrix are, but it knows more than that. Some of this is encoded in our second story, 1 If

this is unfamiliar, you should check that if 𝐴 = (𝑎𝑖𝑗 ), then 𝐴⋆ = (𝑎𝑗𝑖 ).

Chapter 2. The Ellipse Three Ways

17

one that at first blush seems most natural in this situation—geometry. But the theorem we will discuss here is one with which many people are not familiar. It is a beautiful and surprising theorem due to JeanVictor Poncelet. The theorem falls into a general class of results that try to understand when, given two conics, you can find a polygon that circumscribes the smaller and is simultaneously inscribed in the larger. In addition to the examples we saw in Chapter 1, you can imagine that there are many versions of such problems—perhaps you can imagine infinitely many versions! We will discuss just a few of these in Chapter 5. We state only the result we need here; a more general version will appear later. We are interested in ellipses that are inscribed in triangles that are inscribed in the unit circle 𝕋. Not all ellipses can be thus inscribed—you may have seen this when you used the folding method to obtain an ellipse—and one of our goals will be to characterize such ellipses. For now we concentrate on providing two examples. Example 2.4. Consider the circle 𝐸1 of radius 1/2 centered at 0. Show that if 𝛼 lies on the unit circle, then there is a triangle with one vertex at 𝛼 and all vertices on the unit circle circumscribing 𝐸1 ; that is, the edges of the triangle are each tangent to 𝐸1 . For each point of the unit circle, the symmetry of the situation shows that if a triangle exists, it is equilateral, and once we have a triangle corresponding to a particular point, say 𝛼 = 1, then an appropriate rotation of this triangle will work for other values of 𝛼. It is not difficult to check that the triangle with vertices 1, 𝑒𝑖2𝜋/3 , and 𝑒𝑖4𝜋/3 is tangent to the circle at the point 𝑧 = −1/2, and again, the symmetry of the situation makes it clear that this triangle circumscribes this circle. Looking at Figure 2.1 in this context, we are looking at triangles that circumscribe the circle. Example 2.5. Consider the ellipse 𝐸2 with foci 0 and 1/2 and major axis of length 1. Then for each point 𝛼 on the unit circle, there is a triangle circumscribing the ellipse 𝐸2 with vertices on the unit circle and one vertex at 𝛼. Lacking the symmetry present in Example 2.4, the justification of Example 2.5 is a fairly unpleasant computation, but it is an attractive application of a special case of Poncelet’s theorem. Rather than proving

18

Chapter 2. The Ellipse Three Ways

this result directly we apply Poncelet’s theorem, which we state here and prove in Chapter 5. Theorem 2.6 (Poncelet’s closure theorem for triangles). Let 𝐸 denote an ellipse entirely contained in a second ellipse 𝐹. If it is possible to find one triangle that is simultaneously inscribed in 𝐹 and circumscribed about 𝐸, then each point of 𝐹 is a vertex of one such triangle. Before we discuss Example 2.5, let us think about what Poncelet’s theorem is telling us. In Figure 2.3 on the left, the tangent lines we have drawn do not return to their starting point after three steps, so Poncelet’s theorem tells us that no matter where we start, we will not return to our starting point after three steps. On the other hand, we see that in the picture on the right the lines return to the starting point after three steps regardless of where we begin the process.







Figure 2.3. The ellipse 𝐸 on the left is never inscribed in a triangle with vertices on 𝐹, while the ellipse 𝐸 on the right always is.

In Example 2.5, the ellipse has Cartesian equation 1 2 16 4 (𝑥 − ) + 𝑦 2 = 1. 4 3 Consider the triangle with vertices at 𝑤1 = (1, 0), 𝑤2 = (−1/4, √15/4), and 𝑤3 = (−1/4, −√15/4). The line segment 𝑤2 𝑤3 touches the ellipse only at the point (−1/4, 0) and is therefore tangent to it at that point. If we check the line joining 𝑤1 to 𝑤2 we find that this is also tangent to the ellipse at the point (7/12, √15/12) and the final point of tangency follows again by symmetry. Thus, the ellipse is inscribed in a triangle, that triangle is inscribed in the unit circle, and we are looking at triangles that circumscribe the ellipse in Figure 2.2.

Chapter 2. The Ellipse Three Ways

19

Now that we know what happens at one point, Poncelet’s theorem says that at every point on 𝕋 there is a triangle circumscribing the ellipse that is itself inscribed in the unit circle and has one vertex at the given point. It should amaze you that once you have checked the behavior at one point on the outer ellipse, you know something about the behavior about every point on the outer ellipse. We now turn to our final set of examples: The class of finite Blaschke products, which were first introduced by the German mathematician Wilhelm Blaschke. As the word finite suggests, there are also infinite Blaschke products, and though they are very important, we will not discuss them at this time. (A brief discussion of infinite Blaschke products can be found on p. 111.) Thus, we refer only to Blaschke products, which are functions of a complex variable that have the form 𝑛

𝐵(𝑧) = 𝜇 ∏ 𝑗=1

𝑧 − 𝑎𝑗 1 − 𝑎𝑗 𝑧

,

where 𝑛 is a positive integer, 𝑎𝑗 ∈ 𝔻 for 𝑗 = 1, … , 𝑛, and 𝜇 ∈ 𝕋. The number of zeros of 𝐵, counted according to multiplicity, is called the degree of 𝐵; that is, the degree of 𝐵 is 𝑛. Taking 𝑎𝑗 = 0 for all 𝑗 yields special examples of Blaschke products, namely, 𝜇𝑧𝑛 with 𝜇 ∈ 𝕋, and we know how important these are. Now something truly magical will happen. There is nothing about these functions that suggests a relation—even a distant relation—to matrices or ellipses. Or is there? To see what happens we will mention the most basic properties of these functions, all of which can (and will—see Chapter 3 for the proofs) be shown easily. Blaschke products map the open unit disk 𝔻 to itself and the unit circle 𝕋 to itself; they are analytic on an open set containing the closed unit disk and have finitely many zeros in 𝔻. In fact, if an analytic function satisfies the first three of these conditions (which imply the fourth), it must be a Blaschke product—another reason why they are considered an important class of functions. We establish this characterization of finite Blaschke products in Lemma 3.2. Blaschke products have very nice mapping properties, and one such property is that a Blaschke product is an 𝑛-to-1 map of the unit circle

20

Chapter 2. The Ellipse Three Ways

onto itself. As a consequence of this the 𝑛 solutions to 𝐵(𝑧) = 𝜆 are distinct when 𝜆 ∈ 𝕋. What can an ellipse possibly know about a Blaschke product? Here the connection is well hidden but was recently discovered in [35]. Example 2.7. Let 𝐵1 denote the Blaschke product 𝐵1 (𝑧) = 𝑧3 . For each 𝜆 ∈ 𝕋, the three solutions of 𝐵1 (𝑧) = 𝜆 form the vertices of a triangle. Denote the closed region bounded by this triangle by 𝑇𝜆 . Then ⋂𝜆∈𝕋 𝑇𝜆 is a closed circular disk of radius 1/2 centered at the origin. This is easy to see if we think geometrically: For each 𝜆 = 𝑒𝑖𝜃 the three solutions to 𝑧3 = 𝑒𝑖𝜃 are equally spaced on the unit circle. Therefore, we are finding the intersection of all triangular regions, where our triangles are equilateral triangles with vertices on the unit circle. Evidently, this is a closed circular disk centered at the origin. What is the radius? We can choose any triangle to determine the radius, and the triangle with vertices 1, 𝑒𝑖2𝜋/3 , and 𝑒𝑖4𝜋/3 works nicely. The circle is tangent to this triangle on the line joining 𝑒𝑖2𝜋/3 and 𝑒𝑖4𝜋/3 at the point 𝑧 = −1/2. Thus, the radius is 1/2 and we are back to Figure 2.1 again. Example 2.8. Let 𝐵2 denote the Blaschke product 𝐵2 (𝑧) = 𝑧2 (𝑧 − 0.5)/(1 − 0.5𝑧) and 𝜆 ∈ 𝕋. By the discussion above, 𝐵2 is a 3to-1 map of the unit circle onto itself. Therefore, for each 𝜆 ∈ 𝕋, there are three distinct solutions to 𝐵2 (𝑧) = 𝜆, and they are the vertices of a triangle. Let 𝑇𝜆 denote the closed triangular region bounded by the triangle. Then ⋂𝜆∈𝕋 𝑇𝜆 is the region enclosed by the ellipse with foci at the points 0 and 0.5 and major axis of length 1. This second example is more difficult but can be computed with the aid of the following theorem, the proof of which will have to wait until Chapter 3. Theorem 2.9. [35] Let 𝐵 be a Blaschke product of degree 3 with zeros 0, 𝑎, and 𝑏. For 𝜆 ∈ 𝕋, let 𝑧1 , 𝑧2 , 𝑧3 denote the solutions of 𝐵(𝑧) = 𝜆. Then the lines joining 𝑧𝑗 and 𝑧𝑘 , for 𝑗 ≠ 𝑘, are tangent to the ellipse 𝐸 given by the equation |𝑤 − 𝑎| + |𝑤 − 𝑏| = |1 − 𝑎𝑏|. (2.1) Conversely, every point of 𝐸 is the point of tangency of the ellipse with a line segment joining two points on 𝕋 that 𝐵 identifies.

Chapter 2. The Ellipse Three Ways

21












Figure 2.4. A Blaschke ellipse with foci at 𝑎 and 𝑏 illustrating Theorem 2.9.

The equation of the ellipse, (2.1), makes it quite clear that given a point on the ellipse, the sum of the distances from each of the foci, 𝑎 and 𝑏, is a constant (|1 − 𝑎𝑏| in this case). It follows from Theorem 2.9 that ⋂𝜆∈𝕋 𝑇𝜆 is the closed elliptical disk with foci 0 and 1/2 and major axis of length 1. But there is only one ellipse with these foci and this major axis. Therefore, ⋂𝜆∈𝕋 𝑇𝜆 is the region bounded by the ellipse in Figure 2.2. So Examples 2.1, 2.4, and 2.7 produce the same ellipse, though in the first case the foci are the eigenvalues of a matrix, while in the last case the foci are two of the zeros of a Blaschke product. Similarly, Examples 2.2, 2.5, and 2.8 produce the same ellipse with foci at the eigenvalues of a particular matrix in the first case and foci at two of the zeros of the Blaschke product in the last case—yet there seems to be no apparent connection between numerical ranges of 2 × 2 matrices, Poncelet’s theorem, and degree-3 Blaschke products. Our goal is to show that the appearance of the ellipse, in each case, is not a coincidence but rather the result of deep and beautiful mathematical interconnections.

Chapter

3

Blaschke Products Blaschke products are the basic building blocks of analytic functions (holomorphic functions) on the open unit disk, 𝔻. So let us recall our definition of a Blaschke product and isolate it for future reference. Definition 3.1. A finite Blaschke product 𝐵 of degree 𝑛 is a function of the form 𝑛 𝑧 − 𝑎𝑗 𝐵(𝑧) = 𝜇 ∏ , 𝑗=1 1 − 𝑎𝑗 𝑧 where |𝜇| = 1 and 𝑎𝑗 ∈ 𝔻 for 𝑗 = 1, … , 𝑛. In particular, Blaschke products of degree 1 are precisely the disk automorphisms or bijective analytic maps from the disk 𝔻 to itself. (This is a consequence of Schwarz’s lemma; see [118, pp. 143–144].) As we move along in the text, we will be concerned primarily with the sets of points in 𝕋 that the Blaschke product identifies; in other words, the solutions of the equation 𝐵(𝑧) = 𝜆 for 𝜆 ∈ 𝕋. In this case, the constant 𝜇 adds nothing new to the discussion, so we often take 𝜇 = 1 in Definition 3.1. When this is not clear from the context, we will refer to the normalized Blaschke product or state explicitly that we take 𝜇 = 1. Up to the constant 𝜇, Blaschke products are defined by their zeros in 𝔻 and, if an analytic function 𝑓 has the property that |𝑓(𝑧)| ≤ 1 for all 𝑧 ∈ 𝔻, then 𝑓 can be approximated uniformly on compact subsets of 𝔻 by finite Blaschke products.1 Finite (and infinite) Blaschke products capture the zeros of bounded analytic functions thereby encoding a lot of information about such functions. To see what they can tell us in the 1 This

last fact is Carathéodory’s theorem and can be found in [54, p. 6].

23

24

Chapter 3. Blaschke Products

situation we described in Chapter 2, let us look at Example 2.4 more closely. Example 2.4. Consider the circle 𝐸1 of radius 1/2 centered at 0. Show that if 𝛼 lies on the unit circle, then there is a triangle with one vertex at 𝛼 and all vertices on the unit circle circumscribing 𝐸1 ; that is, the edges of the triangle are each tangent to 𝐸1 . This was easy to prove because we realized that the symmetry of the problem implied that we had to divide the unit circle into three equal pieces. Since we know the perimeter of the unit circle, if we start at a point 𝑒𝑖𝜃 the next point that we look for is 𝑒𝑖(𝜃+2𝜋/3) . The point after that is 𝑒𝑖(𝜃+4𝜋/3) and the last one is 𝑒𝑖(𝜃+6𝜋/3) . We are back to where we started as we should be. Simple, right? Now let us take a look at our Example 2.5. Example 2.5. Consider the ellipse 𝐸2 with foci 0 and 1/2 and major axis of length 1. Then for each point 𝛼 on the unit circle, there is a triangle circumscribing the ellipse 𝐸2 with vertices on the unit circle and one vertex at 𝛼. This is closer to Example 2.4 than one might think. The idea is this: We draw one chord of 𝕋 that is tangent to the ellipse, look at where we end up, and then draw another chord from that point that is also tangent to the ellipse. Since we are just repeating this action and hoping to get back to where we started we can imagine that, in some sense, we still want to divide the circle into equal pieces—we just do not want to measure length in the usual way. There are many ways to measure length, so we need to be more creative when we decide how to measure the three arcs of 𝕋 determined by the vertices of the triangle. Given an ellipse 𝐸 with one circumscribing triangle that has all vertices on 𝕋, perhaps a function that maps arcs between vertices to curves of equal length will help us find all triangles circumscribing 𝐸. Is there such a function? The answer is yes (the measure will appear later in Equation (8.5)), and the functions that do this are our Blaschke products. Before turning to an in-depth discussion of these functions, we consider a geometric question about polynomials in the spirit of those we

Chapter 3. Blaschke Products

25

discussed in Chapters 1 and 2. While Blaschke products are not (in general) polynomials, this question will lead to a second one that will show one of the many reasons we study Blaschke products. Suppose we choose 𝑛 equally spaced points 𝑎𝑗 on the unit circle, starting with the point 𝑎0 = 1; see Figure 3.1. For 𝑗 = 1, … , 𝑛 − 1, form the chords that join 𝑎𝑗 to 𝑎0 . What is the product of the lengths of these chords? And what is the connection to our discussion here?






















Figure 3.1. Eight equally spaced points on 𝕋.

The points 𝑎𝑗 , for 𝑗 = 1, … , 𝑛 − 1, together with the point 𝑎0 = 1 are equally spaced on the unit circle, so we have the 𝑛 roots of unity; that is, the points 𝑧 for which 𝑧𝑛 = 1. Thus, 𝑧𝑛 − 1 = (𝑧 − 𝑎0 )(𝑧 − 𝑎1 ) ⋯ (𝑧 − 𝑎𝑛−1 ). If we evaluate at 𝑧 = 1 we see the product we want, namely, (1 − 𝑎1 ) ⋯ (1 − 𝑎𝑛−1 ), multiplied by 0. Of course, multiplying what you want by 0 is never a good idea, so let us divide by 𝑧 −1 before we evaluate at the point 1. We get 𝑧𝑛 − 1 = (𝑧 − 𝑎1 ) ⋯ (𝑧 − 𝑎𝑛−1 ). 𝑧−1

(3.1)

26

Chapter 3. Blaschke Products

The left-hand side of (3.1) is screaming “derivative of 𝑧𝑛 at 𝑧 = 1”, so we take the limit as 𝑧 → 1 (or just do the algebra) and we find out that 𝑛 = (1 − 𝑎1 ) ⋯ (1 − 𝑎𝑛−1 ). We have discovered that the product of the lengths of the chords that we are interested in is 𝑛. (If you found this interesting, we recommend [126] for related results.) That a polynomial with all roots on 𝕋 knows a lot about 𝕋 is not too surprising and suggests that we should invest some time looking at these functions. But polynomials are important for many other reasons, not the least of which are their approximation properties. For example, one form of the Stone–Weierstrass theorem says that any function continuous on the closed unit disk, 𝔻, can be uniformly approximated on 𝔻 by polynomials. What can we approximate with polynomials that have all zeros on the unit circle? A partial answer to this question lies in a result due to Rubinstein [134] that we will soon present in Theorem 3.3. In much of what follows, we will have the opportunity to use the maximum modulus theorem, which we recall says the following: If 𝑓 is an analytic function on a connected open set 𝑈 ⊆ ℂ and there exists 𝑎 ∈ 𝑈 with |𝑓(𝑎)| ≥ |𝑓(𝑧)| for all 𝑧 ∈ 𝑈, then 𝑓 is constant. We now show how the maximum modulus theorem can be used to establish the characterization of Blaschke products that we discussed in Chapter 2. However, in the proof of Theorem 3.3 we need a version of the maximum modulus theorem for ℂ ∪ {∞}; see [29, p. 129]. Lemma 3.2. Let 𝑓 be a nonconstant function. Then 𝑓 is analytic on an open set 𝑈 containing 𝔻 and maps 𝔻 to itself and 𝕋 to itself if and only if 𝑓 is a finite Blaschke product. Proof. We know that if 𝑓 is a finite Blaschke product, it is a product of automorphisms of 𝔻 analytic on an open set containing 𝔻. Thus, 𝑓 is analytic on an open set containing 𝔻 and maps 𝔻 to itself and 𝕋 to itself. So, 𝑓 satisfies the stated conditions. So suppose that we know that 𝑓 is analytic on an open set 𝑈 containing 𝔻, that 𝑓 ∶ 𝔻 → 𝔻 and 𝑓 ∶ 𝕋 → 𝕋. Since the zeros of 𝑓 are isolated in 𝔻 (see [118, p. 139] or consider it an exercise), they cannot cluster in 𝔻. Since |𝑓| = 1 on 𝕋, the zeros cannot cluster on 𝕋, so 𝑓 has finitely many zeros in 𝔻. Shrinking 𝑈 a bit we may assume that 𝔻 ⊂ 𝑈, the zeros 𝑎1 , … , 𝑎𝑛 (listed according to multiplicity) of 𝑓 in 𝑈 are completely contained in 𝔻, and that 1/𝑎𝑗 ∉ 𝑈 for 𝑗 = 1, … , 𝑛. Now

Chapter 3. Blaschke Products

27

we are ready to see how this result will follow from the maximum modulus theorem: Form the Blaschke product 𝐵 with zeros 𝑎1 , … , 𝑎𝑛 (taking 𝜇 = 1 in Definition 3.1). Considering the power series of 𝑓, we see that −1 (𝑧 − 𝑎1 ) ⋯ (𝑧 − 𝑎𝑛 ) is a factor of 𝑓. Further, ((1 − 𝑎1 𝑧) ⋯ (1 − 𝑎𝑛 𝑧)) is analytic on (the modified) 𝑈. Thus, 𝑓/𝐵 and 𝐵/𝑓 are analytic on 𝑈. These functions are continuous on 𝔻, so they must have a maximum on 𝔻. Since |𝑓| = |𝐵| = 1 on 𝕋, the maximum modulus theorem shows that on 𝔻 we must have |𝐵| |𝑓| | | ≤ 1 and | | ≤ 1. |𝐵| |𝑓| Thus, 𝑓 = 𝛾𝐵 for some 𝛾 ∈ 𝕋. This characterization will be particularly useful in the proof of Theorem 3.3 below. Theorem 3.3. Let 𝑞 be a nonconstant polynomial with no zeros in 𝔻. Then there exist polynomials 𝑞𝑘 that approximate 𝑞 uniformly on compact subsets of 𝔻 and have all their zeros on the unit circle. Note that though the statement of Theorem 3.3, which is concerned with approximating polynomials on compact subsets of 𝔻, appears to have nothing to do with Blaschke products, the proof shows that the question has everything to do with them. Proof. Suppose the degree of 𝑞 is 𝑛. Define the polynomial 𝑞 ⋆ by 𝑞 ⋆ (𝑧) = 𝑧𝑛 𝑞(1/𝑧). Since the zeros of 𝑞 lie outside the closed unit disk, the zeros of 𝑞 ⋆ must be inside the disk. Consider the function 𝐵(𝑧) = 𝑞 ⋆ (𝑧)/𝑞(𝑧). ⋆

(3.2)

Since 𝑞 has zeros inside 𝔻 and 𝑞 only has zeros outside 𝔻, we see that 𝐵 is analytic on an open set containing 𝔻. Because the modulus of 𝑞 and 𝑞⋆ agree on the unit circle, 𝐵 also maps 𝕋 to itself. By the maximum modulus theorem 𝐵 cannot assume a value of modulus 1 inside the open unit disk, so 𝐵 ∶ 𝔻 → 𝔻. By considering 1/𝐵 on (ℂ ∪ {∞}) ⧵ 𝔻 and using the extended maximum modulus theorem, we see that |𝐵(𝑧)| > 1 for 𝑧 outside 𝔻. Therefore, 𝐵 assumes values of modulus 1 only on the unit circle. Furthermore, the same holds for 𝑧𝑘 𝐵(𝑧) for every 𝑘 ∈ ℕ. In particular, all solutions to 𝑧𝑘 𝑞 ⋆ (𝑧) = 𝑞(𝑧) must lie on the unit circle. So, if we let

28

Chapter 3. Blaschke Products

𝑞𝑘 (𝑧) = 𝑞(𝑧) − 𝑧𝑘 𝑞⋆ (𝑧), then all roots of 𝑞𝑘 lie on 𝕋 and, as 𝑘 → ∞, we see that 𝑧𝑘 → 0 uniformly on compact subsets of 𝔻. Since 𝑞 ⋆ is fixed in the expression for 𝑞𝑘 , the polynomials 𝑞𝑘 approach 𝑞 uniformly on compact subsets of the unit disk as desired. In Lemma 3.2, we gave the three conditions that characterize finite Blaschke products, and 𝐵, as defined in (3.2), satisfies all of them. So, 𝐵 is a Blaschke product. There is another way to look at this: If we let 𝑝 be the polynomial that appears in the denominator of a Blaschke product 𝐵, then 𝑝 has all its zeros outside of 𝔻. If we let 𝑝⋆ (𝑧) ∶= 𝑧𝑛 𝑝(1/𝑧), which has all of its zeros in 𝔻, a computation shows that 𝑝⋆ is the numerator of 𝐵. Thus, we get 𝐵(𝑧) = 𝑝⋆ (𝑧)/𝑝(𝑧)—the main ingredient in the proof of Theorem 3.3. This suggests one of the natural reasons for investigating Blaschke products. After studying polynomials, it is natural to study rational functions—just as it is natural to study rational numbers once one has studied the integers. But the rational functions appearing in the proof above are special: They are Blaschke products. Yet, in spite of their special nature, they play an important role in investigations involving all bounded analytic functions. The study of Blaschke products is rich and deep and has important connections to function theory as well as operator theory, which we discuss in Chapter 9. The geometry of polynomials has also been well studied; see, for example, Marden’s book entitled, appropriately, The Geometry of Polynomials [108]. The geometry of Blaschke products is not (yet!) well known and is currently an exciting area of research. Blaschke products play an interesting role in many other areas of mathematics, including operator theory, dynamics, interpolation, and complex analysis (e.g., [27], [42], [48], and [128]). A Blaschke product is particularly well behaved on the boundary of 𝔻, in a sense we make precise below. Lemma 3.4. Let 𝐵 be a degree-𝑛 Blaschke product with zeros 𝑎1 , … , 𝑎𝑛 . Then the following hold

Chapter 3. Blaschke Products

29

(1) The logarithmic derivative of 𝐵 is 𝑛 1 − |𝑎𝑗 |2 𝑑 𝐵′ (𝑧) log (𝐵(𝑧)) = =∑ . 𝑑𝑧 𝐵(𝑧) 𝑗=1 (1 − 𝑎𝑗 𝑧)(𝑧 − 𝑎𝑗 )

(2) The argument of 𝐵(𝑒𝑖𝜃 ) is a strictly increasing function of 𝜃.2 (3) For each 𝜆 ∈ 𝕋, there are exactly 𝑛 distinct solutions to 𝐵(𝑧) = 𝜆 and these solutions lie on the unit circle, 𝕋. Proof. The proof of (1) is a straightforward computation and is left to the reader. Using (1) for 𝑧 = 𝑒𝑖𝜃 , we have 𝑛 1 − |𝑎𝑗 |2 𝐵′ (𝑒𝑖𝜃 ) ∑ = . 𝑖𝜃 𝑖𝜃 𝐵(𝑒𝑖𝜃 ) 𝑗=1 (1 − 𝑎𝑗 𝑒 )(𝑒 − 𝑎𝑗 )

If we write 𝐵(𝑒𝑖𝜃 ) = 𝑒𝑖𝜓 and use the chain rule, we see that 𝑑𝜓 𝐵′ (𝑒𝑖𝜃 ) 𝐵′ (𝑒𝑖𝜃 ) = 𝑒𝑖𝜃 = 𝑑𝜃 𝐵(𝑒𝑖𝜃 ) 𝑒−𝑖𝜃 𝐵(𝑒𝑖𝜃 ) 𝑛

=∑ 𝑗=1

1 − |𝑎𝑗 |2 (𝑒−𝑖𝜃 − 𝑎𝑗 )(𝑒𝑖𝜃 − 𝑎𝑗 )

𝑛

=∑ 𝑗=1

1 − |𝑎𝑗 |2 |𝑒𝑖𝜃 − 𝑎𝑗 |2

> 0,

establishing (2). For each 𝜆 ∈ 𝕋, we wish to solve 𝐵(𝑧) = 𝜆 or 𝜇(𝑧 − 𝑎1 )(𝑧 − 𝑎2 ) ⋯ (𝑧 − 𝑎𝑛 ) = 𝜆(1 − 𝑎1 𝑧)(1 − 𝑎2 𝑧) ⋯ (1 − 𝑎𝑛 𝑧). The proof of (3) follows by applying Lemma 3.2 to conclude that 𝐵 can take values of modulus 1 only on the unit circle. So we know that there are 𝑛 solutions to this equation and all of these solutions must lie on the unit circle. By (2), these points are necessarily distinct. The constant 𝜇 played no role in the proof above. This shows why we usually consider normalized Blaschke products. In the future we will often omit 𝜇 without mention. To motivate our main result, we ask that you do the following: Go to the applet ,3 and select the Blaschke Product Explorer. Enter the zeros that you wish to consider and click “Load”. 2 The

continuous branch of the argument at each point is strictly increasing.

3 https://pubapps.bucknell.edu/static/aeshaffer/v1/

30

Chapter 3. Blaschke Products

You will be able to drag zeros out of the disk, and you can also doubleclick in a spot to enter other zeros, but Theorem 2.9 requires that one of the zeros be zero. Always. The reason for this requirement and the consequences of dropping it will be discussed later in the text. (You can experiment to see what happens when this condition is relaxed by going to the box labeled “Load” and removing the zero at zero.) We start by explaining what our applet does. First, you know that if we look at a function 𝑓 ∶ ℝ → ℝ, then you can “picture” that function via a graph; you plot (𝑥, 𝑓(𝑥)). And that is easy to do because the graph of such a function is two dimensional. But, if we have a function 𝐹 ∶ ℂ → ℂ, then it appears that we need four dimensions to “picture” our function, and plotting things in four dimensions is, well, challenging. So the applet does something else—it adds color, and if we wanted a more complete picture of our function, we could also add contours telling us how far we are from the origin. Most of the time we will just add color, and here is how we will do it.4

0

π/2

π

3π/2

2π

Figure 3.2. The colors of the spectrum.

First, color the plane according to the spectrum (if we could straighten the circle out, labeling the angles, we would see the colors in the order depicted in Figure 3.2; red, orange, yellow, green, blue, indigo, violet). We have provided an appropriately colored disk as a handy reference in Figure 3.3. In this picture, you can locate the point 𝑧 = 𝑥 + 𝑖𝑦 if you know both 𝑥 and 𝑦 or, if you prefer, you can locate the point if you know how far it is from the origin (√𝑥 2 + 𝑦 2 ) and its argument or color (a shade of yellow we call hummingbird). We begin with something easy—a Blaschke product with a zero of multiplicity 2 at zero; that is, we start with 𝐵(𝑧) = 𝑧2 . We have “sketched” 4 To learn more about this, we suggest Elias Wegert’s book, Visual Complex Functions

[151].

Chapter 3. Blaschke Products

31

Figure 3.3. Handy reference.

𝐵 in Figure 3.4. Here is what we do: Look at the points in the picture on the left (the domain), and ask where 𝐵 maps them in the picture on the right (the range). So, for example, if we look at the point 𝑧 = 1 on the left and square it, 𝑧2 = 1. So it is colored red in the picture on the left because the point 1 is colored red in the picture on the right. Similarly, if we choose 𝑧 = 𝑖 and square it, 𝑧2 = −1. That is colored sky blue on the right, so we color the new point sky blue in the picture on the left.

Figure 3.4. The phase portrait of 𝐵(𝑧) = 𝑧2 appears on the left.

32

Chapter 3. Blaschke Products

After doing that for every point, we obtain the picture on the left, which is called the phase portrait. And it is so illuminating—we see that zero is the spot where all colors come together, and we see that 𝑓(𝑧) = 𝑧2 takes on every value twice if we count zero twice because of its multiplicity. If we turn our attention to the circle, we see how 𝑧2 wraps the unit circle about itself twice. The contours indicate the modulus of the image of points under 𝐵. Before we move on, you should try to imagine what 𝑔(𝑧) = 𝑧4 should look like and then see what happens using the applet. We are now ready to analyze other complex functions. In this chapter we focus on Blaschke products with three zeros. As an example, we have chosen the normalized Blaschke product with a zero at 0, 0.5 + 0.5𝑖, and −0.5 + 0.5𝑖.

Figure 3.5. One, two, and many triangles.

If you use the “plot” feature, you will see that the colors representing the argument of the Blaschke product repeat three times on 𝕋. These colors depict the 3-to-1 nature of the Blaschke product that we discussed in Lemma 3.4. They also show you that on 𝕋 the argument of the Blaschke product is increasing—the argument is represented by the spectrum of colors, and we also see that they go in the same order (red, orange, yellow, green, blue, indigo, violet) before returning to red. This is what Lemma 3.4, (2) tells you. Finally, we note that the colors all come together at three points, the zeros of 𝐵. What else does the applet show? The applet chooses a value 𝜆1 on 𝕋 and locates the three solutions to 𝐵(𝑧) = 𝜆1 . Then we connect these points to form a triangle. The picture on the left in Figure 3.5 shows the result. The picture in the middle is what we get when we add the triangle associated with a second

Chapter 3. Blaschke Products

33

value, 𝜆2 ∈ 𝕋, and the one on the right is what we get when we choose several distinct values for 𝜆 ∈ 𝕋. You might recognize the picture on the right from previous chapters. It sure looks like an ellipse, and those white dots—the zeros of 𝐵(𝑧)/𝑧—could be the foci. Maybe. But we cannot know whether this really is an ellipse with the aforementioned foci unless we prove it. That is what we will do in the next chapter. We encourage the reader to experiment with this applet ,5 for a while and to form conjectures before moving on. As we mentioned above, it appears that if you form the closed triangular regions 𝑇𝜆 that you get for each point 𝜆 and consider ⋂𝜆∈𝕋 𝑇𝜆 , then the region that you get is an elliptical disk. This was also suggested by Examples 2.7 and 2.8. Our primary goal in the next chapter will be to prove Theorem 2.9, our main theorem on Blaschke products of degree 3. This theorem gives a detailed explanation for the appearance of the ellipse, and we call the ellipses appearing in this theorem Blaschke 3-ellipses. Now perhaps you see how to measure arcs “equally”. Our triangles connect all points on 𝕋 of the same color, say, all points colored red. That means the Blaschke product identifies the three vertices of the triangle and all vertices are red. The Blaschke product mapped the points on the arc of 𝕋 between the two red points onto the unit circle. So, if we measure length in the usual manner, the triangle divided the unit circle into three arcs that appear to have lengths that are not (usually) equal. But the Blaschke product maps each of those arcs onto the entire unit circle, so the Blaschke product thinks all of those arcs have the same length, 2𝜋. Blaschke products are very clever.

5 http://pubapps.bucknell.edu/static/aeshaffer/v1/

Chapter

4

Blaschke Products and Ellipses We are almost ready to present the proof of our main theorem about degree-3 Blaschke products. But before we do so, there are some interesting things happening in the degree-2 case that we would like to consider. Throughout this chapter, Lemma 3.4, which tells us that given 𝜆 ∈ 𝕋 and a Blaschke product 𝐵 of degree 𝑛 there are 𝑛 distinct solutions to the equation 𝐵(𝑧) = 𝜆, will play an important role. We see this already in Theorem 4.1 below. Theorem 4.1. Let 𝐵 be a Blaschke product with zeros at 0 and 𝑎1 ∈ 𝔻. For 𝜆 ∈ 𝕋, let 𝑧1 and 𝑧2 denote the two distinct solutions to 𝐵(𝑧) = 𝜆. Then 𝑎1 lies on the line joining 𝑧1 and 𝑧2 . Conversely, every line passing through 𝑎1 intersects the unit circle at two points 𝑧1′ and 𝑧2′ for which 𝐵(𝑧1′ ) = 𝐵(𝑧2′ ). Since we are interested only in the sets of points that 𝐵 identifies, it is enough to consider normalized Blaschke products as we do below. Proof. We may assume that 𝑧 − 𝑎1 ). 1 − 𝑎1 𝑧 The two solutions to 𝐵(𝑧) = 𝜆 can be written as 𝐵(𝑧) = 𝑧 (

𝑧1 = 𝑎1 + 𝑟1 𝑒𝑖𝜃1 and 𝑧2 = 𝑎1 + 𝑟2 𝑒𝑖𝜃2 for positive real numbers 𝑟1 and 𝑟2 and an appropriate choice of real numbers 𝜃1 and 𝜃2 . We know that 𝐵(𝑧1 ) = 𝐵(𝑧2 ) = 𝜆 and |𝑧1 | = |𝑧2 | = 1. Thus, writing 𝜆 = 𝑒𝑖𝛾 with 𝛾 ∈ ℝ, we have 𝑒𝑖𝛾 = 𝜆 = 𝐵(𝑧𝑗 ) = 𝑧𝑗 (

𝑧𝑗 − 𝑎1 1 − 𝑎1 𝑧𝑗

)= 35

𝑧𝑗 − 𝑎1 𝑧𝑗 − 𝑎1

=

𝑟𝑗 𝑒𝑖𝜃𝑗 𝑟𝑗 𝑒−𝑖𝜃𝑗

= 𝑒2𝑖𝜃𝑗

(4.1)

36

Chapter 4. Blaschke Products and Ellipses

for 𝑗 = 1, 2. So, recalling that |𝑎1 | < 1 and |𝑧1 | = |𝑧2 | = 1, we see that we can choose one of the arguments of 𝑧𝑗 to be 𝛾/2 and the other to be 𝛾/2 + 𝜋. Therefore, 𝑧1 = 𝑎1 + 𝑟1 𝑒𝑖𝛾/2 and 𝑧2 = 𝑎1 − 𝑟2 𝑒𝑖𝛾/2 . Let 𝑡 = 𝑟2 /(𝑟1 + 𝑟2 ). Then 𝑎1 = 𝑡𝑧1 + (1 − 𝑡)𝑧2 , and 𝑎1 lies on the line segment joining 𝑧1 and 𝑧2 . Conversely, if 𝑎1 = 𝑠𝑧1 + (1 − 𝑠)𝑧2 with 0 < 𝑠 < 1, then 𝑧 − 𝑎1 𝑧 − 𝑧2 𝑧 − 𝑎1 𝐵(𝑧2 ) = 2 = 1 = 1 = 𝐵(𝑧1 ). 𝑧1 − 𝑧2 𝑧2 − 𝑎 1 𝑧1 − 𝑎1








Figure 4.1. Connecting points identified by a degree-2 Blaschke product.

So, the degree-2 Blaschke product already provides us with some interesting geometric results. The degree-3 case, as presented in Theorem 2.9, is even more interesting. For the reader’s convenience, we recall the theorem for degree-3 Blaschke products.

Chapter 4. Blaschke Products and Ellipses

37

Theorem 2.9. Let 𝐵 be a Blaschke product of degree 3 with zeros 0, 𝑎1 , and 𝑎2 . For 𝜆 ∈ 𝕋, let 𝑧1 , 𝑧2 , and 𝑧3 denote the three distinct solutions to 𝐵(𝑧) = 𝜆. Then the lines joining 𝑧𝑗 and 𝑧𝑘 , for 𝑗 ≠ 𝑘, are tangent to the ellipse given by |𝑤 − 𝑎1 | + |𝑤 − 𝑎2 | = |1 − 𝑎1 𝑎2 |. Conversely, every point on the ellipse is the point of tangency of a line segment that intersects the unit circle at two points 𝑧1′ and 𝑧2′ for which 𝐵(𝑧1′ ) = 𝐵(𝑧2′ ). The proof hinges on Lemma 4.2 below, which is the same for Blaschke products of arbitrary degree. Since it does not require extra effort on our part (or yours), we include the general result below. We assume 𝐵(0) = 0, so for each 𝜆 ∈ 𝕋 the function 𝐵(𝑧)/𝑧 𝐹(𝑧) ∶= (4.2) 𝐵(𝑧) − 𝜆 is a rational function for which the degree of the numerator is strictly less than the degree of the denominator; furthermore, the denominator has distinct zeros. Therefore, we can use a partial fraction expansion to write 𝐹 in a particularly simple form, and this will aid us in our proof. Lemma 4.2. Let 𝐵 be a Blaschke product of degree 𝑛 ≥ 2 with zeros 0, 𝑎1 , 𝑎2 , …, 𝑎𝑛−1 . Let 𝜆 ∈ 𝕋 and 𝑧1 , … , 𝑧𝑛 denote the 𝑛 distinct solutions to 𝐵(𝑧) = 𝜆. Let the partial fraction expansion of 𝐹 in (4.2) be written as 𝐹(𝑧) =

𝑛 𝑚𝑗 𝐵(𝑧)/𝑧 =∑ . 𝐵(𝑧) − 𝜆 𝑗=1 𝑧 − 𝑧𝑗

Then the following hold. 𝑛

(1) ∑ 𝑚𝑗 = 1. 𝑗=1

(2) For 𝑗 = 1, 2, … , 𝑛, we have 𝑚𝑗 =

𝜆 . 𝑧𝑗 𝐵′ (𝑧𝑗 ) 𝑛−1

(3) For 𝑗 = 1, 2, … , 𝑛, we have

1 − |𝑎𝑘 |2 1 =1+ ∑ . 𝑚𝑗 |𝑧 − 𝑎𝑘 |2 𝑘=1 𝑗

(4) For 𝑗 = 1, 2, … , 𝑛, we have 0 < 𝑚𝑗 < 1.

38

Chapter 4. Blaschke Products and Ellipses

Before we start the proof, note that (1) and (4) tell us that 𝐹 is a convex combination of very simple rational functions. This is really the beauty of partial fractions—it takes a fairly complicated function and writes it as a sum of very simple ones. Proof. Note that we have 𝐵(0) = 0. Using a partial fraction expansion, we obtain 𝑚𝑗 ∈ ℂ with 𝐹(𝑧) =

𝑛 𝑚𝑗 𝐵(𝑧)/𝑧 =∑ . 𝑧 𝐵(𝑧) − 𝜆 𝑗=1 − 𝑧𝑗

To prove (1), consider 𝑛

𝑚𝑗 𝑧 . 𝑧 − 𝑧𝑗 𝑗=1

𝑧𝐹(𝑧) = ∑ Letting 𝑧 → ∞, we have 𝑛

𝑛

𝑧 𝐵(𝑧) = ∑ 𝑚𝑗 lim = ∑ 𝑚𝑗 . 𝑧→∞ 𝐵(𝑧) − 𝜆 𝑧→∞ 𝑧 − 𝑧𝑗 𝑗=1 𝑗=1

1 = lim 𝑛

So 1 = ∑𝑗=1 𝑚𝑗 , which is the desired result. We turn to (2). Since 𝐵(𝑧𝑗 ) = 𝜆, we have (

𝑧 − 𝑧𝑗 (𝑧 − 𝑧𝑗 )𝐵(𝑧) 𝐵(𝑧) = ) 𝑧 𝐵(𝑧) − 𝐵(𝑧𝑗 ) 𝑧(𝐵(𝑧) − 𝜆) = (𝑧 − 𝑧𝑗 )𝐹(𝑧) =∑ 𝑘≠𝑗

𝑚𝑘 (𝑧 − 𝑧𝑗 ) + 𝑚𝑗 . 𝑧 − 𝑧𝑘

The first factor on the left-hand side suggests that we take the derivative. So letting 𝑧 → 𝑧𝑗 , we get 𝑚𝑗 = lim ( 𝑧→𝑧𝑗

𝑧 − 𝑧𝑗 𝐵(𝑧) 𝜆 = , ) lim 𝐵(𝑧) − 𝐵(𝑧𝑗 ) 𝑧→𝑧𝑗 𝑧 𝑧𝑗 𝐵′ (𝑧𝑗 )

which yields (2). For (3), we return to our formula for the derivative from Lemma 3.4, (1). Recalling that 𝐵(0) = 0, we obtain 𝑛−1

(1 − |𝑎𝑘 |2 ) 𝐵′ (𝑧) 1 . = +∑ 𝑧 𝑘=1 (1 − 𝑎𝑘 𝑧)(𝑧 − 𝑎𝑘 ) 𝐵(𝑧)

Chapter 4. Blaschke Products and Ellipses

39

So for 𝑧 ∈ 𝕋, we get 𝑛−1

𝑛−1

(1 − |𝑎𝑘 |2 ) 1 − |𝑎𝑘 |2 𝑧𝐵′ (𝑧) =1+ ∑ =1+ ∑ . |𝑧 − 𝑎𝑘 |2 𝐵(𝑧) 𝑘=1 𝑧(1 − 𝑎𝑘 𝑧)(𝑧 − 𝑎𝑘 ) 𝑘=1 Evaluating at 𝑧𝑗 and using (2), we get 𝑛−1

1 − |𝑎𝑘 |2 1 =1+ ∑ , 𝑚𝑗 |𝑧 − 𝑎𝑘 |2 𝑘=1 𝑗 completing the proof of (3). Finally, (4) follows directly from (3). We return to our degree-3 Blaschke products. The importance of the 𝑚𝑗 , as we will see from the proof of the theorem, is that they tell us where the points of tangency are. If, for example, all 𝑚𝑗 are equal, the proof shows that the points of tangency will occur at the midpoints. In this case, the ellipse is often called the Steiner inellipse of the associated triangle. It is a good exercise to find an example of a Blaschke product and a 𝜆 ∈ 𝕋 for which all 𝑚𝑗 are equal—think about what is happening both algebraically and geometrically. It is an even better exercise to think about necessary and sufficient conditions for a Blaschke product to have all 𝑚𝑗 equal for some 𝜆 ∈ 𝕋; see, for example, [65] and Project 4 in Chapter 15. For the moment, we concentrate on what we can say more generally about degree-3 Blaschke products. We can see how the Blaschke product uses information about the circle to get information about the disk as follows: If 𝜆 ∈ 𝕋 and the three points 𝐵 sends to 𝜆 are 𝑧1 , 𝑧2 , and 𝑧3 , then (𝑧 − 𝑧1 )(𝑧 − 𝑧2 )(𝑧 − 𝑧3 ) . (1 − 𝑎1 𝑧)(1 − 𝑎2 𝑧) Evaluating at 𝑎𝑗 for 𝑗 = 1, 2, we have 𝐵(𝑧) − 𝜆 =

| (𝑎𝑗 − 𝑧1 )(𝑎𝑗 − 𝑧2 )(𝑎𝑗 − 𝑧3 ) | |. 1 = |𝜆| = | | (1 − |𝑎𝑗 |2 )(1 − 𝑎1 𝑎2 ) |

(4.3)

(4.4)

So we have learned that |(𝑎𝑗 − 𝑧1 )(𝑎𝑗 − 𝑧2 )(𝑎𝑗 − 𝑧3 )| = (1 − |𝑎𝑗 |2 )|1 − 𝑎1 𝑎2 |. The function 𝐹 in (4.2) plays a key role in our proof, so we isolate a useful fact here for reference. Let 𝑚 𝑧 + 𝑚 2 𝑧1 . 𝜁3 = 1 2 𝑚1 + 𝑚 2

40

Chapter 4. Blaschke Products and Ellipses

It may look like we pulled 𝜁3 out of thin air, but it gives us two things. First, it is a point on the line segment joining 𝑧1 and 𝑧2 . Second, it is the solution of 𝑚1 𝑚2 + = 0, 𝑧 − 𝑧1 𝑧 − 𝑧2 so it enables us to reduce the three summands in 𝐹 to just two. So we can write 𝐹 as follows: 3

𝑚𝑗 𝑚3 (𝑚1 + 𝑚2 )𝑧 − (𝑚1 𝑧2 + 𝑚2 𝑧1 ) = + 𝑧 − 𝑧𝑗 𝑧 − 𝑧3 (𝑧 − 𝑧1 )(𝑧 − 𝑧2 ) 𝑗=1

𝐹(𝑧) = ∑

𝑧 − 𝜁3 𝑚3 . (4.5) + (𝑚1 + 𝑚2 ) 𝑧 − 𝑧3 (𝑧 − 𝑧1 )(𝑧 − 𝑧2 ) Having said all that, we are ready for our proof. =

Proof of Theorem 2.9. Suppose 𝐵(𝑧1 ) = 𝐵(𝑧2 ) = 𝐵(𝑧3 ) = 𝜆. We focus on just one point of tangency; the others work the same way. Recalling that 𝐹(𝑧) =

3 𝑚𝑗 𝐵(𝑧) =∑ 𝑧(𝐵(𝑧) − 𝜆) 𝑗=1 𝑧 − 𝑧𝑗

and

𝑚1 𝑧2 + 𝑚2 𝑧1 , 𝑚1 + 𝑚 2 it is apparent that 𝜁3 lies on the line segment joining 𝑧1 and 𝑧2 . We will show that 𝜁3 is the point of tangency of the line segment and the ellipse |𝑧 − 𝑎1 | + |𝑧 − 𝑎2 | = |1 − 𝑎1 𝑎2 | in two steps, the first of which is to show that 𝜁3 lies on the ellipse. By definition, 𝐹(𝑎𝑗 ) = 0 for 𝑗 = 1, 2. Thus, (4.5) implies that 𝜁3 =

0 = 𝐹(𝑎𝑗 ) =

(𝑎𝑗 − 𝜁3 ) 𝑚3 (𝑚 + 𝑚2 ). + 𝑎𝑗 − 𝑧3 (𝑎𝑗 − 𝑧1 )(𝑎𝑗 − 𝑧2 ) 1

By Lemma 4.2, 0 < 𝑚𝑗 < 1 and 𝑚1 + 𝑚2 + 𝑚3 = 1, which yields (𝑎𝑗 − 𝜁3 ) | | | 1 | | = (1 − 𝑚3 ) | |. 𝑚3 | | 𝑎𝑗 − 𝑧3 | (𝑎 − 𝑧 )(𝑎 − 𝑧 ) | 𝑗 1 𝑗 2 |

Chapter 4. Blaschke Products and Ellipses

41

Solving for |𝑎𝑗 − 𝜁3 | with 𝑗 = 1, 2, we conclude that 1 1 |𝜁3 − 𝑎1 | + |𝜁3 − 𝑎2 | |1 − 𝑎1 𝑎2 | |1 − 𝑎1 𝑎2 | 𝑚3 | (𝑎 − 𝑧1 )(𝑎1 − 𝑧2 ) | | (𝑎2 − 𝑧1 )(𝑎2 − 𝑧2 ) | |+| |) . (4.6) = (| 1 1 − 𝑚3 | (1 − 𝑎1 𝑎2 )(𝑎1 − 𝑧3 ) | | (1 − 𝑎1 𝑎2 )(𝑎2 − 𝑧3 ) | Combining (4.4) with Equation (4.6), we obtain 1 1 |𝜁3 − 𝑎1 | + |𝜁3 − 𝑎2 | |1 − 𝑎1 𝑎2 | |1 − 𝑎1 𝑎2 | =

1 − |𝑎2 |2 1 − |𝑎1 |2 𝑚3 + ), ( 1 − 𝑚3 |𝑎1 − 𝑧3 |2 |𝑎2 − 𝑧3 |2

and we are almost where we want to be. By Lemma 4.2, (3), we have 1 1 |𝜁3 − 𝑎1 | + |𝜁3 − 𝑎2 | |1 − 𝑎1 𝑎2 | |1 − 𝑎1 𝑎2 | (1 − |𝑎1 |2 ) (1 − |𝑎2 |2 ) 𝑚3 = + ([1 + ] − 1) 1 − 𝑚3 |𝑎1 − 𝑧3 |2 |𝑎2 − 𝑧3 |2 𝑚3 1 =( − 1) = 1. )( 1 − 𝑚 3 𝑚3 Thus, |𝜁3 − 𝑎1 | + |𝜁3 − 𝑎2 | = |1 − 𝑎1 𝑎2 |, whence 𝜁3 lies on the ellipse, as desired. We still need to prove that 𝜁3 is the point of tangency of the line through 𝑧1 and 𝑧2 with the ellipse. To do this, we show that ∠𝑎1 𝜁3 𝑧1 = −∠𝑎2 𝜁3 𝑧2 , so we begin by computing these angles. First, use (4.5) and the definition of 𝐹 to see that 𝐵(𝜁3 )/𝜁3 (𝜁3 − 𝑎1 )(𝜁3 − 𝑎2 ) 𝑚3 (4.7) = 𝐹(𝜁3 ) = = . 𝜁3 − 𝑧3 𝐵(𝜁3 ) − 𝜆 (𝜁3 − 𝑧1 )(𝜁3 − 𝑧2 )(𝜁3 − 𝑧3 ) We now compare the angles made by the line segments joining 𝑎𝑗 and 𝜁3 and joining 𝜁3 and 𝑧𝑗 for 𝑗 = 1, 2. Using Equation (4.7), we calculate (modulo 2𝜋) 𝑎 − 𝜁3 (𝑎 − 𝜁3 )(𝑎2 − 𝜁3 ) 𝑎 − 𝜁3 arg [ 1 ] + arg [ 2 ] = arg [ 1 ] 𝑧1 − 𝜁3 𝑧2 − 𝜁3 (𝑧1 − 𝜁3 )(𝑧2 − 𝜁3 ) 𝑚3 = arg[(𝜁3 − 𝑧3 )𝐹(𝜁3 )] = arg [(𝜁3 − 𝑧3 ) ] = arg 𝑚3 = 0. 𝜁3 − 𝑧3

42

Chapter 4. Blaschke Products and Ellipses

Hence, ∠𝑎1 𝜁3 𝑧1 = −∠𝑎2 𝜁3 𝑧2 , and, by Proposition 1.1, the segment 𝑧1 𝑧2 is tangent to the ellipse at 𝜁3 . What is left to do? We claimed that every point on the ellipse is the point of tangency of the ellipse with a line passing through two points on 𝕋 identified by 𝐵. So let 𝜁 be an arbitrary point on the ellipse and draw the tangent line 𝐿 to the ellipse at 𝜁 (see Figure 4.2). Note that it


=

ζ


ℓ





Figure 4.2. 𝐵(𝑧1′ ) = 𝐵(𝑤𝑘 ) = 𝐵(𝑤ℓ ) and {𝑘, ℓ} = {1, 2}. will intersect the unit circle in two distinct places, which we call 𝑧1′ and 𝑧2′ . From Lemma 3.4, we know that there exist exactly two points on the unit circle, call them 𝑤1 and 𝑤2 , for which 𝐵(𝑧1′ ) = 𝐵(𝑤1 ) = 𝐵(𝑤2 ). But there exist exactly two tangent lines from 𝑧1′ to the ellipse. Since 𝐿 is tangent to the ellipse, it must be one of these two tangent lines. On the other hand, by the first part of our proof, we know that the line through 𝑧1′ and 𝑤1 and the line through 𝑧1′ and 𝑤2 are the two lines tangent to the ellipse. Thus, 𝐿 must be one of these two lines, and 𝑧2′ = 𝑤1 or 𝑧2′ = 𝑤2 , as claimed. This completes the proof of Theorem 2.9. In the proof, we actually obtained a formula for the points of tangency. We isolate this fact for future reference. Note that we always take indices mod 3.

Chapter 4. Blaschke Products and Ellipses

43

Corollary 4.3. Let 𝐵 be a Blaschke product with zeros 𝑎1 , 𝑎2 , and 0. Let 𝜆 ∈ 𝕋 and 𝑧1 , 𝑧2 , 𝑧3 denote the distinct solutions of 𝐵(𝑧) = 𝜆. If 3 𝑚𝑗 𝐵(𝑧)/𝑧 =∑ , 𝐵(𝑧) − 𝜆 𝑗=1 𝑧 − 𝑧𝑗

then the line segments joining 𝑧𝑗 and 𝑧𝑗+1 are tangent to the ellipse |𝑤 − 𝑎1 | + |𝑤 − 𝑎2 | = |1 − 𝑎1 𝑎2 | at the points

𝑚𝑗+1 𝑧𝑗 + 𝑚𝑗 𝑧𝑗+1 for 𝑗 = 1, 2, 3. 𝑚𝑗 + 𝑚𝑗+1

There are many interesting things that follow from this theorem, but we focus on just one here. The result is known as Chapple’s formula or Euler’s identity, and it looks at what we can say when we have a triangle inscribed in one circle and circumscribed about a second circle. (The connection of this work to the Chapple–Euler formula was first noticed by Frantz in [49].) Since there are enough formulas and identities named after Euler to make things confusing and since Chapple was the first to publish a proof of this result (albeit a proof that was difficult to follow; we say more about this later), we call this the Chapple–Euler formula. To establish the Chapple–Euler formula, we first show that if an ellipse 𝐸 is circumscribed by a triangle that is inscribed in the unit circle, then 𝐸 is a Blaschke 3-ellipse. Corollary 4.4. Let 𝐸 be an ellipse with foci 𝑎1 and 𝑎2 . If 𝐸 can be circumscribed by a triangle that is inscribed in 𝕋, then 𝐸 is the Blaschke 3-ellipse with foci at 𝑎1 and 𝑎2 with major axis of length |1 − 𝑎1 𝑎2 |. Proof. By Theorem 2.9, we know that there is an ellipse 𝐸1 with foci at 𝑎1 and 𝑎2 that has the property that given a point 𝑧 ∈ 𝕋, there is a triangle, 𝑇𝑧 , circumscribing it and inscribed in 𝕋 with one vertex at 𝑧. We may suppose that the major axis of 𝐸 is longer than that of 𝐸1 . By our assumption, the ellipses are confocal. Thinking of the nail and string construction of an ellipse, we see that 𝐸 properly contains 𝐸1 . Let 𝑧1 , 𝑧2 , and 𝑧3 be the vertices of the triangle 𝑇 circumscribing 𝐸. Now we attempt to circumscribe 𝐸1 by a triangle 𝑇𝑧1 , inscribed in the unit circle, with one vertex at 𝑧1 . Let 𝑧1 , 𝑤2 , and 𝑤3 be the vertices of 𝑇𝑧1 (see Figure 4.3).

44

Chapter 4. Blaschke Products and Ellipses



















Figure 4.3. Two confocal ellipses.

Because 𝐸 is larger than 𝐸1 and the ellipses are confocal, the argument of 𝑤2 must be greater than that of 𝑧2 ; that is, the line joining 𝑧1 and 𝑤2 lies below that joining 𝑧1 and 𝑧2 . Similarly, the line joining 𝑧1 and 𝑤3 lies above that joining 𝑧1 and 𝑧3 . But then the line joining 𝑤2 and 𝑤3 lies outside 𝐸 and, consequently, cannot be tangent to 𝐸1 . Similarly, we cannot have 𝐸 properly contained in 𝐸1 . Therefore, the only possibility is that 𝐸 = 𝐸1 , which completes the proof. As a consequence of Theorem 2.9 and Corollary 4.4, given 𝑎1 and 𝑎2 in 𝔻 there exists a unique ellipse with foci 𝑎1 and 𝑎2 that can be inscribed in a triangle with all vertices on 𝕋. If the inscribed ellipse is a circle, then we can say more. Corollary 4.5 (The Chapple–Euler formula). If a triangle is inscribed in a circle 𝐶 with center 𝐴 and radius 𝑅 and circumscribes a circle 𝑐 with center 𝑎 and radius 𝑟, then 2𝑟𝑅 = 𝑅2 − 𝑑2 , where 𝑑 = |𝐴 − 𝑎|. Proof. Via scaling and shifting, we can assume that 𝑅 = 1 and 𝐴 = 0. Then 𝑑2 = |𝑎|2 . A point 𝑧𝑐 on 𝑐 satisfies |𝑧𝑐 − 𝑎|2 = 𝑟2 . But the circle is

Chapter 4. Blaschke Products and Ellipses

45

a Blaschke-3 ellipse and there is only one such ellipse, namely, |𝑧𝑐 − 𝑎| + |𝑧𝑐 − 𝑎| = 1 − |𝑎|2 . Thus, 𝑅2 − 𝑑2 = 1 − |𝑎|2 = |𝑧𝑐 − 𝑎| + |𝑧𝑐 − 𝑎| = 2𝑟𝑅, as advertised.





 




Figure 4.4. Chapple–Euler formula: 2𝑟𝑅 = 𝑅2 − 𝑑 2 .

The first known reference to this result appears in a paper of William Chapple in Miscellanea Curiosa Mathematica [25]. Chapple also discussed a closure theorem, in the spirit of Poncelet’s theorem, yet his arguments are difficult to follow and incomplete. “It is not easy to summarise Chapple’s article because almost all the logic in it was wrong … [but] despite his failures in logic, Chapple had grasped essential aspects of the problem”. (See [15].) Euler’s name most likely appears because he studied a related problem and Nicolaus Fuss, who served as Euler’s mathematical assistant, published the formula in a widely circulated journal in 1797. Though Fuss helped Euler prepare papers for publication and was familiar with Euler’s work, Fuss did not attribute the formula to Euler. Nevertheless, the name Euler stuck to this result, and you can find the formula or its consequence (2𝑟 ≤ 𝑅) under the name of Euler’s inequality, Euler’s theorem, or the Chapple–Euler inequality. Fuss also gave a formula that is satisfied when a quadrilateral

46

Chapter 4. Blaschke Products and Ellipses

circumscribes a circle and is inscribed in a second circle [52]. We provide a proof of Fuss’s result in Chapter 13. Incidentally, Fuss married Albertine Euler, the daughter of Leonhard Euler’s eldest son [18, p. 85].

Chapter

5

Poncelet’s Theorem for Triangles In this chapter we consider Poncelet’s theorem as Poncelet did—from a geometric point of view. Projective geometry is a great field to add to one’s toolkit; it provides a different way of looking at many familiar problems, as we will see in a moment. It is not based on a metric, or notion of distance, but rather on the study of points and lines. And while Euclidean geometry is not symmetric—that is, given any two points, there is a unique line passing through them, but given any two lines, there is not necessarily a unique point that lies on both—projective geometry is. One of its beautiful features is the notion of duality, meaning that there is a symmetry between statements about lines and points, a symmetry that we hope to illustrate in this chapter. Jean-Victor Poncelet’s work was essential to the development of the field. Poncelet conceived of the theorem now bearing his name in 1812– 1813 while he was a prisoner in Russia—an imprisonment that was a consequence of his service as an officer in Napoléon’s army. While in prison, without books or notes, he recalled what he had learned from Monge, Carnot, and Brianchon. Poncelet remained in prison until June 1814, and during this time he wrote what he called the Saratov notebook. After his release, Poncelet returned to France. The theorem now named in Poncelet’s honor was part of the Saratov notebook, but the notes were not published until 1862. As a consequence, the first published proof of Poncelet’s theorem appeared in his 1822 paper, Traité des propriétés projectives des figures. There are now many proofs of Poncelet’s theorem (see Chapter 12), but perhaps the most accessible proof (and the one we follow) is due to Halbeisen 47

48

Chapter 5. Poncelet’s Theorem for Triangles

and Hungerbühler [74]. As for Poncelet’s notes, a second edition of Traité des propriétés projectives des figures appeared in 1865–1866 and was reprinted in 1995 [125]. We discuss a special case of Poncelet’s theorem here, leaving a more general treatment for later. In keeping with our desire to make the material in this book be as self-contained as possible, we include a proof of Pascal’s theorem, the main ingredient in Halbeisen’s and Hungerbühler’s derivation. Pascal’s theorem, too, has many proofs. There is a rather short geometric one in [148], but we opt to follow the algebraic proof by Stefanović and Milos̆ević [144]. We state and prove these theorems in real projective 2-space, ℙ2 (ℝ). For 𝑥, 𝑦, and 𝑧 in ℝ we define an equivalence relation on ℝ3 ⧵{(0, 0, 0)} by (𝑥, 𝑦, 𝑧) = (𝑥 ′ , 𝑦 ′ , 𝑧′ ) if 𝑥 ′ = 𝜆𝑥, 𝑦 ′ = 𝜆𝑦, and 𝑧′ = 𝜆𝑧 for some nonzero 𝜆 ∈ ℝ. The equivalence classes are the points of ℙ2 (ℝ), and we will denote them by an arbitrary representative (𝑥, 𝑦, 𝑧) with not all of 𝑥, 𝑦, 𝑧 zero. These are the projective coordinates of a point. We can embed the real plane ℝ2 in ℙ2 (ℝ) with the map (𝑥, 𝑦) ↦ (𝑥, 𝑦, 1). Points that are represented as (𝑥, 𝑦, 0) with at least one of 𝑥 and 𝑦 not 0 are called points at infinity and they make up the line at infinity. Given 𝑎, 𝑏, 𝑐 ∈ ℝ, not all zero, we say the point (𝑥, 𝑦, 𝑧) is on the line (𝑎, 𝑏, 𝑐) if 𝑎𝑥+𝑏𝑦+𝑐𝑧 = 0. Thus, just as we saw for points, lines in ℙ2 (ℝ) are the equivalence classes associated with the equivalence relation on ℝ3 ⧵ {(0, 0, 0)} given by (𝑎, 𝑏, 𝑐) = (𝑎′ , 𝑏′ , 𝑐′ ) if 𝑎′ = 𝛿𝑎, 𝑏′ = 𝛿𝑏, and 𝑐′ = 𝛿𝑐 for some nonzero 𝛿 ∈ ℝ. We denote the lines with an arbitrary representative (𝑎, 𝑏, 𝑐), not all of 𝑎, 𝑏, 𝑐 zero, just as we did for the points. While this may seem odd at first, in fact, things are simplified. Two lines (𝑎, 𝑏, 𝑐1 ) and (𝑎, 𝑏, 𝑐2 ), with 𝑐1 ≠ 𝑐2 and not both of 𝑎 and 𝑏 zero, do not meet in ℝ2 . However, in ℙ2 (ℝ) they intersect at the point (−𝑏, 𝑎, 0), a point at infinity. Every pair of distinct lines meets in exactly one point, and every pair of distinct points determines exactly one line. In other words, when we think about it this way, things work the way we would like them to: points and lines are “dual” objects. To see how projective geometry can help us see things differently, try the following exercise. Exercise 5.1. In the plane, consider four distinct points 𝑝1 = (𝑥1 , 𝑦1 ), 𝑝2 = (𝑥2 , 𝑦2 ), 𝑝3 = (𝑥3 , 𝑦3 ), and 𝑝4 = (𝑥4 , 𝑦4 ),

Chapter 5. Poncelet’s Theorem for Triangles

49

Figure 5.1. Lines meeting at infinity (Creative Commons CC0).

not all of which are collinear. Find the point where the line through 𝑝1 and 𝑝2 intersects the line through 𝑝3 and 𝑝4 (for this exercise you may assume that the two lines intersect). You probably know a way to solve this problem, and we suggest that you solve it now before you read on. We want to present a solution using projective geometry and show how such an approach will simplify things. But first we need to discuss the basic elements of Poncelet’s geometric approach. Our notation will be simplified significantly if we identify a point 𝑃 = (𝑥, 𝑦, 𝑧) with the vector [𝑥 𝑦 𝑧]𝑇 and a line ℓ = (𝑎, 𝑏, 𝑐) with the vector [𝑎 𝑏 𝑐]𝑇 . Using this identification we are able to use three-dimensional vector notation. • The point 𝑃 is on the line ℓ if and only if the dot product 𝑃 ⋅ ℓ = 0. • If 𝑃1 and 𝑃2 are two different points, then the cross product 𝑃1 × 𝑃2 is the line determined by these points. • If ℓ1 and ℓ2 are two different lines, then their intersection is the point ℓ1 × ℓ2 .

50

Chapter 5. Poncelet’s Theorem for Triangles

Alternatively, given two different points 𝑃1 and 𝑃2 , all points on the line through these points are given by 𝛼1 𝑃1 + 𝛼2 𝑃2 , where 𝛼1 , 𝛼2 are real numbers not both zero. Likewise, given two different lines ℓ1 and ℓ2 that intersect at 𝑃, all lines passing through 𝑃 are of the form 𝛼1 ℓ1 + 𝛼2 ℓ2 , where 𝛼1 , 𝛼2 are real, not both zero. Given points 𝑃1 , 𝑃2 , and 𝑃3 , we let [𝑃1 , 𝑃2 , 𝑃3 ] denote the 3 × 3 matrix with 𝑃1 , 𝑃2 , and 𝑃3 as column vectors. With this notation, the points are collinear if and only if det[𝑃1 , 𝑃2 , 𝑃3 ] = 0. Similarly, forming the matrix [ℓ1 , ℓ2 , ℓ3 ] in which the three lines are the column vectors leads to the criterion that the lines intersect in a single point if and only if det[ℓ1 , ℓ2 , ℓ3 ] = 0. So, before moving on, we return to Exercise 5.1. We use projective space to redo the solution you obtained above and we do it two ways. Let 𝑃𝑗 = (𝑥𝑗 , 𝑦𝑗 , 1) for 𝑗 = 1, 2, 3, 4. First, we can find ℓ1 = 𝑃1 × 𝑃2 and ℓ2 = 𝑃3 × 𝑃4 . These represent two lines, so the point we seek is ℓ1 × ℓ2 . For 𝛼 ≠ 0, if we remember to identify (𝛼𝑥, 𝛼𝑦, 𝛼) with (𝑥, 𝑦, 1) and (𝑥, 𝑦, 0) with the point at infinity, we obtain our solution. Or, we can think about it this way: We want to find a point that lies on both ℓ1 = 𝛼1 𝑃1 + 𝛼2 𝑃2 and ℓ2 = 𝛼3 𝑃3 + 𝛼4 𝑃4 . So we want 𝛼1 and 𝛼2 so that the three points 𝛼1 𝑃1 + 𝛼2 𝑃2 , 𝑃3 , and 𝑃4 are collinear; that is, 0 = det[𝛼1 𝑃1 + 𝛼2 𝑃2 , 𝑃3 , 𝑃4 ]. But now we need only use the properties of determinants: 0 = 𝛼1 det[𝑃1 , 𝑃3 , 𝑃4 ] + 𝛼2 det[𝑃2 , 𝑃3 , 𝑃4 ], and we can guess the solution to this—no messy algebraic computations are necessary! Inspection tells us that 𝛼1 = det[𝑃2 , 𝑃3 , 𝑃4 ] and 𝛼2 = − det[𝑃1 , 𝑃3 , 𝑃4 ] works. If we denote the solution by (𝑥0 , 𝑦0 , 𝑧0 ), we see that if 𝑧0 = 0 our lines are parallel; if not, we can divide by it to obtain our solution. While you had to separate out two cases in your solution, one for the case in which the lines are parallel and one when they are not, projective geometry handles them as one case. We turn to the so-called self-duality of the projective plane, which we find to be a useful tool. We will explain the concept briefly below but

Chapter 5. Poncelet’s Theorem for Triangles

51

leave the details to a projective geometry text; see [153, pp. 11–30], for example. A homogeneous polynomial is a polynomial in which each term has the same degree. For instance, 𝑝(𝑥, 𝑦, 𝑧) = 𝜋𝑥 2 𝑧 + 𝑦 3 − 𝑥𝑦𝑧 is a homogenous polynomial while 𝑞(𝑥, 𝑦, 𝑧) = 2𝑥𝑦𝑧 − 3𝑥 2 is not. An algebraic curve is the set of points (𝑥, 𝑦, 𝑧) in ℙ2 (ℝ) that are solutions of a homogeneous polynomial equation 𝑝(𝑥, 𝑦, 𝑧) = 0. Note that this is consistent with the definition of points in ℙ2 (ℝ): If 𝑝 is homogeneous of degree 𝑚, then 𝑝(𝜆𝑥, 𝜆𝑦, 𝜆𝑧) = 𝜆𝑚 𝑝(𝑥, 𝑦, 𝑧) for all 𝜆 ∈ ℝ and thus 𝑝(𝑥, 𝑦, 𝑧) = 0 if and only if 𝑝(𝜆𝑥, 𝜆𝑦, 𝜆𝑧) = 0 for all 𝜆 ≠ 0 in ℝ. The tangent line to the curve 𝑝 = 0 at the point 𝑄 on this curve is the line (𝑝𝑥 (𝑄), 𝑝𝑦 (𝑄), 𝑝𝑧 (𝑄)), where 𝑝𝑥 , 𝑝𝑦 , and 𝑝𝑧 denote the partial derivatives of 𝑝. The dual of an algebraic curve is the set of tangent lines to this curve. It can be shown that the coordinates (𝑎, 𝑏, 𝑐) of these tangent lines are exactly the zeros of another homogeneous polynomial equation. (See Section 15.12 for more information and some examples.) Because of the complete symmetry between points and lines in the projective plane, a reformulation of any theorem obtained by consistently switching points and lines will lead to another theorem, its dual. Let us now see how this duality can be used in the study of conics. A conic in the real projective plane ℙ2 (ℝ) is a curve 𝒞 that consists of all points (𝑥, 𝑦, 𝑧) that satisfy a quadratic equation of the form 𝐴𝑥 2 + 𝐵𝑥𝑦 + 𝐶𝑦 2 + 𝐷𝑥𝑧 + 𝐸𝑦𝑧 + 𝐹𝑧2 = 0,

(5.1)

where 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, and 𝐹 are real numbers not all equal to zero. But you knew that already. Here is the part you may not know: The matrix associated with 𝒞 (determined up to a nonzero multiple) is given by 𝐴 𝑀 = [ 𝐵/2 𝐷/2 𝑇

𝐵/2 𝐷/2 𝐶 𝐸/2 ] . 𝐸/2 𝐹

If you compute [𝑥 𝑦 𝑧] 𝑀 [𝑥 𝑦 𝑧] , you will see why 𝑀 is defined this way. So every conic gives rise to a matrix. There is also a converse to this statement: Every nonzero symmetric real 3 × 3 matrix 𝑀 gives rise to a

52

Chapter 5. Poncelet’s Theorem for Triangles

unique conic 𝒞. With this notation, the points of a conic can be characterized as follows: 𝑇

(𝑥, 𝑦, 𝑧) ∈ 𝒞 if and only if

[𝑥 𝑦 𝑧] 𝑀 [𝑥 𝑦 𝑧] = 0.

A conic is said to be degenerate if det 𝑀 = 0. Geometrically, this corresponds to a pair of straight lines, a line (of multiplicity 2), a point, or the empty set. The theorem below says that five points determine a conic. Theorem 5.2. Let 𝒞1 and 𝒞2 be two conics, at least one of which is nondegenerate. If these conics have at least five different points in common, then they are equal. Proof. We prove the contrapositive statement. So assume 𝒞1 and 𝒞2 are not equal. First, assume that 𝒞1 is nondegenerate and that 𝒞2 is degenerate. Then 𝒞2 ⊆ ℓ1 ∪ ℓ2 for some lines ℓ1 and ℓ2 . A nondegenerate conic intersects a straight line in at most two points. Thus, 𝒞1 ∩ 𝒞2 contains at most four points, and we have completed this case. Now assume that both conics 𝒞1 and 𝒞2 are nondegenerate. We denote the matrices associated with 𝒞1 and 𝒞2 by 𝑀1 and 𝑀2 , respectively, and define a function 𝑓 ∶ [0, 1] → ℝ by 𝑓(𝑡) = det ((1 − 𝑡) det(𝑀2 )𝑀1 − 𝑡 det(𝑀1 )𝑀2 ) . Evidently, 𝑓 is continuous, and since 𝑀1 is 3 × 3 and our curves are nondegenerate, we have 3

𝑓(0) = (det(𝑀2 )) det(𝑀1 ) ≠ 0 and 3

𝑓(1) = − (det(𝑀1 )) det(𝑀2 ) ≠ 0. We see that 𝑓 has opposite signs at the endpoints of the interval [0, 1], so by the intermediate value theorem, there must exist 𝑡0 with 0 < 𝑡0 < 1 and 𝑓(𝑡0 ) = 0. Let 𝛼 = (1 − 𝑡0 ) det(𝑀2 ) and 𝛽 = 𝑡0 det(𝑀1 ) and define 𝑀3 ∶= 𝛼𝑀1 − 𝛽𝑀2 . The matrix 𝑀3 is again a symmetric matrix with det(𝑀3 ) = 0, and therefore it gives rise to a degenerate conic 𝒞3 . If (𝑥, 𝑦, 𝑧) ∈ 𝒞1 ∩ 𝒞2 , then 𝑇

𝑇

[𝑥 𝑦 𝑧] 𝑀3 [𝑥 𝑦 𝑧] = [𝑥 𝑦 𝑧] (𝛼𝑀1 − 𝛽𝑀2 ) [𝑥 𝑦 𝑧] 𝑇

𝑇

= 𝛼 [𝑥 𝑦 𝑧] 𝑀1 [𝑥 𝑦 𝑧] − 𝛽 [𝑥 𝑦 𝑧] 𝑀2 [𝑥 𝑦 𝑧] = 0.

Chapter 5. Poncelet’s Theorem for Triangles

53

Thus, (𝑥, 𝑦, 𝑧) ∈ 𝒞1 ∩ 𝒞3 and we have 𝒞1 ∩ 𝒞2 ⊆ 𝒞1 ∩ 𝒞3 . But, as we saw in the first part of this proof, the set on the right has at most four points. We conclude that 𝒞1 and 𝒞2 intersect in at most four points. Theorem 5.2 tells us that something special happens for five points, so we might expect that something special happens for six points, too. That is indeed the case, and that brings us to Theorem 5.3, or Pascal’s theorem, which tells us what that special consequence is. You may want to explore our applet ,1 labeled Pascal’s Theorem to get a feel for the beauty and importance of this theorem. Before we delve into the technical statement and proof of Theorem 5.3, we want to establish the language that we use throughout this text so that we all have the same understanding of the terminology. A set of points is said to be in general position if the points are distinct and no subset of three points is collinear. By side of a hexagon (or any side of a polygonal chain) we always mean the line containing the side and not just the line segment connecting the vertices. A hexagon (or any polygonal chain) is inscribed in a conic 𝒞 if all of its vertices are on 𝒞. Theorem 5.3 (Pascal). If a hexagon is inscribed in a nondegenerate conic, then the intersection points of the three pairs of opposite sides are collinear and distinct. Conversely, if at least five vertices of a hexagon are in general position and the hexagon has the property that the points of intersection of the three pairs of opposite sides are collinear, then the hexagon is inscribed in a unique nondegenerate conic. It is important to note that the hexagon need not be convex or simple (a hexagon is simple if the line segments between vertices intersect only at vertices). To prove Theorem 5.3 we introduce the following notation. We call the conic 𝒞 and consider a hexagon with the following sequence of vertices: (𝑃1 , 𝑃5 , 𝑃3 , 𝑃6 , 𝑃2 , 𝑃4 ). The intersection points of the three pairs of opposite sides are then denoted by 𝐾 = (𝑃1 × 𝑃5 ) × (𝑃2 × 𝑃6 ), 𝐿 = (𝑃3 × 𝑃6 ) × (𝑃1 × 𝑃4 ), and 𝑀 = (𝑃2 × 𝑃4 ) × (𝑃3 × 𝑃5 ); see Figure 5.2. It is important that we understand what opposite sides are. When the hexagon is convex, this is usually clear, However, in Figure 5.2, it is not so clear. Opposite sides refers to the following: Given a side, count 1 http://pubapps.bucknell.edu/static/aeshaffer/v1/

54

Chapter 5. Poncelet’s Theorem for Triangles









 











Figure 5.2. Illustration of Pascal’s theorem.

three consecutive sides to obtain the opposite side. Thus, in Figure 5.2, ←→ ←→ the side opposite 𝑃1 𝑃5 is 𝑃2 𝑃6 , and the resulting point of intersection is 𝐾. ←→ ←→ Similarly, for 𝑃3 𝑃6 , the opposite side is 𝑃1 𝑃4 and the point of intersection is 𝐿. Now that we have the terminology in place, we can prove Theorem 5.3. Proof. The points 𝐾, 𝐿, and 𝑀 are collinear if they are linearly dependent as vectors. Thus, we need to show that det[𝐾, 𝐿, 𝑀] = 0. We can expect that this determinant is expressible as a function of determinants containing the six points on the conic since 𝐾, 𝐿, and 𝑀 are the intersection points of lines connecting these points. But exactly which function will work is presently unclear. A lengthy calculation using the vector identities (𝑢 × 𝑣) × (𝑤 × 𝑧) = det[𝑢, 𝑤, 𝑧]𝑣 − det[𝑣, 𝑤, 𝑧]𝑢 = det[𝑢, 𝑣, 𝑧]𝑤 − det[𝑢, 𝑣, 𝑤]𝑧 shows that in fact det[𝐾, 𝐿, 𝑀] = det ([𝑃1 , 𝑃3 , 𝑃5 ][𝑃2 , 𝑃4 , 𝑃5 ][𝑃4 , 𝑃6 , 𝑃1 ][𝑃6 , 𝑃2 , 𝑃3 ]) − det ([𝑃2 , 𝑃4 , 𝑃6 ][𝑃1 , 𝑃4 , 𝑃5 ][𝑃3 , 𝑃6 , 𝑃1 ][𝑃5 , 𝑃2 , 𝑃3 ]) .

(5.2)

Chapter 5. Poncelet’s Theorem for Triangles

55 𝑇

Replacing 𝑃6 by the general point 𝑃 = [𝑥 𝑦 𝑧] gives rise to the function 𝑓 defined by 𝑓(𝑥, 𝑦, 𝑧) =𝑓(𝑃) = det ([𝑃1 , 𝑃3 , 𝑃5 ][𝑃2 , 𝑃4 , 𝑃5 ][𝑃4 , 𝑃, 𝑃1 ][𝑃, 𝑃2 , 𝑃3 ]) − det ([𝑃2 , 𝑃4 , 𝑃][𝑃1 , 𝑃4 , 𝑃5 ][𝑃3 , 𝑃, 𝑃1 ][𝑃5 , 𝑃2 , 𝑃3 ]) .

(5.3)

The equation 𝑓(𝑥, 𝑦, 𝑧) = 0 is of the form (5.1) and thus gives rise to a conic 𝒞1 . Using elementary properties of determinants, it follows that 𝑃1 , 𝑃2 , 𝑃3 , 𝑃4 , and 𝑃5 are zeros of 𝑓. Thus, these five points are on the conic 𝒞1 and Theorem 5.2 implies that the two conics 𝒞 and 𝒞1 are identical. Consequently, 𝑃6 is also on 𝒞1 . This means that 𝑃6 is a zero of 𝑓 and hence det[𝐾, 𝐿, 𝑀] = 0, showing that 𝐾, 𝐿, and 𝑀 are collinear. If two of the points 𝐾, 𝐿, and 𝑀 were to coincide, then three of the 𝑃𝑗 would be collinear. Since the 𝑃𝑗 lie on the nondegenerate conic 𝒞, this is impossible. Thus, 𝐾, 𝐿, and 𝑀 are distinct. For the converse, we assume that 𝑃1 , 𝑃2 , 𝑃3 , 𝑃4 , and 𝑃5 are in general position. Then these points are zeros of the function 𝑓 defined above. Thus, they are all on the corresponding conic 𝒞, and that conic must be nondegenerate because it has five points in general position. Since 𝐾, 𝐿, and 𝑀 are collinear, we have det[𝐾, 𝐿, 𝑀] = 0. Now (5.2) and (5.3) imply that 𝑃6 is a zero of 𝑓, so 𝑃6 lies on 𝒞. By Theorem 5.2, this conic is unique. We will now see the duality principle in action—this principle allows us to restate Pascal’s theorem in its dual form. In the same way that points can be in general position, a set of lines is said to be in general position if the lines are distinct and no subset of three of the lines has a common intersection point. A diagonal of a hexagon is a line containing two nonadjacent vertices and not just the segment between these two points. Opposite vertices are two vertices for which there is a path along exactly three consecutive line segments of the hexagon from one vertex to the other. The principal diagonals of a hexagon are the diagonals containing opposite vertices. Finally, a hexagon (or polygonal chain) circumscribes a conic 𝒞 if all of its sides are tangent to 𝒞. Theorem 5.4 (Brianchon). If a hexagon circumscribes a nondegenerate conic, then the three principal diagonals are concurrent and distinct.

56

Chapter 5. Poncelet’s Theorem for Triangles

Conversely, if at least five of the sides of a hexagon are in general position and the hexagon has the property that its three principal diagonals are concurrent, then the six sides are tangent to a unique nondegenerate conic. Again, the hexagon need not be convex or simple. We illustrate the theorem in Figure 5.3 with a nondegenerate conic 𝒞 that is an ellipse and a hexagon with sides ℓ1 , ℓ2 , ℓ3 , ℓ4 , ℓ5 , and ℓ6 , all tangent to 𝒞. Let 𝑘 = (ℓ1 × ℓ2 ) × (ℓ4 × ℓ5 ), that is, the line passing through the point of intersection of ℓ1 and ℓ2 and the point of intersection of ℓ4 and ℓ5 . Similarly, ℓ = (ℓ2 × ℓ3 ) × (ℓ5 × ℓ6 ) and 𝑚 = (ℓ1 × ℓ6 ) × (ℓ3 × ℓ4 ). By Brianchon’s theorem, the lines 𝑘, ℓ, and 𝑚 intersect in one point. You will get an even better appreciation for this theorem if you explore our applet ,2 labeled Brianchon’s Theorem.

ℓ ℓ
ℓ 

ℓ



ℓ

ℓ ℓ Figure 5.3. Illustration of Brianchon’s theorem.

The duality principle can also be applied to Theorem 5.2 to give us the result below. Alternatively, this theorem can be derived directly from Brianchon’s theorem. Theorem 5.5. Let 𝒞1 and 𝒞2 be two conics, at least one of which is nondegenerate. If these conics have at least five different tangent lines in common, then they are equal. 2 http://pubapps.bucknell.edu/static/aeshaffer/v1/

Chapter 5. Poncelet’s Theorem for Triangles

57

Blaise Pascal formulated his theorem in 1639 when he was 16 years old. The theorem was included, without proof, along with other projective geometry theorems in a one-page presentation at a meeting organized in Paris by the French monk and mathematician Marin Mersenne. While it is believed that Pascal did have a proof of his theorem, this proof was lost. Pascal only stated the first part of our Theorem 5.3. The converse is often referred to as the Braikenridge–MacLaurin theorem. It was discovered around 1733 and caused a flurry of heated accusations between the two mathematicians about priority and familiarity with the results; see [113] for information about the controversy. Pascal’s theorem is sometimes called the hexagrammum mysticum theorem; the line through the three points is called the Pascal line. It took 167 years until another Frenchman, Charles Brianchon, formulated, proved, and published the dual theorem in 1806. Brianchon was a rather young student at the École Polytechnique in Paris at the time he did this work. Of course, at this time the duality principle had not yet been developed and Brianchon provided an independent proof. The principle was formulated explicitly for the first time in a series of three papers by a third Frenchman, Joseph Gergonne, in the period from 1824 to 1827 [122]. We are now ready for the main theorem of this chapter. Theorem 5.6 (Poncelet for triangles). Let 𝒞 and 𝒦 be nondegenerate conics in ℙ2 (ℝ) that have no point of intersection. Suppose that there is a triangle inscribed in 𝒦 that also circumscribes 𝒞. Then every point on 𝒦 is the vertex of a triangle that circumscribes 𝒞 and is inscribed in 𝒦. The proof hinges on the following lemma. Lemma 5.7. Let 𝒦 be a nondegenerate conic that has two inscribed triangles, △𝐴1 𝐴2 𝐴3 and △𝐵1 𝐵2 𝐵3 , and none of the six vertices coincide. Then the six sides of the triangles are tangent to a unique nondegenerate conic 𝒞. Proof. Let 𝒦 be the conic containing the two inscribed triangles (see Figure 5.4). The six vertices of the triangles are in general position since they are on a nondegenerate conic and we consider the hexagon with the following sequence of vertices: (𝐴2 , 𝐴1 , 𝐵3 , 𝐵1 , 𝐵2 , 𝐴3 ).

58

Chapter 5. Poncelet’s Theorem for Triangles

By Pascal’s theorem, the three points 𝐾 = (𝐴1 × 𝐴2 ) × (𝐵1 × 𝐵2 ), 𝐿 = (𝐴1 × 𝐵3 ) × (𝐴3 × 𝐵2 ), and 𝑀 = (𝐵1 × 𝐵3 ) × (𝐴2 × 𝐴3 ) are collinear. We enumerate the six sides of the two triangles as follows: ℓ1 = 𝐵1 × 𝐵2 ; ℓ2 = 𝐴1 × 𝐴2 ;

ℓ3 = 𝐵2 × 𝐵3 ; ℓ5 = 𝐵1 × 𝐵3 ; ℓ4 = 𝐴2 × 𝐴3 ; ℓ6 = 𝐴1 × 𝐴3 .

With this notation, 𝐾 = ℓ 1 × ℓ2 , 𝑀 = ℓ4 × ℓ5 , and 𝐿 = ((ℓ2 × ℓ6 ) × (ℓ3 × ℓ5 )) × ((ℓ4 × ℓ6 ) × (ℓ1 × ℓ3 )) . We consider the hexagon with these sides in the following order: (ℓ1 , ℓ3 , ℓ5 , ℓ4 , ℓ6 , ℓ2 ). All six lines are in general position, otherwise three of the vertices of the two triangles would be collinear, which is impossible. The collinearity of 𝐾, 𝐿, and 𝑀 implies that the lines 𝑘 = (ℓ1 × ℓ2 ) × (ℓ4 × ℓ5 ), ℓ = (ℓ2 × ℓ6 ) × (ℓ3 × ℓ5 ), and 𝑚 = (ℓ4 × ℓ6 ) × (ℓ1 × ℓ3 ) all pass through one point, 𝐿. The lines 𝑘, ℓ, and 𝑚 are precisely the three diagonals of the hexagon defined by the six lines. By Brianchon’s theorem, the six lines are tangent to a unique nondegenerate conic 𝒞. In Figure 5.4, the hexagon determined by the lines (ℓ1 , ℓ3 , ℓ5 , ℓ4 , ℓ6 , ℓ2 ) has vertices (𝐵2 , 𝐵3 , 𝑀, 𝐴3 , 𝐴1 , 𝐾) and the conic 𝒞 is inscribed in this hexagon. We now apply this lemma to prove Poncelet’s theorem for triangles. Proof of Theorem 5.6. Let △𝐴1 𝐴2 𝐴3 be a triangle inscribed in a nondegenerate conic 𝒦 and circumscribing a nondegenerate conic 𝒞. Further, let 𝑃 = 𝐵1 be a point on 𝒦, different from the vertices of the triangle. We construct the tangent lines from 𝐵1 to the conic 𝒞 and denote the points of intersection of these tangent lines with 𝒦 by 𝐵2 and 𝐵3 .

Chapter 5. Poncelet’s Theorem for Triangles

59 


ℓ









ℓ

ℓ

ℓ






ℓ





ℓ



 

ℓ


Figure 5.4. Construction in the proof of Lemma 5.7.

A quick case-by-case study of triangles inscribed in one conic and circumscribing another one reveals that the points 𝐵2 and 𝐵3 do in fact exist; it is impossible under the hypothesis of the theorem to have a line through 𝐵1 tangent to both conics. By construction, the six points on 𝒦 are distinct and five of the six sides of the triangles △𝐴1 𝐴2 𝐴3 and △𝐵1 𝐵2 𝐵3 are tangent to the conic 𝒞. By Lemma 5.7, the two triangles circumscribe a conic 𝒞′ . Since 𝒞 and 𝒞′ have five tangent lines in common, they are the same by Theorem 5.5. What does Poncelet’s theorem have to do with Blaschke products? To answer this question we first have to explain how the complex plane of Chapter 3 and the real projective plane of Chapter 5 interact: The points (𝑥, 𝑦, 𝑧) of ℙ2 (ℝ) with 𝑧 ≠ 0 have a unique representation (𝑋, 𝑌) with 𝑋 = 𝑥/𝑧 and 𝑌 = 𝑦/𝑧. The set of these points is the affine part of ℙ2 (ℝ), and this part may be identified with ℂ using the identification 𝑧 = 𝑋+𝑖𝑌. In the context of Chapter 3, the outer conic in Poncelet’s theorem is the unit circle 𝕋. The inner conic is an ellipse, and the natural question to ask is: Which ellipses satisfy the condition for Poncelet’s theorem and thus have the property that every point on 𝕋 is the vertex of a triangle

60

Chapter 5. Poncelet’s Theorem for Triangles

inscribed in 𝕋 that circumscribes the ellipse? We call such ellipses Poncelet 3-ellipses. Actually, we know the answer to this question because it is contained in Theorem 2.9 and Corollary 4.4. Corollary 5.8. The Blaschke 3-ellipses are precisely the Poncelet 3-ellipses.

Chapter

6

The Numerical Range We now return to the numerical range, introduced by Otto Toeplitz in a 1918 paper entitled Das algebraische Analogon zu einem Satz von Fejér (The algebraic analogue of a theorem of Fejér). Toeplitz wanted to associate a complex 𝑛 × 𝑛 matrix with a set in the complex plane, a set that he called the Wertevorrat (which can be translated as supply of values). In English, this same set is called the field of values or, more commonly, the numerical range. The mathematician Eugene Gutkin (who does not like either expression) notes that, “The former adds one more item to the litany of mathematical ‘fields’; the latter is plain awkward. The original name is better in every respect except one: It is German and therefore unacceptable in the English literature” [71]. Following more recent trends, we refer to the numerical range. So recall that given an 𝑛×𝑛 matrix 𝐴, we defined the numerical range 𝑊(𝐴) by 𝑊(𝐴) = {⟨𝐴𝑥, 𝑥⟩ ∶ 𝑥 ∈ ℂ𝑛 , ‖𝑥‖ = 1}. We saw examples of numerical ranges in Chapter 2; one was a circular disk and the other an elliptical disk. Can a set containing just one point be the numerical range of an 𝑛 × 𝑛 matrix with 𝑛 > 1? What about a set containing just two distinct points? The first answer is yes and easy to see, while the second answer is no and not as easy to see—but it will become clear in a moment. Toeplitz had a general idea of what the numerical range might look like: He conjectured that the numerical range of an 𝑛 × 𝑛 matrix is always a convex set, and he showed that the outer boundary is a convex curve. Felix Hausdorff proved Toeplitz’s

61

62

Chapter 6. The Numerical Range

conjecture a year later [78], and this theorem is now called The Toeplitz– Hausdorff theorem. In particular, no matrix can have a set of exactly two points (or 𝑚 > 1 points, for that matter) as its numerical range.1 It will soon be clear to the reader that not every convex set can be the numerical range of an 𝑛 × 𝑛 matrix. Which convex sets can be? This is a difficult question. Toeplitz did show that the numerical range of a 2 × 2 matrix is either a point, a line segment, or an elliptical disk. All of these can be thought of as elliptical disks, though some would be degenerate. Thus, it was Toeplitz who first proved the elliptical range theorem that we dicussed in Chapter 2, and we may think of this as a first step toward describing the sets that can be the numerical range of an 𝑛 × 𝑛 matrix. The Toeplitz–Hausdorff theorem is a relatively easy consequence of Theorem 6.1. In [75], Halmos wrote, “the Toeplitz–Hausdorff theorem says that the numerical range of every operator is a convex subset of the complex plane. It is disappointing that all known proofs of this elegant statement are ugly. The methods are elementary, but the arguments are computational”. We will provide two proofs of the Toeplitz–Hausdorff theorem. The first proof, which Halmos would have been aware of at the time the quote was written, will be as a corollary of Theorem 6.1. Is the proof ugly? We leave that up to you. The second and more recent proof appears at the end of this chapter and is a consequence of the intermediate value theorem. We begin by stating the elliptical range theorem. Then we include the necessary tools to prove it. Finally, we present the proof of the theorem. The Toeplitz–Hausdorff theorem will follow. Theorem 6.1 (Elliptical range theorem). Let 𝐴 be a 2 × 2 matrix with eigenvalues 𝑎 and 𝑏. Then the numerical range of 𝐴 is an elliptical disk with foci at 𝑎 and 𝑏 and minor axis of length (tr(𝐴⋆ 𝐴) − |𝑎|2 − |𝑏|2 )1/2 . To prove this, we need some facts about matrices and their numerical ranges. Below, we let 𝐼𝑛 denote the 𝑛 × 𝑛 identity matrix. Recall that the adjoint, or conjugate transpose, of an 𝑛 × 𝑛 matrix 𝐴 is denoted 𝐴⋆ and we say 𝐴 is self-adjoint if 𝐴 = 𝐴⋆ . It is said to be unitary if it satisfies 𝐴⋆ = 𝐴−1 and normal if 𝐴⋆ 𝐴 = 𝐴𝐴⋆ . Of course, all self-adjoint and 1 It is noted in [71] that this theorem was implicitly proved by Toeplitz in his 1918 paper, though Halmos states [76, p. 110] that Toeplitz proved that the boundary is a convex curve, while Hausdorff showed that the interior had no holes.

Chapter 6. The Numerical Range

63

all unitary matrices are normal. For 𝑗 = 1, … , 𝑛 we let 𝑒𝑗 denote the standard 𝑗th basis vector in ℂ𝑛 ; that is, the coordinates of 𝑒𝑗 are all zero, except the 𝑗th, which is one. Recall that two 𝑛 × 𝑛 matrices 𝐴 and 𝐵 are similar if there exists an invertible matrix 𝑆 such that 𝐴 = 𝑆 −1 𝐵𝑆. If 𝑆 is unitary, we say the matrices 𝐴 and 𝐵 are unitarily equivalent. We see that many properties of matrices are preserved under unitary equivalence; in particular, the numerical range is preserved. Theorem 6.2 is a list of some of the results that we need as well as some that we do not need but that we find too interesting to omit. Proving parts (1)–(3) would be a good exercise for the reader. Schur’s theorem (see (4) below) says that every 𝑛 × 𝑛 matrix 𝐴 is unitarily equivalent to an upper triangular matrix. Since the diagonal entries of an upper triangular matrix are the eigenvalues, Schur’s theorem is often an important tool in problems about eigenvalues. In part (8), we obtain a bound on the size of the numerical range in terms of the norm of the matrix ‖𝐴‖ = sup ‖𝐴𝑥‖. ‖𝑥‖=1

If you are not familiar with matrix norms, you should show that ‖𝐴𝑥‖ ‖𝐴‖ = sup { ∶ 𝑥 ∈ ℂ𝑛 , 𝑥 ≠ 0} ‖𝑥‖ and ‖𝐴𝑥‖ ≤ ‖𝐴‖‖𝑥‖ for all 𝑥 ∈ ℂ𝑛 . In Chapter 2 we mentioned that the numerical range of a matrix is a singleton, {𝜆}, if and only if 𝐴 = 𝜆𝐼 and we promised a proof later. Parts (6) and (7) constitute the promised proof. Theorem 6.2. Let 𝐴 be an 𝑛 × 𝑛 matrix. (1) If 𝑈 is an 𝑛 × 𝑛 unitary matrix, then the numerical range of 𝑈 ⋆ 𝐴𝑈 is equal to the numerical range of 𝐴; that is, 𝑊(𝑈 ⋆ 𝐴𝑈) = 𝑊(𝐴). (2) 𝑊(𝐴⋆ ) = {𝜆 ∶ 𝜆 ∈ 𝑊(𝐴)}. (3) If 𝛼 and 𝛽 are complex numbers, then 𝑊(𝛼𝐴 + 𝛽𝐼) = 𝛼𝑊(𝐴) + 𝛽 ∶= {𝛼𝑧 + 𝛽 ∶ 𝑧 ∈ 𝑊(𝐴)}.

64

Chapter 6. The Numerical Range

(4) (Schur’s theorem) There is a unitary matrix 𝑈 such that 𝑈 ⋆ 𝐴𝑈 is upper triangular. (5) The numerical range of 𝐴 is compact. (6) The numerical range of 𝐴 satisfies 𝑊(𝐴) = {0} if and only if 𝐴 is the 𝑛 × 𝑛 zero matrix. (7) The numerical range of 𝐴 satisfies 𝑊(𝐴) = {𝜆} if and only if 𝐴 = 𝜆𝐼. (8) For 𝑥 ∈ ℂ𝑛 with norm 1, we have |⟨𝐴𝑥, 𝑥⟩| ≤ ‖𝐴‖. Proof. We prove only (4), (5), (6), (7), and (8) here. The proof of Schur’s theorem proceeds by induction. Since the result is clearly true for a 1 × 1 matrix, we show that if, for 𝑛 ∈ ℤ+ , the result holds for 𝑛 × 𝑛 matrices, then for every (𝑛 + 1) × (𝑛 + 1) matrix 𝐴 there is a unitary matrix 𝑈 with 𝑈 ⋆ 𝐴𝑈 upper triangular. So, let 𝐴 be an (𝑛 + 1) × (𝑛 + 1) matrix. Since we work with complex numbers, we may choose an eigenvalue 𝜆1 of 𝐴 and a corresponding unit eigenvector 𝑣1 . Now use the Gram– Schmidt process to obtain an orthonormal basis (𝑣1 , 𝑣2 , … , 𝑣𝑛+1 ). Let 𝑄 denote the (𝑛 + 1) × (𝑛 + 1) matrix for which the 𝑗th column is 𝑣𝑗 and note that because the columns are orthonormal we have 𝑄⋆ 𝑄 = 𝐼; that is, 𝑄 is a unitary matrix. As the reader should check, there exists an 𝑛 × 𝑛 matrix 𝐴𝑛 such that 𝜆 ⎡ 1 0 ⎢ 𝑄⋆ 𝐴𝑄 = ⎢ ⋮ ⎢ ⎣ 0

∗⋯∗ 𝐴𝑛

⎤ ⎥ ⎥. ⎥ ⎦

So our induction hypothesis applied to 𝐴𝑛 implies that there exists 𝑄𝑛 such that 𝑄𝑛 is a unitary matrix and 𝑄𝑛⋆ 𝐴𝑛 𝑄𝑛 is upper triangular. We need the action of 𝑄𝑛 and 𝑄, so we form the following matrix in order to be able to multiply:

Chapter 6. The Numerical Range

65

1 0⋯0 ⎡ ⎤ ⎢ 0 ⎥ 𝐷=⎢ ⎥. ⋮ 𝑄 𝑛 ⎢ ⎥ ⎣ 0 ⎦ Now it is not difficult to see that 𝑈 = 𝑄𝐷 is unitary and 𝑈 ⋆ 𝐴𝑈 = 𝐷 ⋆ 𝑄⋆ 𝐴𝑄𝐷 = [

1 𝟎

𝟎 𝜆 ][ 1 𝑄𝑛⋆ 𝟎

∗⋯∗ 1 𝟎 𝜆 ][ ]=[ 1 𝐴𝑛 𝟎 𝑄𝑛 𝟎

∗⋯∗ ]. 𝑄𝑛⋆ 𝐴𝑛 𝑄𝑛

Since 𝑄𝑛⋆ 𝐴𝑛 𝑄𝑛 is upper triangular, the proof of Schur’s theorem is complete. For (5), we note that ⟨𝐴𝑥, 𝑥⟩ maps the (compact) unit sphere {𝑥 ∈ ℂ𝑛 ∶ ‖𝑥‖ = 1} continuously onto the numerical range, 𝑊(𝐴). Thus, 𝑊(𝐴) is compact. Now consider (6) for the matrix 𝐴. If 𝐴 is the zero matrix, it is clear that 𝑊(𝐴) = {0}, so suppose that the numerical range of 𝐴 is {0}. Then ⟨𝐴𝑥, 𝑥⟩ = 0 for all unit vectors 𝑥. But, for a nonzero vector 𝑥, we have ⟨𝐴𝑥, 𝑥⟩ = 0 if and only if ⟨𝐴(𝑥/‖𝑥‖), 𝑥/‖𝑥‖⟩ = 0, so ⟨𝐴𝑥, 𝑥⟩ = 0 for all 𝑥. Since ⟨𝐴𝑥, 𝑥⟩ = 0 for every vector 𝑥, we have ⟨𝐴⋆ 𝑥, 𝑥⟩ = ⟨𝑥, 𝐴𝑥⟩ = ⟨𝐴𝑥, 𝑥⟩ = 0, and if we let 𝑇 ∶= 𝐴 + 𝐴⋆ , then 𝑇 is self-adjoint and ⟨𝑇𝑥, 𝑥⟩ = 0 for all vectors 𝑥. The following identity, which is true for all self-adjoint operators, is easily checked by expansion: For any two vectors 𝑥 and 𝑦, we have 1 ⟨𝑇𝑥, 𝑦⟩ = (⟨𝑇(𝑥 + 𝑦), (𝑥 + 𝑦)⟩ − ⟨𝑇(𝑥 − 𝑦), (𝑥 − 𝑦)⟩ 4 + 𝑖⟨𝑇(𝑥 + 𝑖𝑦), (𝑥 + 𝑖𝑦)⟩ − 𝑖⟨𝑇(𝑥 − 𝑖𝑦), (𝑥 − 𝑖𝑦)⟩). (6.1) Therefore, ⟨𝑇𝑥, 𝑦⟩ = 0 for all 𝑥 and 𝑦. Thus, if 𝑥 is given and we choose 𝑦 = 𝑇𝑥 we see that ‖𝑇𝑥‖ = 0. So 𝑇 = 0 and 𝐴 + 𝐴⋆ = 0. Consider 𝑆 ∶= 𝑖𝐴 − 𝑖𝐴⋆ . Then 𝑆 ⋆ = −𝑖𝐴⋆ + 𝑖𝐴 = 𝑆, so 𝑆 is self adjoint and ⟨𝑆𝑥, 𝑥⟩ = 0 for all 𝑥. Therefore, (6.1) applies and we see that 𝐴 − 𝐴⋆ = 0. Consequently, 𝐴 = 0, completing the proof of (6).

66

Chapter 6. The Numerical Range

For (7), note that by (3) we know 𝑊(𝐴 − 𝜆𝐼) = 𝑊(𝐴) − 𝜆. Therefore, 𝑊(𝐴) = {𝜆} if and only if 𝑊(𝐴 − 𝜆𝐼) = {0}, and that happens (by (6)) if and only if 𝐴 − 𝜆𝐼 = 0. Finally, (8) relies on the Cauchy–Bunyakovsky–Schwarz inequality [7, p. 172]; that is, the fact that for 𝑢, 𝑣 ∈ ℂ𝑛 , we have |⟨𝑢, 𝑣⟩| ≤ ‖𝑢‖ ‖𝑣‖. Thus, for a unit vector 𝑥, we have |⟨𝐴𝑥, 𝑥⟩| ≤ ‖𝐴𝑥‖ ⋅ ‖𝑥‖ ≤ ‖𝐴‖, completing the proof. Here is an interesting consequence of Schur’s theorem. If we have a 2 × 2 matrix 𝐴, then by Schur’s theorem 𝐴 is unitarily equivalent to ̃ an upper triangular matrix 𝐴.̃ We know three of the entries of 𝐴—the diagonal entries are the eigenvalues and the entry in the (2, 1) position is zero. What, if anything, can we say about the entry in the (1, 2) position? Lemma 6.3. Let 𝐴 be a 2 × 2 matrix. Then 𝐴 is unitarily equivalent to an upper triangular matrix for which the entry in the first row and second column is positive. To familiarize yourself with the techniques in this chapter, we recommend proving this yourself before reading on. Proof. Let 𝜆1 and 𝜆2 denote the eigenvalues of 𝐴. By Schur’s lemma, 𝐴 is unitarily equivalent to an upper triangular matrix 𝜆 𝐴̃ ∶= [ 1 0

𝑐 ] 𝜆2

for some appropriate value of 𝑐 = |𝑐|𝑒𝑖𝜃 . Using the unitary matrix 𝑈 defined by 1 0 𝑈=[ ], 0 𝑒−𝑖𝜃 we see that 𝜆 |𝑐| 𝜆 𝑐 𝐴̃ ∶= 𝑈 ⋆ [ 1 ] ]𝑈 = [ 1 0 𝜆2 0 𝜆2 is unitarily equivalent to 𝐴̃ and therefore to 𝐴.

Chapter 6. The Numerical Range

67

The spectral theorem for normal matrices is particularly beautiful and we are now in a position to prove it. Theorem 6.4 (Spectral theorem for normal matrices). Let 𝐴 be an 𝑛 × 𝑛 matrix with eigenvalues 𝑎1 , … , 𝑎𝑛 . The following are equivalent. (1) The matrix A is normal. (2) There exists a unitary matrix 𝑈 such that 𝑈 ⋆ 𝐴𝑈 is a diagonal matrix. (3) There is an orthonormal set consisting of 𝑛 eigenvectors of 𝐴. Proof. Evidently, (2) and (3) are equivalent. To see that (1) implies (2), suppose that 𝐴 is normal and use Schur’s theorem to choose a unitary matrix 𝑈 for which 𝐴̃ ∶= 𝑈 ⋆ 𝐴𝑈 = (𝑎𝑖𝑗 ) is upper triangular. So, letting 𝑒𝑗 denote the 𝑗th standard basis vector 𝑛 of ℂ𝑛 , we have 𝐴𝑒̃ 1 = 𝑎11 𝑒1 , while 𝐴⋆̃ 𝑒1 = (𝑈 ⋆ 𝐴⋆ 𝑈)𝑒1 = ∑𝑗=1 𝑎1𝑗 𝑒𝑗 . Thus, 𝑛

‖𝐴𝑒̃ 1 ‖2 = |𝑎11 |2 while ‖𝐴⋆̃ 𝑒1 ‖2 = ∑ |𝑎1𝑗 |2 . 𝑗=1

Now, ‖𝐴𝑒̃ 1 ‖2 = ‖𝑈 ⋆ 𝐴𝑈𝑒1 ‖2 = ⟨𝑈 ⋆ 𝐴𝑈𝑒1 , 𝑈 ⋆ 𝐴𝑈𝑒1 ⟩ = ⟨𝑈 ⋆ 𝐴⋆ 𝐴𝑈𝑒1 , 𝑒1 ⟩. But 𝐴 is normal, so ‖𝐴𝑒̃ 1 ‖2 = ⟨𝑈 ⋆ 𝐴𝐴⋆ 𝑈𝑒1 , 𝑒1 ⟩ = ⟨(𝑈 ⋆ 𝐴𝑈)(𝑈 ⋆ 𝐴⋆ 𝑈)𝑒1 , 𝑒1 ⟩ = ⟨𝐴̃𝐴⋆̃ 𝑒1 , 𝑒1 ⟩ = ‖𝐴⋆̃ 𝑒1 ‖2 . 𝑛

Therefore, |𝑎11 |2 = ∑𝑗=1 |𝑎1𝑗 |2 , and we conclude that 𝑎12 = 𝑎13 = ⋯ = 𝑎1𝑛 = 0. Continuing in this manner shows that all off-diagonal entries will be zero and the matrix will be diagonal. To see that all three statements are equivalent, we now show that (2) implies (1): We reduce the problem to diagonal matrices, which we know commute. By assumption, 𝐴 = 𝑈𝐷𝑈 ⋆ for some unitary matrix 𝑈 and diagonal matrix 𝐷, 𝐴⋆ = 𝑈𝐷 ⋆ 𝑈 ⋆ , and 𝐴𝐴⋆ = (𝑈𝐷𝑈 ⋆ )(𝑈𝐷 ⋆ 𝑈 ⋆ ) = 𝑈𝐷𝐷 ⋆ 𝑈 ⋆ = 𝑈𝐷 ⋆ 𝐷𝑈 ⋆ = (𝑈𝐷 ⋆ 𝑈 ⋆ )(𝑈𝐷𝑈 ⋆ ) = 𝐴⋆ 𝐴, and therefore 𝐴 is normal.

68

Chapter 6. The Numerical Range

Exercise 6.5. For 𝑎 > 0, let 𝐴𝑎 = [

1 −1/𝑎

𝑎 ]. −1

Show that 𝐴𝑎 is unitarily equivalent to the matrix 𝑇(1+𝑎2 )/𝑎 = [

0 (1 + 𝑎2 )/𝑎 ]. 0 0

It is easy to check that the eigenvalues of 𝐴𝑎 are both 0. Thus, Schur’s theorem tells us that 𝐴𝑎 is unitarily equivalent to a matrix with zeros on the diagonal. There are several different ways to find the (1, 2) entry. One elementary method is to find a unit eigenvector for 𝐴𝑎 and a second unit vector orthogonal to it. Then use these two vectors to find a unitary matrix 𝑈 satisfying 𝑈 ⋆ 𝐴𝑎 𝑈 = 𝑇(1+𝑎2 )/𝑎 . We leave the details to the reader. Lemma 6.6. Let 𝐴 be an 𝑛 × 𝑛 matrix, 𝛾 ∈ ℝ ⧵ {0}, and (𝐴 + 𝐴⋆ ) + 𝛾(𝐴 − 𝐴⋆ ) . 2 Then, for 𝑥 and 𝑦 in ℝ, we have 𝑥 + 𝑖𝑦 ∈ 𝑊(𝐴) if and only if 𝑥 + 𝑖𝛾𝑦 ∈ 𝑊(𝐶). 𝐶=

This is another good exercise for the reader. We now turn to the proof of the elliptical range theorem (Theorem 6.1). There are many proofs of this theorem. Donoghue, whose 1957 paper is essential reading for those who want to know more, introduced his proof of this theorem by saying, “Most of it is established in [147], but with a rather complicated proof.” We use a proof from the paper A simple proof of the elliptical range theorem by Li [102]. We isolate the special case of the numerical range of a normal matrix for later reference in a lemma. You should think of this lemma as saying that for normal operators the numerical range is the smallest set it can possibly be, once we know that it contains the eigenvalues of the matrix and is convex. Lemma 6.7. Let 𝐴 be a 2×2 normal matrix. Then 𝑊(𝐴) is a line segment with the eigenvalues of 𝐴 as endpoints. Proof. Since 𝐴 is a normal matrix, we know from the spectral theorem that 𝐴 is unitarily equivalent to a diagonal matrix with eigenvalues 𝑎 and

Chapter 6. The Numerical Range

69

𝑏 on the main diagonal. By Theorem 6.2, this diagonal matrix will have the same numerical range as 𝐴. In this case, a computation shows that if we write 𝑥 𝑥 = [ 1 ] , where ‖𝑥‖ = 1, 𝑥2 then 𝑊(𝐴) = {𝑎|𝑥1 |2 + 𝑏|𝑥2 |2 ∶ 𝑥1 , 𝑥2 ∈ ℂ, |𝑥1 |2 + |𝑥2 |2 = 1}. This is a line segment with endpoints 𝑎 and 𝑏. Note that 𝑎 and 𝑏 are the eigenvalues of 𝐴 and that in the event that 𝑎 = 𝑏, the line segment reduces to a point. Thus, we may view the numerical range of a normal matrix as a degenerate ellipse with eigenvalues 𝑎 and 𝑏 and minor axis of length 0. Does this match the formula provided by the elliptical range theorem? Two elementary properties give us the answer. First, for square matrices 𝐴 and 𝐵, we have tr(𝐴𝐵) = tr(𝐵𝐴). Second, we know that our matrix 𝐴 is, in this special case, unitarily equivalent to a diagonal matrix 𝐷 with the eigenvalues of 𝐴, 𝑎 and 𝑏, on the diagonal. Therefore, tr(𝐴⋆ 𝐴) = tr(𝑈 ⋆ 𝐴⋆ 𝑈𝑈 ⋆ 𝐴𝑈) = tr(𝐷 ⋆ 𝐷) = |𝑎|2 + |𝑏|2 , 1/2

so the length of the minor axis is (tr(𝐴⋆ 𝐴) − |𝑎|2 − |𝑏|2 ) = 0, as it should in case the numerical range is a line segment or a point. We turn to the proof of the elliptical range theorem for the general case. Our first step requires replacing 𝐴 by a matrix 𝐴1 ∶= 𝐴−(tr(𝐴)/2)𝐼. It is a really good exercise to derive the length of the minor axis for the ellipse bounding 𝑊(𝐴) from that bounding 𝑊(𝐴1 ). Proof of Theorem 6.1. In view of Lemma 6.7 we assume that 𝐴 is not normal. By (3) in Theorem 6.2, if we replace our matrix 𝐴 with 𝐴1 ∶= 𝐴 − (tr(𝐴)/2)𝐼, we see that 𝑊(𝐴1 )+tr(𝐴)/2 = 𝑊(𝐴) and the numerical range of 𝐴1 is an ellipse if and only if the numerical range of 𝐴 is. Then, by considering 𝐴1 in place of 𝐴, we may assume that the trace of our matrix 𝐴1 is 0. First, suppose that one eigenvalue of 𝐴1 is zero. Since unitarily equivalent matrices have the same trace, it is clear from the upper triangular matrix that if one eigenvalue of 𝐴1 is zero the other must be. By

70

Chapter 6. The Numerical Range

Lemma 6.3, there exists a positive real number 𝑐 such that 𝐴1 is unitarily equivalent to 0 𝑐 [ ]. 0 0 But this is just 𝑐 times the matrix in Example 2.1. It now follows from Theorem 6.2, (3) that 𝑊(𝐴1 ) is a circular disk centered at 0 of radius 𝑐/2. To check that we have obtained the length of the minor axis described in Theorem 6.1, we note that tr(𝐴⋆1 𝐴1 ) = 𝑐2 and therefore 1/2

(tr(𝐴⋆1 𝐴1 ) − |𝑎|2 − |𝑏|2 )

= 𝑐,

as the theorem claims. The result for the original matrix 𝐴 follows. Finally, suppose that 𝐴1 has a nonzero eigenvalue 𝑎. By our assumption that tr(𝐴1 ) = 0, the other eigenvalue must be −𝑎. If we replace 𝐴1 by 𝐴2 = (1/𝑎)𝐴1 , this new matrix has eigenvalues 1 and −1. By Lemma 6.3, we may replace 𝐴2 by a unitarily equivalent matrix 𝐴3 with a positive entry 2𝑐 in the (1, 2) position. By Theorem 6.2, the numerical range of 𝐴3 will be an elliptical disk if and only if the numerical range of 𝐴 was. Thus, we assume 𝐴3 = [

1 2𝑐 ]. 0 −1

If we let 𝐶=

(𝐴3 + 𝐴⋆3 ) +

√1+𝑐2 𝑐

(𝐴3 − 𝐴⋆3 )

2

,

then a computation shows that 𝐶=[

1 −1/(𝑐 + √1 + 𝑐2 )

𝑐 + √1 + 𝑐2 ]. −1

Using Exercise 6.5 with 𝑎 = 𝑐 + √1 + 𝑐2 , we see that 𝐶 is unitarily equivalent to [

0 2√1 + 𝑐2 ]. 0 0

Chapter 6. The Numerical Range

71

But we saw above that 𝑊(𝐶) is a circular disk centered at 0 of radius √1 + 𝑐2 , so 𝑊(𝐶) = {𝑠√1 + 𝑐2 cos 𝑡 + 𝑖𝑠√1 + 𝑐2 sin 𝑡 ∶ 𝑡 ∈ ℝ and 0 ≤ 𝑠 ≤ 1}. By Lemma 6.6 with 𝛾 = √1 + 𝑐2 /𝑐, we have 𝑊(𝐴3 ) = {𝑠√1 + 𝑐2 cos 𝑡 + 𝑖𝑐 𝑠 sin 𝑡 ∶ 𝑡 ∈ ℝ and 0 ≤ 𝑠 ≤ 1}, which is an elliptical disk. Again, we can check that the formula for the minor axis holds in this case. It is time for the proof of the Toeplitz–Hausdorff theorem. The idea is that if you want to show that a set is convex, you have to pick two points and show that the line segment joining them is in your set. Once you pick two points, you have essentially reduced the problem to one of dimension two, and that is what the elliptical range theorem can handle. Before we begin, recall that if 𝑆 is a subspace of ℂ𝑛 , then 𝑆 ⟂ (which we read as “𝑆 perp”) denotes the orthogonal complement of 𝑆; that is, 𝑆 ⟂ = {𝑦 ∈ ℂ𝑛 ∶ ⟨𝑠, 𝑦⟩ = 0 for all 𝑠 ∈ 𝑆}. Every vector 𝑥 ∈ ℂ𝑛 can be written uniquely as the sum of a vector in 𝑆 and a vector in 𝑆 ⟂ ; that is, 𝑥 = 𝑥𝑆 ⊕ 𝑥𝑆 ⟂ , where 𝑥𝑆 ∈ 𝑆 and 𝑥𝑆 ⟂ ∈ 𝑆 ⟂ . Let 𝑃𝑆 denote the orthogonal projection defined on ℂ𝑛 by 𝑃𝑆 (𝑥) = 𝑥𝑆 . Further, for 𝑥 and 𝑦 in ℂ𝑛 , ⟨𝑥, 𝑃𝑆 𝑦⟩ = ⟨𝑥, 𝑦𝑆 ⟩ = ⟨𝑥𝑆 , 𝑦𝑆 ⟩ = ⟨𝑃𝑆 𝑥, 𝑦⟩. 𝑛

(6.2)

𝑛

Given an operator 𝐴 on ℂ , let 𝐴|𝑆 ∶ 𝑆 → ℂ denote the operator restricted to 𝑆 and let 𝐴𝑆 ∶= 𝑃𝑆 𝐴|𝑆 . Then using (6.2), if 𝜆 ∈ 𝑊(𝐴𝑆 ), there exists a unit vector 𝑥 ∈ 𝑆 ⊂ ℂ𝑛 with 𝜆 = ⟨𝐴𝑆 𝑥, 𝑥⟩ = ⟨𝑃𝑆 𝐴|𝑆 𝑥, 𝑥⟩ = ⟨𝑃𝑆 𝐴𝑥, 𝑥⟩ = ⟨𝐴𝑥, 𝑃𝑆 𝑥⟩ = ⟨𝐴𝑥, 𝑥⟩. Therefore, 𝑊(𝐴𝑆 ) ⊆ 𝑊(𝐴).

(6.3)

Note that we have “compressed” the operator 𝐴 to a two-dimensional space and thus we can apply the elliptical range theorem to this compression, 𝐴𝑆 .

72

Chapter 6. The Numerical Range

Theorem 6.8 (The Toeplitz–Hausdorff theorem). The numerical range of an 𝑛 × 𝑛 matrix is convex. Proof. We need to show that the numerical range of a matrix 𝐴 contains the line segment joining any two points in the numerical range. This is only interesting if the points are distinct, so let us choose two points 𝜆1 and 𝜆2 in 𝑊(𝐴) with 𝜆1 ≠ 𝜆2 , and we check that the line segment joining them is in the numerical range. For 𝑗 = 1, 2 there exists a unit vector 𝑥𝑗 such that ⟨𝐴𝑥𝑗 , 𝑥𝑗 ⟩ = 𝜆𝑗 . Note that 𝑥1 and 𝑥2 are linearly independent. Our goal is to figure out how to reduce this problem to the 2 × 2 case so that we can apply the elliptical range theorem. So let 𝑆 = span{𝑥1 , 𝑥2 }. Note that 𝑃𝑆 (𝑥𝑗 ) = 𝑥𝑗 for 𝑗 = 1, 2 and 𝜆𝑗 = ⟨𝐴𝑥𝑗 , 𝑥𝑗 ⟩ = ⟨𝐴|𝑆 𝑥𝑗 , 𝑃𝑆 𝑥𝑗 ⟩ = ⟨𝑃𝑆 𝐴|𝑆 𝑥𝑗 , 𝑥𝑗 ⟩ = ⟨𝐴𝑆 𝑥𝑗 , 𝑥𝑗 ⟩ for 𝑗 = 1, 2. But 𝐴𝑆 ∶ 𝑆 → 𝑆 and therefore we may apply the elliptical range theorem to conclude that 𝑊(𝐴𝑆 ) is an elliptical disk. Since 𝜆1 and 𝜆2 lie in the elliptical disk, the line segment joining 𝜆1 and 𝜆2 does as well. By (6.3), the line segment joining 𝜆1 and 𝜆2 is contained in 𝑊(𝐴), and that completes the proof. Since Halmos’s article appeared, there have been many other proofs of the Toeplitz–Hausdorff theorem; see, for example, [38] and [70]. We include a sketch of a proof we find particularly appealing. We learned of this proof from Crouzeix. One more proof of the Toeplitz–Hausdorff theorem. We wish to show that the numerical range of an 𝑛 × 𝑛 matrix 𝐴 is convex. Let 𝜆1 and 𝜆2 be elements of 𝑊(𝐴). If 𝜆1 = 𝜆2 , the result is clear, so we assume 𝜆1 ≠ 𝜆2 . Shifting and scaling the matrix (see Theorem 6.2) by considering 𝛾𝐴 + 𝛼𝐼 will not change the convexity of the numerical range, so we may assume that 𝜆1 = 0 and 𝜆2 = 1. By the definition of numerical range, there are unit vectors 𝑢 and 𝑣 with ⟨𝐴𝑢, 𝑢⟩ = 1 and ⟨𝐴𝑣, 𝑣⟩ = 0. We claim that we can find 𝜃 ∈ [0, 2𝜋) with 𝑒𝑖𝜃 𝑣 satisfying ⟨𝐴𝑢, 𝑒𝑖𝜃 𝑣⟩ + ⟨𝐴(𝑒𝑖𝜃 𝑣), 𝑢⟩ ∈ ℝ.

(6.4)

Assuming this is true (just for the moment; we return to this below), here is how the rest of the proof goes.

Chapter 6. The Numerical Range

73

Replacing 𝑒𝑖𝜃 𝑣 by 𝑣 in (6.4), we may assume without loss of generality that ⟨𝐴𝑢, 𝑣⟩ + ⟨𝐴𝑣, 𝑢⟩ ∈ ℝ. Define 𝜑 ∶ [0, 1] → 𝑊(𝐴) ∩ ℝ by 𝜑(𝑡) ∶=

⟨𝐴((1 − 𝑡)𝑢 + 𝑡𝑣), (1 − 𝑡)𝑢 + 𝑡𝑣⟩ . ‖(1 − 𝑡)𝑢 + 𝑡𝑣‖2

It is not difficult to check that 𝜑 is a rational function of 𝑡, and it is a good exercise to think about why the denominator cannot vanish on the interval [0, 1]. It follows that 𝜑 is continuous and maps [0, 1] into ℝ. But 𝜑(0) = 1 and 𝜑(1) = 0. So by the intermediate value theorem, the line segment joining 0 and 1 is also in the numerical range, and we have completed the proof, modulo the argument that (6.4) holds. To see that (6.4) is true, note that ⟨𝑣, (𝐴 − 𝐴⋆ )𝑢⟩ = ⟨(𝐴 − 𝐴⋆ )𝑢, 𝑣⟩ and therefore we may choose 𝜃 so that 𝑒𝑖𝜃 ⟨𝑣, (𝐴 − 𝐴⋆ )𝑢⟩ = 𝑒−𝑖𝜃 ⟨(𝐴 − 𝐴⋆ )𝑢, 𝑣⟩. Thus, 𝑒𝑖𝜃 ⟨𝑣, 𝐴𝑢⟩ + 𝑒−𝑖𝜃 ⟨𝑢, 𝐴𝑣⟩ = 𝑒−𝑖𝜃 ⟨𝐴𝑢, 𝑣⟩ + 𝑒𝑖𝜃 ⟨𝐴𝑣, 𝑢⟩, which yields ⟨𝐴𝑢, 𝑒𝑖𝜃 𝑣⟩ + ⟨𝐴(𝑒𝑖𝜃 𝑣), 𝑢⟩ = ⟨𝐴𝑢, 𝑒𝑖𝜃 𝑣⟩ + ⟨𝐴(𝑒𝑖𝜃 𝑣), 𝑢⟩.

It is now time to see how the numerical range, Poncelet’s theorem, and Blaschke ellipses are connected. As we will see in the next chapter, the common thread is that the convex hull of each of the triangles represents the numerical range of a certain unitary matrix that is closely related to our Blaschke product.

Chapter

7

The Connection Revealed We have seen that Blaschke products are naturally associated with Blaschke ellipses and, when the Blaschke product has degree 3, these ellipses are inscribed in triangles. The connection to Poncelet’s theorem is, we hope, also becoming clearer. But what is the relation of the triangles to the numerical range of 2 × 2 matrices? It turns out that the triangles are the boundaries of the numerical ranges of certain unitary 3 × 3 matrices that arise naturally from the 2 × 2 matrix. So let us turn to the numerical range of unitary matrices. In general, the numerical range of an 𝑛 × 𝑛 matrix for 𝑛 > 2 is not easy to compute. However, if the matrix is unitary, the description is accessible and pretty. Is it of use in computing the numerical range of a general 𝑛 × 𝑛 matrix? That is the question we aim to answer below. As we saw in the spectral theorem (Theorem 6.4), every normal matrix, and therefore every unitary matrix, 𝑈, is unitarily equivalent to a diagonal matrix. So, the numerical range is easy to compute—even if the matrix is 𝑛 × 𝑛: 𝑛

𝑊(𝑈) = { ∑ 𝜆𝑗 |𝑥𝑗 |2 ∶ ‖𝑥‖ = 1}, 𝑗=1

where the 𝜆𝑗 denote the eigenvalues of 𝑈. Thus, as a corollary to Theorem 6.4, we obtain a description of the numerical range of a normal matrix. Corollary 7.1. The numerical range of a normal matrix 𝑁 is the convex hull of the eigenvalues of 𝑁.

75

76

Chapter 7. The Connection Revealed

If 𝑈 is unitary and 𝜆 is an eigenvalue of 𝑈 with corresponding unit eigenvector 𝑥, we have 1 = ‖𝑥‖2 = ⟨𝑈 ⋆ 𝑈𝑥, 𝑥⟩ = ⟨𝑈𝑥, 𝑈𝑥⟩ = |𝜆|2 . Thus, a unitary matrix has all its eigenvalues on the unit circle. Some matrices have what is known as a unitary dilation; that is, they sit inside a unitary matrix in a particular way. Our unitary matrices also have a restriction on their size and are therefore called unitary 1-dilations: A matrix 𝐵 is a unitary 1-dilation of an 𝑛 × 𝑛 matrix 𝐴 if 𝐵 is an (𝑛 + 1) × (𝑛 + 1) unitary matrix that is unitarily equivalent to a matrix of the form ∗ ⎡ ⎤ 𝐴 ⋮ ⎥ ⎢ . ⎢ ∗ ⎥ ⎢ ⎥ ⎣ ∗ ⋯ ∗ ∗ ⎦ The unitary dilation of a matrix is closely related to the compression of the matrix that we met in Chapter 6. Halmos realized that every contraction, that is, a matrix 𝐴 with norm ‖𝐴‖ = max{‖𝐴𝑥‖ ∶ 𝑥 ∈ ℂ𝑛 , ‖𝑥‖ = 1} ≤ 1,

(7.1)

has a unitary dilation (though not necessarily a unitary 1-dilation). In 1964, he made a conjecture that we make precise later, but for the time being we think of as saying the following. Halmos’s conjecture ([75]). The numerical range of a contraction is determined by the numerical ranges of its unitary dilations. In fact, Halmos’s conjecture was stated more generally: Recall that an operator 𝑇 from a Hilbert space1 𝐻 to itself is bounded if there exists a constant 𝑀 ∈ ℝ such that ‖𝑇𝑓‖𝐻 ≤ 𝑀‖𝑓‖𝐻 for all 𝑓 ∈ 𝐻. Here, ‖𝑓‖𝐻 denotes the norm in 𝐻 and the subscript is often omitted in the norm notation. In this context, we can define the 1 A Hilbert space is a vector space that has an inner product and is complete with respect to the metric defined by this inner product.

Chapter 7. The Connection Revealed

77

norm of a bounded operator in an analogous manner to the way in which we defined the norm of a matrix in (7.1); that is, ‖𝑇‖ = sup{‖𝑇𝑓‖ ∶ 𝑓 ∈ 𝐻, ‖𝑓‖ = 1}.

(7.2)

Again, 𝑇 is a contraction if ‖𝑇‖ ≤ 1. Halmos showed that every contraction has a unitary dilation, defined on a space containing 𝐻 (actually on 𝐻 ⊕𝐻), and, in this general setting, he stated his conjecture in terms of the closures of the numerical ranges. The conjecture was proved by Choi and Li in 2001 [26]. A precise discussion of this requires familiarity with the fundamentals of Hilbert spaces and operator theory and will have to wait until Chapter 9. Before moving on though, we should note that because unitary operators are much easier to work with than general contractions, Halmos’s conjecture (that is, Choi and Li’s theorem) is a valuable resource for operator theorists. In the representation of the matrices from the class on which we focus, all entries will lie in 𝔻, while the entries of the unitary matrices associated with them lie in 𝔻. Why did Halmos restrict his attention to contractions? Well, if a matrix “sits inside” a unitary matrix 𝑈 like this ⎡ 𝐴 ⎢ 𝑈=⎢ ⎢ ⎣ ∗ ⋯

∗

∗ ⋮ ∗ ∗

⎤ ⎥ ⎥, ⎥ ⎦

then (as the reader should check) 𝐴 has to be a contraction. Now let us start with a (small) contraction. If a 1 × 1 matrix is 𝐴 = [ 𝑎 ], then we can find a unitary dilation of it if |𝑎| < 1. Here is a family of matrices that work, one for each 𝜆 ∈ 𝕋: 𝑈𝜆 = [

𝑎 𝜆(1 − |𝑎|2 )1/2

(1 − |𝑎|2 )1/2 ]. −𝜆𝑎

Note also that things work nicely because |𝑎| < 1. One crucial point is that up to unitary equivalence these are the only 2 × 2 unitary matrices that will work! Because this is so important, we isolate it as an exercise. Exercise 7.2. Show that if |𝑎| < 1 and 𝑈 is a unitary 1-dilation of the 1×1 matrix [ 𝑎 ], then there exists 𝜆 ∈ 𝕋 such that 𝑈 is unitarily equivalent to 𝑈𝜆 .

78

Chapter 7. The Connection Revealed

Further, just as not every ellipse contained in the unit disk is a Poncelet ellipse, not every 2 × 2 matrix will have a Poncelet ellipse as the boundary of its numerical range. As we shall see, those that do are of the form 𝑎 √1 − |𝑎|2 √1 − |𝑏|2 𝐴=[ (7.3) ], 0 𝑏 where 𝑎, 𝑏 ∈ 𝔻. If we take 𝑎 = 𝑏 = 0, we get a 2 × 2 Jordan block. Why is (7.3) a good guess for a matrix that has a Poncelet ellipse bounding its numerical range? We know that a Blaschke 3-ellipse corresponding to a Blaschke product with zeros at 0, 𝑎, and 𝑏 has foci at 𝑎 and 𝑏 and major axis of length |1 − 𝑎𝑏|. From the elliptical range theorem we know that the matrix 𝑎 𝑥 (7.4) 𝐴=[ ] 0 𝑏 has an elliptical numerical range with foci at 𝑎 and 𝑏 and a minor axis of length (tr(𝐴⋆ 𝐴) − |𝑎|2 − |𝑏|2 )1/2 = |𝑥|. If we try to match the lengths of the axes of these confocal ellipses, we see that |𝑥|2 = (1 − |𝑎|2 )(1 − |𝑏|2 ). Consequently, (7.3) is an appropriate matrix to study. Because a unitary matrix must have orthonormal columns and rows, arguing as in Exercise 7.2 we see that a unitary 1-dilation of 𝐴 must be unitarily equivalent to 𝑎 𝑈𝜆 = [ 0 𝜆√1 − |𝑎|2

√1 − |𝑎|2 √1 − |𝑏|2 𝑏 −𝜆𝑎√1 − |𝑏|2

−𝑏√1 − |𝑎|2 √1 − |𝑏|2 ] , 𝜆𝑎𝑏

(7.5)

where 𝜆 ranges over the unit circle. We get our matrix 𝐴 back from every 𝑈𝜆 by compressing it: Let 1 0 𝑃 = [ 0 1 ]. 0 0

(7.6)

Then 𝑃⋆ 𝑈𝜆 𝑃 = 𝐴 for every 𝜆 ∈ 𝕋. In this chapter we show that the numerical range of a matrix of the form described in (7.3) is bounded by a Poncelet 3-ellipse inscribed in

Chapter 7. The Connection Revealed

79

triangles associated with a Blaschke product, and we will show that every Poncelet 3-ellipse is associated with such a matrix. Our matrices 𝐴 are contractions, they have all eigenvalues in the disk, and we can find unitary 1-dilations of them, one for each point on the unit circle. This may seem like an odd class of matrices, but as we will see in Chapter 9, it is actually a nice class associated with a set of operators known as compressions of the shift operator. Now consider the Blaschke product with zeros 𝑎 and 𝑏 and the matrix in (7.3). The eigenvalues of the matrix are the zeros of the Blaschke product. What are the eigenvalues of one of the unitary dilations given by (7.5)? A computation shows that the determinant of 𝑧𝐼 − 𝑈𝜆 is 𝑧(𝑧 − 𝑎)(𝑧 − 𝑏) − 𝜆(1 − 𝑎𝑧)(1 − 𝑏𝑧).

(7.7)

Therefore, the eigenvalues of 𝑈𝜆 are the points on the unit circle that satisfy 𝑧(𝑧 − 𝑎)(𝑧 − 𝑏) = 𝜆. (1 − 𝑎𝑧)(1 − 𝑏𝑧) But the function on the left is a degree-3 Blaschke product that takes 0 to 0. So, letting 𝑧−𝑎 𝑧−𝑏 𝐵(𝑧) = 𝑧 ( ), )( 1 − 𝑎𝑧 1 − 𝑏𝑧 we see that the eigenvalues of 𝑈𝜆 are the three points that 𝐵 sends to 𝜆. In particular, if 𝜆1 ≠ 𝜆2 , then the eigenvalues of 𝑈𝜆1 and 𝑈𝜆2 are all distinct. These observations, together with Theorem 2.9 in Chapter 2, will allow us to prove the main result of this chapter. Before we move on, we mention a fact that is true of all square matrices that are contractions. Lemma 7.3. Let 𝐴 be an 𝑛 × 𝑛 matrix that is a contraction and has all eigenvalues in 𝔻. Then 𝑊(𝐴) ⊆ 𝔻. This follows from the Cauchy–Bunyakovsky–Schwarz inequality or Theorem 6.2, (8), but we provide a detailed proof. Proof. From the Cauchy–Bunyakovsky–Schwarz inequality, for all 𝑥 ∈ ℂ𝑛 with ‖𝑥‖ = 1, we have |⟨𝐴𝑥, 𝑥⟩| ≤ ‖𝐴𝑥‖‖𝑥‖ ≤ 1, so it is clear that 𝑊(𝐴) ⊆ 𝔻. If there is a point 𝜆 ∈ 𝕋 with ⟨𝐴𝑥, 𝑥⟩ = 𝜆, then |⟨𝐴𝑥, 𝑥⟩| = 1 and equality holds in the Cauchy–Bunyakovsky–Schwarz inequality.

80

Chapter 7. The Connection Revealed

This implies that 𝐴𝑥 is a multiple of 𝑥 and therefore 𝐴 has an eigenvalue on the unit circle, contrary to the hypothesis. We are now ready for our main theorem of this chapter. We discuss the geometric significance of Theorem 7.4 once we have completed the proof. Theorem 7.4. Let 𝑎 and 𝑏 be points in 𝔻, and let 𝐴=[

𝑎 0

√1 − |𝑎|2 √1 − |𝑏|2 ]. 𝑏

(7.8)

Then 𝑊(𝐴) is the intersection of the numerical ranges of all unitary 1dilations of 𝐴. Proof. First, we show that 𝑊(𝐴) ⊆ ⋂𝑈 𝑊(𝑈), where the intersection is taken over all unitary 1-dilations. As we discussed above, every unitary 1-dilation of 𝐴 is unitarily equivalent to 𝑈𝜆 for some 𝜆 ∈ 𝕋 and unitarily equivalent matrices have the same numerical range. Thus, we only need to show that 𝑊(𝐴) ⊆ ⋂𝜆∈𝕋 𝑊(𝑈𝜆 ). Recall that if we let 1 0 𝑃 = [ 0 1 ], (7.9) 0 0 then, for 𝑥 ∈ ℂ2 with ‖𝑥‖ = 1 and 𝜆 ∈ 𝕋, we have 𝑃𝑥 ∈ ℂ3 , 𝑃⋆ 𝑈𝜆 𝑃 = 𝐴, and ‖𝑃𝑥‖ = 1. If 𝛼 ∈ 𝑊(𝐴), then there exists 𝑥𝛼 ∈ ℂ2 with ‖𝑥𝛼 ‖ = 1 and for every 𝜆 ∈ 𝕋 we have 𝛼 = ⟨𝐴𝑥𝛼 , 𝑥𝛼 ⟩ = ⟨𝑃⋆ 𝑈𝜆 𝑃𝑥𝛼 , 𝑥𝛼 ⟩ = ⟨𝑈𝜆 𝑃𝑥𝛼 , 𝑃𝑥𝛼 ⟩. Since ‖𝑃𝑥𝛼 ‖=1, we see that 𝛼 ∈ ⋂𝜆∈𝕋 𝑊(𝑈𝜆 ). Thus, 𝑊(𝐴) ⊆ ⋂𝑈 𝑊(𝑈). For the other direction, suppose that 𝛽 ∉ 𝑊(𝐴). If 𝛽 ∉ 𝔻, then 𝛽 will not be in the numerical range of the intersection of two distinct unitary dilations 𝑈𝜆1 and 𝑈𝜆2 of 𝐴; that is, 𝛽 ∉ ⋂𝑈 𝑊(𝑈). So we may assume that 𝛽 ∈ 𝔻. We use Theorem 2.9 to create a unitary dilation 𝑈𝜆 of 𝐴 for which 𝛽 ∉ 𝑊(𝑈𝜆 ). Theorem 6.2 and the Toeplitz–Hausdorff theorem imply that 𝑊(𝐴) is closed and convex. Thus, since 𝛽 ∉ 𝑊(𝐴), there is a half-plane separating 𝛽 from 𝑊(𝐴). (See Figure 7.1.) The boundary line of this half-plane must intersect the unit circle twice at, say, 𝑤1 and 𝑤2 .

Chapter 7. The Connection Revealed

81











β


= Figure 7.1. Separating 𝛽 from 𝑊(𝐴).

Now we use the connection of our matrix to a Blaschke product. Let 𝐵 denote the degree-3 Blaschke product with zeros at 0 and the two eigenvalues of 𝐴, which we recall are 𝑎 and 𝑏. Let 𝑣1 = 𝑤1 , 𝑣2 , and 𝑣3 denote the three points satisfying 𝐵(𝑤1 ) = 𝐵(𝑣1 ) = 𝐵(𝑣2 ) = 𝐵(𝑣3 ). By Theorem 2.9, we know that the triangle with vertices 𝑣1 , 𝑣2 , and 𝑣3 circumscribes a Blaschke ellipse with foci at 𝑎 and 𝑏 and major axis of length |1 − 𝑎𝑏|. On the other hand, Theorem 6.1 implies that 𝑊(𝐴) is bounded by an ellipse with foci at 𝑎 and 𝑏 and minor axis of length (1 − |𝑎|2 − |𝑏|2 + |𝑎𝑏|2 )1/2 . Since the center of the ellipse is at (𝑎 + 𝑏)/2, a computation shows that the major axis has length |1 − 𝑎𝑏|—in other words, the convex hull of the Blaschke ellipse is the numerical range of 𝐴. But our triangle with vertices 𝑣1 , 𝑣2 , 𝑣3 must circumscribe the ellipse and the point 𝛽 cannot lie in or on the triangle. So 𝑈𝐵(𝑤1 ) is a unitary dilation of 𝐴 and its numerical range is, by our discussion above and Corollary 7.1, the convex hull of its eigenvalues, 𝑣1 , 𝑣2 , and 𝑣3 . Since 𝛽 does not lie in or on this triangle, 𝛽 ∉ 𝑊(𝑈𝐵(𝑤1 ) ). Since 𝑈𝐵(𝑤1 ) is a unitary 1-dilation of 𝐴, we see that 𝛽 ∉ ⋂𝑈 𝑊(𝑈), where the intersection is taken over all unitary 1-dilations. Thus, ⋂𝑈 𝑊(𝑈) ⊆ 𝑊(𝐴), completing the proof.

82

Chapter 7. The Connection Revealed

So, given a degree-3 Blaschke product sending 0 to 0, the intersection over all 𝜆 ∈ 𝕋 of the triangular regions formed with the three distinct solutions of 𝐵(𝑧) − 𝜆 = 0 is an ellipse with foci at the zeros of 𝐵(𝑧)/𝑧. This ellipse is the boundary of the numerical range of a 2 × 2 matrix 𝐴 with foci at the eigenvalues of 𝐴. On the other hand, you can start with the 2 × 2 matrix 𝐴 of the form given in (7.3). Its numerical range is an elliptical disk that is the intersection of the numerical ranges of all of the unitary 1-dilations of 𝐴—and those numerical ranges are the convex hulls of triangles with vertices on 𝕋. Then the Blaschke product that sends 0 to 0 and has its other two zeros at the eigenvalues of 𝐴 identifies the vertices of each of the triangular regions obtained from the unitary 1-dilations of 𝐴. What about Poncelet’s theorem for triangles inscribed in 𝕋? Well, we know that the numerical range of a matrix of the special form in (7.3) is an elliptical disk inscribed in triangles, one triangle for each point of the unit circle, and so it is a Poncelet 3-ellipse. But we also know, from Corollary 4.4, that all Poncelet 3-ellipses are Blaschke ellipses. So if we have a matrix 𝐴 for which 𝑊(𝐴) is bounded by a Poncelet 3-ellipse, there is a Blaschke product 𝐵 with zeros at 0 and the eigenvalues of 𝐴 that gives you all of its circumscribing triangles. If you create a matrix of the form (7.3) with the zeros of 𝐵(𝑧)/𝑧 on the diagonal, then the numerical range of that matrix will be bounded by your Poncelet ellipse. So think of it as you wish: You can start with the Blaschke product and look for the matrix that goes with it or start with the matrix and look for the Blaschke product. Either way, you will end up with the same ellipse. Or, if you are interested in Poncelet 3-ellipses, it is enough to consider 2×2 matrices in the form of (7.3) or Blaschke products of degree 3 with zeros at the foci of the ellipse, 𝑎 and 𝑏, and one zero at 0. These unexpected connections are what makes mathematics so beautiful. But it is also valuable from a practical perspective—looking at theorems from three very different perspectives will provide much greater insight. Do these connections appear when we consider 𝑛 × 𝑛 matrices? If we consider the right collection of matrices, much of our work can be extended, though the numerical range will not always be elliptical. Showing this, however, requires us to understand much more about our

Chapter 7. The Connection Revealed

83

Blaschke products, matrices, and the operators hidden behind our matrices. Before turning to that, there is one more short story we would like to tell.

Intermezzo mais apud me omnia fiunt Mathematicè in Natura But in my opinion, everything in nature occurs mathematically –René Descartes, Correspondence with Mersenne, March 11, 1640, p. 36

85

Chapter

8

And Now for Something Completely Different. . . Benford’s Law In 1992, an Arizona state employee by the name of Wayne James Nelson wrote 23 checks to a vendor. We provide the dates of the checks and the amounts1 in Table 8.1. Table 8.1. The 1992 date and check amount. Date 10/09 10/09 10/14 10/14 10/14 10/14 10/19 10/19 10/19 10/19 10/19

Amount in USD 1,927.48 27,902.31 86,241.90 72,117.46 81,321.75 97,473.96 93,249.11 89,658.17 87,776.89 92,105.83 79,949.16

Date 10/19 10/19 10/19 10/19 10/19 10/19 10/19 10/19 10/19 10/19 10/19 10/19

Amount in USD 87,602.93 96,879.27 91,806.47 84,991.67 90,831.83 93,766.67 88,338.72 94,639.49 83,709.28 96,412.21 88,432.86 71,552.16

1 Data from http://alfre.dk/how-to-commit-tax-fraud/ (accessed 9/19/2017).

87

88

Chapter 8. Something Completely Different

Take a moment to study the table. Do you see anything suspicious? No? Well, the State of Arizona did! In 1993, the State of Arizona accused Mr. Nelson of fraud to the tune of almost two million dollars. What might have helped detect this fraud? Read on. We need to start at the beginning. In 1881, the mathematician-astronomer Simon Newcomb noticed that the pages of the logarithm tables for which the first nonzero digit of the number (reading from left to right) was a 1 wore out more quickly than those of the other digits. He provided a table showing the frequency with which the digits 1 through 9 appear as the leftmost digit in a number in the logarithm table; in particular, 1 appears as a first digit about 30.1% of the time and 2 about 17.6% of the time, while the 9 appears only 4.6% of the time as a first digit. In 1938, while working at General Electric, Frank Benford rediscovered this curious fact about the distribution of first digits—a result that is now known as Benford’s law. To understand this law, we introduce some terminology and notation. Given a number 𝑥, we will call the first nonzero digit that we see in the decimal representation of 𝑥 the first digit; although it would be more precise to say “the first significant digit”, that becomes cumbersome. So, the first digit of 25 is 2, which is what we would expect, and the first digit of 0.02 is also 2. Further, we say that an infinite list of positive real numbers satisfies Benford’s law if the first digit 𝑑 occurs with probability log10 ((𝑑 + 1)/𝑑), where 𝑑 = 1, 2, … , 9. Informally, we also say that a long but finite list of positive numbers satisfies Benford’s law if the distribution of first digits is reasonably well approximated by the probabilities above. This informal interpretation covers many familiar examples of lists that satisfy Benford’s law. Of course many sets of data do not follow Benford’s law but many do. Benford made a great effort to collect data to show that this was often the case. He had 20,229 observations from various sets of data, including lengths of rivers, American League baseball statistics, and atomic weights of elements [81]. The 1990s saw a renewed interest in Benford’s

Chapter 8. Something Completely Different

89

Law in part due to its applications to fraud detection; that is, manipulated lists of data tend not to follow Benford’s law, while nonmanipulated lists do tend to follow the law. There is now a website2 devoted to it! In addition, there is a book3 about Benford’s law [120], and [87, pp. 1–12] has a more complete history of it. This is a good time to return to Mr. Nelson. Obviously, 23 data points is not a particularly long list. But there are some curious things about this list that you might have spotted. The first significant digit is a 1 exactly once or 4% of the time—nowhere close to the 30.1% Benford’s law suggests we should see. This should cause us to look at the data more closely, which we now do. We see that all checks were made to the same vendor. This might have a reasonable explanation; perhaps this vendor had prices that, for whatever reason, do not follow Benford’s law. For example, at a certain private liberal arts institution, we happen to know that receipts are not required for meals that are less than $25. Thus, we might expect to see more receipts with a 2 as the first significant digit and very few higher than a 2. The same seems to happen above: all checks are below $100,000. This is a little surprising; a state government would tend to have larger purchases. There are some other surprising features of the data set; for example, the first two checks are much smaller than later ones. In addition, some pairs of digits appear frequently (Nelson seems to have had a fondness for 83, for example). Of course, the main point is that the vendor Nelson wrote the checks to was bogus (actually, Nelson was the vendor), and that was surely the nail in the coffin. Nevertheless, Benford’s law is a quick and easy method that can tell us when we need to look at data more closely. In the rest of this chapter, we connect Benford’s law to Poncelet’s theorem (really!) as well as another story, and we explain some of the mathematics behind Benford’s law. Let us consider an example of a distribution of first digits. We construct Table 8.2 as follows: In the first row, put the integers from 2 to 9. In the second, put the first digit of the integers 22 to 92 . For the third row, put the first digit of 23 to 93 . Thus, in the second position of the third line we would compute the number 33 = 27 and put the number 2 http://www.benfordonline.net

(accessed 12/15/2017)

3 The website http://www.nigrini.com/benfordslaw.htm (accessed 12/15/2017)

has several links to real data sets that follow Benford’s law.

90

Chapter 8. Something Completely Different

2 in that spot. We can continue doing this for as many rows as we like, though it would be handy to have some notation for the first digit, so we will write ⟨⟨𝑥⟩⟩ for the first digit of 𝑥. Thus, ⟨⟨33 ⟩⟩ = 2 and ⟨⟨2/100⟩⟩ = 2. The first ten rows that we get are reproduced in Table 8.2. Table 8.2. Table of first digits. 𝑛 1 2 3 4 5 6 7 8 9 10

⟨⟨2𝑛 ⟩⟩ 2 4 8 1 3 6 1 2 5 1

⟨⟨3𝑛 ⟩⟩ 3 9 2 8 2 7 2 6 1 5

⟨⟨4𝑛 ⟩⟩ ⟨⟨5𝑛 ⟩⟩ 4 5 1 2 6 1 2 6 1 3 4 1 1 7 6 3 2 1 1 9

⟨⟨6𝑛 ⟩⟩ 6 3 2 1 7 4 2 1 1 6

⟨⟨7𝑛 ⟩⟩ ⟨⟨8𝑛 ⟩⟩ ⟨⟨9𝑛 ⟩⟩ 7 8 9 4 6 8 3 5 7 2 4 6 1 3 5 1 2 5 8 2 4 5 1 4 4 1 3 2 1 3

Avez [6, p. 37] attributes the following question about this list to the Russian mathematician Israel M. Gelfand: Is there an 𝑛 such that the first digit of 2𝑛 is 9? This leads naturally to a string of questions about patterns we might see in Table 8.2. There are only 98 different possible rows, so at least one row will appear infinitely often. Will all 98 possible rows appear? Will we ever see the first row, 23456789, again? If so, can we describe the frequency with which it will appear? Will we ever see a row of all twos? threes? You might be thinking that Benford’s law is a catchy, but unrelated, opening. Remember that Benford’s law is about frequencies—how often a number appears. So, we could focus on a single column, say the column with header ⟨⟨2𝑛 ⟩⟩, and look at the frequency with which each digit appears. Now we are ready to define the frequency of a digit 𝑑, where 𝑑 = 1, 2, … , 9, by 1 |{𝑛 ∶ 1 ≤ 𝑛 ≤ 𝑁 and ⟨⟨2𝑛 ⟩⟩ = 𝑑}| , fr(𝑑) = lim 𝑁→∞ 𝑁 assuming this limit exists.4 4 As

noted in [94], “the existence of frequencies is not a foregone conclusion”.

Chapter 8. Something Completely Different

91

If you measured the frequency with which each digit appears in the 2-column, you would find that the digit 1 occurs about 30.1% of the time, the digit 2 about 17.6% of the time, and the digit 3 about 12.5% of the time. Surprisingly, the frequencies are the same no matter which column we look at! (See [45].) Though that might convince you that Gelfand’s question is connected to Benford’s law, you are probably still wondering what these problems are doing in a book about Poncelet’s theorem, ellipses, and Blaschke products. In King’s article, Three problems in search of a measure [94], he claims that Gelfand’s question is “secretly isomorphic” to Poncelet’s theorem. Let us see why. For a positive real number 𝑥, the first digit of 𝑥 and 10𝑥 are the same, and in our first digit problem, we are effectively identifying the intervals … , [0.01, 0.1) , [0.1, 1) , [1, 10) , [10, 100) , … in the following way. Imagine wrapping [1, 10) about the unit circle and breaking it into arcs from [1, 2), [2, 3), and so on. When you get to 10, start wrapping again, adjusting lengths, and breaking the circle into arcs from [10, 20), [20, 30) that “lie on top” of the ones from [1, 2) and [2, 3). So you can “see” Poncelet’s theorem beginning to appear: We may think of the intervals as wrapping around the circle and the lines tangent to our ellipse as having endpoints associated with the highest order digit. To make this more precise, let 𝛼 ∈ ℝ+ and 𝑚 = 10𝛼 and define 𝑇𝑚 ∶ ℝ+ → ℝ+ by 𝑇𝑚 (𝑥) = 𝑚𝑥. Let 𝜓(𝑥) = log10 𝑥 (mod 1) so that 𝜓 maps ℝ+ onto the interval ℐ = [0, 1). Finally, define 𝜌𝛼 (𝑥) = 𝑥 ⊕ 𝛼 = 𝑥 ⊕ log10 (𝑚), where ⊕ denotes addition (mod 1). Then, as one can check, the following diagram commutes: ℝ+

𝑇𝑚

ℝ+

𝜓

𝜓

ℐ

𝜌𝛼

.

ℐ

In other words, 𝜌𝛼 ∘ 𝜓 = 𝜓 ∘ 𝑇𝑚 .

(8.1)

92

Chapter 8. Something Completely Different

Another direct computation shows that things also work well under iteration; that is, if we let (𝑛)

𝑇𝑚 ∶= 𝑇 𝑚 ∘ 𝑇𝑚 ∘ ⋯ ∘ 𝑇𝑚 ⏟⎵⎵⎵⎵⏟⎵⎵⎵⎵⏟ 𝑇𝑚 repeated 𝑛 times

denote the 𝑛th iterate of 𝑇𝑚 , then (𝑛)

𝜌𝑛𝛼 ∘ 𝜓 = 𝜓 ∘ 𝑇𝑚 .

(8.2)

Now for 𝑥 ∈ ℝ+ , we have ⟨⟨𝑥⟩⟩ = 𝑑 if and only if there is an integer 𝑛 with 𝑑10𝑛 ≤ 𝑥 < (𝑑 + 1)10𝑛 ; in other words, 𝑑 ≤ 𝑥 ⋅ 10−𝑛 < 𝑑 + 1. Taking the logarithm (mod 1), we conclude that ⟨⟨𝑥⟩⟩ = 𝑑 if and only if 𝜓(𝑥) ∈ [log10 (𝑑), log10 (𝑑 + 1)), for 𝑑 = 1, … 9. This strengthens one connection: the length of the first interval is log10 2 − log10 1, which is about 0.301, the length of the second is log10 3 − log10 2 ≈ 0.176, and the third has length log10 (10) − log10 9 ≈ 0.0457—the frequencies both Newcomb and Benford observed. And it works in general; that is, the length of the interval [log10 (𝑑), log10 (𝑑 + 1)) is log10 ((𝑑 + 1)/𝑑), which is the same as the frequency that 𝑑 appears as a first digit for 𝑑 = 1, … , 9. There are two questions left to answer: Why are the frequencies the same as the lengths of the intervals? And what is the connection to Poncelet’s theorem? We return to the Poncelet picture with an ellipse 𝐸 entirely contained in the unit circle. Choose a point 𝑧 ∈ 𝕋, construct the right-hand tangent to the ellipse through 𝑧, and intersect it with 𝕋. Denote the point of intersection by 𝑅𝐸 (𝑧). We can do the same construction with the lefthand tangent through 𝑧; in this case, we denote the intersection point of the tangent line with 𝕋 by 𝐿𝐸 (𝑧). If the ellipse is inscribed in a triangle, as in Figure 8.1, and we first find 𝑅𝐸 (𝑧) and then proceed to 𝑅𝐸 (𝑅𝐸 (𝑧)) ∶= (2) (𝑛) 𝑅𝐸 (𝑧), we will end up at 𝐿𝐸 (𝑧). In general, we let 𝑅𝐸 denote the 𝑛th iterate of 𝑅𝐸 . In our new terminology and in our setting, Poncelet’s theorem says that if for some 𝑧 ∈ 𝕋 there exists an integer 𝑛 that is the smallest integer (𝑛) (𝑛) for which 𝑅𝐸 (𝑧) = 𝑧, then for every point 𝑤 on the unit circle 𝑅𝐸 (𝑤) = 𝑤. We will try to give a deeper sense of the connection between the two

Chapter 8. Something Completely Different

93



(
)=
(
)
(
)



(
) Figure 8.1. Poncelet 3-ellipse.

problems without venturing too far into King’s secret weapon—measure theory. So let us go back to Poncelet’s theorem. Suppose, for the moment, that we have an ellipse 𝐸, that 𝑅𝐸 is as above, and that we can find an invertible continuous 𝜑 and a positive real number 𝛼 that make the following diagram commute: 𝕋

𝑅𝐸

𝕋

𝜑

𝜑

ℐ

𝜌𝛼

ℐ

Thus, we have 𝜌𝛼 ∘ 𝜑 = 𝜑 ∘ 𝑅𝐸 .

(8.3)

Again, since 𝜑 is invertible, iterates work well: (𝑛)

𝜌𝑛𝛼 = 𝜑 ∘ 𝑅𝐸 ∘ 𝜑−1 .

(8.4)

𝛼

Using 𝜑 and letting 𝑚 = 10 , we can lift 𝐼𝑑 = [log10 (𝑑), log10 (𝑑 + 1)) onto intervals 𝐽𝑑 of the unit circle for 𝑑 = 1, 2, … , 9. Then we claim that 𝜑−1 ∘ 𝜓 ∶ ℝ+ → 𝕋 makes the Gelfand question, for this choice of 𝑚, “isomorphic” to Poncelet’s theorem. How does it do that?

94

Chapter 8. Something Completely Different

Let us suppose that we are interested in the 2-column of Table 8.2. Then, for a positive integer 𝑛, if we take 𝑚 = 2, then 𝛼 = log10 2 and (𝑛)

(𝑛)

the first digit of 2𝑛 is the first digit of 𝑇2 (1). Thus, 𝜓(2𝑛 ) = 𝜓(𝑇2 (1)). From (8.2) and (8.4), assuming an ellipse 𝐸 exists, we have (𝑛)

(𝑛)

𝑅𝐸 ∘ (𝜑−1 ∘ 𝜓) = (𝜑−1 ∘ 𝜓) ∘ 𝑇2 , and we have moved Gelfand’s question over to the Poncelet system: A (𝑛) question about the first digit of 2𝑛 = 𝑇2 (1) is the same as looking at the behavior of 𝜑−1 ∘ 𝜓(1) under the 𝑛th iterate of 𝑅𝐸 . The time has come5 to produce such a 𝜑: Take 𝑧 and 𝑦 in 𝕋 and use the notation [𝑧, 𝑦) to denote the arc of 𝕋 from 𝑧 to 𝑦, oriented counterclockwise. Suppose we have a finite measure 𝜇, normalized so that 𝜇(𝕋) = 1. Further suppose that 𝜇 does not give points positive measure, gives open intervals positive length, and is 𝑅𝐸 invariant, meaning that 𝜇([𝑥, 𝑦)) = 𝜇([𝑅𝐸 𝑥, 𝑅𝐸 𝑦)). Then for 𝑥, 𝑦, and 𝑧 on 𝕋, we have 𝜇([𝑧, 𝑦)) ⊕ 𝜇([𝑦, 𝑥)) = 𝜇([𝑧, 𝑥)), where we recall that ⊕ denotes addition modulo 1. Fix a point 𝑧0 on the unit circle, let 𝛼 = 𝜇([𝑧0 , 𝑅𝐸 𝑧0 )), and define 𝜑 on 𝕋 by 𝜑(𝑦) = 𝜇([𝑧0 , 𝑦)). So 𝜑(𝑧0 ) = 0 and 𝜑 maps into [0, 1). In addition, 𝜑 is injective and therefore invertible. Though it certainly looks like 𝛼 depends upon the point we choose as 𝑧0 , it turns out that is not the case: For 𝑥 ∈ 𝕋, 𝛼 = 𝜇([𝑧0 , 𝑅𝐸 𝑧0 )) = 𝜇([𝑧0 , 𝑥)) ⊕ 𝜇([𝑥, 𝑅𝐸 𝑧0 )) = 𝜇([𝑅𝐸 𝑧0 , 𝑅𝐸 𝑥)) ⊕ 𝜇([𝑥, 𝑅𝐸 𝑧0 )) = 𝜇([𝑥, 𝑅𝐸 𝑥)). 5 For those who want a more in-depth description of the construction—in particular, how to get the invariant measure in the general case—we recommend King’s original article [94] or Flatto’s book, Poncelet’s Theorem [47].

Chapter 8. Something Completely Different

95

And the diagram commutes: 𝜑 ∘ 𝑅𝐸 (𝑧) = 𝜇([𝑧0 , 𝑅𝐸 𝑧)) = 𝜇([𝑧0 , 𝑅𝐸 𝑧0 )) ⊕ 𝜇([𝑅𝐸 𝑧0 , 𝑅𝐸 𝑧)), while 𝜌𝛼 ∘ 𝜑(𝑧) = 𝜇([𝑧0 , 𝑧)) ⊕ 𝜇([𝑧0 , 𝑅𝐸 𝑧0 )) = 𝜇([𝑧0 , 𝑅𝐸 𝑧0 )) ⊕ 𝜇([𝑅𝐸 𝑧0 , 𝑅𝐸 𝑧)). Assuming the existence of such a measure, we can now provide a concrete example in a simple case. Example 8.1. Let the ellipse 𝐸 be the circle centered at the origin of radius 1/2. We analyzed this circle in Example 2.4, where we showed that the triangles circumscribing 𝐸 are equilateral. Thus, they divide 𝕋 into arcs of equal length. Because our measure is normalized, the length of each arc is 1/3. Therefore, 𝛼 = 1/3 and 𝑚 = 101/3 . Thus, we see that 𝑇𝑚 (𝑥) = 101/3 𝑥 and 𝜓(𝑥) = log10 𝑥 (mod 1). The reader should now check that (8.1) is satisfied. On the other hand, 𝑅𝐸 (𝑒𝑖𝜃 ) = 𝑒𝑖(𝜃+2𝜋/3) and 𝜑(𝑧) = arg(𝑧)/2𝜋 (mod 1). This time the reader should check that (8.3) is satisfied. This gives us the double commutative diagram below. ℝ+

𝑇𝑚

ℝ+

𝜓

𝜓

ℐ

𝜌𝛼

𝜑

𝕋

ℐ 𝜑

𝑅𝐸

𝕋

Where would we get such a measure 𝜇? In the case of Poncelet triangles, we have done enough to produce such a measure: From Corollary 4.4 we know that the Blaschke 3-ellipses are precisely the Poncelet 3-ellipses. So let us begin with an ellipse 𝐸, with foci at points 𝑎 and 𝑏, that is inscribed in a triangle that is itself inscribed in the unit circle. Then there is a Blaschke product 𝐵 with zeros 0, 𝑎, and 𝑏 that is associated with 𝐸. Back in Chapter 3 we argued that one way to get the measure would be to use the Blaschke product: 𝐵 knows where the vertices of the triangles lie and it finds them by identifying the vertices. Between any two vertices, 𝐵 maps the arc around the unit circle, weighing the

96

Chapter 8. Something Completely Different

three arcs equally. Figure 8.2 shows how this might look for a degree-4 Blaschke product. We have now arrived at the point at which we can make this precise. Since 𝐵 is analytic on an open set containing the closed unit disk, we may integrate from one point of the unit circle to another. If we start at 𝑧1 = 𝑒𝑖𝜃1 , we are looking for two other points 𝑧𝑗 = 𝑒𝑖𝜃𝑗 for 𝑗 = 2, 3 that satisfy 𝐵(𝑧1 ) = 𝐵(𝑧2 ) = 𝐵(𝑧3 ). That is, for 𝑗 = 1, 2, 3 we would like to have 6 1 1= (𝑖 arg(𝐵(𝑧𝑗+1 )) − 𝑖 arg(𝐵(𝑧𝑗 ))) 2𝜋𝑖 1 = ( log |𝐵(𝑧𝑗+1 )| + 𝑖 arg(𝐵(𝑧𝑗+1 )) − log |𝐵(𝑧𝑗 )| − 𝑖 arg(𝐵(𝑧𝑗 ))) 2𝜋𝑖 1 = ( log(𝐵(𝑧𝑗+1 )) − log(𝐵(𝑧𝑗 ))) 2𝜋𝑖 𝜃𝑗+1

=∫ 𝜃𝑗

𝑒𝑖𝜃

𝐵′ (𝑒𝑖𝜃 ) 𝑑𝜃 . 𝐵(𝑒𝑖𝜃 ) 2𝜋

(8.5) But, using Lemma 3.4, we can compute the integrand, and we obtain ℎ(𝑒𝑖𝜃 ) ∶= 𝑒𝑖𝜃

𝐵′ (𝑒𝑖𝜃 ) 𝐵(𝑒𝑖𝜃 )

1 − |𝑎|2 1 − |𝑏|2 1 + + ) 𝑒𝑖𝜃 (1 − 𝑎𝑒𝑖𝜃 )(𝑒𝑖𝜃 − 𝑎) (1 − 𝑏𝑒𝑖𝜃 )(𝑒𝑖𝜃 − 𝑏) 1 − |𝑏|2 1 − |𝑎|2 + = 1+ . |1 − 𝑎𝑒𝑖𝜃 |2 |1 − 𝑏𝑒𝑖𝜃 |2

= 𝑒𝑖𝜃 (

It is interesting that ℎ, which certainly does not look real—let alone positive—when we first define it, is always positive. One of the great consequences of this is that we can use it to define a positive measure on 𝕋. For an interval 𝑆 ⊂ 𝕋 we can define 𝑑𝜃 1 𝜇(𝑆) = ∫ ℎ(𝑒𝑖𝜃 ) , 3 𝑆 2𝜋 6 Choose

71].

the branches of the argument and the logarithm appropriately; see [118, p.

Chapter 8. Something Completely Different

97

Figure 8.2. Measuring arc length with Blaschke products.

where ℎ is the function above and, in fact, a function we met earlier in Lemma 4.2 in our study of Blaschke products. It is easy to see that 𝜇 is a finite measure that is appropriately normalized, does not give points positive measure, and gives open intervals positive length. It is just a bit harder to see that 𝜇 is 𝑅𝐸 invariant, but the computations are similar to those that we did above, so we leave this as an exercise for the reader. So, having justified the existence of a measure 𝜇—at least in the case of Poncelet 3-ellipses—we may return to Gelfand’s question and focus on 𝛼. If 𝛼 > 0 is rational, then 𝛼 = 𝑝/𝑞, where 𝑝 and 𝑞 are positive integers, and it is clear that after 𝑞 steps we are back where we started, modulo 1. On the other hand, if 𝛼 is irrational, it turns out that the orbit of the (𝑛) point 0 under 𝜌𝛼 (that is, { 𝜌𝛼 (0) ∶ 𝑛 an integer }) is dense. Why is this so? Well, let us suppose that 𝛼 is irrational, but the orbit is not dense. Then there is a little open interval that the orbit misses, which we suppose is [0, 𝜀) for some positive real number 𝜀. Suppose there were infinitely many distinct values (mod 1) of 𝑟𝛼 as 𝑟 ranges over the integers. Then we must have two for which 𝑟1 𝛼 and 𝑠1 𝛼 are within 𝜀 of each other with, say, 𝑟1 > 𝑠1 . In this case, (𝑟1 − 𝑠1 )𝛼 would be in [0, 𝜀), an interval

98

Chapter 8. Something Completely Different

the orbit misses. Thus, we can have only finitely many points in the orbit and therefore we would eventually return to one of the points; that (𝑟) (𝑠) is, there would exist distinct integers 𝑟 and 𝑠 for which 𝜌𝛼 (0) = 𝜌𝛼 (0). Translating this back into more familiar representation, we would have 𝑟𝛼 − 𝑠𝛼 = 𝑘 ≠ 0 for some integer 𝑘. In particular, 𝛼 would be rational, establishing a contradiction. It is time to put the pieces together. For the 2-column of Table 8.2, we have 𝛼 = log10 2 and that, of course, is an irrational number. Therefore, the orbit of 0 under 𝜌log10 2 is dense in ℐ. This implies that for some positive integer 𝑛 (and, in fact, for infinitely many of them), we get (𝑛)

(𝑛)

𝜓(2𝑛 ) = 𝜓 ∘ 𝑇2 (1) = 𝜌log

10

(𝑛)

2

∘ 𝜓(1) = 𝜌log

10

2 (0)

∈ [log10 (9), 1).

We have answered Gelfand’s question in the positive: For some integer 𝑛, we have ⟨⟨2𝑛 ⟩⟩ = 9. You might be interested to learn that the smallest 𝑛 for which the first digit is a 9 is 𝑛 = 53. To complete the connection to Benford’s law, there is one more theorem that we need. Letting ⌊𝑥⌋ denote the greatest integer less than or equal to 𝑥, we see that 𝑥 − ⌊𝑥⌋ denotes the fractional part of 𝑥. With this notation, we obtain the following special case of Weyl’s theorem. Theorem 8.2. Let 𝛾 be an irrational number, and let [𝑎, 𝑏] ⊂ [0, 1]. Then 1 lim |{𝑘 ∶ 𝑘𝛾 − ⌊𝑘𝛾⌋ ∈ [𝑎, 𝑏], where 1 ≤ 𝑘 ≤ 𝑛}| = 𝑏 − 𝑎. 𝑛→∞ 𝑛 We focus on the 2-column and apply this to log10 (2𝑛 ) = 𝑛 log10 (2). Recall what we learned earlier: The first digit of 𝑥 ∈ ℝ+ is 𝑑 if and only if 𝜓(𝑥) ∈ [log10 (𝑑), log10 (𝑑 + 1)), where 𝜓(𝑥) = log10 𝑥 (mod 1). If we let 𝛾 = log10 2, then 𝛾 is irrational. Putting this together with Theorem 8.2 we see that for 𝑑 = 1, … , 9, the number of times we would expect the first digit of 2𝑛 to be in the interval [log10 𝑑, log10 (𝑑 + 1)) is the length of the interval. For example, when 𝑑 = 2 we get log10 3 − log10 2, which is about 17%—just as Benford’s law claims. When King wrote his paper, he included four questions, some of which were mentioned above: Will 9 occur in the 2-column? Will 23456789 occur again in a row? Will a row consisting of digits that are all the same occur? Will the decimal expansion for an 8-digit prime ever appear? The

Chapter 8. Something Completely Different

99

answer to the first question (that is, the question we just answered) appears in King’s paper as well as the book [5, Appendix 12]. Answers to the other questions can be found in [45], and here is what they tell us. • The only positive integer 𝑛 with ⟨⟨ℓ𝑛 ⟩⟩ = ℓ for all ℓ = 2, … , 9 is 𝑛 = 1. • For a fixed positive integer 𝑛, the eight numbers ⟨⟨2𝑛 ⟩⟩, ⟨⟨3𝑛 ⟩⟩, … , ⟨⟨8𝑛 ⟩⟩, ⟨⟨9𝑛 ⟩⟩ are not all equal; that is, we never have a constant row in our table. • The set

9

{ ∑ ⟨⟨ℓ𝑛 ⟩⟩109−ℓ ∶ 𝑛 ∈ ℤ+ } 𝑙=2

contains precisely 17,596 integers, of which 1,127 are prime numbers. It is now time to look at extensions of Poncelet’s theorem, the elliptical range theorem, and the Blaschke products that connect these.

Part 2 It is only after you have come to know the surface of things, … that you can venture to seek what is underneath. But the surface of things is inexhaustible. –Italiano Calvino, Mr. Palomar And let us not confine our cares To simple circles, planes and spheres –Thorold Gossett, The Kiss Precise (Generalized)

Chapter

9

Compressions of the Shift Operator: The Basics For degree greater than 3, finite Blaschke products are connected to curves that retain some of the properties of the Poncelet curves we have already studied but, in most cases, lose their elliptic nature. The precise operators that provide our matrices are called compressions of the shift operator associated with a finite Blaschke product. They can also be defined on an infinite dimensional space, but in doing so, things become much more complicated (infinitely more complicated). We are primarily concerned with what occurs in finite dimensions, so while we present a brief look at the background, what we are really interested in is a special class of operators: Operators on a particular finite dimensional space (that will be defined in this chapter) that are contractions, have all their eigenvalues in 𝔻, and have the property that rank(𝐼 − 𝐴⋆ 𝐴)1/2 = rank(𝐼 − 𝐴𝐴⋆ )1/2 = 1, where 𝐴 is the matrix representing the operator. This class is often called 𝒮𝑛 , and we come back to it later in this chapter. The special properties of these operators allow them (even in the infinite dimensional case) to serve as a model for all contractions [145]. The matrix representation of these operators includes a very famous class—the 𝑛 × 𝑛 Jordan block with zeros on the main diagonal: 0 1 0 … 0 ⎤ ⎡ ⎢0 0 1 ⋯ 0⎥ 𝐽𝑛 = ⎢⋮ ⋮ ⋮ ⋱ ⋮⎥ . ⎥ ⎢ ⎢0 0 0 ⋯ 1⎥ ⎣0 0 0 ⋯ 0⎦ 103

104

Chapter 9. Compressions of the Shift

If we agree that diagonalization is the best thing that can happen to a matrix, then the Jordan canonical form would have to be the second best thing that can happen; this should give a sense of the importance of these operators. We now turn to a special function space that will allow us to introduce the appropriate spaces and operators that serve as the motivation for our work; motivation that requires some mathematics that is a bit deeper than the results we actually need and use. So, we present the bare minimum here and provide references for those readers who wish to learn more. One really good reference is Young’s book [154]. The reader should also note that we sometimes interchange two limits or an integral and infinite sum; to check that these computations are justified, the reader might consult Royden’s book [133, Chapter 4]. We begin with the Lebesgue space, 𝐿2 (𝕋), named in honor of Henri Lebesgue who introduced the Lebesgue integral in 1904. Let 𝐿2 (𝕋) denote the Hilbert space of square-integrable Lebesgue measurable functions on 𝕋 with the inner product given by 2𝜋

⟨𝑓, 𝑔⟩ = ∫

𝑓(𝑒𝑖𝜃 )𝑔(𝑒𝑖𝜃 )

0

𝑑𝜃 , 2𝜋

where 𝑑𝜃/2𝜋 denotes the normalized Lebesgue measure. In this space, the integral cannot distinguish between functions that are the same except on a set of measure zero, so two functions are identified if they are equal almost everywhere. Thus, elements of 𝐿2 (𝕋) are really equivalence classes of functions, but analysts usually discuss the elements as if they were functions. Letting ‖𝑓‖2𝐿2 (𝕋) = ⟨𝑓, 𝑓⟩, we obtain a norm on 𝐿2 (𝕋). Certain functions are clearly in 𝐿2 (𝕋); for example, all continuous functions are bounded on 𝕋 and Lebesgue measurable and therefore they are members of 𝐿2 (𝕋). ln particular, the functions 𝑒𝑛 defined by 𝑒𝑛 (𝑒𝑖𝜃 ) = 𝑒𝑖𝑛𝜃 for 𝑛 ∈ ℤ are in 𝐿2 (𝕋), and it is easy to compute their norm: 2𝜋

‖𝑒𝑛 ‖2𝐿2 (𝕋) = ⟨𝑒𝑖𝑛𝜃 , 𝑒𝑖𝑛𝜃 ⟩ = ∫ 0

𝑒𝑖𝑛𝜃 ⋅ 𝑒−𝑖𝑛𝜃

𝑑𝜃 = 1. 2𝜋

Note that it was helpful to have divided by 2𝜋. In the same way, it is easy to see that ⟨𝑒𝑖𝑛𝜃 , 𝑒𝑖𝑚𝜃 ⟩ = 0 for 𝑚 ≠ 𝑛 and so (𝑒𝑛 )𝑛∈ℤ is an orthonormal set in 𝐿2 (𝕋). It is also true (see [89, p. 12]) that if 𝑓 ∈ 𝐿2 (𝕋) and

Chapter 9. Compressions of the Shift

105

⟨𝑒𝑛 , 𝑓⟩ = 0 for each 𝑛 ∈ ℤ, then 𝑓 = 0. In this case, (𝑒𝑛 )𝑛∈ℤ is said to be a complete orthonormal basis [89, p. 30]. Thus, given 𝑓 ∈ 𝐿2 (𝕋), we can write ∞

𝑓(𝑒𝑖𝜃 ) = ∑ 𝑎𝑛 𝑒𝑖𝑛𝜃 , 𝑛=−∞

̂ denotes the 𝑛th Fourier coefficient of 𝑓. where 𝑎𝑛 = ⟨𝑓, 𝑒𝑛 ⟩ = 𝑓(𝑛) Parseval’s identity, which you can think of as the Pythagorean theo∞ rem for inner product spaces, says that for 𝑓 ∈ 𝐿2 (𝕋) with 𝑓=∑𝑛=−∞ 𝑎𝑛 𝑒𝑛 , we have ∞

∑ |𝑎𝑛 |2 = ⟨𝑓, 𝑓⟩ = ‖𝑓‖2𝐿2 (𝕋) . 𝑛=−∞

Now we turn to a second important space, the Hardy space 𝐻 2 (𝔻), named in honor of G. H. Hardy. If we have an analytic function 𝑓 de∞ fined by 𝑓(𝑧) = ∑𝑛=0 𝑎𝑛 𝑧𝑛 on 𝔻, then thinking of 𝑓𝑟 ∶ 𝕋 → ℂ defined by 𝑓𝑟 (𝑒𝑖𝜃 ) = 𝑓(𝑟𝑒𝑖𝜃 ) for 0 < 𝑟 < 1, we get a continuous function on 𝕋 and therefore 𝑓𝑟 ∈ 𝐿2 (𝕋). The discussion above and the first of the aforementioned interchangings of an infinite sum and integral imply that 2𝜋

‖𝑓𝑟 ‖2𝐿2 (𝕋)

=∫

2𝜋

𝑖𝜃 2 𝑑𝜃

|𝑓(𝑟𝑒 )|

2𝜋

0

=∫ 0

2𝜋

∞

= ∑ 𝑎𝑛 𝑎𝑚 𝑟 𝑛 𝑟 𝑚 ∫

∞

𝑒𝑖(𝑛−𝑚)𝜃

0

𝑛,𝑚=0

2

| |∞ | ∑ 𝑎 𝑟𝑛 𝑒𝑖𝑛𝜃 | 𝑑𝜃 𝑛 | 2𝜋 | | |𝑛=0 𝑑𝜃 = ∑ |𝑎 |2 𝑟2𝑛 . 2𝜋 𝑛=0 𝑛

(9.1)

These integrals increase as 𝑟 increases, and 𝐻 2 (𝔻) is the set of those analytic functions 𝑓 ∶ 𝔻 → ℂ for which 2𝜋

lim− ∫

𝑟→1

|𝑓(𝑟𝑒𝑖𝜃 )|2

0

𝑑𝜃 < ∞. 2𝜋

The norm on this space is defined by 2𝜋

‖𝑓‖2𝐻 2 (𝔻) = lim− ∫ 𝑟→1

0

|𝑓(𝑟𝑒𝑖𝜃 )|2

𝑑𝜃 . 2𝜋

(9.2)

106

Chapter 9. Compressions of the Shift ∞

If 𝑓 ∈ 𝐻 2 (𝔻) and we write 𝑓(𝑧) = ∑𝑛=0 𝑎𝑛 𝑧𝑛 , then using (9.1) and another of the aforementioned changing of limits, we have 2𝜋

∞

∞

𝑑𝜃 = lim− ∑ |𝑎𝑛 |2 𝑟2𝑛 = ∑ |𝑎𝑛 |2 . 2𝜋 𝑟→1 𝑟→1 0 𝑛=0 𝑛=0 (9.3) Let us look at the functions {𝑓𝑟 ∶ 0 < 𝑟 < 1} associated with 𝑓 ∈ 𝐻 2 (𝔻) more closely. We know that 𝑓𝑟 ∈ 𝐿2 (𝕋) for each 𝑟 satisfying 0 < 𝑟 < 1. But for 𝑟 and 𝑠 satisfying 0 < 𝑟, 𝑠 < 1 we know that ‖𝑓‖2𝐻 2 (𝔻) = lim− ∫

2𝜋

‖𝑓𝑟 −

𝑓𝑠 ‖2𝐿2 (𝕋)

=∫ 0

|𝑓(𝑟𝑒𝑖𝜃 )|2

2

∞ |∞ | | ∑ 𝑎 (𝑟𝑛 − 𝑠𝑛 )𝑒𝑖𝑛𝜃 | 𝑑𝜃 = ∑ |𝑟𝑛 − 𝑠𝑛 |2 |𝑎 |2 . 𝑛 𝑛 | | 2𝜋 |𝑛=0 | 𝑛=0

As the reader should check, it follows from this that every sequence (𝑓𝑟𝑛 ) with 𝑟𝑛 → 1− is a Cauchy sequence. Our estimates plus the fact that 𝐿2 (𝕋) is a Banach space show that every such sequence will, as 𝑟𝑛 → 1− , converge to the same function defined on 𝕋, a function that is usually denoted by 𝑓 ⋆ . What we need is stated precisely in Fatou’s radial limit theorem [54, p. 28]. Theorem 9.1 (Fatou’s radial limit theorem). Let 𝑓 ∈ 𝐻 2 (𝔻). Then the radial limit lim− 𝑓(𝑟𝑒𝑖𝜃 ) = 𝑓 ⋆ (𝑒𝑖𝜃 ) 𝑟→1

exists almost everywhere on 𝕋. Looking back at (9.3), we see that what it says is that ‖𝑓‖2𝐻 2 (𝔻) = ‖𝑓 ⋆ ‖2𝐿2 (𝕋) . Most authors drop the special notation and use 𝑓 to denote the boundary value function of 𝑓. It is also usual to identify 𝐻 2 (𝔻) with the (closed) ̂ subspace of 𝐿2 (𝕋) given by {𝑓 ∈ 𝐿2 (𝕋) ∶ 𝑓(𝑛) = 0 for 𝑛 < 0}. In the fu2 2 ture, we write 𝐻 in place of 𝐻 (𝔻) and it will usually be clear from the context whether the function is considered on the open unit disk or the unit circle; if it is not clear, we promise to remind you of the setting. We always require that our subspaces be closed and 𝐻 2 is, in fact, the closure of the polynomials (where we mean polynomials in 𝑧) [30, p. 120]. It is now time to look at some examples. There are many functions in 𝐻 2 , and we would like to point out that finite Blaschke products are among these functions: They are analytic on 𝔻 and, since they are

Chapter 9. Compressions of the Shift

107

bounded (by 1) on 𝔻, the limit in (9.2) must be finite. All polynomials are in 𝐻 2 , for more or less the same reason finite Blaschke products are. Here are some other functions that have interesting behavior: For 𝑎 ∈ 𝔻, define 1 𝑘𝑎 (𝑧) = . 1 − 𝑎𝑧 The function 𝑘𝑎 is called a reproducing kernel1 for 𝐻 2 at the point 𝑎 ∈ 𝔻, and it is a bounded analytic function on 𝔻 that plays an important role in what follows. We can get the series expansion for 𝑘𝑎 using what we know about geometric series: ∞

𝑛

𝑘𝑎 (𝑧) = ∑ 𝑎 𝑧𝑛 . 𝑛=0

This is quite handy. To see what we mean, let us do a few computations. By Parseval’s identity ∞

∞

‖𝑘𝑎 ‖2𝐿2 (𝕋) = ∑ |𝑎𝑛 |2 = ∑ |𝑎2 |𝑛 = 𝑛=0

𝑛=0

1 . 1 − |𝑎|2

(9.4)

We also frequently use the normalized reproducing kernel defined by √1 − |𝑎|2 . 1 − 𝑎𝑧 Looking at the series expansion for 𝑘𝑎 , we see that for each 𝑛 ≥ 0, 𝑘𝑎̃ (𝑧) =

ˆ𝑎 (𝑛) = 𝑎𝑛 . ⟨𝑒𝑛 , 𝑘𝑎 ⟩ = 𝑘 Since the inner product is linear in the first variable, we know that for each polynomial 𝑝 we have ⟨𝑝, 𝑘𝑎 ⟩ = 𝑝(𝑎). But the polynomials are dense in 𝐻 2 , so this can be extended to all functions in 𝐻 2 . Thus, 𝑘𝑎 allows us to evaluate 𝑓 ∈ 𝐻 2 at 𝑎 via the inner product on our space, reproducing 𝑓. Given this property, the name reproducing kernel is fitting. There are many other functions in 𝐻 2 , but that is another story. Here we focus on 𝐻 2 and “the most important single operator, which plays 1 This is also called the Cauchy kernel because of its appearance in the Cauchy integral formula.

108

Chapter 9. Compressions of the Shift

a vital role in all parts of Hilbert space theory” ([76], p. 40), the shift operator 𝑆 defined on 𝐻 2 by 𝑆(𝑓)(𝑧) = 𝑧𝑓(𝑧). Note that ‖𝑆𝑓‖𝐻 2 = ‖𝑓‖𝐻 2 . Example 9.2. Though this takes us afield from operators acting on a finite dimensional space, we know enough to compute the numerical range of 𝑆. First, by the Cauchy–Bunyakovsky–Schwarz inequality for 𝑓 ∈ 𝐻 2 with norm 1, we have |⟨𝑆𝑓, 𝑓⟩| ≤ ‖𝑆𝑓‖𝐻 2 ‖𝑓‖𝐻 2 = 1. Thus, 𝑊(𝑆) ⊆ 𝔻. Now, let 𝑎 ∈ 𝔻 and consider the normalized reproducing kernel, 𝑘𝑎̃ . Then we have ‖𝑘𝑎̃ ‖𝐻 2 = 1. Using the reproducing property of 𝑘𝑎 , we have ⟨𝑆 𝑘𝑎̃ , 𝑘𝑎̃ ⟩ = (1 − |𝑎|2 )⟨𝑆𝑘𝑎 , 𝑘𝑎 ⟩ = 𝑎. Thus, 𝑎 ∈ 𝑊(𝑆) and we have shown that 𝔻 ⊆ 𝑊(𝑆) ⊆ 𝔻.

(9.5)

Suppose that 𝜆 ∈ 𝑊(𝑆) ∩ 𝕋. Then there exists 𝑔 of norm 1 with 𝜆 = ⟨𝑆𝑔, 𝑔⟩. As before, we have 1 = |𝜆| = |⟨𝑆𝑔, 𝑔⟩| ≤ ‖𝑆𝑔‖𝐻 2 ‖𝑔‖𝐻 2 ≤ 1. In particular, |⟨𝑆𝑔, 𝑔⟩| = ‖𝑆𝑔‖𝐻 2 ‖𝑔‖𝐻 2 and equality holds in the Cauchy–Bunyakovsky–Schwarz inequality. This implies that 𝑆𝑔 and 𝑔 are linearly dependent; that is, there exists a constant 𝜇 with 𝑆𝑔 = 𝜇𝑔. But in this case, we would have 𝑧𝑔(𝑧) = 𝑆𝑔(𝑧) = 𝜇𝑔(𝑧), where 𝑔 is a nonzero analytic function. This is impossible, so we conclude that 𝕋 ∩ 𝑊(𝑆) = ∅. Putting this together with (9.5) we see that 𝑊(𝑆) = 𝔻. In particular, we now have an example that shows that the numerical range of an operator on an infinite dimensional space need not be closed. Here is another way to view 𝑆. If you think about the Fourier coef∞ ficients of 𝑓 as the sequence (𝑎0 , 𝑎1 , …) and 𝑓(𝑧) = ∑𝑛=0 𝑎𝑛 𝑧𝑛 , then 𝑆 shifts the Fourier coefficients of 𝑓 to the right, inserting a zero in the first spot: (𝑎0 , 𝑎1 , …) ↦ (0, 𝑎0 , 𝑎1 , …).

Chapter 9. Compressions of the Shift

109

It is the forward shift. Now from our definition, it is easy to see that 𝐻 2 is invariant under 𝑆; that is, if 𝑓 ∈ 𝐻 2 , then 𝑆(𝑓)(𝑧) = 𝑧𝑓(𝑧) and so 𝑆𝑓 ∈ 𝐻 2 . Are there other subspaces of 𝐻 2 that are invariant under S (where invariant subspace means 𝑆(𝑀) ⊆ 𝑀 and subspace means closed subspace)? Well, sure—there are at least two such subspaces, namely, {0} and 𝐻 2 . But these are not very interesting invariant subspaces, so they are called trivial subspaces. Here is an example of a nontrivial invariant subspace: Let 𝐵 denote any (nonconstant) finite Blaschke product. We claim that 𝐵𝐻 2 = {𝐵ℎ ∶ ℎ ∈ 𝐻 2 } is invariant under the shift operator: If 𝑓 ∈ 𝐵𝐻 2 , then 𝑓 = 𝐵𝑔 for some function 𝑔 ∈ 𝐻 2 and 𝑆(𝑓)(𝑧) = 𝑧𝑓(𝑧) = 𝑧𝐵(𝑧)𝑔(𝑧) = 𝐵(𝑧) (𝑧𝑔(𝑧)) . So 𝑆(𝑓) ∈ 𝐵𝐻 2 and we see that 𝑆(𝐵𝐻 2 ) ⊆ 𝐵𝐻 2 . It is easy to check that 𝐵𝐻 2 is closed under addition and scalar multiplication. But why is it a closed subspace of 𝐻 2 ? Well, if (𝑓𝑛 ) is a sequence in 𝐵𝐻 2 and 𝑓𝑛 → 𝑓 using the norm we defined above, then there exist 𝑔𝑛 ∈ 𝐻 2 such that 𝑓𝑛 = 𝐵𝑔𝑛 . Using the fact that |𝐵| = 1 on the circle, we have ‖𝑓𝑛 − 𝑓𝑚 ‖𝐿2 (𝕋) = ‖𝐵𝑔𝑛 − 𝐵𝑔𝑚 ‖𝐿2 (𝕋) = ‖𝑔𝑛 − 𝑔𝑚 ‖𝐿2 (𝕋) , and we see that the fact that (𝑓𝑛 ) is a Cauchy sequence forces (𝑔𝑛 ) to be one as well. Since 𝐻 2 is complete, every Cauchy sequence converges and so there exists 𝑔 ∈ 𝐻 2 such that 𝑔𝑛 → 𝑔 in 𝐻 2 . It follows that ‖𝑓𝑛 − 𝐵𝑔‖𝐿2 (𝕋) = ‖𝐵𝑔𝑛 − 𝐵𝑔‖𝐿2 (𝕋) = ‖𝑔𝑛 − 𝑔‖𝐿2 (𝕋) → 0. So 𝑓 = 𝐵𝑔 ∈ 𝐵𝐻 2 . Looking over this reasoning, what properties of the Blaschke product did we use? If 𝑢 were a bounded analytic function with radial limits of modulus 1 almost everywhere, the same proof would work. Are there such 𝑢 that are not Blaschke products? Yes, in fact, as the reader can check, the function 𝑧+1 𝑠(𝑧) = exp ( (9.6) ) for 𝑧 ∈ 𝔻, 𝑧−1 is one such function. If we consider the set of bounded analytic functions on 𝔻 that have radial limits of modulus 1 almost everywhere on 𝕋, we get a class of functions called inner functions and this class includes our finite Blaschke products. Since inner functions are bounded, they also

110

Chapter 9. Compressions of the Shift

lie in 𝐻 2 . The example presented in (9.6) shows that the class of inner functions is bigger than just the set of finite Blaschke products. More about such functions can be found in Garnett’s book [54, pp. 71–78], for example. What our argument above then shows is that 𝑢𝐻 2 is an invariant subspace for 𝑆 whenever 𝑢 is an inner function. The next lemma looks at the space 𝐵𝐻 2 with 𝐵 a finite Blaschke product more closely. Lemma 9.3. Let 𝐵 be a finite Blaschke product, and let 𝑓 ∈ 𝐻 2 . Then 𝑓 vanishes on the zeros of 𝐵 (accounting for multiplicity) if and only if 𝑓 ∈ 𝐵𝐻 2 . Proof. It should be clear that if 𝑓 ∈ 𝐵𝐻 2 , then 𝑓 vanishes on the zeros of 𝐵. Now suppose 𝐵 has zeros 𝑎1 , 𝑎2 , … , 𝑎𝑛 (where we repeat zeros if the multiplicity of the zero is greater than 1) and 𝑓 vanishes on the zeros of 𝐵. Since 𝑓 vanishes at 𝑎1 , … , 𝑎𝑛 and 𝑓 is analytic, we know from the power series expansion of 𝑓 that 𝑓(𝑧) = (𝑧 − 𝑎1 ) ⋯ (𝑧 − 𝑎𝑛 )𝑔(𝑧) for some analytic function 𝑔. Since 𝑎𝑗 ∈ 𝔻 and (1 − 𝑎𝑗 𝑧) has one zero at 1/𝑎𝑗 , we see that (1 − 𝑎𝑗 𝑧) never vanishes on 𝔻. Thus, 𝑧 − 𝑎𝑛 𝑧 − 𝑎1 ⋯ 𝑔1 (𝑧), 𝑓(𝑧) = 1 − 𝑎1 𝑧 1 − 𝑎𝑛 𝑧 where the function 𝑔1 is still analytic on 𝔻. What we see is that there is an analytic function 𝑔1 with 𝑓 = 𝐵𝑔1 . All that remains to show is that 𝑔1 ∈ 𝐻 2 . Since the Blaschke product 𝐵 is a finite Blaschke product, there exists 𝜀 > 0 and 𝑠 > 0 with |𝐵(𝑧)| > 𝜀 for 𝑠 < |𝑧| < 1. Thus, on this annulus, 𝑔1 (𝑧) = 𝑓(𝑧)/𝐵(𝑧). It follows that (𝑔1 )𝑟 converges pointwise almost everywhere to the function with boundary values 𝑓/𝐵 and for 𝑟 > 𝑠 the functions (𝑔1 )𝑟 satisfy |(𝑔1 )𝑟 (𝑒𝑖𝜃 )| < 𝜀−1 |𝑓𝑟 (𝑒𝑖𝜃 )|. By the generalized Lebesgue dominated convergence theorem (see [133, p. 270]), we see that the integral in (9.2) is finite. So 𝑔1 ∈ 𝐻 2 and 𝑓 ∈ 𝐵𝐻 2 .

Chapter 9. Compressions of the Shift

111

As an aside, we mention that the function 𝑔1 has the property that |𝑓| = |𝑔1 | almost everywhere on 𝕋 and, consequently, ‖𝑓‖𝐻 2 = ‖𝑔1 ‖𝐻 2 . This integral equality is a case that illustrates the beauty of moving between 𝔻 and 𝕋—that the integrals of |𝑓| and |𝑔1 | are equal is clear on the unit circle and not quite as clear if we are taking the limit of the integrals inside 𝔻. This is one of the many reasons that Blaschke products play such an important role in the study of analytic functions in 𝐻 2 . In fact, given a sequence (𝑎𝑛 ) satisfying the convergence condition ∞ ∑𝑛=1 (1 − |𝑎𝑛 |) < ∞, via an infinite product you obtain a function called an infinite Blaschke product that has zeros precisely at the points 𝑎𝑛 and radial limits of modulus 1 almost everywhere. In this case, you can factor out the zeros of 𝑓, collect them in an infinite Blaschke product 𝐵, divide 𝑓 by 𝐵, and work with a function 𝑔1 with no zeros on 𝔻. This often allows you to do many things you cannot do with functions that have zeros in 𝔻, like taking roots or logarithms. Back to 𝐵𝐻 2 : We now know that the space 𝐵𝐻 2 is precisely the set of functions that vanish on the zeros of 𝐵 and that it is an invariant subspace for the forward shift. The key ingredient that we used was the fact that 𝐵 is a bounded analytic function with the property that |𝐵| = 1 almost everywhere on the unit circle. There are other closed subspaces of 𝐻 2 , and we wish to know which of these are invariant under the shift operator. Beurling’s theorem [54, p. 79] will tell us the answer. Theorem 9.4 (Beurling’s theorem). A nontrivial subspace 𝑀 of 𝐻 2 is invariant under the forward shift 𝑆 if and only if there is a nonconstant inner function 𝑢 such that 𝑀 = 𝑢𝐻 2 . If 𝑆 is the “most important single operator”, the second most important operator would have to be its adjoint, 𝑆 ⋆ . You can show that 𝑆 ⋆ (𝑓)(𝑧) = (𝑓(𝑧) − 𝑓(0)) /𝑧. Thinking of the Fourier coefficients of 𝑓 as an ordered sequence ∞ (𝑎0 , 𝑎1 , …) and 𝑓 = ∑𝑛=0 𝑎𝑛 𝑒𝑛 , we see that 𝑆 ⋆ shifts the Fourier coefficients of 𝑓 to the left: (𝑎0 , 𝑎1 , …) ↦ (𝑎1 , 𝑎2 , …); so it is also called the backward shift. Since the proper invariant subspaces for 𝑆 are 𝑢𝐻 2 , where 𝑢 is inner, the proper invariant subspaces

112

Chapter 9. Compressions of the Shift

for 𝑆 ⋆ are the orthogonal complements of these spaces or ⟂

(𝑢𝐻 2 ) ∶= {𝑓 ∈ 𝐻 2 ∶ ⟨𝑓, 𝑢𝑔⟩ = 0 for all 𝑔 ∈ 𝐻 2 }, where 𝑢 is inner. These important spaces are usually denoted by 𝐾ᵆ or 𝐻 2 ⊖ 𝑢𝐻 2 , rather than (𝑢𝐻 2 )⟂ , and are called model spaces. The reader may find a different characterization of this space more illuminating: we show (in just a moment) that 𝐾ᵆ = 𝐻 2 ∩ 𝑢 (𝑧𝐻 2 ) ;

(9.7)

the bar here denotes complex conjugate (not closure—in fact, we know that 𝑧𝐻 2 is already closed!). Thus, 𝑧𝐻 2 = {𝑧ℎ ∶ ℎ ∈ 𝐻 2 }. To see that (9.7) holds, let 𝑓 ∈ 𝐻 2 . Note that since 𝑢 is inner, 𝑢𝑓 ∈ 2 𝐿 . Now, for all ℎ ∈ 𝐻 2 we have ⟨𝑓, 𝑢ℎ⟩ = 0 if and only if ⟨𝑢𝑓, ℎ⟩ = 0. But this happens if and only if 𝑢𝑓 annihilates all of 𝐻 2 ; that is, the nonnegative Fourier coefficients of 𝑢𝑓 must vanish and, looking at the power series, this happens if and only if 𝑢𝑓 ∈ 𝑧𝐻 2 . Since 𝑢 is an inner function, multiplying both sides by 𝑢 establishes the equality. To sum up our argument, we have shown 𝑓 ⟂ 𝑢𝐻 2 ⟺ 𝑢𝑓 ⟂ 𝐻 2 ⟺ 𝑢𝑓 ∈ 𝑧𝐻 2 ⟺ 𝑓 ∈ 𝑢 (𝑧𝐻 2 ) . At this point, it would be a good exercise for the reader to show that if 𝑎1 , … , 𝑎𝑛 are distinct points in 𝔻, then 𝑘𝑎1 , … , 𝑘𝑎𝑛 are linearly independent. Putting this together with Proposition 9.5, we obtain a basis for 𝐾𝐵 and find that if 𝐵 is a Blaschke product with distinct zeros 𝑎1 , … , 𝑎𝑛 , then 𝑛 is the dimension of 𝐾𝐵 . Proposition 9.5. If 𝐵 is a degree-𝑛 Blaschke product with distinct zeros 𝑎1 , … , 𝑎𝑛 , then 𝐾𝐵 is the span of 𝑘𝑎1 , … , 𝑘𝑎𝑛 .2 Notice that because the span above is finite dimensional, it is also closed. Proof. For each ℎ ∈ 𝐻 2 we know that ⟨𝐵ℎ, 𝑘𝑎𝑗 ⟩ = 𝐵(𝑎𝑗 )ℎ(𝑎𝑗 ) = 0 for 𝑗 = 1, … , 𝑛. It follows that 𝑘𝑎𝑗 ∈ 𝐾𝐵 for all 𝑗. Therefore, span{𝑘𝑎1 , … , 𝑘𝑎𝑛 } ⊆ 𝐾𝐵 . 2 If

𝐵 has a zero of multiplicity 𝑚𝑗 at 𝑎𝑗 , then we take span{𝑘𝑎ℓ 𝑗 , 𝑗 = 1, … , 𝑛, ℓ =

1, … , 𝑚𝑗 }.

Chapter 9. Compressions of the Shift

113

We show that 𝐾𝐵 is the span by showing that 𝐵𝐻 2 = 𝐾𝐵⟂ ⊇ span{𝑘𝑎1 , … , 𝑘𝑎𝑛 }⟂ . Then taking the orthogonal complement one more time will reverse the inclusion. So suppose 𝑓 ∈ span{𝑘𝑎1 , … , 𝑘𝑎𝑛 }⟂ . In particular, ⟨𝑓, 𝑘𝑎𝑗 ⟩ = 0 for 𝑗 = 1, 2, … , 𝑛. So 𝑓 vanishes on the zeros of 𝐵 and, by Lemma 9.3, we know that 𝑓 ∈ 𝐵𝐻 2 , completing the proof. We would like to define a shift operator from 𝐾ᵆ to itself, but 𝐾ᵆ is invariant for 𝑆 ⋆ —not 𝑆. Therefore, when we multiply by 𝑧 the shift operator 𝑆 might take us outside of 𝐾ᵆ . Thus, we have to do one thing that we did not have to do when we considered the shift operating on 𝐻 2 ; we have to project back into the space 𝐾ᵆ . So let 𝑃ᵆ denote the (orthogonal) projection from 𝐻 2 onto 𝐾ᵆ and consider the compressed shift operator 𝑆ᵆ ∶ 𝐾ᵆ → 𝐾ᵆ defined by 𝑆ᵆ (𝑓) = 𝑃ᵆ (𝑆(𝑓)). Thus, 𝑆ᵆ = 𝑃ᵆ 𝑆|𝐾𝑢 . This would be a good time to say more about the operator 𝑃ᵆ , but to do so we need to introduce two more operators. So let 𝑃− denote the orthogonal projection of 𝐿2 (𝕋) onto 𝐿2 (𝕋) ⊖ 𝐻 2 and 𝑃+ ∶ 𝐿2 (𝕋) → 𝐻 2 the orthogonal projection of 𝐿2 (𝕋) onto 𝐻 2 . Recall that 𝐿2 (𝕋) ⊖ 𝐻 2 = 𝑧𝐻 2 and so 𝑃− (𝑓) = 𝑓 for all 𝑓 ∈ 𝑧𝐻 2 . Of course, the identity operator 𝐼 ∶ 𝐿2 (𝕋) → 𝐿2 (𝕋) can be written as 𝐼 = 𝑃+ + 𝑃− . We are ready to study the projection operator 𝑃ᵆ corresponding to an inner function 𝑢: We claim that 𝑃ᵆ (𝑓) = 𝑢𝑃− (𝑢𝑓) for 𝑓 ∈ 𝐻 2 .

(9.8)

First, suppose 𝑓 ∈ 𝐾ᵆ . Since 𝑃ᵆ is the orthogonal projection onto 𝐾ᵆ , we have 𝑃ᵆ (𝑓) = 𝑓. From (9.7) we know that there exists ℎ ∈ 𝐻 2 with 𝑓 = 𝑢𝑧ℎ = 𝑢𝑆ℎ. Using this, the fact that 𝑃− is the orthogonal projection onto 𝑧𝐻 2 , and the fact that 𝑢 ⋅ 𝑢 = 1 on 𝕋, we have 𝑢𝑃− (𝑢𝑓) = 𝑢𝑃− (𝑆ℎ) = 𝑢𝑆ℎ = 𝑓. In particular, 𝑃ᵆ (𝑓) = 𝑢𝑃− (𝑢𝑓) for all 𝑓 ∈ 𝐾ᵆ . Now suppose that 𝑔 ∈ 𝐻 2 satisfies 𝑔 ⟂ 𝐾ᵆ = 𝐻 2 ⊖ 𝑢𝐻 2 . Then 𝑔 ∈ 𝑢𝐻 2 . Therefore, there exists 𝑘 ∈ 𝐻 2 with 𝑔 = 𝑢𝑘. On the unit

114

Chapter 9. Compressions of the Shift

circle, we have 𝑢𝑃− (𝑢𝑔) = 𝑢𝑃− (𝑘) = 𝑢(𝐼 − 𝑃+ )(𝑘) = 0 = 𝑃ᵆ (𝑔). Thus, 𝑃ᵆ (𝑔) = 𝑢𝑃− (𝑢𝑔) for all 𝑔 ∈ 𝑢𝐻 2 . Since 𝐻 2 = 𝑢𝐻 2 ⊕ 𝐾ᵆ , we have now established (9.8). You will certainly find it useful to know the following equivalent ways of writing 𝑃ᵆ (𝑓) for 𝑓 ∈ 𝐻 2 : 𝑃ᵆ (𝑓) = 𝑢𝑃− (𝑢𝑓) = 𝑢(𝐼 − 𝑃+ )(𝑓). It is also good to keep in mind that the operator 𝑃ᵆ is self-adjoint, satisfying the equation ⟨𝑃ᵆ 𝑓, 𝑔⟩ = ⟨𝑓, 𝑃ᵆ 𝑔⟩ for all 𝑓 and 𝑔 in 𝐻 2 . When 𝑢 = 𝐵 is a finite Blaschke product, which is the case we are interested in, we know that 𝐾𝐵 is finite dimensional (and this is precisely when 𝐾ᵆ is finite dimensional) and we can try to find a matrix representation for this operator. But what is a good basis? There are lots of bases—even lots of good ones—but if we find a basis that yields an upper triangular matrix, we can read the eigenvalues off the matrix. In addition, we would also like to have an orthonormal basis for the space. When 𝐵 is a finite Blaschke product with distinct zeros, we saw above that the reproducing kernels form a basis for 𝐾𝐵 , so we might try the Gram–Schmidt process on them to see what we get. That brings us to the so-called Takenaka–Malmquist basis, used by Takenaka in 1925 [146], Malmquist in 1926 [106], and Walsh3 in 1932 [149]. We write 𝑧−𝑎 , where 𝑎 ∈ 𝔻, 𝜑𝑎 (𝑧) = 1 − 𝑎𝑧 for one Blaschke factor, and these will appear in the basis we use. There are a few important things to note about these functions: First, 𝜑𝑎 (𝑎) = 0. Second, |𝜑𝑎 | = 1 on the unit circle and therefore 𝜑𝑎 𝜑𝑎 = 1 on 𝕋. Fi𝑛 nally, if the zeros of 𝐵 are 𝑎1 , … , 𝑎𝑛 , then 𝐵 = 𝜇 ∏𝑗=1 𝜑𝑎𝑗 , where 𝜇 is a unimodular constant. In addition, we need to use the normalized reproducing kernel, 𝑘𝑎̃ . Here is the Takenaka–Malmquist basis in the event that the zeros 𝑎1 , … , 𝑎𝑛 are distinct and ordered so that the matrices we work with are upper triangular. We have 3 Walsh’s

name is usually added when we consider infinite Blaschke products.

Chapter 9. Compressions of the Shift

𝑛

115

𝑛

(𝑘𝑎̃ 1 ∏ 𝜑𝑎𝑗 , 𝑘𝑎̃ 2 ∏ 𝜑𝑎𝑗 , … , 𝑘𝑎̃ 𝑛−1 𝜑𝑎𝑛 , 𝑘𝑎̃ 𝑛 ) . 𝑗=2

(9.9)

𝑗=3

We know that the 𝑘𝑎̃ 𝑗 are in the space 𝐾𝐵 , and we would like to take a moment to convince you that the other functions are too. So, we just need to check that for each 𝑗 = 1, … , 𝑛 − 1 and ℎ ∈ 𝐻 2 we have 𝑛 𝑛 ⟨𝐵ℎ, 𝑘𝑎̃ 𝑗 ∏𝑘=𝑗+1 𝜑𝑎𝑘 ⟩ = 0. On the unit circle, the function ∏𝑘=𝑗+1 𝜑𝑎𝑘 has modulus 1. Therefore, 𝑛

2

𝑛

∏ 𝜑𝑎𝑘 ⋅ ∏ 𝜑𝑎𝑘 𝑘=𝑗+1

𝑘=𝑗+1

| 𝑛 | = || ∏ 𝜑𝑎𝑘 || = 1. |𝑘=𝑗+1 |

So for ℎ ∈ 𝐻 2 , we have 2𝜋

𝑛

⟨𝐵ℎ, 𝑘𝑎̃ 𝑗 ∏ 𝜑𝑎𝑘 ⟩ = ∫ 𝑘=𝑗+1

0

𝑛

𝑛

𝑑𝜃 (𝜇 ∏ 𝜑𝑎ℓ ) ℎ ⋅ (𝑘𝑎̃ 𝑗 ∏ 𝜑𝑎𝑘 ) 2𝜋 𝑘=𝑗+1 ℓ=1

𝑗

= 𝜇⟨(∏ 𝜑𝑎ℓ )ℎ, 𝑘𝑎̃ 𝑗 ⟩ = 0. ℓ=1 𝑛 Thus, the functions 𝑘𝑎̃ 𝑗 ∏𝑘=𝑗+1 𝜑𝑎𝑘 are in 𝐾𝐵 . Since the 𝑘𝑎̃ 𝑗 are normalized reproducing kernels and the 𝜑𝑎𝑗 all have modulus 1 on the unit circle, it is easy to see that this is an orthonormal basis: That the norm is 1 is clear; for orthogonality we check that for 𝑞 > 𝑝 we have 𝑛

𝑛

𝑞

⟨𝑘𝑎̃ 𝑝 ∏ 𝜑𝑎𝑗 , 𝑘𝑎̃ 𝑞 ∏ 𝜑𝑎𝑗 ⟩ = ⟨𝑘𝑎̃ 𝑝 ∏ 𝜑𝑎𝑗 , 𝑘𝑎̃ 𝑞 ⟩ = 0. 𝑗=𝑝+1

𝑗=𝑞+1

𝑗=𝑝+1

We showed that 𝐾𝐵 has dimension 𝑛 in Proposition 9.5 and consequently the Takenaka–Malmquist basis is deserving of the word “basis” in its name. What is the matrix representation for 𝑆𝐵 with respect to this basis? The easiest way to compute it (we think) is to compute the matrix for the adjoint, 𝑆𝐵⋆ on 𝐾𝐵 . Note that since 𝐾𝐵 is invariant for 𝑆 ⋆ , we have 𝑆𝐵⋆ = 𝑆 ⋆ on 𝐾𝐵 . Therefore, the only thing we need is the definition of

116

Chapter 9. Compressions of the Shift

𝑆 ⋆ : For 𝑓 ∈ 𝐾𝐵 , we have (𝑆𝐵⋆ (𝑓))(𝑧) = (𝑆 ⋆ (𝑓))(𝑧) =

𝑓(𝑧) − 𝑓(0) . 𝑧

Now, for 𝑓, 𝑔 ∈ 𝐻 2 , we have 𝑓(𝑧)𝑔(𝑧) − 𝑓(0)𝑔(0) 𝑧 𝑓(𝑧)(𝑔(𝑧) − 𝑔(0)) 𝑔(0)(𝑓(𝑧) − 𝑓(0)) = + 𝑧 𝑧 ⋆ = 𝑓(𝑧)(𝑆 (𝑔))(𝑧) + 𝑔(0)(𝑆 ⋆ (𝑓))(𝑧).

(𝑆 ⋆ (𝑓𝑔))(𝑧) =

(9.10)

We do the 2 × 2 case in detail and provide a sketch of how the 3 × 3 case goes. You can try your hand at higher dimensional cases using the computations below, plus a few more. We first note that using the definition of 𝑆 ⋆ , we get 𝑆 ⋆ (𝑘𝑎̃ ) = 𝑎𝑘𝑎̃ . (9.11) We see that we got more than we bargained for; 𝑘𝑎̃ is an eigenvector for 𝑆 ⋆ with eigenvalue 𝑎. Another computation shows that (𝑆 ⋆ (𝜑𝑎 ))(𝑧) = (1 − |𝑎|2 )/(1 − 𝑎𝑧) = √1 − |𝑎|2 𝑘𝑎̃ (𝑧).

(9.12)

In the 2 × 2 case, the basis is simply (𝑘𝑎̃ 1 𝜑𝑎2 , 𝑘𝑎̃ 2 ). Looking at 𝑘𝑎̃ 1 𝜑𝑎2 and using formula (9.10), we obtain 𝑆 ⋆ (𝑘𝑎̃ 1 𝜑𝑎2 ) = 𝜑𝑎2 𝑆 ⋆ (𝑘𝑎̃ 1 ) + 𝑘𝑎̃ 1 (0)𝑆 ⋆ (𝜑𝑎2 ). Using (9.11) and (9.12), we get 𝑆 ⋆ (𝑘𝑎̃ 1 𝜑𝑎2 ) = 𝑎1 𝑘𝑎̃ 1 𝜑𝑎2 + √1 − |𝑎1 |2 √1 − |𝑎2 |2 𝑘𝑎̃ 2 and

𝑆 ⋆ (𝑘𝑎̃ 2 ) = 𝑎2 𝑘𝑎̃ 2 .

Writing the matrix with respect to our ordered basis, we see that the matrix representation for 𝑆 ⋆ is [

𝑎1 √1 − |𝑎1 |2 √1 − |𝑎2 |2

0 ]. 𝑎2

Taking the adjoint we spot one of our very favorite matrices, [

𝑎1 0

√1 − |𝑎1 |2 √1 − |𝑎2 |2 ], 𝑎2

(9.13)

Chapter 9. Compressions of the Shift

117

which is the matrix we met in Chapter 7 given by (7.3). The reader should check that this works in general. For example, if 𝐵 has three distinct zeros, the same computations apply to the first two vectors and the third basis vector works just the same way. To become more familiar with the concepts and computations here, it would be a really good idea to check that you get the following: 𝑆 ⋆ (𝑘𝑎̃ 1 𝜑𝑎2 𝜑𝑎3 ) = 𝑎1 𝑘𝑎̃ 1 𝜑𝑎2 𝜑𝑎3 + √1 − |𝑎1 |2 √1 − |𝑎2 |2 𝑘𝑎̃ 2 𝜑𝑎3 − 𝑎2 √1 − |𝑎1 |2 √1 − |𝑎3 |2 𝑘𝑎̃ 3 . If we do this calculation in general and then take the adjoint, we get a matrix we refer to as 𝐴 (presented in (9.14) below), representing the compression of the shift. Back in Chapter 7, the matrix given by (9.13) also had a unitary dilation, which looked like 𝑎1 𝑈𝜆 = [ 0 𝜆√1 − |𝑎1 |2

√1 − |𝑎1 |2 √1 − |𝑎2 |2 𝑎2 −𝜆𝑎1 √1 − |𝑎2 |2

−𝑎2 √1 − |𝑎1 |2 √1 − |𝑎2 |2 ] . 𝜆𝑎1 𝑎2

Notice that if we can only add one row and one column to the matrix 𝐴 to get to our unitary matrix, we get one 𝑈𝜆 for each 𝜆 ∈ 𝕋 and, up to unitary equivalent matrices, it turns out that there is nothing else that we can get. This brings us to our matrices corresponding to higher-degree Blaschke products. In what follows, we let 𝑎1 , … , 𝑎𝑛 be points in 𝔻. For 𝑛 ≥ 2, we look at the 𝑛 × 𝑛 matrix 𝐴 = [𝑎𝑖𝑗 ] given by 𝑎𝑖𝑗 =

⎧

𝑎𝑗

𝑗−1 (∏𝑘=𝑖+1 (−𝑎𝑘 )) √1

if 𝑖 = 𝑗, |2 √1

|2

(9.14) − |𝑎𝑖 − |𝑎𝑗 if 𝑖 < 𝑗, ⎨ 0 if 𝑖 > 𝑗. ⎩ We say that the matrix 𝐴 in (9.14) is in standard form, and we have seen above that the matrix 𝐴 represents a compression of the shift in the case when the zeros are distinct. It is possible to adjust the orthonormal basis to show that this representation also works when the zeros are not distinct. And there is one particularly interesting case—the case in which the Blaschke product has all its 𝑛 zeros at zero. In that case, the associated matrix 𝐴 is just the 𝑛 × 𝑛 Jordan block, 𝐽𝑛 . We often say that the

118

Chapter 9. Compressions of the Shift

Blaschke product 𝐵 with zeros 𝑎1 , … , 𝑎𝑛 (distinct or not) is associated with the matrix 𝐴 and write 𝐴 or 𝐴𝐵 , the latter in the event that the association is not clear from the context. If we look at (𝐼 − 𝑆𝐵 𝑆𝐵⋆ ) and (𝐼 − 𝑆𝐵⋆ 𝑆𝐵 ), the reader can check that we get rank-1 operators (you may find (9.7) and (9.8) to be useful tools for the second computation). This turns out to be extremely useful, particularly in combination with the following exercise. Exercise 9.6. If 𝑀 is a self-adjoint matrix, then rank 𝑀 = 1 if and only if rank 𝑀 2 = 1. Sometimes it is convenient to assume that rank(𝐼 − 𝑆𝐵 𝑆𝐵⋆ ) = 1, and sometimes it is easier to consider the rank of (𝐼 − 𝑆𝐵 𝑆𝐵⋆ )1/2 . Exercise 9.6 allows us to move freely between the two. Now as we mentioned in Chapter 7, Halmos showed that all contractions have unitary dilations; if the matrix representing the contraction, 𝑋, is 𝑛 × 𝑛, the dilation can be taken to be 2𝑛 × 2𝑛, and here is what it is [

𝑋 (1 − 𝑋 ⋆ 𝑋)1/2

(𝐼 − 𝑋𝑋 ⋆ )1/2 ]. −𝑋 ⋆

But finding the right 𝑛 rows to add to the matrix is often computationally challenging; if there is a smaller unitary dilation, it would be much more efficient to compute that instead. In the case at hand, the contraction is the 𝑛 × 𝑛 matrix 𝐴 representing 𝑆𝐵 . Looking at the Halmos dilation and recalling that the ranks of both 𝐼 − 𝐴⋆ 𝐴 and 𝐼 − 𝐴𝐴⋆ are 1, you might guess that there is an (𝑛 + 1) × (𝑛 + 1) unitary dilation. And if you guessed this, you would be right and you can see which one works: Since the first column of 𝐴 has only one nonzero element, start there and use the fact that the length of the first column in the unitary dilation must be 1. So, for example, the only thing that can go in the first position is 𝜆√1 − |𝑎1 |2 , where |𝜆| = 1. Then the second column must be orthogonal to it, and the first 𝑛 − 1 entries are given, so the choices for the last entry are limited. You can figure out what works. In any case, that is one way to find unitary 1-dilations of 𝐴. Here is another: just check that the matrices below work.

Chapter 9. Compressions of the Shift

119

Given 𝐴, we define 𝑈𝜆 = [𝑢𝑖𝑗 ] for |𝜆| = 1 by

𝑢𝑖𝑗 =

𝑎 ⎧ 𝑖𝑗 𝑗−1 ⎪ 𝜆( ∏ (−𝑎𝑘 ))√1 − |𝑎𝑗 |2 ⎪ 𝑘=1 ⎨ ⎪ ⎪ ⎩

𝑛 ( ∏𝑘=𝑖+1 (−𝑎𝑘 ))√1 𝑛 𝜆 ∏𝑘=1 (−𝑎𝑘 )

− |𝑎𝑖 |2

if 1 ≤ 𝑖, 𝑗 ≤ 𝑛, if 𝑖 = 𝑛 + 1 and 1 ≤ 𝑗 ≤ 𝑛, if 𝑗 = 𝑛 + 1 and 1 ≤ 𝑖 ≤ 𝑛, if 𝑖 = 𝑗 = 𝑛 + 1.

(9.15) Check that the matrix defined in (9.15) really is unitary; to do this you just need to check that the columns form an orthonormal set. Again, it turns out that up to unitary equivalence, these are all possible unitary 1-dilations of 𝐴 and thus they can be parametrized by 𝜆 ∈ 𝕋. Let 𝑉 ⋆ denote the 𝑛 × (𝑛 + 1) matrix [𝐼𝑛 , 0]. Then 𝑉 ⋆ 𝑈𝜆 𝑉 = 𝐴, telling us that 𝑈𝜆 is a dilation of 𝐴. (It also follows from this that 𝐴 is a contraction: If 𝑥 ∈ ℂ𝑛 , then ‖𝐴𝑥‖ = ‖𝑉 ⋆ 𝑈𝜆 𝑉𝑥‖ ≤ ‖𝑉 ⋆ ‖ ‖𝑈𝜆 (𝑉𝑥)‖ ≤ ‖𝑉𝑥‖ ≤ ‖𝑥‖.) And 𝑈𝜆 is (𝑛 + 1) × (𝑛 + 1), so putting all our information together we see that it is a unitary 1-dilation of 𝐴. Finally, 𝐴 has all eigenvalues inside the unit disk, 𝐴 is a contraction (we have already seen that this is true for any operator with a unitary dilation), and a computation shows that rank(𝐼 − 𝐴⋆ 𝐴) = rank(𝐼 − 𝐴𝐴⋆ ) = 1; that is exactly what matrices have to do to be in the class 𝒮𝑛 ([58, p. 180]). So that connects our matrices to the operators in 𝒮𝑛 . And the connection with Blaschke products? That is the story in the next chapter. For work in operator theory and function spaces at a deeper level than what we have presented here, the reader might consult [50], [121], [135], and [145].

Chapter

10

Higher Dimensions: Not Your Poncelet Ellipse So what is the connection to Blaschke products? We have seen that there is a geometric connection between the triangles formed with vertices at points identified by a degree-3 Blaschke product 𝐵 and Poncelet’s theorem. For 𝑛 > 2, does anything nice happen when we have a degree(𝑛 + 1) Blaschke product 𝐵 and we connect the points on 𝕋 that 𝐵 identifies? The answer is yes. What we see now is more complicated but still beautiful: The (convex) polygons 𝑃𝜆 that we obtain by joining consecutive points that 𝐵 maps to 𝜆 ∈ 𝕋 circumscribe a curve regardless of what value of 𝜆 we choose, just like we have seen before. But, in a new development, the curve we get will not usually be an ellipse. So the curve is not like the curve in Poncelet’s theorem in some ways but it is in other ways—it also has a closure property. It is convenient to make that precise. A smooth curve 𝒞 in 𝔻 has the Poncelet property (or 𝑛-Poncelet property) if for every point 𝑧 on the unit circle, there is an 𝑛-sided (convex) polygon inscribed in 𝕋, circumscribing 𝒞 and having 𝑧 as one of its vertices. We call such a curve a Poncelet curve. So these curves act like Poncelet ellipses, they just do not look like them. As you can see below, they are a sort of generalization of Poncelet ellipses. To see an actual Poncelet curve, we invite you to go to our Blaschke Product Explorer applet ,1 and insert your favorite zeros of a Blaschke product. These will appear in the applet as small white circles. (The critical points of the Blaschke product appear as dark gray circles, but those 1 http://pubapps.bucknell.edu/static/aeshaffer/v1/

121

122

Chapter 10. Not Your Poncelet Ellipse

Figure 10.1. Poncelet curves.

can be removed using the advanced options feature.) Using the “Auto identify points” feature, you can increase the number of line segments you would like to see. Then hit “Plot”, and the line segments joining the points your Blaschke product identifies will appear. You should see something that looks like one of the curves in Figure 10.1. Our goal for this chapter is to show the relation between a class of curves that have the Poncelet property and Blaschke products. Let us consider a normalized finite Blaschke product defined by 𝑛

𝐵(𝑧) = ∏ 𝑗=1

𝑧 − 𝑎𝑗 1 − 𝑎𝑗 𝑧

, where 𝑎𝑗 ∈ 𝔻.

We assume that 𝐵(0) = 0 though this does not make a difference to the curve: If 𝐵(0) ≠ 0, we may compose with an automorphism of 𝔻 and consider the finite Blaschke product 𝐶 ∶= (𝐵 − 𝐵(0))/(1 − 𝐵(0)𝐵). This new Blaschke product identifies the same points on the unit circle that 𝐵 does and thus the Poncelet curve associated with 𝐶 will be the same as that associated with 𝐵. Looking back at Lemma 3.4, we see that the argument of 𝐵 is an increasing function on the unit circle and since the Blaschke product we consider has degree 𝑛, for each point 𝜆 ∈ 𝕋, there are 𝑛 distinct points 𝑧1 , … , 𝑧𝑛 on the unit circle for which 𝐵(𝑧𝑗 ) = 𝜆. As luck would have it, given a matrix 𝐴 of the form (9.14), the unitary dilations of 𝐴 (that is, the 𝑈𝜆 in (9.15) where 𝜆 ranges over all points of

Chapter 10. Not Your Poncelet Ellipse

123

modulus 1) are closely connected to the Blaschke product formed using the eigenvalues of 𝐴. Since the connection between Blaschke products and the matrices is clearest when we use the matrix form given in (9.14) and the unitary 1-dilations 𝑈𝜆 for 𝜆 ∈ 𝕋, we always consider this form of the matrices. Theorem 10.1. Let 𝐴 be the 𝑛 × 𝑛 matrix in standard form (9.14). For each 𝜆 ∈ 𝕋, let 𝑈𝜆 denote the unitary dilation appearing in (9.15). Let 𝐵1 be the degree-𝑛 Blaschke product with the eigenvalues of 𝐴 as zeros. Then the eigenvalues of 𝑈𝜆 are the 𝑛 + 1 values in 𝕋 that are mapped to 𝜆 under the Blaschke product defined by 𝐵(𝑧) ∶= 𝑧𝐵1 (𝑧). A proof by induction for which we provided the first step in (7.7) shows that the eigenvalues of 𝑈𝜆 are what we claim they are. The rest of the proof is computational and can be found in [56, Lemma 2.4]. Since we know that the numerical range of a unitary matrix is the convex hull of its eigenvalues (see Corollary 7.1), we now know that 𝑊(𝑈𝜆 ) is the (closed) region bounded by the polygon with vertices at the points 𝑧1 , … , 𝑧𝑛+1 . What do we get when we “look inside” all the 𝑊(𝑈𝜆 )? In other words, what is ⋂𝜆 𝑊(𝑈𝜆 )? The basic properties of the numerical range can be found in Chapter 6, but we recall the most important relevant facts before turning to some new information. We have seen that the numerical range of an 𝑛 × 𝑛 matrix 𝐸 is a compact, convex set. It contains the set of eigenvalues of 𝐸, which is called the spectrum of 𝐸 and is denoted by 𝜎(𝐸). In addition, if 𝐸 is normal, then 𝑊(𝐸) is the convex hull of 𝜎(𝐸). We have also seen that unitarily equivalent matrices have the same numerical range; that is, the numerical range is a unitary invariant. But we know much more about the numerical range. In 1951, Rudolf Kippenhahn [95] investigated the geometric properties of the numerical range. In his paper, a square matrix 𝐴 is said to have a unitary decomposition if there exists a unitary matrix 𝑈 and square matrices 𝐴1 and 𝐴2 so that 𝐴 can be written as 𝐴 𝑈 ⋆ 𝐴𝑈 = [ 1 𝟎

𝟎 ]. 𝐴2

124

Chapter 10. Not Your Poncelet Ellipse

Some authors refer to such matrices as reducible. (We note, though, that there are other uses for the term reducible matrix, and the reader is advised to be careful when using this terminology.) It is a good exercise to check that the conditions on our matrices (specifically, rank(𝐼 − 𝐴⋆ 𝐴) = 1 and no eigenvalue of modulus 1) imply that our matrix does not have such a unitary decomposition. If 𝐴 does not have a unitary decomposition, Kippenhahn shows that the eigenvalues of 𝐴 must lie in the interior of 𝑊(𝐴). It is not hard to see this for matrices representing operators in the class 𝒮𝑛 : Changing the order of the zeros of the Blaschke product does not change the operator or its numerical range, so let us suppose we want to show that an eigenvalue 𝑎 of 𝐴 lies in the interior of 𝑊(𝐴). We can then write the matrix we wish to consider as 𝑎 √1 − |𝑎|2 √1 − |𝑏|2 ⋆ ⎡ ⎤ 𝑏 ⋆⎥ ⎢0 . ⎢⋮ ⋮ ⋮⎥ ⎢ ⎥ 0 ⋆⎦ ⎣0 By considering unit vectors of the form x = [𝑥1 𝑥2 0 ⋯ 0]𝑇 we see that the numerical range of 𝐴2 = [

𝑎 0

√1 − |𝑎|2 √1 − |𝑏|2 ] 𝑏

is contained in 𝑊(𝐴). From the elliptical range theorem, 𝑊(𝐴2 ) is an elliptical disk and the boundary ellipse has foci at 𝑎 and 𝑏 and the length of the minor axis is √1 − |𝑎|2 √1 − |𝑏|2 , so it is not a degenerate ellipse. Remark 10.2. We see that if 𝐴 is a matrix of the form (9.14), then all eigenvalues of 𝐴 lie in the interior of 𝑊(𝐴). In particular, no eigenvalue of 𝐴 can lie on the boundary of 𝑊(𝐴), just as Kippenhahn claimed.2 Kippenhahn’s key tool is the construction of support lines: Define a support line of 𝑊(𝐴) to be a line that touches the boundary of 𝑊(𝐴) in either one point (possibly with multiplicity greater than 1) or a whole interval. Kippenhahn showed that the numerical ranges we deal with are bounded by differentiable curves. In 1957, Donoghue [41, Theorem 1] provided a different proof that a boundary point of 𝑊(𝐴) at which the 2 Kippenhahn’s argument is completely different.

ter 13.

We consider his approach in Chap-

Chapter 10. Not Your Poncelet Ellipse

125

bounding curve is not differentiable is an eigenvalue of 𝐴.3 We include Donoghue’s proof below. Lemma 10.3. Let 𝑛 > 1, and let 𝐴 be an 𝑛 × 𝑛 matrix. Every boundary point of 𝑊(𝐴) at which the boundary is not differentiable is an eigenvalue of 𝐴. Before we begin the proof of Lemma 10.3, a few comments are in order. We use the same techniques and notation as in the proof of the Toeplitz–Hausdorff theorem on page 72. We also think of 𝐴 as the operator represented by the matrix rather than the matrix itself. Proof. Let 𝑤 be a point of 𝑊(𝐴) at which the boundary curve is not differentiable. Since 𝑊(𝐴) is closed and convex, there must be two different support lines of 𝑊(𝐴) through 𝑤. Since 𝑤 ∈ 𝑊(𝐴), there exists a unit vector 𝑥 ∈ ℂ𝑛 such that 𝑤 = ⟨𝐴𝑥, 𝑥⟩. If 𝑥 and 𝐴𝑥 are linearly dependent, then 𝑤 is an eigenvalue of 𝐴 and we are done. So suppose that 𝑆 = span{𝑥, 𝐴𝑥} is a two-dimensional subspace of ℂ𝑛 and we consider the operator 𝐴𝑆 = 𝑃𝑆 𝐴|𝑆 on 𝑆, where 𝑃𝑆 is the orthogonal projection onto 𝑆. We find that ⟨𝐴𝑆 𝑥, 𝑥⟩ = ⟨𝑃𝑆 𝐴|𝑆 𝑥, 𝑥⟩ = ⟨𝐴𝑥, 𝑥⟩ = 𝑤. Thus, 𝑤 ∈ 𝑊(𝐴𝑆 ). By Theorem 6.1 and (6.3), we conclude that 𝑊(𝐴𝑆 ) is an elliptical disk contained in 𝑊(𝐴). If the elliptical disk is nondegenerate, then it is not possible that two different support lines of 𝑊(𝐴) pass through 𝑤. Thus, the elliptical disk is degenerate and is, therefore, a point or a line segment. In the latter case, 𝑤 must be one of the endpoints because of the two different support lines through this point. In both cases, Theorem 6.1 implies that 𝑤 is an eigenvalue of 𝐴𝑆 . As the reader can check, using the fact that ⟨𝐴𝑆 𝑥, 𝑥⟩ = 𝑤 and that 𝑤 is an eigenvalue of 𝐴𝑆 , we must have 𝐴𝑆 𝑥 = 𝑤𝑥. But 𝑤𝑥 = 𝐴𝑆 𝑥 = 𝑃𝑆 𝐴|𝑆 𝑥 = 𝐴𝑥. Hence, 𝑤 is an eigenvalue of 𝐴 as claimed. Recall that Paul Halmos conjectured that the unitary dilations of a matrix encode a lot of information about the matrix. We stated a more specific version of this conjecture in the special case presented in Theorem 7.4. Theorem 10.4 is the natural extension of Theorem 7.4 to 𝑛 × 𝑛 3 Our

𝑊(𝐴) is closed; Donoghue actually proved this in more generality, assuming 𝑤 ∈ 𝑊(𝐴).

126

Chapter 10. Not Your Poncelet Ellipse

matrices. By invoking Theorem 10.1 and Lemma 10.5, the proof is just a slight modification and therefore would make an excellent exercise for the reader.4 The results below concern the numerical range of matrices of the form (9.14) and apply, therefore, to any unitarily equivalent matrices. Theorem 10.4. Consider a matrix 𝐴 in the form of (9.14) and the family of unitary dilations {𝑈𝜆 }𝜆∈𝕋 in the form of (9.15). Then 𝑊(𝐴) = ⋂𝜆∈𝕋 𝑊(𝑈𝜆 ). In Lemma 10.5 below [55, Lemma 2.2], our goal is to show that given a unitary dilation 𝑈𝜆 of 𝐴, the polygon with vertices at the eigenvalues of 𝑈𝜆 circumscribes 𝑊(𝐴). The proof has three steps: The first step is to show that the eigenvalues of each unitary dilation 𝑈𝜆 , for 𝜆 ∈ 𝕋, are distinct, but this now follows from Lemma 3.4 and Theorem 10.1. The next two steps appear in the proof below. Lemma 10.5. Let 𝐴 be a matrix with representation (9.14) and, for 𝜆 ∈ 𝕋, let 𝑈𝜆 be a unitary 1-dilation of 𝐴. Then the eigenvalues of 𝑈𝜆 are all distinct and are the vertices of a convex (𝑛 + 1)-gon that is inscribed in 𝕋 and circumscribed about 𝑊(𝐴) with each side tangent to the boundary of the numerical range at exactly one point. Proof. We have already shown that the eigenvalues, 𝑤1 , 𝑤2 , … , 𝑤𝑛+1 , of 𝑈𝜆 are distinct. As always, we order these by increasing argument. It is obvious that the polygon with these eigenvalues as vertices is inscribed in 𝕋. Therefore, the heart of the proof is showing that the polygon circumscribes 𝑊(𝐴). We are now ready for the second step of the proof—showing that the line joining 𝑤𝑛 and 𝑤𝑛+1 intersects 𝑊(𝐴). The proof is easily adapted to arbitrary pairs 𝑤𝑗 , 𝑤𝑗+1 (indices are taken modulo 𝑛 + 1). To complete this step, we write 𝐴 as the sum of rank-1 matrices 𝑤𝑗 𝑉1 𝐸𝑗 𝑉1⋆ , and then we try to make the sums involving 𝑤𝑗 for 𝑗 < 𝑛 vanish. Let 𝑉 = [𝐼𝑛 , 0], which we note is an 𝑛 × (𝑛 + 1) matrix. Using the fact that 𝑈𝜆 is unitary with eigenvalues (𝑤1 , 𝑤2 , … , 𝑤𝑛 , 𝑤𝑛+1 ), Theorem 6.4 implies that 𝑈𝜆 is unitarily equivalent to the (unitary) diagonal matrix 4 The proof is a “Blaschke product” version of the proof of [55, Corollary 2.8], where this theorem first appeared.

Chapter 10. Not Your Poncelet Ellipse

127

𝐷 = (𝑢𝑖𝑗 ) with 𝑢𝑗𝑗 = 𝑤𝑗 for 𝑗 = 1, … , (𝑛 + 1). Now 𝐴 = 𝑉𝑋𝐷𝑋 ⋆ 𝑉 ⋆ for some unitary matrix 𝑋, and letting 𝑉1 = 𝑉𝑋, we have 𝐴 = 𝑉1 𝐷𝑉1⋆ . So, letting 𝐸𝑗 = (𝑒ℓ𝑚 ) denote the (𝑛 + 1) × (𝑛 + 1) matrix with 𝑒𝑗𝑗 = 1 and all other entries zero, we have 𝑛+1

𝑛+1

𝐴 = 𝑉1 ( ∑ 𝑤𝑗 𝐸𝑗 ) 𝑉1⋆ = ∑ 𝑤𝑗 𝑉1 𝐸𝑗 𝑉1⋆ . 𝑗=1

We also have

𝑗=1 𝑛+1

∑ 𝐸𝑗 = 𝐼𝑛+1 , 𝑗=1

so

𝑛+1

∑ 𝑉1 𝐸𝑗 𝑉1⋆ = 𝐼𝑛 . 𝑗=1

We wish to obtain information about the line segment joining 𝑤𝑛 and 𝑤𝑛+1 , and we get what we want by considering 𝐾𝑗 = ker(𝑉1 𝐸𝑗 𝑉1⋆ ) and 𝑛−1

producing a nonzero element 𝑥 ∈ ⋂𝑗=1 𝐾𝑗 . Our proof will proceed by induction on 𝑚 for 1 ≤ 𝑚 ≤ 𝑛 − 1, showing that 𝑚

dim (

⋂

𝐾𝑗 ) ≥ 𝑛 − 𝑚.

𝑗=1

We begin by noticing that for every 𝑗, we have rank(𝑉1 𝐸𝑗 𝑉1⋆ ) ≤ 1 and therefore dim ker(𝑉1 𝐸𝑗 𝑉1⋆ ) ≥ 𝑛 − 1. Thus, dim 𝐾𝑗 ≥ 𝑛 − 1 for 𝑗 = 1, … , 𝑛 + 1, yielding the first step of the 𝑚

induction. Suppose that dim (⋂𝑗=1 𝐾𝑗 ) ≥ 𝑛 − 𝑚 for 1 ≤ 𝑚 < 𝑛 − 2. By our induction hypothesis and the fact that, in general, dim(𝑆 ∩ 𝑇) = dim 𝑆 + dim 𝑇 − dim(𝑆 + 𝑇), we have

128

Chapter 10. Not Your Poncelet Ellipse 𝑚+1

dim (

⋂

𝑚

𝐾𝑗 ) = dim (

𝑗=1

⋂

𝐾𝑗 ) + dim 𝐾𝑚+1

𝑗=1 𝑚

− dim ((

⋂

𝐾𝑗 ) + 𝐾𝑚+1 )

𝑗=1

≥ (𝑛 − 𝑚) + (𝑛 − 1) − 𝑛 = 𝑛 − (𝑚 + 1). 𝑚 dim (⋂𝑗=1

Thus, we see that

𝐾𝑗 ) ≥ 𝑛 − 𝑚 for 𝑚 = 1, … , 𝑛 − 1 and letting 𝑚 = 𝑛 − 1, 𝑛−1

dim (

⋂

𝐾𝑗 ) ≥ 1.

𝑗=1 𝑛−1

Let 𝑥0 be a unit vector in the intersection ⋂𝑗=1 𝐾𝑗 . Now we will use 𝑥0 to get a point on 𝑊(𝐴) that also lies on the line segment joining 𝑤𝑛 and 𝑤𝑛+1 as follows: 𝑛+1

⟨𝐴𝑥0 , 𝑥0 ⟩ = ⟨ ∑ 𝑤𝑗 𝑉1 𝐸𝑗 𝑉1⋆ 𝑥0 , 𝑥0 ⟩ 𝑗=1

= 𝑤𝑛 ⟨𝑉1 𝐸𝑛 𝑉1⋆ 𝑥0 , 𝑥0 ⟩ + 𝑤𝑛+1 ⟨𝑉1 𝐸𝑛+1 𝑉1⋆ 𝑥0 , 𝑥0 ⟩. But ⟨𝑉1 𝐸𝑗 𝑉1⋆ 𝑥0 , 𝑥0 ⟩ = ⟨𝐸𝑗 𝑉1⋆ 𝑥0 , 𝑉1⋆ 𝑥0 ⟩ ≥ 0 for every 𝑗 and ⟨𝑉1 𝐸𝑛 𝑉1⋆ 𝑥0 , 𝑥0 ⟩ + ⟨𝑉1 𝐸𝑛+1 𝑉1⋆ 𝑥0 , 𝑥0 ⟩ 𝑛+1

= ⟨ ∑ 𝑉1 𝐸𝑗 𝑉1⋆ 𝑥0 , 𝑥0 ⟩ = ⟨𝐼𝑛 𝑥0 , 𝑥0 ⟩ = 1. 𝑗=1

𝑤𝑛 ⟨𝑉1 𝐸𝑛 𝑉1⋆ 𝑥0 , 𝑥0 ⟩+𝑤𝑛+1 ⟨𝑉1 𝐸𝑛+1 𝑉1⋆ 𝑥0 , 𝑥0 ⟩ lies in the

Thus, ⟨𝐴𝑥0 , 𝑥0 ⟩ = numerical range and on the line segment joining 𝑤𝑛 and 𝑤𝑛+1 . The final step requires showing that this gives us a point of tangency— not just a point of intersection. So suppose that there exist 𝑥 and 𝑦, linearly independent unit vectors, with ⟨𝐴𝑥, 𝑥⟩ and ⟨𝐴𝑦, 𝑦⟩ on the line segment joining 𝑤𝑛 and 𝑤𝑛+1 . Note that 𝑛+1

⟨𝐴𝑥, 𝑥⟩ = ∑ 𝑤𝑗 ⟨𝑉1 𝐸𝑗 𝑉1⋆ 𝑥, 𝑥⟩. 𝑗=1

If ⟨𝑉1 𝐸𝑗 𝑉1⋆ 𝑥, 𝑥⟩ ≠ 0 for some 𝑗 = 1, … , 𝑛 − 1, then ⟨𝐴𝑥, 𝑥⟩ cannot lie on the line segment. Now for 𝑗 = 1, … , 𝑛 − 1, letting 𝑣𝑗 denote the 𝑗th

Chapter 10. Not Your Poncelet Ellipse

129

component of 𝑉1⋆ 𝑥, we have 0 = ⟨𝑉1 𝐸𝑗 𝑉1⋆ 𝑥, 𝑥⟩ = ⟨𝐸𝑗 𝑉1⋆ 𝑥, 𝑉1⋆ 𝑥⟩ = |𝑣𝑗 |2 . 𝑛−1

Thus, 𝑉1 𝐸𝑗 𝑉1⋆ 𝑥 = 0 and we have 𝑥, and similarly 𝑦, in ⋂𝑗=1 𝐾𝑗 . This 𝑛−1

implies that dim (⋂𝑗=1 𝐾𝑗 ) ≥ 2 and thus 𝑛

dim (

⋂

𝑛−1

𝐾𝑗 ) = dim (

𝑗=1

⋂

𝐾𝑗 ) + dim 𝐾𝑛

𝑗=1 𝑛−1

− dim ((

⋂

𝐾𝑗 ) + 𝐾𝑛 )

𝑗=1

≥ 2 + (𝑛 − 1) − 𝑛 = 1. 𝑛

Therefore, we can choose a unit vector 𝑢 in ⋂𝑗=1 𝐾𝑗 . But this would imply that 𝑛+1

⟨𝐴𝑢, 𝑢⟩ = ∑ 𝑤𝑗 ⟨𝑉 ⋆ 𝐸𝑗 𝑉𝑢, 𝑢⟩ = 𝑤𝑛+1 ⟨𝑉1 𝐸𝑛+1 𝑉1⋆ 𝑢, 𝑢⟩ 𝑗=1 𝑛+1

= 𝑤𝑛+1 ⟨ ∑ 𝑉1 𝐸𝑗 𝑉1⋆ 𝑢, 𝑢⟩ = 𝑤𝑛+1 , 𝑗=1

or 𝑤𝑛+1 ∈ 𝑊(𝐴) ∩ 𝕋 = ∅, a contradiction. We have reached the main theorem in this chapter, a Poncelet-like theorem. Theorem 10.6. Let 𝐵 be a Blaschke product of degree 𝑛 + 1 satisfying 𝐵(0) = 0. Then there is a convex curve 𝒞 such that for each 𝜆 ∈ 𝕋 if 𝑧𝑗 (arranged in order of increasing argument) are the points at which 𝐵(𝑧𝑗 ) = 𝜆, then the line joining 𝑧𝑗 to 𝑧𝑗+1 (where the index is taken modulo 𝑛 + 1) is tangent to 𝒞 at precisely one point. Furthermore, every point on the curve 𝒞 is obtained in this manner. A few comments are in order before we begin the proof of Theorem 10.6. We call 𝒞 a Blaschke curve or the Blaschke curve associated with the Blaschke product 𝐵, though we note that different Blaschke products may give rise to the same Blaschke curve. For example, if 𝜑 is an automorphism of the disk and 𝐵 is a Blaschke product, then 𝜑 ∘ 𝐵 and

130

Chapter 10. Not Your Poncelet Ellipse

Figure 10.2. A Blaschke–Poncelet curve.

𝐵 identify the same sets of points and therefore have the same Blaschke curve. In fact, this is the only way to get the same Blaschke curve, as the reader can check (though Theorem 11.12 might be helpful). From among all Blaschke products that yield a given Blaschke curve 𝒞, there is always one that maps zero to zero. The zeros of such a representative Blaschke product have special geometric features, so we will almost always use the Blaschke product associated with 𝒞 that maps zero to zero. By the previous theorem 𝒞 has the property that for each 𝜆 ∈ 𝕋 the curve is inscribed in a (convex) (𝑛 + 1)-gon with one vertex at 𝜆 and all other vertices on 𝕋. Thus, no matter where we begin on 𝕋 there is a (convex) polygon with vertices on 𝕋 circumscribing 𝒞. Therefore, 𝒞 acts like a Poncelet ellipse and it is a Poncelet curve. It is reasonable to ask if all Poncelet curves are Blaschke curves, but they are not, and this is not easy to show. One reference for this is Mirman’s work in [114]. Let us look at Figure 10.2 with our new information in mind. We can now view this picture two ways. From the numerical range point of view, the curve is the boundary of the numerical range of a compression of the shift. The small circles, other than the origin, represent the zeros of the

Chapter 10. Not Your Poncelet Ellipse

131

Blaschke product; that is, the zeros of 𝐵1 (𝑧) ∶= 𝐵(𝑧)/𝑧. The convex hull of a circumscribing polygon represents the numerical range of a corresponding unitary 1-dilation of the operator 𝑆𝐵1 . From the Blaschke product perspective, the circumscribing polygons are the polygons formed by connecting successive points identified by the Blaschke product 𝐵. It is a beautiful fact that this produces a curve with the Poncelet property. We know that the bounding curve is the boundary of the compact, convex set 𝑊(𝐴). Furthermore, we know that if 𝐵 is the Blaschke product above, the zeros of the Blaschke product 𝐵1 (𝑧) = 𝐵(𝑧)/𝑧 are the eigenvalues of 𝐴 and, by the discussion above, are contained in the interior of 𝑊(𝐴). By Lemma 10.3, the curve is differentiable at every point of the boundary, and it makes sense to talk about the tangent to the curve at each point. Proof of Theorem 10.6. By Theorem 10.1, we know that the eigenvalues of 𝑈𝜆 are the points 𝑧𝑗 , for 𝑗 = 1, … , 𝑛 + 1, where 𝐵(𝑧𝑗 ) = 𝜆 and the points are arranged in order of increasing argument. By Lemma 10.5, we know that the line segment 𝑧𝑗 𝑧𝑗+1 is tangent to the boundary of 𝑊(𝐴) at precisely one point, justifying the first two statements in the theorem. To obtain the final statement, choose a point 𝛼0 on the curve 𝒞 and draw the tangent line ℓ to the curve at 𝛼0 . This will intersect the unit circle in two points, 𝑤1 and 𝑤2 . There are 𝑛 + 1 points, 𝑧1 , … , 𝑧𝑛+1 that satisfy 𝐵(𝑧𝑗 ) = 𝐵(𝑤1 ), where 𝑤1 = 𝑧𝑘 for some 𝑘. By Lemma 10.5, the polygon with vertices 𝑧1 , … , 𝑧𝑛+1 circumscribes 𝑊(𝐴). Thus, the line segments 𝑧𝑘−1 𝑤1 and 𝑤1 𝑧𝑘+1 are tangent to the boundary of 𝑊(𝐴). Since 𝑊(𝐴) is a compact, convex set, there are exactly two tangents from 𝑤1 to the boundary of 𝑊(𝐴). Therefore, 𝑤2 = 𝑧𝑘−1 or 𝑤2 = 𝑧𝑘+1 . In either case, 𝐵(𝑤1 ) = 𝐵(𝑤2 ). As an exercise tying these results together, we suggest that the reader show that the numerical range of an 𝑛 × 𝑛 Jordan block with zeros on the diagonal is the disk {𝑧 ∶ |𝑧| ≤ cos(𝜋/(𝑛 + 1))}. Precise descriptions of the curves in Theorem 10.6 are often quite difficult to obtain, and they are rarely familiar curves. But sometimes they are ellipses and special things happen. We will check some of these out in Chapters 13 and 14.

Chapter

11

Interpolation with Blaschke Products It is now time to turn from our focus on geometry to the construction of Blaschke products with desirable properties. The definition of the Blaschke product (see Definition 3.1) gives the zeros explicitly. What is not listed explicitly are the points on the unit circle that the Blaschke product identifies. In this chapter we consider questions about how much freedom we have to prescribe the values of a finite Blaschke product on the unit circle. For the degree-3 case we have already discussed the question of how to prescribe points on the unit circle, though we did not explicitly think of it this way: Pick three distinct points on 𝕋 and let those be the vertices of a triangle, Δ. Now consider any ellipse, 𝐸, circumscribed by Δ. By Corollary 4.4, 𝐸 is a Blaschke ellipse, and the tangents to 𝐸 intersect 𝕋 at points the Blaschke product identifies. When we begin with a triangle, there is a two-parameter family of such ellipses [2]: We can arbitrarily pick a point on each of two of the sides of Δ and get a unique ellipse that is inscribed in Δ and tangent at the selected points. Thus, the triangle alone did not determine a unique Blaschke product. In addition, only certain pairs of points can be the foci of an ellipse circumscribed by Δ. But, given two triangles with interspersed vertices on 𝕋, the Blaschke product is—up to composition with a disk automorphism—determined uniquely; see Theorem 11.12. What can we say about higher-order Blaschke products? How much freedom do we have in specifying values on 𝕋 that will be identified by the Blaschke product? Once we know how many points we want to specify, do we have any control over the degree of the Blaschke product? Can 133

134

Chapter 11. Interpolation with Blaschke Products

we find a sufficiently elementary method that allows us to construct this Blaschke product? These questions lead us to the subject of interpolation, which is usually thought of as the process of finding, if possible, a function of a certain class that maps one set of prescribed values to another. The simplest interpolation problem is seen early in a student’s mathematics education— finding a line through two points. This is later extended to finding a polynomial of lowest possible degree through a set of prescribed points. Some of the work we have done in earlier chapters can even be thought of as interpolation. For example, we proved the existence of an “interpolating conic” given six points in general position. Our work will focus on finite Blaschke products and on trying to prescribe their values on the unit circle. These questions become more difficult to answer when the degree is higher than three because the Blaschke curves are, in general, no longer ellipses. As you may already have noticed, we have solved many different problems by translating a problem from one setting to another. In this chapter we also change our setting to one in which our questions will be easier to answer. We let 𝐇+ = {𝑧 ∈ ℂ ∶ Im(𝑧) > 0} be the upper half-plane, 𝐇− = {𝑧 ∈ ℂ ∶ Im(𝑧) < 0} be the lower half-plane, ℝ∗ = ℝ ∪ {∞} be the extended real line, and ℂ∗ = ℂ ∪ {∞} be the extended complex plane. Recall that a linear fractional transformation is a map of the form 𝑎𝑧 + 𝑏 , where 𝑎, 𝑏, 𝑐, 𝑑 ∈ ℂ and 𝑎𝑑 − 𝑏𝑐 ≠ 0. 𝑓(𝑧) = 𝑐𝑧 + 𝑑 By defining 𝑓(−𝑑/𝑐) = ∞ and 𝑓(∞) = 𝑎/𝑐, we may think of 𝑓 as a map from the extended complex plane to itself. These maps are also referred to as Möbius transformations in the literature. The linear fractional transformations mapping 𝔻 to itself and 𝕋 to itself are the disk automorphisms, which we have seen are precisely the Blaschke products of degree 1. The following lemma uses linear fractional transformations to translate to our new setting. Lemma 11.1. The function 𝑌 ∶ ℂ∗ → ℂ∗ defined by 1+𝑧 𝑌(𝑧) = 𝑖 ( ) 1−𝑧 is a linear fractional transformation that

Chapter 11. Interpolation with Blaschke Products

135

• maps the unit disk, 𝔻, to the upper half-plane, 𝐇+ ; • maps the set ℂ⋆ ⧵ 𝔻 to the lower half-plane, 𝐇− ; • maps the unit circle, 𝕋, to the extended real line, ℝ∗ . Further, 𝑌(𝑒𝑖𝜃 ) increases as 𝜃 increases on the interval 0 < 𝜃 < 2𝜋. Proof. The function 𝑌 is a linear fractional transformation, 𝑌(1) = ∞, 𝑌(𝑖) = −1, 𝑌(−1) = 0, and 𝑌(−𝑖) = 1. Its inverse is given by 𝑌 −1 (𝑧) = (𝑧 − 𝑖)/(𝑧 + 𝑖). Since linear fractional transformations on the extended complex plane map circles to circles (recall that lines are circles passing through infinity), we see that 𝑌(𝕋) = ℝ∗ . Restricted to the unit circle, 𝑌(𝑒𝑖𝜃 ) = − sin 𝜃/(1 − cos 𝜃), so the derivative of 𝑌 with respect to 𝜃 is 𝑑𝑌 1 = . 𝑑𝜃 1 − cos 𝜃 For 0 < 𝜃 < 2𝜋, we have 𝑌 ′ (𝜃) > 0. Lastly, since 𝑌(0) = 𝑖, the disk is mapped to the upper half-plane and the complement of 𝔻 is mapped to the lower half-plane. We consider rational functions 𝐹 = 𝑝/𝑞, where 𝑝 and 𝑞 are polynomials. The degree of the rational function 𝐹 is defined by max{deg 𝑝, deg 𝑞}. Note that this agrees with the definition of the degree of a Blaschke product. We are interested primarily in the mapping properties of rational functions. We say that a rational function, 𝐹 ∶ ℂ∗ → ℂ∗ , is strongly real of positive type if (1) 𝐹(𝐇+ ) = 𝐇+ ; (2) 𝐹(𝐇− ) = 𝐇− ; (3) 𝐹(ℝ∗ ) = ℝ∗ . Strongly real refers to the third property and positive type refers to the first two properties. For example, 𝐹(𝑧) = (𝑧 − 2)/(𝑧 − 1) is strongly real of positive type, and 𝐹(𝑧) = (𝑧 − 1)/(𝑧 − 2) is strongly real but not of positive type. Our interest is in Blaschke products and the following lemma explains how strongly real rational functions of positive type are connected to finite Blaschke products.

136

Chapter 11. Interpolation with Blaschke Products

Lemma 11.2. Given a rational function 𝐹 ∶ ℂ∗ → ℂ∗ that is strongly real of positive type and the linear fractional transformation 𝑌 defined in Lemma 11.1, the function 𝐵 = 𝑌 −1 ∘ 𝐹 ∘ 𝑌 is a finite Blaschke product. Proof. In Lemma 3.2 we gave a three-fold characterization of finite Blaschke products. The requirements that 𝐵 maps 𝔻 to itself and 𝕋 to itself are easily checked in this case. Also, we need to know that 𝐵 is analytic in an open set containing 𝔻. Equivalently, we can ask where 𝐵 = 𝑌 −1 ∘ 𝐹 ∘ 𝑌 is not analytic. Since 𝐵 is a composition of rational functions, it is analytic as long as it has no poles in 𝔻. In order for 𝑌 −1 (𝑧) to be infinite, we need 𝑧 = −𝑖. The rational function 𝐹 is equal to −𝑖 at finitely many points in 𝐇− . Since 𝑌 is one-to-one and 𝑌(𝑧) ∈ 𝐇− if and only if 𝑧 ∈ {𝑧 ∈ ℂ ∶ |𝑧| > 1}, we find that the finite set of points where 𝐵 is not analytic are all in the set {𝑧 ∈ ℂ ∶ |𝑧| > 1}. Hence, 𝐵 is analytic in an open set containing 𝔻. Lemma 11.2 allows us to focus on finding rational functions, 𝐹 ∶ ℂ∗ → ℂ∗ , that are strongly real of positive type and that also possess the right interpolation properties. To begin, let us consider the simplest kind of situation. Lemma 11.3. Let 𝑛 be a positive integer, and let 𝑎1 , 𝑎2 , … , 𝑎𝑛 and 𝑥1 , 𝑥2 , … , 𝑥𝑛 be real numbers satisfying 𝑎1 < 𝑥1 < 𝑎2 < 𝑥2 < … < 𝑎𝑛 < 𝑥𝑛 . ∗

∗

Let 𝐹 ∶ ℂ → ℂ be the rational function defined by 𝑛

𝐹(𝑧) = ∏ 𝑘=1

𝑧 − 𝑥𝑘 . 𝑧 − 𝑎𝑘

Then (1) 𝐹 is strongly real of positive type; (2) 𝐹 has zeros at 𝑥𝑘 for 𝑘 = 1, 2, … , 𝑛; (3) 𝐹 has poles at 𝑎𝑘 for 𝑘 = 1, 2, … , 𝑛.

Chapter 11. Interpolation with Blaschke Products

137


+

π










 








ℝ

𝑛

Figure 11.1. Upper bound for ∑𝑘=1 (arg(𝑧 − 𝑥𝑘 ) − arg(𝑧 − 𝑎𝑘 )).

Proof. The function 𝐹 satisfies the last two requirements by construction. To see that 𝐹 satisfies the first, let 𝑧 ∈ 𝐇+ . Then using the principal argument, we have 𝑛

𝑛

𝑧 − 𝑥𝑘 ) = ∑ (arg(𝑧 − 𝑥𝑘 ) − arg(𝑧 − 𝑎𝑘 )) . 𝑧 − 𝑎𝑘 𝑘=1 𝑘=1 (11.1) As illustrated in Figure 11.1, this expression is bounded by arg(𝐹(𝑧)) = ∑ arg (

arg(𝑧 − 𝑥𝑛 ) − arg(𝑧 − 𝑎1 ) < 𝜋. Since the middle expression in (11.1) is always positive, we see that 0 < arg(𝐹(𝑧)) < 𝜋 and we conclude that 𝐹(𝑧) ∈ 𝐇+ and thus 𝐹(𝐇+ ) ⊆ 𝐇+ . The same reasoning (using Figure 11.1 reflected over the real axis) shows that 𝐹(𝐇− ) ⊆ 𝐇− . That 𝐹(ℝ∗ ) ⊆ ℝ∗ holds is a consequence of the fact that 𝑥𝑗 and 𝑎𝑗 are real. Since the function 𝐹 is surjective, we have equality between the appropriate sets. Therefore, 𝐹 is strongly real of positive type. Using the functions we found in Lemmas 11.1 and 11.3, we can see how to construct a finite Blaschke product that identifies the points in each of two sets of points interspersed on 𝕋.

138

Chapter 11. Interpolation with Blaschke Products

Corollary 11.4. If 𝑧1 , 𝑧2 , … , 𝑧𝑛 and 𝑤1 , 𝑤2 , … , 𝑤𝑛 are two sets of points on 𝕋 for which the principal argument satisfies 0 ≤ arg(𝑧1 ) < arg(𝑤1 ) < arg(𝑧2 ) < arg(𝑤2 ) < ⋯ < arg(𝑧𝑛 ) < arg(𝑤𝑛 ) < 2𝜋, then there exists a Blaschke product of degree 𝑛 such that 𝐵(0) = 0 and 𝐵 identifies the points 𝑧𝑗 for 𝑗 = 1, 2, … , 𝑛 and 𝐵 identifies the points 𝑤𝑗 for 𝑗 = 1, 2, … , 𝑛. We leave it to the reader to write out a formal proof and instead provide an example. You can create your own examples, entering the arguments of the interspersed points of your choice in our applet ,1 Blaschke Product Interpolation Tool. Example 11.5. Suppose we want a degree-3 Blaschke product that is constant on {𝑖, −1, −𝑖}, constant on {(1+𝑖)/√2, (−1+𝑖)/√2, (−1−𝑖)/√2}, and maps zero to zero. We begin by noting that the function 𝑌 of Lemma 11.1 maps these six points to the following real numbers (listed in order on the real line): −1 − √2 < −1 < 1 − √2 < 0 < −1 + √2 < 1. Thus, the function 𝐹 from Lemma 11.3 is 𝐹(𝑧) =

(𝑧 + 1)𝑧(𝑧 − 1) (𝑧 + 1 + √2)(𝑧 − 1 + √2)(𝑧 + 1 − √2)

.

The Blaschke product 𝐵 = 𝑌 −1 ∘ 𝐹 ∘ 𝑌 is given by the formula 𝐵(𝑧) =

𝛼3 𝑧3 + 𝛼2 𝑧2 + 𝛼1 𝑧 + 𝛼0 , 𝛼0 𝑧3 + 𝛼1 𝑧2 + 𝛼2 𝑧 + 𝛼3

where 𝛼3 = −1 − √2 + (2 − √2)𝑖, 𝛼2 = √2 − 3 + √2𝑖, 𝛼1 = 1 − √2 + √2𝑖, and 𝛼0 = √2 − 1 + (2 − √2)𝑖. Since 𝐵 does not map zero to zero, we need to calculate the Blaschke product ˜ = 𝐵 − 𝐵(0) 𝐵 1 − 𝐵(0)𝐵 1 https://pubapps.bucknell.edu/static/aeshaffer/v1/

Chapter 11. Interpolation with Blaschke Products

139

that maps zero to zero, identifies the same sets of points, and is of the same degree as 𝐵. We obtain 𝛽 𝑧2 + 𝛽1 𝑧 + 𝛽0 ̃ , 𝐵(𝑧) = −𝑧 2 𝛽0 𝑧2 + 𝛽1 𝑧 + 𝛽2 where 𝛽2 = 2 − 2√2 + (−2 − √2)𝑖, 𝛽1 = −3 + √2 + (−5 + 2√2)𝑖, and 𝛽0 = 3 − 3√2 + (−5 + 3√2)𝑖. This is a Blaschke product (you should be able to see this without referring to the theorem; try it!), but its zeros are not apparent in this form. Factoring 𝐵̃ numerically gives us the following approximation: (𝑧 − (−0.552 − 0.332𝑖))(𝑧 − (−0.156 + 0.625𝑖)) ̃ 𝐵(𝑧) ≈ 𝜇𝑧 , (1 − (−0.556 + 0.332𝑖)𝑧)(1 − (−0.156 − 0.625𝑖)𝑧) where 𝜇 = (−2 + 2√2 + (2 + √2)𝑖)/(2 − 2√2 + (2 + √2)𝑖) ≈ 0.889 − 0.458𝑖. (Note that |𝜇| = 1.) We mention that because we require 0 to be mapped to 0; it is not (in general) possible to specify the values of the Blaschke product on both of the two sets. To be able to assign the values on the boundary with more freedom than we have done so far, consider a special linear combination of the functions described in Lemma 11.3. Theorem 11.6. For an integer 𝑛 > 1, let 𝑎1 , 𝑎2 , … , 𝑎𝑛 and 𝑥1 , 𝑥2 , … , 𝑥𝑛 be real numbers satisfying 𝑎1 < 𝑥1 < 𝑎2 < 𝑥2 < … < 𝑎𝑛 < 𝑥𝑛 , and let 𝑦1 , 𝑦2 , … , 𝑦𝑛 be real numbers. Then there exists a rational function, 𝐺 ∶ ℂ∗ → ℂ∗ , of degree 𝑛 such that (1) 𝐺 is strongly real of positive type; (2) 𝐺(𝑥𝑗 ) = 𝑦𝑗 for 𝑗 = 1, 2, … , 𝑛; (3) 𝐺 has poles at 𝑎𝑗 for 𝑗 = 1, 2, … , 𝑛. Proof. First, we note that if we pick an 𝑀 ∈ ℝ so that 𝑀 +𝑦𝑗 > 0 for 𝑗 = 1, 2, … 𝑛 and we replace the second condition above by 𝐺(𝑥𝑗 ) = 𝑀 +𝑦𝑗 for all 𝑗 = 1, 2, … , 𝑛, then the values of 𝐺 at all of the 𝑥𝑗 would be positive. If we prove the theorem in this case, then we can recover the function

140

Chapter 11. Interpolation with Blaschke Products

sought after in the theorem as stated by subtracting 𝑀. Thus, we need only to consider the case where all of the 𝑦𝑗 > 0 for 𝑗 = 1, 2, … , 𝑛. For 𝑘 = 1, … , 𝑛, consider the function 𝐺𝑘 defined by 𝑧 − 𝑥𝑗 𝐺𝑘 (𝑧) = ∏ . 𝑧 − 𝑎𝑗 𝑗≠𝑘 This is a function of the type described in Lemma 11.3 and thus is strongly real of positive type. By the ordering of the 𝑎𝑗 and 𝑥𝑗 , each of (𝑥𝑘 − 𝑥𝑗 )/(𝑥𝑘 − 𝑎𝑗 ) is positive for 𝑗 ≠ 𝑘. Therefore, 𝑥𝑘 − 𝑥𝑗 𝐺𝑘 (𝑥𝑘 ) = ∏ > 0. 𝑥 − 𝑎𝑗 𝑗≠𝑘 𝑘 We claim the function 𝐺 we are seeking is defined by 𝑛

𝑦𝑘 𝐺 (𝑧). 𝐺 (𝑥 ) 𝑘 𝑘=1 𝑘 𝑘

𝐺(𝑧) = ∑

(11.2)

It is easy to see that 𝐺 satisfies the first two conditions above. To show that it satisfies the third condition, we note that the only singularities the rational function 𝐺 can possibly have are poles at 𝑎𝑗 . To show that 𝐺 does have a pole at each 𝑎𝑗 , suppose not. Then the rational function 𝐺 is continuous at 𝑎𝑗 and 𝑛

0 = lim (𝑧 − 𝑎𝑗 )𝐺(𝑎𝑗 ) = ∑ ( 𝑧→𝑎𝑗

𝑘=1

𝑎𝑗 − 𝑥𝑚 𝑦𝑘 ∏ ( ) (𝑎𝑗 − 𝑥𝑗 )) . 𝐺𝑘 (𝑥𝑘 ) 𝑚≠𝑘,𝑗 𝑎𝑗 − 𝑎𝑚

Since each of the terms in the sum above is negative, this is a contradiction. The interpolating function (11.2) is said to be written in Lagrange form. In fact, there is an even stronger interpolation result hidden inside this result that we can find by moving the point at infinity to some finite value. Corollary 11.7. For an integer 𝑛 > 1, let 𝑎1 , 𝑎2 , … , 𝑎𝑛 and 𝑥1 , 𝑥2 , … , 𝑥𝑛 be real numbers satisfying 𝑎1 < 𝑥1 < 𝑎2 < 𝑥2 < … < 𝑎𝑛 < 𝑥𝑛 ,

Chapter 11. Interpolation with Blaschke Products

141

and let 𝑦1 , 𝑦2 , … , 𝑦𝑛 be a set of real numbers. Let 𝜆 ∈ ℝ be such that 𝜆 ≠ 𝑦𝑗 for 𝑗 = 1, 2, … , 𝑛. Then there exists a rational function 𝐾 ∶ ℂ∗ → ℂ∗ of degree 𝑛 such that (1) 𝐾 is strongly real of positive type; (2) 𝐾(𝑥𝑗 ) = 𝑦𝑗 for 𝑗 = 1, 2, … , 𝑛; (3) 𝐾(𝑎𝑗 ) = 𝜆 for 𝑗 = 1, 2, … , 𝑛. Proof. Let 𝑇 ∶ ℂ∗ → ℂ∗ be defined by 1 𝑇(𝑧) = 𝜆 − , 𝑧 and let 𝐺 be the function defined in Theorem 11.6 that maps 𝑥𝑗 to 𝑇 −1 (𝑦𝑗 ) for 𝑗 = 1, 2, … , 𝑛 and has poles at each 𝑎𝑗 for 𝑗 = 1, 2, … , 𝑛. Note that since arg(−1/𝐺(𝑧)) = 𝜋 − arg(𝐺(𝑧)), we have that 0 < arg(−1/𝐺(𝑧)) < 𝜋 for 𝑧 ∈ 𝐇+ and −𝜋 < arg(−1/𝐺(𝑧)) < 0 for 𝑧 ∈ 𝐇− . Finally, −1/𝐺(𝑧) ∈ ℝ∗ for 𝑧 ∈ ℝ∗ . Then the function we seek is 𝐾 = 𝑇 ∘ 𝐺. Now let us check our control of the degree of the interpolating function. Theorem 11.8. For an integer 𝑛 > 1, let 𝑏1 , 𝑏2 , … , 𝑏𝑛 be real and distinct, 𝑐1 , 𝑐2 , … , 𝑐𝑛 be real, and 𝑐𝑗 ≠ 𝑐𝑘 for some 𝑗 and 𝑘. Then there exists a rational function, 𝐿 ∶ ℂ∗ → ℂ∗ , of degree at most 𝑛 − 1, strongly real of positive type such that 𝐿(𝑏𝑘 ) = 𝑐𝑘 for 𝑘 = 1, 2, … , 𝑛. To prove this theorem we use Corollary 11.7, which requires that we identify two sets of interspersed points, 𝑎𝑗 and 𝑥𝑗 . This allows us to conclude that the resulting rational function has the same values on the 𝑎𝑗 and arbitrary values on the 𝑥𝑗 . We need to add points to the 𝑏𝑘 in the hypothesis of Theorem 11.8 so that we are able to use Corollary 11.7. Example 11.9, which follows the proof of Theorem 11.8, shows how this is done. Proof. Without loss of generality we may assume that 𝑏1 < 𝑏2 < ⋯ < 𝑏𝑛 . We choose real numbers 𝑢1 , … , 𝑢𝑛−1 such that 𝑏𝑘 < 𝑢𝑘 < 𝑏𝑘+1 , 𝑏𝑛 < 𝑢𝑛 , and 𝑣 ∈ ℝ such that 𝑣 ≠ 𝑐1 and then construct a pair of

142

Chapter 11. Interpolation with Blaschke Products

sequences, (𝑎𝑗 ) and (𝑥𝑗 ), in two steps. In each sequence the terms are indexed in increasing order. For the first step, we choose (𝑎𝑗 ) and (𝑥𝑗 ) so that • (𝑎𝑗 ) consists of the 𝑏𝑘 for all 𝑘 with 𝑐𝑘 = 𝑐1 ; • (𝑥𝑗 ) consists of the 𝑏𝑘 for all 𝑘 with 𝑐𝑘 ≠ 𝑐1 . Note that so far, each sequence has at least one term and, for each 𝑘, the number 𝑏𝑘 is in exactly one of the two sequences. For the second step, we augment (𝑎𝑗 ) by inserting 𝑢𝑘 for each 𝑘 for which 𝑏𝑘 and 𝑏𝑘+1 are terms in (𝑥𝑗 ). Likewise, we augment (𝑥𝑗 ) by inserting 𝑢𝑘 for each 𝑘 for which 𝑏𝑘 and 𝑏𝑘+1 are terms in (𝑎𝑗 ); in addition, add 𝑢𝑛 if 𝑏𝑛 is a term of (𝑎𝑗 ). This construction yields two interspersed sequences of real numbers (𝑎𝑗 ) and (𝑥𝑗 ) each of length 𝑚 ≤ 𝑛 − 1. For 𝑗 = 1, … , 𝑚 we define 𝑦𝑗 = {

𝑐𝑘 𝑣

if 𝑥𝑗 = 𝑏𝑘 for some 𝑘, otherwise.

Taking 𝜆 = 𝑐1 and applying Corollary 11.7, we obtain a rational function 𝐾 ∶ ℂ⋆ → ℂ⋆ of degree 𝑚 satisfying the conclusion of the corollary. In particular, we see that if 𝑏𝑘 = 𝑥𝑗 for some 𝑗, then 𝐾(𝑏𝑘 ) = 𝐾(𝑥𝑗 ) = 𝑦𝑗 = 𝑐𝑘 , and if 𝑏𝑘 = 𝑎𝑗 for some 𝑗, then 𝐾(𝑏𝑘 ) = 𝐾(𝑎𝑗 ) = 𝜆 = 𝑐1 = 𝑐𝑘 . Thus, 𝐿 = 𝐾 is the desired rational function. The proof of Theorem 11.8 provides a rational function with the desired properties that does the required interpolation. However, this need not be the rational function of minimal degree. More information on how to obtain a rational function of minimal degree can be found in the papers [61] and [138]. Before turning to an example of this algorithm in action, we note that if all 𝑐𝑘 are equal, the constant function 𝐹 defined by 𝐹(𝑧) = 𝑐1 will do the interpolation, but it will not be strongly real of positive type. Nevertheless, we are usually after a solution to the interpolation problem that is analytic and of low degree. The constant function certainly does provide that.

Chapter 11. Interpolation with Blaschke Products

143

We are ready to illustrate the proof of Theorem 11.8 with an example that will also show that the rational function obtained using the described algorithm may have degree strictly less than 𝑛 − 1. Example 11.9. Let 𝑏𝑘 = 𝑘 for 𝑘 = 1, 2, … , 7 and 𝑐1 = 1, 𝑐2 = 1, 𝑐3 = 3, 𝑐4 = 3, 𝑐5 = 1, 𝑐6 = 1, and 𝑐7 = 2. A schematic presentation illuminates the construction in the proof above, with the 𝑢𝑘 we must include (because they are between points with consecutive indices in either set) in bold below.

𝑏4

𝑢4

𝑏5

𝐮𝟓

= 𝑏6 =

𝐮𝟑

=

𝑏3

=

𝑢2

=

𝑏2

𝑥5

=

= 𝐮𝟏

𝑥4

=

𝑏1

𝑥3 =

𝑥2 =

𝑥1

𝑎1

𝑎2

𝑎3

𝑎4

𝑎5

𝑢6

𝑏7

𝑢7

The existence of the required function now follows from Corollary 11.7 and the function is of degree 5. As we saw when we considered Corollary 11.4, we can also find statements about Blaschke products that correspond to Theorem 11.6, Corollary 11.7, and Theorem 11.8. We leave it up to the reader to find and prove these statements. We conclude this chapter with the most general interpolation problem we consider. Corollary 11.10. Let 𝑗 = 1, 2, … , 𝑛 and 𝑘 = 1, 2, … , 𝑚. Let 𝐴𝑘 = {𝑑𝑗𝑘 } be 𝑚 interspersed sets of 𝑛 points on ℝ. Then there exists a strongly real function of positive type, 𝑀, such that 𝑀 is constant on 𝐴𝑘 for each 𝑘. We remark that when we say the sets are interspersed, we mean 𝑑11 < 𝑑12 < ⋯ < 𝑑1𝑚 < 𝑑21 < 𝑑22 < ⋯ < 𝑑2𝑚 < ⋯ < 𝑑𝑛1 < 𝑑𝑛2 < ⋯ < 𝑑𝑛𝑚 . 𝑛

Proof. We let 𝑆1 = ⋃𝑘=2 𝐴𝑘 and choose a set 𝑆 of (𝑚 − 2)𝑛 points in ℝ in such a way that the set 𝑆2 ∶= 𝑆 ∪ 𝐴1 has its points interspersed with the points of 𝑆1 . Note that 𝑑11 is the smallest value, and it lies in 𝑆2 . We apply Corollary 11.7 with 𝑆2 containing the 𝑎𝑗 and 𝑆1 the 𝑥𝑗 . Making appropriate choices for 𝑦𝑗 and 𝜆, Corollary 11.7 provides a rational function 𝑀 that is strongly real of positive type and such that 𝑀(𝑑𝑗𝑘 ) = 𝑀(𝑑ℓ𝑘 ) for all 𝑗 and ℓ; that is, 𝑀 is constant on 𝐴𝑘 for each 𝑘.

144

Chapter 11. Interpolation with Blaschke Products

What does this say about Blaschke products and Poncelet curves? Corollary 11.11. Let 𝑗 = 1, 2, … , 𝑛 and 𝑘 = 1, 2, … , 𝑚. Let {𝑧𝑗𝑘 } be 𝑚 sets of 𝑛 interspersed points on 𝕋. Then there exists a Blaschke product such that for each 𝑘, the Blaschke product identifies 𝑧𝑗𝑘 for 𝑗 = 1, 2, … 𝑛, and there exists a Poncelet curve that is circumscribed by polygons 𝑃𝑘 such that for every 𝑘, the vertices of 𝑃𝑘 include 𝑧𝑗𝑘 for 𝑗 = 1, 2, … 𝑛. Proof. Use Corollary 11.10 and Lemma 11.2 to find a Blaschke product 𝐵 with the correct interpolation properties. If 𝐵(0) ≠ 0, let 𝐵̃ =

𝐵 − 𝐵(0) 1 − 𝐵(0)𝐵

.

̃ Then 𝐵̃ has the same interpolation properties as 𝐵 and 𝐵(0) = 0. By Theorem 10.6, we have the corresponding Poncelet curve. We note that the degree of the Blaschke product in Corollary 11.11 could be as large as (𝑚 − 1)𝑛, and, as a consequence, the polygons 𝑃𝑘 will have vertices in addition to the original 𝑧𝑗𝑘 . In the introduction to this chapter, we noted that specifying only one triangle did not determine a unique Blaschke product. Theorem 11.12 will show that two triangles with interspersed vertices determine the Blaschke product, up to composition with a disk automorphism. In fact, the same is true for two convex 𝑛-gons with interspersed vertices on 𝕋. Theorem 11.12. Given two sets of 𝑛 points on the unit circle, 𝑆1 = {𝑧1 , … , 𝑧𝑛 } and 𝑆2 = {𝑤1 , … , 𝑤𝑛 } with 0 ≤ arg(𝑧1 ) < arg(𝑤1 ) < arg(𝑧2 ) < arg(𝑤2 ) < ⋯ < arg(𝑧𝑛 ) < arg(𝑤𝑛 ) < 2𝜋, there is a Blaschke product 𝐵 of degree 𝑛 that is constant on 𝑆1 and constant on 𝑆2 . In addition, if 𝐶 is any other Blaschke product of degree 𝑛 that is constant on 𝑆1 and constant on 𝑆2 , then there is an analytic disk automorphism 𝜑 such that 𝐶 = 𝜑 ∘ 𝐵. Proof. The existence of such a Blaschke product is established in Corollary 11.4. Our goal is to prove the required uniqueness, up to composition with disk automorphisms.

Chapter 11. Interpolation with Blaschke Products

145

So let 𝐵 and 𝐶 be two Blaschke products that identify the two sets of points: 𝐵|𝑆𝑗 = 𝛼𝑗 and 𝐶|𝑆𝑗 = 𝛽𝑗 for 𝑗 = 1, 2. Now choose 𝑢 with arg(𝑤1 ) < arg(𝑢) < arg(𝑧2 ), and let 𝛼3 = 𝐵(𝑢) and 𝛽3 = 𝐶(𝑢). Then 𝛼1 , 𝛼2 , 𝛼3 and 𝛽1 , 𝛽2 , 𝛽3 are ordered in the same sense on the unit circle. Then there is a disk automorphism 𝜑 with the property that 𝜑(𝛼𝑗 ) = 𝛽𝑗 for 𝑗 = 1, 2, 3; see [118, p. 46]. Now 𝜑∘𝐵 and 𝐶 are rational functions of degree 𝑛 that agree on 2𝑛+1 points on the unit circle. Writing 𝜑 ∘ 𝐵 = 𝑝/𝑞 and 𝐶 = 𝑟/𝑠, where 𝑝, 𝑞, 𝑟, and 𝑠 are polynomials of degree at most 𝑛, we have 𝑝𝑠 − 𝑞𝑟 = 0 on 2𝑛 + 1 points. Therefore, 𝑝𝑠 = 𝑞𝑟 everywhere and 𝜑 ∘ 𝐵 = 𝐶. In Example 11.5 we constructed a Blaschke product 𝐵 constant on the set 𝑆1 = {𝑖, −1, −𝑖} and constant on 𝑆2 = {(1 + 𝑖)/√2, (−1 + 𝑖)/√2, (−1−𝑖)/√2}. Theorem 11.12 tells us that if 𝐶 is any other Blaschke product that is constant on 𝑆1 and on 𝑆2 , then there is a disk automorphism 𝜑 with 𝐶 = 𝜑 ∘ 𝐵.

Chapter

12

Poncelet’s Theorem for 𝑛-Gons In Chapter 5 we established Poncelet’s theorem for triangles, and in Chapter 10 we discussed a Poncelet-like theorem for curves that need not be ellipses. When the curve we consider is a conic inscribed in a triangle, the two results match up. In this chapter, we provide a complete proof of the following version of Poncelet’s theorem: Given two nondegenerate conics (see page 52) in the real projective plane that do not intersect and for which there exists a closed 𝑛-sided polygon inscribed in the first conic and circumscribing the second, any (𝑛 − 1)-sided polygonal chain with all its vertices on the first conic and its sides tangent to the second conic can be completed to a closed 𝑛-sided polygon inscribed in the first conic and circumscribing the second. Recall that side of a polygonal chain refers to the line through the vertices (not just the line segment) and a principal diagonal of a hexagon is a line through opposite vertices. We use a surprisingly elementary proof of Poncelet’s theorem, due to Halbeisen and Hungerbühler, that appeared in their 2015 paper [74]. As the authors note, “Poncelet’s treatise was a milestone in the development of projective geometry, and his theorem is widely considered the deepest and most beautiful result about conics”. Halbeisen and Hungerbühler’s idea is, essentially, to use repeated applications of Pascal’s theorem (Theorem 5.3) and its dual, Brianchon’s theorem (Theorem 5.4). We use the following notation throughout this chapter. Let 𝒞 and 𝒟 be two nondegenerate conics in the real projective plane that do not intersect. Let 𝑎1 , … , 𝑎𝑛 (where 𝑛 ≥ 4) denote distinct points on 𝒞 that are the vertices of an 𝑛-sided polygon circumscribing 𝒟. Let 𝑏1 , … , 𝑏𝑛 be distinct points on 𝒞, none of them coinciding with 𝑎ℓ for any ℓ, having the additional property that the lines 𝑏𝑗 × 𝑏𝑗+1 for 1 ≤ 𝑗 < 𝑛 are tangent 147

148

Chapter 12. Poncelet’s Theorem for 𝑛-Gons

to 𝒟. At this point, we have no information about the tangency of the line 𝑏𝑛 × 𝑏1 . We define the following points for integers ℓ, 𝑚, 𝑝, and 𝑞, where 1 ≤ ℓ, 𝑚 ≤ 𝑛, ℓ ≠ 𝑚, and 1 ≤ 𝑝 < 𝑞 ≤ 𝑛: 𝐼ℓ,𝑚 𝑋𝑝,𝑞

= (𝑎ℓ × 𝑎𝑚 ) × (𝑏ℓ × 𝑏𝑚 ), = (𝑎𝑝 × 𝑏𝑞 ) × (𝑎𝑞 × 𝑏𝑝 ).

(12.1)

Note that 𝐼ℓ,𝑚 = 𝐼𝑚,ℓ . If 𝑚 = (ℓ + 1) mod 𝑛, we write 𝐼ℓ for 𝐼ℓ,𝑚 . Finally, we denote the middle index by 𝑐, where 𝑐 = 𝑛/2 if 𝑛 is even or 𝑐 = (𝑛 + 1)/2 if 𝑛 is odd. The idea of the proof is to apply Brianchon’s theorem to five distinct tangent lines to 𝒟 and the line 𝑏𝑛 × 𝑏1 , thus showing that this last line is also tangent to 𝒟. To show that the hypothesis of Brianchon’s theorem is met we need to show that three lines are concurrent. This will be established in a series of lemmas that move us between triples of points. We choose 𝑛 distinct points and list them according to their indices. Working from points with indices in the middle of the list (indices that vary as the index is even or odd), Lemmas 12.1 and 12.2 provide us with three distinct points that are collinear. Lemmas 12.3 and 12.4 show how to move outwards toward the beginning and end of the polygonal chains, respectively, while maintaining the property that we have three collinear points. Lemma 12.5 specifies the end situation of this outward movement along the polygonal vertices. What remains to be shown is that the three collinear points that were migrated carefully are exactly the right ones to ensure that Brianchon’s theorem applies and can be used in the proof of Theorem 12.6. As an example of how these lemmas work, let us consider the case in which 𝑛 = 8. See Figure 12.1, which shows the first and last triple of collinear points. (It is interesting to note that 𝑋4,5 = 𝑋1,8 in the figure; see page 155 for why this is true.) We start with the octagon that is inscribed in 𝒞 and circumscribes 𝒟 and the polygonal chain (𝑏1 , … , 𝑏8 ) that has all its vertices on 𝒞 and seven sides that are tangent to 𝒟. Our goal is to show that the points 𝐼1 , 𝑋1,8 , and 𝐼7 are collinear. Here is how it works. • In this case, 𝑐 = 8/2 = 4 and by Lemma 12.2 the distinct points 𝐼3 , 𝑋4,5 , and 𝐼5 are collinear.

Chapter 12. Poncelet’s Theorem for 𝑛-Gons

149

 





















 





Figure 12.1. The collinear points “move” from the center to the outside.

• By Lemma 12.4, we can shift both indices of 𝑋4,5 away from the central index 𝑐 = 4 and keep collinearity; thus, 𝐼3 , 𝑋3,6 , and 𝐼5 are collinear. • By Lemma 12.3, we can shift the indices of 𝐼3 and 𝐼5 away from the central index 𝑐 = 4 and keep collinearity; thus, 𝐼2 , 𝑋3,6 , and 𝐼6 are collinear. • Again by Lemma 12.4, shifting indices on 𝑋3,6 , the distinct points 𝐼2 , 𝑋2,7 , and 𝐼6 are collinear. • Again by Lemma 12.3, shifting the indices of 𝐼2 and 𝐼6 , the distinct points 𝐼1 , 𝑋2,7 , and 𝐼7 are collinear.

150

Chapter 12. Poncelet’s Theorem for 𝑛-Gons

• Finally, by Lemma 12.4, shifting indices on 𝑋2,7 , the distinct points 𝐼1 , 𝑋1,8 , and 𝐼7 are collinear, which is what we want. The collinearity of these three points and Brianchon’s theorem will enable us to prove Poncelet’s theorem, as we will see at the end of this chapter. We start by showing the existence of three distinct collinear points. For the case in which 𝑛 is odd we use Lemma 12.1, and for the case in which 𝑛 is even we use Lemma 12.2. However, both lemmas hold for all 𝑛 ≥ 4. Lemma 12.1. With the conditions and notation as in the paragraph containing (12.1), the points 𝐼𝑐−1 , 𝑋𝑐−1,𝑐+1 , and 𝐼𝑐 are distinct and collinear.



-



-



-



+












+

-

+


Figure 12.2. By Lemma 12.1, the points 𝐼𝑐−1 , 𝑋𝑐−1,𝑐+1 , and 𝐼𝑐 are collinear.

Proof. We consider the hexagon with vertices on the nondegenerate conic 𝒞 that has the following vertex order: (𝑎𝑐+1 , 𝑎𝑐 , 𝑎𝑐−1 , 𝑏𝑐+1 , 𝑏𝑐 , 𝑏𝑐−1 ).

Chapter 12. Poncelet’s Theorem for 𝑛-Gons

151

By Pascal’s theorem (Theorem 5.3) the points (𝑎𝑐+1 × 𝑎𝑐 ) × (𝑏𝑐+1 × 𝑏𝑐 ) = 𝐼𝑐 , (𝑎𝑐 × 𝑎𝑐−1 ) × (𝑏𝑐 × 𝑏𝑐−1 ) = 𝐼𝑐−1 , (𝑎𝑐−1 × 𝑏𝑐+1 ) × (𝑏𝑐−1 × 𝑎𝑐+1 ) = 𝑋𝑐−1,𝑐+1 are collinear and distinct; see Figure 12.2. Lemma 12.2. With the conditions and notation as in the paragraph containing (12.1), the points 𝐼𝑐−1 , 𝑋𝑐,𝑐+1 , and 𝐼𝑐+1 are distinct and collinear.


-








-



-




+

+

+

+



+


ℓ

 



+


Figure 12.3. By Lemma 12.2 the points 𝐼𝑐−1 , 𝑋𝑐,𝑐+1 , and 𝐼𝑐+1 are collinear.

152

Chapter 12. Poncelet’s Theorem for 𝑛-Gons

Proof. We consider the hexagon determined by the six lines in the order: (𝑎𝑐−1 × 𝑎𝑐 , 𝑏𝑐−1 × 𝑏𝑐 , 𝑏𝑐 × 𝑏𝑐+1 , 𝑏𝑐+1 × 𝑏𝑐+2 , 𝑎𝑐+1 × 𝑎𝑐+2 , 𝑎𝑐 × 𝑎𝑐+1 ). See Figure 12.3 in which the hexagon is emphasized. By assumption, all sides of this hexagon are tangent to the conic 𝒟. By Brianchon’s theorem (Theorem 5.4) the principal diagonals, 𝑘 ℓ 𝑚

= = = = = =

((𝑎𝑐−1 × 𝑎𝑐 ) × (𝑏𝑐−1 × 𝑏𝑐 )) × ((𝑏𝑐+1 × 𝑏𝑐+2 ) × (𝑎𝑐+1 × 𝑎𝑐+2 )) 𝐼𝑐−1 × 𝐼𝑐+1 , ((𝑏𝑐−1 × 𝑏𝑐 ) × (𝑏𝑐 × 𝑏𝑐+1 )) × ((𝑎𝑐+1 × 𝑎𝑐+2 ) × (𝑎𝑐 × 𝑎𝑐+1 )) 𝑏𝑐 × 𝑎𝑐+1 , ((𝑏𝑐 × 𝑏𝑐+1 ) × (𝑏𝑐+1 × 𝑏𝑐+2 )) × ((𝑎𝑐 × 𝑎𝑐+1 ) × (𝑎𝑐−1 × 𝑎𝑐 )) 𝑏𝑐+1 × 𝑎𝑐 ,

are concurrent at a point 𝑃 = ℓ × 𝑚 = (𝑏𝑐 × 𝑎𝑐+1 ) × (𝑏𝑐+1 × 𝑎𝑐 ) = 𝑋𝑐,𝑐+1 . Hence, 𝐼𝑐−1 , 𝑋𝑐,𝑐+1 , and 𝐼𝑐+1 are collinear. Observe that 𝐼𝑐−1 ≠ 𝐼𝑐+1 : Each of these points has two distinct tangent lines to the conic, and if the points were equal, we would have four distinct tangent lines from this point to the conic 𝒟, which is impossible. That 𝑋𝑐,𝑐+1 ≠ 𝐼𝑐−1 and 𝑋𝑐,𝑐+1 ≠ 𝐼𝑐+1 hold is because otherwise we would have three collinear points on the conic 𝒞 (see Figure 12.3). Lemma 12.3. With the conditions and the notation as in the paragraph containing (12.1) and 𝑝 ≤ 𝑞 −2, if the points 𝐼𝑝 , 𝑋𝑝,𝑞 , and 𝐼𝑞−1 are distinct and collinear, then the points 𝐼𝑝−1 , 𝑋𝑝,𝑞 , and 𝐼𝑞 are distinct and collinear. Proof. See Figure 12.4 for an illustration. Since 𝑝 ≤ 𝑞 − 2, the six different tangent lines to the nondegenerate conic 𝒟 form a hexagon with side sequence (𝑏𝑝−1 × 𝑏𝑝 , 𝑏𝑝 × 𝑏𝑝+1 , 𝑎𝑝 × 𝑎𝑝+1 , 𝑎𝑞 × 𝑎𝑞+1 , 𝑎𝑞−1 × 𝑎𝑞 , 𝑏𝑞−1 × 𝑏𝑞 ) . The three principal diagonals are the lines 𝑘 ℓ 𝑚

= = = = =

((𝑏𝑝−1 × 𝑏𝑝 ) × (𝑏𝑝 × 𝑏𝑝+1 )) × ((𝑎𝑞 × 𝑎𝑞+1 ) × (𝑎𝑞−1 × 𝑎𝑞 )) 𝑏𝑝 × 𝑎𝑞 , ((𝑏𝑝 × 𝑏𝑝+1 ) × (𝑎𝑝 × 𝑎𝑝+1 )) × ((𝑎𝑞−1 × 𝑎𝑞 ) × (𝑏𝑞−1 × 𝑏𝑞 )) 𝐼𝑝 × 𝐼𝑞−1 , ((𝑎𝑝 × 𝑎𝑝+1 ) × (𝑎𝑞 × 𝑎𝑞+1 )) × ((𝑏𝑞−1 × 𝑏𝑞 ) × (𝑏𝑝−1 × 𝑏𝑝 )) .

Chapter 12. Poncelet’s Theorem for 𝑛-Gons

153


+

   +






-




-

-

-





ℓ


+

+
 ℓ

  -

 -



 Figure 12.4. Proof of Lemma 12.3.

By Brianchon’s theorem (Theorem 5.4), the lines 𝑘, ℓ, and 𝑚 are distinct and concurrent.

(12.2)

We repeat this construction with a hexagon formed by the following sequence of tangent lines to 𝒟: (𝑏𝑝−1 × 𝑏𝑝 , 𝑎𝑝−1 × 𝑎𝑝 , 𝑎𝑝 × 𝑎𝑝+1 , 𝑎𝑞 × 𝑎𝑞+1 , 𝑏𝑞 × 𝑏𝑞+1 , 𝑏𝑞−1 × 𝑏𝑞 ) .

154

Chapter 12. Poncelet’s Theorem for 𝑛-Gons

The principal diagonals of the hexagon are the lines 𝑘′ ℓ′ 𝑚′

= = = = =

((𝑏𝑝−1 × 𝑏𝑝 ) × (𝑎𝑝−1 × 𝑎𝑝 )) × ((𝑎𝑞 × 𝑎𝑞+1 ) × (𝑏𝑞 × 𝑏𝑞+1 )) 𝐼𝑝−1 × 𝐼𝑞 , ((𝑎𝑝−1 × 𝑎𝑝 ) × (𝑎𝑝 × 𝑎𝑝+1 )) × ((𝑏𝑞 × 𝑏𝑞+1 ) × (𝑏𝑞−1 × 𝑏𝑞 )) 𝑎𝑝 × 𝑏𝑞 , 𝑚.

Again by Theorem 5.4, the lines 𝑘 ′ , ℓ′ , and 𝑚 are distinct and concurrent.

(12.3)

Note that ℓ′ × 𝑘 = (𝑎𝑝 × 𝑏𝑞 ) × (𝑏𝑝 × 𝑎𝑞 ) = 𝑋𝑝,𝑞 . By assumption, 𝐼𝑝 , 𝑋𝑝,𝑞 , and 𝐼𝑞−1 are distinct and collinear and ℓ contains 𝐼𝑝 and 𝐼𝑞−1 . So ℓ also contains the point 𝑋𝑝,𝑞 . Thus, ℓ, ℓ′ , and 𝑘 are concurrent, and since 𝑘 ≠ ℓ, we get the lines ℓ, ℓ′ , and 𝑘 intersect at, and only at, 𝑋𝑝,𝑞 .

(12.4)

′

Statements (12.2) and (12.4) imply that 𝑘, ℓ, 𝑚, and ℓ all intersect in the unique point 𝑋𝑝,𝑞 . Combined with (12.3), we get that 𝑘 ′ also passes through 𝑋𝑝,𝑞 . In particular, 𝐼𝑝−1 , 𝑋𝑝,𝑞 , and 𝐼𝑞 are collinear. The fact that the points 𝐼𝑝−1 , 𝑋𝑝,𝑞 , and 𝐼𝑞 are distinct follows from the same arguments as in the proof of Lemma 12.2. Lemma 12.4. With the conditions and notation as in the paragraph containing (12.1), if the points 𝐼𝑝−1 , 𝑋𝑝,𝑞 , and 𝐼𝑞 are distinct and collinear, then the points 𝐼𝑝−1 , 𝑋𝑝−1,𝑞+1 , and 𝐼𝑞 are distinct and collinear. Proof. Figure 12.5 will guide us through the proof. Since 𝑝 < 𝑞, we get the hexagon with vertices (𝑎𝑝−1 , 𝑎𝑝 , 𝑏𝑞 , 𝑏𝑝−1 , 𝑏𝑝 , 𝑎𝑞 ) on the conic 𝒞. By Pascal’s theorem (Theorem 5.3), the points 𝐼𝑝−1 , 𝑋𝑝,𝑞 , and 𝐼𝑝−1,𝑞 are distinct and collinear.

(12.5)

Similarly, we get a hexagon with vertices (𝑎𝑝−1 , 𝑎𝑞 , 𝑎𝑞+1 , 𝑏𝑝−1 , 𝑏𝑞 , 𝑏𝑞+1 ) on 𝒞. Again by Theorem 5.3, the points 𝐼𝑝−1,𝑞 , 𝑋𝑝−1,𝑞+1 , and 𝐼𝑞 are distinct and collinear.

(12.6)

We obtain three sets of three collinear points from the assumption, (12.5), and (12.6). We conclude that the points 𝐼𝑝−1 , 𝑋𝑝−1,𝑞+1 , and 𝐼𝑞 are collinear. The same arguments we used in the proof of Lemma 12.2 show that the points are also distinct.

Chapter 12. Poncelet’s Theorem for 𝑛-Gons




155


-
+








 +


  +




-
 
-




-

-






Figure 12.5. Proof of Lemma 12.4.

Figure 12.5 suggests that under the hypothesis of Lemma 12.4, we have 𝑋𝑝,𝑞 = 𝑋𝑝−1,𝑞+1 . This fact is irrelevant for our proof, but interesting, and it can be established using Lemma 12.4 together with Brianchon’s theorem applied to the hexagon with side sequence (𝑎𝑞 × 𝑎𝑞+1 , 𝑎𝑞+1 × 𝑎𝑝 , 𝑎𝑝 × 𝑎𝑝−1 , 𝑏𝑝−1 × 𝑏𝑝 , 𝑏𝑝−1 × 𝑏𝑞 , 𝑏𝑞 × 𝑏𝑞+1 ). Lemma 12.5. With the conditions and the notation as in the paragraph containing (12.1), the points 𝐼1 , 𝑋1,𝑛 , and 𝐼𝑛−1 are distinct and collinear. Proof. If 𝑛 ≥ 4 is even, we start with the distinct and collinear points 𝐼𝑐−1 , 𝑋𝑐,𝑐+1 , and 𝐼𝑐+1 that Lemma 12.2 tells us exist. Applying

156

Chapter 12. Poncelet’s Theorem for 𝑛-Gons

Lemma 12.4 followed by Lemma 12.3 and repeating this cycle, we reach the conclusion. If 𝑛 ≥ 4 is odd, we start with the distinct and collinear points 𝐼𝑐−1 , 𝑋𝑐−1,𝑐+1 , and 𝐼𝑐 that exist by Lemma 12.1. Applying Lemma 12.3 followed by Lemma 12.4 and repeating this cycle, we reach the conclusion. Theorem 12.6 (Poncelet). Let 𝒞 and 𝒟 be nondegenerate conics that do not intersect. Suppose that there is an 𝑛-sided polygon inscribed in 𝒞 and circumscribing 𝒟. Consider an (𝑛 − 1)-sided polygonal chain with all vertices on 𝒞 and all sides tangent to 𝒟. Then the side that closes up the polygonal chain (forming a polygon) is also tangent to 𝒟.


-








-



-












ℓ

 





Figure 12.6. Proof of Poncelet’s theorem.

Chapter 12. Poncelet’s Theorem for 𝑛-Gons

157

Proof. If 𝑏𝑗 = 𝑎𝑘 for some 𝑗 and some 𝑘, then the two polygonal chains are the same and the result is trivial. Thus, we assume that 𝑏𝑗 ≠ 𝑎𝑘 for all 𝑗 and all 𝑘. For 𝑛 = 3, the statement follows from Theorem 5.6. So we assume now that 𝑛 ≥ 4 and we use the notation introduced at the beginning of this chapter; that is, we have 𝑛 distinct points 𝑎1 , … 𝑎𝑛 on 𝒞 that are the vertices of an 𝑛-sided polygon circumscribing 𝒟. The points 𝑏1 , … , 𝑏𝑛 , chosen to lie on 𝒞 are distinct and none coincides with 𝑎ℓ for any ℓ. These points have the additional property that the lines 𝑏𝑗 × 𝑏𝑗+1 for 𝑗 = 1, … , 𝑛 − 1 are tangent to 𝒟. In order to prove Poncelet’s theorem we need to show that the line 𝑏𝑛 × 𝑏1 is tangent to 𝒟. We use the symbols 𝐼ℓ,𝑚 , 𝐼ℓ , and 𝑋𝑝,𝑞 that were defined in (12.1). Figure 12.6 should be helpful in understanding this proof. We choose five tangent lines to 𝒟 and the line 𝑏𝑛 × 𝑏1 to form the hexagon (𝑎𝑛−1 × 𝑎𝑛 , 𝑎𝑛 × 𝑎1 , 𝑎1 × 𝑎2 , 𝑏1 × 𝑏2 , 𝑏𝑛 × 𝑏1 , 𝑏𝑛−1 × 𝑏𝑛 ) . Since the conic is nondegenerate, the five tangent lines are in general position, for otherwise there would be a point from which three different tangents could be drawn to 𝒟, which is impossible. We calculate the principal diagonals of the hexagon and get 𝑘 ℓ 𝑚

= = = = = =

((𝑎𝑛−1 × 𝑎𝑛 ) × (𝑎𝑛 × 𝑎1 )) × ((𝑏1 × 𝑏2 ) × (𝑏𝑛 × 𝑏1 )) 𝑎𝑛 × 𝑏1 , ((𝑎𝑛 × 𝑎1 ) × (𝑎1 × 𝑎2 )) × ((𝑏𝑛 × 𝑏1 ) × (𝑏𝑛−1 × 𝑏𝑛 )) 𝑎1 × 𝑏𝑛 , ((𝑎1 × 𝑎2 ) × (𝑏1 × 𝑏2 )) × ((𝑏𝑛−1 × 𝑏𝑛 ) × (𝑎𝑛−1 × 𝑎𝑛 )) 𝐼1 × 𝐼𝑛−1 .

Now 𝑘 × ℓ = (𝑎𝑛 × 𝑏1 ) × (𝑎1 × 𝑏𝑛 ) = 𝑋1,𝑛 and by Lemma 12.5 we see that 𝑋1,𝑛 is also a point on 𝑚. So the three lines 𝑘, ℓ, and 𝑚 intersect in one point. By Brianchon’s theorem (Theorem 5.4), the six lines that we chose originally are tangent to a conic 𝒟′ . Since five of them are already tangent to the conic 𝒟, Theorem 5.5 implies that 𝒟′ = 𝒟. In particular, 𝑏𝑛 × 𝑏1 is tangent to 𝒟, completing the proof of Poncelet’s theorem. An oft-quoted form of Poncelet’s theorem is now immediate.

158

Chapter 12. Poncelet’s Theorem for 𝑛-Gons

Corollary 12.7. Let 𝒞 and 𝒟 be nondegenerate conics that do not intersect. Suppose that there is an 𝑛-sided polygon inscribed in 𝒞 circumscribing 𝒟. Then any point on 𝒞 is the vertex of an 𝑛-sided polygon that is inscribed in 𝒞 and circumscribes 𝒟. Similarly, any point on 𝒟 is the point of tangency of an 𝑛-sided polygon that is inscribed in 𝒞 and circumscribes 𝒟. Be aware that you sometimes find a statement with the conclusion as in the corollary without the condition that the two conics do not intersect. Such a statement is false. However, our original statement in Theorem 12.6 is also true if the condition of nonintersection is replaced by the weaker one of being in general position (at most four points of intersection). The proof is essentially the same with some additional detail work to rule out special cases. With this weaker hypothesis, some points on the conic cannot be used as vertices of circumscribing polygons. We hope that you enjoyed this proof of Poncelet’s theorem. It is elementary, but this should not be confused with trivial—we think Halbeisen and Hungerbühler’s proof is really clever. The points and lines had to be chosen very carefully in order to progress toward the conclusion. The alternating use of Pascal’s theorem and its dual, Brianchon’s theorem, is an excellent example of how to use duality in projective geometry. There are many other proofs of Poncelet’s theorem. In Chapter 5 we discussed Poncelet’s proof. In 1828, Jacobi gave another proof that is reprinted in Volume 1 of his collected works [86]. In 1977, Griffiths and Harris gave a modern proof based on algebraic geometry [69]. Their proof renewed interest in the theorem and resulted in two monographs dedicated to the subject: one by Flatto [47], and the other by Dragović and Radnović [43]. Both of these monographs connect Poncelet’s theorem to dynamical systems and billiards.

Chapter

13

Kippenhahn’s Curve and Blaschke’s Products In this chapter we investigate a surprising connection between the shape of the numerical range of the compression of the shift and when we can write a Blaschke product as the composition of two nontrivial Blaschke products. This question has a long history, starting with the work of Ritt [132] dating back to 1922 and continuing to the present day. In our situation, if we start with a degree-3 Blaschke product 𝐵1 and consider 𝐵(𝑧) = 𝑧𝐵1 (𝑧), then it turns out that the numerical range of 𝑆𝐵1 is an elliptical disk if and only if there exist degree-2 Blaschke products 𝐶 and 𝐷 such that 𝐵 = 𝐶 ∘ 𝐷. You can experiment with our applet ,1 using Compose Tool. Choose Blaschke products 𝐶 and 𝐷, form 𝐵 = 𝐶 ∘ 𝐷, and then look at the Blaschke ellipse associated with 𝐵 that, as we now know, is also the boundary of the numerical range of 𝑆𝐵1 . The important work of Rudolf Kippenhahn is one of our essential tools. So we begin with that as well as a little bit about the man himself. Kippenhahn was born in 1926 in Czechoslovakia but had to leave in 1945. In 1948, he crossed the German border illegally in order to study mathematics in Erlangen. He took a job at the Bamberg Observatory rather than one in mathematics, saying, “I didn’t feel that I would achieve very much in mathematics”. In this chapter we will study one of his major mathematical achievements. In an interview with the American Institute of Physics, Kippenhahn describes how and why he moved from mathematics to physics: 1 https://pubapps.bucknell.edu/static/aeshaffer/v1/

159

160

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products I also first went to Biermann2 and showed him what I’d done, on novae, which was something which he immediately found out was complete nonsense. And so he suggested to me, I should do plasma physics, so I went back to Bamberg and read papers on plasma physics. I got an idea about how magnetic variables would work, I made a theory on magnetic variables, and I sent that theory to Biermann, who didn’t answer for months. Meanwhile, I looked into my theory. I found two mistakes in the theory. And then came Biermann’s letter telling me that there were three mistakes in the theory, each of which made the theory completely wrong. And so, it was all just crackpot; I had sent a theory which was nonsense. They are quite experienced with crackpots. But my advantage was that I had already found two of the mistakes, and after Biermann answered me, I understood also the third mistake. So I sent back a letter saying that I’m really very disappointed that I’d made these three mistakes, and I understand these mistakes, and so I see that my theory is completely nonsense, but I still, although that’s not very encouraging, I don’t want to give up theory of astrophysics, and asked him when I could come and talk to him again. They had plenty of crackpots, but they’ve never had crackpot who immediately gave in that he had made mistakes. Then they allowed me to come, and they asked me to give a talk … From that point on, I got firm connections with the Institute, and two years later, I got a job there.3

Kippenhahn’s article on the numerical range was written in German in 1951 [95] and translated into English in 2008 [96]. To see what he did, we consider a decomposition of square matrices. Every 𝑛 × 𝑛 matrix 𝐴 can be written as the sum of two Hermitian (self-adjoint) matrices: 𝐴 = 𝐻𝐴 + 𝑖𝐾𝐴 , where 𝐻𝐴 = 2 Ludwig

𝐴 + 𝐴⋆ 𝐴 − 𝐴⋆ and 𝐾𝐴 = . 2 2𝑖

(13.1)

Biermann, a physicist at the Max Planck Institute, München. of Rudolf Kippenhahn by Owen Gingerich on June 18, 1978, Niels Bohr Library & Archives, American Institute of Physics, College Park, Maryland, USA, https: //www.aip.org/history-programs/niels-bohr-library/oral-histories/5091 (accessed 9/07/2017). 3 Interview

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

161

Thinking back to the representation of complex numbers, we call 𝐻𝐴 the real part of 𝐴 and 𝐾𝐴 the imaginary part of 𝐴. This decomposition into real and imaginary parts with Hermitian matrices is unique. It is sometimes referred to as the Cartesian decomposition of a matrix. Looking at an example will make certain features of 𝐻𝐴 (and 𝐾𝐴 ) readily apparent. Example 13.1. Let 1+𝑖 𝐴=[ 1 3

𝑖 ]. −𝑖

Then 1−𝑖 𝐴 =[ −𝑖 ⋆

1

2 1 3 ] and a computation yields 𝐻𝐴 = [1 2 −𝑖 𝑖 3

1 3

+𝑖 0

].

As expected, 𝐻𝐴 is Hermitian. It should also be noted that though the diagonal entries of 𝐻𝐴 must be real (why?), the other entries need not be. The Cartesian decomposition of a matrix turns out to be incredibly helpful. Since every Hermitian matrix is normal, the spectral theorem applies to both 𝐻𝐴 and 𝐾𝐴 . But more is true: In Chapters 6 and 7 we looked at the numerical range of normal matrices and found that the numerical range must be the convex hull of the eigenvalues of the matrix. But, as the reader can check, the eigenvalues of a Hermitian matrix must be real and therefore the numerical range is the closed interval that ranges from the minimum eigenvalue to the maximum eigenvalue. We collect these observations below. Theorem 13.2. The numerical range of a Hermitian matrix 𝐻 is a closed interval on the real line. The lower endpoint is the minimum eigenvalue of 𝐻, and the upper endpoint is the maximum eigenvalue. This relatively simple theorem is the second result in Kippenhahn’s paper, and it is what makes everything else work. Using the fact that the numerical range of a matrix 𝐴 is a compact and convex set, Kippenhahn’s idea was to “box in” the numerical range in order to locate the boundary points of 𝑊(𝐴) in the (𝑥, 𝑦)-plane. Once he has the points with maximum and minimum real parts, he rotates the curve to find other points and then he rotates back. Here is how that works [96, Theorem 9].

162

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

Theorem 13.3. If 𝐴 = 𝐻𝐴 + 𝑖𝐾𝐴 and 𝐻𝐴 has eigenvalues 𝛼1 ≤ 𝛼2 ⋯ ≤ 𝛼𝑛 while 𝐾𝐴 has eigenvalues 𝛽1 ≤ 𝛽2 ⋯ ≤ 𝛽𝑛 , then the points 𝑧 = 𝑥 +𝑖𝑦 ∈ 𝑊(𝐴) lie in the interior or on the boundary of the rectangle constructed by the lines 𝑥 = 𝛼1 , 𝑥 = 𝛼𝑛 , 𝑦 = 𝛽1 , and 𝑦 = 𝛽𝑛 . The sides of the rectangle share either one point (with multiplicity that might be greater than 1) or one closed interval with the boundary of 𝑊(𝐴). The sides of the rectangle are thus the support lines we met back in Chapter 10. In case the matrices represent operators in the class 𝒮𝑛 , Lemma 10.5 shows that there cannot be an entire line segment contained in the boundary. As we repeat this procedure, rotating the matrix, we see the curve bounding 𝑊(𝐴). To make that precise, note that we get a vertical support line of 𝑊(𝐴) from 𝑥 = 𝑀𝑒 (𝐴) ∶= the maximum eigenvalue of 𝐻𝐴 . By Theorem 6.2, we know that rotating the matrix has the effect of rotating 𝑊(𝐴). So, if we want other support lines, we rotate the numerical range by multiplying by 𝑒−𝑖𝜑 and then we vary 𝜑. Thus, we are looking for a support line of 𝑒−𝑖𝜑 𝐴, and we rotate that back to get a support line of 𝐴; see Figure 13.1. We know how to get a support line for 𝑊(𝑒−𝑖𝜑 𝐴): 𝑥 = 𝑀𝑒 (𝑒−𝑖𝜑 𝐴) = the maximum eigenvalue of 𝐻𝑒−𝑖𝜑 𝐴 . Since the decomposition (13.1) is unique and 𝑒−𝑖𝜑 𝐴 = ((cos 𝜑)𝐻𝐴 + (sin 𝜑)𝐾𝐴 ) + 𝑖 ((cos 𝜑)𝐾𝐴 − (sin 𝜑)𝐻𝐴 ) , we see that 𝐻𝑒−𝑖𝜑 𝐴 = (cos 𝜑)𝐻𝐴 + (sin 𝜑)𝐾𝐴 . We are looking for the maximum eigenvalue of 𝐻𝑒−𝑖𝜑 𝐴 , so we want the largest solution of 𝑑𝑒𝑡[(cos 𝜑)𝐻𝐴 + (sin 𝜑)𝐾𝐴 − 𝜆𝐼] = 0.

(13.2)

−𝑖𝜑

But now we need to rotate back: Rotating 𝑒 𝐴 back to 𝐴 by multiplying by 𝑒𝑖𝜑 will take care of the matrix, but we also need to rotate the line 𝑥 = 𝑀𝑒 (𝑒−𝑖𝜑 𝐴). To see what happens here, think of the coordinates (𝑥, 𝑦) as 𝑥 = 𝜌 cos 𝜃 and 𝑦 = 𝜌 sin 𝜃 (in polar coordinates) and rotate by 𝑒𝑖𝜑 to see that 𝑥 cos 𝜑 + 𝑦 sin 𝜑 − 𝑀𝑒 (𝑒−𝑖𝜑 𝐴) = 0

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products y

163

y

ϕ

x

x

Figure 13.1. Constructing a support line of 𝑊(𝐴).

is a support line of 𝑊(𝐴). As 𝜑 varies from 0 to 2𝜋, we obtain all support lines of 𝑊(𝐴). For each 𝜑, we see that (13.2) may have (finitely many) other solutions, giving rise to lines parallel to the support lines. Since the support lines are extremal, this yields the boundary of the numerical range. All this leads to Kippenhahn’s Theorem, Theorem 13.4 [96, Theorem 10]. But before we state this theorem, we need to introduce some terminology. We think of ℂ as embedded in ℙ2 (ℝ) by identifying 𝑥+𝑖𝑦 with (𝑥, 𝑦, 1). The latter space is embedded in ℙ2 (ℂ) in the natural way. In the complex plane, the real part of an algebraic curve 𝒞 in ℙ2 (ℂ) is then the part of the curve that lives in ℂ. If a curve 𝒞 is given by the homogeneous polynomial equation 𝑓(𝑥, 𝑦, 𝑧) = 0, the tangent lines 𝑢𝑥 + 𝑣𝑦 + 𝑤𝑧 = 0 to the curve 𝒞 satisfy another homogeneous polynomial equation Φ(𝑢, 𝑣, 𝑤) = 0. This polynomial, Φ, determines the dual curve. The degree of 𝑓 is the degree of the curve 𝒞, and the degree of Φ is the class of 𝒞. See also Section 15.12 for more details and examples. Theorem 13.4 (Kippenhahn). Given an 𝑛 × 𝑛 matrix 𝐴 = 𝐻𝐴 + 𝑖𝐾𝐴 , the degree-𝑛 polynomial 𝐿𝐴 (𝑢, 𝑣, 𝑤) = det[𝑢𝐻𝐴 + 𝑣𝐾𝐴 + 𝑤𝐼] = 0

(13.3)

determines a curve of class 𝑛 in homogeneous line coordinates in the complex plane. The convex hull of this curve is the numerical range of 𝐴. Thus, the dual of the curve defined by 𝐿𝐴 (𝑢, 𝑣, 𝑤) = 0 is an algebraic curve in ℙ2 (ℂ). The real part of this curve is called the boundary generating curve or the Kippenhahn curve and is denoted by 𝐶(𝐴).

164

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

The polynomial 𝐿𝐴 tells us about the spectrum and the numerical range. For example, 𝐿𝐴 (−1, −𝑖, 𝑤) = det[−𝐻𝐴 −𝑖𝐾𝐴 +𝑤𝐼] = det[𝑤𝐼−𝐴] is the characteristic polynomial of 𝐴. Also, for 𝜃 ∈ [0, 2𝜋), the polynomial 𝐿𝐴 (− cos 𝜃, − sin 𝜃, 𝑤) = det[𝑤𝐼 − 𝐻𝑒−𝑖𝜃 𝐴 ] is the one we see in (13.2), and its maximum zero is the signed distance from the origin to the support line it determines. As 𝜃 varies, the support lines yield the numerical range of 𝐴. Before moving on, try the Kippenhahn procedure to obtain support lines for the 3 × 3 Jordan block with zeros on the main diagonal 0 𝐽3 = [0 0

1 0 0

0 1] . 0

From this, you should be able to see that 𝑊(𝐽3 ) is a disk, centered at the origin, of radius 1/√2. The support lines provide one of the many ways to compute the boundary of the numerical range; see [127, Section 6] for the details and [11] for an algorithm that modifies this approach. Because the numerical range appears in many different fields, including differential equations, physics, and quantum computing, it is worth computing the numerical range when it is possible to do so. This leads to various questions: First, when is the numerical range of a matrix circular? Second, when is it elliptical? The elliptical range theorem provides the answers to these questions for 2 × 2 matrices, but what about larger square matrices? Since the second question contains the first, we turn to an answer to the second question in the very next case: a 3 × 3 matrix. Kippenhahn [96, Theorem 26] showed that a 3 × 3 matrix must have one of the following curves as its Kippenhahn curve. (1) The determinant in (13.3), 𝐿𝐴 , factors into three linear factors, and 𝐶(𝐴) consists of three points. (2) The determinant 𝐿𝐴 factors into a linear and a quadratic factor, and 𝐶(𝐴) is a point and an ellipse.

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

165

Figure 13.2. Poncelet 4-curves.

(3) The determinant 𝐿𝐴 is irreducible, and 𝐶(𝐴) is a curve of degree 4 with a double tangent and a cusp, and the boundary of 𝑊(𝐴) contains a flat portion. (4) The determinant 𝐿𝐴 is irreducible, and 𝐶(𝐴) is a curve of degree 6 consisting of an “oval” and a curve with three cusps in its interior.4 For matrices in the class 𝒮𝑛 , the numerical range will never be the convex hull of three points because we know the boundary curve is smooth. But it might look like one of the pictures that you see in Figure 13.2. The numerical range on the right looks elliptical, and you may be wondering where the point described in case 2 is. For the operators we consider, the point lies inside the ellipse, so when we take the convex hull of the boundary, we “cover up” the point. We look at case 2 above more closely in this section. Using Kippenhahn’s classification we are able to give necessary and sufficient conditions for the numerical range of a 3 × 3 matrix to be an 4 An oval 𝒪 has a precise meaning in projective geometry: It is a nonempty set of points in the projective plane with the property that no three points of 𝒪 are collinear and each point of 𝒪 lies on exactly one tangent. See, for example, Projective Geometry: From Foundations to Applications [14].

166

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

elliptical disk. To see how this works, consider the two matrices 𝑎 𝐴 = [0 0

𝑥 𝑏 0

𝑦 𝑎 𝑧 ] and 𝐴1 = [ 0 𝑐 0

𝑠 𝑏 0

0 0] , 𝑐

(13.4)

where we may choose 𝑠 ≥ 0 in the 2 × 2 principal minor by Lemma 6.3. We will now see how these two are connected. The matrix 𝐴1 is a direct sum of two matrices; that is, 𝑎 𝐴1 = [ 0

𝑠 ] ⊕ [𝑐] , 𝑏

(13.5)

and it is a general fact (as well as a good exercise) that the numerical range of a direct sum of matrices is the convex hull of the union of the numerical ranges of the smaller matrices. In this case, the elliptical range theorem tells us that the numerical range of 𝐴1 would be the convex hull of the point 𝑐 and an ellipse 𝐸𝑠 with foci 𝑎 and 𝑏 and minor axis of length 𝑠. So, the Kippenhahn curve 𝐶(𝐴) is an ellipse and a point if and only if 𝐶(𝐴) = 𝐶(𝐴1 ), and (using Kippenhahn’s classification) that happens if and only if 𝐿𝐴 = 𝐿𝐴1 . So, let us see when 𝐿𝐴 = 𝐿𝐴1 . We present a skeleton of the computations here, and the reader is invited to provide the details. First, we compute 𝐻𝐴 and 𝐾𝐴 and form the matrix 𝑢𝐻𝐴 + 𝑣𝐾𝐴 + 𝑤𝐼, obtaining 𝑢ℜ(𝑎) + 𝑣ℑ(𝑎) + 𝑤 ⎡ ⎢ ⎢ 𝑥 ⎢ (𝑢 + 𝑖𝑣) 2 ⎢ ⎢ ⎢ 𝑦 (𝑢 + 𝑖𝑣) ⎣ 2

(𝑢 − 𝑖𝑣)

𝑥 2

𝑢ℜ(𝑏) + 𝑣ℑ(𝑏) + 𝑤 (𝑢 + 𝑖𝑣)

𝑧 2

(𝑢 − 𝑖𝑣) (𝑢 − 𝑖𝑣)

𝑦 2 𝑧 2

𝑢ℜ(𝑐) + 𝑣ℑ(𝑐) + 𝑤

⎤ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎦

Similarly, we form the matrix 𝑢𝐻𝐴1 + 𝑣𝐾𝐴1 + 𝑤𝐼 recalling that we are choosing 𝑠 > 0, and we obtain

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

𝑢ℜ(𝑎) + 𝑣ℑ(𝑎) + 𝑤 ⎡ ⎢ ⎢ 𝑠 (𝑢 + 𝑖𝑣) ⎢ 2 ⎢ ⎢ 0 ⎣

(𝑢 − 𝑖𝑣)

𝑠

167

0

2

𝑢ℜ(𝑏) + 𝑣ℑ(𝑏) + 𝑤

0

0

𝑢ℜ(𝑐) + 𝑣ℑ(𝑐) + 𝑤

⎤ ⎥ ⎥ ⎥. ⎥ ⎥ ⎦

We may write 𝐿𝐴 and 𝐿𝐴1 as polynomials in 𝑤; that is, we write them as 𝐶3 (𝑢, 𝑣) + 𝐶2 (𝑢, 𝑣)𝑤 + 𝐶1 (𝑢, 𝑣)𝑤 2 + 𝑤 3 , where the 𝐶𝑗 (𝑢, 𝑣) denote homogeneous polynomials of degree 𝑗 in 𝑢 and 𝑣 for 𝑗 = 1, 2, 3. Now we compare the coefficients of the polynomials 𝐿𝐴 and 𝐿𝐴1 . A look at the matrices shows that the coefficients of 𝑤 3 and 𝑤 2 in 𝐿𝐴 and 𝐿𝐴1 are equal. Thus, we must consider the coefficient of 𝑤 and the constant terms. In what follows, we ignore terms that are obviously the same. From the coefficient of 𝑤, we obtain the first condition for 𝐿𝐴 to equal 𝐿𝐴1 : 𝑠2 = |𝑥|2 + |𝑦|2 + |𝑧|2 . Now we consider the constant terms, which are linear combinations of 𝑢2 𝑣, 𝑢𝑣 2 , 𝑢3 , and 𝑣3 . From 𝑢2 𝑣, we get 𝑠2 ℑ(𝑐) = ℑ(𝑎)|𝑧|2 + ℑ(𝑏)|𝑦|2 + ℑ(𝑐)|𝑥|2 − ℑ(𝑥𝑦𝑧).

(13.6)

2

Looking at the symmetry of the matrix in 𝑢 and 𝑣 from 𝑢𝑣 , we get 𝑠2 ℜ(𝑐) = ℜ(𝑎)|𝑧|2 + ℜ(𝑏)|𝑦|2 + ℜ(𝑐)|𝑥|2 − ℜ(𝑥𝑦𝑧),

(13.7)

and thus we can write the conditions as 𝑠2 𝑐 = 𝑎|𝑧|2 + 𝑏|𝑦|2 + 𝑐|𝑥|2 − 𝑥𝑦𝑧. Finally, we check the coefficients of 𝑢3 and 𝑣3 : The coefficient of 𝑢3 yields (13.7), and the coefficient of 𝑣3 yields (13.6). Since everything we did above is reversible, these conditions are necessary and sufficient for 𝐿𝐴 to equal 𝐿𝐴1 . Thus, we obtain the following theorem [90].

168

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

Theorem 13.5. Let 𝑎, 𝑏, and 𝑐 be complex numbers, and let 𝑥, 𝑦, and 𝑧 be complex numbers with at least one nonzero. Let 𝑎 𝐴 = [0 0

𝑥 𝑏 0

𝑦 𝑧] . 𝑐

Then the Kippenhahn curve 𝐶(𝐴) is a point that is an eigenvalue of 𝐴 and an ellipse if and only if one of the eigenvalues 𝜆 satisfies (|𝑥|2 + |𝑦|2 + |𝑧|2 )𝜆 = 𝑎|𝑧|2 + 𝑏|𝑦|2 + 𝑐|𝑥|2 − 𝑥𝑦𝑧.

(13.8)

If this condition is satisfied, then 𝐶(𝐴) is the union of 𝜆 and the ellipse having its foci at the other two eigenvalues of 𝐴. The length of the minor axis is 𝑠 = √|𝑥|2 + |𝑦|2 + |𝑧|2 . Earlier, we asked you to use Kippenhahn’s procedure to figure out the numerical range of the Jordan block, 𝐽3 , with zeros on the main diagonal. You can now use Theorem 13.5 to check your answer. In Chapter 10 (see Remark 10.2) we showed that if 𝐴 represents a compression of the shift operator, then no eigenvalue of 𝐴 can lie on the boundary of 𝑊(𝐴). Putting this together with Theorem 13.5 and ordering the zeros of the Blaschke product 𝐵 so that 𝜆 = 𝑐, we get the following corollary. Corollary 13.6. Let 𝑎 𝐴 = [0 0

𝑥 𝑏 0

𝑦 𝑧] 𝑐

with 𝑥 = √1 − |𝑎|2 √1 − |𝑏|2 , 𝑦 = −𝑏√1 − |𝑎|2 √1 − |𝑐|2 and 𝑧 = √1 − |𝑏|2 √1 − |𝑐|2 be the usual representation of a compression of the shift corresponding to a Blaschke product 𝐵1 with zeros at 𝑎, 𝑏, and 𝑐. Then 𝑊(𝐴) is an elliptical disk with foci at 𝑎 and 𝑏 if and only if (|𝑥|2 + |𝑦|2 + |𝑧|2 )𝑐 = 𝑎|𝑧|2 + 𝑏|𝑦|2 + 𝑐|𝑥|2 − 𝑥𝑦𝑧.

(13.9)

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

169

Proof. We know that condition (13.9) is satisfied if and only if 𝐶(𝐴) consists of an ellipse and a point that is an eigenvalue of 𝐴. Thus, (13.9) plus the fact that the eigenvalue must lie inside 𝑊(𝐴) imply that 𝑊(𝐴) is an elliptical disk. Now suppose that the boundary of 𝑊(𝐴), denoted 𝜕𝑊(𝐴) is an ellipse with foci at 𝑎 and 𝑏. Since 𝐴 represents 𝑆𝐵1 , we know (by Remark 10.2) that 𝑐 must lie inside the ellipse. Thus, taking 𝐴1 as in (13.5), we see that 𝑊(𝐴) = 𝑊(𝐴1 ) and 𝐿𝐴 and 𝐿𝐴1 factor. The linear factor is determined by the point 𝑐, and the quadratic factor is determined by the (same) ellipse. Therefore, 𝐿𝐴 = 𝐿𝐴1 and (13.9) must be satisfied. Surprisingly, the conditions in Theorem 13.5 say something about Blaschke products of degree 4 and composition. So what do they say? First, a bit of background is in order. While it is easy to compose two Blaschke products, it is much more difficult to look at a Blaschke product and see if it can be written as a composition of two nontrivial Blaschke products (necessarily of lower degree). When 𝐵 is a Blaschke product that can be written as a composition 𝐵 = 𝐶 ∘ 𝐷 with 𝐶 and 𝐷 Blaschke products of lower degree (but degree greater than 1) we say that 𝐵 is decomposable. Decomposition of polynomials and Blaschke products is closely connected to other interesting questions. The following are good references to learn more about these topics: [33], [36], [119], [131], and [132]. We focus on decomposing Blaschke products of degree 4. So, let 𝐵 be a Blaschke product of degree 4 and define 𝑐−𝑧 , where 𝑐 ∈ 𝔻. (13.10) 𝜓𝑐 (𝑧) = 1 − 𝑐𝑧 As we noted in Chapter 10, the points 𝐵 identifies are the same as those the Blaschke product 𝜓𝐵(0) ∘ 𝐵 =

𝐵(0) − 𝐵

1 − 𝐵(0)𝐵 identifies, and therefore there is no harm in assuming that 𝐵(0) = 0. −1 Further, 𝐵 = 𝐶 ∘ 𝐷 if and only if 𝐵 = 𝐶 ∘ 𝜓𝐷(0) ∘ 𝜓𝐷(0) ∘ 𝐷. Thus, 𝐵 = 𝐶 ∘ 𝐷 if and only if there exist Blaschke products 𝐶1 and 𝐷1 with 𝐷1 (0) = 0, 𝐵 = 𝐶1 ∘ 𝐷1 and the degree of 𝐷1 equal to the degree of 𝐷. Since we assume 𝐵(0) = 0 and 𝐷1 (0) = 0, we also have 𝐶1 (0) = 0. Thus,

170

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

Figure 13.3. The Blaschke ellipse for 𝐵 = 𝐶 ∘ 𝐷.

when we consider decomposable Blaschke products of degree 4 we need only consider those of the form 𝛼−𝑧 𝑐−𝑧 𝐵(𝑧) = (𝑧 ( (13.11) )) ∘ (𝑧 ( )) 1 − 𝛼𝑧 1 − 𝑐𝑧 with 𝛼 and 𝑐 in 𝔻. It will also be useful to look at the Blaschke product 𝐵1 defined by 𝐵(𝑧) = 𝑧𝐵1 (𝑧). Before we begin proving Theorem 13.8, we suggest using the applet 5 , Compose Tool to check the result. In Figure 13.3, we used compose tool to plot 𝐵 = 𝐶∘𝐷 for Blaschke products 𝐶 with zeros at 0 and 0.2−0.1𝑖 and 𝐷 with zeros at 0 and 0.1 − 0.3𝑖. The boundary curve certainly appears to be an ellipse, and we prove that the region it bounds is elliptical and the foci of the boundary ellipse occur at two of the zeros of 𝐵. Here is one more useful fact. The proof is strictly computational and is left as an exercise. 𝑐−𝑎 , Proposition 13.7. Let 𝑎, 𝑐 ∈ 𝔻. For 𝜓𝑐 (𝑎) = 1 − 𝑐𝑎 (1 − |𝑎|2 )(1 − |𝑐|2 ) 2 1 − |𝜓𝑐 (𝑎)| = . |1 − 𝑐𝑎|2 It is now time to show the connection between function theory (when a degree-4 Blaschke product can be decomposed as the composition of 5 https://pubapps.bucknell.edu/static/aeshaffer/v1/

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

171

two degree-2 Blaschke products) and geometry (the geometry of the numerical range). Theorem 13.8. Let 𝐵1 be a degree-3 Blaschke product with zeros 𝑎, 𝑏, and 𝑐, and let 𝐵 be the Blaschke product defined by 𝐵(𝑧) = 𝑧𝐵1 (𝑧). Then 𝐵 is decomposable if and only if the numerical range of the matrix representing the compressed shift operator 𝑆𝐵1 is an elliptical disk. Proof. We use the following computations for both directions of the proof. The matrix 𝐴 that we are using to represent 𝑆𝐵1 was given in (9.14) and satisfies 𝑥 = √1 − |𝑎|2 √1 − |𝑏|2 , 𝑦 = −𝑏√1 − |𝑎|2 √1 − |𝑐|2 , and 𝑧 = √1 − |𝑏|2 √1 − |𝑐|2 . Thus, 𝑐|𝑥|2 + 𝑏|𝑦|2 + 𝑎|𝑧|2 − 𝑥𝑦𝑧 = 𝑐(1 − |𝑎|2 )(1 − |𝑏|2 ) + 𝑏|𝑏|2 (1 − |𝑎|2 )(1 − |𝑐|2 ) +𝑎(1 − |𝑏|2 )(1 − |𝑐|2 ) + 𝑏(1 − |𝑎|2 )(1 − |𝑏|2 )(1 − |𝑐|2 ) = 𝑐(1 − |𝑎|2 )(1 − |𝑏|2 ) +𝑎(1 − |𝑏|2 )(1 − |𝑐|2 ) + 𝑏(1 − |𝑎|2 )(1 − |𝑐|2 ),

(13.12)

and 𝑠2 = |𝑥|2 + |𝑦|2 + |𝑧|2 = (1 − |𝑎|2 )(1 − |𝑏|2 ) + |𝑏|2 (1 − |𝑎|2 )(1 − |𝑐|2 ) + (1 − |𝑏|2 )(1 − |𝑐|2 ). (13.13) First, suppose that 𝐵 can be decomposed to the form given in (13.11). Note that 𝐵(𝑧) = 𝐶(𝑧𝜓𝑐 (𝑧)), where 𝐶(𝑧) = 𝑧𝜓𝛼 (𝑧) for some 𝛼 ∈ 𝔻. Since 𝜓𝑐 is its own inverse, 𝐵(𝜓𝑐 (𝑧)) = 𝐵(𝑧). Thus, 𝜓𝑐 permutes the zeros of 𝐵. Since 𝜓𝑐 obviously permutes the zeros 0 and 𝑐, the zeros of 𝐵 are 0, 𝑐, 𝑎, and 𝑏 = 𝜓𝑐 (𝑎). We claim that the numerical range of 𝑆𝐵1 is bounded by an ellipse with foci at 𝑎 and 𝜓𝑐 (𝑎). So we must show that (|𝑥|2 + |𝑦|2 + |𝑧|2 )𝑐 = 𝑎|𝑧|2 + 𝑏|𝑦|2 + 𝑐|𝑥|2 − 𝑥𝑦𝑧.

172

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

Rotating the zeros by a unimodular constant 𝜆 will change neither the shape of the numerical range nor the equation above. Therefore, we may assume that 𝑐 ∈ ℝ. We first compute 𝑠2 and 𝑐|𝑥|2 + 𝑏|𝑦|2 + 𝑎|𝑧|2 − 𝑥𝑦𝑧 for this matrix. 𝑐−𝑎 Using the fact that 𝑏 = , we see that (13.13) and Proposition 13.7 1 − 𝑐𝑎 yield 𝑠2 = (1 − |𝑎|2 )(1 − |𝑐|2 )

(1 − |𝑎|2 ) + |𝑎 − 𝑐|2 + (1 − |𝑐|2 ) . |1 − 𝑐𝑎|2

(13.14)

Since we assume 𝑐 ∈ ℝ, simplifying yields 𝑠2 =

(1 − |𝑎|2 )(1 − |𝑐|2 ) (2 − 2𝑐ℜ(𝑎)). |1 − 𝑐𝑎|2

(13.15)

Equation (13.12) and Proposition 13.7 yield 𝑐|𝑥|2 + 𝑏|𝑦|2 + 𝑎|𝑧|2 − 𝑥𝑦𝑧 (1 − |𝑎|2 )2 (1 − |𝑐|2 ) (1 − |𝑎|2 )(1 − |𝑐|2 )2 =𝑐 +𝑎 2 |1 − 𝑐𝑎| |1 − 𝑐𝑎|2 +𝑏(1 − |𝑎|2 )(1 − |𝑐|2 ) (1 − |𝑎|2 )(1 − |𝑐|2 ) = (𝑐(1 − |𝑎|2 ) + 𝑎(1 − |𝑐|2 ) + (𝑐 − 𝑎)(1 − 𝑎𝑐)) |1 − 𝑐𝑎|2 (1 − |𝑎|2 )(1 − |𝑐|2 ) =𝑐 (2 − 2𝑐ℜ(𝑎)) = 𝑐𝑠2 . |1 − 𝑐𝑎|2 By Theorem 13.5, the Kippenhahn curve 𝐶(𝐴) is a point—a point that must be an eigenvalue of 𝐴—and an ellipse. In Chapter 10 we saw that the eigenvalues of 𝐴 can never lie on the boundary when 𝐴 represents an operator in 𝒮𝑛 (see Remark 10.2). We conclude that the point must lie in the interior of the numerical range. Thus, if 𝐵 is decomposable, then the numerical range of 𝑆𝐵1 is an elliptical disk. In the other direction, we assume that the numerical range is an elliptical disk and show that this implies that the zeros of 𝐵 are 0, 𝑎, 𝑐, and 𝜓𝑐 (𝑎). If we can show that these are the zeros of 𝐵, then the zeros are the same as those of the Blaschke product (𝑧(

𝑧−𝛽 1 − 𝛽𝑧

)) ∘ (𝑧(

𝑐−𝑧 )), 1 − 𝑐𝑧

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

173

where we take 𝛽 = 𝑎(𝑐 − 𝑎)/(1 − 𝑐𝑎). Therefore, 𝐵 would be decomposable. So we assume the zeros of 𝐵 are 0, 𝑎, 𝑏, and 𝑐 and the numerical range is elliptical. In this case, Theorem 13.5 shows that one of the eigenvalues, 𝜆, must satisfy 𝑠2 𝜆 = 𝑐|𝑥|2 + 𝑏|𝑦|2 + 𝑎|𝑧|2 − 𝑥𝑦𝑧. We note that reordering the zeros of 𝐵1 in the matrix 𝐴 produces a matrix that represents the operator 𝑆𝐵1 and therefore has the same numerical range. Thus, we may assume that the point is 𝑐, and then we show that 𝑏 = 𝜓𝑐 (𝑎). To see this, use (13.12) and (13.13) to check that 𝑐|𝑏|2 (1 − |𝑎|2 ) + 𝑐(1 − |𝑏|2 ) = 𝑎(1 − |𝑏|2 ) + 𝑏(1 − |𝑎|2 ). Solving for 𝑐, we obtain 𝑐=

𝑎(1 − |𝑏|2 ) + 𝑏(1 − |𝑎|2 ) . 1 − |𝑎𝑏|2

(13.16)

Therefore, 𝑏 − 𝑎|𝑏|2 − 𝑏|𝑎|2 + 𝑎|𝑎𝑏|2 𝑐−𝑎 = = 𝑏. 1 − 𝑐𝑎 1 − |𝑎|2 − 𝑎𝑏 + 𝑎𝑏|𝑎|2 This implies that 𝐵 is decomposable, completing the proof. 𝜓𝑐 (𝑎) =

Everything we have done here produces complete information about the ellipse. The proof shows that if the Blaschke product 𝐵 is of the form (13.11), it has zeros 0, 𝑎, 𝑐, and 𝜓𝑐 (𝑎). The associated Blaschke curve is an ellipse with foci 𝑎 and 𝑏 = 𝜓𝑐 (𝑎), and the length of the minor axis is the value 𝑠 given by (13.14). Recall that the length of the minor axis of the (elliptical) numerical range of a 2×2 matrix 𝐴 with eigenvalues 𝑎 and 𝑏 is given by the elliptical range theorem (Theorem 6.1), and it is (tr(𝐴⋆ 𝐴) − |𝑎|2 − |𝑏|2 )

1/2

.

You can check that for the 3 × 3 matrix that we considered in Theorem 13.8 the length of the minor axis is 1/2

𝑠 = (tr(𝐴⋆ 𝐴) − |𝑎|2 − |𝑏|2 − |𝑐|2 )

.

174

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

Recalling that a Blaschke product 𝐵 of degree 𝑛 defines a Blaschke curve that is inscribed in polygons with vertices at the solutions of 𝐵(𝑧) = 𝜆 for each 𝜆 ∈ 𝕋, we obtain a corollary of Theorem 13.8. Corollary 13.9. Let 𝐵 be a Blaschke product of degree 4. Then the Blaschke curve associated with 𝐵 is an ellipse if and only if 𝐵 is decomposable. What happens if the numerical range is a circular disk? In the case under discussion, the boundary of the disk is a Poncelet circle and that circle is inscribed in a quadrilateral. So let us ask a different question: If a circle 𝐶 is inscribed in a quadrilateral that is itself circumscribed by a circle, what (if anything) can we say? First, we may assume that the circumscribing circle is the unit circle— that is just a matter of rescaling. This Poncelet 4-circle 𝐶 is unique: If we can think of “another” circle, 𝐶1 , with the same center as 𝐶 that is also a Poncelet 4-circle, then 𝐶1 = 𝐶. But we can think of another one! We set things up so that we have a 4 × 4 matrix 𝐴 with eigenvalues 𝑎 and 𝑏 = 𝑎, and we choose 𝑐 ∈ 𝔻 so that 𝜓𝑐 (𝑎) = 𝑎. What is the radius of the circle? Since 𝑎 and 𝑏 = 𝜓𝑐 (𝑎) are equal, (13.16) shows that 2𝑎 . 𝑐= 1 + |𝑎|2 (If you are uncomfortable with how we chose 𝑐 here, checking that 𝑐−𝑎 =𝑎 1 − 𝑐𝑎 will make you feel better.) Using (13.13), we obtain the following formula involving the length of the minor axis 1 + |𝑎|2 1 = . 𝑠2 2(1 − |𝑎|2 )2

(13.17)

But 𝑠, in this case, is the diameter of the inscribed circle: 𝑠 = 2𝑟. Thus, we know everything about circles inscribed in quadrilaterals that are themselves inscribed in the unit circle: Letting 𝑟 denote the radius of the inscribed circle, 𝑎 the center, and noting that 𝑅 = 1 is the radius of the circumscribing circle, we obtain 2(1 + |𝑎|2 ) 1 1 1 = = + . 𝑟2 (1 − |𝑎|2 )2 (1 − |𝑎|)2 (1 + |𝑎|)2

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

175

This is a very beautiful result, and we are not the first to notice it. This was proven by Fuss in 1798 and is a natural extension of Chapple’s formula, which we established in Chapter 4. Theorem 13.10 (Fuss’s theorem). If a circle of radius 𝑟 is inscribed in a quadrilateral that is itself inscribed in a circle of radius 𝑅 and if 𝑑 denotes the distance between the centers of the circles, then 1 1 1 + = 2. 𝑟 (𝑅 − 𝑑)2 (𝑅 + 𝑑)2 For further information on closure theorems involving circles, see [28] and [91]. There are so many nice consequences of Theorem 13.8 that it is difficult not to include them all! We mention two more here. If we are given two arbitrary points 𝑎 and 𝑏 in 𝔻, you may be wondering if there is a Poncelet 4-ellipse with 𝑎 and 𝑏 as foci. We can answer that question now: We can always find 𝑐 with 𝜓𝑐 (𝑎) = 𝑏. Now let 𝐵 be the degree-4 Blaschke product with zeros 0, 𝑐, 𝑎, and 𝜓𝑐 (𝑎). As you can check, 𝐵 can be written as a composition of two degree-2 Blaschke products, each vanishing at 0. Then Corollary 13.9 says that the Blaschke curve associated with this Blaschke product is an ellipse with foci at 𝑎 and 𝑏 = 𝜑𝑐 (𝑎). The Poncelet curves that we have considered thus far have all been inscribed in convex polygons. However, Poncelet also considered ellipses inscribed in nonconvex polygons. Thus, it will be useful to have one more definition that clarifies this distinction. Let us call a Poncelet ellipse contained in 𝔻 a convex Poncelet ellipse if one of its circumscribing polygons inscribed in 𝕋 is convex. You should convince yourself that this implies that all such circumscribing polygons are convex. Then we have the following theorem [57], [114]. Theorem 13.11. Every convex Poncelet ellipse is a Blaschke ellipse. This theorem can be proven using Kippenhahn’s work and Bézout’s theorem; see [82] for an interesting discussion of Bézout’s theorem. In particular, the two aforementioned consequences of Theorem 13.8 are these: Every Poncelet 4-ellipse is a Blaschke ellipse, and given two arbitrary points 𝑎 and 𝑏 in 𝔻, there is a unique Poncelet 4-ellipse with foci 𝑎

176

Chapter 13. Kippenhahn’s Curve, Blaschke’s Products

and 𝑏. In any event, it is now clear that Blaschke products and Poncelet ellipses know a lot about each other. For related results we suggest the reader consult Fujimura’s paper [51] where Theorem 13.8 first appeared6 and [66] where the proof of Theorem 13.8 that we have given here appears.

6 While it looks like the length of the minor axis in Fujimura’s theorem is different than ours, a nontrivial computation or the fact that Poncelet 4-ellipses with given foci are unique, shows that the equations are the same.

Chapter

14

Iteration, Ellipses, and Blaschke Products In Chapter 4 we saw that degree-3 Blaschke products give rise to Poncelet ellipses and, in fact, every Poncelet 3-ellipse is a Blaschke ellipse. In Chapter 10 Blaschke products of higher degree were associated with curves that have the Poncelet property (see Theorem 10.6), but the curves were not always ellipses. Finally, in Chapter 13 we saw that Blaschke products of degree 4 give rise to Poncelet ellipses if and only if the Blaschke product is decomposable. Are there classes of Blaschke products of degree greater than 4 for which the associated Blaschke curve, obtained by forming polygons with vertices at the points identified by the Blaschke product, is an ellipse? Try to find Blaschke ellipses inscribed in hexagons or other polygons using the applet ,1 Compose Tool before reading on. It is not easy to find them! However, it turns out that we can find a class of Blaschke products that always gives rise to ellipses; a description of the class is connected to disk automorphisms and an interesting theorem of Frantz that connects these automorphisms to ellipses. We begin by describing these functions and demonstrating Frantz’s theorem. For 𝜙 ∈ ℝ and 𝑎 ∈ 𝔻 we define the disk automorphism 𝑎−𝑧 , (14.1) 𝑀𝑎,𝜙 (𝑧) = 𝑒𝑖𝜙 1 − 𝑎𝑧 which we recognize as one factor of a Blaschke product and a generalization of the function defined in (13.10), namely, 𝜓𝑐 = 𝑀𝑐,0 . Thus, it follows that 𝑀𝑎,𝜙 ∶ 𝔻 → 𝔻, 𝑀𝑎,𝜙 (𝔻) = 𝔻 and 𝑀𝑎,𝜙 (𝕋) = 𝕋. It is also 1 https://pubapps.bucknell.edu/static/aeshaffer/v1/

177

178

Chapter 14. Iteration, Ellipses, and Blaschke Products

easy to check that 𝑀𝑎,𝜙 is an invertible map from 𝔻 onto 𝔻 and its inverse is the disk automorphism 𝑀𝑎𝑒𝑖𝜙 ,2𝜋−𝜙 . Functions of the form (14.1) make up the full class of analytic disk automorphisms, as we saw earlier (see p. 23). Solving 𝑀𝑎,𝜙 (𝑧) = 𝑧 we see that disk automorphisms that are not the identity have at most two fixed points. In fact, if an automorphism is not the identity automorphism, denoted 𝑖𝑑, the possible cases are the following: • There is one fixed point in 𝔻 (and one outside 𝔻), in which case the disk automorphism is called elliptic. • There is exactly one fixed point (of multiplicity 2) on 𝕋, in which case we call the disk automorphism parabolic. • There are exactly two distinct fixed points, both on 𝕋, and the disk automorphism is called hyperbolic. While all three types of disk automorphisms are interesting, we restrict ourselves to elliptic automorphisms in this text. We show how the parameters determine the classification for the elliptic case. Lemma 14.1. For 𝑎 ∈ 𝔻 and 𝜙 ∈ [0, 2𝜋), the nontrivial disk automorphism 𝑀𝑎,𝜙 is elliptic if and only if |𝑎| < | cos(𝜙/2)| or 𝑎 = 0. Proof. First, suppose that 𝑎 ≠ 0. If we let 𝑐 = 𝑖𝑒𝑖𝜙/2 = 𝛼 + 𝑖𝛽 with 𝛼, 𝛽 real and 𝑑 = −𝑎𝑐, then 𝑒𝑖𝜙 = −𝑐/𝑐 and 𝑎 = −𝑑/𝑐. This implies that 𝑐𝑧 + 𝑑 𝑎−𝑧 = 𝑀𝑎,𝜙 (𝑧) = 𝑒𝑖𝜙 with |𝑑| = |𝑎| and 𝛽 = cos(𝜙/2). 1 − 𝑎𝑧 𝑑𝑧 + 𝑐 To find the fixed points we solve 𝑀𝑎,𝜙 (𝑧) = 𝑧 and get 𝑧𝑗 =

𝑖𝛽 ± √|𝑑|2 − 𝛽 2

for 𝑗 = 1, 2. 𝑑 We claim that 𝑀𝑎,𝜙 is elliptic if and only if |𝑑| = |𝑎| < | cos(𝜙/2)| = |𝛽|. As the reader can check, if |𝑑| ≥ |𝛽|, then |𝑧1 | = |𝑧2 | = 1 (and, in addition, 𝑧1 = 𝑧2 when |𝑑| = |𝛽|). Thus, 𝑀𝑎,𝜙 is not elliptic. Conversely, if |𝑑| < |𝛽|, then 𝑧𝑗 = (𝑖/𝑑) (𝛽 ± √𝛽 2 − |𝑑|2 ) for 𝑗 = 1, 2, 𝑧1 ≠ 𝑧2 , and |𝑧1 𝑧2 | = 1. Thus, the two fixed points lie on a ray emanating from the origin and

Chapter 14. Iteration, Ellipses, and Blaschke Products

179

are symmetric with respect to the unit circle. In particular, |𝑧𝑗 | ≠ 1 for 𝑗 = 1, 2. Consequently, 𝑀𝑎,𝜙 has exactly one fixed point in 𝔻. So 𝑀𝑎,𝜙 is elliptic, completing the proof of the claim. Finally, if 𝑎 = 0, then 𝑀0,𝜙 ≠ 𝑖𝑑 is a rotation centered at the origin and thus is always elliptic. Note that this proof essentially handles all three cases; the reader should work out the details for the hyperbolic and parabolic cases. Frantz’s theorem will show that the name “elliptic” is a good choice → denotes the when we discuss elliptic disk automorphisms. Below, ← 𝑢𝑣 line through two distinct points 𝑢 and 𝑣. Theorem 14.2 (Frantz). Let 𝑎 ∈ 𝔻 and 𝜙 ∈ [0, 2𝜋) be such that 𝑀𝑎,𝜙 is ←−−−−−→ an elliptic disk automorphism. Then the set of lines {𝑧𝑀𝑎,𝜙 (𝑧) ∶ 𝑧 ∈ 𝕋} is 𝜙

precisely the set of tangents to the ellipse 𝐸𝑎 with foci 𝑎 and 𝑎𝑒𝑖𝜙 and major axis of length 2 sin(𝜙/2). ←−−−−−→ When the ellipse is degenerate (that is, when 𝜙 = 0), the lines 𝑧𝑀𝑎,𝜙 (𝑧) are concurrent at 𝑎. Frantz’s theorem is actually more general; see [49, p. 780]; he allows for arbitrary functions of the form (14.1) without the restriction that |𝑎| < | cos(𝜙/2)|. In fact, 𝑎 can be outside 𝔻. With this more gen←−−−−−→ eral hypothesis, he interprets a line 𝑧𝑀𝑎,𝜙 (𝑧) to be the tangent line to 𝕋 at 𝑧 in case 𝑧 is a fixed point of 𝑀𝑎,𝜙 , and the conclusion of the theo𝜙

rem is that the lines are tangent to a conic 𝐸𝑎 of eccentricity |𝑎|. For our purposes the restricted statement will suffice. Proof. We consider the degenerate case first; that is, 𝜙 = 0. Since 𝑀𝑎,0 is elliptic, we know there is no fixed point on 𝕋. Therefore, for 𝑧 ∈ 𝕋 the three points 𝑧, 𝑎, and 𝑀𝑎,0 (𝑧) are distinct. We need to show that they are collinear. Since 𝑧 ∈ 𝕋, we have 𝑧 = 1/𝑧, and this yields 𝑧 − 𝑀𝑎,0 (𝑧) 𝑧 𝑧 1 𝑧 = = + +( ) ∈ ℝ. 𝑧−𝑎 𝑧 − 𝑎 1 − 𝑎𝑧 𝑧−𝑎 𝑧−𝑎 Thus, the angle between 𝑧 − 𝑀𝑎,0 (𝑧) and 𝑧 − 𝑎 is 0 or 𝜋 and the points are collinear.

180

Chapter 14. Iteration, Ellipses, and Blaschke Products






 



 

Figure 14.1. Illustration for the proof of Theorem 14.2.

Now assume that 𝜙 ∈ (0, 2𝜋). Note that 𝑀𝑎,𝜙 (𝑧) =

𝑒𝑖𝜙 𝑎 − 𝑒𝑖𝜙 𝑧

.

1 − 𝑎𝑒𝑖𝜙 (𝑒𝑖𝜙 𝑧)

So we again choose 𝑧 ∈ 𝕋 and introduce the notation 𝑢 = 𝑒𝑖𝜙 𝑧, 𝑣 = 𝑀𝑎,0 (𝑧), 𝑤 = 𝑀𝑎,𝜙 (𝑧) = 𝑒𝑖𝜙 𝑣, and 𝑎′ = 𝑒𝑖𝜙 𝑎. As shown in the degener→ by the ate case above, 𝑧, 𝑎, and 𝑣 are collinear. A rotation of the line ← 𝑧𝑣 ← → angle 𝜙 around the origin yields the line 𝑢𝑤 that contains the point 𝑎′ . Thus, (14.2) |𝑣 − 𝑧| = |𝑤 − 𝑢|, |𝑢 − 𝑧| = |𝑤 − 𝑣|, and we have two congruent triangles △𝑧𝑣𝑤 ≅ △𝑤𝑢𝑧 with |𝑤 − 𝑎′ | = |𝑒𝑖𝜙 𝑣 − 𝑒𝑖𝜙 𝑎| = |𝑣 − 𝑎|.

(14.3) → Letting 𝑝 denote the point of intersection of the line parallel to ← 𝑣𝑤 ← → through 𝑎 with the line 𝑧𝑤, we obtain similar triangles, △𝑧𝑎𝑝 ∼ △𝑧𝑣𝑤.

(14.4)

Chapter 14. Iteration, Ellipses, and Blaschke Products

181

To find the last pair of similar triangles we calculate |𝑤−𝑝|

|𝑤 − 𝑝| |𝑤 − 𝑝| = = ′ |𝑤 − 𝑎 | |𝑣 − 𝑎|

|𝑣−𝑧| |𝑣−𝑎| |𝑣−𝑧|

|𝑤−𝑝|

=

|𝑣−𝑧| |𝑤−𝑝|

=

|𝑤 − 𝑧| |𝑤 − 𝑧| = , |𝑣 − 𝑧| |𝑤 − 𝑢|

|𝑤−𝑧|

where the first and fifth equality follow from (14.3) and the third from (14.4). So we also have △𝑤𝑝𝑎′ ∼ △𝑤𝑧𝑢. Now we can show that 𝑝 is on

(14.5)

𝜙 𝐸𝑎 :

|𝑝 − 𝑎| + |𝑝 − 𝑎′ | |𝑧 − 𝑎| |𝑤 − 𝑎′ | = |𝑣 − 𝑤| + |𝑢 − 𝑧| |𝑧 − 𝑣| |𝑤 − 𝑢| |𝑧 − 𝑎| + |𝑎 − 𝑣| = |𝑢 − 𝑧| |𝑧 − 𝑣| = |𝑢 − 𝑧| = 2 sin(𝜙/2).

(by (14.4) and (14.5)) (by (14.3) and (14.2))

Finally, we use (14.5), (14.3), and (14.4) again to argue that ∠𝑤𝑝𝑎′ = ∠𝑤𝑧𝑢 = ∠𝑧𝑤𝑣 = ∠𝑧𝑝𝑎. ←−−−−−→ 𝜙 Thus, we have established that 𝑧𝑀𝑎,𝜙 (𝑧) is tangent to 𝐸𝑎 at 𝑝. It remains to be shown that for 𝜙 ∈ (0, 2𝜋), every tangent line to ←−−−−−→ 𝜙 𝜙 𝐸𝑎 is of the form 𝑧𝑀𝑎,𝜙 (𝑧) for some 𝑧 ∈ 𝕋. First, note that 𝐸𝑎 ⊆ 𝔻: 𝜙

If this were not the case, then 𝐸𝑎 would contain points outside 𝔻 or it would be tangent to 𝕋 in at least one point, say 𝑧 ∈ 𝕋. In the former ←−−−−−→ 𝜙 case we would have points 𝑧 ∈ 𝕋 that are inside 𝐸𝑎 and 𝑧𝑀𝑎,𝜙 (𝑧) could 𝜙

not possibly be tangent to the ellipse. In the latter case, 𝐸𝑎 and 𝕋 would have a common tangent line at the point 𝑧 ∈ 𝕋 of contact. But, at this ←−−−−−→ 𝜙 point no line 𝑧𝑀𝑎,𝜙 (𝑧) can be tangent to 𝐸𝑎 . In either case, we have a 𝜙

contradiction to the first part of this proof. Thus, 𝐸𝑎 ⊆ 𝔻 as claimed. 𝜙 Let ℓ be a tangent line to 𝐸𝑎 . Since the point of tangency is in 𝔻, the line ℓ intersects 𝕋 at two points 𝑧1 and 𝑧2 , neither of which is a fixed point of 𝑀𝑎,𝜙 . From 𝑧1 there are two different tangent lines to the el←−−1 −−−−−−→ ←−−−−−−−→ lipse. By the first part of this proof, the lines 𝑧1 𝑀𝑎,𝜙 (𝑧1 ) and 𝑀𝑎,𝜙 (𝑧1 )𝑧1 𝜙

are both tangent to 𝐸𝑎 and pass through 𝑧1 . We claim that these two lines

182

Chapter 14. Iteration, Ellipses, and Blaschke Products

−1 are distinct. If they were the same, then 𝑀𝑎,𝜙 (𝑧1 ) = 𝑀𝑎,𝜙 (𝑧1 ) and thus (2)

(2)

𝑀𝑎,𝜙 (𝑧1 ) = 𝑧1 . Thus, 𝑀𝑎,𝜙 would have two fixed points: the one associated with 𝑀𝑎,𝜙 and 𝑧1 , which would force it to be the identity. However, for 𝜙 ∈ (0, 2𝜋), the automorphism 𝑀𝑎,𝜙 is not self-inverse, and we have established a contradiction. Thus, the two lines are distinct and one of ←−−−−−−−→ ←−−−−−−−→ the two lines is ℓ; that is, ℓ = 𝑧1 𝑀𝑎,𝜙 (𝑧1 ) or ℓ = 𝑧2 𝑀𝑎,𝜙 (𝑧2 ). The elliptic disk automorphisms hold great promise in determining Blaschke curves that are ellipses. In general, however, the ellipses obtained from Theorem 14.2 are not Poncelet ellipses. So the question arises, when do these curves have the Poncelet property? As it turns out, this is easy to answer because we have a lot of information about disk automorphisms. To this end we call an elliptic disk automorphism canonical if it is of the form 𝑀0,𝜙 . We note that canonical elliptic disk automorphisms are simply counterclockwise rotations about the origin by an angle of 𝜙 − 𝜋 (or, equivalently, 𝜙 + 𝜋). Every elliptic disk automorphism 𝑀𝑎,𝜙 is conjugate to a canonical elliptic automorphism: There exists 𝜓 ∈ ℝ such that 𝑀𝑎,𝜙 = 𝑀𝑏,0 ∘ 𝑀0,𝜓 ∘ 𝑀𝑏,0 ,

(14.6)

where 𝑏 ∈ 𝔻 is the fixed point of 𝑀𝑎,𝜙 . Note that 𝑀𝑏,0 is an elliptic disk automorphism that is its own inverse. (𝑛) The 𝑛th iterate of 𝑀𝑎,𝜙 , denoted 𝑀𝑎,𝜙 , is the identity if and only if (𝑛)

𝑀0,𝜓 = 𝑀0,𝜋+𝑛(𝜋+𝜓) is the identity. The latter can be achieved for a positive integer 𝑛 if and only if 𝜓 is a rational multiple of 𝜋. Thus, we say that 𝑀𝑎,𝜙 has order 𝑛 if 𝜓 − 𝜋 = 2𝜋𝑝/𝑛 for some integers 𝑝 and 𝑛 satisfying 1 ≤ 𝑝 < 𝑛 and (𝑝, 𝑛) = 1. We also say that 𝑀𝑎,𝜙 has convex order 𝑛 if in addition we have 𝑝 = 1 or 𝑝 = 𝑛 − 1. Lemma 14.3. With the notation as above, suppose that 𝑀𝑎,𝜙 has order 𝑛 with 𝜓 − 𝜋 = 2𝜋𝑝/𝑛. Then for an arbitrary but fixed 𝑧 ∈ 𝕋, the polygon (𝑘) with vertices 𝑀𝑎,𝜙 (𝑧) for 𝑘 = 0, … , 𝑛 wraps 𝑝 times around the unit circle if 𝑝 < 𝑛/2 and 𝑛 − 𝑝 times if 𝑝 > 𝑛/2. In particular, the polygon is convex if and only if 𝑝 = 1 or 𝑝 = 𝑛 − 1. (𝑘)

Proof. Suppose that 𝑤 = 𝑀𝑎,𝜙 (𝑧) is the 𝑘th vertex of the polygon. We (𝑘+1)

show how to construct the next vertex, 𝑀𝑎,𝜙 (𝑧) = 𝑀𝑎,𝜙 (𝑤) (see the

Chapter 14. Iteration, Ellipses, and Blaschke Products

183

(-)π/ 



ψ ()
π/




ϕ(
) 

Figure 14.2. Construction of the next vertex.

illustration in Figure 14.2). By the last statement of Theorem 14.2, we ← → see that 𝑢 = 𝑀𝑏,0 (𝑤) is the intersection of 𝑤𝑏 with 𝕋. Thus, 𝑀𝑎,𝜙 (𝑤) = 𝑀𝑏,0 (𝑀0,𝜓 (𝑀𝑏,0 (𝑤))) = 𝑀𝑏,0 (𝑀0,𝜓 (𝑢)) = 𝑀𝑏,0 (𝑒𝑖2𝜋𝑝/𝑛 𝑢) , ←−−−−−→ so the point 𝑀𝑎,𝜙 (𝑤) is the intersection of 𝑀0,𝜓 (𝑢)𝑏 with 𝕋. In the next step, 𝑀0,𝜓 (𝑢) will take the place of 𝑢 and the construction is repeated. To close the polygon in 𝑛 steps, the points 𝑀0,𝜓 (𝑢), and thus 𝑀𝑎,𝜙 (𝑤), must increase their argument by 2𝑝𝜋, so the polygon wraps around the circle 𝑝 times in the positive direction. However, the polygon also closes if the argument is decreased by 2(𝑛 −𝑝)𝜋: The polygon wraps around the circle 𝑛 − 𝑝 times in the negative direction. Since the direction is irrelevant, we can choose a preferred direction. In the statement of the theorem, we have chosen the smaller of the two numbers. To consider specific examples, we can use the fixed point 𝑏 and rotation angle 𝜓 to find 𝑎 and 𝜙: Use (14.6) to see that 𝑎=

𝑏(1 + 𝑒𝑖𝜓 ) |𝑏|2 + 𝑒𝑖𝜓

and

𝜙 = arg (

|𝑏|2 + 𝑒𝑖𝜓 ). 1 + |𝑏|2 𝑒𝑖𝜓

(14.7)

In Figure 14.3 we use 𝑏 = 0.3 + 0.2𝑖 for the fixed point. For the figure

184

Chapter 14. Iteration, Ellipses, and Blaschke Products





ϕ
(
)





ϕ
(
) ()





ϕ
(
)



ϕ
()(
) (𝑘) Figure 14.3. The polygonal chain 𝑀𝑎,𝜙 (𝑧), 𝑘 = 0, … 7 with 𝑝 = 3 for an

elliptic disk automorphism of order 7 on the left and of convex order 7 on the right.

on the left, 𝜓1 = 13𝜋/7 (which leads to a rotation by the factor −𝑒𝑖𝜓1 = 𝑒𝑖(𝜓1 −𝜋) = 𝑒(3/7)2𝜋𝑖 ). For the figure on the right, we use 𝜓2 = 5𝜋/7. Using (14.7) we find the parameters of the two elliptic disk automorphisms 𝑀𝑎𝑗 ,𝜙𝑗 for 𝑗 = 1, 2 to be 𝑎1 = 0.45 + 0.43𝑖, 𝜙1 = −0.46, 𝑎2 = 0.31 − 0.14𝑖, and 𝜙2 = 2.23. In both cases we start with 𝑧 = 𝑒𝑖𝜋/12 . In what follows we consider Blaschke ellipses where we connect nonsuccessive points on 𝕋 that are identified by the Blaschke product, and this requires additional notation. Let 𝐵 be a Blaschke product of degree 𝑛. For 𝜆 ∈ 𝕋 we denote by 𝑧1 , … , 𝑧𝑛 the 𝑛 distinct points on 𝕋 that 𝐵 maps to 𝜆, as always, enumerated by increasing argument. For 1 ≤ 𝑝 < 𝑛 and (𝑛, 𝑝) = 1 we call an ellipse a Blaschke (𝑛, 𝑝)-ellipse associated with 𝐵 if it has the property that for every 𝜆 ∈ 𝕋 the ellipse is inscribed in the (not necessarily convex) polygon formed with the 𝑛 vertices 𝑧1 , … , 𝑧𝑛 by joining 𝑧𝑗 to 𝑧𝑗+𝑝 for 𝑗 = 1, … , 𝑛, where the indices are taken modulo 𝑛. All these ellipses will, of course, be Poncelet ellipses. We are now ready to connect the disk automorphisms to the Blaschke products. Recall that 𝜙 𝐸𝑎 denotes the ellipse appearing in Theorem 14.2. Lemma 14.4. Let 𝑎 = 0 or let 𝑎 ∈ 𝔻 and 𝜙 ∈ [0, 2𝜋) satisfy |𝑎| < | cos(𝜙/2)|. Then the following hold.

Chapter 14. Iteration, Ellipses, and Blaschke Products

185

• If 𝑀𝑎,𝜙 is of order 𝑛 > 1, there exists a Blaschke product 𝐵 of degree 𝑛 such that 𝐵 ∘ 𝑀𝑎,𝜙 = 𝐵. In addition, for every degree-𝑛 Blaschke product 𝐶 with 𝐶 ∘ 𝑀𝑎,𝜙 = 𝐶, 𝜙

there exists an integer 𝑝 with 1 ≤ 𝑝 < 𝑛 and (𝑛, 𝑝) = 1 such that 𝐸𝑎 is a Blaschke (𝑛, 𝑝)-ellipse associated with 𝐶.

• If 𝑀𝑎,𝜙 is of infinite order, there is no finite Blaschke product 𝐶 satisfying 𝐶 ∘ 𝑀𝑎,𝜙 = 𝐶 and no polygon inscribed in 𝕋 circumscribing 𝜙

𝐸𝑎 . Proof. Under the given assumption, Lemma 14.1 implies that 𝑀𝑎,𝜙 is an elliptic disk automorphism. We assume that 𝑀𝑎,𝜙 is of order 𝑛 > 1. If 𝑎 = 0, then 𝑀0,𝜙 (𝑧) = −𝑒𝑖𝜙 𝑧 = 𝑒𝑖(𝜙+𝜋) 𝑧. If we let 𝐵(𝑧) = 𝑧𝑛 , then 𝑛

𝐵 ∘ 𝑀0,𝜙 (𝑧) = (𝑒𝑖(𝜙+𝜋) 𝑧) = 𝐵(𝑧). We now assume that 𝑎 ≠ 0 and construct the sequence 𝑏1 ∶= 𝑀𝑎,𝜙 (𝑎) = 0, 𝑏2 ∶= 𝑀𝑎,𝜙 (𝑏1 ) = 𝑒𝑖𝜙 𝑎, … , (𝑛)

𝑏𝑛 ∶= 𝑀𝑎,𝜙 (𝑏𝑛−1 ) = 𝑀𝑎,𝜙 (𝑎) = 𝑎. We claim that none of the 𝑏𝑗 can be a fixed point of 𝑀𝑎,𝜙 . Suppose to the contrary that 𝑏𝑗 is a fixed point for some 𝑗. Then (𝑛−𝑗)

𝑏𝑗 = 𝑀𝑎,𝜙 (𝑏𝑗 ) = 𝑏𝑛 = 𝑎. Thus, 𝑎 is a fixed point and 𝑎 = 𝑀𝑎,𝜙 (𝑎) = 0, contrary to our assumption. This fact also implies that 𝑏𝑗 ≠ 𝑏𝑘 if 𝑘 ≠ 𝑗, for if this were not the (𝑘−𝑗)

(𝑘−𝑗)

case and, say, 𝑗 < 𝑘, then 𝑏𝑗 = 𝑏𝑘 = 𝑀𝑎,𝜙 (𝑏𝑗 ). Thus, 𝑀𝑎,𝜙 would have two fixed points in 𝔻 and would be the identity, contradicting the fact that the order of 𝑀𝑎,𝜙 is 𝑛 > 1. Define the Blaschke product 𝐵 by 𝑛

𝐵(𝑧) = ∏ 𝑗=1

𝑧 − 𝑏𝑗 1 − 𝑏𝑗 𝑧

.

By construction, 𝐵 and 𝐵 ∘ 𝑀𝑎,𝜙 have the same set of zeros and are both Blaschke products of degree 𝑛. Hence, 𝐵 ∘ 𝑀𝑎,𝜙 = 𝜇𝐵 for some 𝜇 ∈ 𝕋. Evaluating at the fixed point of 𝑀𝑎,𝜙 leads to the conclusion that 𝜇 = 1, and we have constructed a Blaschke product with the required property.

186

Chapter 14. Iteration, Ellipses, and Blaschke Products

Suppose that 𝐶 is a degree-𝑛 Blaschke product satisfying 𝐶∘𝑀𝑎,𝜙 = 𝐶. First, suppose that the disk automorphism is canonical, that is 𝑀𝑎,𝜙 (𝑧) = 𝑒𝑖2𝑝𝜋/𝑛 𝑧 for some integer 𝑝, where 1 ≤ 𝑝 < 𝑛 and (𝑛, 𝑝) = 1. Choose 𝑧0 ∈ 𝕋, and let 𝜆 = 𝐶(𝑧0 ). Now 𝐶(𝑣) = 𝜆 has the 𝑛 solutions 𝑣𝑗 = 𝑧0 𝑒𝑖2𝑗𝜋/𝑛 for 𝑗 = 0, … , 𝑛 − 1. Since 𝑀𝑎,𝜙 (𝑣𝑗 ) = 𝑧0 𝑒𝑖2(𝑗+𝑝)𝜋/𝑛 = 𝑣𝑗+𝑝 , 𝜙

Theorem 14.2 implies that 𝐸𝑎 is a Blaschke (𝑛, 𝑝)-ellipse (which is a circle in this canonical case). If the disk automorphism 𝑀𝑎,𝜙 with fixed point 𝑏 is not canonical, then there exists a disk automorphism 𝑀𝑏,0 with 𝑀𝑎,𝜙 = 𝑀𝑏,0 ∘ 𝑀0,𝜓 ∘ 𝑀𝑏,0 with 𝑀0,𝜓 canonical. Note that if 𝐶 ∘ 𝑀𝑎,𝜙 = 𝐶, then 𝐶 ∘ 𝑀𝑏,0 ∘ 𝑀0,𝜓 ∘ 𝑀𝑏,0 = 𝐶 and so (𝐶 ∘ 𝑀𝑏,0 ) ∘ 𝑀0,𝜓 = 𝐶 ∘ 𝑀𝑏,0 . Thus, we can apply the argument above to 𝐶 ∘ 𝑀𝑏,0 and 𝑀0,𝜓 . So, if (𝑣𝑗 ) is the sequence of points (ordered as usual according to increasing argument) satisfying 𝐶 ∘ 𝑀𝑏,0 (𝑣) = 𝜆, then 𝑀0,𝜓 (𝑣𝑗 ) = 𝑣𝑗+𝑝 for all 𝑗. Letting 𝑀𝑏,0 (𝑣𝑗 ) = 𝑤𝑗 , we see that 𝐶(𝑤𝑗 ) = 𝐶(𝑀𝑏,0 (𝑣𝑗 )) = 𝜆, and (𝑤𝑗 ) is the sequence of points identified by 𝐶. The order is maintained because 𝑀𝑏,0 is a single Blaschke factor. Thus, 𝑀𝑎,𝜙 (𝑤𝑗 ) = 𝑀𝑏,0 ∘ 𝑀0,𝜓 ∘ 𝑀𝑏,0 (𝑤𝑗 ) = 𝑀𝑏,0 ∘ 𝑀0,𝜓 (𝑣𝑗 ) = 𝑀𝑏,0 (𝑣𝑗+𝑝 ) = 𝑤𝑗+𝑝 . 𝜙 𝜙 −−−→ From Theorem 14.2, the line ← 𝑤− 𝑗 𝑤𝑗+𝑝 is tangent to the ellipse 𝐸𝑎 and 𝐸𝑎 is a Blaschke (𝑛, 𝑝)-ellipse, as claimed. Now assume that 𝑀𝑎,𝜙 has infinite order. Suppose to the contrary that there exists a Blaschke product 𝐶 such that 𝐶 ∘ 𝑀𝑎,𝜙 = 𝐶. Let 𝜆 ∈ 𝕋, and (𝑘)

denote one solution of 𝐶(𝑧) = 𝜆 by 𝑧0 . Then 𝑀𝑎,𝜙 (𝑧0 ) for 𝑘 ∈ ℤ+ are (𝑘)

(ℓ)

all solutions of 𝐶(𝑧) = 𝜆. If 𝑀𝑎,𝜙 (𝑧0 ) = 𝑀𝑎,𝜙 (𝑧0 ) for some 𝑘 > ℓ, then (𝑘−ℓ)

𝑀𝑎,𝜙

would have two fixed points—one of them on the unit circle and (𝑘−ℓ)

one in 𝔻, which would force 𝑀𝑎,𝜙 to be the identity and contradict the infinite order of 𝑀𝑎,𝜙 . Thus, no such Blaschke product 𝐶 exists. If for some positive integer 𝑛 there were to exist an 𝑛-gon inscribed in 𝜙 𝕋 and circumscribing 𝐸𝑎 , then by Poncelet’s theorem, every polygonal 𝜙 chain with vertices on 𝕋 circumscribing 𝐸𝑎 would close up in 𝑛 steps. By

Chapter 14. Iteration, Ellipses, and Blaschke Products

187

(𝑘)

Theorem 14.2, (𝑀𝑎,𝜙 (𝑧)), where 𝑧 ∈ 𝕋 and 𝑘 ≥ 0, is a vertex sequence of such a polygonal chain. This would force 𝑀𝑎,𝜙 to have finite order 𝑛, contradicting the assumption. We conclude that no circumscribing polygon with vertices on 𝕋 exists. We can now describe a large class of Blaschke products that gives rise 𝜙 to Blaschke ellipses of the form 𝐸𝑎 . Once we have done this, you will be able to find Blaschke ellipses inscribed in polygons with 𝑛 vertices on 𝕋 for each 𝑛 ≥ 3. Theorem 14.5. Let 𝐵 be a Blaschke product of degree 𝑛 ≥ 2, 𝑎 ∈ 𝔻, and 𝜙 ∈ [0, 2𝜋). The following are equivalent. 𝜙

(1) The ellipse 𝐸𝑎 is a Blaschke (𝑛, 1)-ellipse associated with 𝐵. (2) The disk automorphism 𝑀𝑎,𝜙 is elliptic of convex order 𝑛 and 𝐵 = 𝐵 ∘ 𝑀𝑎,𝜙 . (3) The disk automorphism 𝑀𝑎,𝜙 is elliptic of convex order 𝑛 with fixed point 𝑏 in 𝔻 and 𝑛 𝑏−𝑧 𝐵 =𝜏∘( ) , 1 − 𝑏𝑧 where 𝜏 is a disk automorphism. Proof. Before we begin, we remark that when 𝑛 = 2, we must have 𝜙 = 0 and conversely. In this case, the ellipse is a single point, and the proofs below cover this situation. We assume that (1) holds and will show that this implies (2). Since 𝐸𝑎,𝜙 is a Blaschke ellipse, it is contained in 𝔻. By Theorem 14.2 for 𝑧 ∈ 𝕋, ←−−−−−→ 𝜙 the line 𝑧𝑀𝑎,𝜙 (𝑧) is tangent to 𝐸𝑎 and thus 𝑀𝑎,𝜙 (𝑧) ≠ 𝑧. We conclude that the disk automorphism has no fixed point on 𝕋 and is therefore elliptic. For 𝜆1 ∈ 𝕋 there are 𝑧1 , … , 𝑧𝑛 ∈ 𝕋, ordered by increasing argument, such that 𝐵(𝑧𝑗 ) = 𝜆1 . The polygon 𝑃1 with these vertices circumscribes 𝜙

𝐸𝑎 . By Theorem 14.2, the vertex sequence of 𝑃1 also satisfies 𝑧𝑗+1 = 𝑀𝑎,𝜙 (𝑧𝑗 ) with indices taken modulo 𝑛. Since the polygon is convex and has exactly 𝑛 vertices, the disk automorphism 𝑀𝑎,𝜙 is of convex order 𝑛.

188

Chapter 14. Iteration, Ellipses, and Blaschke Products

Figure 14.4. 𝐵(𝑧) = ((0.5 + 0.5𝑖) − 𝑧)/(1 − (0.5 − 0.5𝑖)𝑧))

5

Now for each 𝑗, we have 𝐵 ∘ 𝑀𝑎,𝜙 (𝑧𝑗 ) = 𝐵(𝑧𝑗+1 ) = 𝜆1 . Using partial fractions, we write 𝑛 𝑚𝑗𝐵 𝐵(𝑧) = 𝛾𝐵 + ∑ 𝑧 − 𝑧𝑗 𝐵(𝑧) − 𝜆1 𝑗=1

for appropriate constants 𝛾𝐵 and 𝑚𝑗𝐵 and 𝑛 𝑚𝑗𝑀 𝐵 ∘ 𝑀𝑎,𝜙 (𝑧) = 𝛾𝑀 + ∑ 𝑧 − 𝑧𝑗 𝐵 ∘ 𝑀𝑎,𝜙 (𝑧) − 𝜆1 𝑗=1

for appropriate constants 𝛾𝑀 and 𝑚𝑗𝑀 . Clearing the denominators, we get two polynomials of degree 𝑛: 𝑛

𝑝(𝑧) = ∏(𝑧 − 𝑧𝑗 ) 𝑗=1

𝑛 𝐵 ∘ 𝑀𝑎,𝜙 (𝑧) 𝐵(𝑧) and 𝑞(𝑧) = ∏(𝑧 − 𝑧𝑗 ) . 𝐵(𝑧) − 𝜆1 𝐵 ∘ 𝑀𝑎,𝜙 (𝑧) − 𝜆1 𝑗=1

Pick 𝑤1 ∈ 𝕋 such that 𝑤1 ≠ 𝑧𝑗 for all 𝑗, and define 𝑤𝑗+1 = 𝑀𝑎,𝜙 (𝑤𝑗 ). Just as above, we conclude that 𝐵(𝑤𝑗 ) = 𝐵 ∘ 𝑀𝑎,𝜙 (𝑤𝑗 ) = 𝜆2 for all 𝑗 and some 𝜆2 ∈ 𝕋. In addition, we also have 𝐵 ∘ 𝑀𝑎,𝜙 (𝑏) = 𝐵(𝑏) for the fixed point 𝑏 ∈ 𝔻 of the disk automorphism. Thus, the polynomials 𝑝 and 𝑞 agree at 𝑛 + 1 points and are therefore identical.

Chapter 14. Iteration, Ellipses, and Blaschke Products

189

𝑛

Since none of ∏𝑗=1 (𝑧 − 𝑧𝑗 ), 𝐵(𝑧) − 𝜆1 , and 𝐵 ∘ 𝑀𝑎,𝜙 (𝑧) − 𝜆2 has a zero in 𝔻, the four functions 𝐵, 𝑝, 𝐵 ∘ 𝑀𝑎,𝜙 , and 𝑞 have exactly the same zeros. Thus, 𝐵 ∘ 𝑀𝑎,𝜙 = 𝜇𝐵 for some 𝜇 ∈ 𝕋. Pick 𝑧0 ∈ 𝔻 with 𝐵(𝑧0 ) ≠ 0. Solving 𝑝(𝑧0 ) = 𝑞(𝑧0 ), we get 𝜆1 (1 − 𝜇)𝐵(𝑧0 ) = 0. Hence, 𝜇 = 1; that is, 𝐵 ∘ 𝑀𝑎,𝜙 = 𝐵 and statement (2) is established. Since 𝑀𝑎,𝜙 is elliptic, by Lemma 14.1 we conclude that 𝑎 = 0 or |𝑎| < | cos(𝜙/2)|. Now Lemma 14.4 shows that (2) implies (1). Note that we can choose 𝑝 = 1 in the statement of Lemma 14.4 because the disk automorphism is of convex order. We now assume statement (3) and show that this implies (2). Since 𝑀𝑎,𝜙 is an elliptic disk automorphism of convex order 𝑛 with fixed point 𝑏, we have 𝑀𝑏,0 ∘ 𝑀𝑎,𝜙 ∘ 𝑀𝑏,0 = 𝑀0,𝜓 , where 𝑀0,𝜓 (𝑧) = 𝑒2𝜋𝑖/𝑛 𝑧 or 𝑀0,𝜓 (𝑧) = 𝑒−2𝜋𝑖/𝑛 𝑧. Then 𝐵 ∘ 𝑀𝑎,𝜙 = 𝜏 ∘ (𝑀𝑏,0 )𝑛 ∘ 𝑀𝑎,𝜙 𝑛

= 𝜏 ∘ (𝑀𝑏,0 ∘ 𝑀𝑎,𝜙 ) = 𝜏 ∘ (𝑀0,𝜓 ∘ 𝑀𝑏,0 )

𝑛

(14.8)

𝑛

= 𝜏 ∘ (𝑀𝑏,0 ) = 𝐵.

Finally, we assume (2) and show that (3) holds. We denote the fixed point of the elliptic disk automorphism 𝑀𝑎,𝜙 by 𝑏 and assume 𝑀𝑎,𝜙 is of 𝑛

convex order 𝑛. We define 𝐷 = (𝑀𝑏,0 ) and note that the calculations in (𝑗)

(14.8) show that 𝐷∘𝑀𝑎,𝜙 = 𝐷. The sets 𝐴𝑧 = {𝑀𝑎,𝜙 (𝑧) ∶ 𝑗 = 0, … , 𝑛 − 1} are identified by 𝐵 and by 𝐷 for every 𝑧 ∈ 𝕋. In particular, we pick 𝑧1 , 𝑧2 , and 𝑧3 in 𝕋 arranged in order of increasing argument from an arc on which both 𝐵 and 𝐷 are injective. Then, 𝜆𝑗 = 𝐵(𝑧𝑗 ) for 𝑗 = 1, 2, 3 are distinct and 𝜇𝑗 = 𝐷(𝑧𝑗 ) for 𝑗 = 1, 2, 3 are also distinct. By Lemma 3.4, the points (𝜆1 , 𝜆2 , 𝜆3 ) and (𝜇1 , 𝜇2 , 𝜇3 ) are ordered (with respect to the argument) in the same way. Using the fact that the cross ratio of four points is invariant under a linear fractional transformation we can find a linear fractional transformation 𝜏 ∶ 𝔻 → 𝔻 with 𝜏(𝜇𝑗 ) = 𝜆𝑗 for 𝑗 = 1, 2, 3. (Construct 𝜏 or consult [118, p. 46].) Since 𝜏 maps circles to

190

Chapter 14. Iteration, Ellipses, and Blaschke Products

circles and it also maps three points on 𝕋 to 𝕋 maintaining order, 𝜏 maps 𝕋 to itself and 𝔻 to itself; that is, 𝜏 is an automorphism of the disk. Write the two Blaschke products 𝐵 and 𝜏 ∘ 𝐷 as 𝑝1 (𝑧) 𝑞 (𝑧) and 𝜏 ∘ 𝐷(𝑧) = 1 , 𝑝2 (𝑧) 𝑞2 (𝑧) where 𝑝1 , 𝑝2 , 𝑞1 , and 𝑞2 are polynomials of degree at most 𝑛. Then 𝑝1 𝑞2 and 𝑝2 𝑞1 are polynomials of degree at most 2𝑛 that are equal on the more (𝑘) than 2𝑛 + 1 distinct points 𝑀𝑎,𝜙 (𝑧𝑗 ) for 𝑘 = 1, … , 𝑛 and 𝑗 = 1, 2, 3. Thus, the polynomials are equal and hence 𝐵(𝑧) =

𝐵 =𝜏∘𝐷 =𝜏∘(

𝑏−𝑧 1 − 𝑏𝑧

𝑛

) .

Figure 14.5. Illustration of Theorem 14.5. In Figure 14.5 we present an example of a Blaschke (6, 1)-ellipse associated with 6

𝐵(𝑧) = (

−0.1 + 0.3𝑖 − 𝑧 ) . 1 − (−0.1 − 0.3𝑖)𝑧

(14.9)

Chapter 14. Iteration, Ellipses, and Blaschke Products

191

For the corresponding 𝑀𝑎,𝜙 to be of convex order 6 we find that the canonical disk automorphism 𝑀0,𝜓 requires 𝜓 = 4𝜋/3. Using the equations in (14.7), we calculate 𝑎 = −0.317+0.096𝑖, 𝜙 = 4.37061, and 𝑎𝑒𝑖𝜙 = 0.197 + 0.267𝑖. The length of the major axis is then 2 sin(𝜙/2) = 1.634. It would be an excellent exercise for the reader to use Theorem 14.5 and (14.7) to prove the following corollary. Corollary 14.6. Let 𝑏 ∈ 𝔻, 𝜇 ∈ 𝕋, and 𝑘, 𝑚, 𝑛 be positive integers such that 𝑛 = 𝑚𝑘 and 𝑘 ≥ 2. Let 𝑏−𝑧

𝑛

) . 1 − 𝑏𝑧 For each 𝜆 ∈ 𝕋, let (𝑧1 , … , 𝑧𝑛 ) denote the ordered points satisfying 𝐵(𝑧) = −→ 𝜆. Then each of the 𝑚 closed polygons with sides ← 𝑧𝑗− 𝑧− 𝑗+𝑚 circumscribes the ellipse with foci 𝐵(𝑧) = 𝜇 (

𝑎1 = 𝑏

𝑒2𝜋𝑖/𝑘 − 1 𝑒2𝜋𝑖/𝑘 − 1 and 𝑎 = 𝑏 2 𝑒2𝜋𝑖/𝑘 − |𝑏|2 |𝑏|2 𝑒2𝜋𝑖/𝑘 − 1

and major axis of length 2 sin(𝜙/2), where 𝜙 = arg (

𝑒2𝜋𝑖/𝑘 − |𝑏|2 ). |𝑏|2 𝑒2𝜋𝑖/𝑘 − 1

If 𝑘 = 2, the ellipse degenerates to the point 2𝑏 𝑎= . 1 + |𝑏|2 The ellipses of the last corollary are (𝑘, 1)-Blaschke curves, and we can get one of them for each factor 𝑘 of the degree 𝑛 of the Blaschke product. Thus, we get a whole family of ellipses. (The reader might be interested to compare this with Mirman’s “package of Poncelet curves” in [115].) In addition, the foci of these ellipses are all on a circle. This is summarized in the following corollary. Corollary 14.7. Let 𝑎 ∈ 𝔻 and 𝜙 ∈ [0, 2𝜋) be such that 𝑀𝑎,𝜙 is an elliptic disk automorphism of finite order 𝑛 > 1 with fixed point 𝑏 ∈ 𝔻. Let 𝐵 be a Blaschke product satisfying 𝐵 ∘ 𝑀𝑎,𝜙 = 𝐵. If 𝑛1 , 𝑛2 , … , 𝑛ℓ are all the divisors of 𝑛 that are greater than 1, then 𝐵 gives rise to ℓ Poncelet ellipses, one of them possibly degenerate. The foci of all of these ellipses lie on a circle centered at 𝑏/(1 + |𝑏|2 ) that passes through the origin.

192

Chapter 14. Iteration, Ellipses, and Blaschke Products

We again leave it as an exercise for the reader to establish this corollary. The proof of this corollary reinforces all of the material in this chapter. So let us continue our investigation of the Blaschke product defined in (14.9) with a look at what the previous two corollaries say in this particular case. The factors of 6 are 𝑛1 = 6, 𝑛2 = 3, and 𝑛3 = 2. The ellipse corresponding to 𝑛1 was shown in Figure 14.5. The ellipses corresponding to 𝑛2 and 𝑛3 are in Figure 14.6. (Note that the curve corresponding to 𝑛3 is degenerate, as it should be.) Finally, in Figure 14.7 we combine

Figure 14.6. Ellipses corresponding to the factors 𝑛2 = 3 and 𝑛3 = 2.

the graphs of the three ellipses and show the circle on which the foci of all three ellipses are located. We can also express our results using operator theoretic terminology: The composition operator 𝐶𝑀𝑎,𝜙 ∶ 𝐻 2 → 𝐻 2 is defined by 𝐶𝑀𝑎,𝜙 𝑓 = 𝑓 ∘ 𝑀𝑎,𝜙 . Composition operators can be defined in more generality but they must, of course, be well defined. To see that 𝐶𝑀𝑎,𝜙 really maps 𝐻 2 to itself, the reader is referred to [140, p. 16] or [34, Section 3.1]; here we need only consider the action of 𝐶𝑀𝑎,𝜙 on the set of finite Blaschke products ℬ, where it is clear that the map is well defined. Thus, we consider 𝐶𝑀𝑎,𝜙 ∶ ℬ → ℬ on the set ℬ and with respect to a disk automorphism 𝑀𝑎,𝜙 . The condition 𝐵 ∘ 𝑀𝑎,𝜙 = 𝐵 is then equivalent to the requirement that 𝐵 is an eigenvector of the composition operator 𝐶𝑀𝑎,𝜙 with eigenvalue 1. Using

Chapter 14. Iteration, Ellipses, and Blaschke Products

193

Figure 14.7. The family of ellipses for the Blaschke product of (14.9).

this language and combining Corollary 14.7 with Lemma 14.4, we get our final result for this particular class of Blaschke products. Corollary 14.8. Let 𝑎 ∈ 𝔻 and 𝜙 ∈ [0, 2𝜋) be such that 𝑀𝑎,𝜙 is an elliptic disk automorphism of finite order 𝑛 > 1. Suppose 𝐵 is a Blaschke product of degree 𝑛 that is an eigenvector with eigenvalue 1 of the composition operator 𝐶𝑀𝑎,𝜙 . Let 𝑛1 , 𝑛2 , … , 𝑛ℓ be the factors of 𝑛 that are greater than 1. For each integer 𝑝𝑘 with 1 ≤ 𝑝𝑘 < 𝑛𝑘 /2 and (𝑝𝑘 , 𝑛𝑘 ) = 1 there exists a Blaschke (𝑛𝑘 , 𝑝𝑘 )-ellipse. Putting the results in Chapters 9, 10, and 14 together, we obtain a corollary that ties things back to compressions of the shift. We again leave the proof of this to the reader. Corollary 14.9. Let 𝑎 ∈ 𝔻 and 𝜙 ∈ [0, 2𝜋) be such that 𝑀𝑎,𝜙 is an elliptic disk automorphism of convex order 𝑛 > 1, and let 𝐵 be a Blaschke product of degree 𝑛 − 1. Suppose that the Blaschke product 𝐶(𝑧) = 𝑧𝐵(𝑧) satisfies 𝐶 ∘ 𝑀𝑎,𝜙 = 𝐶. Then the numerical range of 𝑆𝐵 is an elliptical disk.

194

Chapter 14. Iteration, Ellipses, and Blaschke Products

We conclude this chapter with an example that shows the abundance of Blaschke ellipses. We choose 15

(−0.1 + 0.3𝑖) − 𝑧 ) . 1 − (−0.1 − 0.3𝑖)𝑧 Equations (14.7) allow us to find the corresponding elliptic disk automorphism of order 15. By Theorem 14.5, the Blaschke product 𝐵 is an eigenvector with eigenvalue 1 of the composition operator. We thus expect (𝑛, 𝑝)-Blaschke ellipses for (𝑛, 𝑝) equal 𝐵(𝑧) = (

(15, 1), (15, 2), (15, 4), (15, 7), (5, 1), (5, 2), and (3, 1). Figure 14.8 shows the seven ellipses associated with 𝐵. The four black envelopes correspond to the Poncelet ellipses circumscribed by 15-gons, the two blue ones to the Poncelet ellipses circumscribed by 5-gons, and the red one is a Poncelet 3-ellipse.

Figure 14.8. The “super family” of seven Blaschke ellipses associated with 𝐵.

On Surprising Connections It is a good rule of thumb that if in the course of mathematical research an ellipse appears, there is likely to be an interesting result nearby. In the stories we have shared in this book, the ellipse has repeatedly shown up as the object connecting disparate parts of mathematics. Each of the ideas that we have discussed is worthy of study as an object in its own right, but our goal has not been to study Poncelet’s theorem, Blaschke products, and the numerical range. Rather, our goal has been to weave these threads together into one story. The historical development of each of these threads is distinct: Blaschke products and the numerical range received the most attention in the twentieth century, and Poncelet’s theorem first appeared in the nineteenth century. In addition, the mathematical history of each is not connected in any direct way to the others. Now, in the 21st century, we bring this material together. In the first part of the book we showed the interconnections between these three threads when the Blaschke curve is associated with a Blaschke product of degree 3, the matrices are 2×2, and the inner conics in Poncelet’s theorem are ellipses inscribed in triangles that are inscribed in the unit circle. These connections are of interest in their own right, but they also set the stage for the more general setting. What happens if the degree of the Blaschke product is greater than 3? What happens if the matrices are 𝑛 × 𝑛 with 𝑛 > 2? And what about Poncelet ellipses inscribed in 𝑛-gons with 𝑛 > 3? In the second part of the book we saw that we were able to obtain similar results in the general case, but it came at a cost—we had to give up on the elliptical nature of the numerical range and therefore on the elliptic nature of the Blaschke curve determined by the Blaschke product. Though we lost the ellipse, we retained the Poncelet property of the curve. And, of course, we saw that there 195

196

On Surprising Connections

were special instances in which the higher-degree Blaschke curve was an ellipse, the numerical range of the matrix was elliptical, and the Poncelet ellipse was inscribed in a polygon with more than three sides. In Chapters 12, 13, and 14 we saw that the appearance of an ellipse told us something unexpected. In the third part of this book (Chapter 15), we will share more surprising connections between the objects that have appeared here. To encourage active participation on the part of the reader, this final part has been written as a collection of projects, many openended. We hope the reader will continue to think about the way these seemingly disparate objects are interrelated. We leave the reader with one last intriguing connection to another branch of mathematics that we have not spent much time on thus far — number theory. Just as Poncelet’s theorem is a key idea bringing analytic objects like Blaschke products and (linear) algebraic objects like the numerical range together, we can use ideas related to Poncelet’s theorem to build a connection to number theory. We have already seen a bit of this in Chapter 8, but here is one more question for you to think about before you delve enthusiastically into the projects. Consider the sequence of positive integers 1, 2, 3, 4, 6, 8, 9, 12, 15, 16, 21, 24, 24, 32, 36, … . Assuming this describes a sequence that has something to do with the results in this book, can you figure out what that relation is? And, if so, does your relation tell you what we think the next terms will be? (Right now, we are asking you to read our minds. We tell you what we think in a moment.) In fact, this sequence has a interesting relation to the topics we have presented. A simple form of Poncelet’s theorem occurs when both conics are circles. We have examined this twice in this book: The first time was in Chapter 4 when we looked at the Chapple–Euler formula, which deals with triangles that are simultaneously inscribed in one circle of radius 𝑅 and circumscribe a second circle of radius 𝑟. The second time was in Chapter 13 when we studied Fuss’s theorem. If we let 𝑑 denote the distance between the centers of the circles and we define 1 1 1 , 𝑏= , 𝑐= , 𝑎= 𝑅+𝑑 𝑅−𝑑 𝑟

On Surprising Connections

197

then we see that the Chapple–Euler formula can be rewritten as 𝑎 + 𝑏 = 𝑐, a degree 1 equation in 𝑎, 𝑏, and 𝑐, and Fuss’s theorem can be rewritten as 𝑎2 + 𝑏2 = 𝑐2 , a degree 2 equation in 𝑎, 𝑏, and 𝑐. If we consider the case of an 𝑛-gon instead of just a triangle or quadrilateral, we can ask for the degree, 𝑑𝑛 , that yields a necessary and sufficient formula in terms of 𝑎, 𝑏, and 𝑐 for an 𝑛-gon to be inscribed in a circle and circumscribed by another. It turns out that there are such formulas, and you can write them down. The degree, 𝑑𝑛 , of the formulas for 𝑛 = 3, 4, 5, … are 1, 2, 3, 4, 6, 8, 9, 12, 15, 16, 21, 24, 24, 32, 36, … , where the first is the degree we obtained from the Chapple–Euler formula and the second is the degree from Fuss’s theorem. Even more can be said: It turns out that this sequence has a closed form expression. Given 𝑛, we can take its prime factorization 𝑞𝑗

𝑛 = 2𝑞0 ∏ 𝑝𝑗 , 𝑗

where 𝑝𝑗 is an odd prime and 𝑞𝑗 ∈ ℕ. Then 𝑑𝑛 =

2(𝑞𝑗 −1) 2 4𝑞0 ∏𝑝 (𝑝𝑗 − 1). 8 𝑗 𝑗

This is surely an unexpected result and further evidence of the intriguing mathematics one can find related to the ideas we have discussed. To learn more about this, visit the Wolfram page on Poncelet’s porism [152] and the original paper by Kerawala [91]. We hope the reader will find more interesting and, perhaps, surprising connections in the projects.

Part 3 “Why,” said the Dodo, “the best way to explain it is to do it.” –Lewis Carroll, Alice in Wonderland

Chapter

15

Fourteen Projects for Fourteen Chapters We provide a project for each of the first fourteen chapters in this book. The projects range from standard exercises to open research problems. Once you have completed Chapter 𝑋, you are ready to work the project labeled 15.𝑋. In all projects below, a literature and web search should include MathSciNet or zbMATH (Zentralblatt) if those are available to you.

15.1 Constructing Great Ellipses In Chapter 1 we discussed two ways to construct an ellipse: One is the familiar pin and string method and the other, in which we fold the paper and obtain a family of lines, is often referred to as the envelope method, but there are many other ways. Ideas can be found online or in books; see, for example, [104, Chapter 2]. Project 1. After a literature and web search, choose two methods and a team and construct a large ellipse in a visible place on your campus. Be sure to get permission for the construction first.

15.2 What’s in the Envelope? The envelope method for constructing an ellipse described in Chapter 1 (see Figure 1.3) is a particular example of how to find an envelope of a family of curves. Chapter 4 shows how Blaschke products provide the tangent lines to an ellipse and thus generate the envelope of the tangent lines. But what is the precise definition of an envelope? For the moment, 201

202

Chapter 15. Fourteen Projects for Fourteen Chapters

think of an envelope 𝐸 of a family of curves ℱ as a curve such that every point of 𝐸 is a point of tangency to a curve in ℱ; some definitions require that 𝐸 touch each curve. We give a precise definition soon (with some caveats), but first we give an example of how to compute an envelope. We emphasize that care must be taken to ensure that whatever method you use does in fact yield what you want. Why and when our method works can be found in the references, which we encourage you to read as part of this project. See [31, pp. 170–179] and, for a modern view of envelopes, see [88]. So suppose that a family of curves is defined by 𝐹(𝑥, 𝑦, 𝜃) = 0,

(15.1)

where 𝜃 runs through an interval. We assume that the curve 𝐸 can be parametrized locally (and smoothly) by the variable 𝜃 and that points (𝑥, 𝑦) on 𝐸 have the property that 𝑥 and 𝑦 are continuously differentiable functions of 𝜃. Thus, assuming the partial 𝐹𝜃 exists, we see that for (𝑥, 𝑦) to be on the envelope we must also have 𝐹𝜃 (𝑥, 𝑦, 𝜃) = 0.

(15.2)

We then (if possible) eliminate the parameter 𝜃, replacing that variable in (15.1) to obtain the equation of the curve in terms of 𝑥 and 𝑦. This procedure is sometimes referred to as the envelope algorithm. For us, the envelope of a family of curves is the curve 𝐸 that we obtain from the envelope algorithm. It turns out that there are cases in which (15.1) and (15.2) are satisfied, but the curve we obtain is not tangent to each member of ℱ. And other things that “feel right” are not true; for example, the envelope is not always the boundary of the region filled in by the family. Thus, once you get what you think is the envelope, you need to check that you have what you think you have. The references provide much more detailed information to guide you. Before we state the specific project for this chapter, we provide a well-known example of how to compute the envelope of a family of lines. This example appears in a problem known as the ladder problem [88], which seems to have first appeared in a book entitled Recreations in Mathematics [103, p. 163]. The Ladder Problem. How long a ladder can you carry horizontally around a (right-angled) corner of a corridor? See Figure 15.1.

15.2. What’s in the Envelope?

203

Figure 15.1. Ladder in a corridor.

Here is the solution: Assuming a ladder of fixed length 𝐿 is moved around a corner, let us find the envelope of the lines we get by picturing the ladder sliding along the outside wall (the 𝑦- and 𝑥-axes in the first quadrant), always keeping the tips of the ladder on the axes. So, we think of the ladder as forming a right triangle with legs on the 𝑥- and 𝑦-axes and hypotenuse of length 𝐿. Denote the corner of the inside wall by (𝑎, 𝑏). Parametrize the lines via the angle the line makes with the 𝑥axis, calling the angle 𝜃. We can assume that 𝜃 ∈ (0, 𝜋/2). Thus, the equation of each line (or position for a ladder) is 𝑦 𝑥 + = 𝐿. (15.3) sin 𝜃 cos 𝜃 Now differentiate with respect to 𝜃 to obtain 𝑦

cos 𝜃 2

=𝑥

sin 𝜃 cos2 𝜃

3

or

𝑦=𝑥

sin 𝜃 . cos3 𝜃

sin 𝜃 Substituting this into (15.3) and recalling that the curve appears in the first quadrant, we see that the curve must satisfy 𝑥 2/3 + 𝑦 2/3 = 𝐿2/3 .

(15.4)

204

Chapter 15. Fourteen Projects for Fourteen Chapters

(

)

Figure 15.2. Enveloping curve for the ladder problem.

Thus, what we have here is a piece (a quarter, actually) of an astroid, which is a familiar curve. Now we can answer the ladder problem: As long as the point (𝑎, 𝑏) stays outside the region bounded by the (first quadrant) of the astroid, the ladder can be turned about the corner. So the critical case is when (𝑎, 𝑏) satisfies 𝑎2/3 + 𝑏2/3 = 𝐿2/3 , and the longest the ladder can be is (𝑎2/3 + 𝑏2/3 )3/2 . See [88] for further discussion of envelopes, this problem, and how the envelope and boundary of the region are connected. So, if we have a family of curves defined by 𝐹(𝑥, 𝑦, 𝜃) = 0 for which 𝐹𝜃 exists and is continuous, 𝑥 and 𝑦 are continuously differentiable functions of 𝜃; the envelope of the family will be the set of points (𝑥, 𝑦) for which 𝐹(𝑥, 𝑦, 𝜃) = 0 and 𝐹𝜃 (𝑥, 𝑦, 𝜃) = 0. In our situation, the partials satisfy various conditions that allow us to find this curve in other ways, but we omit that discussion here.

15.2. What’s in the Envelope?

205

There are many well-known examples of envelopes of curves, and a rigorous search will turn up many articles, websites, and books that include such information. However, most of the books are quite old, so we suggest you begin learning more about envelopes via a web search and the following references (and the references therein): [31, pp. 170–179], [68], [98], [111], and [116]. Project 2. Make up a family of curves with an interesting envelope. Create an applet to go with your family of curves.1 Of course, you can modify the ladder problem in a number of ways as well. It is not always easy to find the envelope of a family, and you will not always end up with a familiar curve. Just for reference, here are some well-known examples (see [104] for a more complete list). (1) An applet for envelopes of lines2 can currently be found online. (2) The envelope of the family of curves 𝑦 = (𝑥 − 𝑡)3 for 𝑡 ∈ ℝ is discussed in [98], where it is shown that the envelope is the 𝑥-axis. This is a particularly interesting example; think about what happens when you try to find the envelope by looking at intersections of nearby curves! (3) Choose a circle 𝐶1 and a fixed point 𝐴 on the circle. Now consider the family of circles with center 𝐴𝐶 on 𝐶1 passing through the point 𝐴. The envelope of this family is a cardioid. (4) Keep the fixed circle 𝐶1 of the previous example. If we now consider the family of circles with center 𝐴𝐶 on 𝐶1 that are tangent to the horizontal line segment through the center of 𝐶1 , this produces a nephroid, an object that is kidney shaped. As you can see, the possibilities are endless. You can also create objects using these techniques; this is called string art (see [129]). 1 Here is a nice example of what you can do: https://www.geogebra.org/m/ c2Aqbj8D (accessed 12/15/2017). 2 http://www.dankalman.net/ume/env/envelopes.html (accessed 3/14/2018)

206

Chapter 15. Fourteen Projects for Fourteen Chapters

15.3 Sendov’s Conjecture We began Chapter 3 with a discussion of polynomials, the first step in understanding rational functions. Though it may seem that we know or should know everything about polynomials, that is emphatically not the case.3 Here is one example of an interesting open question about complex polynomials concerning the relationship between the zeros of a polynomial and its critical points (zeros of its derivative). Sendov’s conjecture. Let 𝑝 denote a complex polynomial having all of its zeros in the closed unit disk. Then every zero of 𝑝 is at most a distance 1 from a critical point of 𝑝. One result along these lines is the Gauss–Lucas theorem. Theorem 15.1 (Gauss–Lucas theorem). Let 𝑝 be a nonconstant polynomial with derivative 𝑝′ . Then the roots of 𝑝′ lie in the convex hull of the roots of 𝑝. Three different proofs of Theorem 15.1 can be found in [20] as well as a bit of history on the theorem. There are other classical theorems of this type, such as Walsh’s two-circle theorem about polynomials with zeros in two disks and Jensen’s theorem about critical points of polynomials with real coefficients (see [108] and [141]). Knowing these results, it is reasonable to imagine that one can say more about the location of the critical points of polynomials and it is easy to imagine how such a conjecture might be made; lots of computer experimentation might suggest what is true. But that is not what happened in this case: Sendov’s conjecture goes back to at least 1959 and simply came from good intuition. Who was it that had such good intuition? The first thing to do, normally, would be a search for “Sendov’s conjecture”. That is still a good starting point, but the reader will soon learn that Blagovest Sendov proposed this conjecture to Obreschkoff in 1958 or 1959, to Marden in 1962 and, independently, to Illiev. Illiev spoke informally about the conjecture, and it eventually made its way to Walter Hayman’s 1967 book, Research Problems in Function Theory, where it was attributed to Illiev ([79]); see [130, Section 7.3]. Thus, the earliest references often refer to 3 In fact, there is a whole book [9] devoted to results on polynomials approached through problem solving; a corrected reprint is also available.

15.3. Sendov’s Conjecture

207

Illiev’s conjecture (or, to make matters worse, Illieff’s, Illief’s, or Illyeff’s conjecture). There is currently an extensive bibliography4 for the conjecture online that is useful as a starting point. The conjecture has been established in several special cases; for example, Schmeisser proved the conjecture for the cases in which the degree 𝑛 of the polynomial is 𝑛 = 3 or 𝑛 = 4 as early as 1969. Other proofs for 𝑛 = 3 were presented in [19]. These proofs are accessible to those with a background in complex analysis. Sendov’s conjecture is known to be true if the polynomial has at most eight distinct zeros [21], when all zeros are on the unit circle ([62], [136]), and if the polynomial vanishes at 0 [136]. In 2014 Dégot [39] showed that for a zero 𝑎 of a polynomial 𝑝, there is an integer 𝑁 such that if the degree of 𝑝 is larger than 𝑁, then the closed disk of radius 1 about 𝑎 contains a critical point of 𝑝. At the time of this writing, Sendov’s conjecture remains a conjecture. However, restricting to low-degree polynomials or adding conditions make the problem solvable. It is interesting to note that we know the conjecture is true for small degree and we know the conjecture is true for large degree (a degree that depends on the point 𝑎), though we do not know how large “large” really is. This conjecture sounds easy but is most likely very hard. Nevertheless, as Sheil-Small notes: Let not age deter the bright young newcomer. Simple problems do sometimes have simple solutions, but nevertheless last a long time. . . However, a word of warning: simple solutions are rarely simply found and can cost a great deal of time and effort. Most (all?) mathematicians spend 99% of their time failing to solve the problems in which they are interested. The author would be delighted with a 1% success rate. . . What about Blaschke products? Earlier it was convenient to use the fact that Blaschke products can be written as 𝑞 ⋆ /𝑞, where 𝑞 and 𝑞⋆ are the polynomials defined in Chapter 3. Thus, it is not surprising that the location of the critical points of Blaschke products is an interesting area to explore. To say more about this, we need to introduce hyperbolic 4 http://parallel.bas.bg/~pencho/sendov/pap0bs.html

(accessed 3/5/2018)

208

Chapter 15. Fourteen Projects for Fourteen Chapters

geometry—an introduction that requires a brief mention of Euclidean geometry.

ℓ ℓ


ℓ
ℓ

Figure 15.3. Lines ℓ1 , ℓ2 , and ℓ3 through 𝑃 are parallel to ℓ in this Poincaré model.

Euclid worked with five postulates, the fifth of which is called the parallel postulate and states that given a line ℓ and a point 𝑃 not on ℓ, there exists a unique line ℓ′ through 𝑃 and parallel to ℓ. Hyperbolic geometry is a non-Euclidean geometry in which one assumes all postulates except the fifth, replacing it by its negation. As a consequence, in hyperbolic geometry there might be many lines through a point parallel to a given line. The Poincaré disk is one way to model hyperbolic geometry: The points in the geometry lie in the unit disk, and the arcs of circles that intersect the unit circle at right angles (including line segments through the origin) are the geodesics. Just like in Euclidean geometry, there is a unique geodesic through an arbitrary pair of distinct points. However, given a geodesic, 𝒢, that does not contain the origin, there are infinitely many geodesics through the origin that do not intersect 𝒢 and hence are parallel to it. Every other point not on a given geodesic displays this “non-Euclidean” behavior; see Figure 15.3. Walsh [150] noticed the connection between the study of polynomials with Euclidean geometry and the study of Blaschke products with non-Euclidean geometry. When we

15.3. Sendov’s Conjecture

209

use hyperbolic geometry, a hyperbolic convex set is a set 𝑆 such that for any two points in the set, the geodesic arc joining the two points is also contained in 𝑆. The hyperbolic convex hull of a set 𝑇 is the smallest hyperbolic convex set containing 𝑇. When 𝑇 is a finite set of points, in the Poincaré model, the hyperbolic convex hull of 𝑇 is a set for which the boundary is a “polygon” of circular arcs that lie on geodesics. Blaschke products also satisfy a hyperbolic Gauss–Lucas theorem: The critical points of a Blaschke product 𝐵 inside 𝔻 lie in the hyperbolic convex hull of the zeros of 𝐵; see [150], [141, p. 377], or [110, Chapter 3], where other basic information can be found. It follows that the critical points in 𝔻 of a Blaschke product 𝐵 lie in the Euclidean convex hull of 0 and the zeros of 𝐵. For a proof that covers infinite Blaschke products, see [24, Theorem 2.1]. For the location of zeros using hyperbolic geometry [142] is a good reference. The applet ,5 Blaschke Product Explorer helps to visualize these relationships. Remember that the white circles denote zeros, while the gray circles denote critical points. Exercise 15.2. Choose a Blaschke product of degree 3. With a computer or by hand using a straight edge and a compass, construct the hyperbolic convex hull of the zeros of the Blaschke product (similar to our Figure 15.3). Verify that the critical points of your Blaschke product are in this hyperbolic convex hull. Project 3. (Variants of Sendov’s conjecture, more advanced knowledge probably required.) There are many interesting variations of Sendov’s conjecture and many different possible approaches to the proofs of these conjectures. Some of these ideas are outlined in a paper [92] dedicated to Julius Borcea, a mathematician who made important contributions to the study of Sendov’s conjecture. This paper gives an idea of what is possible— though heeding Sheil-Small’s warning, we note that attempting a full solution to these problems will be both difficult and time consuming. In addition, Sendov’s 2002 paper discusses the Sendov conjecture and other related conjectures [139]. (Blaschke products) Before exploring the situation for Blaschke products, it is helpful to understand the Gauss–Lucas theorem (Theorem 15.1) for polynomials. Working through the project in [20] is an excellent first 5 https://pubapps.bucknell.edu/static/aeshaffer/v1/

210

Chapter 15. Fourteen Projects for Fourteen Chapters

step. Following that, investigate Rolle’s theorem in the complex plane to see how the situation for real polynomials differs from that for complex polynomials. Begin with Marden’s survey [109]; see also [137] (be sure to read Marden’s MathSciNet review of this paper!). Marden’s paper mentions two conjectures, one of which is Sendov’s. Once the reader is familiar with the way the critical points of complex polynomials behave and is aware of some of the differences between the real and complex settings, it would be useful to play with the applet (or create your own algorithm) to look at where the critical points of a Blaschke product are relative to its zeros. You can ask where the critical points are once the zeros are in place, or you can turn the question around and try to get the critical points where you want them and see where the zeros end up. Then formulate conjectures, examples, and counterexamples, starting with low-degree Blaschke products. In fact, finite Blaschke products can be used to show that if the zeros of a polynomial lie on a circle, then the polynomial satisfies Sendov’s conjecture. You will know enough to solve this after Chapter 4, or you can read the proof in Section 4 of [37].

15.4 Generalizing Steiner Inellipses In Chapter 4 we saw that for each Blaschke product 𝐵 of degree 3 with 𝐵(0) = 0, there is a unique Blaschke ellipse associated with 𝐵: For each 𝜆 ∈ 𝕋 the ellipse is inscribed in the triangle formed with the three (distinct) solutions of 𝐵(𝑧) = 𝜆. The foci of the ellipse are the zeros of the Blaschke product 𝐵1 (𝑧) = 𝐵(𝑧)/𝑧. Some ellipses have particularly interesting properties: For example, when the ellipse is inscribed in a triangle with points of tangency at the midpoints of each side, something special happens. Two results along these lines are Steiner’s theorem and Siebeck’s theorem. The first of these theorems is due to the Swiss mathematician, Jakob Steiner, and the second is due to Jörg Siebeck. Theorem 15.3 (Steiner’s theorem). Given a triangle 𝑇 with vertices 𝑧1 , 𝑧2 , and 𝑧3 , there is a unique inscribed ellipse tangent to 𝑇 at the midpoints of each of its sides. The foci of the ellipse are

15.4. Generalizing Steiner Inellipses

211

𝑧1 + 𝑧2 + 𝑧3 𝑧 + 𝑧 2 + 𝑧 3 2 𝑧1 𝑧2 + 𝑧 1 𝑧3 + 𝑧 2 𝑧3 ± ( 1 . ) − √ 3 3 3 This ellipse is called the Steiner inellipse. Siebeck presented a connection between these ellipses and critical points of degree-3 polynomials. Theorem 15.4 (Siebeck’s theorem). Given a triangle 𝑇 with vertices 𝑧1 , 𝑧2 , and 𝑧3 , consider the degree-3 polynomial 𝑝 with zeros 𝑧1 , 𝑧2 , and 𝑧3 . Then the critical points of 𝑝 are the foci of the Steiner inellipse of 𝑇. As a warmup, find an ellipse inscribed in the triangle with vertices (−𝑎, 0), (𝑎, 0), and (𝑐, ℎ), where 𝑎, 𝑐, and ℎ are positive real numbers using Steiner’s theorem and then using Siebeck’s theorem. A simple proof of Theorem 15.4 appears in [8]. Here is an outline of how it goes. Proof. Shifting the points in an appropriate way, we may assume that the zeros 𝑧1 , 𝑧2 , 𝑧3 satisfy 𝑧1 + 𝑧2 + 𝑧3 = 0. Show that this implies that the critical points, 𝑎 and 𝑏, satisfy 𝑎 + 𝑏 = 0; that is 𝑎 = −𝑏. Now let 𝑤 denote the point of tangency to the Steiner ellipse on the line segment joining 𝑧1 and 𝑧2 so that 𝑤 is the average of 𝑧1 and 𝑧2 . Write the derivative of the polynomial 𝑝 two ways (maybe three, depending on how you count it) to conclude that 𝑧1 − 𝑧 2 2 ) . 2 Now use the parallelogram identity to show that 3(𝑤 + 𝑎)(𝑤 − 𝑎) = − (

2(|𝑤 + 𝑎| + |𝑤 − 𝑎|)2 = 2|𝑤 + 𝑎|2 + 2|𝑤 − 𝑎|2 + 4|(𝑤 + 𝑎)(𝑤 − 𝑎)| = 4|𝑤|2 + 4|𝑎|2 + 4|(𝑤 + 𝑎)(𝑤 − 𝑎)| 1 = |𝑧1 + 𝑧2 |2 + 4|𝑎|2 + |𝑧1 − 𝑧2 |2 3 1 1 2 2 = |𝑧1 + 𝑧2 | + |𝑧1 + 𝑧2 |2 + |𝑧1 − 𝑧2 |2 + 4|𝑎|2 3 3 3 2 = (|𝑧1 |2 + |𝑧2 |2 + |𝑧3 |2 ) + 4|𝑎|2 . 3

(15.5)

212

Chapter 15. Fourteen Projects for Fourteen Chapters

This same result will hold for the midpoint of the line segments joining any two of the points 𝑧1 , 𝑧2 , or 𝑧3 . In particular, we see that there is a constant 𝐶 so that |𝑤 + 𝑎| + |𝑤 − 𝑎| = 𝐶 for all three midpoints of the sides of the triangle. Thus, these three midpoints lie on an ellipse 𝐸 with foci 𝑎 and −𝑎. To complete the proof, considering arguments in (15.5) we see that arg(𝑤 + 𝑎) + arg(𝑤 − 𝑎) = 𝜋 + 2 arg(𝑧1 − 𝑧2 ). This ensures that the sides of the triangle are tangent to 𝐸 at 𝑤 and 𝐸 must be the Steiner inellipse of the triangle. Which Blaschke ellipses have a circumscribing triangle for which they are Steiner inellipses? Here is the answer. Theorem 15.5. Let 𝑎 and 𝑏 be points in 𝔻, and let 𝐵 be the degree-3 Blaschke product 𝐵(𝑧) = 𝑧(

𝑧−𝑎 𝑧−𝑏 )( ). 1 − 𝑎𝑧 1 − 𝑏𝑧

Then the following are equivalent: (1) There is a cubic polynomial 𝑝 with zeros on the unit circle and critical points at 𝑎 and 𝑏. (2) There exists a triangle inscribed in the unit circle such that the Blaschke ellipse associated with 𝐵 is the Steiner inellipse of the triangle. (3) The points 𝑎 and 𝑏 satisfy |𝑎 + 𝑏| |. |𝑎𝑏| = || 2 | Prove this theorem or read the proof in [65]. There are many ways to modify this result. You can increase the degree of the polynomial, and you will get conditions on the zeros (that are somewhat unattractive). But you can also try moving the points of tangency around to see what else you can say. That brings us to our project. Project 4. We have discussed Steiner inellipses (the case in which the ellipse is inscribed in a triangle at the midpoints) thoroughly above. The Mandart inellipse is an example of another well-studied inellipse. While

15.5. Steiner’s Porism and Inversion

213

we will not discuss the Mandart inellipse here, we encourage you to look for information about it before completing this project. Modify the ellipse so that it is still inscribed in a triangle but the points of tangency are not necessarily the midpoints. When can you say something interesting? In the event that you can say something interesting about the ellipse, given one focus, where must the other focus lie? What happens if the ellipse is inscribed in a quadrilateral that is itself inscribed in 𝕋?

15.5 Steiner’s Porism and Inversion Poncelet’s theorem is sometimes called Poncelet’s porism, from the Greek word πόρισμα (pórisma). The meaning of this term has changed over time. It is often used to mean a proposition that yields conditions under which a problem has either no solution or infinitely many solutions. However, Boyer [17, Chapter 5] states that according to Pappus, a porism is “intermediate between a theorem, in which something is proposed for demonstration, and a problem, in which something is proposed for construction.” (For a longer discussion of what a porism is in mathematical literature, see Boyer’s book.) Regardless of what we think a porism is, we now present a result that sounds a lot like Poncelet’s theorem and is called Steiner’s porism. This result6 goes like this: Consider two circles, 𝒞1 and 𝒞2 with 𝒞1 entirely contained in 𝒞2 . Suppose we draw a circle 𝐴1 that is in the closed region 𝑆 bounded by 𝒞1 and 𝒞2 , and 𝐴1 is tangent to both 𝒞1 and 𝒞2 . Now draw a circle 𝐴2 in 𝑆 tangent to 𝐴1 , 𝒞1 , and 𝒞2 . Continue in this way, drawing circles 𝐴𝑗+1 ≠ 𝐴𝑗−1 in 𝑆 tangent to 𝐴𝑗 , 𝒞1 , and 𝒞2 . Let us agree to call this sequence of tangent circles, (𝐴𝑗 ), a chain of circles between 𝒞1 and 𝒞2 . One of two things will happen to this chain: Either you will not return to the circle 𝐴1 , or you will return as in Figure 15.4. If you return to 𝐴1 this will happen no matter what the position is of the first circle, 𝐴1 . The chain of circles between 𝒞1 and 𝒞2 can even wind around 𝒞1 a few times before it closes and Steiner’s porism still applies. This remarkable result sounds a lot like Poncelet’s theorem—but it is much easier to prove. 6 Due

to Jakob Steiner; that is, the same Steiner for whom Theorem 15.3 is named.

214

Chapter 15. Fourteen Projects for Fourteen Chapters

 


Figure 15.4. A chain of seven circles between 𝒞1 and 𝒞2 that closes. Theorem 15.6 (Steiner’s porism). Let 𝒞1 and 𝒞2 be two circles with 𝒞1 entirely contained in 𝒞2 . If one chain of circles between 𝒞1 and 𝒞2 closes, then all such chains close. When we considered Example 2.4 we noted that Poncelet’s theorem is easy to prove in the special case that the two ellipses are concentric circles. The same is true for Steiner’s porism: If the circles are concentric, the picture remains the same under rotation. The difference is that understanding this special case is not enough to prove Poncelet’s theorem, but it is enough to prove Steiner’s porism! Exercise 15.7. Prove Steiner’s porism using the outline below. Recall that automorphisms of the closed unit disk that map the unit circle to itself and the unit disk to itself are of the form 𝑧−𝑎 𝜑(𝑧) = 𝜇 with |𝜇| = 1 and 𝑎 ∈ 𝔻. 1 − 𝑎𝑧 These map circles to circles, if we agree that lines are circles, and they preserve angles. Explain why we can assume that the larger circle is 𝕋. Then show that by using an automorphism of the disk we may reduce the problem to the case in which 𝒞2 = 𝕋, and we may further assume that 𝕋 and 𝒞1 are concentric; see Figure 15.5.

15.5. Steiner’s Porism and Inversion

215

Now finish the proof of Steiner’s porism. Once you have done that, write and implement a computer program that will produce a figure like Figure 15.4.






Figure 15.5. A transformation of Figure 15.4 so that 𝒞1 and 𝒞2 are concentric.

As we will see later, the centers of the circles in the chain lie on an ellipse [123, p. 57]. Other proofs of Steiner’s porism use inversion, which is mainly a tool for transforming the situation you are given to a related one that preserves important properties. Steiner is often credited as having discovered the concept of inversion, but the history on the subject is complicated. In his article, The origins of the geometric principle of inversion [124], Patterson says, “When one seeks to find to whom we are indebted for the invention of this fruitful method of geometry he is confronted only by meagre footnotes and confusing references in the literature.” What is certain is that Steiner played an important role. You are familiar with many transformation techniques. For example, reflecting across a line is a transformation, but it is not particularly useful as far as the study of geometric properties of figures is concerned. Linear fractional transformations change problems in a useful way; you might

216

Chapter 15. Fourteen Projects for Fourteen Chapters

use it to switch from functions acting on a disk to functions defined on a half-plane. (This is precisely what happens in Chapter 11.) Linear fractional transformations are examples of transformations that preserve circles and angles.












Figure 15.6. Finding the inverse point.

The inversion technique is more complicated than either of these two transformations. To describe inversion we begin with a circle, called the reference circle, with center 𝑂 and radius 𝑅. First, suppose 𝐴 lies outside the circle. To find the inverse of 𝐴 with respect to this circle find a point 𝐵 on the circle such that 𝑂𝐵 is perpendicular to 𝐴𝐵; see Figure 15.6. Let 𝐴′ denote the foot of the altitude of the triangle △𝑂𝐵𝐴 on 𝑂𝐴. Then 𝐴′ is called the inverse of the point 𝐴 with respect to the reference circle. Though we have only defined the inverse in the event that 𝐴 lies outside the circle, the other possibilities can be handled similarly. Note that the two triangles △𝑂𝐵𝐴 and △𝑂𝐴′ 𝐵 are similar and |𝑂𝐴| 𝑅 = or |𝑂𝐴| ⋅ |𝑂𝐴′ | = 𝑅2 . 𝑅 |𝑂𝐴′ | In fact, the inverse of a point 𝐴 is often defined to be the point 𝐴′ such that 𝐴′ lies on the line joining 𝑂 and 𝐴 and satisfies |𝑂𝐴|⋅|𝑂𝐴′ | = 𝑅2 . The

15.5. Steiner’s Porism and Inversion

217

inverse of the point 𝑂 is taken to be the point at infinity, and, conversely, the inverse of the point at infinity is 𝑂. Before moving on, you should do a few straight edge and compass constructions to see where points end up. Exercise 15.8. To become familiar with inversion, experiment with the applet ,7 Circle Inversion. Then construct (with a straight edge and a compass) the inverse 𝐴′ of a point 𝐴 with respect to a circle 𝐶. Consider the cases when 𝐴 is inside 𝐶 and when 𝐴 is not inside 𝐶 separately. When we discuss inverses of points below, assume that the points are not the center of the reference circle. You should show that the inverse of a point 𝑧 ∈ ℂ with respect to a circle with radius 𝑅 and center 𝑧0 is given by 𝑅2 (𝑧 − 𝑧0 ) 𝑧 ′ = 𝑧0 + (15.6) . |𝑧 − 𝑧0 |2 In addition, work out the special case in which 𝐶 is the unit circle 𝕋. For each point 𝑧 write its inverse in terms of 𝑧 (or 𝑧). The inverse of a point inside the reference circle must lie outside the circle, a point outside will have an inverse inside the circle, and it should be clear that a point on the reference circle goes to itself. Other basics, of which we prove only the first few, are the following. Under inversion, • lines through 𝑂 go to lines through 𝑂; • lines not through 𝑂 go to circles through 𝑂; • circles that do not pass through 𝑂 go to circles that do not pass through 𝑂; • a circle orthogonal to the reference circle goes to itself; • angles are preserved (inversion is a conformal map). Let us now see why a few of these are true. We have seen that the reference circle goes to itself, and you should convince yourself that lines through 𝑂 go to lines through 𝑂. So consider a line ℓ that does not go through 𝑂. We give an argument in the event that the line intersects the reference circle in at most one point; the case in which ℓ intersects the 7 https://pubapps.bucknell.edu/static/aeshaffer/v1/

218

Chapter 15. Fourteen Projects for Fourteen Chapters

circle in two points can be handled similarly. Let 𝐴 be the point on ℓ such that 𝑂𝐴 is perpendicular to ℓ. Let 𝐵 be another point on the line ℓ. Consider 𝐴′ and 𝐵′ , the inverted points of 𝐴 and 𝐵. These lie inside











ℓ Figure 15.7. Similar triangles. or on the circle. So we have the triangle △𝑂𝐵′ 𝐴′ inside the circle and a second triangle △𝑂𝐴𝐵; see Figure 15.7. These share one angle and, from our discussion above, we have |𝑂𝐴′ ||𝑂𝐴| = 𝑅2 = |𝑂𝐵′ ||𝑂𝐵|, where 𝑅 is the radius of the reference circle. Thus, the two triangles are similar and ∠𝑂𝐵′ 𝐴′ is also a right angle. Now consider the circle 𝒞′ with diameter 𝑂𝐴′ . The angle ∠𝑂𝐵′ 𝐴′ is a right angle and therefore 𝐵′ lies on 𝒞′ , completing the proof that the inverse of the line ℓ is a circle through 𝑂 containing 𝐴′ and 𝐵′ . Details of the other properties are left to the reader and can be found in the charming book, Gems of Geometry [10], which has motivated much of the discussion here.

15.5. Steiner’s Porism and Inversion

219

Before turning to the proof of Steiner’s porism, we need one more definition. The radical axis of two circles consists of the locus of points from which the line segments tangent to 𝒞1 and 𝒞2 are of equal length. Given a point 𝑃 on the radical axis, there is a unique circle centered at 𝑃 that intersects both of the circles at right angles. Conversely, the center of a circle that intersects both of the circles at right angles lies on the radical axis. Exercise 15.9. Given two nonintersecting circles such that neither lies inside the other, show that the radical axis is a straight line perpendicular to the line containing the centers of the circles. We now have all the ingredients except the most important property. Proposition 15.10. If we choose two circles that do not intersect, we can invert them with respect to a third circle to obtain a pair of concentric circles. This is precisely what we need to prove Steiner’s theorem. Why is it true? If one circle lies inside the other, we may invert with respect to the larger one and the two circles will then lie outside each other. So we assume that neither circle lies inside the other. A circle centered on the radical axis with radius the common length of the tangents to the circles 𝒞1 and 𝒞2 intersects both circles at right angles. Thus, we can find two intersecting circles, 𝑐1 and 𝑐2 , that are orthogonal to 𝒞1 and 𝒞2 . Make an inversion with respect to a circle with the center at one of the points of intersection of 𝑐1 and 𝑐2 . Then 𝑐1 and 𝑐2 will map onto two straight lines, intersecting at a point 𝑁, and 𝒞1 and 𝒞2 will map onto two circles orthogonal to those lines and therefore both centered at the point 𝑁. We are ready to see how the proof of Steiner’s theorem goes if we use inversion. Proof of Steiner’s porism. We are given 𝒞1 and 𝒞2 with 𝒞1 entirely contained in 𝒞2 . Let 𝒜 be a chain of circles that returns to the starting circle, say 𝐴1 = 𝐴𝑛 , where 𝑛 > 1 is the first integer for which this happens. Invert the circles 𝒞1 and 𝒞2 to obtain concentric circles. When we do this, the chain of circles will also be mapped to circles tangent to the inverted circles and, therefore, the circles will have equal radii. The setup is now invariant under rotations, completing the proof.

220

Chapter 15. Fourteen Projects for Fourteen Chapters

Exercise 15.11. Prove the following: The centers of the circles in the chain lie on an ellipse. Figure 15.8 will help you prove this.










 

Figure 15.8. Three circles and three centers, 𝑃, 𝑂1 , and 𝑂2 .

Project 5. To give you an idea of what can be done with inversion, here are two examples of theorems for which you should find a proof using inversion. Once you have completed these exercises, locate other theorems (of which there are many) that can be proved using inversion,8 and understand their statements and proofs. Your work should include specific nontrivial examples that satisfy the hypotheses of the theorem; for example, if you were to illustrate Steiner’s theorem, you would produce a specific pair of nonconcentric circles that have a closed chain.9 Here are the two examples we promised to present. Either look for proofs of these (using inversion) that you rewrite in your own words or try to prove them on your own. 8 These and others can be found in an entertaining article http://www.ams.org/ samplings/feature-column/fcarc-kissing (accessed 12/15/2017). 9 Some suggestions for use in Mathematica appear in http://www. mathematica-journal.com/2014/05/the-arbelos/ (accessed 12/15/2017).

15.5. Steiner’s Porism and Inversion

221

Theorem 15.12 (Ptolemy’s theorem). Let 𝑄 be a quadrilateral with vertices on a circle. Denoting the four ordered vertices 𝐴, 𝐵, 𝐶, and 𝐷, |𝐴𝐶| ⋅ |𝐵𝐷| = |𝐴𝐵| ⋅ |𝐶𝐷| + |𝐵𝐶| ⋅ |𝐴𝐷|. This can be proved in many different ways and has a number of interesting corollaries that we encourage you to find and prove. We remark that a quadrilateral with all four vertices on a circle is said to be a cyclic quadrilateral. For the next one, recall that the curvature of a circle is the reciprocal of the radius. Theorem 15.13 (Descartes’s circle theorem). Given four mutually tangent circles with curvatures 𝑏𝑗 , then 𝑏12 + 𝑏22 + 𝑏32 + 𝑏42 = (𝑏1 + 𝑏2 + 𝑏3 + 𝑏4 )2 /2. Descartes’s theorem is often called Soddy’s formula, but Descartes described this result to Princess Elizabeth of Bohemia in 1643. The proof was rediscovered (independently) by many others, including the NobelPrize-winning chemist, Frederick Soddy. Soddy’s 1936 proof [143] was published as a poem, beginning with the lines For a pair of lips to kiss maybe Involves no trigonometry. ’Tis not so when four circles kiss Each one the other three. To bring this off the four must be As three in one or one in three. It seems only fair that Soddy should get some credit for this proof—at least for originality of presentation. Poetic extensions of the circle theorem followed Soddy’s proof and include The Kiss Precise (Generalized) by Thorold Gosset and The Kiss Precise (Further Generalized) by Fred Lunnon.10 Gosset’s article appears with the following footnote [67]: “The final stanza of Soddy’s poem announces Soddy’s discovery of an analogous formula for spheres. After Soddy’s verses appeared, Thorold Gosset wrote additional lines to describe the more general case for tangency, or ‘kissing,’ of 𝑛 + 2 hyperspheres in 𝑛 dimensions”. For more about Descartes’s theorem, see [99]. 10 http://pballew.net/soddy.html

(accessed 12/15/2017)

222

Chapter 15. Fourteen Projects for Fourteen Chapters

15.6 The Numerical Range and Radius In Chapter 6 we saw that the numerical range of a 2 × 2 matrix is always an elliptical disk and we provided a formula for the boundary ellipse in the elliptical range theorem. In general, it is not easy to give a formula for the boundary curve for the numerical range of a matrix. But when we consider special classes of matrices we can say more; for example, if 𝐴 is 𝑛 × 𝑛 and normal, it follows from results in Chapter 6 that the numerical range of 𝐴 is the convex hull of the eigenvalues of 𝐴. (This is proved in Corollary 7.1, but you can prove it using only tools from Chapter 6.) Here is a second example, starting with Hermitian matrices. Proposition 15.14. Let 𝐴 be an 𝑛 × 𝑛 matrix. Then 𝑊(𝐴) ⊂ ℝ if and only if 𝐴 is Hermitian. Proof. Suppose 𝐴 = 𝐴⋆ . Then for a unit vector 𝑥 ∈ ℂ𝑛 , we have ⟨𝐴𝑥, 𝑥⟩ = ⟨𝑥, 𝐴⋆ 𝑥⟩ = ⟨𝑥, 𝐴𝑥⟩ = ⟨𝐴𝑥, 𝑥⟩, so 𝑊(𝐴) ⊆ ℝ. Since 𝑊(𝐴) is bounded, we see that 𝑊(𝐴) ⊂ ℝ. Now suppose that 𝑊(𝐴) ⊂ ℝ. Then a similar computation shows that for all unit vectors 𝑥 ∈ ℂ𝑛 , we have ⟨(𝐴 − 𝐴⋆ )𝑥, 𝑥⟩ = 0. In particular, 𝑊(𝐴 − 𝐴⋆ ) = {0}. The conclusion follows from Theorem 6.2, (6). We say that a matrix 𝐴 is positive definite if 𝐴 = 𝐴⋆ and ⟨𝐴𝑥, 𝑥⟩ > 0 for all 𝑥 ∈ ℂ𝑛 with 𝑥 ≠ 0. When ⟨𝐴𝑥, 𝑥⟩ > 0 for all 𝑥 ∈ ℂ𝑛 ⧵ {0}, we can say something special about 𝑊(𝐴) (what?). We would, therefore, like to know when a matrix is positive definite. Let us begin with a simple example. Is 1 2 𝐴0 = [ ] 2 1 positive definite? Look up some equivalent ways of defining positive definiteness, and you will find that there are many ways to see if a matrix is positive definite. Here are some other interesting results for classes of matrices and operators. If 𝐴 is nilpotent, that is, 𝐴𝑛 = 0 for some positive integer 𝑛, then the numerical range of 𝐴 is a circular disk; if 𝐴 is an 𝑛 × 𝑛 Jordan block with zeros on the main diagonal, then the numerical range is a disk of

15.6. The Numerical Range and Radius

223

radius cos(𝜋/(𝑛 + 1)). This will follow (relatively) easily from the results in Chapter 10, and it appears as an exercise in that chapter. A completely different proof of this fact appears in Haagerup and de la Harpe’s paper [72], where they show that for any contraction 𝑇 on a Hilbert space 𝐻, if 𝑇 𝑛 = 0 for 𝑛 ≥ 2, then the numerical radius, 𝑤(𝑇), where 𝑤(𝑇) = sup{|⟨𝑇𝑥, 𝑥⟩| ∶ ‖𝑥‖ = 1, 𝑥 ∈ 𝐻},

(15.7)

satisfies the inequality 𝑤(𝑇) ≤ cos(𝜋/(𝑛 + 1)). It is easy to see that this inequality is correct for a 2 × 2 Jordan block with zeros on the main diagonal—in fact, this was one of our very early examples, namely, Example 2.1. (Note that this, together with some of the numerical range basics, tells us what the numerical range of a general Jordan block is. What is it?) This study has been extended in various ways: More is known about the numerical range of quasi-nilpotent operators, that is, those operators for which the spectrum is the singleton {0}, and so-called 0 − 1 matrices; that is, matrices in which every entry is either a 0 or a 1. For results on those matrices with at most one 1 in each row and column and zeros elsewhere, see [107]. You should begin this project by checking the results for specific matrices. Here is an example of what you might do. Consider the two matrices 0 1 1 0 ] and 𝐴2 = [ ]. 1 0 0 −1 Show that these two matrices are unitarily equivalent. By Theorem 6.2, these two should have the same numerical range. Do they? Is 𝑊(𝐴1 𝐴2 ) = 𝑊(𝐴2 𝐴1 )? Now pick two 2 × 2 other matrices, 𝐴3 and 𝐴4 , with 𝑊(𝐴3 ) = 𝑊(𝐴4 ). Are they unitarily equivalent? Is 𝑊(𝐴3 𝐴4 ) = 𝑊(𝐴4 𝐴3 )? 𝐴1 = [

Project 6. In Chapter 6 we considered only 2 × 2 matrices, yet even for these matrices there are many interesting questions one might think about. In addition, we have shown that an 𝑛 × 𝑛 matrix is Hermitian if and only if 𝑊(𝐴) ⊂ ℝ. Is it true that a 2 × 2 matrix is normal if and only if the numerical range is the convex hull of the eigenvalues? What if the matrix is 𝑛 × 𝑛? The above discussion suggests other questions: If two 𝑛 × 𝑛 matrices 𝐴 and 𝐵 have the same numerical range, are they unitarily equivalent? Is it

224

Chapter 15. Fourteen Projects for Fourteen Chapters

always the case that 𝑊(𝐴𝐵) = 𝑊(𝐵𝐴)? Of course, if an answer is negative the next question is when (if ever) is it the case that the answer is positive. Pick a special class of matrices and investigate the behavior of the class with regard to the numerical range. You should also consider the behavior of its numerical radius. When 𝐴 is an 𝑛 × 𝑛 matrix, rather than an operator on a general Hilbert space, the supremum in the numerical radius definition becomes a maximum; that is, the numerical radius of 𝐴 is 𝑤(𝐴) = 𝑚𝑎𝑥{|⟨𝐴𝑥, 𝑥⟩| ∶ ‖𝑥‖ = 1}.

(15.8)

For other ideas, see the paper [85]. We mention two more examples of classes of matrices for which more is known: 3 × 3 matrices (see [90] for a focus on when the numerical range is elliptical) and companion matrices (see, for example, [60] and Theorem 15.31 below).

15.7 Pedal Curves and Foci In Project 2 we saw how to compute the envelope of a family of curves. In Project 4 we saw that Blaschke ellipses corresponding to Blaschke products of degree 3 are the envelopes of a family of lines and we singled out a special ellipse, the Steiner inellipse. Also, in Chapter 4 we learned that the numerical range of a 2 × 2 matrix is always elliptical. Putting all these ideas together, we can ask the following question: Does something interesting happen when you move an ellipse along a prescribed path? When a curve is moved so that it always touches two fixed curves, the locus of a point or the envelope of an associated line or curve is a glissette. For example, if an ellipse slides along the 𝑥- and 𝑦-axes (sort of like the ladder problem with ellipses) the locus of the center of the ellipse is an arc of a circle. This relies on the following lemma. Lemma 15.15. The set of points of intersection between any two perpendicular tangents to an ellipse lie on a circle with the same center as the ellipse. Proof. We may assume that the ellipse is 𝑦 2 𝑥 2 ( ) + ( ) = 1. 𝐴 𝐵

15.7. Pedal Curves and Foci

225

Figure 15.9. Intersection points of perpendicular tangent lines lie on a circle.

Suppose that 𝑦 = 𝑚𝑥 + 𝑏 is tangent to the ellipse, and find the point at which this line intersects the ellipse. To this end, we have 𝐵2 𝑥 2 + 𝐴2 (𝑚𝑥 + 𝑏)2 = (𝐴𝐵)2 . Solving, we have (𝐴2 𝑚2 + 𝐵2 )𝑥 2 + 2𝐴2 𝑚𝑏𝑥 + (𝐴2 𝑏2 − (𝐴𝐵)2 ) = 0. Since we have a tangent line, this line intersects the ellipse at one point; that is, the discriminant must be zero. So 4𝐴4 𝑚2 𝑏2 = 4(𝐴2 𝑚2 + 𝐵2 )(𝐴2 𝑏2 − 𝐴2 𝐵2 ). Solving for 𝑏 > 0, we get 𝑏 = √𝐴2 𝑚2 + 𝐵2 . Thus, the point of intersection (𝑥0 , 𝑦0 ) of the two (perpendicular) tangent lines must satisfy 𝑦0 − 𝑚𝑥0 = √(𝐴𝑚)2 + 𝐵2 ,

226

Chapter 15. Fourteen Projects for Fourteen Chapters

and since the perpendicular tangent line must satisfy the corresponding equation with slope −1/𝑚, we have 𝑦0 +

1 𝐴2 𝑥0 = √ 2 + 𝐵 2 . 𝑚 𝑚

This leads to the equation (𝑦0 − 𝑚𝑥0 )2 + (𝑚𝑦0 + 𝑥0 )2 = (𝐴𝑚)2 + 𝐵2 + 𝐴2 + (𝐵𝑚)2 . We get the same equation for 𝑏 < 0. Thus, 𝑥02 + 𝑦02 = 𝐴2 + 𝐵2 . Consequently, the set of points of intersection between two perpendicular tangent lines is a circle of radius √𝐴2 + 𝐵2 . Now let us try something different. Choose a circle and a point inside the circle. Draw a line ℓ through the point. At the points at which ℓ intersects the circle, draw lines perpendicular to the line ℓ. If you repeat this several times (we repeated it five times at evenly spaced points) you will see Figure 15.10.

Figure 15.10. A circle, a point, and some perpendicular lines.

15.7. Pedal Curves and Foci

227

Perhaps it is not clear what is happening, so repeat this more often. Looking at Figure 15.11 we see that, once again, we end up with an ellipse. Try constructing an ellipse in this way. You can do it using a computer, string art, or some other method of construction.

Figure 15.11. A circle, a point, and more perpendicular lines. When we do this, we end up with a curve called the negative pedal of the original; that is, start with a curve 𝐶 and a fixed point 𝑃0 = (𝑎, 𝑏) called the pedal point. For a point 𝑃 on the curve, draw a line through 𝑃 perpendicular to 𝑃0 𝑃. The envelope of these lines is the negative pedal curve of 𝐶 with respect to 𝑃0 . It appears that the ellipse is the negative pedal of the circle with respect to the point we chose. What is special about the point? It appears to be a focus. This suggests possibilities for a project. Project 7. (Pedal and negative pedal curves.) The fact that the procedure we have just described produces a curve called the negative pedal suggests that there is also a pedal curve. Find out what a pedal curve is, and then give some examples and some constructions of both pedal curves and negative pedal curves. (Foci and Blaschke products.) Let Δ be a triangle with vertices on the unit circle. Show that there is a nondegenerate ellipse inscribed in Δ with

228

Chapter 15. Fourteen Projects for Fourteen Chapters

one focus at a point 𝑎 ∈ 𝔻 if and only if 𝑎 lies (strictly) inside Δ. If, in addition, 𝑎 ∈ ℝ, what are the possible values of the second focus of the ellipse? Give, explicitly, a Blaschke product 𝐵 that identifies the vertices of the triangle and satisfies 𝐵(0) = 𝐵(𝑎) = 0. After investigating these cases, there are many directions to go, none of which will be straightforward. We suggest considering the case in which the focus 𝑎 ∉ ℝ. You might also consider cases in which the ellipse is inscribed in 𝑛-sided polygons for 𝑛 > 3. The existence of a Blaschke product 𝐵 with zeros at 0 and 𝑎 that identifies the vertices of the triangle can be found in [63].

15.8 The Power of Positivity In our work on the connection between Benford’s law and Poncelet’s theorem we considered powers of integers and iteration. This project investigates what happens when we consider matrices in place of real numbers. Recall that an 𝑛 ×𝑛 matrix is said to be a nonnegative matrix if all of its entries are nonnegative and a positive matrix if all of the entries are positive. Nonnegative matrices occur naturally; for example, if the entries of a matrix denote the time it takes to complete a task, they will be nonnegative. Thus, it is not an accident that such matrices are well studied. In addition, the notion of iteration is an important tool in many fields, as in Newton’s method or finding equilibria in market models. One of the most important theorems about positive matrices and iteration is the Perron–Frobenius theorem. Theorem 15.16 (Perron–Frobenius theorem). Let 𝐴 = [𝑎𝑖𝑗 ] be a positive 𝑛 × 𝑛 matrix. Then the matrix 𝐴 has a positive eigenvalue 𝑟 such that (1) the eigenvalue 𝑟 has multiplicity 1; (2) any other eigenvalue, 𝑠, of 𝐴 satisfies |𝑠| < 𝑟; (3) there are right and left eigenvectors, 𝑣 and 𝑤, associated with 𝑟 that have all entries positive (𝐴𝑣 = 𝑟𝑣 and 𝑤 𝑇 𝐴 = 𝑟𝑤 𝑇 );

15.8. The Power of Positivity

229

(4) if the entries in each row of 𝐴 sum to one, then 𝑟 = 1, its associated eigenvector is 𝑒 ∶= [1 1 … 1]𝑇 , and lim 𝐴𝑛 = 𝑒𝑢𝑇 ,

𝑛→∞

where 𝑢𝑇 is the unique left eigenvector with positive entries summing to 1 that corresponds to the eigenvalue 1. A vector with nonnegative entries in which the sum of the entries is 1 is said to be a stochastic vector. There are many different proofs of Theorem 15.16, and we leave it to the reader to find one he or she can read. Instead, we focus on a curious application. Here is the idea. For 𝑗 ∈ ℤ+ , let 𝑃𝑗 = (𝑎𝑗 , 𝑏𝑗 , 𝑐𝑗 ) ∈ ℂ3 . Suppose that 𝑃1 , 𝑃2 , and 𝑃3 are arbitrary but fixed. The remaining points in the sequence (𝑃𝑗 ) are constructed by taking the centroid of the previous three points. For example, 𝑎 + 𝑎2 + 𝑎3 𝑏1 + 𝑏2 + 𝑏3 𝑐1 + 𝑐2 + 𝑐3 𝑃4 = ( 1 , , ), 3 3 3 and the remaining points are constructed in this manner. Does the sequence (𝑃𝑗 ) converge? If so, how does the limit depend on the initial three points? To find the answers, we introduce the matrix 𝑎𝑗 𝐴𝑗 = [ 𝑎𝑗+1 𝑎𝑗+2

𝑏𝑗 𝑏𝑗+1 𝑏𝑗+2

𝑐𝑗 𝑐𝑗+1 ] . 𝑐𝑗+2

Thus, the rows of 𝐴𝑗 are the coordinates of the points 𝑃𝑗 , 𝑃𝑗+1 , and 𝑃𝑗+2 . The matrix 𝐴1 is fixed, and you should convince yourself that 0 1 0 ⎡ ⎤ 𝐴𝑗+1 = 𝐵𝐴𝑗 , where 𝐵 = ⎢ 0 0 1 ⎥ . ⎢ 1 1 1 ⎥ ⎣ 3 3 3 ⎦ To avoid repeated rows, it suffices to consider the matrices 𝐴1 , 𝐴4 , 𝐴7 , …. 𝑛 This leads to the sequence (𝐴3𝑛+1 ) with 𝐴3𝑛+1 = (𝐵3 ) 𝐴1 for 𝑛 ≥ 0. A

230

Chapter 15. Fourteen Projects for Fourteen Chapters

calculation shows that 1

1

1

⎡3 ⎢ ⎢1 3 𝐵 =⎢ ⎢9 ⎢ ⎢4 ⎣ 27

3

⎤ ⎥ 4 ⎥ ⎥, 9 ⎥ ⎥ 16 ⎥ 27 ⎦ 3

4 9 7 27

which is a positive matrix with rows summing to 1. Thus, we may apply the Perron–Frobenius theorem to compute 𝑛

lim (𝐵3 ) .

𝑛→∞

A computation shows that the left eigenvector 𝑢𝑇 in Theorem 15.16, (4) is 1 1 1 𝑢𝑇 = [ ]. 6 3 2 Thus, 1

1

1

⎡6 3 2⎤ ⎢ ⎥ ⎢1 1 1⎥ 3 𝑛 𝑇 ⎢ ⎥𝐴 . lim (𝐵 ) 𝐴1 = (𝑒𝑢 ) 𝐴1 = ⎢6 3 2⎥ 1 𝑛→∞ ⎥ ⎢ ⎢1 1 1⎥ ⎣6 3 2⎦ From this we can conclude that (𝑃𝑛 ) converges and its limit is the point 𝑎 𝑏 𝑏 𝑐 𝑐 𝑎 𝑎 𝑏 𝑐 ( 1 + 2 + 3, 1 + 2 + 3, 1 + 2 + 3). 6 3 2 6 3 2 6 3 2 In this case, we can diagonalize the matrix 𝐵3 . Thus, an alternate approach would be to write 𝐵3 = 𝑃𝐷𝑃−1 with 𝐷 diagonal and to use this to compute the limit. For more on this see [40]. Suppose that 𝐴 is a nonnegative matrix and consider the sequence of powers of 𝐴; that is, (𝐴𝑛 ). Then several questions arise. Project 8. Let 𝐴 be a 2 × 2 matrix with eigenvalues 𝑎1 and 𝑏1 in 𝔻. Let 𝑎𝑛 and 𝑏𝑛 denote the eigenvalues of 𝐴𝑛 , and let 𝑠𝑛 denote the length of the minor axis of the ellipse bounding 𝑊(𝐴𝑛 ). Does the limit of (𝑎𝑛 ), (𝑏𝑛 ), or (𝑠𝑛 ) exist? If so, what are the limits? If not, explain why they do not exist. Consider the question in the event that 𝑎1 and 𝑏1 are arbitrary complex numbers.

15.9. Similarity and the Numerical Range

231

For further investigation, consider 𝑘 × 𝑘 matrices. While this will not be easy, you may obtain answers for special classes of matrices. Since the elliptical range theorem no longer applies for 𝑘 > 2, you might replace the question about length of the minor axis by the question of what happens to the spectral radius (see Section 15.9) and the numerical radius. For further research on Benford’s law, see [13, Chapter 7], [12], and [112].

15.9 Similarity and the Numerical Range In this project, we consider the numerical radius introduced in (15.7) and a new object called the spectral radius. Given a matrix 𝐴, the spectral radius of 𝐴 is 𝜌(𝐴) = max{|𝜆| ∶ 𝜆 an eigenvalue of 𝐴}. Let us examine this definition. First, you should explain why 𝜌(𝐴) ≤ ‖𝐴‖. Can you give an example of a matrix 𝐴 with 𝜌(𝐴) < ‖𝐴‖? This inequality gives a rough bound on 𝜌(𝐴). We already know that if two matrices 𝐴 and 𝐵 are unitarily equivalent, then they have the same numerical range. In this project we explore some of the following questions. Are the spectral radii of 𝐴 and 𝐵 equal? What happens if we assume only that 𝐴 and 𝐵 are similar; in other words, if there exists an invertible matrix 𝑆 such that 𝑆 −1 𝐴𝑆 = 𝐵: Is 𝜌(𝐴) = 𝜌(𝐵)? Is 𝑊(𝐴) = 𝑊(𝐵)? We begin this investigation with some concepts for you to explore before digging into the project. First, look back at Section 15.8. What does the Perron–Frobenius theorem (Theorem 15.16) say about the spectral radius of positive matrices? We turn our attention to some particular examples. Consider the matrix 2 1 𝐴=[ ]. 1 2 Find the eigenvalues of 𝐴, and then find 𝜌(𝐴). Choosing three different invertible 2 × 2 matrices 𝑆 (do not choose the identity!) investigate the eigenvalues of 𝑆 −1 𝐴𝑆. Now answer the following: If 𝐴1 and 𝐴2 are unitarily equivalent matrices, must they have the same eigenvalues? Still assuming 𝐴1 and 𝐴2 are unitarily equivalent, must they have

232

Chapter 15. Fourteen Projects for Fourteen Chapters

the same spectral radius? In this project, we investigate (among other things) what, if anything, changes if 𝐴1 and 𝐴2 are assumed to be similar rather than unitarily equivalent, and we consider the numerical radius rather than the spectral radius. We have the following theorem about 𝜌(𝐴) that provides a little more insight. Theorem 15.17. Let 𝐴 be a 𝑘 × 𝑘 matrix. Then 𝜌(𝐴) = lim𝑛→∞ ‖𝐴𝑛 ‖1/𝑛 . We prove this in the special case that 𝐴 is Hermitian; that is, 𝐴 = 𝐴⋆ . Proof in the case that 𝐴 is Hermitian. Since 𝐴 is Hermitian, we can find an orthonormal basis for ℂ𝑛 consisting of eigenvectors of 𝐴. Denote this basis by {𝑣1 , … , 𝑣𝑛 }. Let 𝜆0 denote an eigenvalue of 𝐴 of maximum 𝑛 modulus. Now let 𝑥 ∈ ℂ𝑛 with ‖𝑥‖ = 1 and write 𝑥 = ∑𝑗=1 𝛼𝑗 𝑣𝑗 for 𝑛

an appropriate choice of 𝛼1 , … , 𝛼𝑛 with ∑𝑗=1 |𝛼𝑗 |2 = 1. Then, writing 𝐴𝑣𝑗 = 𝜆𝑗 𝑣𝑗 , we have 2

𝑛 ‖ 𝑛 ‖ ‖𝐴𝑥‖2 = ‖‖ ∑ 𝛼𝑗 𝜆𝑗 𝑣𝑗 ‖‖ = ∑ |𝛼𝑗 |2 |𝜆𝑗 |2 ≤ |𝜆0 |2 . ‖𝑗=1 ‖ 𝑗=1

So ‖𝐴‖ ≤ |𝜆0 |. On the other hand, if 𝑥0 is a unit eigenvector corresponding to 𝜆0 , then ‖𝐴‖ ≥ ‖𝐴𝑥0 ‖ = |𝜆0 |. Thus, ‖𝐴‖ = |𝜆0 | = 𝜌(𝐴). For each integer 𝑛 ∈ ℤ+ , we get ‖𝐴𝑛 ‖ ≤ ‖𝐴‖𝑛 = |𝜆0 |𝑛 and ‖𝐴𝑛 ‖ ≥ ‖𝐴 𝑥0 ‖ = |𝜆0 |𝑛 , so ‖𝐴𝑛 ‖ = |𝜆0 |𝑛 and ‖𝐴𝑛 ‖1/𝑛 = |𝜆0 |. Thus, 𝑛

lim ‖𝐴𝑛 ‖1/𝑛 = |𝜆0 | = ‖𝐴‖ = 𝜌(𝐴).

𝑛→∞

The proof for general matrices is more difficult. One accessible proof can be found online.11 Recall that we denote the set of eigenvalues or spectrum of 𝐴 by 𝜎(𝐴) and that 𝑤(𝐴) denotes the numerical radius defined in (15.8). We can now begin investigating 𝑤(𝐴). 11 http://users.cms.caltech.edu/ jtropp/notes/Tro01-Spectral-Radius. ̃ pdf (accessed 12/15/2017)

15.9. Similarity and the Numerical Range

233

Exercise 15.18. Let 𝐴 and 𝐵 be two 𝑛 × 𝑛 matrices. Prove statements (1)–(6) and answer parts (7) and (8). (1) The numerical radius satisfies 𝑤(𝐴) = 0 if and only if 𝐴 = 0. (2) For every constant 𝛼, we have 𝑤(𝛼𝐴) = |𝛼|𝑤(𝐴). (3) The numerical radius satisfies 𝑤(𝐴 + 𝐵) ≤ 𝑤(𝐴) + 𝑤(𝐵). (4) If 𝐴 is a normal matrix, then 𝑤(𝐴) = 𝜌(𝐴). (5) If 𝐴 is a Hermitian matrix with 𝑤(𝐴) ≤ 1, then ‖𝐴‖ = 𝑤(𝐴) and thus ‖𝐴‖ ≤ 1. (6) The following string of inequalities holds: 𝜌(𝐴) ≤ 𝑤(𝐴) ≤ ‖𝐴‖ ≤ 2𝑤(𝐴). (7) Is 𝑤(𝐴𝐵) = 𝑤(𝐴)𝑤(𝐵)? If not, can you find examples of when these two are equal? (Looking for a more challenging question? Try giving conditions for the two to be equal.) (8) Do similar matrices have the same numerical radius? As in Exercise 15.18 (6), it is possible to show that the equivalent statement is true for bounded operators. Thus, if 𝑇 is a bounded operator, then ‖𝑇𝑥‖ ≤ 2 whenever 𝑤(𝑇) ≤ 1 and ‖𝑥‖ ≤ 1. In fact, Theorem 15.20 is true (and you can prove it with the tools you have thus far!). The proof for general bounded operators will be precisely the same, so we state it in its most general form; see [97, Theorem 3.1]. Another interesting related result can be found in [44]. Exercise 15.19. Prove the following theorem, using the rough outline provided. Theorem 15.20. Let 𝑇 be a bounded operator on a Hilbert space 𝐻, and let 𝑥 ∈ 𝐻. If 𝑤(𝑇) ≤ 1 and ‖𝑥‖ ≤ 1, then ‖𝑇𝑥‖2 ≤ 2 + 2√1 − |⟨𝑇𝑥, 𝑥⟩|2 .

234

Chapter 15. Fourteen Projects for Fourteen Chapters

Proof Outline. Given 𝑥 ∈ 𝐻, multiply 𝑇 by a unimodular constant so that you may assume ⟨𝑇𝑥, 𝑥⟩ ≥ 0. Consider the operators 𝐻𝑇 = (𝑇 + 𝑇 ⋆ )/2 and 𝐾𝑇 = (𝑇 − 𝑇 ⋆ )/(2𝑖), which play an important role in Chapter 13. Then the key inequality is ‖𝑇𝑥 − ⟨𝑇𝑥, 𝑥⟩𝑥‖ ≤ ‖𝐻𝑇 𝑥 − ⟨𝐻𝑇 𝑥, 𝑥⟩𝑥‖ + ‖𝐾𝑇 𝑥 − ⟨𝐾𝑇 𝑥, 𝑥⟩𝑥‖. A computation shows that ‖𝑇𝑥 − ⟨𝑇𝑥, 𝑥⟩𝑥‖2 = ‖𝑇𝑥‖2 − |⟨𝑇𝑥, 𝑥⟩|2 . Similarly, you should show that ‖𝐻𝑇 𝑥 − ⟨𝐻𝑇 𝑥, 𝑥⟩𝑥‖2 = ‖𝐻𝑇 𝑥‖2 − |⟨𝐻𝑇 𝑥, 𝑥⟩|2 ≤ 1 − |⟨𝑇𝑥, 𝑥⟩|2 and ‖𝐾𝑇 𝑥 − ⟨𝐾𝑇 𝑥, 𝑥⟩𝑥‖2 = ‖𝐾𝑇 𝑥‖2 − |⟨𝐾𝑇 𝑥, 𝑥⟩|2 ≤ 1. Then use these inequalities to complete the proof. You should have shown that the numerical radius and spectral radius are both unitary invariants, but the numerical radius is not invariant under similarity while the spectral radius is. What if we restrict ourselves to special matrices? What happens if we restrict ourselves to operators in the class 𝒮𝑛 ? Project 9. Sometimes similar matrices have the same numerical range and sometimes they do not. Investigate this, beginning with 2 × 2 matrices. While you work on this question, think about the distinction between the numerical radius and the spectral radius: When do two 2 × 2 matrices have the property that 𝜌(𝐴) = 𝑤(𝐴)? When is 𝑤(𝐴) = ‖𝐴‖? What about the 𝑛 × 𝑛 matrices that represent compressions of the shift operator? And, as direction for further work, what happens for general 𝑛 × 𝑛 matrices?

15.10 The Importance of Being Zero Sometimes the fact that a particular value lies in the numerical range tells you something about your matrix. For example, suppose you have a matrix 𝐴 and you know that the trace of 𝐴, denoted tr(𝐴), lies in the numerical range. It turns out that this implies that 𝐴 is unitarily equivalent to a matrix with tr(𝐴) as the (1, 1) entry and zeros in every other position on the diagonal. The proof of this (taken from [46]) is surprisingly elementary.

15.10. The Importance of Being Zero

235

Theorem 15.21. Let 𝐴 be an 𝑛 ×𝑛 matrix with complex entries. Then 𝐴 is unitarily equivalent to a matrix with main diagonal (tr(𝐴), 0, … , 0) if and only if tr(𝐴) ∈ 𝑊(𝐴). Proof. Suppose 𝐴 has a main diagonal given by (tr(𝐴), 0, … , 0). By considering the unit vector 𝑥 = [𝑥1 ⋯ 𝑥𝑛 ]𝑇 with 𝑥1 = 1 and all other entries 0, we see that tr(𝐴) ∈ 𝑊(𝐴). From here, the full result follows, so it is the other direction that is interesting. We turn to proving it. Suppose that tr(𝐴) ∈ 𝑊(𝐴). The proof will be by induction on 𝑛. The case 𝑛 = 1 is clear, so we assume 𝑛 > 1. Since we assume tr(𝐴) ∈ 𝑊(𝐴), there is a unit vector 𝑥 with ⟨𝐴𝑥, 𝑥⟩ = tr(𝐴). Let (𝑥1 , … , 𝑥𝑛 ) be an ordered orthonormal basis for ℂ𝑛 with 𝑥1 = 𝑥. Thus, 𝐴 is unitarily equivalent to a matrix of the form tr(𝐴) 𝐵 𝑀=[ ], 𝐶 𝐷 where 𝐷 is (𝑛 − 1) × (𝑛 − 1). So, the trace of 𝐴 and 𝑀 are equal and therefore 𝑡𝑟(𝐷) = 0. If 𝑛 = 2, then 𝐷 is the zero matrix and the result is true in this case. If 𝑛 > 2, we need to show that 𝑡𝑟(𝐷) = 0 ∈ 𝑊(𝐷), and then we can apply the induction hypothesis to 𝐷. Let 𝜆1 , … , 𝜆𝑛−1 be the eigenvalues of 𝐷, repeated according to multiplicity. Then 1 1 (𝜆 + ⋯ + 𝜆𝑛−1 ) = 𝑡𝑟(𝐷) = 0. 𝑛−1 1 𝑛−1 But then we know several things: First, 𝜆𝑗 ∈ 𝑊(𝐷) for all 𝑗; second, 0 is in the convex hull of the 𝜆𝑗 ; and third, 𝑊(𝐷) is convex. Thus, 0 ∈ 𝑊(𝐷). Applying the induction hypothesis to 𝐷, we obtain a unitary matrix 𝑈 such that 𝑈 ⋆ 𝐷𝑈 has a main diagonal consisting of all zeros. Letting 1 𝑉=[ 0

0 ] 𝑈

and noting that 𝑉 is unitary and that 𝑉 ⋆ 𝑀𝑉 has main diagonal (tr(𝐴), 0, … , 0), the desired conclusion follows.

236

Chapter 15. Fourteen Projects for Fourteen Chapters

Certain things follow easily from this. For example, since the matrix 𝐴 − (tr(𝐴)/𝑛)𝐼 has trace zero, Theorem 15.21 shows that every matrix is unitarily equivalent to a matrix with constant main diagonal. Interesting things also happen when 0 ∈ 𝑊(𝐴). In Theorem 15.21 we saw that if the trace of 𝐴 is 0 and lies in 𝑊(𝐴), then 𝐴 is unitarily equivalent to a matrix with zeros on the main diagonal. For bounded operators on a Hilbert space, there are many interesting related results. For example, given a compact operator 𝑇 acting on a separable infinite dimensional Hilbert space, the numerical range of 𝑇 is closed if and only if 0 is in the numerical range; see [100]. Following this, Agler showed that for a bounded operator 𝑇 on a Hilbert space, if 𝑇 is compact and 0 lies in the interior of the numerical range, then the boundary of 𝑊(𝑇) is particularly nice; see [3]. Since the location of zero in (or not in) the numerical range of a matrix (or operator) can give you information about your matrix (or operator), people study what is known as the zero inclusion question. For matrices this is a natural question. We study eigenvectors corresponding to nonzero eigenvalues, which we can think of as those vectors 𝑥 such that 𝐴𝑥 and 𝑥 are parallel. The zero inclusion question studies vectors 𝑥 for which 𝐴𝑥 and 𝑥 are orthogonal. Project 10. Investigate the zero inclusion question: Given a matrix 𝐴, is 0 ∈ 𝑊(𝐴)? One class of operators for which one can say quite a bit is the class of composition operators on the Hardy space 𝐻 2 . For this, let 𝜑 ∶ 𝔻 → 𝔻 be an analytic map. The composition operator 𝐶𝜑 ∶ 𝐻 2 → 𝐻 2 is defined by 𝐶𝜑 (𝑓) = 𝑓 ∘ 𝜑. A discussion of the boundedness of these operators as well as other elementary properties can be found in [34, Chapter 3] and [140, Chapter 1]. Here we concentrate only on a brief discussion of the zero inclusion question. It is known, and not difficult to show, that if 𝜑 is not one-to-one, then 0 lies in the interior of 𝑊(𝐶𝜑 ). Thus, the interesting case is the one in which 𝜑 is injective. In [16], the authors show that if 𝜑 ≠ 𝑖𝑑 is an arbitrary self-map of 𝔻, then 0 ∈ 𝑊(𝐶𝜑 ). More recently, Higdon [80] showed that if 𝜑 is not a linear fractional self-map of 𝔻 and has its attractive fixed point on 𝕋, then 0 lies in the interior of 𝑊(𝐶𝜑 ). In fact, taken together, the results in these two papers give a complete description of when 0 is in the numerical range of a composition operator. We learn more about fixed points in Chapter 14 and Section 15.14.

15.11. Building a Better Interpolant

237

15.11 Building a Better Interpolant We studied one way to approach interpolation in Chapter 11. There is another natural way to handle interpolation problems (and it may have been the one you thought of first). As before, we have real numbers 𝑦1 , … , 𝑦𝑛 as well as 𝑎𝑗 and 𝑥𝑗 satisfying 𝑎1 < 𝑥1 < 𝑎2 < 𝑥2 < ⋯ < 𝑎𝑛 < 𝑥𝑛 . We are interested only in interpolating the 𝑥𝑗 to 𝑦𝑗 , and we relax the condition on the poles. We require only that the poles lie among the 𝑎𝑗 ; that is, the poles may be a proper subset of {𝑎1 , … , 𝑎𝑛 }. We try a step-by-step approach. Getting the interpolation right at one point, 𝑥1 , is easy. We can start with the constant function 𝐾1 defined by 𝐾1 (𝑧) = 𝑦1 = 𝜆1 . To satisfy the interpolation condition at 𝑥2 and possibly create a pole at 𝑎1 and nowhere else, we can consider the degree-1 function 𝐾2 defined by 𝑧 − 𝑥1 𝐾2 (𝑧) = 𝜆1 + 𝜆2 𝑧 − 𝑎1 for an appropriate choice of 𝜆2 . (Really? When would we not create a pole at 𝑎1 ?) This suggests a general approach. Define 𝐾ℓ as 𝐾ℓ (𝑧) = 𝜆1 + 𝜆2

2 ℓ−1 𝑧 − 𝑥𝑗 𝑧 − 𝑥𝑗 𝑧 − 𝑥1 + 𝜆3 ∏ + ⋯ + 𝜆ℓ ∏ . 𝑧 − 𝑎1 𝑧 − 𝑎𝑗 𝑧 − 𝑎𝑗 𝑗=1 𝑗=1

For an appropriate choice of 𝜆1 , … , 𝜆ℓ , we will see that 𝐾ℓ has degree at most ℓ − 1, maps 𝑥𝑗 to 𝑦𝑗 for 𝑗 = 1, … , ℓ, and has its poles among 𝑎1 , … , 𝑎ℓ−1 (if it has any at all). Thus, 𝐾𝑛 would suffice for our interpolation; it is a rational function of degree at most 𝑛 − 1. We can buy a bit of freedom that will come in handy below by adding a term and considering a degree-𝑛 rational function—the same degree we needed in Chapter 11 for the interpolation. Thus, we obtain as a solution to our problem the function 𝑛+1

𝑘−1

𝐾𝑛+1 (𝑧) = ∑ (𝜆𝑘 ∏ 𝑘=1

𝑗=1

𝑧 − 𝑥𝑗 ), 𝑧 − 𝑎𝑗

(15.9)

238

Chapter 15. Fourteen Projects for Fourteen Chapters

where 𝜆𝑗 for 𝑗 = 1, … , 𝑛 are determined as explained above and 𝜆𝑛+1 ≠ 0 but is otherwise arbitrary. We refer to the interpolating function written in this form as the Newton form. One of the great advantages of writing the function that does the interpolation in Newton form is that it is relatively easy to extend the interpolation to additional points. For example, suppose you already have a function 𝐾ℓ and you want to preserve the interpolation of 𝑥𝑗 to 𝑦𝑗 for 𝑗 = 1, … , ℓ, have all of its poles among 𝑎𝑗 for 𝑗 = 1, … , ℓ − 1, add possibly an additional pole at 𝑎ℓ , and add an interpolation pair mapping 𝑥ℓ+1 to 𝑦ℓ+1 (satisfying the appropriate conditions). Then we need only add one term and determine one coefficient to ensure that 𝐾ℓ+1 does the interpolation. In Theorem 11.6 we found a function 𝐺 ∶ ℂ∗ → ℂ∗ that produced the same interpolation results under the same hypotheses as the function 𝐾𝑛+1 . We called 𝐺 the Lagrange form of the interpolation function. How do the two interpolating functions, 𝐺 and 𝐾𝑛+1 , compare? To a large extent the two functions are the same but may nevertheless differ in a crucial aspect. To make this statement precise we first ask you to work the following exercise. Exercise 15.22. Show that if 𝑥1 , … , 𝑥𝑛 and 𝑎1 , … , 𝑎𝑛 are 2𝑛 distinct real numbers, then the set 𝑛

𝑛

{∏(𝑧 − 𝑎𝑗 ), (𝑧 − 𝑥1 ) ∏(𝑧 − 𝑎𝑗 ), … , 𝑗=1

𝑗=2 𝑘

𝑛

𝑛

∏(𝑧 − 𝑥𝑗 ) ∏ (𝑧 − 𝑎𝑗 ), … , ∏(𝑧 − 𝑥𝑗 )} 𝑗=1

𝑗=𝑘+1

𝑗=1

forms a basis for the space of complex polynomials of degree at most 𝑛. This exercise will help us establish the following theorem [64, Theorem 5]. Theorem 15.23. Let 𝑎1 , … , 𝑎𝑛 and 𝑥1 , … , 𝑥𝑛 be distinct real numbers. Let 𝐹 be a rational function of degree 𝑛 such that 𝐹 has poles at 𝑎1 , … , 𝑎𝑛 . Then there exist uniquely determined constants 𝛼1 , … , 𝛼𝑛+1 such that 𝐹(𝑧) = 𝛼1 + 𝛼2 (

𝑛 𝑧 − 𝑥𝑗 𝑧 − 𝑥1 ) + ⋯ + 𝛼𝑛+1 ∏ ( ). 𝑧 − 𝑎1 𝑧 − 𝑎𝑗 𝑗=1

15.11. Building a Better Interpolant

239

Further, if 𝐻(𝑧) = 𝛽1 + 𝛽2 (

𝑛 𝑧 − 𝑥𝑗 𝑧 − 𝑥1 ) + ⋯ + 𝛽𝑛+1 ∏ ( ) 𝑧 − 𝑎1 𝑧 − 𝑎𝑗 𝑗=1

and 𝐻(𝑥𝑗 ) = 𝐹(𝑥𝑗 ) for 𝑗 = 1, … , 𝑛, then 𝛼𝑗 = 𝛽𝑗 for 𝑗 = 1, … , 𝑛. If, in addition, there is another point, 𝑧0 , with 𝑧0 ≠ 𝑥𝑗 and 𝑧0 ≠ 𝑎𝑗 for 𝑗 = 1, … , 𝑛 and 𝐹(𝑧0 ) = 𝐻(𝑧0 ), then 𝐹 = 𝐻. Exercise 15.24. Prove Theorem 15.23. For the first part, notice that 𝑛

𝐹(𝑧) ∏ (𝑧 − 𝑎𝑗 ) 𝑗=1

is a polynomial of degree 𝑛 and thus has a unique representation using the basis given in Exercise 15.22. We now compare the Lagrange form of the interpolation function 𝐺 of (11.2) and the Newton form 𝐾𝑛+1 of (15.9). The function 𝐺 is a rational function of degree 𝑛 satisfying the hypotheses of Theorem 15.23. Thus, 𝐺(𝑧) = 𝛾1 + 𝛾2 (

𝑛 𝑧 − 𝑥𝑗 𝑧 − 𝑥1 ) + ⋯ + 𝛾𝑛+1 ∏ ( ), 𝑧 − 𝑎1 𝑧 − 𝑎𝑗 𝑗=1

where 𝛾𝑗 for 𝑗 = 1, … , 𝑛 + 1 are uniquely determined coefficients. The function 𝐾𝑛+1 in (15.9) is already given in this form. Since 𝐺(𝑥𝑗 ) = 𝑦𝑗 = 𝐾𝑛+1 (𝑥𝑗 ) for 𝑗 = 1, … , 𝑛, by Theorem 15.23 we conclude that 𝜆𝑗 = 𝛾𝑗 for 𝑗 = 1, … , 𝑛. Thus, the two interpolation functions 𝐺 and 𝐾𝑛+1 presented in the form given by Theorem 15.23 differ only in the last coefficient. Is this difference significant? Recall from Theorem 11.6 that the function 𝐺 is also strongly real of positive type—a crucial property in translating this to interpolation on 𝔻. (The function 𝐺 also has a pole at each 𝑎𝑗 for 𝑗 = 1, … , 𝑛, while 𝐾𝑛+1 ’s poles are only among these values; it might have fewer poles.) But 𝐾𝑛+1 is not necessarily strongly real of positive type. We can fix that! In the construction of 𝐾𝑛+1 , the last coefficient, 𝜆𝑛+1 , was arbitrary with the exception of the requirement that 𝜆𝑛+1 ≠ 0. It turns out that a proper choice of this coefficient guarantees that 𝐾𝑛+1 will be strongly real of positive type. Such a choice is given in

240

Chapter 15. Fourteen Projects for Fourteen Chapters

[64, Theorem 8]. We encourage you to read the proof showing that 𝑛

𝜆𝑛+1 = (

𝑛

𝑥𝑛 − 𝑎1 𝑛−1 ∑ 𝑦𝑗 − ∑ 𝜆𝑗 , ) 𝑚 𝑗=1 𝑗=1

(15.10)

where 𝑚 = min{|𝑥𝑗 − 𝑥𝑘 | ∶ 𝑗, 𝑘 = 1, … , 𝑛 and 𝑗 ≠ 𝑘} will make 𝐾𝑛+1 strongly real of positive type. We turn to the question of which of the two interpolation functions requires less work to obtain and which is faster to evaluate. To reduce the number of computations necessary to produce 𝐾𝑛+1 , it is often useful to write the Newton form in a slightly different way, called the nested form of 𝐾𝑛+1 . We indicate how to do this for the case 𝐾4 . From (15.9), we have 𝐾4 (𝑧) = 𝜆1 + 𝜆2

2 3 𝑧 − 𝑥𝑗 𝑧 − 𝑥𝑗 𝑧 − 𝑥1 + 𝜆3 ∏ + 𝜆4 ∏ . 𝑧 − 𝑎1 𝑧 − 𝑎𝑗 𝑧 − 𝑎𝑗 𝑗=1 𝑗=1

Factoring, we get 𝐾4 (𝑧) = 𝜆1 +

𝑧 − 𝑥3 𝑧 − 𝑥1 𝑧 − 𝑥2 𝜆 )) , (𝜆2 + (𝜆3 + 𝑧 − 𝑎1 𝑧 − 𝑎2 𝑧 − 𝑎3 4

where 𝜆𝑗 is obtained from 𝐾4 (𝑥𝑗 ) = 𝑦𝑗 for 𝑗 = 1, 2, 3, and using (15.10), 𝑥3 − 𝑎1 2 ) (𝑦1 + 𝑦2 + 𝑦3 ) − (𝜆1 + 𝜆2 + 𝜆3 ) 𝑚 with 𝑚 = min{𝑥3 − 𝑥2 , 𝑥3 − 𝑥1 , 𝑥2 − 𝑥1 }. Writing 𝐾4 in its nested form reduces the number of multiplications required to find the coefficients 𝜆𝑗 and evaluate 𝐾4 at a point 𝑧. 𝜆4 = (

Exercise 15.25. For 𝐾𝑛+1 as in (15.9), write the coefficients 𝜆𝑘 in terms of 𝑥𝑗 , 𝑦𝑗 , and 𝑎𝑗 for appropriate choices of 𝑗. Be sure to choose the last coefficient so as to satisfy the interpolation properties. Exercise 15.26. Find the nested form of 𝐾𝑛+1 for an arbitrary positive integer 𝑛. Project 11. We have now provided two ways of writing a rational function that does the desired interpolation. Analyze the number of computations needed to compute the rational function in each form, Lagrange and Newton (in nested form). Now, assuming you have the rational function, analyze the number of computations needed to evaluate the function at a

15.12. Foci of Algebraic Curves

241

point 𝑧 in the complex plane. Using the techniques we have learned, use each algorithm to create a Blaschke product that satisfies an appropriate interpolation problem. Implement the more efficient algorithm, making sure your algorithm allows you to evaluate the Blaschke product at an arbitrary point 𝑧 ∈ 𝔻. It is possible to produce an algorithm on the open unit disk rather than transforming the problem to one on the upper half-plane. For more on this, see [32]. Project 11 is closely tied to Project 13 below.

15.12 Foci of Algebraic Curves In Chapter 5 we introduced projective coordinates and discussed duality. The geometric space we work in is the complex plane ℂ, and we think of it as being embedded in the two-dimensional real projective plane ℙ2 (ℝ) by 𝑧 = 𝑥 + 𝑖𝑦 ↦ (𝑥, 𝑦, 1). The real projective plane is, in turn, embedded in the complex projective plane ℙ2 (ℂ) in the natural way. We first consider the following question: Given an equation for an algebraic curve 𝒞 in the projective plane, how do we find the equation for the dual curve? Suppose that 𝒞 is given by the homogeneous polynomial equation 𝑓(𝑥, 𝑦, 𝑧) = 0. The tangent lines 𝜆𝑥 + 𝜇𝑦 + 𝜈𝑧 = 0 to the curve 𝒞 satisfy another homogeneous polynomial equation Φ(𝜆, 𝜇, 𝜈) = 0, which is referred to as the dual or tangential equation. This latter polynomial determines the dual curve. The degree of 𝑓 is the degree of the curve 𝒞, and the degree of Φ, which is the degree of the dual curve, is the class of 𝒞. We follow [83] to find the dual Φ to a homogenous polynomial 𝑓. If 𝑓 determines the curve 𝒞 and 𝑃 = (𝑥0 , 𝑦0 , 𝑧0 ) is a point at which the partial derivatives of 𝑓 exist and are not all zero, then 𝑓𝑥 (𝑥0 , 𝑦0 , 𝑧0 )𝑥 + 𝑓𝑦 (𝑥0 , 𝑦0 , 𝑧0 )𝑦 + 𝑓𝑧 (𝑥0 , 𝑦0 , 𝑧0 )𝑧 = 0 is the tangent line to 𝒞 at 𝑃. The line 𝜆𝑥 + 𝜇𝑦 + 𝜈𝑧 = 0 determines the same tangent line if and only if 𝜆 = 𝑓𝑥 (𝑥0 , 𝑦0 , 𝑧0 )𝑡, 𝜇 = 𝑓𝑦 (𝑥0 , 𝑦0 , 𝑧0 )𝑡, and 𝜈 = 𝑓𝑧 (𝑥0 , 𝑦0 , 𝑧0 )𝑡 for some 𝑡 ≠ 0. Finally, we eliminate 𝑥0 , 𝑦0 , 𝑧0 , and 𝑡 to get the homogenous equation Φ(𝜆, 𝜇, 𝜈) = 0.

242

Chapter 15. Fourteen Projects for Fourteen Chapters

Example 15.27. Consider the curve 𝒞 given by 𝑓(𝑥, 𝑦, 𝑧) = 𝛼𝑦 2 𝑧 −𝑥(𝑥 2 + 𝑦 2 ) for some 𝛼 ∈ ℝ. This gives rise to the following equations: 𝜇 𝜆 𝑓𝑥 (𝑥, 𝑦, 𝑧) = −3𝑥 2 − 𝑦 2 = , 𝑓𝑦 (𝑥, 𝑦, 𝑧) = 2𝛼𝑦𝑧 − 2𝑥𝑦 = , 𝑡 𝑡 and 𝜈 𝑓𝑧 (𝑥, 𝑦, 𝑧) = 𝛼𝑦 2 = (15.11) 𝑡 for some 𝑡 ≠ 0. Starting with 𝑓(𝑥, 𝑦, 𝑧) = 0, we get 𝑦(𝛼𝑦𝑧 − 𝑥𝑦) = 𝑥 3 . After squaring both sides, we obtain 2

𝑦 2 (𝛼𝑦𝑧 − 𝑥𝑦) = 𝑥 6 , and using (15.11) this yields 3 𝜈 𝜇2 1 𝜈 + 𝜆) . = − ( 𝛼𝑡 4𝑡 2 27𝑡3 𝛼 Since 𝑡 ≠ 0, we obtain the dual equation

Φ(𝜆, 𝜇, 𝜈) = 27𝛼 2 𝜈𝜇2 + 4(𝜈 + 𝛼𝜆)3 = 0. Note that the curve 𝒞 in Example 15.27 is of degree 3 and class 3. It is time to try one of your own. Exercise 15.28. Find the equation of the dual curve to the cubic 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 3 + 𝑦 3 + 𝑧3 = 0. What are the degree and class of this curve? You should also check that the unit circle 𝑥 2 + 𝑦 2 − 𝑧2 = 0 has the dual equation 𝜆2 + 𝜇2 − 𝜈 2 = 0; that is, it is self-dual. The tangent lines to the unit circle that pass through the origin, (0, 0, 1), are of particular interest, and you should check that these lines are tangent to the unit circle at the points 𝜔+ = (1, 𝑖, 0) and 𝜔− = (1, −𝑖, 0). These two points are called the circular points. A point 𝑆 is called a focus of a curve 𝒞 if the tangent lines through 𝜔+ and 𝜔− intersect at 𝑆. If 𝑆 is real, that is, it has coordinates (𝑥, 𝑦, 1) with 𝑥 and 𝑦 real, then we call it a real focus. In general, a curve 𝒞 has (deg 𝒞) real foci and (deg 𝒞)2 − (deg 𝒞) nonreal foci; see [83, p. 69]. If a curve passes through the line at infinity, then the number of foci is reduced. Also, if 𝜔+ and 𝜔− are on the curve, then the

15.12. Foci of Algebraic Curves

243

two tangent lines through 𝜔+ and 𝜔− , respectively, intersect in a point that is called a singular focus. Exercise 15.29. Suppose that 𝑎 and 𝑏 are real with 𝑏 > 𝑎. Using the definition above, find all four foci of the ellipse 𝑥 2 /𝑎2 + 𝑦 2 /𝑏2 = 1. Show that the real foci are what you would expect the foci of this ellipse to be. (Note that you first need to turn the ellipse into a projective curve by homogenizing the equation.) A generalization of Theorem 15.4 will be of use in this project. Theorem 15.30 (Siebeck). Let 𝑧1 , … , 𝑧𝑛 be complex numbers, no three of which are collinear, and let 𝑚1 , … , 𝑚𝑛 be real numbers with 𝑚𝑗 + 𝑚𝑘 ≠ 0 for all 𝑗, 𝑘 with 1 ≤ 𝑗, 𝑘 ≤ 𝑛. Then the zeros of the function 𝑛

𝑚𝑗 𝑧 − 𝑧𝑗 𝑗=1

𝐹(𝑧) = ∑

(15.12)

are the foci of an algebraic curve of class 𝑛 − 1 that is tangent to each line segment 𝑧𝑗 𝑧𝑘 at a point that divides the line segment in the ratio 𝑚𝑗 ∶ 𝑚𝑘 , for 1 ≤ 𝑗, 𝑘 ≤ 𝑛. Proof (adapted from [108]). Let 𝑧 = 𝑥+𝑖𝑦 and 𝑧𝑗 = 𝑥𝑗 +𝑖𝑦𝑗 with 𝑥, 𝑦, 𝑥𝑗 , and 𝑦𝑗 real. Further, we let 𝑡 = 1/𝑧, which allows us to rewrite (15.12) as 𝑛

𝑚𝑗 . 𝑡𝑥𝑗 + 𝑖𝑡𝑦𝑗 − 1 𝑗=1

𝐹(1/𝑡) = −𝑡 ∑

(15.13)

Taking our motivation from (15.13), we let ℒ𝑗 (𝜆, 𝜇, 𝜈) = 𝜆𝑥𝑗 + 𝜇𝑦𝑗 + 𝜈 and consider the equation 𝑛

𝑛

𝑚𝑗 ℒ (𝜆, 𝜇, 𝜈) 𝑗=1 𝑗

Φ(𝜆, 𝜇, 𝜈) = (∏ ℒ𝑘 (𝜆, 𝜇, 𝜈)) ∑ 𝑘=1 𝑛

= ∑ 𝑚𝑗 ∏ ℒ𝑘 (𝜆, 𝜇, 𝜈) = 0. 𝑗=1

(15.14)

𝑘≠𝑗

Now (15.14) is the dual of an equation of an algebraic curve of class 𝑛−1, which we denote by 𝒞. If 𝑧 = 𝑥 + 𝑖𝑦 ≠ 0 is a zero of 𝐹, then 𝑛

𝑚𝑗 = −𝑧𝐹(𝑧) = 0 ℒ (1/(𝑥 + 𝑖𝑦), 𝑖/(𝑥 + 𝑖𝑦), −1) 𝑗=1 𝑗 ∑

244

Chapter 15. Fourteen Projects for Fourteen Chapters

and

𝑛

𝑚𝑗 = 𝑧𝐹(𝑧) = 0. ℒ (−1/(𝑥 − 𝑖𝑦), 𝑖/(𝑥 − 𝑖𝑦), 1) 𝑗=1 𝑗 ∑

We conclude that (1/(𝑥 + 𝑖𝑦), 𝑖/(𝑥 + 𝑖𝑦), −1) and (−1/(𝑥 − 𝑖𝑦), 𝑖/(𝑥 − 𝑖𝑦), 1) are both solutions of Φ(𝜆, 𝜇, 𝜈) = 0 and hence both are tangent lines to 𝒞. We can check that both lines contain the point 𝑧, which we are identifying with the point (𝑥, 𝑦, 1). In addition, the first tangent line contains 𝜔+ and the second one contains 𝜔− . Thus, if 𝑧 ≠ 0 is a zero of 𝐹, then 𝑧 is a focus of the curve 𝒞. If 𝐹(0) = 0, then 𝑛

𝑚𝑗 = −𝐹(0) = 0 𝑥 + 𝑖𝑦𝑗 𝑗=1 𝑗 ∑

𝑛

and

𝑚𝑗 = −𝐹(0) = 0. 𝑥 − 𝑖𝑦𝑗 𝑗=1 𝑗 ∑

This implies that the line (1, 𝑖, 0), which contains 𝜔+ , and the line (1, −𝑖, 0), which contains 𝜔− , are both tangents to 𝒞. Their intersection point 𝑧 = 0, which we identify with (0, 0, 1), is thus also a focus of 𝒞. Taken together, these two cases show that all zeros of 𝐹 are foci of the curve 𝒞. Since 𝐹 has 𝑛 − 1 zeros (where 𝑛 − 1 is the class of 𝒞), they are all the real foci of 𝒞. Denote the line that passes through the points (𝑥𝑗 , 𝑦𝑗 , 1) and (𝑥𝑘 , 𝑦𝑘 , 1) for fixed 𝑗, 𝑘 with 𝑗 ≠ 𝑘 by (𝜆0 , 𝜇0 , 𝜈0 ). Then ℒ𝑗 (𝜆0 , 𝜇0 , 𝜈0 ) = 0 = ℒ𝑘 (𝜆0 , 𝜇0 , 𝜈0 ) and hence, by (15.14), we see that Φ(𝜆0 , 𝜇0 , 𝜈0 ) = 0. That is, the line through 𝑧𝑗 and 𝑧𝑘 is tangent to 𝒞. Finally, we calculate the point of tangency on the line (𝜆0 , 𝜇0 , 𝜈0 ). Using duality, this point must be 𝜕Φ 𝜕Φ 𝜕Φ (𝜆 , 𝜇 , 𝜈 ), (𝜆 , 𝜇 , 𝜈 ), (𝜆 , 𝜇 , 𝜈 )) . 𝜕𝜆 0 0 0 𝜕𝜇 0 0 0 𝜕𝜈 0 0 0 The partials are 𝜕Φ (𝜆 , 𝜇 , 𝜈 ) = (𝑚𝑗 𝑥𝑘 + 𝑚𝑘 𝑥𝑗 ) ∏ ℒℓ (𝜆0 , 𝜇0 , 𝜈0 ), 𝜕𝜆 0 0 0 ℓ≠𝑗,𝑘 𝑃=(

𝜕Φ (𝜆 , 𝜇 , 𝜈 ) = (𝑚𝑗 𝑦𝑘 + 𝑚𝑘 𝑦𝑗 ) ∏ ℒℓ (𝜆0 , 𝜇0 , 𝜈0 ), and 𝜕𝜇 0 0 0 ℓ≠𝑗,𝑘 𝜕Φ (𝜆 , 𝜇 , 𝜈 ) = (𝑚𝑗 + 𝑚𝑘 ) ∏ ℒℓ (𝜆0 , 𝜇0 , 𝜈0 ). 𝜕𝜈 0 0 0 ℓ≠𝑗,𝑘

15.13. Companion Matrices and Kippenhahn

245

Since (𝑚𝑗 + 𝑚𝑘 ) ∏ℓ≠𝑗,𝑘 ℒℓ (𝜆0 , 𝜇0 , 𝜈0 ) ≠ 0, we may divide each coordinate by this expression, completing the calculation for the point of tangency: 𝑚𝑗 𝑥𝑘 + 𝑚𝑘 𝑥𝑗 𝑚𝑗 𝑦𝑘 + 𝑚𝑘 𝑦𝑗 𝑃=( , , 1) . 𝑚𝑗 + 𝑚𝑘 𝑚𝑗 + 𝑚𝑘 Project 12. To investigate the foci of algebraic curves, start by looking at conics. Find their class and show that if the foci are calculated using the general definition as given in this project, then you obtain the usual foci of conics. Then look at Blaschke products and show that the zeros of the Blaschke products are in fact often (when?) the foci of the Blaschke curve. It might be helpful to use Lemma 4.2 in the process. When we consider algebraic curves that are not conics, the foci do not have a nice geometric meaning that can be used to construct the curve, which is in contrast to the case for ellipses, hyperbolas, and parabolas. Can we say anything about the foci? Where are they located with respect to the Blaschke curves? If you want to read about general foci, you probably need to look at older texts. For a more recent view of things, some interesting ideas can be found in [101].

15.13 Companion Matrices and Kippenhahn To start this project off, we look at some special 3×3 matrices and obtain more information about the numerical range. First, a bit of notation: Let 𝑧1 , 𝑧2 , and 𝑧3 be complex numbers, and let 𝑝 be a polynomial defined by 𝑝(𝑧) = (𝑧 − 𝑧1 )(𝑧 − 𝑧2 )(𝑧 − 𝑧3 ) = 𝑧3 + 𝑎𝑧2 + 𝑏𝑧 + 𝑐 with 𝑎 = −(𝑧1 + 𝑧2 + 𝑧3 ), 𝑏 = 𝑧1 𝑧2 + 𝑧1 𝑧3 + 𝑧2 𝑧3 , and 𝑐 = −𝑧1 𝑧2 𝑧3 . The companion matrix associated with 𝑝 is 0 𝑃=[0 −𝑐

1 0 −𝑏

0 1 ]. −𝑎

(15.15)

For higher-degree polynomials, companion matrices are defined in an analogous way. The polynomial 𝑝 is both the characteristic polynomial as well as the minimal polynomial of 𝑃, which explains why it makes sense to think of the matrix as a companion to 𝑝; see [84, p. 230]. Note

246

Chapter 15. Fourteen Projects for Fourteen Chapters

that the eigenvalues of 𝑃 are 𝑧1 , 𝑧2 , and 𝑧3 . What is the numerical range of 𝑃? Theorem 15.31. Given nonzero complex numbers 𝑧1 and 𝑧2 with |𝑧1 | = |𝑧2 |, there exists at least one and at most three complex numbers, one of which we denote by 𝑧3 , such that the companion matrix of 𝑝, where 𝑝(𝑧) = (𝑧 − 𝑧1 )(𝑧 − 𝑧2 )(𝑧 − 𝑧3 ), has an elliptical numerical range (possibly degenerate) with foci at 𝑧1 and 𝑧2 . The proof of Theorem 15.31 uses the material in Chapter 13 and can be found in [23]. We provide a proof of this theorem using Theorem 15.32, which is a more general form of Theorem 13.5. Theorem 15.32. Let 𝐴 be a 3 × 3 matrix with eigenvalues 𝑧1 , 𝑧2 , and 𝑧3 . Then 𝑊(𝐴) is an elliptical disk with minor axis of length √𝑑, where 3

𝑑 = tr(𝐴⋆ 𝐴) − ∑ |𝑧𝑗 |2 𝑗=1

and foci at 𝑧1 and 𝑧2 if and only if 3

(1) the constant 𝑑 = tr(𝐴⋆ 𝐴) − ∑𝑗=1 |𝑧𝑗 |2 > 0, (2) one of the eigenvalues (which we choose to be 𝑧3 ) satisfies 3

𝑧3 = tr(𝐴) + (1/𝑑) ( ∑ |𝑧𝑗 |2 𝑧𝑗 − tr(𝐴⋆ 𝐴2 )) ,

(15.16)

𝑗=1

and (3) the eigenvalue 𝑧3 lies inside the ellipse with foci 𝑧1 , 𝑧2 and minor axis of length √𝑑. Recall that a matrix is said to be reducible if it is unitarily equivalent to the direct sum of two other matrices. While our proof of Theorem 15.31 focuses on irreducible matrices, there is a proposition about reducible companion matrices that can be used to complete the proof in the reducible case. This proposition (that is, Proposition 15.33) says that a 3 × 3 companion matrix is reducible when its eigenvalues are equally distributed in a sense that will be made precise below. The next two propositions are due to Gau and Wu [59].

15.13. Companion Matrices and Kippenhahn

247

Proposition 15.33. A 3 × 3 companion matrix is reducible if and only if its eigenvalues can be written as 𝑎𝜔1 , 𝑎𝜔2 , and (1/𝑎)𝜔3 , where 𝑎 ≠ 0 is a complex number and 𝜔1 , 𝜔2 , and 𝜔3 are the three cube roots of unity. Proposition 15.34 will allow us to move the eigenvalues of the irreducible 3 × 3 companion matrices for which we compute the numerical range, thus simplifying the computation in the proof of Theorem 15.31. Proposition 15.34. If 𝐴 is a companion matrix, then for every 𝜆 ∈ 𝕋 the matrix 𝜆𝐴 is unitarily equivalent to a companion matrix. As indicated above, we provide the proof in the case when the matrix is irreducible. You should convince yourself that the fact that the matrix is not reducible implies that it is not normal. This implies that 𝑑 as defined in Theorem 15.32 must be positive. In addition, considering the irreducible case has special implications for the third condition in Theorem 15.32. If the point 𝑧3 did not lie inside the boundary ellipse, it would follow that 𝐴 is unitarily equivalent to the direct sum [𝑧0 ] ⊕ 𝐵 for some matrix 𝐵. This, in turn, would imply that the matrix is reducible; see p. 123. Thus, we need not check the third condition above if the matrix under consideration is irreducible. The special case (a reducible matrix or normal matrix) can be handled via Proposition 15.33 as in [23]. Proof of Theorem 15.31 for an irreducible matrix. We are assuming that the matrix 𝑃 is irreducible, and therefore 𝑑 > 0 and the point 𝑧3 must lie in the interior of the ellipse. Thus, the first and third conditions in Theorem 15.32 hold. Turning to the second condition, we may assume that 𝑧1 = 𝑧2 . (Why? This is your chance to use Proposition 15.34.) Let 𝑡 = 𝑧1 𝑧2 = |𝑧1 |2 , 𝑠 = 𝑧1 + 𝑧2 = 2ℜ(𝑧1 ), and 𝑧 = 𝑧3 = 𝑥 + 𝑖𝑦. With 𝑎, 𝑏, and 𝑐 as in (15.15), we have 𝑎 = −(𝑠 + 𝑧), 𝑏 = 𝑡 + 𝑠𝑧, and 𝑐 = −𝑡𝑧. A calculation shows that 𝑑 = 2 + |𝑎|2 + |𝑏|2 + |𝑐|2 − (2𝑡 + |𝑧|2 ).

(15.17)

To establish the second condition, we claim that 𝑧 must be real. To this end, write (15.16) in a slightly different form. We have 𝑑𝑧 = −𝑎𝑑 + 𝑠𝑡 + |𝑧|2 𝑧 − tr(𝑃⋆ 𝑃2 ).

248

Chapter 15. Fourteen Projects for Fourteen Chapters

Since 𝑑𝑧 + 𝑎𝑑 = −𝑠𝑑, the second condition becomes 0 = 𝑠𝑑 + 𝑠𝑡 + |𝑧|2 𝑧 − tr(𝑃⋆ 𝑃2 ).

(15.18)

Note that tr(𝑃⋆ 𝑃2 ) = −𝑎 (1 + |𝑎|2 + |𝑏|2 + |𝑐|2 ) + 𝑎𝑏 + 𝑏𝑐 = (𝑠 + 𝑧)(1 + |𝑠 + 𝑧|2 + |𝑡 + 𝑠𝑧|2 + 𝑡 2 |𝑧|2 )− (𝑠𝑡 + 𝑡𝑧 + 𝑠2 𝑧 + 𝑠|𝑧|2 + 𝑡 2 𝑧 + 𝑠𝑡|𝑧|2 ). (15.19) Taking the imaginary part of (15.18), we get that ℑ(𝑧) = 0 and therefore 𝑧 is real, establishing our claim, or ℑ(𝑧) ≠ 0 and therefore |𝑧|2 = 1 + |𝑠 + 𝑧|2 + |𝑡 + 𝑠𝑧|2 + 𝑡 2 |𝑧|2 + 𝑡 − 𝑠2 − 𝑡 2 .

(15.20)

We note that (15.20) can be simplified and is equal to (𝑠2 + 𝑡 2 )|𝑧|2 = − (1 + 𝑡 + (2𝑠)(1 + 𝑡)𝑥) .

(15.21)

So we assume that (15.20) holds and show that, under our assumptions, this leads to a contradiction. Substituting (15.20) into (15.17) yields 𝑑 = 2 + (𝑠2 + 𝑡 2 ) + 2𝑠(1 + 𝑡)𝑥 + (𝑠2 + 𝑡 2 )|𝑧|2 − 2𝑡 = 1 + 𝑠2 + 𝑡 2 − 3𝑡.

(15.22)

Continuing this super-fun computation and using (15.20) and (15.19), we get tr(𝑃⋆ 𝑃2 ) = (𝑠 + 𝑧) (𝑠2 + 𝑡 2 − 𝑡 + |𝑧|2 ) − (𝑠𝑡 + 𝑡𝑧 + (𝑠2 + 𝑡 2 )𝑧 + (𝑠 + 𝑠𝑡)|𝑧|2 ). (15.23) Substituting (15.22) and (15.23) into (15.18) yields 𝑠 2𝑥 = −𝑠|𝑧|2 − . 𝑡 Substituting this back into (15.21) and solving for |𝑧|2 shows that we must have |𝑧|2 < 0, which is silly. Therefore, 𝑊(𝑃) can only be elliptical if 𝑧 is real. Writing 𝑧 = 𝑥, the equation we need to solve (namely, (15.18)) is a cubic in 𝑥. It is easy to check that (15.18) defines a polynomial of degree

15.13. Companion Matrices and Kippenhahn

249

at most three, that the coefficients are real, and the coefficient of 𝑧3 is nonzero. (The intrepid reader can show that (15.18) becomes (𝑡 2 + 𝑠2 )𝑥 3 + (𝑠𝑡 + 2𝑠)𝑥 2 + (1 − 𝑡)𝑥 − 𝑠 = 0 after some additional computations.) Therefore, there are at most three choices for 𝑧. For these choices of 𝑧, the second condition in Theorem 15.32 will be satisfied. Since the coefficients of the polynomial are real, there must be at least one real solution; that is, there is at least one solution for 𝑥. In fact, the more general result is true [23]. Theorem 15.35. Given two complex numbers 𝑧1 and 𝑧2 there exists at least one complex number 𝑧3 , and at most five, such that the corresponding companion matrix has elliptic numerical range with foci at 𝑧1 and 𝑧2 . The proof is rather lengthy and computational, and we will not present it here. Instead, we note some other facts about the numerical range of 3 × 3 matrices. And though we know less about 3 × 3 matrices than we do about 2 × 2 matrices, we do know a lot. The following special case of 3 × 3 matrices in the form (15.24) can be analyzed fully. We note that Schur’s theorem and the unitary invariance of the numerical range tell us something about all 3×3 matrices with one eigenvalue of multiplicity 3. Writing 𝑎 𝐴 = [0 0

𝑥 𝑎 0

𝑦 𝑧] , 𝑎

(15.24)

it turns out that • 𝑊(𝐴) is a disk if and only if 𝑥𝑦𝑧 = 0; • 𝑊(𝐴) has a flat portion on its boundary if and only if |𝑥| = |𝑦| = |𝑧| > 0. In this case, 𝐶(𝐴) is a cardioid; • 𝑊(𝐴) has an oval shape if 𝑥𝑦𝑧 ≠ 0 and |𝑥|, |𝑦|, |𝑧| are not all equal. Exercise 15.36. Either prove this or read the proof in [90, Theorem 4.1]. We now have examples of matrices that fall into each of the four classes in Kippenhahn’s classification appearing on page 164. (Information in

250

Chapter 15. Fourteen Projects for Fourteen Chapters

[77] might also be useful.) Kippenhahn provides other examples and shows that it is possible to compute the boundary generating curve of the matrix 1 0 ⎤ ⎡0 − 2 ⎢ ⎥ ⎢1 1 1 ⎥ 𝐴 = ⎢ 2 √2 − 2 ⎥ . ⎢ ⎥ ⎢ ⎥ ⎢ 1 1 ⎥ − ⎦ ⎣0 2

√2

Do so: To investigate the curve you might try using a Monte Carlo simulation to approximate the curve, choosing vectors 𝑥 randomly on the unit sphere and calculating ⟨𝐴𝑥, 𝑥⟩. Other algorithms to compute the numerical range are available12 and can be found via a web search. We suggest new ways in the project. Project 13. (Kippenhahn) Find other examples of 3 × 3 matrices, and determine which of the cases in Kippenhahn’s classification they fall into. (Blaschke) In Chapter 11 we gave an algorithm using the Lagrange form of a rational function on the upper half-plane to compute a Blaschke product identifying two sets of points on the unit circle and taking zero to zero. Call this Blaschke product 𝐵, and let 𝐵(𝑧) = 𝑧𝐵1 (𝑧). In Project 11 we presented a second way to construct the Blaschke product using the (nested) Newton form of the corresponding function on the upper half-plane. Using each algorithm, you can create another algorithm to approximate the curve bounding 𝑊(𝑆𝐵1 ), the numerical range of the compression of the shift corresponding to 𝐵1 . This, like Kippenhahn’s algorithm, gives an “outer” estimate for the numerical range since it always circumscribes the curve. How would you find an inner estimate? (There is enough material in this book to tell you how to do this!) Can you find other algorithms? If so, how can you evaluate which is best? Be sure to try your algorithm out on several examples. (See Project 11 for some ideas of where to start.) 12 An older method appears at https://www.math.iupui.edu/~ccowen/ Downloads/33NumRange.pdf (accessed 2/8/2017).

15.14. Denjoy–Wolff Points and Blaschke Products

251

15.14 Denjoy–Wolff Points and Blaschke Products Blaschke products also have fixed points in 𝔻, just as disk automorphisms do. This is a consequence of Brouwer’s fixed point theorem [117, p. 351]. How many fixed points are there and where are they? To begin this project, we suggest that you go to the Blaschke Product Explorer applet ,.13 Working with the “show fixed points” option in the extras, experiment with different Blaschke products. You will see more interesting things happening if you remove the zero at zero. Think about why that is so before reading on. Exercise 15.37. Prove the first three statements below, and find the required examples in (4) and (5). (1) A Blaschke product of degree 𝑛 that is not the identity has at most 𝑛 + 1 fixed points in ℂ∗ . (2) If 𝑧0 is a fixed point of a Blaschke product 𝐵, then 1/𝑧0 is also a fixed point of 𝐵. (If 𝑧0 = 0, then 1/𝑧0 = ∞.) (3) If a Blaschke product has no fixed point in 𝔻, then all its fixed points are in 𝕋. (4) Give an example of a Blaschke product of degree at least 2 that has a fixed point other than 0 in 𝔻. (5) Give an example of a Blaschke product of degree at least 2 that has no fixed point in 𝔻. The following theorem shows that there is a very special fixed point associated with a Blaschke product. Theorem 15.38 (Denjoy–Wolff). Let 𝐵 be a Blaschke product of degree at least 2. Then there is a unique point 𝑤 in 𝔻 such that the iterates of 𝐵 tend to 𝑤 uniformly on compact subsets of 𝔻. This unique point is called the Denjoy–Wolff point of the Blaschke product. 13 https://pubapps.bucknell.edu/static/aeshaffer/v1/

252

Chapter 15. Fourteen Projects for Fourteen Chapters

A proof of Theorem 15.38 can be found in [34, Theorem 2.51] or [22]. Below we follow the proof in [140, p. 79] for the special case in which 𝐵 has a fixed point in 𝔻. The proof will depend heavily on Schwarz’s lemma; see [118, p.143]. Here is the exact statement of what we prove. Proposition 15.39. Suppose that 𝐵 is a Blaschke product of degree at least 2 that has a fixed point 𝑧0 in 𝔻. Then 𝐵(𝑛) converges uniformly to the constant 𝑧0 on compact subsets of 𝔻. Proof. If 𝑧0 ≠ 0, we consider the new Blaschke product 𝐶 = 𝜓𝑧0 ∘𝐵∘𝜓𝑧0 , where the notation is as defined in (13.10). Then 𝐵 and 𝐶 are Blaschke products of the same degree, 𝐶 has 0 as its fixed point, and 𝐵(𝑛) tends to 𝑧0 uniformly on compact subsets of 𝔻 if and only if 𝐶 (𝑛) tends to 0 uniformly on compact subsets of 𝔻. Thus, it suffices to consider the case when 𝐵 has a fixed point at 0. Fix 𝑟 so that 0 < 𝑟 < 1, and let 𝑀(𝑟) = max{|𝐵(𝑧)| ∶ |𝑧| ≤ 𝑟}. Since the degree of 𝐵 is at least 2, we know that 𝐵 is not a rotation. By Schwarz’s lemma, |𝐵(𝑧)| < |𝑧| for all 𝑧 ∈ 𝔻. This implies that 𝛿 ∶= 𝑀(𝑟)/𝑟 < 1. We now consider the new function 𝐵(𝑟𝑧) 𝑓(𝑧) = for 𝑧 ∈ 𝔻. 𝑀(𝑟) Again by Schwarz’s lemma, |𝑓(𝑧)| ≤ |𝑧| for all 𝑧 ∈ 𝔻. Since 𝑓 is continuous on the closed unit disk, we can extend the inequality to 𝑧 ∈ 𝔻 and get 𝑀(𝑟)|𝑟𝑧| |𝐵(𝑟𝑧)| ≤ for all 𝑧 ∈ 𝔻. 𝑟 This implies that 𝑀(𝑟) |𝑤| = 𝛿|𝑤| for all 𝑤 ∈ 𝑟𝔻. 𝑟 Iterating (15.25), as below, we get |𝐵(𝑤)| ≤

(15.25)

||𝐵(𝑛) (𝑤)|| ≤ 𝛿 ||𝐵(𝑛−1) (𝑤)|| ≤ 𝛿 2 ||𝐵(𝑛−2) (𝑤)|| ≤ ⋯ ≤ 𝛿 𝑛 |𝑤| ≤ 𝛿 𝑛 𝑟 < 𝛿 𝑛 for all 𝑤 ∈ 𝑟𝔻. This implies that (𝐵(𝑛) ) converges uniformly to 0 on compact subsets. Thus, in general, (𝐵(𝑛) ) converges uniformly to the constant 𝑧0 on any compact subset of 𝔻. It follows from Proposition 15.39, as you should check, that a Blaschke product of degree at least 2 has at most one fixed point in 𝔻, at least one on 𝕋, and at most 𝑛 − 1 on 𝕋.

15.14. Denjoy–Wolff Points and Blaschke Products

253

Project 14. For a disk automorphism, 𝑀𝑎,𝜙 , we know that we can put conditions on 𝑎 and 𝜙 that ensure that the automorphism is elliptic. Can we also tell from the form of a Blaschke product whether it has a fixed point in 𝔻? Classify Blaschke products as we have classified disk automorphisms. When is a Blaschke product in each class? Do not expect a complete answer. Begin small: Start with Blaschke products of degree 2 or 3 and consider spe𝑛

cial cases, like our Blaschke products ((𝑏 − 𝑧)/(1 − 𝑏𝑧)) of Chapter 14 (sometimes called unicritical Blaschke products). Investigate the dynamics of a Blaschke product: Look up the definitions of the Fatou set and the Julia set. What are these sets for Blaschke products of degree at least 2 with a fixed point in 𝔻? Can you also say something if there is no fixed point in 𝔻? You will find some guidance and partial answers in [48].

Bibliography [1] S. Adlaj, An eloquent formula for the perimeter of an ellipse, Notices Amer. Math. Soc. 59 (2012), no. 8, 1094–1099, DOI 10.1090/noti879. MR2985810 [2] M. Agarwal, J. Clifford, and M. Lachance, Duality and inscribed ellipses, Comput. Methods Funct. Theory 15 (2015), no. 4, 635–644, DOI 10.1007/s40315-015-0124-0. MR3428821 [3] J. Agler, Geometric and topological properties of the numerical range, Indiana Univ. Math. J. 31 (1982), no. 6, 767–777, DOI 10.1512/iumj.1982.31.31053. MR674866 [4] G. Almkvist and B. Berndt, Gauss, Landen, Ramanujan, the arithmetic-geometric mean, ellipses, 𝜋, and the Ladies diary, Amer. Math. Monthly 95 (1988), no. 7, 585– 608, DOI 10.2307/2323302. MR966232 [5] V. I. Arnol′d and A. Avez, Ergodic problems of classical mechanics, W. A. Benjamin, Inc., New York-Amsterdam, 1968. Translated from the French by A. Avez. MR0232910 [6] A. Avez, Ergodic theory of dynamical systems, vol. 1, University of Minnesota, Institute of Technology, 1966. [7] S. Axler, Linear algebra done right, 3rd ed., Undergraduate Texts in Mathematics, Springer, Cham, 2015. MR3308468 [8] E. Badertscher, A simple direct proof of Marden’s theorem, Amer. Math. Monthly 121 (2014), no. 6, 547–548, DOI 10.4169/amer.math.monthly.121.06.547. MR3225469 [9] E. J. Barbeau, Polynomials, Problem Books in Mathematics, Springer-Verlag, New York, 1989. MR987938 [10] J. Barnes, Gems of geometry, 2nd ed., Springer, Heidelberg, 2012. MR2963305 [11] N. Bebiano, J. da Providência, A. Nata, and J. P. da Providência, Revisiting the inverse field of values problem, Electron. Trans. Numer. Anal. 42 (2014), 1–12. MR3183614 [12] A. Berger and T. P. Hill, A basic theory of Benford’s law, Probab. Surv. 8 (2011), 1–126, DOI 10.1214/11-PS175. MR2846899 [13] A. Berger and T. P. Hill, An introduction to Benford’s law, Princeton University Press, Princeton, NJ, 2015. MR3242822 [14] A. Beutelspacher and U. Rosenbaum, Projective geometry: from foundations to applications, Cambridge University Press, Cambridge, 1998. MR1629468 [15] H. J. M. Bos, C. Kers, F. Oort, and D. W. Raven, Poncelet’s closure theorem, Exposition. Math. 5 (1987), no. 4, 289–364. MR917349 [16] P. S. Bourdon and J. H. Shapiro, When is zero in the numerical range of a composition operator?, Integral Equations Operator Theory 44 (2002), no. 4, 410–441, DOI 10.1007/BF01193669. MR1942033 [17] C. B. Boyer and U. C. Merzbach, A history of mathematics, John Wiley & Sons, 2011.

255

256

Bibliography

[18] R. E. Bradley and E. Sandifer, Leonhard Euler: Life, work and legacy, vol. 5, Elsevier, 2007. [19] D. A. Brannan, On a conjecture of Ilieff, Proc. Cambridge Philos. Soc. 64 (1968), 83– 85. MR0220906 [20] M. A. Brilleslyper and B. Schaubroeck, Explorations of the Gauss–Lucas theorem, PRIMUS 27 (2017), no. 8-9, 766–777. [21] J. E. Brown and G. Xiang, Proof of the Sendov conjecture for polynomials of degree at most eight, J. Math. Anal. Appl. 232 (1999), no. 2, 272–292, DOI 10.1006/jmaa.1999.6267. MR1683144 [22] R. B. Burckel, Iterating analytic self-maps of discs, Amer. Math. Monthly 88 (1981), no. 6, 396–407, DOI 10.2307/2321822. MR622955 [23] W. Calbeck, Elliptic numerical ranges of 3 × 3 companion matrices, Linear Algebra Appl. 428 (2008), no. 11-12, 2715–2722, DOI 10.1016/j.laa.2007.12.018. MR2416583 [24] G. Cassier and I. Chalendar, The group of the invariants of a finite Blaschke product, Complex Variables Theory Appl. 42 (2000), no. 3, 193–206, DOI 10.1080/17476930008815283. MR1788126 [25] W. Chapple, An essay on the properties of triangles inscribed in and circumscribed about two given circles, Miscellanea Curiosa Mathematica 4 (1746), 117–124. [26] M.-D. Choi and C.-K. Li, Constrained unitary dilations and numerical ranges, J. Operator Theory 46 (2001), no. 2, 435–447. MR1870416 [27] M. Chuaqui and C. Pommerenke, On Schwarz–Christoffel mappings, Pacific J. Math. 270 (2014), no. 2, 319–334, DOI 10.2140/pjm.2014.270.319. MR3253684 [28] W. Cieślak and E. Szczygielska, On Poncelet’s porism, Ann. Univ. Mariae CurieSkłodowska Sect. A 64 (2010), no. 2, 21–28, DOI 10.2478/v10062-010-0011-0. MR2771117 [29] J. B. Conway, Functions of one complex variable, 2nd ed., Graduate Texts in Mathematics, vol. 11, Springer-Verlag, New York-Berlin, 1978. MR503901 [30] J. B. Conway, The theory of subnormal operators, Mathematical Surveys and Monographs, vol. 36, American Mathematical Society, Providence, RI, 1991. MR1112128 [31] R. Courant, Differential and integral calculus. Vol. II, Wiley Classics Library, John Wiley & Sons, Inc., New York, 1988. Translated from the German by E. J. McShane; Reprint of the 1936 original; A Wiley-Interscience Publication. MR1009559 [32] D. Courtney and D. Sarason, A mini-max problem for self-adjoint Toeplitz matrices, Math. Scand. 110 (2012), no. 1, 82–98, DOI 10.7146/math.scand.a-15198. MR2900072 [33] C. C. Cowen, Finite Blaschke products as compositions of other finite Blaschke products, arXiv:1207.4010, 2012. [34] C. C. Cowen and B. D. MacCluer, Composition operators on spaces of analytic functions, Studies in Advanced Mathematics, CRC Press, Boca Raton, FL, 1995. MR1397026 [35] U. Daepp, P. Gorkin, and R. Mortini, Ellipses and finite Blaschke products, Amer. Math. Monthly 109 (2002), no. 9, 785–795, DOI 10.2307/3072367. MR1933701 [36] U. Daepp, P. Gorkin, A. Shaffer, B. Sokolowsky, and K. Voss, Decomposing finite Blaschke products, J. Math. Anal. Appl. 426 (2015), no. 2, 1201–1216, DOI 10.1016/j.jmaa.2015.01.039. MR3314888

Bibliography

257

[37] U. Daepp, P. Gorkin, and K. Voss, Poncelet’s theorem, Sendov’s conjecture, and Blaschke products, J. Math. Anal. Appl. 365 (2010), no. 1, 93–102, DOI 10.1016/j.jmaa.2009.09.058. MR2585079 [38] C. Davis, The Toeplitz–Hausdorff theorem explained, Canad. Math. Bull. 14 (1971), 245–246, DOI 10.4153/CMB-1971-042-7. MR0312288 [39] J. Dégot, Sendov conjecture for high degree polynomials, Proc. Amer. Math. Soc. 142 (2014), no. 4, 1337–1349, DOI 10.1090/S0002-9939-2014-11888-0. MR3162254 [40] J. Ding and T. H. Fay, The Perron–Frobenius theorem and limits in geometry, Amer. Math. Monthly 112 (2005), no. 2, 171–175, DOI 10.2307/30037416. MR2121328 [41] W. F. Donoghue Jr., On the numerical range of a bounded operator, Michigan Math. J. 4 (1957), 261–263. MR0096127 [42] R. G. Douglas, S. Sun, and D. Zheng, Multiplication operators on the Bergman space via analytic continuation, Adv. Math. 226 (2011), no. 1, 541–583, DOI 10.1016/j.aim.2010.07.001. MR2735768 [43] V. Dragović and M. Radnović, Poncelet porisms and beyond, Frontiers in Mathematics, Birkhäuser/Springer Basel AG, Basel, 2011. Integrable billiards, hyperelliptic Jacobians and pencils of quadrics. MR2798784 [44] S. W. Drury, Symbolic calculus of operators with unit numerical radius, Linear Algebra Appl. 428 (2008), no. 8-9, 2061–2069, DOI 10.1016/j.laa.2007.11.007. MR2401640 [45] J. Eising, D. Radcliffe, and J. Top, A simple answer to Gelfand’s question, Amer. Math. Monthly 122 (2015), no. 3, 234–245, DOI 10.4169/amer.math.monthly.122.03.234. MR3327713 [46] P. A. Fillmore, On similarity and the diagonal of a matrix, Amer. Math. Monthly 76 (1969), 167–169, DOI 10.2307/2317264. MR0237526 [47] L. Flatto, Poncelet’s theorem, American Mathematical Society, Providence, RI, 2009. Chapter 15 by S. Tabachnikov. MR2465164 [48] A. Fletcher, Unicritical Blaschke products and domains of ellipticity, Qual. Theory Dyn. Syst. 14 (2015), no. 1, 25–38, DOI 10.1007/s12346-015-0133-4. MR3326210 [49] M. Frantz, How conics govern Möbius transformations, Amer. Math. Monthly 111 (2004), no. 9, 779–790, DOI 10.2307/4145189. MR2104049 [50] E. Fricain and J. Mashreghi, The theory of ℋ(𝑏) spaces. Vol. 2, New Mathematical Monographs, vol. 21, Cambridge University Press, Cambridge, 2016. MR3617311 [51] M. Fujimura, Inscribed ellipses and Blaschke products, Comput. Methods Funct. Theory 13 (2013), no. 4, 557–573, DOI 10.1007/s40315-013-0037-8. MR3138353 [52] N. Fuss, De quadrilateris quibus circulum tam inscribere quam circumscribere licet, Nova Acta Acad. Sci. Petrop. 10 (1797), 103–125. [53] M. Gardner, New mathematical diversions, revised edition, MAA Spectrum, Mathematical Association of America, Washington, DC, 1995. MR1335231 [54] J. B. Garnett, Bounded analytic functions, 1st ed., Graduate Texts in Mathematics, vol. 236, Springer, New York, 2007. MR2261424 [55] H.-L. Gau and P. Y. Wu, Numerical range of 𝑆(𝜙), Linear Multilinear Algebra 45 (1998), no. 1, 49–73. [56] H.-L. Gau and P. Y. Wu, Lucas’ theorem refined, Linear and Multilinear Algebra 45 (1999), no. 4, 359–373, DOI 10.1080/03081089908818600. MR1684719

258

Bibliography

[57] H.-L. Gau and P. Y. Wu, Condition for the numerical range to contain an elliptic disc, Linear Algebra Appl. 364 (2003), 213–222, DOI 10.1016/S0024-3795(02)005487. MR1971096 [58] H.-L. Gau and P. Y. Wu, Numerical range and Poncelet property, Taiwanese J. Math. 7 (2003), no. 2, 173–193, DOI 10.11650/twjm/1500575056. MR1978008 [59] H.-L. Gau and P. Y. Wu, Companion matrices: reducibility, numerical ranges and similarity to contractions, Linear Algebra Appl. 383 (2004), 127–142, DOI 10.1016/j.laa.2003.11.027. MR2073899 [60] H.-L. Gau and P. Y. Wu, Numerical ranges of companion matrices, Linear Algebra Appl. 421 (2007), no. 2-3, 202–218, DOI 10.1016/j.laa.2006.03.037. MR2294336 [61] C. Glader, Minimal degree rational unimodular interpolation on the unit circle, Electron. Trans. Numer. Anal. 30 (2008), 88–106. MR2480071 [62] A. W. Goodman, Q. I. Rahman, and J. S. Ratti, On the zeros of a polynomial and its derivative, Proc. Amer. Math. Soc. 21 (1969), 273–274, DOI 10.2307/2036982. MR0239062 [63] P. Gorkin, L. Laroco, R. Mortini, and R. Rupp, Composition of inner functions, Results Math. 25 (1994), no. 3-4, 252–269, DOI 10.1007/BF03323410. MR1273115 [64] P. Gorkin and R. C. Rhoades, Boundary interpolation by finite Blaschke products, Constr. Approx. 27 (2008), no. 1, 75–98, DOI 10.1007/s00365-006-0646-3. MR2336418 [65] P. Gorkin and E. Skubak, Polynomials, ellipses, and matrices: two questions, one answer, Amer. Math. Monthly 118 (2011), no. 6, 522–533, DOI 10.4169/amer.math.monthly.118.06.522. MR2812283 [66] P. Gorkin and N. Wagner, Ellipses and compositions of finite Blaschke products, J. Math. Anal. Appl. 445 (2017), no. 2, 1354–1366, DOI 10.1016/j.jmaa.2016.01.067. MR3545246 [67] T. Gosset, “The Kiss Precise (Generalized)”, in Strange attractors: Poems of love and mathematics (S. Glaz and J. Growney, editors), A K Peters, Ltd., Wellesley, MA, 2008, 189. MR 2490399 [68] J. W. Green, Classroom notes: On the envelope of curves given in parametric form, Amer. Math. Monthly 59 (1952), no. 9, 626–628, DOI 10.2307/2306769. MR1528268 [69] P. Griffiths and J. Harris, A Poncelet theorem in space, Comment. Math. Helv. 52 (1977), no. 2, 145–160, DOI 10.1007/BF02567361. MR0498606 [70] K. Gustafson, The Toeplitz–Hausdorff theorem for linear operators, Proc. Amer. Math. Soc. 25 (1970), 203–204, DOI 10.2307/2036559. MR0262849 [71] E. Gutkin, The Toeplitz–Hausdorff theorem revisited: relating linear algebra and geometry, Math. Intelligencer 26 (2004), no. 1, 8–14, DOI 10.1007/BF02985393. MR2034035 [72] U. Haagerup and P. de la Harpe, The numerical radius of a nilpotent operator on a Hilbert space, Proc. Amer. Math. Soc. 115 (1992), no. 2, 371–379, DOI 10.2307/2159255. MR1072339 [73] A. J. Hahn, Mathematical excursions to the world’s great buildings, Princeton University Press, Princeton, NJ, 2012. MR2962336 [74] L. Halbeisen and N. Hungerbühler, A simple proof of Poncelet’s theorem (on the occasion of its bicentennial), Amer. Math. Monthly 122 (2015), no. 6, 537–551, DOI 10.4169/amer.math.monthly.122.6.537. MR3361732

Bibliography

259

[75] P. R. Halmos, Numerical ranges and normal dilations, Acta Sci. Math. (Szeged) 25 (1964), 1–5. MR0171168 [76] P. R. Halmos, A Hilbert space problem book, 2nd ed., Graduate Texts in Mathematics, vol. 19, Springer-Verlag, New York-Berlin, 1982. Encyclopedia of Mathematics and its Applications, 17. MR675952 [77] T. R. Harris, M. Mazzella, L. J. Patton, D. Renfrew, and I. M. Spitkovsky, Numerical ranges of cube roots of the identity, Linear Algebra Appl. 435 (2011), no. 11, 2639– 2657, DOI 10.1016/j.laa.2011.03.020. MR2825272 [78] F. Hausdorff, Der Wertvorrat einer Bilinearform (German), Math. Z. 3 (1919), no. 1, 314–316, DOI 10.1007/BF01292610. MR1544350 [79] W. K. Hayman, Research problems in function theory, The Athlone Press University of London, London, 1967. MR0217268 [80] W. M. Higdon, On the numerical ranges of composition operators induced by mappings with the Denjoy–Wolff point on the boundary, Integral Equations Operator Theory 85 (2016), no. 1, 127–135, DOI 10.1007/s00020-016-2287-0. MR3503182 [81] T. P. Hill, A statistical derivation of the significant-digit law, Statist. Sci. 10 (1995), no. 4, 354–363. MR1421567 [82] J. Hilmar and C. Smyth, Euclid meets Bézout: intersecting algebraic plane curves with the Euclidean algorithm, Amer. Math. Monthly 117 (2010), no. 3, 250–260, DOI 10.4169/000298910X480090. MR2640851 [83] H. Hilton, Plane algebraic curves, Oxford University Press, 1920. [84] K. Hoffman and R. Kunze, Linear algebra, 2nd ed., Prentice-Hall, Inc., Englewood Cliffs, N.J., 1971. MR0276251 [85] J. Holbrook and J.-P. Schoch, Theory vs. experiment: multiplicative inequalities for the numerical radius of commuting matrices, Topics in operator theory. Volume 1. Operators, matrices and analytic functions, Oper. Theory Adv. Appl., vol. 202, Birkhäuser Verlag, Basel, 2010, pp. 273–284, DOI 10.1007/978-3-0346-0158-0_14. MR2723281 [86] C. G. J. Jacobi, Gesammelte Werke. Bände I–VIII (German), Herausgegeben auf Veranlassung der Königlich Preussischen Akademie der Wissenschaften. Zweite Ausgabe, Chelsea Publishing Co., New York, 1969. MR0260557 [87] A. Jamain, Benford’s law, Master’s thesis, Imperial College of London (2001). [88] D. Kalman, Solving the ladder problem on the back of an envelope, Math. Mag. 80 (2007), no. 3, 163–182. MR2322082 [89] Y. Katznelson, An introduction to harmonic analysis, 3rd ed., Cambridge Mathematical Library, Cambridge University Press, Cambridge, 2004. MR2039503 [90] D. S. Keeler, L. Rodman, and I. M. Spitkovsky, The numerical range of 3 × 3 matrices, Linear Algebra Appl. 252 (1997), 115–139, DOI 10.1016/0024-3795(95)00674-5. MR1428632 [91] S. M. Kerawala, Poncelet porism in two circles, Bull. Calcutta Math. Soc. 39 (1947), 85–105. MR0026339 [92] D. Khavinson, R. Pereira, M. Putinar, E. B. Saff, and S. Shimorin, Borcea’s variance conjectures on the critical points of polynomials, Notions of positivity and the geometry of polynomials, Trends Math., Birkhäuser/Springer Basel AG, Basel, 2011, pp. 283–309, DOI 10.1007/978-3-0348-0142-3_16. MR3051172 [93] C. Kimberling, The shape and history of the ellipse in Washington, D.C., Department of Mathematics, University of Evansville, http://faculty.evansville.edu/ck6/ellipse.pdf. (Accessed October 2016).

260

Bibliography

[94] J. L. King, Three problems in search of a measure, Amer. Math. Monthly 101 (1994), no. 7, 609–628, DOI 10.2307/2974690. MR1289271 [95] R. Kippenhahn, Über den Wertevorrat einer Matrix (German), Math. Nachr. 6 (1951), 193–228, DOI 10.1002/mana.19510060306. MR0059242 [96] R. Kippenhahn, On the numerical range of a matrix, Linear Multilinear Algebra 56 (2008), no. 1-2, 185–225, DOI 10.1080/03081080701553768. Translated from the German by Paul F. Zachlin and Michiel E. Hochstenbach [MR0059242]. MR2378310 [97] H. Klaja, J. Mashreghi, and T. Ransford, On mapping theorems for numerical range, Proc. Amer. Math. Soc. 144 (2016), no. 7, 3009–3018, DOI 10.1090/proc/12955. MR3487232 [98] A. Kock, Envelopes—notion and definiteness, Beiträge Algebra Geom. 48 (2007), no. 2, 345–350. MR2364794 [99] J. C. Lagarias, C. L. Mallows, and A. R. Wilks, Beyond the Descartes circle theorem, Amer. Math. Monthly 109 (2002), no. 4, 338–361, DOI 10.2307/2695498. MR1903421 [100] J. S. Lancaster, The boundary of the numerical range, Proc. Amer. Math. Soc. 49 (1975), 393–398, DOI 10.2307/2040652. MR0372644 [101] J. C. Langer and D. A. Singer, Foci and foliations of real algebraic curves, Milan J. Math. 75 (2007), 225–271, DOI 10.1007/s00032-007-0078-4. MR2371544 [102] C.-K. Li, A simple proof of the elliptical range theorem, Proc. Amer. Math. Soc. 124 (1996), no. 7, 1985–1986, DOI 10.1090/S0002-9939-96-03307-2. MR1322932 [103] H. Licks, Recreations in mathematics, D. Van Nostrand Company, New York, 1917. Available online in public domain. [104] E. H. Lockwood, A book of curves, Cambridge University Press, New York, 1961. MR0126191 [105] C. Maclaurin, A treatise on fluxions, Ruddimans, Edinburgh, 1742. [106] F. Malmquist, Sur la détermination d’une classe de fonctions analytiques par leurs valeurs dans un ensemble donné de poits, in C.R. 6ième Cong. Math. Scand. Kopenhagen, 1925. Gjellerups, Copenhagen, 1926, pp. 253–259. [107] M. Marcus and B. N. Shure, The numerical range of certain 0, 1-matrices, Linear and Multilinear Algebra 7 (1979), no. 2, 111–120, DOI 10.1080/03081087908817266. MR529878 [108] M. Marden, Geometry of polynomials, 2nd ed., Mathematical Surveys, No. 3, American Mathematical Society, Providence, R.I., 1966. MR0225972 [109] M. Marden, The search for a Rolle’s theorem in the complex domain, Amer. Math. Monthly 92 (1985), no. 9, 643–650, DOI 10.2307/2323710. MR810661 [110] J. Mashreghi, Derivatives of inner functions, Fields Institute Monographs, vol. 31, Springer, New York; Fields Institute for Research in Mathematical Sciences, Toronto, ON, 2013. MR2986324 [111] V. J. Matsko, Generic ellipses as envelopes, Math. Mag. 86 (2013), no. 5, 358–365, DOI 10.4169/math.mag.86.5.358. MR3141737 [112] S. J. Miller, A quick introduction to Benford’s law, Benford’s law: Theory and applications, Princeton Univ. Press, Princeton, NJ, 2015, pp. 3–22, DOI 10.1515/9781400866595. MR3411056 [113] S. Mills, Note on the Braikenridge–Maclaurin theorem, Notes and Records Roy. Soc. London 38 (1984), no. 2, 235–240, DOI 10.1098/rsnr.1984.0014. MR783589 [114] B. Mirman, Numerical ranges and Poncelet curves, Linear Algebra Appl. 281 (1998), no. 1-3, 59–85, DOI 10.1016/S0024-3795(98)10037-X. MR1645335

Bibliography

261

[115] B. Mirman, V. Borovikov, L. Ladyzhensky, and R. Vinograd, Numerical ranges, Poncelet curves, invariant measures, Linear Algebra Appl. 329 (2001), no. 1-3, 61–75, DOI 10.1016/S0024-3795(01)00233-6. MR1822222 [116] H. F. Montague, Envelopes associated with a one-parameter family of straight Lines, Natl. Math. Mag. 13 (1938), no. 2, 73–75. MR1569600 [117] J. Munkres, Topology, 2nd ed., Prentice-Hall, Inc., Englewood Cliffs, N.J., 2000. [118] R. Nevanlinna and V. Paatero, Introduction to complex analysis, Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., 1969. Translated from the German by T. Kövari and G. S. Goodman. MR0239056 [119] T. W. Ng and C. Y. Tsang, Chebyshev-Blaschke products: solutions to certain approximation problems and differential equations, J. Comput. Appl. Math. 277 (2015), 106– 114, DOI 10.1016/j.cam.2014.08.028. MR3272168 [120] M. Nigrini, Benford’s law: Applications for forensic accounting, auditing and fraud detection, vol. 586, John Wiley & Sons, 2012. [121] N. K. Nikolski, Operators, functions, and systems: an easy reading. Vol. 2, Mathematical Surveys and Monographs, vol. 93, American Mathematical Society, Providence, RI, 2002. Model operators and systems; Translated from the French by Andreas Hartmann and revised by the author. MR1892647 [122] J. J. O’Connor and E. F. Robertson, The MacTutor history of mathematics archive, School of Mathematics and Statistics, University of St. Andrews, Scotland, http://www-history.mcs.st-and.ac.uk/. (Accessed 12/15/2017). [123] C. S. Ogilvy, Excursions in geometry, Courier Corporation, 1990. [124] B. C. Patterson, The origins of the geometric principle of inversion, Isis 19 (1933), no. 1, 154–180. [125] J. V. Poncelet, Traité des propriétés projectives des figures: ouvrage utile à ceux qui s’ occupent des applications de la géométrie descriptive et d’opérations géométriques sur le terrain. T. 2, Gauthier-Villars, Imprimeur-Libraire, 1866. [126] T. E. Price, Products of Chord Lengths of an Ellipse, Math. Mag. 75 (2002), no. 4, 300– 307. MR1573631 [127] P. Psarrakos and M. Tsatsomeros, Numerical range: (in) a matrix nutshell, No. Bd. 1 in Mathematics Notes from Washington State University, Department of Mathematics, Washington State University, 2002. [128] H. Queffélec and K. Seip, Decay rates for approximation numbers of composition operators, J. Anal. Math. 125 (2015), 371–399, DOI 10.1007/s11854-015-0012-6. MR3317907 [129] G. Quenell, Envelopes and string art, Math. Mag. 82 (2009), no. 3, 174–185, DOI 10.4169/193009809X468779. MR2522910 [130] Q. I. Rahman and G. Schmeisser, Analytic theory of polynomials, London Mathematical Society Monographs. New Series, vol. 26, The Clarendon Press, Oxford University Press, Oxford, 2002. MR1954841 [131] J. Rickards, When is a polynomial a composition of other polynomials?, Amer. Math. Monthly 118 (2011), no. 4, 358–363, DOI 10.4169/amer.math.monthly.118.04.358. MR2800347 [132] J. F. Ritt, Prime and composite polynomials, Trans. Amer. Math. Soc. 23 (1922), no. 1, 51–66, DOI 10.2307/1988911. MR1501189 [133] H. L. Royden, Real analysis, 3rd ed., Macmillan Publishing Company, New York, 1988. MR1013117

262

Bibliography

[134] Z. Rubinstein, On the approximation by 𝐶-polynomials, Bull. Amer. Math. Soc. 74 (1968), 1091–1093, DOI 10.1090/S0002-9904-1968-12057-9. MR0232003 [135] D. Sarason, Sub-Hardy Hilbert spaces in the unit disk, University of Arkansas Lecture Notes in the Mathematical Sciences, vol. 10, John Wiley & Sons, Inc., New York, 1994. A Wiley-Interscience Publication. MR1289670 [136] G. Schmeisser, Bemerkungen zu einer Vermutung von Ilieff (German), Math. Z. 111 (1969), 121–125, DOI 10.1007/BF01111192. MR0264040 [137] I. J. Schoenberg, A conjectured analogue of Rolle’s theorem for polynomials with real or complex coefficients, Amer. Math. Monthly 93 (1986), no. 1, 8–13, DOI 10.2307/2322536. MR824585 [138] G. Semmler and E. Wegert, Boundary interpolation with Blaschke products of minimal degree, Comput. Methods Funct. Theory 6 (2006), no. 2, 493–511, DOI 10.1007/BF03321626. MR2291147 [139] Bl. Sendov, Generalization of a conjecture in the geometry of polynomials, Serdica Math. J. 28 (2002), no. 4, 283–304. MR1965232 [140] J. H. Shapiro, Composition operators and classical function theory, Universitext: Tracts in Mathematics, Springer-Verlag, New York, 1993. MR1237406 [141] T. Sheil-Small, Complex polynomials, Cambridge Studies in Advanced Mathematics, vol. 75, Cambridge University Press, Cambridge, 2002. MR1962935 [142] D. A. Singer, The location of critical points of finite Blaschke products, Conform. Geom. Dyn. 10 (2006), 117–124, DOI 10.1090/S1088-4173-06-00145-7. MR2223044 [143] F. Soddy, The Kiss Precise, Nature 137 (1936), 1021. [144] N. Stefanović and M. Milošević, A very simple proof of Pascal’s hexagon theorem and some applications, Proc. Indian Acad. Sci. Math. Sci. 120 (2010), no. 5, 619–629, DOI 10.1007/s12044-010-0047-7. MR2779392 [145] B. Sz.-Nagy, C. Foias, H. Bercovici, and L. Kérchy, Harmonic analysis of operators on Hilbert space, revised and enlarged edition, Universitext, Springer, New York, 2010. MR2760647 [146] S. Takenaka, On the orthogonal functions and a new formula of interpolation, Jap. J. Math. 2 (1925), 129–145. [147] O. Toeplitz, Das algebraische Analogon zu einem Satze von Fejér (German), Math. Z. 2 (1918), no. 1-2, 187–197, DOI 10.1007/BF01212904. MR1544315 [148] J. van Yzeren, A simple proof of Pascal’s hexagon theorem, Amer. Math. Monthly 100 (1993), no. 10, 930–931, DOI 10.2307/2324214. MR1252929 [149] J. L. Walsh, Interpolation and functions analytic interior to the unit circle, Trans. Amer. Math. Soc. 34 (1932), no. 3, 523–556, DOI 10.2307/1989366. MR1501650 [150] J. L. Walsh, Note on the location of zeros of the derivative of a rational function whose zeros and poles are symmetric in a circle, Bull. Amer. Math. Soc. 45 (1939), no. 6, 462–470, DOI 10.1090/S0002-9904-1939-07012-2. MR1564005 [151] E. Wegert, Visual complex functions, Birkhäuser/Springer Basel AG, Basel, 2012. An introduction with phase portraits. MR3024399 [152] E. W. Weisstein, Poncelet’s porism, from MathWorld–A Wolfram Web Resource, http://mathworld.wolfram.com/PonceletsPorism.html. (Accessed July 2017). [153] J. W. Young, Projective geometry, The Carus Mathematical Monographs, The Mathematical Association of America, Chicago, 1930. [154] N. Young, An introduction to Hilbert space, Cambridge Mathematical Textbooks, Cambridge University Press, Cambridge, 1988. MR949693

Index (𝑢𝐻 2 )⟂ , 112 𝐴, 117 𝐴⋆ , 62 𝐵𝐻 2 , 109 𝐶(𝐴), 163 𝐶𝜑 , 236 𝐶𝑀𝑎,𝜙 ∶ ℬ → ℬ, 192 𝐻 2 , 106 𝐻 2 (𝔻), 105 𝐻𝐴 + 𝑖𝐾𝐴 , 161 𝐼ℓ , 148 𝐼ℓ,𝑚 , 148 𝐽3 , 164 𝐽𝑛 , 103 𝐾ᵆ , 112 𝐿2 (𝕋), 104 𝑀𝑒 , 162 𝑀𝑎,𝜙 , 177 𝑃+ , 113 𝑃− , 113 𝑃ᵆ , 113 𝑆, 108 𝑆 ⟂ , 71 𝑆 ⋆ , 111 𝑆ᵆ , 113 𝑈𝜆 , 119 𝑊(𝐴), 14 𝑋𝑝,𝑞 , 148 ̂ 𝑓(𝑛), 105

⟨⟨𝑥⟩⟩, 90 ⟨𝑥, 𝑦⟩, 13 ⌊𝑥⌋, 98 ℂ, 14 ℂ∗ , 134 ℂ𝑛 , 13 ℙ2 (ℂ), 163 ℙ2 (ℝ), 48 ℝ∗ , 134 𝕋, 17 𝐇+ , 134 𝐇− , 134 ℬ, 192 𝒮𝑛 , 103 ‖𝐴‖, 63, 76 ‖𝑥‖, 13 𝜔+ , 242 𝜔− , 242 ← → 54, 179 𝑢𝑣, 𝜓𝑐 , 169, 177 𝜎(𝐸), 123, 232 𝑘𝑎̃ , 107 tr(𝐴), 16, 234 𝜑𝑎 , 114 𝑒𝑗 , standard basis 𝐿2 (𝕋), 104 𝑒𝑗 , standard basis ℂ𝑛 , 63 𝑓 (𝑛) , 92 𝑓𝑟 , 105 𝑘𝑎 , 107 𝑤(𝐴), 232 263

264 𝑤(𝑇), 223

Index Braikenridge–MacLaurin theorem, 57 Brianchon’s theorem, 55 Brianchon, Charles, 57

adjoint, 16 affine part, 59 algebraic curve, 51 applets ,, x Cauchy kernel, 107 Blaschke product tools, 29–33, 121, Cauchy–Bunyakovsky–Schwarz in159, 170, 177, 209, 251 equality, 66, 79, 108 experimental tools, 138 chain of circles between 𝒞1 and 𝒞2 , geometry tools, 8, 53, 56, 217 213 astroid, 204 Chapple, William, 45 Chapple–Euler formula, 43–46, 175, Bedford, Frank, 88 197 Bell, Eric Temple, 1 circular point, 242 Benford’s Law, 88 circumscribing a conic, 55 Beurling’s theorem, 111 class 𝒮𝑛 , 103, 162 Blaschke Colosseum, 3 curve, 129, 174 companion matrix, 245 ellipse complete orthonormal basis, 105 (𝑛, 𝑝)-ellipse, 184 composition operator, 192, 236 3-ellipse, 33, 43, 60, 82 compressed shift operator, 113 product, 19, 23, 28 compression, 71 characterization, 19, 26 matrix, 117 composition, 159, 169, 171 conic, 51 decomposable, 169 associated matrix, 51 degree, 19 degenerate, 52 degree-2, 35 constructing an ellipse degree-3, 37–44, 82 envelope method, 201 degree-4, 169–174 folding, 7–10 finite, 19 pin and string method, 201 infinite, 111 contraction, 76 normalized, 23 curve properties, 28, 37–38 algebraic, 51 unicritical, 253 class, 163, 241 Blaschke, Wilhelm, 19 degree, 163, 241 Borcea, J., 209 dual, 51, 163, 241 boundary generating curve, 163

Index real part, 163 tangent line, 51 cyclic quadrilateral, 221 Dégot, J., 207 degenerate, 52 Denjoy–Wolff point, 251 Denjoy–Wolff theorem, 251 Descartes’s circle theorem, 221 Descartes, René, 85 diagonal, 55 direct sum of matrices, 166 disk automorphism, 23, 134, 177 elliptic, 178 canonical, 182 characterization, 178 convex order, 182 order, 182 hyperbolic, 178 parabolic, 178 dual, 51, 241–242 dual curve, 163 dual equation, 241 duality, 47, 57 eigenvalue, 14, 232 Hermitian matrix, 161 eigenvector, 14 ellipse, 3 area enclosed, 5 construction, 8–10 diameter, 3 equation, 4 focus, 3 major axis, 3 minor axis, 4

265 perimeter, 5–7 Ellipse, President’s Park South, 3 elliptical range theorem, 16, 62, 68– 71, 164, 173, 222 envelope, 202, 204 Euclidean norm, 13 Euler, Leonhard, 45 Fatou’s radial limit theorem, 106 field of values, 61 first digit, 88 focus, 242 ellipse, 3 real, 242 singular, 243 Fourier coefficient, 105 Frantz’s theorem, 179 frequency, 90 Fuss’s theorem, 175, 197 Fuss, Nicolaus, 45 Gauss–Lucas theorem, 206 Gelfand’s questions, 90, 98 general position of lines, 55 of points, 53 Gergonne, Joseph, 57 glissette, 224 half-plane lower, 134 upper, 134 Halmos, Paul, vii, 125 Hardy space, 105 Hermitian matrix eigenvalue, 161 numerical range, 161

266 hexagon diagonal, 55 inscribed in conic, 53 opposite sides, 53 opposite vertices, 55 principal diagonal, 55 side, 53 simple, 53 hexagrammum mysticum theorem, 57 Hilbert space, 76 homogeneous polynomial, 51 hyperbolic convex hull, 209 convex set, 209 hyperbolic geometry, 208

Index Kippenhahn, Rudolf, 123, 159–160 Lagrange form, 140 Lebesgue space, 104 Lebesgue, Henri, 104 line at infinity, 48 linear fractional transformation, 134

Möbius transformations, 134 Maclaurin expansion, 7 Maclaurin, C., 7 major axis, 3 Mandart inellipse, 212 Marden, M., 210 matrix adjoint, 16 associated to conic, 51 Cartesian decomposition, 161 Illiev, L., 206 companion, 245 inner function, 109 direct sum, 166 inner product Hermitian, 160, 222 2 𝐿 (𝕋), 104 imaginary part, 161 standard, 13 nilpotent, 222 inscribed in a conic, 53 nonnegative, 228 interpolation, 134, 237–241 norm, 63, 76 invariant subspace, 109, 111 normal, 62 inverse of a point, 216 normal, numerical range, 68, 75 inversion, 215–221 positive, 228 iterate, 92, 182, 251 positive definite, 222 Jensen’s theorem, 206 real part, 161 Jordan block, 78, 103, 131, 164, 168, reducible, 124, 246 222 self-adjoint, 62, 160 similar, 63 Kepler, Johannes, 6 trace, 16, 69, 234 King, Jonathan, 91 unitarily equivalent, 63 Kippenhahn curve, 163 unitary, 62, 75 Kippenhahn’s theorem, 163 unitary decomposition, 123

Index maximum modulus theorem, 26 measure, 95–97 Mersenne, Marin, 57 minor axis, 4 model space, 112 negative pedal curve, 227 Nelson, Wayne James, 87 nephroid, 205 nested form, 240 Newcomb, Simon, 88 Newton form, 238 nilpotent, 222 nonnegative matrix, 228 norm Euclidean, 13 matrix, 63 vector, 13 normalized Blaschke product, 23 numerical radius matrix, 224, 232 operator, 223 numerical range, 14, 61 circular disk, 174 elliptical disk, 159, 168, 171 Hermitian matrix, 161, 222 normal matrix, 75 properties, 63, 68 operator quasi-nilpotent, 223 bounded, 76 contraction, 77 nilpotent, 222 norm, 77 shift, 108 opposite sides, 53

267 opposite vertices, 55 orthogonal complement, 71 orthogonal projection, 71 oval, 165 parallel postulate, 208 Parseval’s identity, 105 Pascal line, 57 Pascal’s theorem, 53 Pascal, Blaise, 57 pedal curve, 227 pedal point, 227 Perron–Frobenius theorem, 228 phase portrait, 30–32 Poincaré model, 208 points at infinity, 48 polygonal chain inscribed in conic, 53 polynomial homogenous, 51 Poncelet curve, 121, 130 ellipse 3-ellipse, 60, 82 4-ellipse, 175 convex, 175 property, 121 Poncelet’s theorem, 92 alternate version, 157 false statements, 158 for triangles, 18, 57 general, 156 Poncelet, Jean-Victor, 47 porism, 213 positive matrix, 228 principal diagonal, 55

268 projection, 71, 113 projective geometry, 47 projective coordinates, 48 projective space, 48 Ptolemy’s theorem, 221 quasi-nilpotent, 223 radial limit, 106 radical axis, 219 rational function, 135 degree, 135 reference circle, 216 reproducing kernel, 107 normalized, 107, 114 Saratov notebook, 47 Schmeisser, G., 207 Schur’s theorem, 64 Sendov’s conjecture, 206 shift backward, 111 forward, 109 shift operator, 108 adjoint, 111 compressed, 113 side of a polygonal chain, 53 Siebeck’s theorem general, 243 triangles, 211 Soddy’s formula, 221 spectral radius, 231 spectral theorem for normal matrices, 67 spectrum, 123, 232 standard form, 117 standard inner product, 13

Index Steiner inellipse, 39, 211 Steiner’s porism, 213 Steiner’s theorem, 210 Stone–Weierstrass theorem, 26 strongly real of positive type, 135, 239 support line, 124, 162–164 Takenaka–Malmquist basis, 114 tangent line, 51 tangential equation, 241 Toeplitz, Otto, 61 Toeplitz–Hausdorff theorem, 61–62, 71–73 trace, 16, 234 Traité des propriété projectives des figures, 48 unit circle, 17 unitary 1-dilation, 76–82, 119 unitary dilation, 76 vector, 49 cross product, 49 dot product, 49 stochastic, 229 Walsh’s two-circle theorem, 206 Weyl’s theorem, 98 zero inclusion question, 236

AMS / MAA

THE CARUS MATHEMATICAL MONOGRAPHS

Mathematicians delight in finding surprising connections between seemingly disparate areas of mathematics. Whole domains of modern mathematics have arisen from exploration of such connections—consider analytic number theory or algebraic topology. Finding Ellipses is a delight-filled romp across a three-way unexpected connection between complex analysis, linear algebra, and projective geometry. The book begins with Blaschke products, complex-analytic functions that are generalizations of disk automorphisms. In the analysis of Blaschke products, we encounter, in a quite natural way, an ellipse inside the unit disk. The story continues by introducing the reader to Poncelet’s theorem—a beautiful result in projective geometry that ties together two conics and, in particular, two ellipses, one circumscribed by a polygon that is inscribed in the second. The Blaschke ellipse and the Poncelet ellipse turn out to be the same ellipse, and the connection is illuminated by considering the numerical range of a 2 × 2 matrix. The numerical range is a convex subset of the complex plane that contains information about the geometry of the transformation represented by a matrix. Through the numerical range of n × n matrices, we learn more about the interplay between Poncelet’s theorem and Blaschke products. The story ranges widely over analysis, algebra, and geometry, and the exposition of the deep and surprising connections is lucid and compelling. Written for advanced undergraduates or beginning graduate students, this book would be the perfect vehicle for an invigorating and enlightening capstone exploration. The exercises and collection of extensive projects could be used as an embarkation point for a satisfying and rich research project. You are invited to read actively using the accompanying interactive website, which allows you to visualize the concepts in the book, experiment, and develop original conjectures.

For additional information and updates on this book, visit www.ams.org/bookpages/car-34

CAR/34