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Content: Rational pencils of cubics and configurations of six or seven points in RP(2)Points, lines and conics in the planeConfigurations of six pointsConfigurations of seven pointsPencils of cubics with eight base points lying in convex position in RP(2)Pencils of cubicsList of conicsLink between lists and pencilsPencils with reducible cubicsClassification of the pencils of cubicsTabularsApplication to an interpolation problemAlgebraic curvesHilbert's 16th problemMcurves of degree 9Mcurves of degree 9 with deep nestsMcurves of degree 9 with four or three nestsMcurves of degree 9 or 11 with nonempty ovalCurves of degree 11 with many nestsTotally real pencils of curves
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
Séverine FiedlerLe Touzé
Université Toulouse III, Toulouse, France
CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 334872742 © 2019 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acidfree paper Version Date: 20180719 International Standard Book Number13: 9781138322578 (Hardback) International Standard Book Number13: 9781138590519 (Paperback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 9787508400. CCC is a notforprofit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com
To my parents, Thomas, Delphine, Etienne.
Contents
Preface
xi
List of Figures
xvii
List of Tables
xxi
Acknowledgments Contributors Symbols
I
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Rational cubics and configurations of six or seven points in RP 2 1
1 Points, lines and conics in the plane 1.1 Configurations of points . . . . . . . . . . . . . . . . . . . . . 1.2 Definitions and results . . . . . . . . . . . . . . . . . . . . . .
3 3 5
2 Configurations of six points 2.1 Rational pencils of cubics . . . . . . . . . . . . . . . . . . . . 2.2 Diagrams and codes . . . . . . . . . . . . . . . . . . . . . . .
7 7 17
3 Configurations of seven points 3.1 Fourteen configurations . . . . . . . . . . . . . . . . . . . . . 3.2 Linewalls and conicwalls . . . . . . . . . . . . . . . . . . . . 3.3 Refined linewalls . . . . . . . . . . . . . . . . . . . . . . . .
31 31 40 50
II
Pencils of cubics with eight base points lying in convex position in RP 2 57
4 Pencils of cubics 4.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Singular pencils . . . . . . . . . . . . . . . . . . . . . . . . .
59 59 60
5 Lists 5.1 Points in convex position and conics . . . . . . . . . . . . . . 5.2 Admissible lists . . . . . . . . . . . . . . . . . . . . . . . . .
65 65 68 vii
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Contents 5.3 5.4 5.5 5.6
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71 75 76 78
6 Link between lists and pencils 6.1 Nodal lists . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Pairs of distinguished cubics . . . . . . . . . . . . . . . . . . 6.3 Changes of lists and of pencils . . . . . . . . . . . . . . . . .
83 83 85 88
7 Pencils with reducible cubics 7.1 Two nongeneric lists . . . . . . . . . . . . . . . . . . . . . . 7.2 Pencil with six reducible cubics . . . . . . . . . . . . . . . . . 7.3 Symmetric lists . . . . . . . . . . . . . . . . . . . . . . . . .
91 91 93 97
8 Classification of the pencils of cubics 8.1 Nodal pencils . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Inductive constructions . . . . . . . . . . . . . . . . . . . . .
103 103 106
9 Tables
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10 Application to an interpolation problem 10.1 A nongeneric pencil of cubics . . . . . . . . . . . . . . . . . 10.2 Solution to the interpolation problem . . . . . . . . . . . . .
123 123 126
III
Extremal lists . . . . . . . . . . . . Distances between points . . . . . . Isotopies of octuples of points . . . Elementary changes . . . . . . . . .
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Algebraic curves
11 Hilbert’s 16th problem 11.1 Real and complex schemes . . . . . . . . 11.2 Classical restriction method and degree 7 11.3 Orevkov’s method . . . . . . . . . . . . . 11.4 M curves of degree 9 . . . . . . . . . . .
129 . . . .
131 131 136 140 145
12 M curves of degree 9 with deep nests 12.1 Results and rigid isotopy invariants . . . . . . . . . . . . . . 12.2 Curves without O1 jumps . . . . . . . . . . . . . . . . . . . .
155 155 158
13 M curves of degree 9 with four or three nests 13.1 Statement of the results and first proofs . . . . 13.2 Inequalities . . . . . . . . . . . . . . . . . . . . 13.3 M curves with three nests and a jump . . . . . 13.4 End of the proof, using two Orevkov formulas
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163 163 173 183 194
14 More restrictions 14.1 M curves of degree 9 or 11 with one nonempty oval . . . . . 14.2 Curves of degree 11 with many nests . . . . . . . . . . . . .
203 203 208
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Contents
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15 Totally real pencils of cubics 15.1 Two real schemes of sextics . . . . . . . . . . . . . . . . . . . 15.2 Nodal pencil again . . . . . . . . . . . . . . . . . . . . . . . .
211 211 213
Bibliography
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Index
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Preface
“C’est a ` la fois tr`es simple et tr`es compliqu´e.” Capitaine Haddock, Tintin au Pays de L’or Noir Two points in the real projective plane determine a line, five points determine a conic, and nine points a cubic. What objects do n ≤ 9 generic points determine? In addition to nonsingular curves, let us allow rational curves with nodes at some of the chosen points, and consider not only single curves but also pencils. The space of real algebraic curves of degree m is a projective space RP N with N = m(m + 3)/2, the condition that a curve F (x : y : z) passes through a given point p is an equation F (p), linear in the N +1 coefficients of F . The curve F is singular at p if it fulfills two further linear equations ∂F ∂x (p) = 0 and ∂F (p) = 0 (the third partial derivative vanishes automatically by Euler’s ∂y identity). In the generic situation, a singular point is a node, and a rational curve has g = (m − 1)(m − 2)/2 nodes. For m = 3, 4, 5 one has 3g < N , hence n = N − 2g points determine ng rational curves of degree m with nodes at g of them. Thus, seven points determine seven rational cubics, and eight points determine 56 rational quartics and 28 rational quintics. A generic pencil of curves of degree m has m2 base points, any N − 1 of them determine the pencil. The discriminantal hypersurface ∆ ⊂ RP N , formed by the singular curves, has degree 3(m − 1)2 . Hence, a real pencil has s ≤ 3(m − 1)2 singular real curves, where s has the same parity as 3(m − 1)2 . A pencil of conics has four base points and three singular conics (double lines). For m = 3, 4, 5, n = N − 1 − 2g points determine ng rational pencils of curves of degree m with nodes at g of them. Table 0.1 completes the answer to the question above. Let us focus now our attention on the degree 3 and try to go further. Is it possible to describe the possible positions of the cubics (single curves or pencils) with respect to the n points? The cases n = 6 and 7 are studied in Part 1 of this research note. Let six points 1, . . . 6 lie in general position in the real projective plane and consider the rational pencil of cubics based at these points, with a node at one of them, say 1. This pencil has exactly five reducible cubics, dividing it in five portions. At most one of these portions may contain a pair of cuspidal cubics, dividing it in three subportions; the middle one is formed by cubics with an isolated node, the other two contain cubics with a loop passing through no base point apart from the node. We call a combinatorial cubic a topological type (cubic, points) and a combinatorial pencil the cyclic sequence of five combinatorial reducible cubics. Up to the action of the symmetric group S5 on {2, . . . 6}, there are exactly seven possible xi
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pencil of lines line three reducible conics (double lines) pencil of conics conic six pencils of rational cubics seven rational cubics, 35 pencils of rational quartics, seven pencils of rational quintics 8 pencil of cubics, 56 rational quartics, 28 rational quintics 9 cubic TABLE 0.1 Curves determined by n = 1 . . . 9 generic points combinatorial pencils with a node at 1. Consider now the set of six pencils obtained, making the node to be 1, . . . 6. Up to the action of S6 on {1, . . . 6}, there are exactly four possible sets of six combinatorial pencils (Theorem 1.1). Let seven points 1, . . . 7 lie in general position in the plane. Up to the action of S7 on {1, . . . 7}, there are exactly fourteen possible sets of seven nodal combinatorial cubics passing through the seven points, with respective nodes at 1, . . . 7 (Theorem 1.2 and Figures 3.83.10). Let X = (P1 , . . . Pn ) be an element of (RP 2 )n with n ≥ 5, treated as a configuration of n points in one single real projective plane. The (combinatorial) ordered L or Qconfiguration of X is the data describing the mutual positions of each point with respect to the set of lines through any two others, and (for Q) the set of conics through any five others. Quotient by Sn yields the (combinatorial) unordered L or Qconfiguration. An element X is generic if no three of the points are aligned, and (for Q) no six points lie on the same conic. Two generic elements X and X 0 are deformation equivalent if there exists a path in (RP 2 )n containing only generic configurations, that connects one to the other up to the action of Sn . In each setting L or Q, the deformation class of X determines its unordered configuration, but is the converse true? The answer is no in general, but yes for n = 6 and 7, see [27], [29], [90]. In order to encode the ordered Qconfigurations, we shall use the set of six combinatorial pencils for n = 6, and the set of seven combinatorial cubics for n = 7. Each setting L and Q defines a stratification of the space (RP 2 )n , the chambers correspond to the generic configurations, the walls to the first degenerations of the generic configurations. Each of these stratifications induces a stratification of the quotient space (RP 2 )n /Sn . This space may thus be endowed with an adjacency graph, the vertices (resp. the edges) correspond to the chambers (resp. the walls). For n = 6 and 7 and for each setting L and Q, we find the adjacency graph. (The case n = 7, Q is new, while the other three had previously been obtained by Sergey Finashin [28] with a different method.) Part 2 deals with the case n = 8. We classify the pencils of cubics with eight
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base points lying in convex position in RP 2 . (The general case is too complicated.) A generic pencil of cubics has nine base points, and is determined by any eight of them. Exactly eight of the s ≤ 12 singular cubics are distinguished cubics, that is to say real nodal cubics with a loop passing through some base points. The circle parameterizing the real part of the pencil is divided in eight portions by the distinguished cubics; four pairwise nonconsecutive portions contain only smooth cubics with an oval passing through an even number of base points. Each of the other four portions contains connected cubics, plus possibly one subportion bounded by a pair of singular cubics, one with an isolated node, the other with a loop passing through no base point. The inside of the subportion consists of cubics with an oval passing through no base point. (The total number of such subportions is 0, 1 or 2.) Consider eight generic points 1, . . . 8 lying in convex position in the plane; one may associate to these points two combinatorial objects. The combinatorial pencil P(1, . . . 8) is the sequence of topological types (cubic, base points) realized by the eight successive distinguished cubics of the actual pencil. The list L(1, . . . , 8) is the ordered Qconfiguration of the eight points. As the points lie in convex position, the list enumerates the 56 conics passing through five of them plus, for each conic, the position of each of the remaining three points, inside or outside. Written in full extent, the lists are full of redundancies. A simple encoding will allow us to get rid of those and to classify the lists. Let us start with a configuration of seven points 1, . . . 7 in convex position; it determines a set of seven rational cubics. By abuse of language, we call a cubic a topological type (cubic, seven points). There are exactly 14 possible sets of seven cubics, denoted by N ±, N = 1, . . . 7; these sets are all equivalent up to cyclic permutation and symmetries. (The cubics of N ± with node at N have a loop passing through all other six points.) A configuration of eight points gives rise ˆ , N = 1, . . . 8 of seven rational cubics passing through all of to eight sets N the points but N . The data L(1, . . . 8) may then be encoded with the octuple (ˆ 1, . . . ˆ 8). Up to the action of the dihedral group D8 , eight points 1, . . . 8 lying in convex position may realize exactly 47 lists (Theorem 5.1). Move the eight points preserving the strict convexity. An elementary change is the change induced on a list by a move letting one point cross a conic through five others. Up to the action of D8 , there are exactly 17 elementary changes. Four of the 47 lists are nodal , that is to say realizable by eight points on a nodal cubic, one of them being the node. What are the connections between lists and pencils? If one moves the eight points preserving the strict convexity, a degeneration of the (actual) pencil occurs when one of the points 1, . . . 8 comes together with the ninth base point 9, or when six of the points 1, . . . 8 become coconic, 9 is then aligned with the other two. At this moment, two singular cubics of the pencil come together to yield one singular cubic with a node at one of the points 1, . . . 8, or a reducible cubic. Bezout’s theorem allows us to recover the initial pair of (combinatorial) singular cubics. Among the 17 elementary changes, only one is inessential , that is to say it does not alter the combinatorial pencil. If a combinatorial pencil fits to two different lists, then these
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lists differ by an inessential change. Any nonnodal list corresponds to one single combinatorial pencil, and a nodal list gives rise to several combinatorial pencils that are obtained successively from one another by swaps of 9 with other base points. Up to the action of D8 , eight points 1, . . . 8 lying in convex position may realize exactly 43 combinatorial pencils (Theorem 4.1). Parts 1 and 2 may be read independently. Part 3 presents several results about the topology of real algebraic curves in RP 2 , all of them in connection with Hilbert’s 16th problem. Let A be a nonsingular real algebraic curve of degree m. Its complex part CA ⊂ CP 2 is a Riemannian surface of genus g = (m − 1)(m − 2)/2, and its real part RA is a collection of L ≤ g + 1 circles embedded in RP 2 . If L = g + 1, we say that A is an M curve. The complex conjugation conj of CP 2 acts on CA with RA as the fixed points set; thus, CA \ RA is connected or splits in two homeomorphic halves that are exchanged by conj. In the latter case, we say that A is dividing, for example the M curves are dividing. A circle embedded in RP 2 is called oval or pseudoline depending on whether it realizes the class 0 or 1 of H1 (RP 2 ). If m is even, the L components of RA are ovals; if m is odd, RA contains exactly one pseudoline. In 1876, Harnack proved that for each degree m = 2k (resp. m = 2k + 1), there exist curves realizing all values of L ranging between 0 (resp. 1) and g +1. The topological classification is thus completed. What about the isotopy classification? The isotopy type of RA ⊂ RP 2 , describing the mutual positions of the ovals in the plane, is called real scheme of A. The first part of Hilbert’s 16th problem deals with the following question: what are the real schemes realizable by curves of given degree m? This problem requires two complementary approaches, constructions and restrictions. The classification is completed up to m = 7. For higher degrees, one restricts the study to the M curves because of the large number of cases. The classification in degree 8 is almost completed, with 83 realized isotopy types, and only six open cases left. This research note deals only with the restriction side of the problem. The classical method, combining Bezout’s theorem with auxiliary lines, conics (single curves or pencils), and complex orientation results, is presented in Section 11.2, along with examples in degree 7. The first person who studied thoroughly the degree 9 was Anatoly Korchagin. After systematic constructions, this author stated in 1986 three conjectures predicting the nonrealizability of several series of isotopy types. His survey [52] from 1997 contained the complete state of knowledge at that time. We give an update in Section 11.4, with 583 realized isotopy types and 290 open cases left. On the construction side, the main contribution is due to Stepan Orevkov. He found in particular counterexemples to the second half of the third Korchagin conjecture. Some of the restrictions are presented in the next chapters. In Chapters 1214, we exclude in total 266 real schemes of degree 9 (we do not count those that were forbidden previously by other authors), proving parts of the Korchagin conjectures. The proofs are based on the classical restriction method, improved with the following additional tools: auxiliary curves of higher degrees (rational cubics and quartics) and
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Orevkov’s complex orientation formulas. Chapter 14 presents also restrictions for M curves of degree 11, obtained with a generic pencil of cubics. A real pencil of curves is totally real with respect to some real algebraic curve A if it has only real intersections with CA. A curve A may be endowed with a totally real pencil if and only if it is dividing, see [53]. But the proof of this fact does not give explicit pencils. In Chapter 15, we consider the case of M − 2sextics with real scheme h2 q 1h6ii or h6 q 1h2ii. It is known that such sextics are dividing. The results from Part 2 are applied to prove that these sextics may be endowed with totally real pencils of cubics. One can explicitly describe the combinatorial pencil for the first real scheme. When I was a graduate student in Rennes, my director Ilia Itenberg proposed me to work on various questions around Hilbert’s 16th problem. I chose to specialize on the restriction side of the problem, and liked particularly the results on complex orientation. This is how I had the luck to meet my husband, and I followed him to Toulouse in 1997. By chance, several people working on Hilbert’s 16th problem lived in this very place at that time. The research on the subject, that had come to a a standstill after the golden era in Russia, was revived, especially with the breakthrough of Stepan Orevkov (who lived across the road from us), and the constructions of Benoˆıt Chevallier. This was the best possible environment to write my PhD thesis. Since then, I have been working mostly alone, even if I keep contact to fellow researchers. The present book gathers my main results obtained in a time range of twenty years. This is only the visible part of the iceberg, as there were many detours and trails leading nowhere. For instance, I have been trying for years to prove Korchagin’s third conjecture. My first attempt dates back to 2004, when I was a postdoc student in Berkeley. I managed to prove the first half in 2010, and Orevkov found counterexamples disproving the second half only lately! Also, I had the idea to use rational or generic pencils of cubics as restriction tool. In the generic case, the restriction is, theoretically, purely algebraic. Orevkov and myself raised the hope to find thus a nonsingular pseudoholomorphic curve that is not realizable algebraically. Unfortunately, it didn’t work. But I had the idea to classify the pencils so as to make the method more effective. This was the genesis of two big articles that form now Parts 1 and 2 of the present book. The study of the rational cubics lead me to configurations of points, and I came across Sergey Finashin and Arzu Zabun who were working on the same topic. Their results are widely cited and used in Part 1. Now I would like to apologize by the reader that some notations and definitions are perhaps not exactly standard, especially in the first two parts of the book. If I may venture also an advice: It may help to have a pencil and a sheet of paper at hand. Toulouse, June 2018 [email protected]
List of Figures
2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10
2.17 2.18 2.19 2.20 2.21 2.22
The 36 zones . . . . . . . . . . . . . . . . . . . . . . . . . . . Cremona transformations for zones A and B . . . . . . . . . Cremona transformations for zones C and D . . . . . . . . . Cremona transformations for zones E and F . . . . . . . . . Cremona transformation for zone G . . . . . . . . . . . . . . Pencils with node in 1 for zones A and B . . . . . . . . . . . Pencils with node in 1 for zones C and D . . . . . . . . . . . Pencils with node in 1 for zones E and F . . . . . . . . . . . Pencil with node in 1 for zone G . . . . . . . . . . . . . . . . Marked diagrams for 1 ∈ A, . . . G, the circle represents the conic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Refined 1diagrams . . . . . . . . . . . . . . . . . . . . . . . Four configurations β, δ, γ, α encoded each by six equivalent refined diagrams . . . . . . . . . . . . . . . . . . . . . . . . Crossing of walls starting from an αconfiguration, using diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Crossings of walls starting from a βconfiguration, using diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Crossings of walls starting from a γconfiguration, using diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Crossings of walls starting from a δconfiguration, using diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Codes for the four configurations β, δ, γ, α . . . . . . . . . . Crossings of walls starting from a βconfiguration, using codes Crossings of walls starting from a δconfiguration, using codes Crossings of walls starting from a γconfiguration, using codes Crossing of walls starting from an αconfiguration, using codes Adjacency graph of (RP 2 )6 /S6 . . . . . . . . . . . . . . . . .
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3.1 3.2 3.3 3.4 3.5 3.6 3.7
The eleven conicwalls A, . . . K . . . . . . . . . . . . . . The three nonrealizable unmarked conicdiagrams . . Conicdiagrams and codes ˆ1, . . . ˆ6 for A, . . . K . . . . . Codes of (E, 6), (D, 6), (C, 6), (B, 6), (A, 6) and (C 0 , 6) Codes of (G, 6), (H, 6), (K, 6), (I, 6) and (J, 6) . . . . . Three new configurations: 7 ∈ R, 7 ∈ T , 7 ∈ V . . . . . Codes of R, T , V . . . . . . . . . . . . . . . . . . . . .
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2.11 2.12 2.13 2.14 2.15 2.16
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3.10 3.11 3.12 3.13 3.14 3.15 3.16
Sets of seven nodal cubics for (E, 6), (D, 6), (C, 6), (B, 6), (A, 6) and (C 0 , 6) . . . . . . . . . . . . . . . . . . . . . . . . Sets of seven nodal cubics for (G, 6), (H, 6), (K, 6), (I, 6) and (J, 6) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sets of seven nodal cubics for R, T , V . . . . . . . . . . . . The two mutual positions of one line and four points . . . . Moving the line . . . . . . . . . . . . . . . . . . . . . . . . . The 27 linewalls W 1, . . . W 27, the 14 lineconicsubwalls . . Adjacency graph of (RP 2 )7 /S7 in the linear setting . . . . . Linewalls W 4, W 16 and W 12 . . . . . . . . . . . . . . . . . Adjacency graph of (RP 2 )7 /S7 in the quadratic setting . . .
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The degeneration of P into Psing . . . . . . . . . . . . . . .
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The lists 1+ and 1− . . . Construction of the list 1+ ˆ Degenerations of the list G Elementary changes . . . .
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The cubics (1±, 1)n , (1−, 8)n , . . . (1−, 2)n . . . . . . . . . . . Pair of cubics: C31 , C32 if (A, B) = (1, 9), C33 , C34 if (A, B) = (9, 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cubics (1−, 7), (7, 1−), (1−, 67) and (1−, E) . . . . . . . . . Cubics obtained with the close pairs (ˆ8 = 1±, ˆ1 = 8±), and (ˆ 8=ˆ 1 = 7+) . . . . . . . . . . . . . . . . . . . . . . . . . . . Cubics obtained with the elementary pair (ˆ1 : 5+, ˆ5 : 8−) . .
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7.1 7.2 7.3 7.4 7.5 7.6 7.7
First configuration l of points with four reducible cubics . Second configuration l0 of points with four reducible cubics Pencil of cubics realized with the configuration l . . . . . . Pencil of cubics realized with the configuration l0 . . . . . Pascal and Pappus theorems . . . . . . . . . . . . . . . . . Configuration with six reducible cubics . . . . . . . . . . . Pencil with six reducible cubics . . . . . . . . . . . . . . .
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The path p . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nodal pencils obtained from the list max(ˆ1 = 8−) . . . . . .
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The pencil Psing . . . . . . . . . . . . . . . . . . . . . . . . . Interpolation of C . . . . . . . . . . . . . . . . . . . . . . . .
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11.1 11.2 11.3 11.4 11.5
Two quintics hJ q 4i with different types . . . Fiedler chain with a pencil of lines . . . . . . . Folds . . . . . . . . . . . . . . . . . . . . . . . The real scheme hJ q 1h14ii is not realizable . Curve of degree 7 with jump . . . . . . . . . .
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xix
11.6 11.7 11.8 11.9
The complex scheme hJ q 3+ q 1+ h6+ q 5− ii is not realizable Braid b (trivial) and surface N in the case of a conic . . . . Proof of Fiedler’s theorem using a piece of braid . . . . . . . ˆ . . . . . . . . . . . . . . . . . . . . . . . . Isotopy of ˆb in N
141 144 144 145
12.1 12.2 12.3 12.4 12.5
O1 jump . . . C9 with J = 1 Case 1 . . . . Case 2 . . . . Case 3 . . . .
156 157 160 160 161
13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 13.10 13.11 13.12 13.13 13.14 13.15 13.16 13.17 13.18 13.19 13.20 13.21 13.22 13.23 13.24 13.25 13.26
Ovals of a ninth degree curve giving rise to a jump . Three nests of C9 . . . . . . . . . . . . . . . . . . . . Zone containing the oval D . . . . . . . . . . . . . . . C9 with four nests . . . . . . . . . . . . . . . . . . . . Curve C9 with three nests and several jumps . . . . . Jump in O3 , D front, D back . . . . . . . . . . . . . Main part of C18 . . . . . . . . . . . . . . . . . . . . Contradiction with the conic C2 (A1 ) . . . . . . . . . The six ovals in convex position, the three conics . . The hexagon H and the six triangles Z1 , . . . Z6 . . . . The three zones containing the base lines . . . . . . . The hexagon H0 . . . . . . . . . . . . . . . . . . . . . Curve C18 along with the pencil of lines FC , D front, E ∈ T0 . . . . . . . . . . . . . . . . . . . . . . . . . . E ∈ T1 . . . . . . . . . . . . . . . . . . . . . . . . . . F ∈ T3 . . . . . . . . . . . . . . . . . . . . . . . . . . E ∈ T0 . . . . . . . . . . . . . . . . . . . . . . . . . . Missing portion, ED ∩ A1 C ∈ T1 , ED ∩ A1 C ∈ T0 . . E ∈ T1 case 1 . . . . . . . . . . . . . . . . . . . . . . E ∈ T1 , case 2 . . . . . . . . . . . . . . . . . . . . . . E ∈ T1 , case 3 . . . . . . . . . . . . . . . . . . . . . . E ∈ T1 , case 4 . . . . . . . . . . . . . . . . . . . . . . F ∈ T3 . . . . . . . . . . . . . . . . . . . . . . . . . . O1 up, O1 down . . . . . . . . . . . . . . . . . . . . . The double lines . . . . . . . . . . . . . . . . . . . . . The possible cubics C3 . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D back . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
164 165 166 167 170 172 174 177 179 180 181 183 185 186 186 186 187 188 189 190 191 192 193 197 200 201
14.1
The dotted arcs in the triangles are pieces of O, the plain arcs are pieces of J . . . . . . . . . . . . . . . . . . . . . . . . . . The last oval of one chain is connected with O . . . . . . . . M curves hJ q 1+ h6+ q 8− ii and hJ q 1− h15+ q 12− ii . . . The pencil of cubics sweeping out C11 . . . . . . . . . . . . .
205 205 207 209
Pencil of cubics with eight base points distributed in the eight empty ovals of h2 q 1h6ii . . . . . . . . . . . . . . . . . . . .
212
14.2 14.3 14.4 15.1
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xx
List of Figures 15.2 15.3
The cubic C3 . . . . . . . . . . . . . . . . . . . . . . . . . . The next singular cubic in the pencil leads to a contradiction
214 215
List of Tables
0.1
Curves determined by n = 1 . . . 9 generic points . . . . . . .
xii
2.1 2.2 2.3
The seven pencils of conics . . . . . . . . . . . . . . . . . . . The seven pencils of cubics . . . . . . . . . . . . . . . . . . . The four sets of six pencils . . . . . . . . . . . . . . . . . . .
8 8 16
3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9
The 14 generic configurations and their monodromy Adjacencies via linewalls . . . . . . . . . . . . . . Adjacencies via linewalls, continued . . . . . . . . Adjacencies via linewalls, end . . . . . . . . . . . Adjacencies via conicwalls . . . . . . . . . . . . . Admissible triples for the 11 conicwalls . . . . . . . The 14 lineconicwalls . . . . . . . . . . . . . . . . Adjacencies via the 22 strict refined linewalls . . . Adjacencies between unordered configurations . . .
groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41 47 48 49 49 51 52 54 55
5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14
The two lists L(1, . . . 6) . . . . . . . . . . . . . . . . . . . . . Action of D8 . . . . . . . . . . . . . . . . . . . . . . . . . . . Possible values of ˆ8 for the chains n ˆ± = n ˆ ± (8), n = 1, . . . , 7 ˆ 8 = 1+, lists L1 , . . . , L32 . . . . . . . . . . . . . . . . . . . . ˆ 8 = 1+, lists L33 , . . . L64 . . . . . . . . . . . . . . . . . . . . ˆ 8 = 1−, lists L65 , . . . , L70 . . . . . . . . . . . . . . . . . . . . ˆ 8 = 2+, lists L71 , . . . , L74 . . . . . . . . . . . . . . . . . . . . ˆ 8 = 2−, lists L75 , . . . , L87 . . . . . . . . . . . . . . . . . . . . ˆ 8 = 3−, lists L88 , . . . , L95 . . . . . . . . . . . . . . . . . . . . The chains ˆ 1 + (8) and ˆ8 + (1) . . . . . . . . . . . . . . . . . Distances A → B for the principal lists with ˆ8 = 1+ . . . . . Distances A → B for the principal lists with ˆ8 6= 1+ . . . . . Representatives of the 17 orbits of elementary changes . . . Inductive construction of the principal lists . . . . . . . . . .
66 69 70 72 73 74 74 74 75 76 77 78 80 82
6.1
Changes of pairs of distinguished cubics . . . . . . . . . . . .
89
7.1 7.2 7.3
Almost generic symmetric lists, first . . . . . . . . . . . . . . Almost generic symmetric lists, continued . . . . . . . . . . Almost generic symmetric lists, end . . . . . . . . . . . . . .
99 100 101
xxi
xxii
List of Tables
8.2 8.3
Pencils max(ˆ 1 = 8−), max(ˆ1 = 8+), max(ˆ1 = 6−), max(ˆ1 = 6+) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nonnodal pencils with ˆ8 = 1+ . . . . . . . . . . . . . . . . Nonnodal pencils with ˆ8 6= 1+ . . . . . . . . . . . . . . . .
108 109 110
9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11
The lists ˆ 8 = L(1, . . . 7) . . . . . . The lists ˆ 8 = L(1, . . . 7), continued ˆ 8 = 1+ . . . . . . . . . . . . . . . ˆ 8 = 1− . . . . . . . . . . . . . . . ˆ 8 = 2+ . . . . . . . . . . . . . . . ˆ 8 = 2− . . . . . . . . . . . . . . . ˆ 8 = 3+ . . . . . . . . . . . . . . . ˆ 8 = 3− . . . . . . . . . . . . . . . ˆ 8 = 4+ . . . . . . . . . . . . . . . Elementary changes . . . . . . . . Elementary changes, continued . .
. . . . . . . . . . .
112 113 114 115 116 117 118 119 120 121 122
Real schemes of degrees m = 1, . . . 5, complete classification Complex schemes of degrees m = 1, . . . 5, complete classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Complex M schemes of degree 7 with and without jump, complete classification . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Admissible real schemes for the M curves of degree 9 . . . . 11.5 Two nests . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Three nests odd, odd, odd . . . . . . . . . . . . . . . . . . . 11.7 Three nests even, odd, odd . . . . . . . . . . . . . . . . . . . 11.8 Three nests odd, even, even . . . . . . . . . . . . . . . . . . 11.9 Four nests, complete classification . . . . . . . . . . . . . . . 11.10 Deep nest . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.11 Pseudoholomorphic curves whose algebraic realizability is unknown . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
133
13.1 13.2 13.3
197 197
8.1
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11.1 11.2
13.4 13.5 13.6 13.7
Parameters that depend on Sl . . . . . . . . . . . . . . . . . Fi depends on S¯i . . . . . . . . . . . . . . . . . . . . . . . . Short complex schemes even, even, even, along with Ei , i = 0, . . . 3 and Z . . . . . . . . . . . . . . . . . . . . . . . . . . The cases with Fi − Gj − Gk 6= 0 are forbidden . . . . . . . Admissible short complex types even, even, even, along with Z Admissible short complex schemes along with E0 . . . . . . Short complex schemes left, along with E1 , E2 , E3 . . . . . .
135 142 145 148 149 150 151 151 152 153
198 198 198 199 199
Acknowledgments
I am grateful to Sergey Finashin and Arzu Zabun for useful discussions about Part 1. Part 2 contains an uptodate (2018) state of knowledge of the isotopy types realizable by ninth degree M curves. I would not have been able to establish this survey without the precious help of Stepan Orevkov and Benoˆıt Chevallier. They gave me lists with all of the isotopy types they have realized, and went as far as to rerun their computer programs to make sure that no case was forgotten. I wish to thank them very heartily. Finally, I am very grateful to Thomas for his constant and kind support. He was also a source of inspiration, as I use his theorem about complex orientations many times in Part 3. This book would never have existed without him.
xxiii
Contributors
Stepan Orevkov Universit´e Paul Sabatier Toulouse, France Benoˆıt Chevallier Universit´e Paul Sabatier Toulouse, France
Stepan Orevkov and Benoˆıt Chevallier have kindly authorized me to include their unpublished results about ninth degree curves in Section 11.4.
xxv
Symbols
Symbol Description 1, 2, . . . Points in the plane 12345 Conic passing through the points 1, . . . 5 in this ordering A, . . . G (Chapter 2) seven zones in Figure 2.1 Sn Symmetric group on n points hg1 , g2 i (Part 1) subgroup of Sn generated by the elements g1 and g2 kˆ Subconfiguration of six points obtained from the configuration of {1, . . . 7} removing k from this set, or sublist of seven points obtained from the list of {1, . . . 8} removing k from this set cr Cremona transformation α, . . . δ (Part 1) The four unordered configurations of six points or their chosen representatives αγ Wall between α and γ G Monodromy group in the quadratic setting G0 Monodromy group in the linear setting Dn Dihedral group on n points a, σ (Part 2) generators of D8 (+1) Generator a of order 8 of D8 15 Generator σ of order 2 of D8
A5
Alternating group on five points a, . . . d0 (Part 1) the seven pairs (configuration, orbit) A, . . . K (Chapter 3) the 11 conicwalls of seven points ' Equivalence between ordered configurations: an element of S7 maps one onto the other (E, 6) An ordered configuration of seven points with six of them in convex position R, T, V Three nonequivalent ordered configurations of seven points with no six of them in convex position (7, 0, 0, 0) A quadruple 6 : (6b) (Chapter 3) orbit data of 6 in (E, 6) W1 (Chapter 3) A linewall of seven points τ A projective transformation A367 (Chapter 3) lettertriple encoding a lineconic configuration of seven points W 41 (Chapter 3) A strict refined linewall P Pencil of cubics Psing Singular pencil of cubics RP Real part of a real pencil of cubics ∆ Discriminantal hypersurface xxvii
xxviii
J O χ C3 A, B, . . . k < C2 k > C2 \ 1+ C3 (k) Ln L
max(L)
min(L)
1→2
Σn ˆ B) (A, (1−, 7)n
(12, L)
Symbols (Chapters 4 and 11), elsewhere convex hull Pseudoline Oval or loop Euler characteristic Cubic (actual or combinatorial) (Parts 2 and 3) points or empty ovals The point k is interior to the conic C2 The point k is exterior to the conic C2 Number of One of the 14 lists of seven points 1, . . . 7 Cubic with node at k List of eight points number n List (Part 2), number of real components of a real algebraic curve (Chapter 11), line everywhere else The maximal list of eight points realizable with a given sublist L of seven points The minimal list of eight points realizable with a given sublist L of seven points Distance between the points 1 and and 2 for a list L of eight points 1, . . . 8 Sum 1 → 2 + · · · + 8 → 1 for the list Ln Cubic from the list Aˆ with node at B A combinatorial cubic through 1, . . . 8 with node at 7 A combinatorial distinguished cubic in a pencil P (the position of the ninth
(12, C) (1−, 7) (1−, 67) (1−, E) ˆ B) ˆ (A, l, l0
li , li0 λi , λ0i RA CA g conj CA+ Π+ , Π−
Λ+ , Λ−
Lt Lt FX FABCD FS
Cm
base point 9 on it is not given) A combinatorial distinguished cubic A combinatorial distinguished cubic A combinatorial distinguished cubic A combinatorial distinguished cubic Elementary pair, close pair, or elementary change (Chapter 7) nongeneric lists with four sets of six coconic points (Chapter 7) almost generic lists (Chapter 7) other nongeneric lists Real part of a real algebraic curve A Complex part of a real algebraic curve A Genus Complex conjugation Onehalf of CA \ RA for A dividing Numbers of positive and negative injective pairs for a dividing curve Numbers of positive and negative ovals for a dividing curve of odd degree Pencil of lines Line of Lt Pencil of lines based at the point X Pencil of conics based at A, B, C, D S stands for AABCDEF : rational pencil of cubics based at these six points with node at A (Real part of a) real algebraic curve of degree m
Symbols Oi
Nonempty ovals and by extension, nest that they surround α, β Numbers of empty interior and exterior ovals (Part 3 except Chapter 12) hJ q 1i Real scheme of a dividing cubic hJ q 1− i Complex scheme of a dividing cubic XY Z (Part 3) principal triangle determined by three points X, Y, Z chosen in three empty ovals of a curve with odd degree [XY ] (Part 3) principal segment [XY ]0 (Part 3) nonprincipal segment J Almost complex structure ω Symplectic form b (Section 11.3 only) braid ˆb Closure of b σi Artin generator of the braid group Bm e(b) Exponent sum of b l (Section 11.3 only) link + π− Number of injective pairs with positive outer oval and negative empty inner oval l+ , l− Numbers of positive and negative nonempty ovals of an M curve with deep nest λ+ , λ− Numbers of positive and negative empty ovals of an M curve with deep nest
xxix Ai i ηi αi αi+ , αi− a, b, c
Ti , Qi λi
µi
O Si S¯i (+, −) (+, u) πi , πi0 Ni , Mi Gi , Fi Z
(Chapter 13) base ovals Contribution +1 or −1 of Oi to Λ+ − Λ− Contribution +1 or −1 of Ai to Λ+ − Λ− Number of empty ovals in the nest Oi Numbers of positive and negative ovals in the nest Oi (Chapters 11 and 12 only) numbers of outer, median and inner ovals Triangles and quadrangles (Chapter 13) contribution of the ovals in the triangle Ti to Λ+ − Λ− (Chapter 13) contribution of the ovals in the quadrangle Qi to Λ+ − Λ− (Chapter 13) image of J by the Cremona transformation (Chapter 13) short complex scheme of the nest Oi (Chapter 13) short complex type of the nest Oi A short complex scheme A short complex type (Chapter 13) parameters of the nests Parameters of the nests, continued Parameters of the nests, end (Chapter 13) zone (union of triangles) that may contain nonprincipal exterior ovals
Part I
Rational cubics and configurations of six or seven points in RP 2
1 Points, lines and conics in the plane
CONTENTS 1.1 1.2
Configurations of points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Definitions and results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1
Configurations of points
3 5
Let (P1 , . . . Pn ) be an element of (RP 2 )n with n ≥ 5, treated as a configuration of n points in one single real projective plane. The (combinatorial) ordered Lconfiguration of (P1 , . . . Pn ) is the combinatorial data describing the mutual positions of each point with respect to the set of lines through any two others, and the (combinatorial) ordered Qconfiguration is the data describing the mutual positions of each point with respect to the lines through two others and the conics through five others (L stands for linear , Q for quadratic). These data will be described with some encodings involving the indices 1, . . . n. Let us fix a setting L or Q. Two configurations of points (P1 , . . . Pn ) and (P10 , . . . Pn0 ) are combinatorially equivalent if their ordered configurations coincide up to the action of the symmetric group Sn ; the equivalence classes are called (combinatorial) unordered configurations. A configuration is generic if no three points are aligned, and (for Q only) no six points are coconic (i.e. on the same conic). The almost generic configurations have three aligned points or (for Q only) six coconic points, and are otherwise generic. An L or Qdeformation or rigid isotopy is a path in some stratum of (RP 2 )n , that does not cross a substratum of higher codimension. One may define similarly an L or Qdeformation of the whole plane. Two configurations of n points are deformation equivalent if they may be deformed one into the other up to the action of Sn . For n = 5, there is only one deformation class. There exist two other obvious equivalence relations: Two configurations (P1 , . . . Pn ) and (P10 , . . . Pn0 ) are point by point combinatorially equivalent if their ordered configurations coincide, and they are point by point deformation equivalent if they may be deformed one into the other. Each setting L and Q gives rise to a stratification of (RP 2 )n . The chambers (resp. the walls) are the point by point deformation classes of the generic (resp. the almost generic) configurations. The quotient space (RP 2 )n /Sn has an algebraic variety structure as the orbit 3
4
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
space of a finite group action. The two stratifications of (RP 2 )n induce stratifications of (RP 2 )n /Sn . The chambers (resp. the walls) are the deformation classes of the generic (resp. the almost generic) configurations. The adjacency graph of (RP 2 )n /Sn is obtained assigning a vertex to each chamber and an edge to each wall. For n = 5, the generic combinatorial L and Qconfigurations coincide and are given by the cyclic ordering of the points on the conic they determine. Part 1 of this book deals with n = 6 and 7 in both settings L and Q. The configurations considered hereafter are always supposed to be generic or almost generic. For each of the four cases, consider a configuration (P1 , . . . Pn ). Its ordered configuration will be represented by several equivalent codes (diagrams whose vertices are marked with 1, . . . n). It turns out that two configurations are point by point deformation equivalent if (and only if) they are point by point combinatorially equivalent. They are thus deformation equivalent if and only if they are combinatorially equivalent (see [27], [29] and Proposition 3.2). So, the chambers (resp. the walls) in (RP 2 )n are in bijection with the generic (resp. almost generic) ordered configurations. In the quotient space (RP 2 )n /Sn , the chambers (resp. the walls) are in bijection with the unordered generic (resp. almost generic) configurations. It will be convenient to call the almost generic combinatorial configurations walls for short. A purely combinatorial study allows us to obtain the four adjacency graphs (the case n = 7, Q is new, whereas the other three were previously obtained by S. Finashin [28]). The stratified space (RP 2 )6 /S6 has four chambers and four linewalls for both L and Q. In the quadratic setting, a supplementary conicwall appears inside of one chamber. For n = 7, there exist configurations with three aligned points and six coconic points. The (actual) linewalls of the linear setting split thus in several refined linewalls for the quadratic setting. Let (P1 , . . . Pn ) be a generic element of (RP 2 )n , the subgroup of Sn formed by all permutations that may be realized with point by point deformations is the monodromy group of the configuration. For n = 6 or 7, it is also the subgroup of permutations of Sn that preserves the ordered configuration of (P1 , . . . Pn ). (The orbit of a code under the action of the monodromy group is the set of all equivalent codes.) For all orderings of the n points, the associated monodromy groups are isomorphic, so one can also speak of the monodromy group of an unordered configuration. The monodromy group will be denoted by G in the quadratic setting, and by G0 in the linear setting. How many different combinatorial classes may be realized by generic configurations of n points? Do they coincide with the deformation classes? This problem has a long history. Usually, the objects under consideration are not configurations of points, but dual arrangements of lines. In the linear setting, results have been obtained with various methods; some come from algebraic geometry (Geometric Invariant Theory, Hilbert schemes, del Pezzo surfaces), others from combinatorics (Matroid Theory). In the modern approach, a line arrangement is endowed with an oriented matroid [5], which encodes the combinatorial type of line arrangement in terms of mutual position of the partition
Points, lines and conics in the plane
5
polygons. The first description of the line arrangements for n = 6 and 7 dates back to 1932, with the papers [10], [87] of Cummings and White. In 1988, Mn¨ev found configurations of n points that are combinatorially equivalent but not rigidly isotopic, with n ≥ 19. What is the lowest number n such that the combinatorial and deformation classifications do not coincide? Y. Tsukamoto found the best bound up to now with n = 13. On the other hand, S. Finashin [27] proved that the classifications coincide for n = 6 and 7. The numbers of deformation classes (or equivalently, of unordered configurations) are 4 for n = 6 and 11 for n = 7. For n = 8, a computeraided enumeration gave 135 different combinatorial types of generic arrangements (R. Canham, E. Halsey, 1971). For n = 9, there are 4381 types (J. RichterGebert, G. GonzalesSpringer, G. Laffaille, 1989). Let us consider now the quadratic setting. For n = 6, the Qdeformation classes coincide with the Ldeformation classes, and there are 4 of them [27]. For n = 7, there are exactly 14 unordered configurations [72]. (We will give an alternative proof of this fact.) To find them, J. Sekiguchi worked with dual line arrangements (with no three concurrent lines and no six lines tangent to a common conic) and established a link between such arrangements and the root system E7 . S. Finashin and A. Zabun [29], [90] proved that the combinatorial and the deformation classifications coincide. They obtained also the classification of the Aronhold sets of seven bitangents to real plane quartics with four ovals: They realize 14 types, corresponding via the del Pezzo surfaces of degree 2 to the 14 unordered combinatorial Qconfigurations. For n = 8 there are 1629 unordered configurations, obtained via the dual arrangements with a computeraided enumeration, based on an analysis of the root system E8 [31].
1.2
Definitions and results
Throughout Part 1, the word configuration written without further precision will stand for a combinatorial configuration associated to an actual configuration of 6 or 7 points. Unless otherwise stated, it will be by default supposed to be generic, ordered, and quadratic. Consider six points 1, . . . 6 lying in generic position in the plane; five of them determine 10 lines and one conic, dividing RP 2 in 36 zones. The configuration they determine may be described by any one of the six equivalent pieces of information: cyclic ordering of five chosen points on their conic and zone containing the sixth one. Consider the rational pencil of cubics through these points, with a node at one of them, say 1: the generic cubics all have a node at 1, the nongeneric cubics are five reducible cubics plus, possibly, some cuspidal cubics. We ignore the latter and consider the sequence of reducible cubics. Let us call a combinatorial cubic a topological type (cubic, points), up to the following identification: a loop passing through no other point than the node will be assimilated to an isolated node. Let us
6
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
call a combinatorial pencil the cyclic sequence of five combinatorial reducible cubics; they are of the form 1m ∪ 1ijkl, where {i, j, k, l, m} = {2, 3, 4, 5, 6}. Definition 1.1 The ordered configuration associated to six points 1, . . . 6 is the set of six combinatorial rational pencils of cubics based at 1, . . . 6, with respective nodes at 1, . . . 6. Let us consider now seven points 1, . . . 7. For a point n among the seven, we shall denote by n ˆ the (Q or L)subconfiguration of six points, obtained by removing n from the set {1, . . . 7}. The (Q or L)configuration determined by the seven points may then be described with the set ˆ1, . . . ˆ7 of seven subconfigurations. One may alternatively use the following: Definition 1.2 The ordered configuration associated to seven points 1, . . . 7 is the set of seven combinatorial nodal cubics through the points 1, . . . 7, with respective nodes at 1, . . . 7. Theorem 1.1 Let 1, . . . , 6 be six generic points in RP 2 . Up to the action of S5 on {2, . . . 6}, there are exactly seven possible combinatorial rational pencils of cubics based at these six points with a node at 1. Up to the action of S6 on {1, . . . 6}, there are exactly four possible sets of six combinatorial rational pencils of cubics based at these six points with respective nodes at 1, . . . 6. Otherwise stated, six generic points may realize exactly four different unordered configurations (Table 2.3). Theorem 1.2 Seven generic points 1, . . . 7 in RP 2 may realize exactly fourteen different unordered configurations. Up to the action of S7 , there are correspondingly exactly fourteen possible sets of seven combinatorial nodal cubics through the seven points, with respective nodes at 1, . . . 7 (Figures 3.83.10). The definitions above give temporary encodings of the combinatorial configurations. The codes mentioned in the previous section will be introduced later. Theorem 1.1 will be proved in Chapter 2, Theorem 1.2 in Chapter 3.
2 Configurations of six points
CONTENTS 2.1 2.2
Rational pencils of cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Diagrams and codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1
Rational pencils of cubics
7 17
Proof of Theorem 1.1: Up to the action of S5 on {2, . . . 6}, we may assume that the six points are disposed as shown in Figure 2.1, where 2, 5, 3, 6, 4 lie in this ordering on a conic, and 1 is in one of the zones A, . . . G. Let 1 ∈ A. We perform a Cremona transformation cr : (x0 ; x1 ; x2 ) → (x1 x2 ; x0 x2 ; x0 x1 ) with base points 1, 4, 5. Let us denote the respective images of the lines 14, 15, 45 by 5, 4, 1. For the other points, we keep the same notation as before cr. The three conics 14365, 31452, and 61542 are mapped onto three lines 36, 23 and 26; see the upper part of Figure 2.2. After cr, let us consider the pencil of conics 1236, and five particular conics of this pencil: the three double lines and the two conics passing respectively through 4 and 5. The cyclic ordering of these conics in the pencil is easily determined. Perform the Cremona transformation back. The pencil of conics 1236 is mapped onto the rational pencil of cubics based at 1, . . . 6 with node at 1. The five particular conics are mapped onto the five reducible cubics. We repeat this procedure for the various positions of 1 ∈ B, . . . G, with a Cremona transformation based either in 145 or in 136, see Figures 2.22.5. (The cuspidal cubics are images by cr of conics tangent to the base line that does not contain 1.) The sequences of five particular conics are displayed in Table 2.1. The sequences of five reducible cubics for each case A, . . . G are shown in Table 2.2, the pencils are drawn in Figures 2.62.9. More precisely, we have represented the combinatorial cubics in the five successive portions bounded by the reducible cubics. The upper middle pictures of each pencil B, C, G are not quite correct: in the actual pencils, the portion starts and finishes with cubics having a loop containing no base point other than 1. In between, there is a pair of cuspidal cubics, and in the middle, cubics with an isolated node. We have represented only the latter. Consider any one of these seven pencils. We observe the following proper7
8
Pencils of Cubics and Algebraic Curves in the Real Projective Plane F B 3 6
D
A C
5
G
4
2
E
FIGURE 2.1 The 36 zones 1 ∈ A, cr(145) 12 ∪ 36 13462 1 ∈ B, cr(145) 12 ∪ 36 31426 1 ∈ C, cr(136) 12 ∪ 45 15 ∪ 42 1 ∈ D, cr(145) 12 ∪ 36 31246 1 ∈ E, cr(136) 12 ∪ 45 15 ∪ 24 1 ∈ F, cr(145) 12 ∪ 36 34126 1 ∈ G, cr(136) 12 ∪ 45 15 ∪ 24
13 ∪ 26 13 ∪ 26 41562 13 ∪ 26 41562 13 ∪ 26 41562
16 ∪ 23 16532 16 ∪ 32 25163 41523 14 ∪ 25 16 ∪ 32 16352 41523 14 ∪ 25 16 ∪ 23 32156 41532 14 ∪ 25
TABLE 2.1 The seven pencils of conics 1∈A 1∈B 1∈C 1∈D 1∈E 1∈F 1∈G
12 ∪ 51436 12 ∪ 31645 12 ∪ 15364 12 ∪ 31654 12 ∪ 15364 12 ∪ 16453 12 ∪ 14635
15 ∪ 31264 15 ∪ 31642 15 ∪ 31264 15 ∪ 31624 15 ∪ 12346 15 ∪ 31642 15 ∪ 12436
TABLE 2.2 The seven pencils of cubics
13 ∪ 51624 13 ∪ 51642 13 ∪ 51264 13 ∪ 51642 13 ∪ 12546 13 ∪ 16425 13 ∪ 12456
16 ∪ 31452 16 ∪ 31425 16 ∪ 21453 16 ∪ 31425 16 ∪ 12453 16 ∪ 13524 16 ∪ 12453
14 ∪ 21653 14 ∪ 31625 14 ∪ 21653 14 ∪ 31652 14 ∪ 12653 14 ∪ 13526 14 ∪ 12635
Configurations of six points
9
3 1
6 1
5
4
5
2 6
3
1 4
5
4
2
3
6
2
1 5
4 3
6
2
1
1
3
6
3
5
4
5
6
4 2
2 6
3 1 5
6
5
4
2
1
3
4
2
FIGURE 2.2 Cremona transformations for zones A and B
10
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
4
1
2
2
2
1
5
1 5
4 6
6
3
3
5
3
2
1
6
3
1 4
2
4 2
1 6 5
6
5
4
3
1
3
6
5
3
5
1
6
2
6
3
4
5
4
4
2
FIGURE 2.3 Cremona transformations for zones C and D
Configurations of six points
11
1 4
1
5
2
6
4 3
2
6
2
1
5
2
2 6
3
3
1 6
3
4
5
1
5
4 2
4 2
6
2
1 5
6
3
6
3
5
5
1
6
1
4
3
4
3
5
4
FIGURE 2.4 Cremona transformations for zones E and F
12
Pencils of Cubics and Algebraic Curves in the Real Projective Plane 1 4 6
4 5
2
2
1
5
1
6
3
6
3
4
3
5
1
6
5
2
4
3
2
FIGURE 2.5 Cremona transformation for zone G ties: The cyclic ordering of the points 2, . . . 6 is the same on all of the generic cubics; it is given by the pencil of lines based at 1 (this fact follows from Bezout’s theorem). The cyclic ordering of 2, . . . 6 given by the lines of the successive reducible cubics is the same as the cyclic ordering of these points on the conic that they determine. The mutual cyclic orderings of 2, . . . 6 given on one hand by the conic 25364, and on the other hand by the pencil of lines based at 1, are displayed in the marked diagrams of Figure 2.10, where the circles represent the conics. Let us now consider together all six pencils of cubics with nodes at 1, . . . 6. If we take 1 in one of the zones E, F, G exterior to the conic 25364, then we note that 2 lies inside of the conic determined by the other five points. So, up to the action of S6 , we may assume that 1 is interior to a conic 25364. There are exactly four sets of six pencils, see Table 2.3 where k < C2 (resp. k > C2 ) means k interior to C2 (resp. k exterior to C2 ). Hence, six generic points may realize exactly four different unordered Qconfigurations α, β, γ, δ. This finishes the proof of Theorem 1.1. For a given set of pencils, each choice of a point 1, . . . 6 gives rise to a marked diagram. Let us remove the markings (the names of the points); it turns out that all six unmarked diagrams of a set are identical, and each set corresponds to a different unmarked diagram. Note that the unmarked diagrams also remain unchanged if the roles of the circle and of the polygonal line are swapped.
Configurations of six points
3
13
3
6 1
13
16
5 4 3
6
3
31264
1
12
5 2
2
15
2
14
13
1
51642
6
3 5
4
31642
3
1
6
2
12
31425
5
2
FIGURE 2.6 Pencils with node in 1 for zones A and B
1
3
31645 5
2
4
14
6 4
2
6
3
4
4
5
16
21653
4
1 6
5
6
1
51436
4
1
4
2
2
5
3
1
31452
5
2 15
6
3
1
51624
4
5
6
31625
14
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
2
2 16
1 4
21453
5
14
1
6
21653
6
1
3
51642
4
3
6 16
1
31624 3
6
31425
12
2
31654
2
FIGURE 2.7 Pencils with node in 1 for zones C and D
6
4
4
4
5
1
3 5
2
31264
5
5 2
15
4
3
5
15
1
15364
5
13
3
2 12
6
6
5 6
2
4
1
1 4
3
2
1
3
51264
5
4
3
6
13
1
3
6
5
4 2
14
31652
Configurations of six points 16
1 4
12453
5
2
6
2
15 2
4
3
13
6
3
16425
14
1
3
6
5
12
6
2
31645
3
4
5
4
4
2
16253
1
3
5 4
2
12346
5
6
1
5
3
1
15364
5
14253
5
2 6
3
16
1 4
3
12
6
1
2
1 4
52164
5
6
12653
3
13
1
4
3
6
14
15
15
1
16423
6
5
4
2
2
FIGURE 2.8 Pencils with node in 1 for zones E and F 4
1 5
2 6
12
14
12635 4
15364
4 2 6
1
5
15
3
FIGURE 2.9 Pencil with node in 1 for zone G
12453 4
5
2 6
3
16
1
1 4
5 3
6
3
12436
1
2
2
6
13 5
3
12456
12 ∪ 31645 23 ∪ 51642 35 ∪ 31642 43 ∪ 51642 53 ∪ 51642 63 ∪ 51642 12 ∪ 15364 21 ∪ 25364 35 ∪ 31264 42 ∪ 15364 53 ∪ 51264 62 ∪ 15364 12 ∪ 31654 23 ∪ 51642 35 ∪ 31624 43 ∪ 51642 53 ∪ 51642 63 ∪ 51642
β:1∈B 1 < 25364 2 > 31645 3 > 51642 4 < 31625 5 < 31642 6 > 31425 γ:1∈C 1 < 25364 2 > 15364 3 > 51264 4 < 21653 5 < 31264 6 > 21453 δ:1∈D 1 < 25364 2 < 31654 3 > 51642 4 > 31652 5 > 31624 6 > 31425
TABLE 2.3 The four sets of six pencils
12 ∪ 51436 25 ∪ 31264 35 ∪ 31264 42 ∪ 51436 53 ∪ 51624 63 ∪ 51624
α:1∈A 1 < 25364 2 < 51436 3 < 51624 4 < 21653 5 < 31264 6 < 31452
15 ∪ 31624 21 ∪ 25364 31 ∪ 25364 41 ∪ 25364 51 ∪ 25364 61 ∪ 25364
15 ∪ 31264 25 ∪ 31264 31 ∪ 25364 41 ∪ 25364 51 ∪ 25364 61 ∪ 25364
15 ∪ 31642 21 ∪ 25364 31 ∪ 25364 41 ∪ 25364 51 ∪ 25364 61 ∪ 25364
15 ∪ 31264 21 ∪ 25364 31 ∪ 25364 41 ∪ 25364 51 ∪ 25364 61 ∪ 25364
13 ∪ 51642 26 ∪ 31425 36 ∪ 31425 46 ∪ 31425 56 ∪ 31425 64 ∪ 31652
13 ∪ 51264 23 ∪ 51264 32 ∪ 15364 46 ∪ 21453 52 ∪ 15364 64 ∪ 21653
13 ∪ 51642 26 ∪ 31425 36 ∪ 31425 46 ∪ 31425 56 ∪ 31425 64 ∪ 31625
13 ∪ 51624 24 ∪ 21653 36 ∪ 31452 46 ∪ 31452 52 ∪ 51436 64 ∪ 21653
16 ∪ 31425 25 ∪ 31624 34 ∪ 31652 45 ∪ 31624 52 ∪ 31654 62 ∪ 31654
16 ∪ 21453 26 ∪ 21453 36 ∪ 21453 45 ∪ 31264 56 ∪ 21453 65 ∪ 31264
16 ∪ 31425 24 ∪ 31625 34 ∪ 31625 42 ∪ 31645 54 ∪ 31625 62 ∪ 31645
16 ∪ 31452 23 ∪ 51624 32 ∪ 51436 45 ∪ 31264 56 ∪ 31452 65 ∪ 31264
14 ∪ 31652 24 ∪ 31652 32 ∪ 31654 42 ∪ 31654 54 ∪ 31652 65 ∪ 31624
14 ∪ 21653 24 ∪ 21653 34 ∪ 21653 43 ∪ 51264 54 ∪ 21653 63 ∪ 51264
14 ∪ 31625 25 ∪ 31642 32 ∪ 31645 45 ∪ 31642 52 ∪ 31645 65 ∪ 31642
14 ∪ 21653 26 ∪ 31452 34 ∪ 21653 43 ∪ 51624 54 ∪ 21653 62 ∪ 51436
16 Pencils of Cubics and Algebraic Curves in the Real Projective Plane
Configurations of six points
3
17
6
5
3 4
6 4
5
2 A
3
2 C, E
6
5
4 2 D
4
5
2 B, F 3
6
4
2
6
5 3 G
FIGURE 2.10 Marked diagrams for 1 ∈ A, . . . G, the circle represents the conic
2.2
Diagrams and codes
The seven configurations from Figure 2.1 corresponding to the choices A, . . . G of the zone containing 1 may be encoded by refining the marked diagrams from Figure 2.10 as shown in Figure 2.11. First, we indicate whether 1 lies inside or outside of the conic 25364 using a dotted polygonal line if 1 is inside, and a plain polygonal line if 1 is outside. Once this is done, the marked diagrams of B, F and G still correspond to several zones. The diagram of B fits to the other four Blike zones, and each of these five zones may be characterized by the point of the conic situated opposite to it; for B this point is 2. Let us add to the diagram of B a dot at the point 2. The same argument applies to F . Let us now consider the zone G; it is a triangle having a vertex at 2 and whose sides are supported by the lines 24, 26, 35. There are in total 10 Glike zones and the diagram of G also fits to the one having 4 as vertex and whose sides are supported by the lines 24, 54, and 36. In order to differentiate these two zones, we provide the edge 24 in the diagram of G with an arrow from 2 to 4. The extra point 1 is indicated inside of the diagrams in Figure 2.11. Let us call an ndiagram a diagram whose extra point is n. A configuration may be encoded by any one of its six ndiagrams (n = 1, . . . 6). See Figure 2.12, where all six ndiagrams in a row are equivalent. For each of the (underlying actual) four configurations of points, let one single point n move, until it crosses a line through two others or a conic through five others (the actual configuration follows a path that crosses a wall in the stratified space (RP 2 )6 ). Figures 2.132.16 show the corresponding changes of the ndiagrams for n = 1, . . . 6. We deduce from Figures 2.132.16 the following:
18
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
Proposition 2.1 Each βconfiguration is adjacent to another βconfiguration via a conicwall, and to three δconfigurations via linewalls. Each δconfiguration is adjacent to two βconfigurations and to four γconfigurations via linewalls. Each γconfiguration is adjacent to six δconfigurations and to one αconfiguration via linewalls. Each αconfiguration is adjacent to ten γconfigurations via linewalls. Our next concern is to define new kinds of diagrams, or codes, replacing the set of six equivalent ndiagrams. It is of course impossible to define a unique code describing a configuration. However, the codes will turn out to be more convenient than the ndiagrams. Let us say briefly that a point n is interior for the configuration if n is interior to the conic determined by the other five (the ndiagram is dotted). The upper part of Figure 2.17 shows four codes corresponding to the four configurations of Figure 2.12. Let us explain how we defined them. In the case β, we represent the six points with six dots disposed on a circle in the natural cyclic ordering given by the convex position. The dots are colored alternately in black and white, and the white dots correspond to the interior points. The case δ offers no such evident solution, so one has to make a choice that is not entirely satisfactory. The interior points in the δconfiguration of Figure 2.12 are 1 and 2. The two polygonal lines of their diagrams may be seen as closed paths intercepting successively five points. Let us embed these two paths in a different way, drawing one of them with a dotted and the other with a plain line. We get a graph with vertices 1, . . . 6, having two different kinds of edges. To recover the remaining four ndiagrams from this graph, note that the polygonal lines of these diagrams are oriented. Let us choose a starting point: 1 for the 3 and 6diagrams, 2 for the 4 and 5diagrams. The orderings with which the five points are met by the polygonal lines of these four diagrams are shown with arrows describing paths in the graph, as indicated in Figure 2.17. Finally, note that this encoding is not unique: the second graph drawn in the bottom of the figure represents the same configuration. Two graphs obtained from one another swapping pairs of points with a vertical symmetry represent, of course, the same configuration. For the case γ, note that the 1diagram is dotted and has 2 as its bottom point, whereas the 2diagram is plain and has 1 as its bottom point. We say that 1 and 2 form a pair. The other four points may be similarly distributed in two pairs. Let us encode this with a (nonconnected) graph having the six points as vertices, and three edges endowed with arrows as shown in Figure 2.17; each arrow connects an interior point to its associated exterior point. The original ndiagrams may be deduced easily from this graph. For the case α, let us simply observe that the ndiagrams may be deduced easily from one another. For example, start with the 1diagram. In this diagram, 1 and 2 are separated by branch 54 of the star. To get the 2diagram, it suffices to swap 1 with 2 and 5 with 4. To encode this configuration, we will simply use any one of the ndiagrams, with the circle removed. For convenience we will speak of the code of an ordered generic configuration.
Configurations of six points
19
In reality, there are always several equivalent codes. This is also true for the almost generic configurations; see hereafter. Let us call a (combinatorial) wall a combinatorial configuration obtained as first degenerations of a generic configuration, with three points on a line (linewall) or six points on a conic (conicwall). An ordered linewall is the data of the cyclic ordering with which the line L meets: the three aligned points P4 , P5 , P6 of the configuration, and the three lines L12 , L23 , L13 determined by the other three points. An ordered conicwall is the data of the cyclic ordering of the six points 1, 2, 3, 4, 5, 6 on the conic. Figures 2.182.21 show all possible crossings of walls, starting from the four ordered configurations α, β, γ, δ, using now codes instead of diagrams. In Figure 2.21, the set of ten lines is written, but for space reasons, we drew the codes of only two adjacent γconfigurations. Note that one may encode any unordered generic configuration of six points with its unmarked code (obtained by removing the indices). This will no longer be true in general for n = 7, as each index 1, . . . 7 appears several times in the codes. The monodromy group of a configurations is the subgroup of permutations of S6 mapping one of its codes onto the set of equivalent codes. With the help of Figure 2.17, we find thus: G(β) = {id, (36)(45), (14)(23), (15)(26), (145)(236), (154)(263)} = D3 , G(δ) = {id, (36)(45), (12)(3465), (12)(3564)} = Z/4, G(γ) = {id, (36)(45), (32)(15), (14)(26), (154)(236), (263)(145)} = D3 . The group G(α) is a icosahedral group (isomorphic to A5 ), it has 60 elements: id, 24 elements of order two (conjugacy class of (12)(45)), 20 of order five (conjugacy class of (25364)), and 15 of order three (conjugacy class of (142)(356)). The action of G(α) on the points is transitive. For each of the other three configurations, the points are distributed in two orbits, so there are in total seven pairs (configuration, orbit): a (for α); b:{2, 3, 6}, b0 :{1, 4, 5} (β); c:{2, 3, 6}, c0 :{1, 5, 4} (γ); d:{1, 2}, d0 :{3, 4, 5, 6} (δ). Consider an ordered configuration along with its monodromy group. The unordered linewalls adjacent to this configuration correspond to the orbits of the adjacent triples under the action of the group. For a βconfiguration, the unordered conicwall ββ corresponds to the unique orbit of the adjacent sextuple under the action of the group. For example, G(δ) gives rise to two orbits of triples: {136, 245} (δβ) and {246, 134, 156, 235} (δγ), see Figure 2.19. Let us encode α, δ, γ, β now as Lconfigurations. The codes for the first three stay unchanged. For β, the dots must now have all the same color, say black. Let us look at the monodromy groups of the Lconfigurations. One has G0 (β) = D6 ; note that this group acts transitively on the points. For each of the other three config
20
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
3 5
6 1
3 4
5
1
3 4
5
2 B
2 A 3 5
6
1 2 C
1
3 4
5
3 5
6 1 2 D
4
1
6 4
2 E
2 F
6 4
6
4 6
2 1
5
3 G
FIGURE 2.11 Refined 1diagrams urations, one has G0 = G. In the linear setting, there are in total six pairs (configuration, orbit): a, b, c, c0 , d, d0 . The monodromy groups of the walls are easily found: G(αγ) = {id, (13)(45), (23)(56), (12)(46), (123)(465), (132)(456)}, G(γδ) = {id}, G(δβ) = {id, (13)(46)}, G(ββ) = D6 = h(26)(35), (123456)i. The conicwall connects two βconfigurations that have the same monodromy group G = {id, (26)(35), (15)(24), (13)(46), (135)(246), (153)(264)}. The monodromy group of a linewall is the biggest common subgroup of the groups associated to the two adjacent configurations. (In the case of βδ, this is true with both G(β) and G0 (β).) Lemma 2.1 (S. Finashin) In each setting L and Q, two generic or almost generic configurations (P1 , . . . P6 ) and (P10 , . . . P60 ) are point by point deformation equivalent if (and only if ) they are point by point combinatorially equivalent. Thus, the deformation classes coincide with the combinatorial classes. See [27] or [29]. Combining this lemma with Prop 2.1, one gets the following proposition, that had previously been proved by Finashin [27], [28]: Proposition 2.2 (S. Finashin) In the linear setting, the stratified space (RP 2 )6 /S6 has four chambers and three linewalls. In the quadratic setting, one supplementary conicwall appears inside of one chamber. The adjacency graphs are as shown in Figure 2.22 (the dotted loop represents the conicwall).
Configurations of six points
3
6
4
1
5
4
5
3
1
4
2
1 4
5
6
5 6
4
3
5
4
2 6
1
2
2
3
1
5
2 6
1 5
3
3 4
4
2 6
5
1
4 2
5
4
6
3 6
4
3
3
5
1
2
5
5 2
4 6
3
1 2
1
1
4
6 6
3
3
4
5
1
6
4 3
3 5
4
1 6
1
5
2
5 4
2
4
3
1
1 6
1
4
2
3
3 2
2 4
1
2 6
2
5
2
5
5
1 3
6
6
6
1 3
2
2
3
3
2
4
6
6
4
5
4
1
5
3
1
1 6
3
5
2
6
2
5
21
1 6
3
4 6
5
2
FIGURE 2.12 Four configurations β, δ, γ, α encoded each by six equivalent refined diagrams
22
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
3 5
2
6 3
14 5
3
123
1
5
4
2
2
6
5
4
1
6
5 6
3
2
5 1
4
5 1
4 3
3 2
1
1
26 1 6
3
5
4
6 1
2
2 3
1 34
3
5
2
4
4
3 6
5
2
2
6
2
4 3
2
5
6
5 1
5 4
1
134
1
3
4
6
5
1
5
4 1
245
2 5
3
6
4
4 6 2
4 5 1
3
1 5
1
5
3 4
1
6
2
3
2
4 6
3 2
5
2
6
6
5 2
5
4
3
4
1
1
6
1 6
4
5 356
156
5 3 2
3
2 4
1
6
2 145
3
145
246
6
3
356
5
3
3
5
2 5
346
6 4
2
2
6
6
5 24 5
1 3 2
1 4
3
5
1 1
5
3
4 3
6
1
4
6 4
2
346
6
1
1 3
4 1
3
1 5
23 5
2
6 2
4
5
6
123
2 5 4
5
6 3
3
6
3 1
246
4 2
5
5
6
4
3 23 5
134
1 2
6
2 1
5
6
4 2
4
1 26
6
3 1
2
4
4 1
5
6 1
126
1 6
3
246
4 5
2
2
346
3
5 4
2 6
2
15 6
3
1
FIGURE 2.13 Crossing of walls starting from an αconfiguration, using diagrams
5
3 6 4
1
Configurations of six points
23
4 3 5
3
6 1
4
2
1
5
136
6
3 4
2
1
5
253164
6
6 4
1 6 4 2 53
2
4
6 1
1 3
2
3
1
4
5 2
6
5
5 2
3
1
246
5
23 5
5
2
4
3
6
3 1
2 6
4 164 253 5
1 3
2 2 4
5 3
4
2 35
6
6
1 1
136
1
5
6
6 3
6
2 4
5
46 2
3
2 4
1
6 5
3
1
253164
2 4
5
3
4 1
2
4 6
3
2
5 1 6 42 53 2 4
2
3 5 6
1
253
4
3 5 6
2 1
253164
4
3 5
4
1 1
3
6
3 5
1 6
136
6 5
2 4
246
4
2
FIGURE 2.14 Crossings of walls starting from a βconfiguration, using diagrams
1
2 6 3
5
24
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
4
2 1
6
4
5 6
3
1 5
2
145 6
3
1 4
6 12 6
3
4
5
123
2
5
1
4 1
2
126
2
4
5
6 3
1
3
3
246
5
5
1
1
3
4
2
6 4
2
2
6
5
3
6 356
6
4
1 3
3 2
5
2 6
5
1
2
6
235
4
1 4
123
6 2
5 4
3
2
3
1
6
5
235
145 2
3
3 1
4
4
5
1
5 6
123
4
2
346
1
5
3
5
1 6
3
1
6
2 2
4
6 1
5 3
4
3
2
3 1 5
6
5
4
4
346
246
6
1
2
4 6
4 5
1
5 6
3
2 3
2
145 6
2
246 5 4
356
3
2 235
3
5
1
1
4
356
1
126
5
5
6
4
4
6
2 5
2
3
1 3
6
34 6
4
3 4
2 6
5 1
3
6
FIGURE 2.15 Crossings of walls starting from a γconfiguration, using diagrams
1
2
Configurations of six points
3
6
5
1
5
25
4
4 2
6
2
24 5
136 3 5 4
2
134
6
5
1
4
2
156
2
5
1
3
1
6
3
2
3
235
1
24 6
6
2 6
3 4
5
136
245
6
5
3
134
2
5 4
235
5
2
3 6
6
2
1
4
3
1
24 6
13 4
5
1
2 1
4 6
3
4
1
3
235
1
136 3
3
5 2
15 6
5 6
6
1
5 4
2
2
5
4 5
5
6
24 5
6
6
3
3
2
1 4
2 5
1
3 4
4 6
2
6 1
3 5
1
6
1
4
5 4
4
2
6
3
5
1 3
1 2
4
6 5
1
4
6
2
4
5 3
2
3
1
4
246
4
2
156
2
6 1
3
4
1 6
3
5 6
2
4
5
FIGURE 2.16 Crossings of walls starting from a δconfiguration, using diagrams
3 1
26
Pencils of Cubics and Algebraic Curves in the Real Projective Plane 3
1
1
6
5
4 2
6
3
5
4
1 2
5 4
2
6
3
3
1
5
6
4
2
2 5
4
3
6 1
FIGURE 2.17 Codes for the four configurations β, δ, γ, α 3 5
1
2
6 4 164253 1
3
6
5
4 2
24 6
13 6
6 4 2 35 15 13
3 1 6
4
235
2 4 45 2 5
2 5 3 16 14 46
1
5
2
6
6
3
3
2
1
5
5
4
4
1
3
2
6
FIGURE 2.18 Crossings of walls starting from a βconfiguration, using codes
Configurations of six points
27
2 5
4
6
3
1 3
6
5
4
2
1
1
4
3
24 5
2
5
6
4 2 5 1 6 3 6 13 13 6
24 6
6 1 3 25 45 24
2 4 13 6 15 35
1 6
3
5
4
2 35
1 34
2
2 5 1 6 3 1 4 46 2
1 3 2 5 4 26 56 1 56
3
1
2
1 6 24 5 23 34 5
4 6
3 1
1
5
6
5 6
3 4
2
FIGURE 2.19 Crossings of walls starting from a δconfiguration, using codes
4
28
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
5 1
3
4
2
1
2
6
5
2
4
3 6
23 5 3 5 14 16 2 46
123 1 2 46 3 56 45
5 4
3
6
5
4 346 4 6 12 3 25 15
1 3 2
1 26 1 2 35 6 34 45
1
35 6 3 5
6
14
1
24 6 12
2
4
2
5
6 3
246 4 6 35 2 3
5
2
6
1 4
13 15
145 1 26 4 36 5 4 6
23
6
1
4
3 5
2 1
5
3
FIGURE 2.20 Crossings of walls starting from a γconfiguration, using codes
Configurations of six points
29
4 6
2 1
5
3 1 45 1 23
23 5
5
36
1
3
2
4
26
16
5 2
5 4
2
1
6
6 1 23 12 6
14 3 5
3 134 15 6 24 5
246 3 46
4 356
FIGURE 2.21 Crossing of walls starting from an αconfiguration, using codes
46
30
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
γ
α
α
αγ
FIGURE 2.22 Adjacency graph of (RP 2 )6 /S6
δ
γ
δ
γδ
β
β
ββ
δβ
3 Configurations of seven points
CONTENTS 3.1 3.2 3.3
Fourteen configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Linewalls and conicwalls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Refined linewalls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1
Fourteen configurations
31 40 50
A (combinatorial ordered Q)configuration of seven points 1, . . . 7 will be encoded with the set of seven subcodes describing the subconfigurations ˆ1, . . . ˆ7. As for the case n = 6, we shall call linewalls (resp. conicwalls) the almost generic configurations with three aligned points (resp. six coconic points). Consider an ordered conicwall, with 1, . . . 6 disposed in this ordering on the conic. The mutual cyclic orderings of 1, . . . , 6 given respectively by the conic and the pencil of lines based at 7 may be described with a conicdiagram: a closed polygonal line with six vertices, inscribed in a circle (where a vanishing triangle of the polygonal line is assimilated to a triple point). It is easily seen that there are exactly eleven admissible unmarked conicdiagrams. Proposition 3.1 Seven points 1, . . . 7 in RP 2 with six of them on a conic, but otherwise generic, may realize exactly eleven different unordered conicwalls. Seven generic points 1, . . . 7 in RP 2 , with six of them lying in convex position, may realize exactly eleven different unordered configurations. Up to the action of S7 , there are correspondingly exactly eleven possible sets of seven combinatorial nodal cubics. Proof: Let us consider for a start seven points 1, . . . 7 such that 1, . . . 6 lie in this ordering on a conic, and draw this conic as an ellipse in some affine plane. The point 7 lies in a zone bounded by some of the 15 lines determined by 1, . . . 6. The lines 36, 14 and 25 give rise to six sectors, each containing one edge of the hexagonal convex hull of the six points. Up to cyclic permutation of 1, . . . 6, we may assume that 7 lies in the sector containing the edge 61. Let us move the six points, keeping them coconic, until three triplets of lines become concurrent, as shown in the upper part of Figure 3.1. Move 7 along so that it does not cross any of the 15 lines nor the conic. Note 31
32
Pencils of Cubics and Algebraic Curves in the Real Projective Plane I J H
G
E
F
6
C
D B
1
C’
5
2 4
6
B
1
5
6
4
4
3 G 6 1
5
6
4
A
6
2
6 2
3
2
I
4
4 6
4
6
K
3
1 2
5
3
4 1 2
5 4
1 2
3
2
J
5
1
5
3
6 2
4
3
FIGURE 3.1 The eleven conicwalls A, . . . K
5 4
6
1 2 3
1 2
1
5
E, F
5
3
4
1
6
1
5
H
5
3
D 1
3 6
2 4
C
5
2
3
3
Configurations of seven points
33
FIGURE 3.2 The three nonrealizable unmarked conicdiagrams that the zone containing 7 may be a triangle that shrinks in the end to a triple point. If 7 is not in such a vanishing zone, one may assume up to the symmetry (61)(52)(43) that 7 lies in the end in one of the nine zones B, . . . J of Figure 3.1. Note that the (ordered) configuration realized by the seven points is preserved all along the motion. If 7 is in a vanishing triangle, this triangle is either A or K, see the bottom part of Figure 3.1 (up to the action of (61)(52)(43) for K). There are thus eleven unordered conicwalls; name them A, . . . K after the zone containing 7. Note that these eleven zones give rise in total to eight unmarked conicdiagrams, the other three that turn out to be unrealizable are shown in Figure 3.2. Figure 3.3 shows for each zone A, . . . K: the unmarked conicdiagram (refined with either a dotted or a plain polygonal line depending on whether 7 lies inside or outside of the conic), and the six subcodes ˆ 1, . . . ˆ 6. Let now 1, . . . 7 be seven generic points, such that six of them lie in convex position. We may assume that the seventh point (extra point) is either inside or outside of all six conics determined by the first six. Let indeed 1, . . . 6 lie in convex position, 6 being outside of the conic 12345, and let 7 lie between two of the six conics. If the seven points lie in convex position (the position of 7 in the cyclic ordering is arbitrary), consider one of the two conics adjacent to 7; let n ∈ 1, . . . 6 be the point that does not lie on this conic, n may be taken as extra point. Otherwise, up to the action of D3 on 1, . . . , 6, one of the following two situations occurs: 1, 2, 3, 4, 5, 7 lie in convex position and 6 is outside of all six conics they determine, or 2, 3, 4, 5, 6, 7 lie in convex position and 1 lies inside of all six conics they determine. Let now 1, . . . 6 lie in convex position, and 7 be either inside or outside of all six conics determined by 1, . . . 6. One may move the seven points until the first six become coconic, preserving the configuration all along. So, up to the action of S7 , any configuration of seven points with six of them in convex position may be obtained from one of the eleven nongeneric configurations A, . . . K by moving 6 away from the conic 12345. Let X be one of the zones in Figure 3.1. Denote by (X, 6) the configuration obtained from X moving 6 to the outside of the conic 12345, and (X, 60 ) the configuration obtained moving 6 to the inside. Let us enumerate the pairs of equivalent configurations, along with the elements of S7 mapping one onto the other: (X, 6) and (X, 60 ) with X ∈ {A, B, D, E, F, J} are swapped by the symmetry (61)(52)(43), this symmetry swaps also (C 0 , 6) with (C, 60 ). The symmetry (42)(76)(15) swaps (D, 6) with (G, 60 ), and (C, 6) with (H, 60 ). The symmetry (63)(14)(52) swaps (I, 6) with (I, 60 ), and (K, 6) with (K, 60 ). The cyclic permutation (1234567)
34
Pencils of Cubics and Algebraic Curves in the Real Projective Plane ^ 1 6
1
4
3
6
1
3
5
2
^ 2 5
4
6 7
2
4
1
1
6
2
7
4
7
3
6
1
6
1
^ 6 3
2
1 4
7
3
2
6
7 2
5
1 4
4
3
5
3
5
2
6
1
6
7
3 4
1
3
1
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3 5
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2
5 7
7 4
7
1
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3
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6
4
1
5 3
5 3
2
5
1
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4 7
2
6
7
5 6
7 4
4
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6 7
3
2 1
1
^ 5
5
5
3
6
1
6
1
7
^ 4
5 6
7
4
2 4
5
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2
3 5
^ 3
3
6
2
3
5
3
1
2
5
6
7
6
4
1
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2
4 7
2
6
3
3
3
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2
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2
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5
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1
6
1
6
1
6
4
1
2 3
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6
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7
1
7
7
7 7
2 3
7
7
7
7
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2
6
1
6
1
6
1
6
1
5
5
3
5
3
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2
5
2
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2
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1
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1
4
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3
6
2
3
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3
6
5
5
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5
7
5
7
5
7
4
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2
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1
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1
3
1
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1
3
7
2
5 4
3 6
3 3
1
5
3
2
3
2
2
1
2
2
2
6
5
4
2
4
1
4
1
3
1
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7
4
5
7
5
7
5
7
5
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3
1
3
7
2 4
3 6
6
2
1
5
6 4
1
3
5
6 4
1
3
5
6 2
4
2
6
3
7
6
7
2
2
5
3
1
3
1
4
5
7
4
7
3
7
5
2
4
1
2
6
1
7 6
2
6
1 6
1
2
4
1
4
4
5 6
5
2
3
5
3
6
3
7
6
4
1 3
5 5
4
1
2 6
7
7
2 3
6
7
2 3
2
3
1 4
5
3 2
4
5
7
2
5
1
1
6 7
3
2
3
5
4
5
5
7 3
7
6
5
1 7
2
6
7 4
1
5
2
3 6
6
1
6 5
4
1
1
2
7
7
4 6
5
2
4 2
2
6
4
6
4
6
4
3
1
3
1
3
2
7
1
7
1
7
5
7
4
7
4
1
2
5 4
7
3 3
3
2
FIGURE 3.3 Conicdiagrams and codes ˆ1, . . . ˆ6 for A, . . . K
6
6
5
Configurations of seven points
35
maps (E, 6) onto (F, 60 ). There are thus exactly eleven different unordered configurations of seven points with six of them in convex position. Let us name each of them after some ordered configuration representing it. Later on, we will introduce a more canonical encoding. Each of the zones X ∈ {A, B, I, J, K} gives rise to one unordered configuration, say (X, 6), the pair E, F gives rise to one, say (E, 6), the pair D, G gives rise to two, say (D, 6) and (G, 6), and the pair C, H gives rise to three, say (C, 6), (C 0 , 6) and (H, 6). Figure 3.4 and Figure 3.8 display respectively the codes and the sets of seven nodal cubics with nodes at 1, . . . 7 for the configurations (E, 6), (D, 6), (C, 6), (B, 6), (A, 6), (C 0 , 6). Figure 3.5 and Figure 3.9 display the codes and the sets of cubics for the configurations (G, 6), (H, 6), (K, 6), (I, 6), (J, 6). Let us explain how to find these sets of cubics. To get a cubic with node at a given point, say 1, it suffices to perturb the reducible cubic 17 ∪ 123456, moving one point, say 6, away from the conic 12345 in the appropriate direction. Note that crossing a line lnm induces the change of the four subconfigurations pˆ, p 6= l, n, m, and of the three cubics with respective nodes at l, n, m. Three configurations with no six points in convex position may be obtained taking 7 in one of the zones R, T and V of Figure 3.6. The codes of these configurations are shown in Figure 3.7. We have thus constructed 14 unordered generic configurations; to complete the proof of Theorem 1.2, we need to grant that there exist no others. This will be done in the next section. The sets of seven nodal cubics corresponding to the three new configurations are shown in Figure 3.10. Each cubic may be obtained in several ways perturbing reducible cubics. For example, the cubic with a node at 7 of the first configuration may be obtained from 176 ∪ 57423 moving 6 to the left of the line 17, from 274 ∪ 56173 moving 4 to the top from the line 24, etc. For a given configuration, let us count the numbers of types β, δ, γ and α realized by the subconfigurations ˆ1, . . . ˆ7; we get a quadruple (nβ , nδ , nγ , nα ). There are exactly eleven possible quadruples. Note that the 14 unordered Qconfigurations give rise to 11 unordered Lconfigurations, corresponding to the eleven quadruples. Nine of the quadruples correspond each to a unique unordered Qconfiguration The quadruple (3, 4, 0, 0) fits to two unordered Qconfigurations, the quadruple (2, 2, 3, 0) fits to three unordered Qconfigurations, we use indices to differentiate them, see Table 3.1.1 Two unordered generic configurations have the same quadruple if and only if they are adjacent via a conicwall. Let us consider some ordered configuration of seven points 1, . . . 7. Each point n appears in six subcodes; for each subcode ˆ For example, ˆ note the orbit type of n for the action of G(k) ˆ (or G0 (k)). k, 1 in (B, 6) realizes a twice and each of the four types b0 , d0 , c, c0 once; we encode this as follows: 1 : (2a, 1b0 , 1d0 , 1c, 1c0 ). This is the orbit data of 1. Each configuration may thus be endowed with its set of seven orbit data for the points 1, . . . 7. Note that some part of this set may suffice to determine the 1 A similar notation is used by A. Zabun and S. Finashin [90], [29], but with different choices of indices. Theirs may be obtained from mine swapping 1 and 2 for (3, 4, 0, 0), and performing the cyclic permutation (123) for (2, 2, 3, 0).
36
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
^ 1 7
^ 2 7
^ 3 7
^ 4 7
^ 5 7
^ 6 7
^ 7 1 6
6
2 6
1
6
1 6
1 6
1 5
1
5
3 5
3 5
2 5
2 4
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2 5
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6 1
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1 1
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4 7
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3
2 7
1 7
2 5
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1
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6 7
6
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1 3
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3 4
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5
5
3 4
3
1
4
2
4
2
6
2
6
1
5
1
5
3
7
FIGURE 3.4 Codes of (E, 6), (D, 6), (C, 6), (B, 6), (A, 6) and (C 0 , 6)
7
4
Configurations of seven points ^ 1
^ 2
3
37
^ 3
^ 4 2
2
3
^ 5
^ 6
6
^ 7
5
1
4
2
4
1
4
1
3
1
4
7 4
7 6
2
5
7
5
7
5
7
5
7
3
1 3
1 5
3
6
6
2
4
1
3
5 6
1
4
3
7
6
5 6
7
7
2
2
4
5
1
3
1
3
1
4
7 6
5
7
4
6
3
1 5
6
5
5
2 2
4
2
5
5
6
6
2
7
2
1
6
4
6
4
6
4
3
1
3
1
3
1
6
2
7
1
7
1
7
5
7
4
7
4
7
5
3
2
3
2
4
3
1 5
6
7
6 5
6
3
1
4
3 7
5
2
5
7
4
2
4
1
5
7
5
7
5
1
6 4
6
5 2
7
1
4 2
3
3
2
3
1
3
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6
2
2
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5
3
7
1
1
1
7
2
2
4 4
1
6
7
6 3
5
6
6
4
3
6
6
3 4
2
2
2
4
1
1
2
7
2
7
6
6
3
5
3
5
4
4
2 3 4
FIGURE 3.5 Codes of (G, 6), (H, 6), (K, 6), (I, 6) and (J, 6)
2
6
2
R
5
3
3
6 V
T 4
1
FIGURE 3.6 Three new configurations: 7 ∈ R, 7 ∈ T , 7 ∈ V
5
1
4
38
Pencils of Cubics and Algebraic Curves in the Real Projective Plane ^ 1
^ 2
6 2
3
5
4
4 7
3
7 4 2
6
6
7
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5
1
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1 7
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7
^ 7
3
1
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^ 6
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^ 5
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7
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^ 4
^ 3
5
4
2
3 6
7 7
4
6
2
3
1 3
5
1
2
4 3
5
6
FIGURE 3.7 Codes of R, T , V 6− 4
6
5 3
7
3
6 7
4 3
1
5 3
3
5
1
7
4 3
5
7
1 2
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6 7 2
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1 6 1 6 1 6
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2 5
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1
5
4 3
6 7
5
7 2
6 7
6
7
3
1
6
5
4
7
2
3
6
5
3
1 5
4
1
2
4
7
3
1
2
6
5
4
4
3
2
FIGURE 3.8 Sets of seven nodal cubics for (E, 6), (D, 6), (C, 6), (B, 6), (A, 6) and (C 0 , 6)
Configurations of seven points
7
2 3
6 5 7
1
4
3
7
7 5
1 6
4
6
5
6
5
4
7
3
2 3
1
4
5
2
4 7
7
1
6
5
3 1
6
5
2
4
3
3
5
4 7
1
6
5
3 1
6
2
6
5
2
3
4
4
3
4
1
6
2
4
7 1
4
2
3
4
7
6
1
5
2
5
2
3
2
1 2
7 6
5 4
7
3
1 2 2
7 1
1
6
2
3
5
7
4 2
7 1
1
4
2
3
1
3
6
1
5 4
2
3
4
7
1
1
7 6
7
6
5
2
3 6
5
1
3
7
3
7
3 6
5
4
5 4
2
6
5
2
3
4
3
1
1
3
7
6
5
4
7 6
7
6
5
2
6
4
2
1
3
5
1
5
3
2
3
4
7
2
4
7
5
4
1
6
1
5 4
3
7
6
5
2
6
4
7 3
7
2 3
4
5
5
1
7
2
6 2
4
1
6
7
6
5
3
4 7
1
6
2
5
3
7
2
3
2
4
1
6
1
6
5
7
2
4 7
2
1
1
6
5
4
6
5
7
1
39
6
2
3
3
5
4
FIGURE 3.9 Sets of seven nodal cubics for (G, 6), (H, 6), (K, 6), (I, 6) and (J, 6)
6 2
7
4
5
2
3
7
5
4
5
3
6 5
1
7
7
4
2 5 1
4
6
3 7
5
2
4
6 7
5
3 4
2
3 7
4
2
7
7
3 4
2
5
4
6
5
1
FIGURE 3.10 Sets of seven nodal cubics for R, T , V
7 1
4
2 5
3
7
5
4
1
6
1
7
5
3 6
5
2 4
5
2
4
1 2
3
7
4
7 1
3
6 7
4
1 2
4
3 4
1
5
3 6
6
7
5
2
1 3
7
6
2 4
1 2
3
6
3 7
1 6
5
4
3
6
6
2
3 1
5
1 6
7
5
1
2
3
1
1
2
6
6
2
4
1 2
3
3
7
5
1 6
6
2
5 1
6
3 7 4
40
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
configuration, for example (E, 6) is determined by the two orbit data 6 : (6b), 7 : (6b0 ). Consider seven generic points 1, . . . 7 lying in convex position (their unordered configuration is (7, 0, 0, 0)), they may realize exactly fourteen different ordered configurations, encoded 1±, . . . 7±, where 1+ is determined by 1 : (6b), 7 : (6b0 ). (Thus, E(6) is 6−.) See also Proposition 5.1 in Part 2. The monodromy group of a configuration may be easily found using its code. An element of G (or G0 in the linear setting) must map n onto a point m such that the subcodes n ˆ and m ˆ have the same type α, β, γ or δ, and n, m realize the same orbit data. Once we have the image m of one point n, it is easy to find the images of the other points using the correspondence n ˆ → m. ˆ This procedure allows us to get the groups G and G0 , see Table 3.1. The list of groups G0 was found originally by S. Finashin [27], and we rediscovered it here with another method.
3.2
Linewalls and conicwalls
In the linear setting, the stratified space (RP 2 )7 /S7 has 11 chambers and 27 walls, see [27], [28] (where S. Finashin considered actually dual arrangements of lines). The chambers correspond to the 11 unordered Lconfigurations, they are described by the quadruples. Let us explain how to find and encode the linewalls. Consider first the possible positions of a line L with respect to four points 1, 2, 3, 4. Choose three points among 1, . . . 4; they give rise to four triangles in the plane. We call principal triangle the one containing the fourth point. Add a line L passing through none of the four points. This line cuts either three or four principal triangles, see upper and lower part of Figure 3.11. The corresponding symmetry groups are S3 = h(12), (123)i and Z/2 = {id, (23)}. The four points give rise to six lines; let us look at the six intersection points of these lines with L. The cyclic ordering of these intersections on L allows us to recover the mutual position of L with 1, 2, 3, 4. This information will be encoded using a circle with six marked points, two points have the same color if they are in the same orbit for the action of the symmetry group. We need two colors (red and blue) in the first case, and four colors (yellow, blue, red and green) in the second case 2 . To get the walls in (RP 2 )7 /S7 , distribute three unmarked points (colored black) in all possible ways on the two circles from Figure 3.11, that stay for first marked. For the second circle, if an interval between a blue point, say 24, and a red point, say 13, contains no black point, we may move the line L (or equivalently the point 4) until L crosses the intersection 24 ∩ 13. On the circle, the positions of 13 and 24 are swapped, and the colorings of all six points change, see the second circle in Figure 3.12. The 2 As this book is printed in black and white, we use letters to indicate the colors: b for blue, r for red, g for green and y for yellow.
unordered configuration (7, 0, 0, 0) (3, 4, 0, 0)1 (3, 4, 0, 0)2 (2, 2, 3, 0)1 (2, 2, 3, 0)2 (2, 2, 3, 0)3 (1, 2, 2, 2) (1, 0, 6, 0) (1, 6, 0, 0) (1, 2, 4, 0) (1, 4, 2, 0) (0, 4, 3, 0) (0, 3, 3, 1) (0, 6, 1, 0)
G0 D7 = h(12)(37)(46), (1234567)i Z/2 = h(34)(25)(16)i Z/2 = h(14)(23)(57)i {id} {id} {id} Z/2 = h(16)(25)(34)i S3 = h(16)(25)(34), (135)(246)i Z/2 = h(14)(36)(25)i Z/2 = h(14)(36)(25)i Z/2 = h(16)(25)(34)i {id} Z/3 = h(137)(456)i Z/3 = h(137)(456)i
G {id} {id} {id} {id} {id} {id} {id} Z/3 = h(135)(246)i {id} {id} {id} {id} Z/3 = h(137)(456)i Z/3 = h(137)(456)i
TABLE 3.1 The 14 generic configurations and their monodromy groups
ordered configuration (E, 6) (D, 6) (G, 6) (C, 6) (C 0 , 6) (H, 6) (B, 6) (A, 6) (K, 6) (I, 6) (J, 6) R T V
Configurations of seven points 41
42
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
1 2
4
1 3
2
1
3
4
3
2
2
4
3
13 r
1 4 2
3
14
b
12
r
b
b
1
4
34
r 23
24
2
1 2
4
4 3
3
3
1
4
2
2 4
1
1 3
14 y
1 12 b
b 13
4 2
3
34
r
g
r 24
23
FIGURE 3.11 The two mutual positions of one line and four points
third and fourth circles show the colorings obtained with the last two possible positions of 34, 12, 13, 24. In other words, the group G acting on the four points is D4 = {id, (23), (14), (12)(34), (13)(24), (14)(23), (1243), (1342)}. We can thus reduce the number of colors to two (red and green), see the bottom circle in Figure 3.12. (We may see the pairs 2, 3 and 1, 4 as opposite edges of a square, the red points on the circle are the vertices of the square, and the green points are the intersections of the pairs of opposite lines supporting the edges.) Remove now the markings on the circles; one gets in total 27 unordered linewalls W 1, . . . W 27, splitting in three groups according to the distribution of colors for the six intersections with the line L, see the 27 circles in Figure 3.13 (where the markings should be ignored, they will be used in the next section). For n = 6, the linear and quadratic linewalls coincide. This is no longer true for n = 7, as there exist nongeneric configurations with three aligned points and six points on a nonsingular conic. Such a configuration (otherwise
Configurations of seven points
12
b
34 r
14 y g
b
13
12
r
r
24
34
b
23
14 g y
43
r
24
34
b
b
13
12
r
23
14 y g 23
12 r 34
r
14 g g
b
24
34
r
r
13
12
b
14 g y
r
13
b
24
23
r 13 r 24
23
FIGURE 3.12 Moving the line
generic) is called lineconic. The lineconic strata divide the Llinewalls in several Qlinewalls. In the sequel, we shall call the latter refined linewalls. Proposition 3.2 (S. Finashin, A. Zabun) In each setting L and Q, two generic configurations (P1 , . . . P7 ) and (P10 , . . . P70 ) are point by point deformation equivalent if (and only if ) they are point by point combinatorially equivalent. Thus, the generic deformation classes coincide with the generic combinatorial classes. The proofs may be found in [27] (Lsetting) and [90] or [29] (Qsetting). Proposition 3.3 In each setting L and Q, two almost generic configurations (P1 , . . . P7 ) and (P10 , . . . P70 ) are point by point deformation equivalent if (and only if ) they are point by point combinatorially equivalent. Thus, the almost generic deformation classes coincide with the almost generic combinatorial classes. Proof: Let first (P1 , . . . P7 ) and (P10 , . . . P70 ) be two Lconfigurations realizing the same ordered linewall, with P4 , P5 , P6 on a line L and P40 , P50 , P60 on a line L0 . A projective transformation τ maps (L012 , L013 , L023 , L0 ) onto (L12 , L13 , L23 , L). Note that τ is realizable with a point by point Qdeformation of the whole plane, as P GL(3, R) is connected. One has τ (Pi0 ) = Pi , i = 1, 2, 3. Moreover, Qj = τ (Pj0 ), j = 5, 6, 7 lie on L, and Q4 = τ (P40 ) lies in the zone of RP 2 \(L12 ∪L13 ∪L23 ∪L) containing P4 . The line L is divided in six segments by its intersections with L12 , L13 , L23 , P1 Q4 , P2 Q4 , P3 Q4 . Let us perform a further point by point deformation, moving Q4 along a segment of
44
14
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
13 r b
b
34
14
13 r b
W1
12
r
b
r 23
12
r
b
r
b
34
14
12 r r
g
r
23
12
r
r 13 r 24
34
r 23
r W12 r 5 12 5 7 4 3 2 4 6 12 5 6 4 3 7 r 12 6 7 4 3 1 3 r 34 14 g
14 g
14 g 12 r
r 13
r
g
r 24
23 14g
7 5 1 2 r W21 r 1 3 15 4 2 6 3 34 r r 24 g 6 23
5 1 2 r W22 r 1 3 7 15 43 6 2 34 r g r 24 6 23
14 y
14 y b 13
12 b
W25
r
g 23
14 12
24
34
r
b 13 g 23
r 23
14 g
r 13
b W4 b 34 14 25 36 1 2 14 25 37 23 r r b 5 24 7 14 6 12
r
g
r 24
14
13 r b
12
r
b
12 r
14
13 r b
24
34
r
g
23
23
14 g
14 g
1 2 r W18 r 1 3 6 12 4 356 5 r r 24 g 34 23
12
r
b
6 1 2 r W23 r 1 3 16 4 352 7 34 r r 24 g 5 23
r 13
12 r
r 24
r 13 g 23
1 2 W27 b b 13 6 12 4 35 6 7 5 r r 24 g 34 23
FIGURE 3.13 The 27 linewalls W 1, . . . W 27, the 14 lineconicsubwalls
34
r
g
r
24
14 g 12 r
r 13 W20
34
r
g 23
W24
r
r 13
23
14g
34
14 g W15
23
12 r
r 23
24
7 1 2 W19 r r 13 6 124356 5 r r 24 g 34
14g
34
b W10
W14
r
r 23
24
24 14 g
34
b W5
13 r
b W9 b 34 14 3 5 2 6 5 r r 23 b
W13
34
13 r
14y
W26
r
12 r
7
W17
34
b
13 7 r b W8 b 3 4 14 2 536 r r 23 b 5 24
12
1 2 r W16 r 1 3 164352 3 4 1 74 3 5 2 2 4 r r g 5 23
34
r
7
14
6 b
b
b 34
24
24
7 14g
12 b
12
23g
23 6
r 23
W7
W11
34
b
b
24 14 g
b
6
W3
13 r
W6
12
14
24
13 r b
34
W2
24
14
b
13 r
r 24
r 24
Configurations of seven points
45
line onto P4 . During this deformation, one or two segments among the six mobile segments on L may collapse (inducing changes as shown in Figure 3.12). But such a collapse may take place only if the segment does not contain any one of the points Qk , k = 5, 6, 7. So one can move these three points along so that they stay on their respective mobile segments. Finally, shift Qk , k = 5, 6, 7 along L onto Pk , we are done. (See also [27] for an alternative proof.) To prove the statement for the refined linewalls, it suffices to achieve, additionally, that during the motion, none of the points Qk crosses a conic C through P1 , P2 , P3 , Q4 , Ql , with k, l ∈ {5, 6, 7}. Assume that Qk must cross a conic C. This conic cuts L at Ql and a second point S. The point S divides one of the six segments in two segments. By hypothesis, one of these segments contains Qk and collapses during the deformation (at some moment, C is tangent to L), reappearing afterwards on the other side of Ql . During the process, Qk has crossed C from the inside to the outside, passing through S, and then Qk crosses C again, from the outside to the inside, passing through Ql . Contradiction. Let now (P1 , . . . P7 ) and (P10 , . . . P70 ) be two configurations realizing the same ordered conicwall, with P1 , . . . P6 (resp. P10 , . . . P60 ) lying in this cyclic ordering on a conic C (resp. C 0 ). A projective transformation τ (realizable with a point by point Qdeformation of the whole plane) maps C 0 onto C. Let Qi = τ (Pi0 ). The points Q1 , . . . Q6 lie in this cyclic ordering on C. The 15 lines passing through pairs of points Qi , Qj divide the plane in Qzones. One shifts now {Q1 , . . . Q6 } onto {P1 , . . . P6 } along C. Assume that the Qzone containing Q7 does not collapse in the process, so one may move Q7 along. The action of the monodromy group D6 extends to the Qzones, and the Qzones containing respectively P7 and Q7 are in the same orbit, we are done. Assume now that the Qzones containing P7 and P70 are collapsible (vanishing triangles). The unordered conicwall is A or K. Up to the action of D6 , we may assume that (P1 , . . . P6 ) lie on C as shown in one of the two lower pictures in Figure 3.1. Let L∞ be a line at infinity and L0ij be the line through Pi0 , Pj0 . Let τ be the projective transformation that maps (L014 , L063 , L025 , L∞ ) onto (L14 , L63 , L25 , L∞ ) in the case A, or (L045 , L063 , L012 , L∞ ) onto (L45 , L63 , L12 , L∞ ) in the case K. In both cases, the triangle containing P70 is mapped onto the one containing P7 . Let Qi = τ (Pi0 ); these six points are distributed pairwise on the three lines supporting the triangle. One may deform now the conic τ (C 0 ) onto C. The points Qi , i = 1, . . . 6 move along their respective lines onto Pi , and Q7 moves in the triangle onto P7 . Proposition 3.4 For each of the 14 configurations (E, 6), . . . V , the unordered configurations realized by the adjacent configurations are as shown in Tables 3.23.5, along with the corresponding unordered linewalls (for the Lsetting) and conicwalls. Proof: For each of the 14 configurations (E, 6), . . . V , we proceed as follows. Using the rule from Figures 2.182.21 we get the set of adjacent triples for each subconfiguration ˆ1, . . . ˆ7. There are thus seven sets of triples. The
46
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
adjacent triples lnm for the configuration are those appearing four times in total (in all sets except for ˆl, n ˆ , m). ˆ Draw the code of the configuration obtained after the crossing. Each crossing induces the change of four subcodes. To determine the unordered linewall among W 1, . . . W 27, we need to find the cyclic ordering of nine points on the line L. Write for each of the four subcodes the cyclic ordering of the six relevant points on L, using the rule of Figures 2.182.21. We get thus four circles marked with six points each. The partial cyclic orderings given by these four circles allow us to recover the cyclic ordering of the nine points without ambiguity, except for pairs of adjacent red points. For example, start from (D, 6) and take the crossing 456, the partial cyclic orderings are: 4, 5, 6, 27, 37, 23 (change of ˆ1), 4, 5, 17, 6, 37, 13 (ˆ 2), 4, 5, 17, 6, 27, 12 (ˆ 3), 4, 5, 6, 12, 13, 23 (ˆ7). The resulting cyclic ordering is 4, 5, 17, 6, 27,{37, 12}, 13, 23, the wall is W 18. The information obtained is gathered in Tables 3.23.4. By Proposition 3.1, there are exactly eleven unordered conicwalls, and they have representatives A, . . . K. Each ordered configuration (X, 6) is adjacent to (X, 60 ) via the conicwall X. The equivalences between ordered configuration are: (X, 6) ' (X, 60 ) for X ∈ {A, B, I, J, K}; (D, 6) ' (D, 60 ) ' (G, 60 ); (E, 6) ' (E, 60 ) ' (F, 6) ' (F, 60 ); and (C, 6) ' (H, 60 ). We get thus the information in Table 3.5, and Proposition 3.4 is proved. It follows from this proposition that there exist no other unordered configurations than the 14, this finishes the proof of Theorem 1.2. Proposition 3.4 with Tables 3.23.4 implies immediately: Proposition 3.5 In the linear setting, the adjacency graph of the stratified space (RP 2 )7 /S7 is as shown in Figure 3.14. This graph was first given in [27], [28], using dual arrangements of lines. For each arrangement, S. Finashin obtained the adjacent linewalls as orbits of triangles under the action of its group of symmetries. In our setting, the arrangements of lines and triangles are replaced by configurations of points and triples. We choose representatives of the eleven unordered Lconfigurations, for example, by removing C 0 (6), (G, 6) and (H, 6) from our canonical set of fourteen Qconfigurations. Let us encode each of the eleven representatives now as an Lconfiguration, removing the colors black and white from the dots on the circles representing the βsubcodes. We determine the set of adjacent triples for each subconfiguration ˆ1, . . . ˆ7 (each βsubcode has now six adjacent triples instead of three). There are thus seven sets of triples. The adjacent triples lnm for the configuration are those appearing four times in total (in all sets except for ˆl, n ˆ , m). ˆ Consider an Lconfiguration, with monodromy group G0 , see Table 3.1. The adjacent (unordered) linewalls correspond to the orbits of the adjacent triples under the action of G0 . For example (A, 6) has two adjacent linewalls: {367, 257, 147} (W 23) and {234, 456, 126, 345, 123, 156} (W 14). There is one exception: the configuration (0, 4, 3, 0), represented by R, has monodromy group {id}, but the adjacent wall W 26 is represented by two triples, 125 and 357. As a matter of fact, W 26 is a twosided inner (ac
Configurations of seven points
47
(E, 6)
(7, 0, 0, 0)
167 W 12 456 W 12
(3, 4, 0, 0)1 (3, 4, 0, 0)2
(D, 6)
(3, 4, 0, 0)1
267 W 4 157 W 4 456 W 18 167 W 12
(2, 2, 3, 0)1 (2, 2, 3, 0)2 (1, 4, 2, 0) (7, 0, 0, 0)
(C, 6)
(2, 2, 3, 0)1
267 W 4 (3, 4, 0, 0)1 157 W 8 (1, 2, 2, 2) 367 W 16 (2, 2, 3, 0)1 126 W 27 (1, 6, 0, 0) 456 W 19 (1, 2, 4, 0)
(B, 6)
(1, 2, 2, 2)
157 267 257 367 234 456 147 126
W8 W8 W 23 W 21 W 13 W 20 W 21 W 25
(2, 2, 3, 0)1 (2, 2, 3, 0)2 (1, 0, 6, 0) (1, 2, 2, 2) (1, 2, 2, 2) (0, 3, 3, 1) (1, 2, 2, 2) (0, 4, 3, 0)
(A, 6)
(1, 0, 6, 0)
257 367 234 456 147 126
W 23 W 23 W 14 W 14 W 23 W 14
(1, 2, 2, 2) (1, 2, 2, 2) (0, 4, 3, 0) (0, 4, 3, 0) (1, 2, 2, 2) (0, 4, 3, 0)
C 0 (6)
(2, 2, 3, 0)2
157 W 4 (3, 4, 0, 0)1 267 W 8 (1, 2, 2, 2) 234 W 11 (2, 2, 3, 0)2 456 W 17 (0, 6, 1, 0) 147 W 16 (2, 2, 3, 0)2
TABLE 3.2 Adjacencies via linewalls
48
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
(G, 6)
(3, 4, 0, 0)2
234 567 467 127
(H, 6)
(2, 2, 3, 0)3
467 W 4 (3, 4, 0, 0)2 457 W 27 (1, 6, 0, 0) 234 W 6 (0, 4, 3, 0) 367 W 16 (2, 2, 3, 0)3 127 W 19 (1, 2, 4, 0)
(K, 6)
(1, 6, 0, 0)
456 W 1 234 W 5 367 W 22 457 W 27 127 W 27
(I, 6)
(1, 2, 4, 0)
457 W 19 (2, 2, 3, 0)1 234 W 7 (0, 3, 3, 1) 467 W 9 (1, 4, 2, 0) 367 W 22 (1, 6, 0, 0) 137 W 9 (1, 4, 2, 0) 126 W 2 (1, 4, 2, 0) 127 W 19 (2, 2, 3, 0)3
(J, 6)
(1, 4, 2, 0)
467 W 9 (1, 2, 4, 0) 234 W 2 (1, 2, 4, 0) 567 W 18 (3, 4, 0, 0)1 137 W 9 (1, 2, 4, 0) 126 W 3 (0, 6, 1, 0) 127 W 18 (3, 4, 0, 0)2
TABLE 3.3 Adjacencies via linewalls, continued
W1 W 12 W4 W 18
(1, 6, 0, 0) (7, 0, 0, 0) (2, 2, 3, 0)3 (1, 4, 2, 0)
(3, 4, 0, 0)2 (0, 4, 3, 0) (1, 2, 4, 0) (2, 2, 3, 0)3 (2, 2, 3, 0)1
Configurations of seven points
49
R
(0, 4, 3, 0)
236 457 247 357 167 156 134 125
W 25 (1, 2, 2, 2) W 6 (2, 2, 3, 0)3 W 14 (1, 0, 6, 0) W 26 (0, 4, 3, 0) W 15 (0, 4, 3, 0) W 5 (1, 6, 0, 0) W 24 (0, 3, 3, 1) W 26 (0, 4, 3, 0)
T
(0, 3, 3, 1)
247 356 236 357 467 145 137 167 125 134
W 20 W7 W 20 W 24 W7 W7 W 10 W 24 W 20 W 24
V
(0, 6, 1, 0)
247 356 236 467 145 137 125
W 17 (2, 2, 3, 0)2 W 3 (1, 4, 2, 0) W 17 (2, 2, 3, 0)2 W 3 (1, 4, 2, 0) W 3 (1, 4, 2, 0) W 10 (0, 3, 3, 1) W 17 (2, 2, 3, 0)2
(1, 2, 2, 2) (1, 2, 4, 0) (1, 2, 2, 2) (0, 4, 3, 0) (1, 2, 4, 0) (1, 2, 4, 0) (0, 6, 1, 0) (0, 4, 3, 0) (1, 2, 2, 2) (0, 4, 3, 0)
TABLE 3.4 Adjacencies via linewalls, end (E, 6)
(7, 0, 0, 0)
(D, 6)
(3, 4, 0, 0)1
(C, 6)
(2, 2, 3, 0)1
(B, 6) (1, 2, 2, 2) (A, 6) (1, 0, 6, 0) (C 0 , 6) (2, 2, 3, 0)2 (G, 6) (3, 4, 0, 0)2 (H, 6) (2, 2, 3, 0)3 (K, 6) (1, 6, 0, 0) (I, 6) (1, 2, 4, 0) (J, 6) (1, 4, 2, 0) TABLE 3.5 Adjacencies via conicwalls
E F D G C H B A C G H K I J
(7, 0, 0, 0) (7, 0, 0, 0) (3, 4, 0, 0)1 (3, 4, 0, 0)2 (2, 2, 3, 0)2 (2, 2, 3, 0)3 (1, 2, 2, 2) (1, 0, 6, 0) (2, 2, 3, 0)1 (3, 4, 0, 0)1 (2, 2, 3, 0)1 (1, 6, 0, 0) (1, 2, 4, 0) (1, 4, 2, 0)
50
Pencils of Cubics and Algebraic Curves in the Real Projective Plane W10
(0, 3, 3, 1) W24
(1, 0, 6, 0) W7
W21
W20
W14
(0, 4, 3, 0) W15
W26
W23 W25
W13
(1, 2, 2, 2)
W17
W11 W8
W6
(2, 2, 3, 0) W27
W5 W19
(0, 6, 1, 0) W16
W4
(3, 4, 0, 0)
W1
(7, 0, 0, 0) W3
(1, 6, 0, 0)
W22
(1, 2, 4, 0)
W12
W18 W2 W9
(1, 4, 2, 0)
FIGURE 3.14 Adjacency graph of (RP 2 )7 /S7 in the linear setting tual) wall. The other five inner walls W 15, W 13, W 21, W 11 and W 16 are onesided, see [27].
3.3
Refined linewalls
Let us now determine the adjacency graph of the stratified space (RP 2 )7 /S7 in the quadratic setting. Proposition 3.6 In the quadratic setting, the stratified space (RP 2 )7 /S7 has exactly 38 refined linewalls. Proof: To describe the refined linewalls, we need first to determine, for each of the 27 linewalls, all sets of six points that may be coconic. For each linewall with some admissible conic, we mark the black points with names 5, 6, 7, and indicate the conic(s) inside of the circle, see Figure 3.13. Let us explain how to spot these conics. The eleven unordered conicwalls are represented by A, . . . K, with conic 123456. Table 3.6 displays all triples of points with 7 that may become aligned for A, . . . K. Most of these triples are directly visible in Figure 3.1, but the safest way to find them without forgetting one is the following: write the set of all admissible triples for each subconfiguration ˆ 1, . . . ˆ 6 using the rules shown in Figures 2.182.21; the relevant triples nm7 are those appearing four times in total (in all sets but n ˆ , m). ˆ Each of the lineconic configurations obtained may be denoted by a letter followed by a triple, and there are in total 36 combinations lettertriple. Note that a lineconic configuration may be encoded with two lettertriples, for example, B157 = C157, see Figure 3.1. Note also that some are mapped onto others by elements of S7 : The cyclic permutation (165432) sends A367 onto A257 and A257 onto A147; the symmetry s = (16)(25)(34) maps Xij7 onto Xs(i)s(j)7 for X ∈ {A, B, D, E, F, J}. The symmetry (14)(25)(36) swaps the elements of
Configurations of seven points A B C D E F G H I J K
367, 367, 367, 157, 167 127, 467, 367, 367, 467, 457,
51 257, 267, 267, 267,
147 257, 147, 157 157 167
567 567, 457, 457, 567, 367,
127 467, 127 467, 137, 127 137, 127 127
TABLE 3.6 Admissible triples for the 11 conicwalls the pairs (I457, H127), (K127, H457) and (I137, I467). One finds in total 14 different lineconic unordered configurations; for each of them, choose a lettertriple as representative. See then Figure 3.1 to get the cyclic ordering with which the line meets the three aligned points, and the six lines determined by the other four points. The left column of Table 3.7 displays the symmetries mapping the 14 lettertriples onto the lineconic subwalls of the linewalls in Figure 3.13. The right column displays the lists of equivalent lettertriples. Looking at Figure 3.13, we see that there are actually 15 conics in total, distributed in 11 linewalls. But W 12, 125643 and W 12, 126743 are equivalent as they are swapped by the symmetry (13)(24)(75). (Note also that this symmetry maps W 12, 125743 onto itself.) Eight of the linewalls are marked each with one single conic. Each of these linewalls has two refined linewalls, obtained moving one of the six points away from the conic either to the inside or to the outside, so as to realize a βcode. We shall encode it as follows. Say 123456 lie in convex position and each of the points 1, 3, 5 is exterior to the conic through the other five, we write briefly (123456, 135). The symmetry (13)(24)(56) maps W 22, 154362 onto itself, and swaps W 22, (154362, 235) with W 22, (154362, 146). This symmetry maps also W 23, 164352 onto itself, and swaps W 23, (164352, 236) with W 23, (164352, 145). The three linewalls W 4, W 12 and W 16, that are each marked with several conics, are represented in Figure 3.15. Let n be a point and C2 be a conic; we write n < C2 (n > C2 ) if n lies inside (outside) of C2 . For W 4, one has: 6, 7 < 14253 or 6 > 14253 and 7 < 14253, or 6, 7 > 14253. For W 16, one has: 6, 7 < 14352 or 6 > 14352 and 7 < 14352, or 6, 7 > 14352. Each of the two linewalls W 4 and W 16 has three refined linewalls, see the encoding with βcodes in Table 3.8. The wall W 12 gives rise to four cases: 6, 7 < 12543 and 6 < 12743; 6, 7 < 12543 and 6 > 12743; 6 < 12543 and 7 > 12543; 6, 7 > 12543. The triples of βcodes ˆ5, ˆ 6 and ˆ 7 for these four cases are:
52
Pencils of Cubics and Algebraic Curves in the Real Projective Plane C267 G467 B157 I467 D167
(1325) (2453)(67) (16342) (245) (176524) (1543)(67) F 127 (16275)(34) (164)(253) C367 (354) (14235) H367 (354)(67) (14235)(67) G127 (16)(25) H127 (16)(25) B367 (23654) (14365) I367 (13524) (3654) A367 (136524) (14235) H457 (12456)
W 4, 142536 W 4, 142537 W 8, 142536 W 9, 143526 W 12, 125743
D267, D157 H467 B267, C157 J467, J137, I137 E167
W 12, 126743 F 567, G567 W 12, 125643 W 16, 164352 W 16, 174352 W 18, 124356 J127, J567 W 19, 124356 I127, I457 W 21, 154263 B147 W 22, 154362 K367 W 23, 164352 A257, A147, B257 W 27, 124356 K127, K457
TABLE 3.7 The 14 lineconicwalls ˆ 5 = 126743, 273, ˆ 6 = 125743, 154, ˆ7 = 125643, 154, (125743, 126743) ˆ 5 = 126743, 164, ˆ6 = 125743, 154, ˆ7 = 125643, 154, (126743) ˆ 5 = 126743, 273, ˆ 6 = 125743, 273, ˆ7 = 125643, 154, (125643, 125743), ˆ 5 = 126743, 273, ˆ6 = 125743, 273, ˆ7 = 125643, 263, (125643) (We have indicated the adjacent conicwalls for each of them.) The symmetry (13)(24)(57) swaps the first with the third, and the second with the fourth. So W 12 gives rise to only two refined linewalls. To characterize them, let us consider the middle point in the group of three aligned points (6 in Figure 3.13). Either 6 is interior for only one of the codes ˆ5, ˆ7, or 6 is interior for both codes ˆ 5 and ˆ 7. The 11 linewalls with conic(s) give rise together to exactly 22 refined linewalls, let us call them the strict refined linewalls. A representative for each of them is obtained adding some βcodes after the name W N of the wall. For N = 12, we need three βcodes, for N = 4, 16 we need two, for the other values of N , we need only one, see Table 3.8. We have yet to find out, for each refined linewall, the adjacent pair of unordered configurations. Appropriate symmetries mapping our 22 representatives onto walls between configurations of types (X, 6) or (X, 60 ) allow us to achieve this, see Tables 3.8, 3.9. The total number of refined linewalls is 22+(27−11) = 38. Combining Proposition 3.4 with Table 3.5 and Proposition 3.6 with Table 3.9, one gets immediately:
Configurations of seven points
53
1 4
2
3 6
5
2
24
14
1
7
23 34
13
1 5
4
2
3 2
16 7 14 4
24
13
23
3
1
2 23 2
1 5 67
4
3 14 4
24
13 3
FIGURE 3.15 Linewalls W 4, W 16 and W 12 Proposition 3.7 The adjacency graph for the stratified space (RP 2 )7 /S7 in the quadratic setting is as shown in Figure 3.16. Here again, the walls adjacent to a configuration correspond to the orbits of the adjacent triples and eventual adjacent cyclically ordered sextuples under the action of its monodromy group G. There is one single exception: the twosided inner linewall W 26, see end of Section 3.2. All of the other inner walls are onesided.
54
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
W 41 W 42 W 43 W 81 W 82 W 91 W 92 W 121 W 122 W 161 W 162 W 163 W 181 W 182 W 191 W 192 W 211 W 212 W 22 W 23 W 271 W 272
(142536, 456) (1523) (C, 6) (142537, 123) (142536, 123) (1523) (C, 60 ) (142537, 123) (142536, 456) (2354)(67) (H, 6) (142537, 457) (142536, 123) (12436) (B, 6) (142536, 456) (12436) (B, 60 ) (143526, 456) (254) (I, 6) (143526, 132) (254) (I, 60 ) (126743, 273) (1345)(67) (E, 6) (125743, 154) (125643, 154) (126743, 164) (1324) (F, 6) (125743, 154) (125643, 154) (164352, 236) (345) (C, 6) (174352, 145) (164352, 145) (345) (C, 60 ) (174352, 145) (164352, 236) (67)(345) (H, 6) (174352, 237) (124356, 236) (16)(25) (G, 60 ) (124356, 145) (16)(25) (G, 6) (124356, 236) (16)(25) (H, 60 ) (124356, 145) (16)(25) (H, 6) (154263, 235) (24563) (B, 6) (154263, 146) (24563) (B, 60 ) (154362, 146) (14253) (I, 6) (164352, 236) (15324) (A, 6) (124356, 236) (16542) (H, 60 ) (124356, 145) (16542) (H, 6)
TABLE 3.8 Adjacencies via the 22 strict refined linewalls
267
(D, 6)
267
(D, 60 )
467
(G, 6)
157 157 467 467 167
(C, 6) (C, 60 ) (J, 6) (J, 60 ) (D, 6)
567
(G, 6)
367
(15)(24)(C, 6)
367
(15)(24)(C, 60 )
367
(15)(24)(H, 6)
127 (J, 60 ) 127 (J, 6) 127 (I, 60 ) 127 (I, 6) 367 (15)(24)(B, 6) 367 (15)(24)(B, 60 ) 367 (K, 6) 367 (B, 6) 457 (K, 60 ) 457 (K, 6)
Configurations of seven points
55
(3, 4, 0, 0)1
W 41 W 42 W 43 W 81 W 82 W 91 W 92 W 121 W 122 W 161 W 162 W 163 W 181 W 182 W 191 W 192 W 211 W 212 W 22 W 23 W 271 W 272
(3, 4, 0, 0)2 (1, 2, 2, 2) (1, 2, 4, 0) (7, 0, 0, 0) (2, 2, 3, 0)1 (2, 2, 3, 0)2 (2, 2, 3, 0)3 (1, 4, 2, 0) (1, 2, 4, 0) (1, 2, 2, 2) (1, 6, 0, 0) (1, 0, 6, 0) (1, 6, 0, 0)
(2, 2, 3, 0)1 (2, 2, 3, 0)2 (2, 2, 3, 0)3 (2, 2, 3, 0)1 (2, 2, 3, 0)2 (1, 4, 2, 0) (1, 4, 2, 0) (3, 4, 0, 0)1 (3, 4, 0, 0)2 (2, 2, 3, 0)1 (2, 2, 3, 0)2 (2, 2, 3, 0)3 (3, 4, 0, 0)1 (3, 4, 0, 0)2 (2, 2, 3, 0)1 (2, 2, 3, 0)3 (1, 2, 2, 2) (1, 2, 2, 2) (1, 2, 4, 0) (1, 2, 2, 2) (2, 2, 3, 0)1 (2, 2, 3, 0)3
TABLE 3.9 Adjacencies between unordered configurations
W10
(0, 3, 3, 1)
(0, 6, 1, 0) W161
W20 W24
W212 A
(1, 0, 6, 0) W7
W23
W25
W14
(0, 4, 3, 0)
(1, 2, 2, 2) W8 B
2
(2, 2, 3, 0) 3 W272
W26
(1, 2, 4, 0)
W163
H
W6
W191
(3, 4, 0, 0) 1
W42
(3, 4, 0, 0) 2
W43
W271
W22
W2
W92
W91
(7, 0, 0, 0)
FIGURE 3.16 Adjacency graph of (RP 2 )7 /S7 in the quadratic setting
W12 2
F W181
W1
K
E
W121
G
(1, 6, 0, 0)
W5 W19 2
D
W41
C W11 W162
(2, 2, 3, 0) 2
W15
I
(2, 2, 3, 0) 1
W81
W211 W13
W17
W182 J
(1, 4, 2, 0)
W3
Part II
Pencils of cubics with eight base points lying in convex position in RP 2
4 Pencils of cubics
CONTENTS 4.1 4.2
Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Singular pencils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1
Preliminaries
59 60
Given nine generic points in CP 2 , there exists one single cubic passing through them. Given eight generic points in CP 2 , there exists a oneparameter family of cubics passing through them. We will call such a family a pencil of cubics. Let F0 and F1 be two cubics of a pencil P. They intersect at a ninth point. As the other cubics of P are linear combinations of F0 and F1 , they all pass through this ninth point. We call these nine points the base points of P. If eight of the base points are real, the pencil is real, and hence the ninth base point is also real. A pencil of cubics is a line in the space CP 9 of complex cubics. Let ∆ be the discriminantal hypersurface of CP 9 , formed by the singular cubics; this hypersurface is of degree 12. Hence, a generic pencil of cubics intersects ∆ transversally at 12 regular points. Otherwise stated, a generic pencil P has exactly 12 singular (nodal) cubics. A nongeneric pencil will be called singular pencil . Let P be a real pencil with nine real base points and denote the real part of P by RP. Recall that a circle embedded in RP 2 is called oval (resp. pseudoline or odd component) if it realizes the class 0 (resp. 1) of H1 (RP 2 ). The real part of a generic real cubic consists either of one pseudoline, or of one pseudoline plus one oval. Let n ≤ 12 be the number of real singular cubics of P. Let C3 be one of these cubics. The node (double point) P of C3 is isolated if the tangents to C3 at P are nonreal, otherwise P is nonisolated . If P is nonisolated, C3 \ P = J ∪ O, where [J ∪ P ] 6= 0 and [O ∪ P ] = 0 in H1 (RP 2 ). We say that O is the loop and J is the odd component of C3 . Notice that the loop O is convex. The estimation of n presented hereafter is due to V. Kharlamov, see [11]. One has: n = n1 +n2 +n3 , where n1 is the number of cubics with an isolated node, n2 is the number of cubics with a loop containing no base points, and n3 is the number of cubics with a loop containing some base points. To evaluate n, one recalculates the Euler characteristic of RP 2 , fibering RP 2 with the 59
60
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
cubics of RP. Each isolated node, and each base point contributes by +1; and each nonisolated node contributes by −1, so that one gets: 1 = χ(RP 2 ) = 9 + n1 − (n − n1 ). So, n − 2n1 = 8. Thus, n = 8, 10 or 12, correspondingly, n1 = 0, 1 or 2. Consider a motion in RP starting from a cubic with an isolated node. If we choose the direction of the motion properly, an oval appears, grows, and attaches itself to the pseudoline, forming a loop that contains no base point. Conversely, starting from a cubic with a loop containing no base point, one can move in the pencil so that there appears a cubic with an oval. As this oval lies inside of the loop, it shrinks when one moves farther, and degenerates into an isolated node. Thus, n2 = n1 and n3 = 8 independently of n. Let us call the eight cubics of the third type the distinguished cubics of P. We shall picture RP by a circle, divided in eight portions by the eight distinguished cubics. Let us remark that this number n3 = 8 is the Welschinger invariant W3 , see [85]. The number of real rational plane curves of degree d going through 3d − 1 generic points of RP 2 is always finite. Let c1 be the number of such curves with an even number of isolated nodes, and c2 be the number of such curves with an odd number of isolated nodes. Welschinger proved that the difference Wd = c1 − c2 does not depend on the choice of the 3d − 1 points. Let 1, . . . 8 be eight generic points in RP 2 , determining a pencil of cubics P. Let 9 be the ninth base point, and C3 be a real cubic of P. We call combinatorial cubic C3 the topological type of (C3 , 1, . . . , 9), and combinatorial pencil the cyclic sequence of the successive eight combinatorial distinguished cubics. From now on, the same notation P = P(1, . . . 8) will refer to the (real part of the) pencil of cubics determined by 1, . . . 8, and to the combinatorial pencil. It will always be clear from the context whether we speak of a combinatorial or of an actual pencil. Theorem 4.1 Up to the action of D8 , eight generic points 1, . . . 8 lying in convex position in RP 2 may realize exactly 43 combinatorial pencils P(1, . . . 8).
4.2
Singular pencils
Let P be an (actual) pencil of cubics with only real base points. Move these points till P degenerates into a singular pencil Psing . The degeneration is generic if and only if Psing intersects ∆ transversally at 10 regular points and 1. Psing is tangent to ∆ at one regular point, or 2. Psing crosses transversally a stratum of codimension 1 of ∆.
Pencils of cubics
61
In both cases, two singular cubics C31 and C32 of P come together to yield one singular cubic C30 of Psing . The cubic C30 is necessarily real, the cubics C31 and C32 are either both real or complex conjugated. Let F be a generic cubic of ∆ with a node at some point p, then TF ∆ = {F + GG(p) = 0}. Therefore, Psing satisfies the condition 1) if and only if Psing has a double base point, at p. Otherwise stated, Psing is obtained from P by letting two base points A and B of P come together. Move A towards B along the line (AB). For simplicity, we assume that B is the origin (0, 0) of the plane, and the direction is the xaxis. The condition that some cubic H passes through A and B becomes at the limit: H(0, 0) = 0 and ∂H ∂x (0, 0) = 0. All of the cubics of Psing but one are tangent to the xaxis. A unique cubic C30 is singular at A = B (for this cubic, the other partial derivative ∂H ∂y (0, 0) is also equal to 0). In case 2), the cubic C30 must be reducible (product of a line and a conic), or have a cusp. If C30 is reducible, the genericity imposes that three of the base points lie on the line, and the other six lie on the conic. If C30 has a cusp, the cubics C31 and C32 are either nondistinguished real cubics, one with a loop, the other with an isolated double point, or complex conjugated. See Figure 4.1, where the various possible types for the degeneration P → Psing have been denoted by 1a, 1b, 2a, 2b, 2c, 2d. In the righthand part of the figure, the dotted crosses symbolize the nonreal nodes of the cubics C31 and C32 . Moving further, one gets a new pencil P 0 . The cubic C30 is replaced by a new pair of nodal cubics C33 , C34 . If P → Psing realizes one of the types 1a, 1b, 2a or 2b, P 0 → Psing is of the same type, whereas the types 2c and 2d are swapped. Let us describe more precisely the case 1). We call elementary arc AB an arc connecting A to B and containing no other base point. As P can degenerate into Psing letting A and B come together, some cubics of P must have an elementary arc AB. Start from such a cubic and move in any direction in RP. At some moment, the mobile arc AB must glue to another arc, and then disappear. Thus, P has two distinguished cubics corresponding to the openings of the arc AB. We call singular elementary arc AB the nonsmooth arc AB of either of these cubics. For 1a, these two cubics are C31 and C32 ; they come together to yield the cubic C30 , which has a nonisolated double point at A = B. All three combinatorial cubics C30 , C31 and C32 are identical outside of a neighborhood of A ∪ B. Note that the pair of cubics C31 , C32 divide the pencil RP in two portions; in one portion, the cubics have an elementary arc AB, in the other they have none. The latter portion disappears in the motion P → Psing . For 1b, the two cubics corresponding to the opening of AB divide the pencil in two portions; the cubics in each portion have respectively one and two elementary arc(s) AB. The cubics of the latter portion all have an oval passing through A, B, and no other base point. When the motion reaches Psing , a nodal cubic C30 appears inside of this portion. Following this portion, one sees the oval shrink till it becomes an isolated node. Then the oval reappears on the other side and grows. Let us finally mention that eight points in RP 2 determine a pencil of cubics, with a ninth base point, if and only if no four of the eight points
62
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
are aligned, and no seven are coconic. If three of the points are aligned, the ninth base point lies on the conic determined by the five others; if six points are coconic, the ninth base point is on the line determined by the other two: the pencil is singular. If four points are aligned, one gets a pencil of reducible cubics, the product of the line and the pencil of conics determined by the other four points. If seven points are coconic, one gets again a pencil of reducible cubics, the product of the conic and the pencil of lines through the remaining point. If two points coincide, say A and B, one gets a oneparameter family of singular pencils: consider the cubic C30 through 1, . . . 8, with node at A = B. The ninth point 9 may be chosen anywhere on this cubic. Finally, six generic points 1, . . . 6 in RP 2 determine six rational pencils, with respective nodes at 1, . . . 6. Each pencil has exactly five reducible cubics. Let us call combinatorial cubic C3 a topological type (C3 , 1, . . . 6) and combinatorial pencil the cyclic sequence of five combinatorial reducible cubics. Up to the action of S6 on 1, . . . 6, there are exactly four lists of six combinatorial pencils, see Part 1.
Pencils of cubics
63
1a
1
A
A=B
B
C3
C30
1b 1
C3 A=B
Psing
P
A
B
2a
C30
2 C3
Psing
2
C3
P
2b
C30
1
C3
C30
2
Psing
C3
1
C3
2
P
2c
Psing
C3
P
2d
C30
1
C3
C30
2
Psing
C3
1
C3
2
P
FIGURE 4.1 The degeneration of P into Psing
Psing
C3
P
5 Lists
CONTENTS 5.1 5.2 5.3 5.4 5.5 5.6
Points in convex position and conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . Admissible lists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Extremal lists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Distances between points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Isotopies of octuples of points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Elementary changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1
Points in convex position and conics
65 68 71 75 76 78
We say that n points lie in convex position in RP 2 if there exists a line L ⊂ RP 2 such that the n points lie in convex position in the affine plane RP 2 \ L. Note that any set of n ≤ 5 points in RP 2 lies in convex position. Consider n ≥ 5 generic points lying in strictly convex position in RP 2 , say 1, . . . , n. Their ordered combinatorial Qconfiguration will be called list L(1, . . . , n) for short. So a generic list, written in full extent, consists of (n − 5) × n5 elementary data of the type k < C2 (k lies inside of the conic C2 ) or k > C2 (k lies outside of C2 ). In Chapter 7, we will consider also nongeneric configuration of points 1, . . . , 8 with six of them on a conic, and the corresponding nongeneric lists. How many different lists L(1, . . . , n) may be realized by n generic points 1, . . . , n lying in convex position in the plane? For n = 6, note that the condition 6 > 12345 is equivalent to 5 < 12346. One single elementary data determines L(1, . . . 6). There are only two lists: 6 < 12345 and 6 > 12345. We write briefly: \L(1, . . . , 6) = 2. The two lists l1 , l2 , written in full extent are displayed in Table 5.1. Let now n = 7. Let C3 be a cubic passing through the points 1, . . . , 7. We will call (combinatorial) cubic C3 the topological type of (C3 , 1, . . . , 7). (Whenever the context allows no ambiguity, we shall omit the word combinatorial .) Let F be the equation of C3 and p be a point among 1, . . . , 7. The condition F (p) = 0 is a linear equation in the coefficients of F . If p is a singular point of C3 , one gets two supplementary linear equations: ∂F ∂x (p) = 0 65
66
Pencils of Cubics and Algebraic Curves in the Real Projective Plane L(1, . . . 6) C2 in out in out 12345 6 6 12346 5 5 12356 4 4 12456 3 3 13456 2 2 23456 1 1
TABLE 5.1 The two lists L(1, . . . 6) and ∂F ∂y (p) = 0. Thus, as 1, . . . , 7 are generic in the affine plane, there exists exactly one real nodal cubic passing through 1, . . . , 7 with the double point at p. Let S(1, . . . , 7) be the set of seven combinatorial nodal cubics passing through the points 1, . . . , 7, one of them being the double point. Denote by C3 (k) the cubic with node at k. We may ask the same question as above, replacing L(1, . . . 7) by S(1, . . . 7). Proposition 5.1 \L(1, . . . , 7) = \S(1, . . . , 7) = 14. The fourteen sets S(1, . . . 7) are denoted by 1±, 2±, . . . , 7±, and they are all equivalent up to the action of the dihedral group D7 . The combinatorial data S(1, . . . 7) and L(1, . . . 7) are equivalent, and determined by only two elementary data: S(1, . . . 7) = 1+ if and only if 7 < 23456 and 1 > 23456. So, we may speak of the list S(1, . . . 7). The lists 1+ and 1− are shown in Figure 5.1; the other lists n± are obtained from 1± performing on 1, . . . , 7 the cyclic permutation that replaces 1 by n. (The double point of the last cubic in each list may be an isolated node or a crossing. If it is a crossing, the loop is attached to arc 61 of the odd branch, as shown in the figure, or to arc 12.) Note that any of the five nonextremal cubics of the list S(1, . . . , 7) determines the whole of this list. Let us denote by C3 (k) the cubic with node at k. Proof: The first six points 1, . . . 6 may realize two different lists L(1, . . . 6). For either of them, there are seven possible positions of the point 7 with respect to the set of conics passing through five points among 1, . . . , 6. One checks easily that the data L(1, . . . , 6), position of 7 determines the list L(1, . . . , 7). Thus, \L(1, . . . , 7) = 14. Let 1, . . . 7 be such that 7 < 23456 and 1 > 23456, see Figure 5.2. One may move 1, preserving the convex position, towards 23456 till 1 reaches this conic. The cubic C3 (k) with k = 2, 3, 4, 5 or 6 is obtained as a perturbation of the reducible cubic k7 ∪ 23456 letting 1 move back to the outside of 23456 (in Figure 5.2, we show the case k = 3). The condition 1 moves to the outside of 23456 is equivalent to 6 moves to the inside of 12345. The cubic C3 (1) is obtained by perturbing the reducible cubic 17 ∪ 12345. Assume that the cubic
Lists
67
1+
1
2
7 6
1−
3
7
3 4
5
7
2
2 3
5
4
4
2
7 6
1
6
3
6 5
7
1
2
6
4
5
1
1
6 5
5
1 3 4
4
3
5
6
1
7 2
3
7
. . .
1
7
2
6
. . .
4
2 5
4
3
FIGURE 5.1 The lists 1+ and 1− C3 (7) has a loop, passing through some other of the seven points. This cubic cuts then C3 (6) at more than nine points, contradicting Bezout’s theorem. Applying now Bezout’s theorem with lines, we see that if C3 (7) has a loop, it is attached to one of the arcs 61 or 12 of the odd component. We have constructed 1+, and cyclic permutations and symmetries allow us to obtain thirteen other sets. Until now, we have proved that: \L(1, . . . 7) = 14, two elementary data of L(1, . . . 7) determine S(1, . . . 7), and \S(1, . . . 7) ≥ 14. Hence, \S(1, . . . 7) = 14, the data L(1, . . . 7) and S(1, . . . 7) are equivalent. The correspondences between the fourteen lists S(1, . . . 7) and L(1, . . . 7) (written in full extent) are displayed in Tables 9.19.2. (For convenience, we have gathered some of the tables in Chapter 9.) With the notations of Part 1, the lists L(1, . . . 7) are of type (7, 0, 0, 0), and one has E(6) = 6− Let A, B be two consecutive points in a configuration 1, . . . , 7 realizing a generic list, and let C2 be the conic through the other five points. By definition, the distance A → B is 0 if A, B lie both inside or both outside of C2 ; +1 if A < C2 and B > C2 ; −1 if A > C2 and B < C2 . For the list 1+, one has thus 1 → 2 = · · · = 6 → 7 = 0, and 7 → 1 = +1. The first nontrivial case is n = 8. It will turn out that the least number of elementary data necessary to determine a list L(1, . . . 8) depends on the list. Let us denote by ˆi the list of seven points obtained by removing i from the set {1, . . . 8}. The list L(1, . . . , 8) may be encoded by the octuple of sublists (ˆ 1, ˆ 2, . . . , ˆ 8) with ˆi = k±, k ∈ {1, . . . , ˆi, . . . 8}. Consider the dihedral group D8 of symmetries of the octagon, generated by the cyclic permutation a = +1 and the symmetry with respect to the axis σ = 15, see Table 5.2. D8 = {a, σa8 = id, σ 2 = id, aσ = σa−1 } Theorem 5.1 Up to the action of D8 , eight generic points 1, . . . 8 lying in
68
Pencils of Cubics and Algebraic Curves in the Real Projective Plane 1
1
7
6
2 3
5
3 5
4
4
1 7
7
1
2
6 5
2
7
6
4
3
6
2 5
4
3
FIGURE 5.2 Construction of the list 1+ convex position may realize exactly 47 different lists L(1, . . . 8). The total number of generic lists is 752 = 47 × 16. This theorem will be proved in two steps, in Sections 5.2 (restriction part) and 5.6 (construction part).
5.2
Admissible lists
Let 1, . . . 8 lie in convex position and {A, . . . G} = {1, . . . 7}. Move 8, leaving the other points fixed and preserving the convex position. Consider an event ˆ The of the form 8 crosses a conic ABCDE, which induces a change of G. remaining point F may lie inside or outside of the conic ABCDE, depending on the list L(1, . . . , 7) (see upper and lower parts of Figure 5.3). So there are: 21 choices for the conic ABCDE, two choices for G, and two possible positions of the last point F with respect to ABCDE. Hence, in total, 84 possibilities. Figure 5.3 gathers all of these possibilities, showing how the cubic ˆ with double point at C changes when 8 crosses ABCDE. This cubic of G ˆ The point 8 (not represented) may be placed in six entirely determines G. different ways on each cubic, according to the cyclic ordering of the set of points {8, A, B, C, D, E, F }. Assume for example that 8 is situated between A ˆ : E− → A+ and B in the cyclic ordering. When 8 enters ABCDE, one has: G ˆ : F − → F + (if F outside of ABCDE). (if F inside of ABCDE), and G
Lists
69 15 1↔1 2↔8 3↔7 4↔6 5↔5
26 1↔3 2↔2 4↔8 5↔7 6↔6
37 1↔5 2↔4 3↔3 6↔8 7↔7
48 1↔7 2↔6 3↔5 4↔4 8↔8
(+1)(15) 1↔2 3↔8 4↔7 5↔6
(+1)(26) 1↔4 2↔3 5↔8 6↔7
(+1)(37) 1↔6 2↔5 3↔4 7↔8
(+1)(48) 1↔8 2↔7 3↔6 4↔5
TABLE 5.2 Action of D8 ˆ For each G ∈ {1, . . . , 7}, we find two possible chains of degenerations of G while letting 8 cross successively all six conics determined by A, . . . F from the ˆ outside to the inside. The chain starting with 8± will be denoted by G±(8). For ˆ each 8 ∈ {1±, . . . 7±} and each G ∈ {1, . . . , 7} one watches which of the two possible chains is realized, see Table 5.3. For each list ˆ8 ∈ {1±, 2±, 3±, 4+}, we draw a diagram whose rows are the chains of degenerations of ˆ1, . . . ˆ7. (We drop the other cases ˆ 8 ∈ {7±, 6±, 5±, 4−}, that can be deduced from the first ones by the action of D8 .) Let C2 and C20 be two adjacent conics in a column; we add a vertical arrow from C2 to C20 if the following holds: 8 outside of C2 implies 8 outside of C20 . See Tables 9.39.9. Chasing in the diagrams, we may find all of the admissible orbits, for the action of D8 , realizable by the lists L(1, . . . 8). The explicit lists L(1, . . . 8) obtained in this procedure are gathered in Tables 5.45.9. We denote these lists by L1 , . . . L95 , according to their appearance order. Note that we cannot completely rule out redundancies: We get sometimes several representatives of the same orbit. So, we choose for each orbit one representative that we write with normal fonts, the equivalent lists are written in bold. The lists written in normal fonts will be called for convenience principal lists, even if their choice is not canonical. Tables 5.45.5 show the 64 admissible lists with ˆ8 = 1+. The first and the last are deduced one from the other by ±1. The 15 lists with ˆ5 = 6+ are mapped onto the 15 lists with ˆ3 = 4+ by +3. The set of six lists with ˆ2 = 1− splits into two subsets that are mapped one onto the other by the symmetry 15. The set of 20 lists with ˆ4 = 3− splits into two subsets that are mapped one onto the other by 26. The set of six lists with ˆ6 = 5− splits into two subsets that are mapped one onto the other by 37. Up to the action of D8 , there are exactly 32 admissible lists with ˆ8 = 1+. Let now ˆ8 = 1−. First thing, we rule out the orbit of ˆ 8 = 1+, that is to say, any list which is mapped by some element of D8 onto a list with ˆ8 = 1+. In other words, we set: ˆ1 6= 2+, 8−,
70
Pencils of Cubics and Algebraic Curves in the Real Projective Plane A F
F B
E
A
F B
E C
D
C
D
A
F
B
E
C
D
A
B
E
D
FIGURE 5.3 ˆ Degenerations of the list G
ˆ 1+ ˆ 1− ˆ 2+ ˆ 2− ˆ 3+ ˆ 3− ˆ 4+ ˆ 4− ˆ 5+ ˆ 5− ˆ 6+ ˆ 6− ˆ 7+ ˆ 7−
ˆ8 1+, 2+, 2−, 4+, 4−, 6+, 6− 1−, 3+, 3−, 5+, 5−, 7+, 7− 1+, 1−, 2−, 4+, 4−, 6+, 6− 2+, 3+, 3−, 5+, 5−, 7+, 7− 1+, 1−, 3+, 4+, 4−, 6+, 6− 2+, 2−, 3−, 5+, 5−, 7+, 7− 1+, 1−, 3+, 3−, 4−, 6+, 6− 2+, 2−, 4+, 5+, 5−, 7+, 7− 1+, 1−, 3+, 3−, 5+, 6+, 6− 2+, 2−, 4+, 4−, 5−, 7+, 7− 1+, 1−, 3+, 3−, 5+, 5−, 6− 2+, 2−, 4+, 4−, 6+, 7+, 7− 1+, 1−, 3+, 3−, 5+, 5−, 7+ 2+, 2−, 4+, 4−, 6+, 6−, 7−
TABLE 5.3 Possible values of ˆ 8 for the chains n ˆ± = n ˆ ± (8), n = 1, . . . , 7
C
Lists
71
ˆ 2 6= 3+, 1−, ˆ 3 6= 4+, 2−, ˆ 4 6= 5+, 3−, ˆ5 6= 6+, 4−, ˆ6 6= 7+, 5− and ˆ7 6= 8+, 6−. Table 5.6 shows the six new admissible lists obtained. They split into two sets that are mapped one onto the other by 15, so we are left with only three new orbits. For ˆ 8 = 2+ we rule out the orbits of ˆ8 = 1±, that is we set: ˆ1 6= 2±, 8±, ˆ 2 6= 3±, 1±, ˆ 3 6= 4±, 2±, ˆ 4 6= 5±, 3±, ˆ5 6= 6±, 4±, ˆ6 6= 7±, 5± and ˆ7 6= 8±, 6±. Table 5.7 shows the four new admissible lists with ˆ8 = 2+. They split into two sets that are mapped one onto the other by 15, we get two new orbits. Let now ˆ 8 = 2−. Once we have ruled out the orbits of ˆ8 = 1±, 2+, we get the 13 new admissible lists shown in Table 5.8. Some of them are deduced from each other by ±2, 15 or 26 so that there are only eight new orbits. Let ˆ8 = 3+; once we have ruled out the orbits of ˆ8 = 1±, 2±, we find no new admissible lists. For ˆ 8 = 3−, after excluding the orbits of ˆ8 = 1±, 2±, 3+, we get the eight new admissible lists shown in Table 5.9. They split into two sets that are mapped one onto the other by ±2, we select one of them. This set splits in two subsets that are mapped one onto the other by 15. We get thus two new orbits. At last, let ˆ 8 = 4+; once we have excluded the orbits of ˆ8 = 1±, 2±, 3±, we find no new admissible lists. The total number of admissible orbits is 47. One checks easily that the 16 lists in each orbit are all distinct. (For example, assume that a principal list with ˆ 8 = 1+ is invariant for (+1). This list should verify: ˆ8 = 1+ and ˆ 1 = 2+. But there is no such list.) The total number of admissible lists is thus 47 × 16 = 752. The nongeneric lists invariant for some elements of D8 will be studied in Section 7.3.
5.3
Extremal lists
Consider a configuration of eight points 1, . . . 8 lying in convex position in the plane. Any piece of information Pˆ = n±, with P ∈ {1, . . . 8} is equivalent to a statement of the form F < C2 and G > C2 , where F, G are two points among 1, . . . , 8, different from P . For example, ˆ8 = 1+ if and only if 7 < 23456 and 1 > 23456. The correspondences for P = 8 are indicated with bold fonts in Tables 9.19.2, and the other cases are easily deduced from these by the action of D8 . We say that a list is maximal (or minimal ) for some F ∈ 1, . . . , 8 if F lies outside (or inside) of all the conics determined by five of the other seven points. Say F = 8, then for each possible ˆ8 = n±, there is one maximal and one minimal list. The maximal list max(ˆ8 = n±) is obtained from the given diagram ˆ 8 = n± taking for all m ˆ with m = 1, . . . 7 the labels above the first column of horizontal arrows. The minimal list min(ˆ8 = n±) is obtained from the given diagram ˆ 8 = n± taking for all m ˆ with m = 1, . . . 7 the labels above the last column of horizontal arrows. These notations are consistent with the action of D8 . The extremal lists are easily realizable as follows: start from a configuration of eight points lying on the same conic C2 . Move first
72
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
8+ 8+ 8+ 8+ 8+ 8+ 8+ 8+ 8+ 8+ 8+ 5− 8+ 5−
8+ 8+ 8+ 8+ 6− 5− 5+
8+ 8+ 8+ 8+ 6+ 5+ 5+
8+ 8+ 8+ 6− 6− 5− 3−
8+ 8+ 8+ 6− 6+ 5+ 3−
8+ 8+ 8+ 6+ 6+ 3− 3−
8+ 8+ 8+ 3− 3− 3− 3−
ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
8+ 8+ 6− 6− 6− 5− 3+
8+ 8+ 6− 6− 6+ 5+ 3+
8+ 8+ 6− 6+ 6+ 3− 3+
8+ 8+ 6− 3− 3− 3− 3+
8+ 8+ 6+ 6+ 6+ 3+ 3+
8+ 8+ 6+ 3− 3− 3+ 3+
8+ 8+ 4− 3− 3+ 3+ 3+
8+ 8+ 4+ 3+ 3+ 3+ 3+
ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
8+ 6− 6− 6− 6− 5− 1−
8+ 6− 6− 6− 6+ 5+ 1−
8+ 6− 6− 6+ 6+ 3− 1−
8+ 6− 6− 3− 3− 3− 1−
8+ 6− 6+ 6+ 6+ 3+ 1−
8+ 6− 6+ 3− 3− 3+ 1−
8+ 6− 4− 3− 3+ 3+ 1−
8+ 6− 4+ 3+ 3+ 3+ 1−
ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
8+ 6+ 6+ 6+ 6+ 1− 1−
8+ 6+ 6+ 3− 3− 1− 1−
8+ 6+ 4− 3− 3+ 1− 1−
8+ 6+ 4+ 3+ 3+ 1− 1−
8+ 4− 4− 3− 1− 1− 1−
8+ 4− 4+ 3+ 1− 1− 1−
8+ 4+ 4+ 1− 1− 1− 1−
8+ 1− 1− 1− 1− 1− 1−
TABLE 5.4 ˆ 8 = 1+, lists L1 , . . . , L32
Lists
73
ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
6− 6− 6− 6− 6− 5− 1+
6− 6− 6− 6− 6+ 5+ 1+
6− 6− 6− 6+ 6+ 3− 1+
6− 6− 4− 3− 3+ 3+ 1+
6− 6− 4+ 3+ 3+ 3+ 1+
ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
6− 6+ 6+ 6+ 6+ 1− 1+
6− 6+ 6+ 3− 3− 1− 1+
6− 6− 6− 6− 6− 6+ 6+ 4− 4− 4+ 4− 4+ 4− 4+ 4+ 3− 3+ 3− 3+ 1− 3+ 3+ 1− 1− 1− 1− 1− 1− 1− 1− 1+ 1+ 1+ 1+ 1+
6− 1− 1− 1− 1− 1− 1+
ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
6+ 6+ 6+ 6+ 6+ 1+ 1+
6+ 6+ 6+ 3− 3− 1+ 1+
6+ 6+ 4− 3− 3+ 1+ 1+
6+ 6+ 4+ 3+ 3+ 1+ 1+
6+ 6+ 6+ 4− 4− 4+ 4− 4+ 4+ 3− 3+ 1− 1− 1− 1− 1+ 1+ 1+ 1+ 1+ 1+
6+ 1− 1− 1− 1− 1+ 1+
ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
4− 4− 4− 3− 1+ 1+ 1+
4− 4− 4+ 3+ 1+ 1+ 1+
4− 4+ 4+ 1− 1+ 1+ 1+
4− 1− 1− 1− 1+ 1+ 1+
4+ 4+ 4+ 1+ 1+ 1+ 1+
2+ 1+ 1+ 1+ 1+ 1+ 1+
TABLE 5.5 ˆ 8 = 1+, lists L33 , . . . L64
6− 6− 6− 3− 3− 3− 1+
6− 6− 6+ 6+ 6+ 3+ 1+
6− 6− 6+ 3− 3− 3+ 1+
4+ 1− 1− 1+ 1+ 1+ 1+
2− 1− 1+ 1+ 1+ 1+ 1+
74
Pencils of Cubics and Algebraic Curves in the Real Projective Plane ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
7− 1+ 1+ 1+ 1+ 1+ 1+
7+ 1+ 1+ 1+ 1+ 1+ 1−
5− 1+ 1+ 1+ 1+ 1− 1−
5+ 1+ 1+ 1+ 1− 1− 1−
3− 1+ 1+ 1− 1− 1− 1−
3+ 1+ 1− 1− 1− 1− 1−
TABLE 5.6 ˆ 8 = 1−, lists L65 , . . . , L70 ˆ1 4+ 4− 6+ 6− ˆ2 8− 8− 8− 8− ˆ3 8− 8− 8− 8− ˆ4 2+ 8− 8− 8− ˆ5 2+ 2+ 8− 8− ˆ6 2+ 2+ 2+ 8− ˆ7 2+ 2+ 2+ 2+ TABLE 5.7 ˆ 8 = 2+, lists L71 , . . . , L74 ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
4+ 8+ 8− 2+ 2+ 2+ 2+
4+ 4− 8− 2+ 2− 2− 2−
6− 8+ 8− 8− 8− 8− 2+
4+ 6− 8− 2+ 2+ 2+ 2−
4+ 6+ 8− 2+ 2+ 2− 2−
4− 8+ 8− 8− 2+ 2+ 2+
ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
6+ 6− 8− 8− 8− 2+ 2−
6+ 6+ 8− 8− 8− 2− 2−
4− 4− 8− 8− 2− 2− 2−
6− 6− 8− 8− 8− 8− 2−
4− 6− 8− 8− 2+ 2+ 2−
4− 6+ 8− 8− 2+ 2− 2−
TABLE 5.8 ˆ 8 = 2−, lists L75 , . . . , L87
6+ 8+ 8− 8− 8− 2+ 2+
Lists
75 ˆ 1 ˆ 2 ˆ 3 ˆ 4 ˆ 5 ˆ 6 ˆ 7
5+ 7+ 7− 1+ 1− 1− 3+
5+ 5− 7+ 1+ 1− 3+ 3−
5− 7+ 7− 1+ 1+ 1− 3+
5+ 5− 5− 5+ 5− 5− 7+ 5− 7+ 5− 7− 7+ 7− 7+ 7+ 1+ 1+ 1+ 1+ 1+ 1− 1+ 1+ 1− 1+ 3+ 1− 3+ 1− 3+ 3+ 3− 3+ 3− 3−
TABLE 5.9 ˆ 8 = 3−, lists L88 , . . . , L95 8, either to the interior or to the exterior of C2 . Then, move two further points F and G, one to the interior and the other to the exterior of C2 , so as to obtain the desired list ˆ8. The extremal lists are distributed in 12 orbits: two are both maximal and minimal, five are only maximal, five are only minimal. Note that max(ˆ1 = n±) and max(ˆ1 = (10 − n)∓) are deduced one from the other by 15. We denote the orbit of these two lists briefly by (n±, (10 − n)∓). The maximal orbits (2+, 8−) and (3+, 7−) are also minimal: max(ˆ 1 = 8−) = min(ˆ 2 = 1−) and max(ˆ1 = 3+) = min(ˆ2 = 1+). The principal extremal lists are: L2 = min(ˆ7 = 5−), L3 = min(ˆ7 = 5+), L5 = min(ˆ 7 = 3−), L32 = max(ˆ1 = 8+), L48 = max(ˆ1 = 6−), L56 = max(ˆ1 = 6+), L64 = max(ˆ 1 = 2+) = min(ˆ8 = 1+), L65 = max(ˆ1 = 7−) = min(ˆ8 = 1−), L66 = max(ˆ 1 = 7+), L67 = max(ˆ1 = 5−), L71 = min(ˆ1 = 4+), and L72 = min(ˆ 1 = 4−).
5.4
Distances between points
For the lists L(1, . . . , 8), we define the distance between consecutive points A, B in the cyclic ordering. Let (A, . . . H) = (1, . . . 8) up to some cyclic permutation and reversion. The distance A → B is the number of conics passing through five points among C, D, E, F, G, H that separate A from B, multiplied by −1 if A is the outermost of the two points. To determine a distance A → B, ˆ of L(1, . . . 8), as explained in one needs only to look at the sublists Aˆ and B the following example. Let us find out the distance 8 → 1 in the two cases: ˆ 1 = 8+, ˆ 8 = 1+ and ˆ 1 = 8+, ˆ8 = 6+. One has: 7 < 23456, letting 8 (resp. 1) cross the sequence of six conics determined by 2, . . . 7 yields the chain ˆ1 + (8) (resp. ˆ 8 + (1)), see Table 5.10. For ˆ1 = 8+, ˆ8 = 1+, both points 8 and 1 lie outside of all six conics, 8 → 1 = 0. For ˆ1 = 8+, ˆ8 = 6+, one has 8 > 23456, and 23467 < 1 < 23457, thus 8 → 1 = −2. ˆ = A Note that the distance A → B is equal to 0 if and only if: Aˆ = B, B ˆ ˆ or A = B = N , with N ∈ {C, . . . H}.
76
Pencils of Cubics and Algebraic Curves in the Real Projective Plane 8+
23456 /
6−
23457 /
6+
23467 /
4−
23567 /
4+
24567 /
2−
34567 /
2+
1+
23456 /
6−
23457 /
6+
23467 /
4−
23567 /
4+
24567 /
2−
34567 /
2+
TABLE 5.10 The chains ˆ 1 + (8) and ˆ 8 + (1) Tables 5.115.12 give the values of these invariants for all of the 47 principal lists Ln , and their sum Σn . Note that the 47 octuples (1 → 2, . . . , 8 → 1) listed in Tables 5.115.12 are all different. Furthermore, their orbits under the action of D8 are also all distinct. Observing the tables allows us to derive the following: Proposition 5.2 Any list L(1, . . . , 8) is determined by the octuple (1 → 2, 2 → 3, . . . , 8 → 1). Each orbit is determined by an octuple of integer numbers (ranging between −6 and +6), defined up to cyclic permutation, and reversion with change of all signs. The absolute value Σn  is an invariant of the orbits, that takes all of the even values between 0 and 8. The numbers of orbits realizing Σn  = 0, 2, 4, 6, 8 are respectively: 14, 12, 16, 4 and 1. We have thus a new encoding for the lists and the orbits.
5.5
Isotopies of octuples of points
Consider the space (RP 2 )8 endowed with the quadratic stratification (see Section 1.1). Proposition 5.3 The octuples of points realizing a given list L(1, . . . 8) lie all in the same chamber of (RP 2 )8 . This chamber is adjacent to the deep stratum, formed by the configurations of eight coconic points. Proof: Let 1, . . . 8 realize some list L. Denote by ht a homothety centered at any point P different from 1, . . . 8, with rate t. The homotheties ht , t ∈ [1, ∞] give rise to an isotopy of 1, . . . 8. During this isotopy, the list L is preserved, except for the endpoint when the mobile points 1, . . . 8 all become aligned. They are also on some reducible conic. The configuration is thus on the deep stratum. Move in this stratum so that the conic becomes nonreducible. With a slight perturbation, we can make the isotopy generic, in other words, it is completely contained in the same chamber except for the endpoint. Consider now two configurations realizing the same list. Perform for each of them a generic isotopy till all of the eight points lie on the same
Lists
n 1→2 2 0 3 0 4 0 5 0 6 0 7 0 8 0 10 0 11 0 12 0 13 0 14 0 15 0 18 −1 19 −1 20 −1 21 −1 22 −1 23 −1 25 −2 26 −2 32 −5 34 0 35 0 36 0 37 0 38 0 41 −1 48 −4 49 0 56 −3 64 0
77
2→3 0 0 0 0 0 0 0 1 1 1 2 2 3 0 0 0 1 1 2 0 0 0 0 0 0 1 1 0 0 0 0 0
3→4 0 0 0 −1 −1 −2 −3 0 −1 −2 0 −1 0 0 −1 −2 0 −1 0 0 −1 0 0 −1 −2 0 −1 0 0 0 0 0
4→5 5→6 0 −1 1 0 2 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 2 0 1 −1 0 1 0 0 1 0 0 0 2 0 1 −1 0 0 3 0 2 0 0 1 0 0 1 0 0 0 2 0 1 0 3 0 0 0 4 0 1 0 0
6→7 0 −1 0 −2 −1 0 0 −2 −1 −1 0 0 0 −3 −2 −2 −1 −1 −1 0 0 0 −4 −3 −3 −2 −2 −1 −1 0 0 0
TABLE 5.11 Distances A → B for the principal lists with ˆ8 = 1+
7→8 5 4 4 3 3 3 3 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0
8→1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 2 2 6
Σn 4 4 6 0 2 2 0 2 2 0 6 4 4 −2 −2 −4 2 0 0 2 0 −4 −2 −2 −4 2 0 2 −4 6 0 6
78
Pencils of Cubics and Algebraic Curves in the Real Projective Plane n 1→2 2→3 3→4 4→5 5→6 6→7 7→8 8→1 65 −1 0 0 0 0 0 −1 6 66 −2 0 0 0 0 1 0 5 67 −3 0 0 0 −1 0 0 4 71 5 0 1 0 0 0 0 −2 72 4 0 0 −1 0 0 0 −3 75 4 −1 1 0 0 0 1 −1 78 3 −2 1 0 0 −1 0 −1 80 3 −1 0 −1 0 0 1 −2 82 1 −2 0 0 1 −1 0 −3 83 0 −3 0 0 2 0 0 −3 84 0 −4 0 −2 0 0 0 −2 86 2 −2 0 −1 0 −1 0 −2 87 1 −3 0 −1 1 0 0 −2 88 −2 1 −1 1 0 1 −1 1 92 −1 0 −2 0 −1 2 0 2
Σn 4 4 0 4 0 4 0 0 −4 −4 −8 −4 −4 0 0
TABLE 5.12 Distances A → B for the principal lists with ˆ8 6= 1+ conic. Up to an affine transformation mapping one conic onto the other and an isotopy along this conic, we may assume that both isotopies have the same endpoint. So we have a path connecting the two configurations and having one single nongeneric point. Perturb the path in a neighborhood of this point. In this neighborhood, all of the walls intersect pairwise transversally. There are two possible perturbations, one crosses all of the walls twice, the other crosses no wall at all.
5.6
Elementary changes
In this section, we finish the proof of Theorem 5.1. Let us call elementary change the change induced on a list L = L(1, . . . 8) letting one point among 1, . . . , 8 cross a conic determined by five others, in some direction. Up to the action of D8 , there are exactly 17 elementary changes, see Figure 5.4 and ˆ = Q±, M ˆ = R±) involved in some elementary change Table 5.13. A pair (N is called elementary pair . Four of the 17 elementary changes are invariant under the action of 26 ∈ D8 , and two are invariant under the action of 37. The orbits of these six elementary changes each contain eight elements; the other eleven orbits each contain 16 elements. There are in total 224 elementary changes and correspondingly 224 elementary pairs, see Tables 9.109.11. Proposition 5.4 Let 1, . . . , 8 be eight points in convex position, realizing a list L. Let A, B be two of these points, consecutive for the cyclic ordering.
Lists
79
1
2
8
7
4
6 88
6
4
3
6
4
6
5
3
6
4
7 6
7
8
7
5
6
FIGURE 5.4 Elementary changes
4
3 5
6
4
4
5 2
3
8 6
5
1
7
7
3
8
1
3
2
8
2
6
4
6 3
7
5
1
5 2
4
4
5
1
3
2
8
3
6
2
6 3
2
1
7
5 1
7
4
8
5
1 8
4
3
4
2
8
5
1
1
7
2
7
2 8
6
1
3
2
7
5
8
5
8
4
6
4
1
3
8
7
7 6
2
3
2
5 1
3
2
8
4
6
5
1 7
7
1
8
3
2
8
5 1 2
7
1
3
4 5
80
Pencils of Cubics and Algebraic Curves in the Real Projective Plane ˆ 1 : 2+ → 2− ˆ 2 : 1+ → 1−
ˆ1 : 3+ → 8− ˆ2 : 1+ → 1−
ˆ1 : 8− → 3+ ˆ2 : 1− → 1+
ˆ 1 : 8− → 3+ ˆ 2 : 8− → 3+
ˆ1 : 3− → 3+ ˆ3 : 1+ → 1−
ˆ1 : 3+ → 3− ˆ3 : 1− → 1+
ˆ 1 : 2− → 4+ ˆ 3 : 1+ → 1−
ˆ1 : 4+ → 2− ˆ3 : 1− → 1+
ˆ1 : 4+ → 2− ˆ3 : 8− → 2+
ˆ 1 : 2− → 4+ ˆ 3 : 2+ → 8−
ˆ1 : 4+ → 4− ˆ4 : 1+ → 1−
ˆ1 : 5+ → 3− ˆ4 : 1+ → 1−
ˆ 1 : 3− → 5+ ˆ 4 : 1− → 1+
ˆ1 : 3− → 5+ ˆ4 : 8− → 2+
ˆ1 : 5− → 5+ ˆ5 : 1+ → 1−
ˆ 1 : 4− → 6+ ˆ 5 : 1+ → 1−
ˆ1 : 6+ → 4− ˆ5 : 8− → 2+
TABLE 5.13 Representatives of the 17 orbits of elementary changes 1. If A → B = 0, one may move the set of points, shifting A towards B, in such a way that the generic list L(1, . . . 8) is preserved until A = B. 2. If A → B = ±n, with n > 0, the points A and B are separated from each other by n of the six conics determined by C, D, E, F, G, H. One may move the set of points, shifting A towards B, till A = B, in such a way that the list L(1, . . . , 8) undergoes exactly n elementary changes, corresponding to the n conics. Proof: For any pair of consecutive points P, Q among 1, . . . 8, denote by [P Q] the affine segment P Q belonging to the convex hull of the eight points. Move A towards B as prescribed, leaving the other points fixed. The point A moves inside of a triangle BHR supported by the lines (BC), (BH), (GH), where R = (BC) ∩ (GH). In most cases, one may find a path connecting the starting point A to B inside of the triangle, that does not cut any one of the 15 conics passing through B and four other points among C, . . . H. The exception is the following: let C2 be one of the conics BDEF G, BDEF H, BDEGH, BDF GH or BEF GH. If A, C both lie outside of C2 , and the second intersection of C2 with the line BC is on the edge BR of the triangle, then any path A → B must cross C2 . Assume now that we may move all of the points at the same time. We perform an isotopy in (RP 2 )8 inside of the chamber of L, towards the deep stratum. When the octuple of points gets close enough to this stratum, the exceptional case can no longer occur: let
Lists
81
C2 be any one of the five conics, and assume that A, C both lie outside of C2 , then the second intersection of C2 with the line (BC) lies on the segment [BC]. . Proposition 5.5 The elementary changes are always realizable: for each elementary pair appearing in an existing list L, one may perform the corresponding elementary change to realize a new list. Proof: Recall that a statement of the form 6 crosses 12345 from the inside to the outside is equivalent to 5 crosses 12346 from the outside to the inside. All of the 17 elementary changes but one may be thus interpreted as motions of a point A towards a consecutive point B such that A → B 6= 0, till A crosses the first conic separating it from B. The only exception is the third change: ˆ 1 : 8− → 3+, ˆ 2 : 1− → 1+. So, for any elementary pair of L that is not in the orbit of (ˆ 1 = 8−, ˆ2 = 1−), the corresponding elementary change is realizable. Let now L have the elementary pair (ˆ1 = 8−, ˆ2 = 1−). Note that ˆ 1 = 8− belongs to the chain ˆ1 − (8), and ˆ2 = 1− belongs to the chain ˆ 2 + (8). Using Table 5.3, we deduce that ˆ8 = 1−; using Table 9.4, we see that ˆ = 1− for N = 3, . . . 7. The list is max(ˆ1 = 8−). The point 8 lies inside N of all the conics determined by 1 and four points among 2, . . . 7. Up to some isotopy towards the deep stratum, we may assume that the conic 34567 has its second intersection point with the line (17) on the affine segment [17]. Denote by T the triangle supported by the lines (67), (71), (12) containing 8. One may move 8 towards [17] in T , preserving the list, till 8 reaches 34567. Starting from a list that is already realized, we will say that we perform the ˆ, M ˆ ) without further precision, as there is no ambiguity possible. change (N In Table 5.14, the number n stands for a list Ln (as in Tables 5.115.12). The rows represent sequences of elementary changes and actions of elements of D8 . Three rows start with an extremal list (n = 1, 65, 71), each of the other rows starts with a list that was already realized in a previous row. All of the principal lists appear in this table: each of them may be obtained from an extremal list performing successive elementary changes and symmetries. The sequences have been chosen so as to reach all of the principal lists with the least possible number of starting lists (note that some intermediate lists are also extremal). This finishes the proof of Theorem 5.1.
(ˆ 6, ˆ 7) (ˆ 5, ˆ 6) (ˆ 5, ˆ 6) (ˆ 3, ˆ 5) (37) (ˆ 4, ˆ 5) (37) (ˆ 3, ˆ 6) (ˆ 1, ˆ 7) (ˆ 1, ˆ 4) (ˆ 2, ˆ 8) (ˆ 1, ˆ 4) (ˆ 1, ˆ 5) (ˆ 3, ˆ 8)
2 (ˆ5, ˆ7) 3 5 (37) 9 10 (ˆ4, ˆ6) 11 15 17 (ˆ5, ˆ6) 18 ˆ 21 (ˆ2, 6) 25 33 (ˆ5, ˆ6) 34 ˆ ˆ 37 (2, 6) 41 66 (ˆ1, ˆ6) 67 72 75 (ˆ2, ˆ7) 78 80 (ˆ2, ˆ7) 86 83 (ˆ2, ˆ6) 82 (−1) 88 (15) or (ˆ1, ˆ5)
TABLE 5.14 Inductive construction of the principal lists
1 6 9 14 3 22 2 35 65 71 71 75 87 87
4 12 19 26 35 49
87 (ˆ3, ˆ7)
(ˆ5, ˆ6) (ˆ4, ˆ5) (ˆ4, ˆ6) (ˆ4, ˆ5) (ˆ4, ˆ6) (ˆ1, ˆ6)
(ˆ2, ˆ6) 90
6
7
(ˆ3, ˆ6) 38
(ˆ4, ˆ5) 36
92
(ˆ2, ˆ5) 84
(ˆ3, ˆ6) 22
(ˆ4, ˆ5) 13
(ˆ4, ˆ6)
(ˆ4, ˆ5) 20
(ˆ3, ˆ6) 14
(ˆ4, ˆ7) 8
(ˆ3, ˆ 5) 23
(ˆ4, ˆ 5)
82 Pencils of Cubics and Algebraic Curves in the Real Projective Plane
6 Link between lists and pencils
CONTENTS 6.1 6.2 6.3
Nodal lists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pairs of distinguished cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Changes of lists and of pencils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1
Nodal lists
83 85 88
A configuration of eight points lying on a nodal cubic C3 , one of them being the node, is not generic. Up to the action of D8 , there are exactly eight possible combinatorial cubics C3 ; see Figure 6.1 where the successive cubics are denoted by (1±, 1)n , (1−, k)n for k = 8, . . . 2. Note that the encoding is consistent with the action of D8 . 1 8
3
7 6
5
4
1
8
2
2
7
3 6
5
1
8
4
7 6 5
1
8 2
7
3
6
4
5
4
3
2
FIGURE 6.1 The cubics (1±, 1)n , (1−, 8)n , . . . (1−, 2)n Definition 6.1 A list L(1, . . . , 8) = (ˆ1, ˆ2, . . . , ˆ8) is nodal if it is realizable by 1, . . . , 8 on a nodal cubic with its double point at one of these points. Proposition 6.1 Up to the action of the group D8 , \L(1, . . . , 8)nodal = 4. The nodal orbits are the maximal orbits (8−, 2+), (8+, 2−), (6−, 4+), (6+, 4−). Here are representatives of each orbit along with the corresponding nodal cubics. L(1, . . . , 8) = (ˆ 1, ˆ 2, . . . , ˆ8) = max(ˆ 1 = 8−), realizable with (1±, 1)n , (1−, k)n , k = 2, . . . 8, 83
84
Pencils of Cubics and Algebraic Curves in the Real Projective Plane max(ˆ 1 = 8+), realizable with (1±, 1)n and (1−, 8)n , max(ˆ 1 = 6−) and max(ˆ1 = 6+), both realizable with (1±, 1)n .
Proof: Let 1, . . . , 8 lie in convex position on some nodal cubic C3 , one of these points being the node. Up to the action of D8 , we may assume that C3 is one of the cubics (1±, 1)n , (1−, k)n , k = 2 . . . 8. If C3 = (1±, 1)n , one has ˆi ∈ {1±} for all i ∈ {2, . . . , 8}. Using the two diagrams ˆ8 = 1± of Tables 9.39.4 , one finds 14 possibilities for the list L(1, . . . , 8), namely the maximal lists max(ˆ 1 = N ±), N = 2, . . . , 8. One has ˆ1 = 8+ ⇐⇒ 7 < 23456 and 8 > 23456. One can choose the points 2, . . . , 8 on the loop of C3 so that this condition is achieved. One has ˆ 1 = 7− ⇐⇒ 8 < 23456 and 7 > 23456. By Bezout’s theorem between 23456 and (1±, 1)n , one cannot choose the points 1, . . . , 8 on the loop of this cubic verifying this condition. Finishing this argument with the other possible values of ˆ1, one finds that 1, . . . , 8 may be chosen on (1±, 1)n so as to realize the eight lists max(ˆ1 = N ±), with N = 2, 4, 6, 8. Similarly, one proves that 1, . . . , 8 on the cubic C3 = (1−, 8)n can realize exactly the first two lists max(ˆ1 = 8±); and points 1, . . . , 8 on any cubic C3 = (1−, k)n , k ∈ {2, . . . , 7} must realize the first list max(ˆ1 = 8−). Given two points P, Q among 1, . . . , 8, we denote by (Pˆ , Q) the cubic of the list Pˆ with a double point at Q. Proposition 6.2 The three conditions hereafter are equivalent: 1. The list L(1, . . . , 8) determines the (combinatorial) pencil P(1, . . . , 8). 2. The list L(1, . . . , 8) is not nodal. ˆ the position of G with respect 3. ∀G ∈ {1, . . . , 8}, ∀C3 cubic of G, to C3 is determined by L(1, . . . , 8). Proof: ¬2 ⇒ ¬1. Let L(1, . . . , 8) be a nodal list. Up to the action of D8 , we may assume it is one of the four lists in Proposition 6.1. Any one of these lists is realizable with the eight points on a nodal cubic C3 = (1±, 1)n . The points 1, . . . , 8 give rise to a singular pencil Psing with 1 = 9. Perturb the pencil moving 1 away from the node onto the odd component or onto the loop of C3 (leaving the other seven points fixed). In the generic pencil obtained, C3 is replaced by a pair of distinguished cubics C31 , C32 or C33 , C34 , see Figure 6.2. Using Bezout’s theorem, we see easily that these two pairs of cubics cannot belong to the same pencil. 2 ⇒ 3. According to Proposition 5.3, any two configurations of points realizing the same list may be connected by a path inside of their common chamber of (RP 2 )8 . As L(1, . . . , 8) is not nodal, none of the points G may ˆ cross any cubic of the list G. 3 ⇒ 1. Consider a configuration of points realizing a list L(1, . . . 8) and giving rise to some pencil P. Move the configuration preserving the list; the combinatorial pencil degenerates only if some base point A among 1, . . . 8
Link between lists and pencils
85 B
B 2
A
8
3
7 6
5
4
A
8
2 3
7 6
5
4
FIGURE 6.2 Pair of cubics: C31 , C32 if (A, B) = (1, 9), C33 , C34 if (A, B) = (9, 1) comes together with 9. But this amounts to saying that each point G 6= A ∈ ˆ A). {1, . . . , 8} comes onto the cubic (G, Propositions 5.4 and 6.2 imply immediately: Proposition 6.3 Let 1, . . . , 8 be eight points in convex position, realizing a nonnodal list L. Let A, B be two of these points, consecutive for the cyclic ordering. We may move the set of points, shifting A towards B, in such a way that the combinatorial pencil does not degenerate till A = B (if A → B = 0) or till A reaches the first of the n conics separating A from B (if A → B 6= 0).
6.2
Pairs of distinguished cubics
In what follows, we call (combinatorial) cubic C3 a topological type (C3 , 1, . . . 8) (the position of 9 is not yet specified). We define an encoding for distinguished cubics that is consistent with the action of D8 . Let (1−, N ), N = 3, . . . 8 be the cubic obtained from (1−, N )n , shifting N away from the node X onto the arc X2 of the loop, and let (N, 1−), N = 3, . . . 7 be the cubic obtained from (1−, N )n , shifting N away from the node onto the arc X1 of the odd component, see Figure 6.3. For (P, Q) = (3, 4), (4, 5), . . . , (8, 1), denote by (1−, P Q), the cubic obtained from (1−, P )n (or (1−, Q)n ) shifting the node X away from P (or Q) into the interior of the arc P Q. Let (1−, E) (E stands for end ) be the cubic that could be defined as (1−, N ), with N = 2 or as (1−, P Q), with (P, Q) = (2, 3). We chose this specific notation to avoid double notation for one single combinatorial cubic. The cubics encoded hereabove are called cubics of the family 1−. At last, denote by (81, L) and (81, C) the first and the second cubic in Figure 6.4. The letters L and C stand respectively for loop and odd component. In Chapter 8 where we classify the pencils, we will consider combinatorial cubics (C3 , 1, . . . 9). To encode them, we use the notation defined here for (C3 , 1, . . . , 8), enhanced with the position of 9. The cubic C3 = (C3 , 1, . . . , 8) is divided into nine successive arcs by the points 1, . . . 8, X. If C3 = (81, L), then 9 lies on the arc XX. If C3 is of the family
86
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
N , follow C3 starting from N in the direction , and write which oriented arc contains 9, see Tables 8.18.3. 1
8
2
7 6
3 5
4
1
8
7
7
1
8
2
6
3 5
4
2 6
3 5
4
1
8 7 6
2 5
4
3
FIGURE 6.3 Cubics (1−, 7), (7, 1−), (1−, 67) and (1−, E) Let 1, . . . 8 realize a list L and a pencil P. Assume we know L and not P. We explain hereafter how to determine pairs of distinguished cubics of P. The position of 9 is not specified; it can sometimes be found using Bezout’s theorem. Leave 1, . . . 7 fixed and move 8 towards 1, preserving the convex position, till 8 reaches 1 (if 8 → 1 = 0) or 8 reaches the first of the six conics separating it from 1 (if 8 → 1 6= 0). Letting 8 cross this conic induces an elementary change on the list, see Proposition 5.4. The only other base point of the pencil that moves is 9. Assume that in the motion, the eight points are never on a cubic with node at one of them. This condition is always achieved if the list is nonnodal. More generally, if the condition is achieved for a given set of points 1, . . . 8 with 8 → 1 6= 0, we say that the corresponding elementary change is realizable starting from this set of points. The mobile pencil P(1, . . . 8) is preserved all along, except in the end, when it becomes a singular pencil Psing . Let C30 , C31 and C32 be the three singular cubics involved, ˆ = A or Aˆ = B ˆ = N , with see Section 4.2. A close pair is a pair Aˆ = B, B A, B consecutive. Up to the action of D8 there are exactly seven close pairs: (ˆ 8 = 1+, ˆ 1 = 8+), (ˆ 8=ˆ 1 = N +), N = 2, . . . 7. 1. If 8 → 1 = 0, the singular cubic C30 , with a double point at 8 = 1, is identical to both (ˆ8, 1) and (ˆ1, 8) (auxiliary cubics). The close pairs ˆ 8=ˆ 1 = 2+ or 7− (deduced one from the other by (+1)(48)) are inessential : one does not know a priori whether the double points of the auxiliary cubics are isolated or not. Therefore, the cubics C31 , C32 may be nonreal. There are in total 16 inessential close pairs (one orbit). For the other close pair (ˆ1, ˆ8) (essential close pairs), the auxiliary cubics have each a nonisolated double point. All along the motion 8 → 1, the position of 1 with respect to (ˆ1, 8) and the position of 8 with respect to (ˆ8, 1) are preserved. Using Bezout’s theorem with these cubics, one finds out the corresponding pair C31 , C32 . In all of the cases, any one of the four combinatorial data: cubic (ˆ 1, 8) enhanced with the position of 1, cubic (ˆ8, 1) enhanced with the position of 8, cubic C31 , and cubic C32 determines the other three. The close pairs (ˆ8 = 1+, ˆ1 = 8+) and (ˆ8 = 1−, ˆ1 = 8−) give
Link between lists and pencils
87
rise to the same three admissible pairs C31 , C32 ; they are shown in the upper part of Figure 6.4, along with the auxiliary cubics. These three pairs are: (81, L), (81, C) (8+, 1), (8−, 78) (1+, 12), (1−, 8) The other close pairs (ˆ8, ˆ1) each give rise to two admissible pairs C31 , C32 , see the table below, where N ranges from 3 to 6. Note that all of these cubics C31 , C32 are distinguished. The case ˆ8 = ˆ1 = 7+ is shown in the lower part of Figure 6.4. ˆ ˆ ˆ8 = ˆ1 = N + ˆ8 = ˆ1 = N − 8=ˆ 1 = 7+ 8 = ˆ1 = 2− (7+, 8), (1, 7+) (2−, 8), (12, C) (N +, 1), (8, N +) (N −, 8), (1, N −) (78, C), (7+, 1) (2−, 1), (8, 2−) (N +, 8), (1, N +) (N −, 1), (8, N −) 2. If 8 → 1 6= 0, one has C30 = C2 ∪ (1P ), P ∈ {2, . . . , 7}. If P = 2 and both 1 and 2 are outside of C2 = 34567, then one does not know a priori whether the double points of the reducible cubic C30 are real or complex conjugated. Therefore, the cubics C31 , C32 may be nonreal. Let 8 go to the outside of C2 = 34567, 1, 2 > 34567, the elementary change is: ˆ1 : 2− → 2+, ˆ2 : 1− → 1+; it is inessential . There are in total 16 inessential elementary changes (one orbit). Even if the cubics C31 , C32 are real, the combinatorial pencil is the same before and after performing an inessential elementary change. The other elementary changes are essential . Perform one of them, then either cubic C31 , C32 may be obtained from the reducible cubic C30 moving 8 away from C2 and perturbing one of the double points of C30 . All of the pairs obtained are distinguished cubics. Indeed, let s be the intersection of the line (1P ) with the interior of C2 , and let s1 , s2 be the two arcs of C2 on either side of (1P ). The loop of each cubic C31 , C32 is obtained by perturbing one of the si ∪ s, i ∈ {1, 2}, and both si ∪ s contain some points among 1, . . . , 8. We give an example in Figure 6.5; the elementary pair is (ˆ1 : 5+, ˆ5 : 8−), the corresponding pair of distinguished cubics is (5, 8−), (5+, 1). The first elementary change ˆ1 : 2− → 2+, ˆ2 : 1− → 1+ in Figure 5.4 and Table 5.13 is inessential. The changes of combinatorial distinguished cubics corresponding to the 17 elementary changes from Table 5.13 are displayed in Table 6.1. An elementary pair appearing in some (in)essential elementary change will be called an (in)essential elementary pair. Up to the action of D8 , two pairs are both elementary and close: (ˆ1 = 8+, ˆ8 = 1+) (essential only as a close pair), and (ˆ1 = ˆ8 = 2+) (essential only as an elementary pair). If a list L is nonnodal, each essential elementary pair of L gives a pair
88
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
of distinguished cubics, see Table 6.1, and each essential close pair of L gives two or three admissible pairs of distinguished cubics. The rest of the list allows us to determine which pair of cubics is actually realized. 8
7
8
1 7
2 3
6
3 5
8
7
1
6
3
4
5
6
2
8
6
4
5
7
7 8
6 5
4
1 3
7 6 4
3
5
3
4
2
6 5
2 4
3
6 5
2
5
4
1
8 1
6 5
3
4 7
3
8
2
8 2
3
4
6
3
1
1
7
7 2
3
6
1
4 7
1
5
5
6
7 8
8 1
5
2
3
4 7 8
1
6 5
2
8
2 4
8 2
5
4
1
6
3
4
8
6
3
5
8
1
7
2
7
4
77
2 5
1
3
5
1
6
2
6
7
3
4 8
3 6
8 1
2
7
3 4
8
1
7
2
5
2
1
8
6
4
7
5
1
8
7
2
6
4
5
1
2 4
3
FIGURE 6.4 Cubics obtained with the close pairs (ˆ8 = 1±, ˆ1 = 8±), and (ˆ8 = ˆ1 = 7+)
6.3
Changes of lists and of pencils
Let 1, . . . , 8 be eight generic points lying in strictly convex position in RP 2 , realizing a list L(1, . . . 8) and a pencil P = P(1, . . . 8). Move the eight points, keeping them distinct and strictly convex. At some moment, a degeneration P → Psing occurs, after which one gets a new pencil P 0 , See Section 4.2. We are interested only in the degenerations that affect the combinatorial pencil, so we leave aside the cases 1b, 2b, 2c, 2d. When the list undergoes an elementary change, six of the eight chosen base points become coconic, and 9 becomes aligned with the other two. If the change is essential, the new pencil P 0 is
Link between lists and pencils
89
8
8
8
1
1
7
1
7 2
6 5
3
4
7 2
6 5
4
2
6
3
5
4
3
FIGURE 6.5 Cubics obtained with the elementary pair (ˆ1 : 5+, ˆ5 : 8−)

(1+, 2) → (1−, E) (1, 3+) → (1+, 12)
(1−, E) → (1+, 2) (1+, 12) → (1, 3+)
(8−, 2) → (23, L) (81, L) → (3+, 1)
(3, 1+) → (3, 1−) (1, 3−) → (1, 3+)
(3, 1−) → (3, 1+) (1, 3+) → (1, 3−)
(1+, 3) → (1−, 3) (12, C) → (1, 4+)
(1−, 3) → (1+, 3) (1, 4+) → (12, C)
(8−, 3) → (2+, 3) (4+, 1) → (2−, 1)
(2+, 3) → (8−, 3) (2−, 1) → (4+, 1)
(4, 1+) → (4, 1−) (1, 4+) → (1, 4−)
(1+, 4) → (1−, 4) (1, 5+) → (1, 3−)
(1−, 4) → (1+, 4) (1, 3−) → (1, 5+)
(8−, 4) → (2+, 4) (3−, 1) → (5+, 1)
(5, 1+) → (5, 1−) (1, 5−) → (1, 5+)
(1+, 5) → (1−, 5) (1, 4−) → (1, 6+)
(8−, 5) → (2+, 5) (6+, 1) → (4−, 1)
TABLE 6.1 Changes of pairs of distinguished cubics
90
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
obtained from P by replacing the pair of distinguished cubics C31 , C32 by a new pair C33 , C34 (case 2a). If the change is inessential, we are in case 2a or 2b. In case 2a, only one of the two cubics C31 , C32 is distinguished, and one recovers the same pair of combinatorial cubics after the degeneration (left to the reader). Performing an inessential change on the list does not change the pencil. Conversely, one may sometimes change the pencil without changing the list. The degeneration occurs letting one of the first eight base points A come together with 9 (1a). The cubic C30 of Psing passes through 1, . . . 8 and has a node at A, the corresponding list L(1, . . . 8) is nodal. Move farther, then the new generic pencil P 0 is deduced from P swapping the positions of A and 9 on all of the eight (combinatorial) distinguished cubics. Performing an inessential change on a nodal list again yields a nodal list. So, if a pencil is realizable by two different lists, these lists are both nonnodal or both nodal. In the latter case, we say that the pencil is nodal .
7 Pencils with reducible cubics
CONTENTS 7.1 7.2 7.3
Two nongeneric lists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pencil with six reducible cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Symmetric lists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1
Two nongeneric lists
91 93 97
Among the principal lists, exactly ten have a set of four disjoint elementary pairs. The lists Ln with n = 11, 12, 13, 14, 19, 20, 21, 22 each have disjoint elementary pairs (ˆ 1, ˆ 8), (ˆ2, ˆ7), (ˆ3, ˆ6), (ˆ4, ˆ5). Each of the two lists Ln with n = 88, 92 has disjoint elementary pairs (ˆ1, ˆ5), (ˆ2, ˆ6), (ˆ3, ˆ7), (ˆ4, ˆ8). We explain hereafter how to realize these lists directly. Consider a pair of ellipses intersecting at four points, 1, 3, 5, 7, see Figure 7.1. Denote by 9 the intersection of the diagonal lines 15 and 37. Draw a vertical and a horizontal line, both passing through 9. Let 4, 8, 2, 6 be four supplementary points chosen such as: 4, 8 are the intersections of the vertical line with one ellipse; 2, 6 are the intersections of the horizontal line with the other ellipse; each pair of points lying on one ellipse is in the interior of the other. By construction, the points 1, . . . 8 lie in convex position. Moreover, there exist two supplementary conics passing through six points: 234678 and 124568. As a matter of fact, the pencil of cubics P determined by 1, . . . , 8 has 9 as the ninth base point and four distinguished cubics, all of them reducible: 498 ∪ 123567, 296 ∪ 134578, 397 ∪ 124568 and 195 ∪ 234678. Let us perturb the pencil, moving slightly the points 8 and 9. As 1, 2, 3, 5, 6, 7 are on a conic, 8, 4 and 9 must stay aligned. The point 8 leaves the three conics 23467, 12456, 13457. Letting 8 cross each conic from the inside to the outside yields the following elementary changes: Conic 23467: (ˆ1 = 5−, ˆ5 = 1+) → (ˆ1 = 5+, ˆ5 = 1−). Conic 12456: (ˆ 3 = 7−, ˆ7 = 3+) → (ˆ3 = 7+, ˆ7 = 3−). Conic 13457: ˆ ˆ (2 = 7+, 6 = 1−) → (ˆ 2 = 5−, ˆ6 = 3+). Let us now move 3 away from the conic 12567. Letting 3 cross 12567 from the inside to the outside yields the following elementary change: (ˆ4 = 7−, ˆ8 = 5+) → (ˆ4 = 1+, ˆ8 = 3−). The first perturbation may be done so as to realize six different positions of 8 with respect to the set of conics 23467, 12456, 13457. Then, move 3 to the outside 91
92
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
of 12567. We obtain six lists from Table 5.9 (the first five and the last one), among which we find L88 and L92 . Consider a pair of ellipses intersecting at four points 2, 4, 5, 7, see Figure 7.2. Denote by 9 the intersection of the lines 27 and 45. Draw a line passing through 9 and cutting the ellipses on their arcs 57 and 24. Let 6 and 3 be the intersections of this line with one ellipse, chosen so that these points lie outside of the second ellipse. Draw another line, passing through 9 and cutting the second ellipse at two points 8, 1 on the arc 72. One may choose this second line in such a way that the points 1, . . . 8 lie in convex position. By construction, there exist two supplementary conics passing through six of the points: 123678 and 134568. The pencil of cubics P determined by 1, . . . , 8 has 9 as ninth base point and four distinguished cubics, all of them reducible: 189 ∪ 234567, 369∪124578, 459∪123678 and 279∪134568. Let us perturb the pencil, moving the points 8 and 9. The point 8 leaves the three conics 13456, 12457, 12367. Letting 8 cross each conic from the inside to the outside yields the following elementary moves: Conic 13456: (ˆ2 = 6−, ˆ7 = 1−) → (ˆ2 = 8+, ˆ7 = 3+). Conic 12457: (ˆ 3 = 6+, ˆ6 = 3+) → (ˆ3 = 6−, ˆ6 = 3−). Conic 12367: (ˆ 4 = 3−, ˆ 5 = 3−) → (ˆ 4 = 6+, ˆ5 = 6+). Let us now move 7 away from the conic 23456. Letting 7 cross 23456 from the inside to the outside yields the following elementary move: (ˆ1 = 8+, ˆ8 = 1+) → (ˆ1 = 8−, ˆ8 = 1−). The first perturbation may be done so as to realize six different positions of 8 with respect to the set of conics 13456, 12457, 12367. Then, move 7 to the inside of 23456. We obtain the lists L11 , L12 , L14 , L19 , L21 , L22 . Starting again from the pencil with four reducible cubics we realize the two missing lists as follows. List L13 : move first 7 to the outside of 12458 and the inside of 12368; then, move 8 to the outside of 13456. List L20 : move first 6 to the outside of 23457 and the inside of 12378; then move 8 to the outside of 12457. Proposition 7.1 Let 1, . . . 8 be eight points lying in convex position in the plane and let k be the number of conics passing through exactly six of them. One has k ≤ 4. If k = 4, then the points realize, up to the action of D8 , one of the two nongeneric lists l and l0 shown in Figures 7.17.2. The orbit of l has two elements, the orbit of l0 has eight elements. Proof: Perturbing slightly the configuration 1, . . . 8 must yield a generic list with four distinct elementary pairs; otherwise stated, a list that is in the orbit of some of the 10 lists Ln . Up to the action of D8 , the original configuration 1, . . . 8 must be as shown in Figure 7.1 or 7.2. The list l is encoded by the data: 2, 6 < 134578, 4, 8 < 123567, 1, 5 > 234678, 3, 7 > 124568. This list is invariant by the action of id, 15, 26, 37, 48, +2, −2, +4. The list l0 is encoded by the data: 1, 8 > 234567, 5, 4 < 123678, 2, 7 < 134568, 3, 6 > 124578. This list is invariant by the action of (+1)(48). A pencil of cubics with eight base points lying in convex position in the real plane (no seven of them being coconic) has at most four reducible cubics, and the corresponding four lines all pass through the ninth base point. Up to the action of D8 , the points realize one of the lists l, l0 . Each of these lists
Pencils with reducible cubics
93
gives rise to a unique pencil that is easily constructed by applying Bezout’s theorem with the reducible cubics, see Figures 7.37.4. In the next section, we drop the condition of convexity and search for more singular pencils.
8 7 6 5
1 2
9
4
3
FIGURE 7.1 First configuration l of points with four reducible cubics
7.2
Pencil with six reducible cubics
Let us say that a cubic is completely reducible if it is the product of three lines. A complex pencil contains at most four such cubics [74], [89]; the corresponding twelve lines and the nine base points are such that: each point lies on four lines and each line passes through three points. Recall that the nine inflection points of a complex cubic realize such a configuration. The Hessian 2 F of a cubic F (x1 ; x2 ; x3 ) is the cubic H(x1 ; x2 ; x3 ) = det( ∂x∂i ∂x ) (determinant j of the matrix formed with the second partial derivatives). The Hessian pencil associated to F is the pencil generated by F and its Hessian H. The base points of this pencil are the nine inflection points of C3 . The Hessian pencils realize the upper bound of four completely reducible cubics. Let us now go back to pencils with only real base points. A pencil with nine real base points cannot have four completely reducible cubics: the Sylvester
94
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
8
1
7
9
2
6
3 4
5
FIGURE 7.2 Second configuration l0 of points with four reducible cubics
8 7
9
6 5
4
8
7
1 2
6 5
4
FIGURE 7.3 Pencil of cubics realized with the configuration l
5
2 3
4
8
7
2
1 9
5
1
3
8
6
3
9
6
7
9
1 2
4
3
Pencils with reducible cubics 9
6 5
9
7
6
7 4
8
8
5
4
8
5
4
2 3
5
4 1
8
9 7
2
6
3
6
1
7
1
7
1 2 3
4
9 2
5
8
7
6
1
8
9
9
1 2
3
95
6
3
2
5
4
3
FIGURE 7.4 Pencil of cubics realized with the configuration l0 Gallai theorem states that given n points in the real plane, they are either all collinear or there exists a line containing exactly two of them (see e.g.,[4]). Consider the pencil generated by the two completely reducible cubics (BF ) ∪ (AD) ∪ (CE) and (DF ) ∪ (AC) ∪ (BE) in Figure 7.5, the base points are A, . . . I. We recover an elementary proof of the Pascal theorem: assume that G, H, I are on a line, and let C2 be a conic passing through five of the other base points. Then, the cubic (GHI) ∪ C2 belongs to the pencil, otherwise stated, A, B, C, D, E, F are coconic. Conversely, if A, . . . F are coconic, then G, H, I are aligned. The particular case where C2 is the product of two lines is the Pappus theorem, the pencil here has three completely reducible cubics.
A
A G
F
H E
B
G F
C
I D
FIGURE 7.5 Pascal and Pappus theorems
E
B
H C
I D
96
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
B
E
A
D
G
F H
I
C FIGURE 7.6 Configuration with six reducible cubics
A
E D
G I
G I
E
F
B D
C
E
A
F H
FIGURE 7.7 Pencil with six reducible cubics
F H
I
A
E
G I
D C
B F H
C
A
E
G
D I
B
D
G
H
C A
B
C
F H
B
E
A G I
C
D
B F H
Pencils with reducible cubics
97
Let us now search for a pencil having the maximal number of six reducible cubics. Each of the corresponding six lines must pass through exactly three base points. If one point lies on four of these lines, then a fifth line intersects these four at four base points, which is a contradiction. If each base point lies on at most two lines, draw five lines, each of them passes through four base points, a contradiction again. Thus, one point lies on exactly three lines, and six other base points are distributed pairwise on these lines. Draw two further lines; they must intersect at one of these six points, otherwise they would each pass through four base points. We get thus the following distribution of the base points on the lines and the conics: four points A, B, C, D each lie on three lines and three conics, three points D, E, F each lie on two lines and four conics and the other two points H, I each lie on six conics. Assume that H, I are also real. Up to the action of the symmetric group S4 on the first chosen base points A, B, C, D, the configuration is as shown in Figure 7.6. The sequence of six reducible cubics of the pencil is represented in Figure 7.7. Note that the nonsingular cubics of the pencil are all disconnected.
7.3
Symmetric lists
We proved in Section 5.2 that a generic list is preserved by no element of D8 ; otherwise stated, each orbit contains 16 lists. Let us now search for the symmetric lists that are invariant for some element of D8 ; we met two of them, l and l0 , in Section 7.1. Let λ0 be the list 12345678 consisting of eight coconic points. We say that a list is almost generic if it has one unique configuration of six coconic points. Proposition 7.2 Each element of D8 leaves some nongeneric lists invariant: (±1) and (±3): λ0 ; (±2) and (±4): λ0 , l and (+1)l; (+1)(48): λ0 , l0 , (+4)l0 = (+1)(26)l0 ; 15: λ0 , four lists with seven coconic points; two lists, each with four sets of six coconic points; ten lists, each with three sets of six coconic points; two lists, each with two sets of six coconic points; and 24 almost generic lists, divided in two sets of 12 that are mapped one onto the other by (+4) (or equivalently 37). All of these lists have a conic 234678. See description ahead and Tables 7.17.3. In what follows, a stands for a cyclic permutation, and σ for a symmetry 15, 26, 37 or 48. Recall that any two cyclic permutations commute, whereas aσ = σ(−a). Let m be a cyclic permutation, and L be a list such that m·L = L. Any list L0 in the orbit of L is also invariant under the action of m and −m. Up to the action of D8 , we may assume that L contains a conic C2 = 123456, 123457, 123467 or 123567. Moreover, L contains the images of this conic by m, m ◦ m . . . and if P < C2 , then m · P < m · C2 . One finds out easily that (±1), (±3) preserve only λ0 , whereas (±2), (±4) each preserve three lists:
98
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
λ0 , l and (+1)l. A list invariant by (+1)(48) must contain the conics 124578, 134568, 123678 and 234567. Thus, (+1)(48) preserves three lists: λ0 , l0 and (+4)l0 = (+1)(26)l0 . A list invariant by 15 must contain a conic 234678. Let us first look for the almost generic lists. Let L be such a list; the orbit of L contains eight elements, each symmetry σ leaves two of them invariant. If σ · L0 = L0 , one may perturb L0 in two ways to get generic lists that are deduced from one another by σ. The 16 generic lists thus obtained form an orbit of D8 . We select all of the principal lists having an elementary pair ˆ, M ˆ ), with (N, M ) = (1, 5), (2, 6), (3, 7) or (4, 8), such that the nongeneric (N list obtained, making the six points different from N, M coconic, is invariant by the symmetry N M . Up to cyclic permutations, we get thus 24 almost generic lists invariant by 15, and they split in two sets of 12 that are deduced from one another by the action of +4 (or 37). These lists may be obtained also as perturbations of more singular lists invariant by 15. The symmetric lists invariant by 15 are enumerated hereafter: 1. If 1 or 5 lies on 234678 (a) One list with eight coconic points: λ0 = 12345678 (b) Four lists with seven coconic points: 5 < 1234678, 5 > 1234678, 1 < 2345678, 1 > 2345678. 2. If 1 > 234678 and 5 < 234678 (a) Four almost generic lists l1 , l2 , l3 , l4 (b) Three lists λ1 , λ2 , λ3 each having two more configurations of six coconic points: λ1 : 123456, 145678; λ2 : 123457, 135678; λ3 : 123458, 125678 3. If 1 < 234678 and 5 > 234678 (a) Four almost generic lists li0 = (+4)(li ) (b) Three lists λ0i = (+4)(λi ) each having two more configurations of six coconic points: λ01 : 123458, 125678; λ02 : 123457, 135678; 0 λ3 : 123456, 145678 4. If 1 > 234678 and 5 > 234678 (a) Eight almost generic lists l5 , l6 , l7 , l8 , l50 , l60 , l70 , l80 , (b) Four lists each having one more configuration of six coconic points: λ4 : 123578; λ5 = (+4)(L4 ) : 134567; λ6 , λ7 = (+4)(L6 ) : 124568 (c) Two lists each having two more configurations of six coconic points: λ8 , λ9 = (+4)(λ8 ): 123567, 134578
Pencils with reducible cubics l1 l2 l3 l4 ˆ 1 6+ ↔ 4− 6+ ↔ 4− 6+ ↔ 4− 6+ ↔ 4− ˆ 2 1− 4− 4+ 4− ˆ 3 1− 4+ 4+ 4− ˆ 4 1− 3+ 1− 3− ˆ 5 1− ↔ 1+ 1− ↔ 1+ 1− ↔ 1+ 1− ↔ 1+ ˆ 6 1+ 7− 1+ 7+ ˆ 7 1+ 6− 6− 6+ ˆ 8 1+ 6+ 6− 6+ l10 l20 l30 l40 ˆ 1 5− ↔ 5+ 5− ↔ 5+ 5− ↔ 5+ 5− ↔ 5+ ˆ 2 5+ 3− 5+ 3+ ˆ 3 5+ 2− 2− 2+ ˆ 4 5+ 2+ 2− 2+ ˆ 5 2+ ↔ 8− 2+ ↔ 8− 2+ ↔ 8− 2+ ↔ 8− ˆ 6 5− 8− 8+ 8− ˆ 7 5− 8+ 8+ 8− ˆ 8 5− 7+ 5− 7− TABLE 7.1 Almost generic symmetric lists, first (d) One list with three more configurations of six coconic points: λ10 = l : 124568, 134578, 123567 5. 1 < 234678 and 5 < 234678 0 0 0 (a) Eight almost generic lists l9 , l10 , l11 , l12 , l90 , l10 , l11 , l12 (b) Four lists each having one more configuration of six coconic points: λ11 : 123578; λ12 = (+4)(λ11 ) : 134567; λ13 , λ14 = (+4)(λ13 ) : 124568 (c) Two lists each having two more configurations of six coconic points: λ15 , λ16 = (+4)(λ15 ) : 123567, 134578 (d) One list having three more configurations of six coconic points: λ17 = (+1)(λ10 ) = (+1)(l) : 124568, 123567, 134578
99
100
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
l5 l6 l7 l8 ˆ 1 5− ↔ 5+ 5− ↔ 5+ 5− ↔ 5+ 5− ↔ 5+ ˆ 2 1+ 7+ 7+ 7− ˆ 3 1+ 7− 7+ 1+ ˆ 4 1+ 1+ 1+ 1+ ˆ 5 1+ ↔ 1− 1+ ↔ 1− 1+ ↔ 1− 1+ ↔ 1− ˆ 6 1− 1− 1− 1− ˆ 7 1− 3+ 3− 1− ˆ 8 1− 3− 3− 3+ l50 l60 l70 l80 ˆ 1 5− ↔ 5+ 5− ↔ 5+ 5− ↔ 5+ 5− ↔ 5+ ˆ 2 5− 5− 5− 5− ˆ 3 5− 7+ 7− 5− ˆ 4 5− 7− 7− 7+ ˆ 5 1+ ↔ 1− 1+ ↔ 1− 1+ ↔ 1− 1+ ↔ 1− ˆ 6 5+ 3+ 3+ 3− ˆ 7 5+ 3− 3+ 5+ ˆ 8 5+ 5+ 5+ 5+ TABLE 7.2 Almost generic symmetric lists, continued
Pencils with reducible cubics
l9 l10 l11 l12 ˆ 1 6+ ↔ 4− 6+ ↔ 4− 6+ ↔ 4− 6+ ↔ 4− ˆ 2 8+ 6− 6+ 8− ˆ 3 8− 8+ 8+ 8− ˆ 4 8− 8− 8+ 8− ˆ 5 8− ↔ 2+ 8− ↔ 2+ 8− ↔ 2+ 8− ↔ 2+ ˆ 6 2+ 2+ 2− 2+ ˆ 7 2+ 2− 2− 2+ ˆ 8 2− 4+ 4− 2+ 0 0 0 l90 l10 l11 l12 ˆ 1 6+ ↔ 4− 6+ ↔ 4− 6+ ↔ 4− 6+ ↔ 4− ˆ 2 6+ 6+ 6− 6+ ˆ 3 6+ 6− 6− 6+ ˆ 4 6− 8+ 8− 6+ ˆ 5 8− ↔ 2+ 8− ↔ 2+ 8− ↔ 2+ 8− ↔ 2+ ˆ 6 4+ 2− 2+ 4− ˆ 7 4− 4+ 4+ 4− ˆ 8 4− 4− 4+ 4−
TABLE 7.3 Almost generic symmetric lists, end
101
8 Classification of the pencils of cubics
CONTENTS 8.1 8.2
Nodal pencils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Inductive constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1
Nodal pencils
103 106
We will use the method exposed in Section 6.2 to construct the nodal pencils. Let T be the triangle (12), (67), (17) containing 8. The condition that 1, . . . , 8 realizes the list max(ˆ 1 = 8−) splits into eight disjoint subconditions. There is an ordering 14567 > (ˆ 8, 1) > (ˆ8, 7) > (ˆ8, 6) > · · · > (ˆ8, 2) > 34567 in T : the point 8 lies between two consecutive of these curves. When 8 lies on a cubic (ˆ 8, N ) (otherwise stated, the eight chosen base points are on a cubic (1−, N )n ), the pencil is singular, with 9 = N . By Bezout’s theorem with the cubics (ˆ 8, N ), the degeneration 9 = 8 may occur only if 8 lies between (ˆ8, 1) ˆ and (8, 7). Start with 8 between 14567 and (ˆ8, 1); the elementary change 8 goes to the outside of 14567, 2, 3 < 14567 writes: ˆ2 : 1− → 4+, ˆ3 : 1− → 4+, it is realizable. The essential elementary pair (ˆ2 = 1−, ˆ3 = 1−) gives thus two cubics: (1−, 3), (12, L). The ninth base point 9 must lie on the odd component of (12, L). By Bezout’s theorem, 9 lies then on the arc 1X of (1−, 3). Move 3 toward 4 along the loop of (12, L), leaving the other seven chosen base points fixed. The ninth point 9 moves along the cubic but cannot cross X (otherwise 9 should come together with 1 or 2 at X). So 9 stays on the odd component of (12, L). So the first degeneration of the pencil occurs when 3 reaches 4 (as 3 → 4 = 0), and the close pair (ˆ3 = 1−, ˆ4 = 1−) gives two admissible pairs of distinguished cubics: (3, 1−), (1−, 4) or (1−, 3), (4, 1−). We know already that (1−, 3) belongs to the pencil, so the correct pair is the first one. Using the other pairs of consecutive points A, B with A → B = 0, we get other cubics: the pair 4, 5 gives again (4, 1−) and a new cubic (1−, 5), the pairs 5, 6 and 6, 7 give (6, 1−) and (1−, 7). Note that the points 2, 3 verify 2 → 3 = 0 but the corresponding close pair (ˆ2 = 1−, ˆ3 = 1−) is inessential. The position of 9 on each cubic is again obtained with Bezout’s theorem. Until now we have found six cubics. Note that all of them have an arc 81, so the missing cubics correspond to the openings of this arc. Starting from (12, L), move into 103
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
the portion formed of cubics with ovals. The next distinguished cubic will be (81, L), then it will be (81, C). The complete pencil is shown in the first row of Table 8.1. Let now 8 follow a path p crossing successively (ˆ8, 1) and (ˆ8, 7). We will prove that at some moment, 8 must indeed come together with 9. Letting 8 cross (ˆ 8, 1) changes the pencil swapping the positions of 1 and 9 on all combinatorial cubics, see the second row of Table 8.1. The point 8 lies now in the zone between the arcs 17 of (ˆ8, 1) and (ˆ8, 7), and this zone is divided in an upper and a lower part by the path p. Applying Bezout’s theorem with (ˆ 8, 1) and (1−, 81), 2X on one hand, and with (ˆ8, 7) and (1−, 7), 18 on the other hand, we get that 9 lies also in this zone. When 8 follows the path p from (ˆ 8, 1) and (ˆ 8, 7), the point 9 moves from 1 to 7. Let 9 cross p at some point P and assume that at this moment, 8 has the position Q. Assume first that Q lies ahead of P , see Figure 8.1. When 8 previously had the position P , the point 9 was still in the upper zone. But on the other hand, 9 must have been placed at Q (recall that the positions of eight base points determine the position of the ninth one). This is a contradiction. One gets a similar contradiction taking Q behind P on the path. So, at some moment, one must have 9 = 8. Letting 8 successively cross (ˆ8, 1), . . . (ˆ8, 2), we obtain in total the first nine pencils of Table 8.1, they are drawn in Figure 8.2 where (G, H, I, A, B, C, D, E, F ) = (9, 2, 3, 4, 5, 6, 7, 8, 1), (1, 2, 3, 4, 5, 6, 7, 8, 9), (1, 2, 3, 4, 5, 6, 7, 9, 8), (1, 2, 3, 4, 5, 6, 9, 7, 8), (1, 2, 3, 4, 5, 9, 6, 7, 8), (1, 2, 3, 4, 9, 5, 6, 7, 8), (1, 2, 3, 9, 4, 5, 6, 7, 8), (1, 2, 9, 3, 4, 5, 6, 7, 8), (1, 9, 2, 3, 4, 5, 6, 7, 8). If 8 lies outside of the loop of (ˆ8, 1), then one gets the pencil with G = 9. If 8 lies between (ˆ 8, 1) and (ˆ 8, 7), one gets the next two pencils with (E, F ) = (8, 9) and (9, 8). The other positions of 8 give rise to the other pencils, switching successively 9 with 7, 6, . . . , 2. 1 8 P
Q 2
7 3
6 5
4
FIGURE 8.1 The path p Let 1, . . . 8 realize max(ˆ1 = 8−). The inessential elementary change ˆ1 :
Classification of the pencils of cubics G
G
E D
D C B A
H
F
E
I A
C
C
B
D I
H
B A
I
E
G
F
E C
C
H I
D
I
B
D
G H
B A
I
C
H B
C
A F
G
E F H
D C
A
B
G
F
E
H
F
D
F G
E
105
D
E
B
I
A F
G I
H
A
FIGURE 8.2 Nodal pencils obtained from the list max(ˆ1 = 8−) 8− → 8+, ˆ 8 : 1− → 1+ (7 enters the conic 23456, 1, 8 outside of 23456) is realizable if and only if the points realize one of the first three nodal pencils. Let 1, . . . 8 realize one of these pencils; we may move 8 leaving the other points fixed, till the eight points lie on a cubic (1±, 1)n or (1−, 8)n , see Figure 6.1. The conic 23456 has a sixth intersection point P with the loop of the cubic. The point P lies on some arc AB where A and B are two consecutive points among 1, . . . 7. If we move B towards A on the cubic, P moves in the opposite direction; at some moment one has P = B (the cubic and the conic have an ordinary tangency point), then the positions of the two points are swapped. So, up to some swaps like this, we may arrange that P lies on arc 67 of the cubic. Move 7 towards 6 along the cubic till 7 crosses 23456, and we are done. Let now 1, . . . 8 realize one of the other six pencils. The sixth intersection point of 23456 with the cubic (7, 1−) is on the loop of this cubic, whereas 7 is on the odd component. The elementary change is not realizable. The elementary change ˆ1 : 8− → 8+, ˆ8 : 1− → 1+ replaces max(ˆ1 = 8−) by max(ˆ 1 = 8+), and leaves the combinatorial pencil unchanged. The nodal list max(ˆ 1 = 8+) is thus realizable by the three pencils of Figure 8.2 with (G, H, I, A, B, C, D, E, F ) = (9, 2, 3, 4, 5, 6, 7, 8, 1), (1, 2, 3, 4, 5, 6, 7, 8, 9), (1, 2, 3, 4, 5, 6, 7, 9, 8)
The elementary change 1ˆ : 8+ → 6−, 7ˆ : 1− → 1+ (8 enters the conic 23456, 7 inside and 1 outside of 23456) is realizable only for the first two pencils. For both, perform the change of pairs: (1−, 7), (81, C) → (1+, 7), (1, 6−), one gets the two pencils corresponding to the nodal list max(ˆ1 = 6−). The elementary change ˆ 1 : 6− → 6+, ˆ6 : 1− → 1+ (8 enters the conic 23457, 1, 6 outside of 23457) may be performed on both previous pencils replacing
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
the pair (6, 1−), (1, 6−) by the pair (6, 1+), (1, 6+); the new pencils obtained realize the nodal list max(ˆ1 = 6+).
8.2
Inductive constructions
We will now classify the pencils of cubics with eight base points lying in convex position, up to the action of D8 on these points. Representatives of each of the 43 equivalence classes obtained are shown in Tables 8.18.3. The upper nine pencils in Table 8.1 are the nodal pencils obtained from the list max(ˆ 1 = 8−) = (15) · L64 . The first three of them correspond also to the list max(ˆ 1 = 8+) = L32 . The next two pencils after the blank line correspond to the list max(ˆ 1 = 6−) = L48 . The last two pencils after the second blank line correspond to the list max(ˆ1 = 6+) = L56 . The nodal lists give rise in total to 13 orbits of pencils, see Table 8.1. Recall that two lists that are obtained from one another by an inessential change must be both nodal or both nonnodal. For any nonnodal list Ln , denote by Pn the corresponding pencil of cubics. Hereafter, (ˆ 5, ˆ 6) always stands for the same elementary change (ˆ5 : 6− → 6+, ˆ 6 : 5− → 5+); and (ˆ1, ˆ8) stands for (ˆ1 : 8+ → 8−, ˆ8 : 1+ → 1−). They are both inessential. One has: (ˆ 5, ˆ 6) · L3 = L4 , 26 ◦ (+1)(48) ◦ (ˆ 1, ˆ 8) · L4 = (+3) ◦ (ˆ1, ˆ8) · L4 = L15 , (ˆ 5, ˆ 6) ◦ 37 · L3 = (ˆ 5, ˆ 6) · L17 = L18 . Thus, P4 = P3 ; P15 = (+3) · P4 ; P18 = 37 · P3 . One has: (ˆ 5, ˆ 6) · L5 = L6 , ˆ (5, ˆ 6) ◦ 37 · L5 = (ˆ 5, ˆ 6) · L9 = L10 , ˆ 26 ◦ (+1)(48) ◦ (1, ˆ 8) · L6 = (+3) ◦ (ˆ1, ˆ8) · L6 = L23 . Thus, P6 = P5 ; P10 = 37·P5 ; P23 = (+3)·P6 . One has (ˆ5, ˆ6)◦37·L2 = L34 . Thus, P34 = 37 · P2 . Finally, (+1)(48) ◦ (ˆ1, ˆ8) · Ln = Lm for (n, m) = (7, 26), (8, 25), (11, 22), (12, 21), (13, 20), (14, 19). Thus, Pm = (+1)(48) · Pn . Let us say that two nonnodal lists are similar if the pencils that they determine coincide up to the action of D8 . Hereafter we denote briefly by n the list Ln . The 43 principal nonnodal lists are distributed in 30 equivalence classes for the similarity: two classes contain four elements each, (3, 4, 15, 18) and (5, 6, 10, 23), seven classes contain two elements (2, 34), (7, 26), (8, 25), (11, 22), (12, 21), (13, 20), (14, 19) each, and 21 classes each contain one single element. So the nonnodal lists give rise to exactly 30 equivalence classes of pencils, see Tables 8.28.3. To construct the nonnodal pencils in the easiest way, we may follow pieces of the sequences from Table 5.14, with four starting lists: L2 , L65 , L71 , L88 . Construct directly the starting pencils corresponding to L65 , L71 , L88 using
Classification of the pencils of cubics
107
again the method exposed in Section 6.2. To get the starting pencil corresponding to L2 in the shortest way, observe that the list L2 is obtained from the (nonprincipal and nodal) list L1 by an elementary change (ˆ6, ˆ7). The list L1 = max(ˆ 8 = 1+) = (+1)(48) max(ˆ1 = 8−) corresponds to nine pencils. However, the elementary change (ˆ6, ˆ7) is realizable only for the first of them, shown in the first row of Table 8.2.
(81, L) XX (1−, 18) 2X (81, L) XX (1−, 18) 2X
(12, L) XX (1+, 12) 8X (12, L) XX (1+, 12) 8X
(1+, 7) X1 (1+, 7) 12
(1+, 7) X1 (1+, 7) 12
(81, C) 18 (81, C) X1 (1−, 78) X1 (7, 1−) X6 (7, 1−) 65 (7, 1−) 54 (7, 1−) 43 (7, 1−) 32 (7, 1−) 2X
(6, 1+) 61 (6, 1+) 12
(1, 6−) 16 (1, 6−) X1
(1−, 7) X1 (1−, 7) 18 (1−, 7) 8X (1−, 67) 2X (1−, 6) X5 (1−, 6) 54 (1−, 6) 43 (1−, 6) 32 (1−, 6) 26
(1, 6+) 16 (1, 6+) X1
(6, 1−) 61 (6, 1−) 18
(6, 1−) 61 (6, 1−) 18 (6, 1−) 87 (6, 1−) 7X (1−, 56) X1 (5, 1−) X4 (5, 1−) 43 (5, 1−) 32 (5, 1−) 2X
TABLE 8.1 Pencils max(ˆ 1 = 8−), max(ˆ 1 = 8+), max(ˆ1 = 6−), max(ˆ1 = 6+)
(81, L) XX (1−, 81) 2X (1−, 8) X7 (1−, 8) 76 (1−, 8) 65 (1−, 8) 54 (1−, 8) 43 (1−, 8) 32 (1−, 8) 28
(12, L) XX (1+, 12) 8X (1+, 12) 78 (1+, 12) 67 (1+, 12) 56 (1+, 12) 45 (1+, 12) 34 (1+, 12) 23 (1+, 12) X2
(1−, 5) X1 (1−, 5) 18
(1−, 5) X1 (1−, 5) 18
(1−, 5) X1 (1−, 5) 18 (1−, 5) 87 (1−, 5) 76 (1−, 5) 6X (1−, 45) 2X (1−, 4) X3 (1−, 4) 32 (1−, 4) 24
(4, 1−) 41 (4, 1−) 18
(4, 1−) 41 (4, 1−) 18
(4, 1−) 41 (4, 1−) 18 (4, 1−) 87 (4, 1−) 76 (4, 1−) 65 (4, 1−) 5X (1−, 34) X1 (3, 1−) X2 (3, 1−) 2X
(1−, 3) X1 (1−, 3) 18
(1−, 3) X1 (1−, 3) 18
(1−, 3) X1 (1−, 3) 18 (1−, 3) 87 (1−, 3) 76 (1−, 3) 65 (1−, 3) 54 (1−, 3) 4X (1−, E) 2X (1−, E) X2
108 Pencils of Cubics and Algebraic Curves in the Real Projective Plane
TABLE 8.2 Nonnodal pencils with ˆ 8 = 1+
(78, L) XX (56, L) XX (56, L) XX (56, L) XX (56, L) XX (34, L) XX (18, L) XX (18, L) XX (18, L) XX (18, L) XX (18, L) XX (18, L) XX (18, L) XX (18, L) XX (18, L) XX (18, L) XX
(81, L) XX (81, L) XX (81, L) XX (81, L) XX (81, L) XX (81, L) XX (65, L) XX (34, L) XX (34, L) XX (65, L) XX (65, L) XX (34, L) XX (34, L) XX (65, L) XX (65, L) XX (65, L) XX
(81, C) 18 (81, C) 18 (81, C) 18 (81, C) 18 (81, C) 18 (81, C) 18 (6+, 4) X6 (3−, 5) X3 (3−, 5) X3 (6+, 4) X6 (6+, 4) X6 (3−, 5) X3 (3−, 5) X3 (6+, 4) X6 (6+, 4) X6 (6+, 4) X6
(8+, 2) X8 (8+, 2) X8 (8+, 2) X8 (8+, 2) X8 (8+, 2) X8 (8+, 2) X8 (6, 3−) 63 (6, 3−) 63 (3, 6+) 36 (3, 6+) 36 (6, 3−) 63 (6, 3−) 63 (3, 6+) 36 (3, 6+) 36 (3, 6+) 36 (3, 6+) 36
(3, 8+) 38 (3, 8+) 38 (3, 8+) 38 (3, 8+) 38 (3, 8+) 38 (3, 8+) 38 (3, 6−) 36 (3, 6−) 36 (6, 3+) 63 (6, 3+) 63 (3, 6−) 36 (3, 6−) 36 (6, 3+) 63 (6, 3+) 63 (6+, 2) X6 (6+, 2) X6
(8+, 4) X8 (8+, 4) X8 (8+, 4) X8 (3−, 7) X3 (3−, 7) X3 (3−, 7) X3 (3+, 7) X3 (3+, 7) X3 (3+, 7) X3 (3+, 7) X3 (6−, 2) X6 (6−, 2) X6 (6−, 2) X6 (6−, 2) X6 (6, 1−) 61 (1, 6+) 16
(5, 8+) 58 (5, 8+) 58 (5+, 7) X5 (6−, 4) X6 (6, 3−) 63 (6, 3−) 63 (8+, 2) X8 (8+, 2) X8 (8+, 2) X8 (8+, 2) X8 (1, 6−) 16 (1, 6−) 16 (1, 6−) 16 (1, 6−) 16 (1, 6−) 16 (6, 1+) 61
(8+, 6) 1 X8 (5−, 7) 2 X5 (56, C) 3, 4 65 (56, C) 6, 5 65 (6+, 4) 7 X6 (3−, 5) 8 X3 (18, C) 11 81 (18, C) 12 81 (18, C) 14 81 (18, C) 13 81 (1+, 7) 35 X1 (1+, 7) 36 X1 (1+, 7) 38 X1 (1+, 7) 37 X1 (1+, 7) 41 X1 (1+, 7) 49 X1 Classification of the pencils of cubics 109
TABLE 8.3 Nonnodal pencils with ˆ 8= 6 1+
(1+, 2) 82 (1+, 2) 82 (1+, 2) 82 (4, 2+) 1X (4−, 1) 51 (4, 2+) 1X (4, 2+) 1X (4−, 1) 51 (4−, 1) 51 (4−, 1) 51 (4−, 1) 51 (8−, 5) 15 (8−, 5) 15 (5, 1−) 2X (1, 5−) 6X
(3, 1+) 8X (3, 1+) 8X (3, 1+) 8X (2+, 5) 15 (2+, 5) 15 (2+, 5) 15 (2+, 5) 15 (2+, 5) 15 (2+, 5) 15 (2+, 5) 15 (2, 4−) 5X (6+, 1) 51 (6+, 1) 51 (1, 5+) 4X (5, 1+) 8X
(1+, 4) 84 (1+, 4) 84 (1+, 4) 84 (6, 2+) 1X (6, 2+) 1X (6, 2+) 1X (6, 2+) 1X (6, 2+) 1X (6, 2+) 1X (2, 6+) 5X (2−, 5) 35 (2, 6+) 5X (6, 2+) 1X (1+, 4) 84 (1+, 4) 84
(5, 1+) 8X (5, 1+) 8X (5, 1+) 8X (2+, 7) 17 (2+, 7) 17 (2+, 7) 17 (2, 6−) 7X (2+, 7) 17 (2, 6−) 7X (6, 2−) 3X (6, 2−) 3X (6, 2−) 3X (2, 6−) 7X (3−, 8) 48 (3−, 8) 48
(1+, 6) 86 (1+, 6) 86 (1, 5−) 6X (8, 2+) 1X (8, 2+) 1X (2, 8+) 7X (2−, 7) 37 (2, 8+) 7X (2−, 7) 37 (2−, 7) 37 (2−, 7) 37 (2−, 7) 37 (2−, 7) 37 (3, 7−) 8X (7, 3−) 4X
(7, 1+) 8X (1, 7+) 6X (1−, 6) 26 (2, 8−) 1X (2, 8−) 1X (8, 2−) 3X (8, 2−) 3X (8, 2−) 3X (8, 2−) 3X (8, 2−) 3X (8, 2−) 3X (8, 2−) 3X (8, 2−) 3X (7, 3+) 2X (3, 7+) 6X
(1, 7−) 8X (7, 1−) 2X (7, 1−) 2X (8−, 3) 13 (8−, 3) 13 (8−, 3) 13 (8−, 3) 13 (8−, 3) 13 (8−, 3) 13 (8−, 3) 13 (8−, 3) 13 (8−, 3) 13 (8−, 3) 13 (7+, 2) 62 (7+, 2) 62
(1−, 8) 28 (1−, 8) 28 (1−, 8) 28 (4+, 1) 31 (4, 8−) 1X (4+, 1) 31 (4+, 1) 31 (4, 8−) 1X (4, 8−) 1X (4, 8−) 1X (4, 8−) 1X (4, 8−) 1X (4, 8−) 1X (1−, 6) 26 (1−, 6) 26 92
88
82
83
84
87
86
80
78
75
72
71
67
66
65 110 Pencils of Cubics and Algebraic Curves in the Real Projective Plane
9 Tables
111
TABLE 9.1 The lists ˆ 8 = L(1, . . . 7)
ˆ 8 C2 23456 23457 23467 23567 24567 34567 13456 13457 13467 13567 14567 12456 12457 12467 12567 12356 12357 12367 12346 12347 12345 1+ out 1 1, 6 5 1 1, 4 3 1 1, 2 2, 7 2 6 2, 5 2 4 2, 3 7 3 3, 6 5 3 3, 4 4, 7 4 6 4, 5 7 5 5, 6 6, 7 in 7
1− out 1, 7 6 1 1, 5 4 1 1, 3 2 1 2, 7 2 6 2, 5 2 4 2, 3 7 3 3, 6 5 3 3, 4 4, 7 4 6 4, 5 7 5 5, 6 6, 7 in
2+ in out 1, 7 1 6 1, 5 1 4 1, 3 1 2 2, 7 6 2 2, 5 4 2 2, 3 3 7 3, 6 3 5 3, 4 4, 7 6 4 4, 5 5 7 5, 6 6, 7
2− in out 1, 7 1 6 1, 5 1 4 1, 3 1, 2 7 2 2, 6 5 2 2, 4 3 2 3 7 3, 6 3 5 3, 4 4, 7 6 4 4, 5 5 7 5, 6 6, 7
3+ out 1, 7 6 1 1, 5 4 1 1, 3 1, 2 2 7 2, 6 2 5 2, 4 2 3 7 3 3, 6 5 3 3, 4 4, 7 4 6 4, 5 7 5 5, 6 6, 7 in
3− out 1, 7 6 1 1, 5 4 1 1 3 1, 2 2 7 2, 6 2 5 2, 4 2, 3 3, 7 6 3 3, 5 4 3 4, 7 4 6 4, 5 7 5 5, 6 6, 7 in
4+ in out 1, 7 1 6 1, 5 1 4 3 1 1, 2 7 2 2, 6 5 2 2, 4 2, 3 3, 7 3 6 3, 5 3 4 4, 7 6 4 4, 5 5 7 5, 6 6, 7
112 Pencils of Cubics and Algebraic Curves in the Real Projective Plane
4− in out 1, 7 1 6 1, 5 1, 4 3 1 1, 2 7 2 2, 6 5 2 2 4 2, 3 3, 7 3 6 3, 5 3, 4 7 4 4, 6 5 4 5 7 5, 6 6, 7
TABLE 9.2 The lists ˆ 8 = L(1, . . . 7), continued
ˆ 8 C2 23456 23457 23467 23567 24567 34567 13456 13457 13467 13567 14567 12456 12457 12467 12567 12356 12357 12367 12346 12347 12345 5+ out 1, 7 6 1 1, 5 1, 4 1 3 1, 2 2 7 2, 6 2 5 4 2 2, 3 3, 7 6 3 3, 5 3, 4 4 7 4, 6 4 5 7 5 5, 6 6, 7 in
5− out 1, 7 6 1 1 5 1, 4 1 3 1, 2 2 7 2, 6 2, 5 4 2 2, 3 3, 7 6 3 3 5 3, 4 4 7 4, 6 4, 5 5, 7 6 5 6, 7 in
6+ in out 1, 7 1 6 5 1 1, 4 3 1 1, 2 7 2 2, 6 2, 5 2 4 2, 3 3, 7 3 6 5 3 3, 4 7 4 4, 6 4, 5 5, 7 5 6 6, 7
6− in out 1, 7 1, 6 5 1 1, 4 3 1 1, 2 7 2 2 6 2, 5 2 4 2, 3 3, 7 3, 6 5 3 3, 4 7 4 4 6 4, 5 5, 7 5, 6 7 6
7+ out 1, 7 1, 6 1 5 1, 4 1 3 1, 2 2 7 6 2 2, 5 4 2 2, 3 3, 7 3, 6 3 5 3, 4 4 7 6 4 4, 5 5, 7 5, 6 6 7 in
7− in out 1 7 1, 6 1 5 1, 4 1 3 1, 2 2, 7 6 2 2, 5 4 2 2, 3 3 7 3, 6 3 5 3, 4 4, 7 6 4 4, 5 5 7 5, 6 6, 7
Tables 113
TABLE 9.3 ˆ 8 = 1+
ˆ 7
ˆ 6
ˆ 5
ˆ 4
ˆ 3
ˆ 2
ˆ 1
6−
6−
6−
6−
6−
5−
5−
/ 23456 O / 13456 O / 12456 O / 12356 O / 12346 O / 12345
/ 12345
8+
8+
8+
8+
8+
8+
8+
/ 12346
/ 12347 O
/ 12347
/ 12357 O
/ 12457 O
/ 13457 O
/ 23457 O
5+
5+
6+
6+
6+
6+
6+
/ 12356
/ 12357 O
/ 12367 O
/ 12367
/ 12467 O
/ 13467 O
/ 23467 O
3−
3−
3−
3−
4−
4−
4−
/ 12456
/ 12457 O
/ 12467 O
/ 12567 O
/ 12567
/ 13567 O
/ 23567 O
3+
3+
3+
3+
4+
4+
4+
/ 13456
/ 13457 O
/ 13467 O
/ 13567 O
/ 14567 O
/ 14567
/ 24567 O
1−
1−
1−
1−
1−
1−
2−
/ 23456
/ 23457 O
/ 23467 O
/ 23567 O
/ 24567 O
/ 34567 O
/ 34567
1+
1+
1+
1+
1+
1+
2+
/ˆ 7
/ˆ 6
/ˆ 5
/ˆ 4
/ˆ 3
/ˆ 2
/ˆ 1
114 Pencils of Cubics and Algebraic Curves in the Real Projective Plane
TABLE 9.4 ˆ 8 = 1−
ˆ 7
ˆ 6
ˆ 5
ˆ 4
ˆ 3
ˆ 2
ˆ 1
3+
6−
6−
6−
6−
5−
5−
/ 34567 O / 13456 O / 12456 O / 12356 O / 12346 O / 12345
/ 12345
8−
8+
8+
8+
8+
8+
8+
/ 12346
/ 12347 O
/ 12347
/ 12357 O
/ 12457 O
/ 13457 O
/ 24567 O
5+
5+
6+
6+
6+
6+
3−
/ 12356
/ 12357 O
/ 12367 O
/ 12367
/ 12467 O
/ 13467 O
/ 23567 O
3−
3−
3−
3−
4−
4−
5+
/ 12456
/ 12457 O
/ 12467 O
/ 12567 O
/ 12567
/ 13567 O
/ 23467 O
3+
3+
3+
3+
4+
4+
5−
/ 13456
/ 13457 O
/ 13467 O
/ 13567 O
/ 14567 O
/ 14567
/ 23457 O
1−
1−
1−
1−
1−
1−
7+
1+
1+
1+
1+
1+
/ 23567 / 23467 / 23457 / 23456
1+
7−
/ 24567
/ 34567
/ 23456 O
/ˆ 7
/ˆ 6
/ˆ 5
/ˆ 4
/ˆ 3
/ˆ 2
/ˆ 1
Tables 115
TABLE 9.5 ˆ 8 = 2+
ˆ 7
ˆ 6
ˆ 5
ˆ 4
ˆ 3
ˆ 2
ˆ 1
2+
2+
2+
2+
2+
/ 23567 O / 23467 O / 23457 O / 23456
8−
8−
8−
8−
3+
6−
/ 24567 O
/ 34567 O
/ 23456
8−
8−
8+
2−
2−
2−
2−
/ 13467 / 13457 / 13456
2−
3−
6+
/ 13567
/ 14567
/ 14567
/ 23457
/ 12456
/ 12457
/ 12467
4+
4+
4+
5+
5+
/ 12567
/ 12567
5+
4−
/ 13567
/ 23467
/ 12356
/ 12357
/ 12367
/ 12367
/ 12467
/ 13467
/ 23567
4−
4−
4−
5−
5−
5−
4+
/ 12346
/ 12347
/ 12347
/ 12357
/ 12457
/ 13457
/ 24567
6+
7+
7+
7+
7+
7+
2−
/ 12345
/ 12345
/ 12346
/ 12356
/ 12456
/ 13456
/ 34567
6−
7−
7−
7−
7−
7−
2+
/ˆ 7
/ˆ 6
/ˆ 5
/ˆ 4
/ˆ 3
/ˆ 2
/ˆ 1
116 Pencils of Cubics and Algebraic Curves in the Real Projective Plane
TABLE 9.6 ˆ 8 = 2−
ˆ 7
ˆ 6
ˆ 5
ˆ 4
ˆ 3
ˆ 2
ˆ 1
2+
2+
2+
2+
2+
/ 23567 O / 23467 O / 23457 O / 23456
8−
8−
8−
8−
6−
6−
/ 24567 O
/ 13456
/ 23456
8−
8+
8+
/ 13456
/ 13457 O
/ 13467 O
2−
2−
2−
2−
2−
/ 14567 O / 13567 O
6+
6+
/ 13457
/ 23457
4+
4+
4+
/ 12457 / 12456
5+
5+
4−
4−
/ 12467
/ 12567
/ 12567
/ 13467
/ 23467
/ 12356
/ 12357
4−
4−
4−
5−
/ 12367
/ 12367
5−
4+
4+
/ 12467
/ 13567
/ 23567
/ 12346
/ 12347
/ 12347
/ 12357
/ 12457
/ 14567
/ 24567
6+
7+
7+
7+
7+
1−
2−
/ 12345
/ 12345
/ 12346
/ 12356
/ 12456
/ 34567
/ 34567
6−
7−
7−
7−
7−
1+
2+
/ˆ 7
/ˆ 6
/ˆ 5
/ˆ 4
/ˆ 3
/ˆ 2
/ˆ 1
Tables 117
TABLE 9.7 ˆ 8 = 3+
ˆ 7
ˆ 6
ˆ 5
ˆ 4
ˆ 3
ˆ 2
ˆ 1
3+
3+
6−
6−
6−
5−
5−
/ 34567
/ 34567
/ 12456 O / 12356 O / 12346 O / 12345
/ 12345
8−
8−
8+
8+
8+
8+
8+
/ 12346
/ 12347 O
/ 12347
/ 12357 O
/ 12457 O
/ 14567 O
/ 24567 O
5+
5+
6+
6+
6+
3−
3−
/ 12356
/ 12357 O
/ 12367 O
/ 12367
/ 12467 O
/ 13567 O
/ 23567 O
3−
3−
3−
3−
4−
5+
5+
/ 12456
/ 12457 O
/ 12467 O
/ 12567 O
/ 12567
/ 13467 O
/ 23467 O
3+
3+
3+
3+
4+
5−
5−
1−
1−
1−
1−
/ 13467 / 13457 / 13456
1−
7+
7+
/ 13567
/ 14567
/ 13457 O
/ 23457 O
/ 23456
/ 23457
/ 23467
/ 23567
/ 24567
/ 13456
/ 23456 O
1+
1+
1+
1+
1+
7−
7−
/ˆ 7
/ˆ 6
/ˆ 5
/ˆ 4
/ˆ 3
/ˆ 2
/ˆ 1
118 Pencils of Cubics and Algebraic Curves in the Real Projective Plane
TABLE 9.8 ˆ 8 = 3−
ˆ 7
ˆ 6
ˆ 5
ˆ 4
ˆ 3
ˆ 2
ˆ 1
6−
6−
5−
5−
/ 12356 O / 12346 O / 12345
/ 12345
8+
8+
8+
2+
8+
/ 24567
3+
/ 34567
8−
8−
3+
/ 34567
8−
/ 12346
/ 12347 O
/ 12347
/ 12357 O
/ 14567
/ 14567
/ 24567
5+
5+
6+
6+
2−
3−
3−
/ 12356
/ 12357 O
/ 12367 O
/ 12367
/ 12567 O
/ 13567 O
/ 23567 O
3−
3−
3−
3−
5+
5+
5+
3+
3+
3+
/ 12457 / 12456
3+
5−
5−
5−
/ 12467
/ 12567
/ 12467 O
/ 13467 O
/ 23467 O
/ 13456
/ 13457
/ 13467
/ 13567
/ 12457
/ 13457 O
/ 23457 O
1−
1−
1−
1−
7+
7+
7+
/ 23456
/ 23457
/ 23467
/ 23567
/ 12456
/ 13456 O
/ 23456 O
1+
1+
1+
1+
7−
7−
7−
/ˆ 7
/ˆ 6
/ˆ 5
/ˆ 4
/ˆ 3
/ˆ 2
/ˆ 1
Tables 119
TABLE 9.9 ˆ 8 = 4+
ˆ 7
ˆ 6
ˆ 5
ˆ 4
ˆ 3
ˆ 2
ˆ 1
2+
2+
2+
2+
/ 23567 O / 23467 O / 23457 O / 23456
8−
8−
8−
6−
/ 12456
8+
8−
6−
6−
/ 13456
/ 23456
8+
8+
/ 13456
/ 13457 O
/ 13467 O
/ 13567 O
/ 12457
/ 13457
/ 23457
2−
2−
2−
2−
6+
6+
6+
/ 12456
/ 12457 O 4+
4+
4+
5+
/ 12567 O / 12467 O
4−
4−
4−
/ 12467
/ 13467
/ 23467
4−
4−
/ 12356
4−
5−
4+
4+
4+
/ 12357
/ 12367
/ 12367
/ 12567
/ 13567
/ 23567
/ 12346
/ 12347
/ 12347
/ 12357
/ 14567
/ 14567
/ 24567 O
6+
7+
7+
7+
1−
1−
2−
/ 12345
/ 12345
/ 12346
/ 12356
/ 24567
/ 34567 O
/ 34567
6−
7−
7−
7−
1+
1+
2+
/ˆ 7
/ˆ 6
/ˆ 5
/ˆ 4
/ˆ 3
/ˆ 2
/ˆ 1
120 Pencils of Cubics and Algebraic Curves in the Real Projective Plane
Tables
121 ˆ 1 3+ ↔ 8− 3+ ↔ 8− 2+ ↔ 2− 2+ ↔ 2− ˆ 2 3+ ↔ 8− 1+ ↔ 1− 3+ ↔ 8− 1+ ↔ 1− ˆ 1 3− ↔ 3+ 3− ↔ 3+ 2− ↔ 4+ ˆ 2+ ↔ 8− 1+ ↔ 1− 2+ ↔ 8− 3
2− ↔ 4+ 1+ ↔ 1−
ˆ 1 5+ ↔ 3− 5+ ↔ 3− 4+ ↔ 4− ˆ 4 2+ ↔ 8− 1+ ↔ 1− 2+ ↔ 8−
4+ ↔ 4− 1+ ↔ 1−
ˆ 1 5− ↔ 5+ 5− ↔ 5+ ˆ 2+ ↔ 8− 1+ ↔ 1− 5
4− ↔ 6+ 2+ ↔ 8−
4− ↔ 6+ 1+ ↔ 1−
ˆ 1 7+ ↔ 5− ˆ 6 2+ ↔ 8−
7+ ↔ 5− 1+ ↔ 1−
6+ ↔ 6− 2+ ↔ 8−
6+ ↔ 6− 1+ ↔ 1−
ˆ 1 7− ↔ 7+ ˆ 7 2+ ↔ 8−
7− ↔ 7+ 1+ ↔ 1−
6− ↔ 8+ 6− ↔ 8+ 2+ ↔ 8− 1+ ↔ 1−
ˆ 1 2+ ↔ 7− ˆ 8 2+ ↔ 7−
2+ ↔ 7− 1+ ↔ 1−
8+ ↔ 8− 8+ ↔ 8− 2+ ↔ 7− 1+ ↔ 1−
ˆ 2 3− ↔ 3+ 3− ↔ 3+ 1− ↔ 4+ 1− ↔ 4+ ˆ 3 1− ↔ 4+ 2− ↔ 2+ 1− ↔ 4+ 2− ↔ 2+ ˆ 2 5+ ↔ 3− ˆ 4 1− ↔ 3+
5+ ↔ 3− 2− ↔ 2+
4+ ↔ 4− 4+ ↔ 4− 1− ↔ 3+ 2− ↔ 2+
ˆ 2 5− ↔ 5+ 5− ↔ 5+ 4− ↔ 6+ 4− ↔ 6+ ˆ 5 1− ↔ 3+ 2− ↔ 2+ 1− ↔ 3+ 2− ↔ 2+ ˆ 2 7+ ↔ 5− ˆ 6 1− ↔ 3+
7+ ↔ 5− 2− ↔ 2+
6+ ↔ 6− 6+ ↔ 6− 1− ↔ 3+ 2− ↔ 2+
ˆ 2 7− ↔ 7+ 7− ↔ 7+ 6− ↔ 8+ 6− ↔ 8+ ˆ 7 1− ↔ 3+ 2− ↔ 2+ 1− ↔ 3+ 2− ↔ 2+ ˆ 2 1+ ↔ 7− ˆ 8 1− ↔ 3+
1+ ↔ 7− 2− ↔ 2+
8+ ↔ 8− 8+ ↔ 8− 1− ↔ 3+ 2− ↔ 2+
ˆ 3 4− ↔ 4+ 4− ↔ 4+ 2− ↔ 5+ 2− ↔ 5+ ˆ 2− ↔ 5+ 3− ↔ 3+ 2− ↔ 5+ 3− ↔ 3+ 4 TABLE 9.10 Elementary changes
122
Pencils of Cubics and Algebraic Curves in the Real Projective Plane ˆ 3 6+ ↔ 4− 6+ ↔ 4− 5+ ↔ 5− 5+ ↔ 5− ˆ 5 2− ↔ 4+ 3− ↔ 3+ 2− ↔ 4+ 3− ↔ 3+ ˆ 3 6− ↔ 6+ 6− ↔ 6+ 5− ↔ 7+ 5− ↔ 7+ ˆ 2− ↔ 4+ 3− ↔ 3+ 2− ↔ 4+ 3− ↔ 3+ 6 ˆ 3 8+ ↔ 6− 8+ ↔ 6− 7+ ↔ 7− 7+ ↔ 7− ˆ 7 2− ↔ 4+ 3− ↔ 3+ 2− ↔ 4+ 3− ↔ 3+ ˆ 3 8− ↔ 8+ 8− ↔ 8+ 7− ↔ 1+ 7− ↔ 1+ ˆ 2− ↔ 4+ 3− ↔ 3+ 2− ↔ 4+ 3− ↔ 3+ 8 ˆ 4 5− ↔ 5+ 5− ↔ 5+ 3− ↔ 6+ 3− ↔ 6+ ˆ 3− ↔ 6+ 4− ↔ 4+ 3− ↔ 6+ 4− ↔ 4+ 5 ˆ 4 7+ ↔ 5− 7+ ↔ 5− ˆ 6 3− ↔ 5+ 4− ↔ 4+
6+ ↔ 6− 3− ↔ 5+
6+ ↔ 6− 4− ↔ 4+
ˆ 4 7− ↔ 7+ 7− ↔ 7+ 6− ↔ 8+ 6− ↔ 8+ ˆ 7 3− ↔ 5+ 4− ↔ 4+ 3− ↔ 5+ 4− ↔ 4+ ˆ 4 1+ ↔ 7− ˆ 8 3− ↔ 5+
1+ ↔ 7− 4− ↔ 4+
8+ ↔ 8− 3− ↔ 5+
8+ ↔ 8− 4− ↔ 4+
ˆ 5 6− ↔ 6+ 6− ↔ 6+ 4− ↔ 7+ 4− ↔ 7+ ˆ 6 4− ↔ 7+ 5− ↔ 5+ 4− ↔ 7+ 5− ↔ 5+ ˆ 5 8+ ↔ 6− ˆ 7 4− ↔ 6+
8+ ↔ 6− 5− ↔ 5+
7+ ↔ 7− 4− ↔ 6+
7+ ↔ 7− 5− ↔ 5+
ˆ 5 8− ↔ 8+ 8− ↔ 8+ 7− ↔ 1+ 7− ↔ 1+ ˆ 8 4− ↔ 6+ 5− ↔ 5+ 4− ↔ 6+ 5− ↔ 5+ ˆ 6 7− ↔ 7+ 7− ↔ 7+ 5− ↔ 8+ 5− ↔ 8+ ˆ 7 5− ↔ 8+ 6− ↔ 6+ 5− ↔ 8+ 6− ↔ 6+ ˆ 6 1+ ↔ 7− ˆ 8 5− ↔ 7+
1+ ↔ 7− 6− ↔ 6+
8+ ↔ 8− 8+ ↔ 8− 5− ↔ 7+ 6− ↔ 6+
ˆ 7 8− ↔ 8+ 8− ↔ 8+ 6− ↔ 1+ 6− ↔ 1+ ˆ 6− ↔ 1+ 7− ↔ 7+ 6− ↔ 1+ 7− ↔ 7+ 8 TABLE 9.11 Elementary changes, continued
10 Application to an interpolation problem
CONTENTS 10.1 10.2
10.1
A nongeneric pencil of cubics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solution to the interpolation problem . . . . . . . . . . . . . . . . . . . . . . . . . . .
123 126
A nongeneric pencil of cubics
Piecewise rational curves are widely used in computeraided geometric design and geometric modeling. This has various reasons, for instance that parametric representations by rational functions comply with industrial standards. In [33], [34], the authors describe a method to approximate an (not necessarily rational) algebraic curve by pieces of rational cubics. By approximation theory, the best approximation by rational curves is one which intersects the given curve in the maximal number of points. The idea in [33], [34] is to take four points in the interior of the curve which is to be approximated, and to search for an arc of a rational cubic that interpolates the curve segment at these four points and at the tangents of the two ends. If a solution exists, it can be easily found by solving a linear system. Problem: Consider a smooth curve in the plane. Cut this curve into convex pieces and find on each piece C four points such that there exists a rational cubic C3 verifying that some parametrized arc of C3 is isotopic to C, tangent to C at both extremities, and interpolates the four points. See also [18]. Let us consider eight points 1, . . . 8 lying in strictly convex position on a piece of curve C, in such a way that 2 is the first point and 1 is the last. The point number n lies in a triangle Tn supported by three lines determined by the pairs (n − 1, n − 2), (n − 1, n + 1) and (n + 1, n + 2) (where n ∈ Z/8, so 8 = 0). Move 3 towards 2 and 8 towards 1, so that the pairs of points 2, 3 and 1, 8 are replaced by tangencies. Let Psing be the singular pencil of cubics having 4, 5, 6, 7 as base points, and such that the generic cubics of Psing are tangent to C at 2 = 3 and 1 = 8. This pencil is tangent to ∆ at two regular points. Proposition 10.1 Either condition 1) or 2) implies 3). 123
124
Pencils of Cubics and Algebraic Curves in the Real Projective Plane 1. The distinct points 1, . . . , 8 verify: 2 < 34567, 8 > 34567, 8 < 14567, and the point 7 is not separated from the side [68] of T7 by the cubic (ˆ 7, 3). 2. The distinct points 1, . . . , 8 verify: 1 < 45678, 3 > 45678, 3 < 24567, and the point 4 is not separated from the side [53] of T4 by the cubic (ˆ 4, 8). 3. The pencil Psing contains a cubic C3 such that some parametrized arc of C3 is isotopic to C, tangent to C at 2 = 3 and at 1 = 8, and interpolates successively the points 2 = 3, 4, 5, 6, 7, 8 = 1.
By symmetry, it suffices to prove that 1) implies 3). Proof: Eight points 1, . . . 8 lying in convex position realize max(ˆ8 = 1+) if and only if 7 < 23456, 1 > 23456 and 8 > 12345: the first two conditions ˆ = 8+ for N = grant that ˆ 8 = 1+ (Table 9.1), the third grants then that N 1, . . . 7 (Table 9.3). Performing the symmetry (+1)(48) yields the first three conditions of 1), hence the eight points realize the list max(ˆ1 = 8−). The last condition of 1) may be rewritten as 8 is separated from the side [17] of T8 by the cubic (ˆ 8, 3). The pencil P is as shown in Figure 8.2, with 9 = G, F, E, D, C, B or A. The case 9 = F is excluded by the hypothesis that 8 is very close to 1. Let us denote by PN the pencil with 9 = N . Letting 8 move to 1, the two successive nodal cubics of PN corresponding to the opening of the elementary arc 18 come together to yield a cubic with a nonisolated node at 1 = 8, and the portion between them disappears (case 1a of Figure 4.1). These cubics are (81, L), (81, C) for N = G, and (1+, 12), (1−, 8) for the other five choices of N . Letting 3 move to 2, two possibilities arise, according to whether (ˆ2, 3) and (ˆ3, 2) have each an isolated node (case 1b), or a crossing (case 1a). In the first case, two cubics with nonreal nodes come together to yield a cubic with an isolated node at 3 = 2 inside of the portion formed by cubics with ovals through the two base points 2, 3. Following this portion, one sees the oval shrink till it becomes an isolated node, then it reappears on the other side and glues to the pseudoline. In the second case, the cubic (1−, 3) comes together with a cubic with a loop passing through no base point to yield a cubic with a crossing at 2 = 3 and a loop passing through no base point. Moving further in the pencil, one sees cubics with an oval passing through no base point; this oval shrinks till it becomes an isolated node and disappears. The singular pencil Psing realizes one of the six possibilities shown in Figure 10.1, where (P, Q, R, S, T ) take values (4, 5, 6, 7, 9), (4, 5, 6, 9, 7), (4, 5, 9, 6, 7), (4, 9, 5, 6, 7) and (9, 4, 5, 6, 7). Denote by C(1), . . . C(2) . . . the successive cubics of the pencil, starting from the upper left one. There are two choices for the last two cubics, C(7), C(8) or C 0 (7), C 0 (8). For all six values of N , C(8) and C 0 (7) are solutions to the interpolation problem. There is in each case another solution C(k) with k = (1, 2, 3, 4, 5, 6) for N = (G, E, D, C, B, A). These cubics are obtained from the cubics (12, L), (1−, 78), (1−, 67), (1−, 56), (1−, 45) and (1−, 34) of PG , replacing the pairs 8, 1 and 3, 2 by tangencies.
Application to an interpolation problem
9
9
1, 8 7
7 5
6
6
9
1, 8
7
5
4
5
1, 8
7
6 3, 2
3, 2 4
9
7 1, 8 6
5
9
5
3, 2
3, 2
4
4
9
1, 8
9
9
1, 8
7
6 5
4
5
7 6
3, 2
7
6
5
1, 8
3, 2
7
4 6
9
1, 8
3, 2
1, 8
4
9
1, 8 7
6
6
5
4
5
3, 2
4
T Q
R
P
R
3, 2
Q P T
S
P
Q
R
T
Q 3, 2
FIGURE 10.1 The pencil Psing
P
P 1, 8
S R
1, 8
3, 2
1, 8
P
Q
3, 2 P
R P
T
Q
P
S
Q
T
S
R
3, 2
R
S
1, 8
R
3, 2
Q
S
1, 8
T
S
1, 8
R
T 3, 2
R
T 1, 8
S
1, 8
S
3, 2
S
3, 2
T
1, 8
Q
125
3, 2 1, 8
T
Q
3, 2 P
126
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
10.2
Solution to the interpolation problem
In this section, we assume that the curve we want to approximate is algebraic or analytic. Let P be the starting point of the piece C (whose endpoint is not yet known). At each step, we have to find one point Q satisfying some condition, which is always achieved if Q is chosen close enough to P . The idea is to let all of the points Q coincide with P first, and then to free these points one after the other. Start: make P = 2 = 3 = 4 = 5 = 6 = 7 = 8 = 1 Let C2 be the conic that is tangent to C with order 5 at P . 1. First case: the germ of C near P lies outside of the conic C2 . •Q = 8 = 1, P = 2 = 3 = 4 = 5 = 6 = 7 Find 8 = 1 such that the arc 21 of C is outside of C2 = 34567; and the conic 14567 meets C only at P (tangency of order 4) and at 1. •Q = 7, P = 2 = 3 = 4 = 5 = 6 Find 7 such that the conic 34567 meets C only at P (tangency of order 4) and 7; and the conic 14567 meets C only at P (tangency of order 3), 7 and 1. •Q = 6, P = 2 = 3 = 4 = 5 Find 6 such that the conic 34567 meets C only at P (tangency of order 3), 6 and 7; and the conic 14567 meets C only at P (tangency of order 2), 6, 7 and 1. •Q = 5, P = 2 = 3 = 4 Find 5 such that the conic 34567 meets C only at P (tangency of order 2), 5, 6 and 7; and the conic 14567 meets C only at P = 4, 5, 6, 7 and 1. •Q = 4, P = 2 = 3 Find 4 such that the conic 34567 meets C only at P = 3 4, 5, 6 and 7; and the conic 14567 meets C only at P = 4, 5, 6, 7 and 1. •Let C3 be the cubic tangent to C at 8 = 1, having an ordinary cusp tangent to C at 2 = 3, and passing through the points 4, 5, 6. If C3 has no supplementary intersection with C, leave 4, 5, 6, 7 unchanged. Otherwise, let 4, 5, 6, 7 be any successive four intersection points of C3 with the interior of C. 2. Second case: the germ of C near P lies inside of the conic C2 .
Application to an interpolation problem
127
•Q = 8 = 1, P = 2 = 3 = 4 = 5 = 6 = 7 Find 8 = 1 such that the arc 21 of C is contained in the conic C2 = 24567; and the conic 45678 meets C only at P (tangency of order 4) and 8. •Q = 7, P = 2 = 3 = 4 = 5 = 6 Find 7 such that the conic 24567 meets C only at P (tangency of order 4) and 7; and the conic 45678 meets C only at P (tangency of order 3), 7 and 8. •Q = 6, P = 2 = 3 = 4 = 5 Find 6 such that the conic 24567 meets C only at P (tangency of order 3), 6 and 7; and the conic 45678 meets C only at P (tangency of order 2), 6, 7 and 8. •Q = 5, P = 2 = 3 = 4 Find 5 such that the conic 24567 meets C only at P (tangency of order 2), 5, 6 and 7; and the conic 45678 meets C only at P = 4, 5, 6, 7 and 8. •Q = 4, P = 2 = 3 Find 4 such that the conic 24567 meets C only at P = 2 4, 5, 6 and 7; and the conic 45678 meets C only at P = 4, 5, 6, 7 and 8. •Let C3 be the cubic tangent to C at 2 = 3, having an ordinary cusp tangent to C at 8 = 1, and passing through the points 5, 6, 7. If C3 has no supplementary intersection with C, leave 4, 5, 6, 7 unchanged. Otherwise, let 4, 5, 6, 7 be any four successive intersection points of C3 with the interior of C. If the points 4, 5, 6, 7 lie on the cubic C3 , this cubic is a solution to the interpolation problem. Otherwise, the distinct points 1, . . . , 8 satisfy by construction one of the conditions 1) and 2) of Proposition 10.1. See Figure 10.2
128
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
34567
7
1
6 1
P
P 34567
P
14567
6
5
4
6
1
34567
P
1 34567
14567
5
4
7
7 P
1
34567
14567
14567
5
7
6 7 8, 1
2, 3
14567
5
4
6 7 8, 1
2, 3
7
6
8 P
8
P
6
8
24567
5
4
7
45678
45678
P
45678
24567
6
8
5
6
5
6
4
7
7
2, 3
8, 1
24567 4 2, 3
FIGURE 10.2 Interpolation of C
8
24567
24567
P
P
45678
45678
5
7
7
8, 1
Part III
Algebraic curves
11 Hilbert’s 16th problem
CONTENTS 11.1 11.2 11.3 11.4
11.1
Real and complex schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Classical restriction method and degree 7 . . . . . . . . . . . . . . . . . . . . . . . Orevkov’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M curves of degree 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
131 136 140 145
Real and complex schemes
The first part of Hilbert’s sixteenth problem deals with the classification of the isotopy types realizable by real plane algebraic curves of given degree in RP 2 . Let A be a real algebraic nonsingular plane curve of degree m. Its complex part CA ⊂ CP 2 is a Riemannian surface of genus g = (m − 1)(m − 2)/2, and its real part RA ⊂ RP 2 is a collection of L ≤ g + 1 circles embedded in RP 2 . If L = g + 1, we say that A is a maximal curve or M curve. A circle embedded in RP 2 is called an oval or a pseudoline depending on whether it realizes the class 0 or 1 of H1 (RP 2 ). If m is even, the L circles of RA are ovals, if m is odd, one single circle is a pseudoline. An oval separates RP 2 into a M¨ obius band and a disc. The latter is called the interior of the oval. An oval of RA is empty if its interior contains no other oval. An oval is exterior if it is surrounded by no other oval. Two ovals form an injective pair if one of them lies in the interior of the other one. A nest of depth d + 1 is a configuration of ovals (O0 , O1 , . . . , Od ) such that Oi lies in the interior of Oj for all pairs i, j, with j > i. An M curve of degree m = 2k or m = 2k + 1 is said to have a deep nest if it has a nest of depth k − 1. Notice that for a curve with a deep nest, all of the other ovals are empty by Bezout’s theorem with an auxiliary line. All of the deep nests of the curve have the same set O1 , . . . Ok−1 of nonempty ovals. The real scheme of A is the isotopy type of RA ⊂ RP 2 , encoded as follows. The symbol hJ i stands for a curve consisting of one single pseudoline; hni stands for a curve consisting of n empty ovals. If X is the symbol for a curve without a pseudoline, 1hXi is the curve obtained by adding a new oval, containing all of the others in its interior. Finally, a curve which is the union
131
132
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
of two disjoint curves hAi and hBi, having the property that none of the ovals of one curve is contained in an oval of the other curve, is denoted by hA q Bi. The complex conjugation conj of CP 2 acts on CA with RA as fixed points set. Thus, CA \ RA is connected, or splits in two homeomorphic halves which are exchanged by conj. In the latter case, we say that A is dividing. Following Klein’s terminology, we say also that a curve is of type I if it is dividing, and of type II otherwise. A real scheme is of type I, type II , or indefinite, according to whether it is realizable only by dividing curves, only by nondividing curves, or by curves of both types. Assume that CA is oriented canonically. For A dividing, we choose a half CA+ of CA \ RA. The orientation of CA+ induces an orientation on its boundary RA. This orientation, which is defined up to complete reversion, is called complex orientation of A. One can provide all the injective pairs of RA with a sign as follows: such a pair is positive if and only if the orientations of its two ovals induce an orientation of the annulus that they bound in RP 2 . Let Π+ and Π− be the numbers of positive and negative injective pairs of A. If A has odd degree, each oval of RA can be provided with a sign: given an oval O of RA, consider the M¨obius band M obtained by cutting away the interior of O from RP 2 . The classes [O] and [2J ] of H1 (M) coincide or are opposite. In the first case, O is negative, otherwise O is positive. Let Λ+ and Λ− be respectively the numbers of positive and negative ovals of RA. The complex scheme of A is obtained by enriching the real scheme with the complex orientation. This may be done by adding arrows on the circles in a picture of the curve. For m odd, let A have the real scheme hJ q 1hαi q βi. The complex scheme of A is then encoded by hJ q 1 hα+ q α− i q β+ q β− i where ∈ {+, −} is the sign of the nonempty oval; α+ , α− are the numbers of positive and negative ovals among the α; β+ , β− are the numbers of positive and negative ovals among the β. By convention, we say that the complex scheme of a nondividing curve coincides with its real scheme. Consider the space RP N , N = m(m + 3)/2 of real algebraic curves of degree m, and denote by ∆ the discriminantal hypersurface formed by the singular curves. The connected components of RP N \ ∆ are called chambers. Two algebraic nonsingular curves of given degree are rigidly isotopic if they lie in the same chamber: a rigid isotopy between the two curves is a path connecting them inside of this chamber. There are thus three classification problems, each one being a refinement of the previous one: real schemes, complex schemes, and rigid isotopy. These problems require two complementary approaches, constructions and restrictions. In 1876, Harnack obtained the complete classification of the real schemes for m ≤ 5, using elementary tools. Moreover, his construction method allowed him to get series of curves of arbitrary degree with all of the admissible numbers of components (0 ≤ L ≤ g + 1 for m even, 1 ≤ L ≤ g + 1 for m odd), with iterated applications of the small perturbation theorem. Different such series of curves were obtained similarly by other mathematicians (Hilbert, Brusotti, Wiman). Let A have even degree m = 2k; an oval is even if it lies inside of an even number of other ovals, otherwise it is odd . Let p be the number of even ovals, and n be the number of odd ones.
Hilbert’s 16th problem
133
After systematic constructions, Ragsdale stated in 1906 a conjecture giving upper bounds for the numbers p and n. This conjecture was disproved about eighty years later [81], [42], but it is still open for the M curves. In 1900, Hilbert included the classification of the real schemes in his 16th problem, and asked specifically about the degree 6. Bezout’s theorem with an auxiliary line implies that a sixth degree M curve can have at most one nonempty oval. At the beginning of the 20th century, two different real schemes had been realized: h9 q 1h1ii (Harnack) and h1 q 1h9ii (Hilbert). In the statement of his 16th problem, Hilbert included the question whether other isotopy types were possible. The answer was found only in 1969, when Gudkov managed with repeated quadratic transformations to construct an M curve of degree 6 with five interior ovals h5 q 1h5ii. Between the 1930s and the 1990s of the last century, great progress was achieved by the Russian school (Petrovski, Rokhlin, Arnold, Kharlamov, and others). Connecting the problem to manifolds of dimension 4 (with branched coverings and the theory of Smith in the particular case of involutions) allowed proof of many inequalities and congruences for curves of even degree, see for example the survey [88]. We state here only two emblematic results. Let A be a dividing curve of degree 2k, one has: Petrovskii inequalities −3/2k(k − 1) ≤ p − n ≤ 3/2k(k − 1) + 1 Rokhlin congruence If A is an M curve, p − n ≡ k 2 (mod 8) Rokhlin’s congruence, proved in 1972, grants that there are no other possible real schemes for the M curves of degree 6. The study of complex orientations [70], [13] gave further restrictions for both the real and the complex schemes. On the construction side of the problem, a breakthrough was achieved around 1980 with Viro’s method: a powerful tool that allows us to obtain nonsingular curves with a prescribed topology, based on polyhedral subdivisions of Newton polygons, see [80] or the survey [52].1 Viro [81] [83] thus constructed curves of degree 6 and 7 realizing all of the admissible real schemes. Around 2000, Orevkov [55] developed a new powerful restriction method based on link theory, with computations of signatures that may be performed by a computer program. The idea underlying this method is exposed briefly in Section 11.3. 1 2 3 4 5
hJ i h1i, h0i hJ q 1i, J h4i, h3i, h2i, h1h1ii, h1i, h0i hJ q 6i, hJ q 5i, hJ q 4i, hJ q 3i, hJ q 1h1ii hJ q 2i, hJ q 1i, hJ i
TABLE 11.1 Real schemes of degrees m = 1, . . . 5, complete classification 1 A combinatorial version of this method, called Viro patchworking, arises when the subdivisions are triangular.
134
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
What is the state of knowledge now? Up to m = 6, the three classifications are entirely completed. They coincide for m ≤ 4 (Klein theorem). The classification of real and complex schemes is completed for degree 7. For m ≤ 5 and m = 7, the real schemes that are allowed by Bezout’s theorem with auxiliary lines and conics are all realized. The classification of the real and complex schemes realizable by curves of degrees m ≤ 5 are shown in Tables 11.1, 11.22 Nikulin and Kharlamov [54], [44] proved that the rigid isotopy class of a curve of degree m = 5 or 6 is determined by its real scheme, along with its type I or II.3 Note that for m = 5, the real scheme hJ q 4i fits to two rigid isotopy classes. Figure 11.1 shows two quintics with this real scheme, obtained by perturbing the union of two ellipses and a line in two different ways. A simple argument shows that these curves are not rigidly isotopic. Choose points inside of each oval and consider the set of lines connecting these points pairwise. Three lines determine four triangles in the plane; the principal triangle is the one whose edges do not cut J . The first quintic presents one oval lying in the principal triangle determined by the other three, the second quintic has no such oval. Along a rigid isotopy between the two curves, one oval should cross the line through two others; this is impossible by Bezout’s theorem. The two quintics also have different types. A complex orientation of the dividing quintic may be found with the complex version of the small perturbation theorem: Consider a set of dividing curves Ai intersecting only at real points, transversally, and let A be a curve obtained by perturbing each crossing in one of the two possible ways. This curve is dividing if and only if one can endow the real parts RAi with complex orientations in such a way that all perturbations are compatible with the local orientations, and RA has thus the induced complex orientation. We give particular attention to the case of the M curves (for m ≥ 8, this choice is made necessary by the large number of cases). For m ≤ 6, two M curves with the same real scheme are rigidly isotopic. We have already mentioned the three real schemes realizable by M sextics: h9 q 1h1ii, h1 q 1h9ii, h5 q 1h5ii. The M curves of degree 7 realize 14 real schemes hJ q 15i, hJ q β q 1hαii, α + β = 14, 1 ≤ α ≤ 13. Curves with real scheme hJ q 15i and hJ q 13 q 1h1ii are realizable with the methods of Harnack and Hilbert.4 Two series of curves constructed by Viro [81] and Korchagin [48], complete the list. We will see in the next section that the three classifications are distinct for the M curves of degree 7. For m = 8, curves realizing 83 real schemes have been constructed, and there remain six open cases. These six real schemes are realizable by real pseudoholomorphic curves, [64]. Viro’s method amounts to perturbing singular curves, whose singularities satisfy certain conditions, in a controlled way. Korchagin applied this method to obtain M curves of degree 2 The notation h1h1−ii for the curve of degree 4 with two nested ovals means that these ovals form a negative injective pair. 3 The doublesheeted covering of CP 2 branched over a sextic is a K3surface; this fact is used in the proof for m = 6. 4 Hilbert’s method also gives M curves of degree 7 with α = 2, 3 and 12.
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9 realizing 404 different real schemes. Shustin later improved Viro’s method, allowing more degenerated singularities. Implementing this improved method with computeraided systematic constructions, Orevkov and Chevallier have realized 179 more cases, see Section 11.4.
FIGURE 11.1 Two quintics hJ q 4i with different types
degree dividing 1 2
hJ i h1i
3
hJ q 1− i
4
h4i
nondividing
h0i hJ i h1h1− ii 5
h3i h2i h1i h0i
hJ q 3+ q 3− i hJ q 1+ q 3− i hJ q 1− h1− ii
hJ q 5i hJ q 4i hJ q 3i hJ q 2i hJ q 1i hJ i
TABLE 11.2 Complex schemes of degrees m = 1, . . . 5, complete classification
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11.2
Classical restriction method and degree 7
The classical restriction method combines Bezout’s theorem with auxiliary lines or conics, and the following theorems about complex orientations: Rokhlin formula: If m = 2k, then 2(Π+ − Π− ) = L − k 2 RokhlinMishachev formula: If m = 2k + 1, then 2(Π+ − Π− ) + (Λ+ − Λ− ) = L − 1 − k(k + 1) Fiedler theorem: Let Lt = {Lt , t ∈ [0, 1]} be a pencil of real nonsingular dividing curves. Consider two curves Lt1 and Lt2 of Lt , which are tangent to RA at two points P1 and P2 , such S that P1 and P2 are related by a pair of conjugated imaginary arcs in CA ∩ ( CLt ). Orient Lt1 coherently to RA in P1 , and transport this orientation through Lt to Lt2 . Then this orientation of Lt2 is compatible with that of RA in P2 . (See also [13], [70]). In most cases, the pencils used are pencils of lines, see Figures 11.2 and 11.3. Note that if one considers a complete pencil Lt , the parameter t describes a circle RP 1 instead of an interval. The choice of a direction for the pencil allows us to define two types of tangency points: t1 if two real points are replaced by two complex conjugated points, t2 for the converse situation. A sequence of ovals that are connected one to the next by pairs of conjugated imaginary arcs will be called a Fiedler chain, see Figure 11.2. To form a chain, one allows actually two consecutive ovals to be connected indirectly via a fold due to some component of the curve as shown in Figure 11.3. A fold arises with a pair of tangency points of each type on the component. To apply the theorem as a restriction tool, we consider some pencil sweeping out a piece of the curve under consideration, and try to determine all possible ways to distribute the ovals in Fiedler chains. We can safely ignore the folds, as they only reduce the number of possible distributions. The pencil is maximal if none of the curves in it has more than two nonreal intersection points with CA. With a maximal pencil, the Fiedler chains are easily spotted. The definitions hereafter will be useful in the sequel. Let Cm be an M curve of odd degree m. Given an empty oval X of Cm , we shall also call X a point chosen in the interior of the oval and FX a pencil of lines based at X. Let us fix an orientation of RP 2 \ J , then FX : Y → Y , ∈ {±} are the complete pencils based at X, starting and ending at Y , that rotate in either direction. Given two empty ovals X, Y , the principal segment [XY ] is the segment of line XY that cuts J an even number of times, and the other
Hilbert’s 16th problem
FIGURE 11.2 Fiedler chain with a pencil of lines
FIGURE 11.3 Folds
137
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
segment is denoted by [XY ]0 . Three points distributed in three empty ovals X, Y, Z determine four triangles in RP 2 . The principal triangle XY Z is the triangle bounded by the segments [XY ], [Y Z] and [XZ]. Definition 11.1 A pencil of lines FX : Y → Z has a J jump if it sweeps out the nonprincipal segment [Y Z]. Note that the total number of J jumps of a complete pencil FX : Y1 → Y1 over a sequence of ovals Y1 , . . . , Yn is always odd and does not depend on the choices of and Y1 . Assume that Y1 , . . . , Yn form a Fiedler chain, then the signs of Yi and Yi+1 coincide if and only if the pencil FX : Yi → Yi+1 has a J jump.
Definition 11.2 An ordered set of empty ovals F1 , . . . , Fn of Cm lies in a convex position if for each triple Fi , Fj , Fk , the principal triangle Fi Fj Fk does not containSany other oval of the group and F1 , . . . , Fn are the successive vertices of Fi Fj Fk (the convex hull of the group). Let us give now simple examples of restrictions obtained with the classical method, in the case of degree 7. The M curves of degree 7 have 15 ovals, and at most one of them is nonempty. (Otherwise, a line through two ovals A and B interior to different ovals, would have nine intersection points with the curve.) Viro proved that the empty ovals cannot be all interior, in other words, the real scheme hJ q 1h14ii is not realizable. Assume the contrary; let O be the nonempty oval, and A, B, C be three other ovals. By Bezout’s theorem with the lines AB, AC, BC, the principal triangle ABC lies entirely in the interior of O. Suppose that ABC contains an oval D. Then, for any choice of a fifth oval E, the conic through A, B, C, D, and E cuts the curve at 16 points, a contradiction. Thus, ABC contains no ovals. This is true for all choices of A, B, C, hence the empty ovals lie in convex position in the interior of O. Let us fix now a choice of ovals A, B, C, and consider three pencils of lines based at points distributed on these three ovals. These pencils give rise to three Fiedler chains that form together a cyclic chain, containing the 14 interior ovals. (The cyclic ordering of the ovals in the chain is given by the convex position.) The 14 ovals have alternating orientations with respect to both O and J . Thus, Π+ − Π− = 0 and Λ+ − Λ− = , where = ±1 is the contribution of O. On the other hand, the RokhlinMishachev formula reads: 2(Π+ − Π− ) + (Λ+ − Λ− ) = 3, a contradiction. Definition 11.3 Let C7 be a curve of degree 7 with a nonempty oval O. We say that C7 has a jump if there exist two exterior ovals B, C and two interior ovals A, D such that A and D are separated inside of O by any line passing through B and C. Let C7 have a jump, then the lines AB, AC, BD, CD and J divide the plane in zones. The other ten empty ovals are distributed in four of them: two triangles T1 , T2 and two quadrangles Q1 , Q2 , see Figure 11.5. As a matter of
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139
fact, if there was an oval E in another zone, then the conic through A, B, C, D, and E would cut C7 at 16 points, a contradiction. With the help of conics, one proves easily the following: Lemma 11.1 Let C7 be an M curve of degree 7 with a nonempty oval O. Then the interior ovals lie in convex position inside of O. 1. If C7 has a jump, then the empty ovals have a natural cyclic ordering B1 , . . . Bβ1 , A1 , . . . Aα1 , C1 , . . . Cβ2 , D1 , . . . Dα2 , where the Bi , Cj are exterior, Ak , Dl are interior, α1 + α2 = α, β1 + β2 = β. This ordering is given by the complete pencils of lines FAk , FDl , and the pencils of lines FBi , FCj sweeping out O from the first interior oval to the last one. 2. If C7 has no jump, then the empty ovals have a natural cyclic ordering B1 , . . . Bβ , A1 , . . . Aα , where the Bi are exterior and the Aj interior. This ordering is given by the complete pencils of lines FAj , and the pencils of lines FBi sweeping out O from the first interior oval to the last one. The empty ovals of C7 with a jump are thus distributed in four groups, and the distribution α1 , β1 , α2 , β2 of ovals in each group is a rigid isotopy invariant of the curve. Two pencils of lines FC : D → B → A and FB : A → C → D give rise to two Fiedler chains, forming together a cyclic chain involving all empty ovals. As these pencils have no J jumps, the ovals in the chain have alternating signs, hence Λ+ − Λ− = where = ±1 is the contribution of the nonempty oval O. Combining with the RokhlinMishachev formula 2(Π+ −Π− )+(Λ+ −Λ− ) = 3, one gets that Π+ −Π− = 1 if O is positive, 2 if O is negative. For C7 without a jump, the pencils FAj sweeping out the exterior (empty) ovals all have the same number j of J jumps, dividing the sequence of ovals B1 , . . . Bβ in j +1 consecutive groups. The distribution β1 , . . . βj+1 (with β1 + . . . βj+1 = β) of the ovals in the groups is a rigid isotopy invariant of the curve. The interior ovals swept out by the pencils FBi form one single Fiedler chain. They have alternating orientations with respect to both O and J , hence Π+ − Π− = 0, 1 or −1. The arguments above (originally from [13]) allow us to establish two temporary lists of 11 admissible complex schemes with a jump, and 36 without a jump. The complex scheme hJ q 3+ q 1+ h6+ q 5− ii appears in the list without a jump, we prove hereafter with the classical method that it is not realizable.5 Assume the contrary. As the three exterior ovals have the same sign, one must have j = 2. The oval B2 lies in a quadrangle Q whose edges are supported by the lines A1 B1 , A1 B3 , A11 B1 , A11 B3 (see Figure 11.6), this quadrangle is divided in two triangles by the principal segment [B1 B3 ]. If B2 is in the upper triangle, the conic B1 B2 B3 A1 A11 cuts: The set of five chosen ovals at 10 points, O at two points and J at four points, hence in 5 This complex scheme is also excluded by Orevkov’s formula for curves with a deep nest, see the next section and Chapter 12.
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
total 16 points, in contradiction with Bezout’s theorem. Let B2 lie in the lower triangle (as shown in the figure) and apply Fiedler’s theorem with the pencil of lines FB2 : A11 → B3 . The lines in this pencil have at most four nonreal intersections with C7 ; there are thus two chains, starting at O and A11 respectively. One of these ovals should be connected with B3 , but this is impossible as the orientations are incompatible, see Figure 11.6. All of the 11 admissible complex schemes with a jump are realizable, see [81], [48], [57]. There exist curves realizing 14 of the admissible complex schemes without a jump, constructed in [40], [81], [48] and [59]. The other 22 cases of the list have been forbidden in [15], [55], [57], [58] and [30].6 The classification of the complex schemes realizable by M curves of degree 7 is thus completed, see Table 11.3.
A
C
B
FIGURE 11.4 The real scheme hJ q 1h14ii is not realizable
11.3
Orevkov’s method
An almost complex structure J on CP 2 is a linear operator Jx : Tx (CP 2 ) → Tx (CP 2 ) such that Jx2 = −id and J depends smoothly on x. We assume that J is tame: there exists a symplectic form ω such that ∀x ∈ CP 2 , ∀v ∈ Tx (CP 2 ), 6 Eight of the nine restrictions in [15] were obtained with the help of an auxiliary pencil of cubics, constructed step by step using Bezout’s theorem. We don’t reproduce the proof here, as a simpler way to exclude these complex schemes was then found by Orevkov [55], [58].
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141
T1
T2
C
A Q1
D Q2
B
FIGURE 11.5 Curve of degree 7 with jump
Q B1 B2
A11
B3
A1
FIGURE 11.6 The complex scheme hJ q 3+ q 1+ h6+ q 5− ii is not realizable
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
real scheme
without jump
hJ q 1 q 1h13ii hJ hJ hJ q 2 q 1h12ii hJ q 3 q 1h11ii hJ hJ q 4 q 1h10ii hJ hJ q 5 q 1h9ii hJ hJ q 6 q 1h8ii hJ hJ q 7 q 1h7ii hJ hJ q 8 q 1h6ii hJ hJ q 9 q 1h5ii hJ hJ q 10 q 1h4ii hJ q 11 q 1h3ii hJ hJ hJ q 12 q 1h2ii hJ q 13 q 1h1ii hJ hJ hJ hJ q 15i
with jump
q 1+ q 1− h7+ q 6− ii q 1+ q 1+ h6+ q 7− ii hJ hJ hJ q 3+ q 2− q 1− h5+ q 4− ii hJ q 3+ q 2− q 1+ h4+ q 5− ii hJ q 4+ q 3− q 1− h4+ q 3− ii hJ q 4+ q 3− q 1+ h3+ q 4− ii hJ q 5+ q 4− q 1− h3+ q 2− ii hJ q 5+ q 4− q 1+ h2+ q 3− ii hJ q 6+ q 5− q 1− h2+ q 1− ii hJ q 6+ q 5− q 1+ h1+ q 2− ii hJ q 7+ q 6− q 1− h1+ ii q 7+ q 6− q 1+ h1− ii q 9+ q 6− i q 2+ q 1− q 1+ h5+ q 6− ii
q 2− q 1− h7+ q 5− ii q 2+ q 1− q 1+ h5+ q 6− ii q 1+ q 3− q 1− h6+ q 4− ii q 3+ q 2− q 1+ h4+ q 5− ii q 2+ q 4− q 1− h5+ q 3− ii q 4+ q 3− q 1+ h3+ q 4− ii q 3+ q 5− q 1− h4+ q 2− ii q 5+ q 4− q 1+ h2+ q 3− ii q 4+ q 6− q 1− h3+ q 1− ii q 6+ q 5− q 1+ h1+ q 2− ii q 5+ q 7− q 1− h2+ ii
TABLE 11.3 Complex M schemes of degree 7 with and without jump, complete classification
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143
ω(v, Jv) > 0. A surface realizing m[CP 1 ] ∈ H2 (CP 2 ) is a Jholomorphic curve of degree m if its tangent planes at each point are invariants by J. When J is not specified, we speak also of a pseudoholomorphic curve Denote by conj the complex conjugation x 7→ x ¯ in CP 2 . Let J be also real : −1 conj∗ ◦ J = J ◦ conj∗ : Tx → Tx¯ for all x ∈ CP 2 . A Jholomorphic curve is then real if it is invariant by conj. Real algebraic curves are thus as special case of pseudoholomorphic curves, J being the multiplication by i. According to Gromov’s theory, two points determine a unique Jholomorphic line and five points a unique conic. One point determines a pencil of lines, four points a pencil of conics, eight points a pencil of cubics. If the points are real, the curve or pencil of curves they determine is also real. The classical method combining complex orientations and Bezout’s theorem with auxiliary lines, conics or rational cubics (single or pencil) is also valid for real pseudoholomorphic curves. All restrictions involving purely topological tools (such as double coverings) apply to pseudoholomorphic curves. Orevkov’s approach with quasipositive braids [55] provides both a restriction and a construction method for such curves. Let us give hereafter an idea of how it works. Consider a real algebraic curve A of degree m and denote by CA its complexification. Let P be a generic point in RP 2 , L be a real line that doesn’t contain P , and LP be the pencil of lines based at P . The projection πP : CA → CL is an mfold branched covering, and the ramification points are all positive, as tangency points of complex lines with CA. One has πP−1 (RL) = RA ∪ S, where S is the union of all pairs of imaginary arcs connecting two by two the points of tangencies of LP with RA. Consider a circle L , obtained from RL by a small shift into the upper halfsphere CL+ . Let then L+ ⊂ CL+ be the halfsphere bounded by L . The preimage πP−1 (L ) is a closed braid ˆb with m strings. Consider the union of the lines of LP through all points Q ∈ L ; remove P from it, then one gets a solid torus V . Let N ⊂ CA be the surface πP−1 (L+ ). One has ˆb ⊂ V ⊂ S 3 = ∂B 4 ' L+ × C, and N ⊂ B 4 . Let b ∈ Bm be a braid whose closure is ˆb. To get such a braid b in practice, one needs to know the LP scheme of A, which is its real scheme, along with the mutual position of RA and RLP . Place P at infinity and choose a starting line to the left of the picture. Ruling towards right, each double point of RA∪S of type t1 (see Section 11.2) is replaced by some negative twist σi−1 whereas the double points of type t2 yield straight branches. Finally, one has to add at the end of the braid the Garside element ∆m = (σ1 σ2 . . . σm−1 ) . . . (σ1 σ2 σ3 )(σ1 σ2 )(σ1 ) (because RA is on a M¨ obius band RP 2 \ P ). For A dividing, endow b with an orientation, obtained from that of ˆb = ∂N ⊂ CA, then one recovers Fiedler’s theorem (with pencils of lines), see Figure 11.8. Let N ± = N ∩ A± , l± = ∂N ± . One has lk(l+ , l− ) = N + .N − = 0. This yields a new proof of the Rokhlin and RokhlinMishachev formulas. Orevkov also found refinements of these formulas for M curves with deep nests [55] and with four nests [60]. Let Ass12 = As1 ∩ πp−1 (CLs2 ), Figure 11.7 illustrates what happens with a conic. For any curve A, let us perform an isotopy of ˆb in N ; the only obstructions
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
are the ramification points, hence b = Πj aj σij a−1 (b is quasipositive), see j Figure 11.9. Let e : Bm → Z, σi 7→ 1, then e(b) is the number of ramification points. The RiemannHurwitz formula reads χ(N ) = m − e(b) = 2µ(N ) − 2g(N ) − µ(ˆb) where µ(.) stands for the number of connected components. On the other hand, a link l ⊂ S 3 that is the boundary of a surface N ⊂ B 4 must fulfill the TristramMurasugi inequalities. This yields a necessary condition for a braid to be quasipositive, and this condition is usable for restrictions. Consider an LP scheme C, such that the lines of the pencil have all at least m−4 intersections with C. If there are never less than m−2 intersection points with the real scheme, one knows how the points of tangency are connected two by two. So one may construct the braid and test the quasipositivity condition. If some lines have only m−4 intersections with the real scheme, one must consider all of the possibilities, taking account that there is an unknown number n of twists between two imaginary arcs of S (and −n between their conjugates). One gets a set B of admissible braids with m strings. Orevkov [60] proved also that given an LP scheme, if one of the braids b ∈ B is quasipositive, then the LP scheme is realizable by a Jholomorphic curve of degree m, along with a pencil of Jholomorphic lines. Several cases of pseudoholomorphic curves whose isotopy type is not realizable algebraically are known [56], [61], [26], see also the first paragraph of [17]. However, those curves are all singular. CA
CA A+
A+ +
N
+
RA A b
∆2
A+
N
S
FIGURE 11.7 Braid b (trivial) and surface N in the case of a conic
FIGURE 11.8 Proof of Fiedler’s theorem using a piece of braid Remark: Most of the restriction methods are also valid for real pseudoholomorphic curves. But a restriction involving a generic pencil of cubics is, theoretically, purely algebraic. This is due to the fact that an algebraic pencil
Hilbert’s 16th problem
145
N
b
FIGURE 11.9 ˆ Isotopy of ˆb in N has a fixed ninth base point, whereas a pseudoholomorphic pencil has not. In a joint paper with S. Orevkov [17], we present a pseudoholomorphic affine sextic, whose isotopy type is not realizable algebraically (Theorem 1). The restriction part of the theorem used a pencil of cubics, with eight base points distributed in ovals of the sextic. Unfortunately, there is one fatal error (last statement in 5.4). A correct proof of Theorem 1, based on the HilbertRohn method, is given in the recently issued corrigendum [26].
11.4
M curves of degree 9
In this section, we call a nest a configuration 1hαi formed by one oval containing α nonempty ovals in its interior, whereas a deep nest is as defined in Section 11.1. Table 11.4 displays the admissible real schemes for M curves of degree 9, obtained applying Bezout’s theorem with auxiliary lines and conics. (In this table, the numbers αi are all ≥ 1.) hJ hJ hJ hJ hJ hJ
q 28i q β q 1hα1 ii, β + α1 = 27 q β q 1hα1 i q 1hα2 ii, β + α1 + α2 = 26 q β q 1hα1 i q 1hα2 i q 1hα3 ii, β + α1 + α2 + α3 = 25 q β q 1hα1 i q 1hα2 i q 1hα3 i q 1hα4 ii, β + α1 + α2 + α3 + α4 = 24 q a q 1hb q 1hciii, c ≥ 1, a + b + c = 26
TABLE 11.4 Admissible real schemes for the M curves of degree 9 The isotopy type with 28 empty ovals is realized with Harnack’s construction method. After systematic constructions, Korchagin [49], [52] stated the following three conjectures:
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
Conjecture 1 (Korchagin) Let C9 be an M curve of degree 9 with real scheme hJ q a q 1hb q 1hciii. Then a > 0. Conjecture 2 (Korchagin) Let C9 be an M curve of degree 9 with real scheme hJ q β q 1hα1 i q 1hα2 i q 1hα3 i q 1hα4 ii. Then: 1. The numbers αi , i = 1, 2, 3, 4 are all odd, 2. β ≡ 0 (mod 4) Conjecture 3 (Korchagin) Let C9 be an M curve of degree 9 with real scheme hJ qβ q1hα1 iq1hα2 iq1hα3 ii. Then two of the numbers αi , i = 1, 2, 3 are odd. Let us say that a real scheme is (still) admissible if it is not forbidden. Conjecture 1 is true only for odd b and c, see Theorem 12.1, and [63] for the counterexamples. Conjecture 2 is entirely proved, point 1 with Theorem 13.1 and point 2 in [61]. Korchagin has constructed M curves with four nests realizing 33 of the 55 real schemes left, see Table 11.9, and the other 22 cases have been forbidden by Orevkov (private communication). Theorem 13.2 proves onehalf of Conjecture 3: One at least of the numbers αi , i = 1, 2, 3 must be odd. Orevkov has constructed lately 38 algebraic curves (and 60 pseudoholomorphic curves) even, even, odd, disproving thus the other half of the conjecture. Will the classification of the real schemes realizable by ninth degree M curves be completed some day? Much progress has been achieved since Korchagin’s survey [52] in 1997. We present hereafter the uptodate (June 2018) state of knowledge. M curves of degree 9 1. hJ q 28i is realizable. 2. M curves of degree 9 with one nest: hJ q β q 1hα1 ii α1 ≥ 1 and α1 + β = 27. The two real scheme with (α1 , β) = (27, 0) or (26, 1) are forbidden, see Section 14.1 or [24], hence β ≥ 2. There are thus 25 admissible cases. Korchagin constructed the 21 cases with 1 ≤ α1 ≤ 19 and α1 = 22, 23. Orevkov [59] realized those with α1 = 20 and 21, the last two cases α1 = 24 and 25 are open. 3. M curves of degree 9 with two nests: hJ q β q 1hα1 i q 1hα2 i α1 + α2 + β = 26 and α1 ≤ α2 . The cases with β = 0 are not realizable (see [13], [46]), and the cases with β = 1 have been excluded by Orevkov (private communication). This leaves 144 still admissible cases, among which 127 have been realized, see Table 11.5.
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147
4. M curves of degree 9 with three nests: hJ qβq1hα1 iq1hα2 iq1hα3 ii 1 ≤ α1 ≤ α2 ≤ α3 , α1 + α2 + α3 + β = 25. The cases with β = 0 are forbidden, see [13], [47]. Moreover, one of the αi , i = 1, 2, 3 must be odd, see Section 13.1 or [21]. The total number of admissible cases is thus 353, among which 172 have been realized (46/67 for odd, odd, odd; 88/161 for even, odd, odd; 38/125 for even, even, odd), see Tables 11.6, 11.7, 11.8. 5. M curves of degree 9 with four nests: hJ q β q 1hα1 i q 1hα2 i q 1hα3 i q 1hα4 ii α1 + α2 + α3 + α4 + β = 24 The numbers αi , i = 1, . . . 4 are odd, see Section 13.1 or [16]. Moreover, β ≡ 0 mod 4, see [61]. Among the 55 cases left, 22 have been excluded by Orevkov (private communication). The other 33 cases have been constructed by Korchagin. The classification is completed, see Table 11.9. 6. M curves of degree 9 with deep nest: hJ q a q 1hb q 1hciii a + b + c = 26, c ≥ 1. These conditions give a set of 351 admissible cases. Korchagin [47] excluded the cases (b, c) = (0, 26), (0, 25), (1, 25), (1, 24), (2, 24), (3, 23) and (25, 1). If a = 0, then b, c must be even, and b ≥ 4, see Chapter 12 or [20], this excludes ten more cases. Orevkov [60] has proved that if b = 0, then c is odd and c ≤ 19, this excludes another 14 cases. He has also forbidden the cases (b, c) = (1, 23), (1, 22), (24, 2), see [60] and [63]. Thus, 34 of the 351 cases have been excluded, see right column of Table 11.10. This leaves 317 still admissible cases, among which 227 have been realized, see Table 11.10. Orevkov obtained all of the realized cases by computeraided systematic constructions, perturbing curves with given singularities. Parts of them had been constructed previously by Korchagin [49], [50], [51], Polotovskii [66], Orevkov [59], [63], or Chevallier (private communication). We deal here with algebraic curves, but all of the restrictions obtained also apply to the wider set of pseudoholomorphic curves. The admissible cases are the same in both settings. Do there exist nonsingular pseudoholomorphic curves whose real scheme is not realizable algebraically? This question is still open. Orevkov has constructed 87 pseudoholomorphic M curves of degree 9 whose algebraic realizability is unknown: 11 curves with two nests, 38 with three nests, and 38 with a deep nest, see Table 11.11. The state of knowledge is now (June 2018): 873 still admissible cases, among which 583 have been realized by algebraic curves, and 670 by pseudoholomorphic curves. The lists of constructed cases contain all of the pseudoholomorphic M curves that may be obtained perturbing the curves having the following singularities:
148
Pencils of Cubics and Algebraic Curves in the Real Projective Plane 1. N28 : four smooth branches pairwise tangent in the order two, and a smooth branch cutting them transversally (case partially dealt with by Korchagin), 2. X33 : four smooth branches pairwise tangent with order 3, 3. X45 : four smooth branches pairwise tangent with order 4, 4. Six smooth branches intersecting pairwise transversally (case partially dealt with by Korchagin), 5. Three smooth branches pairwise tangent with the order 6, and a smooth branch tangent to each of them with the order two, 6. Three smooth branches pairwise tangent with the order 7, and a smooth branch tangent to each of them with the order two.
Note however that some of the real schemes cannot be constructed by perturbing singularities of these six types. They are obtained with a perturbation of N40 : four smooth branches pairwise tangent with the order 3, and a smooth branch cutting them transversally; and one of the schemes in Table 11.11 is obtained perturbing N52 : four smooth branches two by tangent with the order 4, and a smooth branch cutting them transversally. Many of the algebraic curves had previously been obtained by Chevallier, also with a perturbation of N52 . α1 1 2 3 4 5 6 7 8 9 10 11 12 TABLE 11.5 Two nests
α2 realized 1, . . . 22 2, . . . 21 3, . . . 18; 20 4, . . . 18 5, . . . 18 6, . . . 15; 17 7, . . . 16 8, . . . 15 9, . . . 12; 14 10, 12, 13 11, 12
α2 open 23 22 19, 21 19, 20 19 16, 18 17 16 13, 15 11, 14 13 12
Hilbert’s 16th problem
(α1 , α2 ) (1, 1) (1, 3) (1, 5) (1, 7) (1, 9) (1, 11) (3, 3) (3, 5) (3, 7) (3, 9) (5, 5) (5, 7) (5, 9) (7, 7)
149
α3 realized 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21 3, 5, 7, 9, 11, 13, 15 5, 7, 9, 11, 13, 15 7, 9, 11, 15 9, 11 3, 5, 7, 9, 11 5, 7, 9, 11 7, 9 5, 7 7, 9 9
TABLE 11.6 Three nests odd, odd, odd
α3 open 17, 19 17 13 13 11 13, 15, 17 13, 15 11, 13 9, 11 9, 11, 13 11 9 7
150
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
(α1 , α2 ) (2, 1) (2, 3) (2, 5) (2, 7) (2, 9) (2, 11) (4, 1) (4, 3) (4, 5) (4, 7) (4, 9) (6, 1) (6, 3) (6, 5) (6, 7) (6, 9) (8, 1) (8, 3) (8, 5) (8, 7) (10, 1) (10, 3) (10, 5) (10, 7) (12, 1) (12, 3) (12, 5) (14, 1) (14, 3) (14, 5) (16, 1) (16, 3) (18, 1) (18, 3) (20, 1) (22, 1)
α3 realized 1, 3, 5, 7, 9, 11, 13, 17 3, 5, 7, 9, 11, 15 5, 7, 9, 13 7, 11 1, 3, 5, 7, 9, 11, 15 3, 5, 7, 9 5, 7, 11 9 1, 3, 5, 7, 9, 11, 13 3, 5, 7, 11 5, 7 7 1, 3, 5, 7, 9, 11 3, 5, 7, 9 5, 7 1, 3, 5, 7, 9, 11, 13 3, 5, 7 5, 7 7 1, 3, 5, 7, 9 5 1, 3, 5, 9 1, 3 1 1
TABLE 11.7 Three nests even, odd, odd
α3 open 15, 19, 21 13, 17, 19 11, 15, 17 9, 13, 15 9, 11, 13 11 13, 17, 19 11, 13, 15, 17 9, 13, 15 7, 11, 13 9, 11 15, 17 9, 13, 15 9, 11, 13 9, 11 9 13, 15 11, 13 9, 11 7, 9 9, 11 9 11 3, 7, 9 5, 7 7 3, 5, 7 5 5, 7 3, 5 3, 5 3 1, 3
Hilbert’s 16th problem (α1 , α2 ) (1, 2) (1, 4) (1, 6) (1, 8) (1, 10) (3, 2) (3, 4) (3, 6) (3, 8) (3, 10) (5, 2) (5, 4) (5, 6) (5, 8) (7, 2) (7, 4) (7, 6) (7, 8) (9, 2) (9, 4) (9, 6) (11, 2) (11, 4) (11, 6) (13, 2) (13, 4) (15, 2) (15, 4) (17, 2) (19, 2)
151 α3 realized 2, 6, 10, 14, 18 4, 8, 12, 16 6, 10 8, 12 4, 8, 12 6, 10 8 2, 6, 10 4, 8, 12 4, 8, 12 6 8 2, 6 4, 8 4 6 2 4
α3 open 4, 8, 12, 16, 20 6, 10, 14, 18 8, 12, 14, 16 10, 14 10, 12 2, 6, 10, 14, 16, 18 4, 8, 12, 14, 16 6, 10, 12, 14 8, 10, 12 10 4, 8, 12, 14, 16 6, 10, 14 6, 8, 10, 12 8, 10 2, 6, 10, 14 4, 8, 10, 12 6, 10 8 4, 8, 10, 12 6, 10 6, 8 2, 6, 8, 10 4, 8 6 4, 6, 8 4, 6 2, 6 4 2, 4 2
TABLE 11.8 Three nests odd, even, even β 20 16 12 8 4 0
(α1 , α2 , α3 , α4 ) (1, 1, 1, 1) (1, 1, 1, 5), (1, 1, 3, 3) (1, 1, 1, 9), (1, 1, 3, 7), (1, 1, 5, 5), (1, 3, 3, 5), (3, 3, 3, 3) (1, 1, 1, 13), (1, 1, 3, 11), (1, 1, 5, 9), (1, 1, 7, 7), (1, 3, 3, 9) (1, 3, 5, 7), (1, 5, 5, 5), (3, 3, 3, 7), (3, 3, 5, 5) (1, 1, 1, 17), (1, 1, 3, 15), (1, 1, 5, 13), (1, 1, 7, 11), (1, 1, 9, 9) (1, 3, 5, 11), (1, 3, 7, 9), (1, 5, 7, 7), (3, 3, 7, 7),(3, 5, 5, 7) (1, 1, 1, 21), (1, 1, 7, 15), (1, 1, 11, 11), (1, 5, 7, 11), (1, 7, 7, 9) (5, 5, 7, 7)
TABLE 11.9 Four nests, complete classification
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
c 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
b realized 0, . . . 20; 22 2, . . . 20 1, . . . 20 2, 3, 4; 6, . . . 22 3, 4, 6, 7, 8; 10, . . . 18; 20 2, 3, 4; 6, . . . 18; 20 4, 6, 7, 8, 10, 11, 12; 14, . . . 17 4, 5, 6, 8, 9, 10; 12, . . . 18 2, 4, 6, 8, 9, 10; 12, . . . 16 2, . . . 6; 8, 9; 11, . . . 16 2, 4, 6, 8; 10, . . . 13 2, 4, 6, 8; 10, . . . 14 2, 4, 6, 8, 10, 11, 12 4, 5; 7, . . . 12 4, 7, 8, 9 4; 6, . . . 10 4, 6, 7, 8 3, 4, 5, 6, 7, 8 3, 4, 5 2, . . . 6 2, 3, 4 2, 3, 4
TABLE 11.10 Deep nest
b open 21, 23, 24 1, 21, 22, 23 0, 21, 22 1, 5 0, 1, 2, 5, 9, 19 1, 5, 19 0, 1, 2, 3, 5, 9, 13, 18 1, 2, 3, 7, 11 0, 1, 3, 5, 7, 11 1, 7, 10 0, 1, 3, 5, 7, 9, 14 1, 3, 5, 7, 9 0, 1, 3, 5, 7, 9 1, 2, 3, 6 0, 1, 2, 3, 5, 6, 10 1, 2, 3, 5 0, 1, 2, 3, 5 1, 2 0, 1, 2, 6 1 1 2
b forbidden 25 0, 24 23 0 21 0 19 0 17 0 15 0 13 0 11 0 9 0 7 0 0, 5 0, 1 0, 1, 3 0, 1, 2 0, 1 0
Hilbert’s 16th problem
Two nests α1 α2 1 23 22 2 3 21 5 19 6 18 7 17 16 8 9 15 10 11 11 13 12 12
153
Three nests (α1 , α2 ) α3 (1, 2) 4, 8, 12, 15, 21 (1, 4) 6, 10 (1, 5) 18 (1, 6) 8, 12 (1, 7) 13, 16 (1, 8) 10, 15 (1, 10) 12 (1, 11) 11, 12 (2, 3) 6, 10 (2, 4) 5 (2, 5) 8, 12 (2, 6) 7 (2, 7) 10, 15 (2, 9) 9 (3, 3) 14 (3, 6) 6, 10 (4, 5) 6, 10 (5, 5) 9 (5, 7) 11, 12 (6, 6) 7 (6, 7) 10, 11 (7, 8) 9
c 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Deep nest b 1, 5 2, 5, 9 5 1, 2, 3, 9, 13 1, 3, 7, 11 3, 5, 7, 11 7 3, 5, 7, 9 3, 5, 7, 9 3, 7, 9 3 2, 3 3, 5 3, 5
TABLE 11.11 Pseudoholomorphic curves whose algebraic realizability is unknown
12 M curves of degree 9 with deep nests
CONTENTS 12.1 12.2
12.1
Results and rigid isotopy invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Curves without O1 jumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
155 158
Results and rigid isotopy invariants
Let C9 be an M curve with real scheme hJ q a q 1hb q 1hciii. We shall call the ovals of the groups a, b and c respectively outer , median and inner ovals. Let O1 and O2 be the nonempty ovals, with O1 ⊂ Int(O2 ). Theorem 12.1 The real schemes hJ q 1hb q 1hciii, with b + c = 26 and b, c odd are not realizable by M curves of degree 9. Definition 12.1 Let C9 be an M curve of degree 9 with a deep nest. We say that C9 has an O1 jump if there exist two median ovals A, C and two inner ovals B, D, such that a line passing through A and C separates B from D in the interior of O1 . Theorem 12.2 Let C9 be an M curve with real scheme hJ q 1hb q 1hciii, with even b, c: 1. If C9 has O1 jumps, then the complex scheme of C9 is b+6 c+6 c−6 hJ q 1− h( b−6 2 )+ q ( 2 )− q 1− h( 2 )+ q ( 2 )− iii or b−4 c−4 c+4 hJ q 1+ h( b+4 2 )+ q ( 2 )− q 1+ h( 2 )+ q ( 2 )− iii.
2. Otherwise, the complex scheme of C9 is b−4 c c hJ q 1− h( b+4 2 )+ q ( 2 )− q 1− h( 2 )+ q ( 2 )− iii.
Consider the ninth degree M curves with real scheme hJ q 1hb q 1hciii, b + c = 26, c ≥ 1. There were initially 26 admissible such real schemes. Theorem 12.1 excludes the 13 ones with b, c odd, Theorem 12.2 additionally rules out the cases b = 0 and 2. The cases b = 0, 1, 2, 3 and 25 had previously been forbidden in [47]. S. Orevkov [63] constructed M curves realizing the 10 real schemes with b, c even and 4 ≤ b ≤ 22. Each of these curves has three 155
156
Pencils of Cubics and Algebraic Curves in the Real Projective Plane A B
O1
O2
D
C J
FIGURE 12.1 O1 jump O1 jumps (see Definition 12.2 ahead), and they realize the complex schemes b−4 c−4 c+4 hJ q 1+ h( b+4 2 )+ q ( 2 )− q 1+ h( 2 )+ q ( 2 )− iii. He also excluded the case (b, c) = (24, 2). The classification of the ninth degree M curves with a deep nest and no outer ovals is thus completed. To prove the theorems, we shall need the classical complex orientation results, plus two Orevkov formulas. Let A be an Mcurve of degree m = 2k +1, with a deep nest. Let l+ and l− be respectively the numbers of positive and negative nonempty ovals, and let λ+ and λ− be respectively the numbers of positive and negative empty ovals. Let πsS , S, s ∈ {+, −} be the number of pairs (O, o) where o is an empty oval surrounded by O and (S, s) are the signs of (O, o). Then: + + Formula 1 (Orevkov) π− − π+ = (l+ )2 − − + (λ+ − λ− )/2 = (l− )2 + l− − π− Formula 2 (Orevkov) π+
Notice that l+ + l− = k − 2, and the injective pairs formed of two nonempty ovals bring a contribution of l+ l− − l− (l−2 −1) − l+ (l+2 −1) to Π+ − Π− . Adding the two Orevkov formulas yields the RokhlinMishachev formula. Let C9 be an M curve with real scheme hJ q a q 1hb q 1hciii. The next two lemmas are easily proved with auxiliary conics. Lemma 12.1 The c inner ovals lie in convex position. Lemma 12.2 If C9 has an O1 jump determined by A, B, C, D, then the principal segment [AC] cuts O1 , see Figure 12.1. The lines (AB), (BC), (CD), (DA) give rise to three quadrangles and four triangles. All of the remaining empty ovals lie in the union of the four triangles. Proposition 12.1 Let C9 have a median oval A1 such that: the pencil FA1 sweeping out O1 meets successively empty ovals A2 , A3 , . . . , A2J+2 , where the Aj , j even (resp. odd) are inner (resp. median) ovals. If J ≥ 1, then:
M curves of degree 9 with deep nests
A1
T1 A T2
2
157
T4 A4
A3
O1
O2
T3
FIGURE 12.2 C9 with J = 1 1. The ovals A1 , A2 , . . . A2J+2 lie in convex position; let ∆ be their convex hull. 2. Given a sequence of consecutive ovals Ai , Aj , Ak , Al , consider the triangle T[Aj Ak ] determined by the lines (Ai Aj ), (Aj Ak ), (Ak Al ), such that T[Aj Ak ] ∩ ∆ = [Aj Ak ]. Consider the union Z of the 2J + 2 triangles of this form. All of the remaining empty ovals lie in Z; if an oval A lies in T[Aj Ak ] ∩ Int(O2 ), A is not separated in T[Aj Ak ] from the edge [Aj Ak ] by J . 3. Let C1 , . . . , Cc and B1 , . . . , Bb be respectively the sets of inner and of median ovals. LetSBi be a median oval and Z(i) = Z \ {Ti1 , Ti2 }, where: if Bi ∈ Aj , j odd then Ti1 , Ti2 are the triangles having Bi as vertex; otherwise Ti1 is the triangle containing Bi and Ti2 is the triangle having a common median vertex with Ti1 . Consider: the complete pencils FCi (i = 1, . . . c) and the partial pencils FBi (i = 1, . . . b) sweeping out Z(i). The partial orderings of the empty ovals arising from these pencils are all consistent, and they give rise to a natural cyclic ordering of the empty ovals. 4. If a = 0, then λ+ − λ− = 0. Proof: The points 1), 2), 3) are easily proved using Bezout’s theorem with conics. Notice that if a = 0, the pencils of lines FAi : Aj → Tj → Ak have no J jumps over the sequences of ovals they sweep out, and give rise to a closed Fiedler chain involving all of the empty ovals, hence 4) follows. Definition 12.2 Let C9 be an M curve with a deep nest, and no outer ovals. The number of O1 jumps of C9 is J. Let J ≥ 1, the distribution of the jumps is the data (l1 , . . . , l2J+2 ), where li , i = 1, . . . , 2J + 2 are the cardinals of the successive groups of inner and median ovals in the cyclic ordering.
158
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
The case J = 1 is illustrated in Figure 12.2. Note that the number of O1 jumps and their distribution is a rigid isotopy invariant of C9 . Let C9 be an M curve with real scheme hJ q 1hb q 1hciii. We reproduce first the arguments from [47] to prove that C9 must have some median ovals. Suppose namely that b = 0. Let A, B, C be three of the 26 inner ovals and consider the pencils of lines FA : B → C, FB : C → A, and FC : A → B. These pencils have no J jumps over the sequences of inner ovals they sweep out, and they give rise to a closed Fiedler chain involving all of the inner ovals. Therefore, λ+ − λ− = 0, Π+ − Π− = ±1 and Λ+ − Λ− ∈ {0, 2, −2}. This contradicts the RokhlinMishachev formula. The real scheme hJ q 1h1h26iii is not realizable. Let now b > 0. Assume C9 has O1 jumps. Let 1 , 2 and n be respectively the contributions of O1 , O2 and of the inner ovals to Λ+ − Λ− (where 1 , 2 ∈ {+1, −1}, and n ∈ Z). By Proposition 12.1 (4), the contribution of the median ovals to λ+ − λ− is −n. Hence, by the RokhlinMishachev formula, one must have: 2(−1 2 − n1 ) + 1 + 2 = 8. There are four possible solutions: 1 = 2 = −1, n = 6; 1 = −1, 2 = 1, n = 3; 1 = 1, 2 = −1, n = −3; 1 = 2 = 1, n = −4. + + In the second case, one has l+ = 1, π− − π+ = 0; in the third case, one has + + l+ = 1, π− − π+ = 3. This contradicts the first Orevkov formula. Either of the other two cases verifies both Orevkov formulas. The numbers n, b, c have the same parity. If C9 has O1 jumps, these numbers are even, and Theorem 12.2 (1) is proved.
12.2
Curves without O1 jumps
Lemma 12.3 Let C9 be an M curve with real scheme hJ q1hbq1hciii without O1 jumps and such that c ≥ 2 and b ≥ 1. Let A be a median oval, and B, C be the extreme inner ovals met by the pencil FA sweeping out O1 . The complete pencil FC gives rise to a cyclic Fiedler chain involving all other empty ovals. In the corresponding cyclic ordering, all of the inner ovals are consecutive. Proof: Let T1 , . . . , T4 be the four triangles determined by A, B and C, T1 and T2 being the two ones with edge [BC]. The inner ovals lie in Z10 ∪Z20 , where Zi0 = Ti ∩ Int(O1 ), i = 1, 2. One of the zones Zi0 , i = 1, 2 is empty. Indeed, assume either zone contains an oval Ei . Then the conic through A, B, C, E1 , E2 cuts C9 at more than 18 points, a contradiction. Consider the pencil of lines FC starting at B and sweeping out the nonempty zone Zi0 , i ∈ {1, 2}. Let
M curves of degree 9 with deep nests
159
D be the last inner oval met by this pencil. The pencil FC : B → D meets no median oval. Indeed, assume that a median oval E is met by this pencil. Then, the conic EBADC cuts C9 at more than 18 points, a contradiction. The complete pencil FC gives rise to a cyclic Fiedler chain involving all other empty ovals. This chain splits into two consecutive subchains formed respectively by the inner and the median ovals. Proposition 12.2 Let C9 be an M curve of degree 9 with a deep nest, and denote by 1 any one of the inner ovals. The complete pencil of lines F1 has at most three J jumps over the sequence of median ovals. Note that the number of J jumps of a pencil based at an inner oval stays unchanged if one performs a rigid isotopy. Lemma 12.4 Let C9 be an M curve of degree 9 with a deep nest, and 1 be any one of the inner ovals. Assume there exist five other empty ovals 2, 3, 4, 5, 6 met successively by the pencil of lines based in 1, and such that there is a J jump between any two successive ovals in the cyclic ordering. Denote by F1123456 the rational pencil of cubics through 1, . . . , 6 with a double point at 1. Then, up to the action of the dihedral group D5 on 2, . . . , 6, there are three possible sequence of five reducible cubics for the pencil F1123456 : 16 ∪ 14523, 14 ∪ 12356, 12 ∪ 14365, 15 ∪ 12643, 13 ∪ 15426, 12 ∪ 14365, 15 ∪ 12643, 16 ∪ 12543, 13 ∪ 12456, 14 ∪ 12356, 16 ∪ 15234, 14 ∪ 15326, 15 ∪ 13264, 13 ∪ 14265, 12 ∪ 14365. Proof Assume there exists an inner oval 1 of C9 such that F1 has five J jumps over the median ovals. Let 2, 3, 4, 5, 6 be five median ovals met successively by the pencil, with a J jump between any two successive ovals in the cyclic ordering. Notice first that if F1 : X, Y, Z, X has a J jump between X and Y and between Y and Z, then: F1 has also a J jump between Z and X, and 1 lies in the principal triangle XY Z. Thus, 1 ∈ 234 ∩ 345 ∩ 456 ∩ 562 ∩ 623. As no three points among 1, . . . , 6 can be on a line, these five principal triangles must have a 2dimensional intersection. Let X, Y be two median ovals. The principal segment [XY ] cannot cut J by Bezout’s theorem. Thus, the convex hull of the points 1, . . . , 6 lies in RP 2 \ J . Fix an orientation of RP 2 \ J , to get rid of the symmetries, and assume that the positive pencil F1+ meets successively 2, 3, 4, 5, 6. Case 1: The five points lie in convex position. Consider a pair of points, that are consecutive for F1+ , say 2, 3, and assume these points are also consecutive for the convex cyclic ordering. This cyclic ordering is 2, 3, X, Y, Z, with X, Y, Z ∈ {4, 5, 6}. If X, Y, Z = 6, 5, 4 or 5, 6, 4, then 234 ∩ 345 = [34]; if X, Y, Z = 6, 4, 5 or 5, 4, 6, then 632 ∩ 456 = 6; if X, Y, Z = 4, 6, 5 or 4, 5, 6, then 234 ∩ 456 = 4. This is a contradiction. Then, the only possible convex cyclic ordering of the five points is 2, 4, 6, 3, 5. In the oriented RP 2 \ J , there are a priori two possibilities for the positive cyclic convex ordering of the five
160
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
points: 2, 4, 6, 3, 5 and 2, 5, 3, 6, 4. As F1+ sweeps out successively 2, 3, 4, 5, 6, the first of these possibilities is realized, see Figure 12.3. Case 2 : One of the points, say 2, lies in the convex hull of the other four. Assume 3 and 4 are consecutive in the cyclic convex ordering. Then, 234 ∩ 562 = 2. Contradiction. There are a priori two possibilities for the positive cyclic convex ordering of the four points: 3, 6, 4, 5 or 3, 5, 4, 6. In the first case, the quadrangle 3645 is divided into four triangles: T1 = 346 ∩ 356, T2 = 356 ∩ 345, T3 = 345 ∩ 456, T4 = 346 ∩ 456. If 2 ∈ T1 ∪ T4 , then 632 ∩ 345 = 3; if 2 ∈ T2 , then 623 ∩ 456 = 6. If 2 ∈ T3 , then 1 ∈ 234 ∩ 562 and the pencil F1+ sweeps out successively 6, 5, 4, 3, 2. This is a contradiction. In the second case, the quadrangle 3546 is divided into four triangles: T1 = 345∩356, T2 = 356 ∩ 346, T3 = 346 ∩ 456 and T4 = 345 ∩ 456. If 2 ∈ T1 ∪ T2 , then 623 ∩ 456 = 6; if 2 ∈ T3 , then 623 ∩ 345 = 3. This is a contradiction. Hence, 2 ∈ T4 , see Figure 12.4.
6
3 1 5
4 2
FIGURE 12.3 Case 1
3
6 1
5
FIGURE 12.4 Case 2
2
4
M curves of degree 9 with deep nests
161
6
3 5
1
4
2
FIGURE 12.5 Case 3 Case 3 : Two of the points lie in the principal triangle determined by the other three. Assume the two points are not consecutive for F1+ , say these points are 4, 6. Then 234 ∩ 345 = [34]. This is a contradiction. Thus, up to cyclic permutation of 2, 3, . . . , 6, one can choose 4, 5 as interior points. There are a priori two possibilities for the positive convex ordering of the other three points: 2, 6, 3 and 2, 3, 6. As 1 ∈ 632 and F1+ sweeps out successively 6, 2, 3, the latter possibility is excluded. The triangle 632 is divided in six triangles Ti , i = 1, . . . , 6 by the lines 42, 43 and 46, such that: T1 ∪ T2 = 346, T3 ∪ T4 = 234, T5 ∪ T6 = 246; and T1 , T6 have 4, 6 as common vertices, T2 , T3 have 4, 3 as common vertices, T4 , T5 have 4, 2 as common vertices. If 5 ∈ T1 ∪ T2 ∪ T4 ∪ T5 , then 345 ∩ 456 = [45]; if 5 ∈ T6 , then 456 ∩ 562 = [56]. This is a contradiction. One has 5 ∈ T3 , see Figure 12.5. Six points in generic position may realize four unordered Qconfigurations, see Theorem 1.1. Let us consider their four representatives α, β, γ, δ in Figure 2.17, corresponding to the positions 1 ∈ A, B, C and D in Figure 2.1. Now back to our curve. The points 1, . . . 6 realize the configuration α in case 1, (12)(36)γ in case 2 and (35)(46)δ in case 3. The three rational pencils of cubics are found immediately with help of Table 2.3. Proof of Proposition 12.2: Assume that there exists an inner oval 1 such that F1 has five J jumps over the set of median ovals. There exist median ovals 2, . . . 6 realizing the conditions of Lemma 12.4. Any cubic C3 from one of the three rational pencils of cubics has 27 intersection points with the union of the six base ovals, O1 , O2 and J . So the other ovals of C9 cannot be swept out, a contradiction. We can prove this without even drawing the pencil. The cubic C3 is obtained by perturbing a reducible cubic C2 ∪L. One has C2 ∩L = {1, P } (P is not a base point). Let C2 be any one of the admissible conics listed in Lemma 12.4. Both arcs 1P of C2 contain base points other than 1. Hence, C3 has a loop passing through 1 and some other base points. So C3 has four intersection points with each of the ovals O1 and 1. Morevover, C3 cuts each of the other five base ovals twice. Note also that C3 is divided in seven arcs by the base points. The conic C2 is divided in five arcs by the base points on
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
it; let s be any one of the four arcs that do not contain P . One arc of C3 is isotopic to s with fixed extremities. Two such arcs s1 and s2 of C2 cut each J once and O2 twice. The corresponding arcs of C3 cut each J at least once (an odd number of times), and O2 at least twice. A third arc of C3 must cut J . As it connects two median ovals or a median oval to the inner oval 1, it cuts O2 twice. We have found the 27 intersections. Proof of Theorems 12.1 and 12.2(2): Let C9 be an M curve of degree 9 with real scheme hJ q 1hb q 1hciii and without O1 jumps. Let 1 , 2 and n be respectively the contributions of O1 , O2 and of the median ovals to Λ+ − Λ− ; if c is odd, let 3 be the contribution of the inner ovals to Λ+ − Λ− (where 1 , 2 , 3 ∈ {+1, −1} and n ∈ Z). Let A be a median oval and C be one of the extreme inner ovals met by FA . Applying Lemma 12.3 and Proposition 12.2 with the complete pencil of lines FC , we prove that if c is odd, then n ∈ {±1, ±3} and if c is even, n ∈ {0, ±2, ±4}. The RokhlinMishachev formula yields respectively for odd and for even c: 2(−1 2 − 3 2 − 3 1 − n2 ) + 1 + 2 + n + 3 = 8 2(−1 2 − n2 ) + 1 + 2 + n = 8 For c odd, there is no solution. For c even, there are two solutions: 1 = 2 = −1, n = 4; 1 = 1, 2 = −1, n = 2. + + In the latter case, one has: l+ = 1, π− − π+ = 0. This contradicts the first Orevkov formula. In the first case, both Orevkov formulas are verified. The numbers n, b, c have the same parity, if C9 has no O1 jump, these numbers are even.
13 M curves of degree 9 with four or three nests
CONTENTS 13.1 13.2 13.3 13.4
13.1
Statement of the results and first proofs . . . . . . . . . . . . . . . . . . . . . . . . . Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M curves with three nests and a jump . . . . . . . . . . . . . . . . . . . . . . . . . . End of the proof, using two Orevkov formulas . . . . . . . . . . . . . . . . . .
163 173 183 194
Statement of the results and first proofs
We consider now M curve of degree 9 with real scheme hJ q 1hα1 i q 1hα2 i q 1hα3 i q 1hα4 i q βi or hJ q 1hα1 i q 1hα2 i q 1hα3 i q βi. In this chapter, we call a nest a piece of curve 1hαi i, and the curves under consideration have either four or three nests. Let Oi , i = 1, . . . 3 or 4 be the nonempty ovals. For convenience, we shall sometimes name a nest after its exterior oval Oi . Definition 13.1 Let C9 be a curve of degree 9 and let O be a nonempty oval of C9 . The curve C9 has a jump in O determined by A, B, C, D if there exist two empty ovals B, C inside of O, an oval A inside of some other nonempty oval O0 , and an empty oval D exterior to O, such that B and C are separated inside of O by any line passing through A and D (see Figure 13.1). Lemma 13.1 Let C9 be a curve of degree 9 with three or four nests, and a jump determined by A, B, C, D. Then D is exterior. Proof: Bezout’s theorem with C9 and the line AD implies that D is either exterior, or interior to the same oval O0 as A. In the latter case, the conic through A, B, C, D and a fifth oval interior to a nonempty oval different from O and O0 cuts C9 at more than 18 points, a contradiction. Remark: The jumps are rigid isotopy invariants. Lemma 13.2 Let C9 be a curve of degree 9 with a jump in O determined by A, B, C, D. Let A0 lie together with A inside of O0 . Then, A0 , B, C, D also give rise to a jump. 163
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
O’
O
A
B
C
D
FIGURE 13.1 Ovals of a ninth degree curve giving rise to a jump Proof: Assume this is not true, so A0 lies in the sector (DB, DC) that contains the nonprincipal segment [BC]0 . By Lemma 13.1, the line BC cannot separate A from A0 inside of O0 , so the conic through A, A0 , B, C, D is A0 BDAC or ABA0 CD. This conic cuts C9 at 20 points, a contradiction. Theorem 13.1 Let C9 be an M curve of degree 9 with real scheme hJ q 1hα1 i q 1hα2 i q 1hα3 i q 1hα4 i q βi. Then 1. All of the αi are odd. 2. The complex scheme of C9 is of the form: hJ q 11 h( α12+1 )−1 q ( α12−1 )1 i q 12 h( α22+1 )−2 q ( α22−1 )2 iq 13 h( α32+1 )−3 q ( α32−1 )3 i q 14 h( α42+1 )−4 q ( α42−1 )4 i q βi, with i ∈ {+, −} Korchagin [47] had previously proved that β must be even, and if β = 0, then α1 , α2 , α3 , α4 are all odd. Theorem 13.1 excludes another 213 real schemes. Theorem 13.2 Let C9 be an M curve of degree 9 with real scheme hJ q 1hα1 i q 1hα2 i q 1hα3 i q βi. At least one of the αi , i = 1, 2, 3 is odd. One may reformulate this, saying that the case even, even, even is nonrealizable. Theorem 13.2 excludes the 53 real schemes hJ q 1hα1 i q 1hα2 i q 1hα3 i q βi (α1 + α2 + α3 + β = 25, α1 ≤ α2 ≤ α3 ) with α1 , α2 , α3 even. Among them, the 12 ones with β = 1 had already been excluded by A. Korchagin in [46]. In May 2018, S. Orevkov constructed M curves with three nests even,
M curves of degree 9 with four or three nests A1 Q2 A3 O3
µ2
T0
J
O1
Q3
λ0
Q1
µ1
T1
λ1
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A2
µ3
T3
O2 T2
λ3
λ2
FIGURE 13.2 Three nests of C9 even, odd, disproving thus Conjecture 3 (see Section 11.4). We proved in [25] that the conjecture is true for the curves without a jump. (As the proof is very long, we do not include it in this book.) Let C9 have three or four nests, we choose three empty ovals Ai , i = 1, 2, 3 distributed in three nests with nonempty ovals Oi , i = 1, 2, 3. The baselines A1 A2 , A2 A3 , A3 A1 , and the pseudoline J determine four triangles T0 , T1 , T2 , T3 and three quadrangles Q1 , Q2 , Q3 in RP 2 , where T0 is the principal triangle A1 A2 A3 . Note that J does not cut T0 , by Bezout’s theorem, see Figure 13.2 (where the λi and µi should be ignored, they will be defined later). For C9 with three nests, we shall call the six ovals O1 , O2 , O3 , A1 , A2 , A3 principal ovals. Lemma 13.3 Let C9 be a curve of degree 9 with three or four nests, and consider any triple of nonempty ovals O1 , O2 , O3 . Assume that C9 has a jump in O3 , determined by A1 , B, C, D. Up to permutation of A1 , A2 , the ovals A1 , A2 , C, D, B lie in convex position. So, A2 , B, C, D also determine a jump. Proof: Each of the nonempty ovals Oi is divided in four zones by the lines Ai Aj , Ai Ak , lying respectively in the two triangles T0 , Ti and the two quadrangles Qj , Qk (see Figure 13.2). If C is in a triangle, then the conic through A1 , A2 , C, D, B cuts C9 at 20 points, a contradiction. Up to permutation of A1 , A2 , we may assume that C is in Q1 , see Figure 13.3. Hence, A1 , A2 , C, B lie in convex position. By hypothesis, D lies in the sector (A1 B, A1 C) containing the principal segment [BC]. Using again Bezout’s theorem with the conic through A1 , A2 , C, D, B, we get that D must lie in the sector (A2 , B, A2 C) containing [BC], outside of the convex hull A1 A2 CB. So, C lies in the triangle T ⊂ (Q1 ∪ T1 ) determined by the lines A1 B, A2 C and the segment [BC].
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
Moreover, D is in T ∩ Q1 . (Note that [BC] is not necessarily entirely interior to O3 , and O3 may cut any one of the segments [A1 D], [A1 D]0 . The choice made in Figure 13.3 is arbitrary.)
A1 Q2 O3
T3
Q3 T0
B C
D
J
O1
A2 Q1
O2 T2
T1
FIGURE 13.3 Zone containing the oval D Definition 13.2 Let C9 be a curve of degree 9 with at least two nests. The curve C9 has nk jumps in Ok with repartition (l1 , . . . , l2nk +1 ) if the pencil of lines FA sweeping out Ok meets successively 2nk + 1 groups of ovals, which are situated alternatively in, out, . . . , in the interior of Ok and have cardinals l1 , . . . , l2nk +1 , for any choice of A interior to another nonempty oval Oi . It follows from Lemmas 13.2 and 13.3 that the number of jumps in Ok and their repartition does not depend on the choice of A. Thus, Definition 13.2 is correct. In Definition 13.1, we may omit the mention of A, and speak of a jump in O determined by B, C, D. Lemma 13.4 Let C9 have four nests. Then, for any choice of A1 , A2 , A3 distributed in the other nonempty ovals O1 , O2 , O3 , the oval O4 lies in one of the quadrangles Q1 , Q2 , Q3 . Proof: Assume that O4 lies in T0 ∪ T1 ∪ T2 ∪ T3 , and let A4 be an empty oval interior to O4 . The conic through A1 , A2 , A3 , A4 and a fifth empty oval cuts C9 at 20 points, a contradiction. Lemma 13.5 (Korchagin) Let C9 have four nests. Then C9 verifies the following properties (see Figure 13.4): 1. The lines A1 A2 , A2 A3 , A3 A4 , A4 A1 divide RP 2 in three quadrangles and four triangles: A1 A2 A3 A4 , A1 M A3 N , A2 M A4 N , A1 A2 M , A2 A3 N , A3 A4 M , A4 A1 N , where M = A1 A4 ∩ A2 A3 and
M curves of degree 9 with four or three nests
167
N
O1
Z3 O4
A1
A4
Z2
Z4
M
J
A3
A2
O3 Z1
O2
FIGURE 13.4 C9 with four nests N = A1 A2 ∩ A3 A4 . The first of these quadrangles is not cut by J but the other three are; exactly two successive triangles are not cut by J , say A1 A4 N and A3 A4 M . 2. All of the remaining ovals of C9 lie in Z1 ∪ Z2 ∪ Z3 ∪ Z4 , where the Zi are four zones of RP 2 defined respectively as: the zone of the triangle A2 A3 N that is separated from N by J , the zone of the triangle A1 A2 M that is separated from M by J , the triangle A1 A4 N and the triangle A3 A4 M . Proof: By Lemma 13.4, O4 lies in Q1 ∪ Q2 ∪ Q3 , up to permutation of the indices, we may assume that O4 lies in Q2 , 1) follows immediately. To prove 2), assume there exists some empty oval E of C9 outside of Z1 ∪ Z2 ∪ Z3 ∪ Z4 . The conic through A1 , A2 , A3 , A4 , E cuts C9 at 20 points, a contradiction. Proposition 13.1 Let C9 be a curve of degree 9 with four nests. Then C9 has no jump. Proof: Let A4 = B and assume that C9 has a jump in O4 determined by B, C, D. By Lemmas 13.3 and 13.5, the ovals C, D lie in the same zone Z3 or Z4 , say it is Z4 , and A1 , A2 , A3 , C, D, B lie in convex position. By Bezout’s theorem, D lies outside of the conic A1 A2 A3 CB. The six points realize a βconfiguration, the orbits of G(β) are b : {A1 , A3 , D} and b0 : {A2 , B, C}. Each point in the orbit b (resp. b0 ) lies outside (resp. inside) of the conic passing through the other five, see Section 2.2. Let us make (A1 , A2 , A3 , C, D, B) =
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
(5, 3, 1, 6, 4, 2). The point 1 lies in the zone F determined by the other five points, see Figure 2.1. The rational pencil of cubics with node at 1 is shown in the lower part of Figure 2.8. All of the cubics of this pencil cut: the set of ovals A1 , A2 , A3 , B, C, D, O1 , O2 , O3 , O4 plus J at 27 points. The other ovals of C9 cannot be swept out, a contradiction. Lemma 13.6 Let C9 be an M curve of degree 9 with four nests. One has: Λ+ − Λ− = 0 and Π+ − Π− = 4. Proof: The two pencils of lines FA1 = A2 → A3 → A4 and FA3 = A4 → A1 → A2 sweep out together all of the 28 ovals. Each pencil gives rise to two Fiedler chains, as the number of nonreal intersection points of the lines with C9 is at most four. (Gluing these chains yields either two cyclic chains or one double cyclic chain.) As there are no J jumps between two consecutive ovals in a chain, one has Λ+ − Λ− = 0. The RokhlinMishachev formula reads 2(Π+ − Π− ) + Λ+ − Λ− = 8, hence Π+ − Π− = 4. Proof of Theorem 13.1: Consider the maximal pencil of lines FAj sweeping out Oi . As C9 has no jump, the αi ovals interior to Oi form a Fiedler chain and have thus alternating orientations (with respect to both Oi and J ). So the contribution of the nest Oi to Π+ − Π− is 0 if αi is even and 1 or −1 if αi is odd. As Π+ − Π− = 4, the contribution of each nest must be +1. Till the end of this chapter, C9 stands for an M curve of degree 9 with three nests. Lemma 13.7 The curve C9 has at most one jump. Otherwise stated, n1 + n2 + n3 = 0 or 1. By convention, we fix that if C9 has a jump, it is in O3 (n1 = n2 = 0 and n3 = 1). Proof: Let C9 have at least two jumps, and assume first that there are jumps in two different ovals, see the upper part of Figure 13.5. Let C3 (A) be the cubic passing through A, G, F, D, C, E, B, with node at A. The line AE (resp. AF ) cuts C3 (A) at A with multiplicity two, and at E (resp. F ). Moreover, AE separates B from C inside of O3 , and AF separates D from G inside of O2 . Thus, C3 (A) cannot contain an arc BC interior to O3 or an arc DG interior to O2 . This cubic cuts the set of seven empty ovals (with multiplicity) at 16 points, and O2 ∪ O3 at 8 points. So C3 (A) must have either an isolated node at A, or a loop contained in the interior of O1 . Let us make (A, G, F, D, C, E, B) = (1, 2, 3, 4, 5, 6, 7). These seven points lie in convex position and realize the list 2+ or 7− (see Section 5.1). In both cases, the cubic with node at 2 cuts C9 at more than 27 points, a contradiction. Assume now that there are at least two jumps in the same oval O3 , see the lower part of Figure 13.5. The ovals (A, G, C, E, B) on one hand, (A, G, F, D, C) on the other hand, lie in convex position. Consider the cubics C3 (A) and C3 (G) passing each through A, G, F, D, C, E, B, with nodes at A and G respectively. With similar arguments as above, we get that none of these cubics can contain arcs BC, BF or CF interior to O3 . Thus, each of these cubics cuts the set of
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seven empty ovals at 16 points, and O3 at 6 points. Moreover, C3 (A) cuts O2 twice and C3 (G) cuts O1 twice. So C3 (A) has either an isolated node at A, or a loop contained in the interior of O1 and C3 (G) has an isolated node at G, or a loop contained in the interior of O2 . Moreover, the odd component J3 (A) of C3 (A) meets successively B, E, C, D, F, G and the odd component J3 (G) of C3 (G) meets successively A, B, E, C, D, F . The cyclic ordering B, E, C, D, F is the same on the two odd components. Let us make (A, G, F, D, C, E, B) = (1, 2, 3, 4, 5, 6, 7). Up to the action of S7 , these seven points must realize one of the 14 Qconfigurations in Figures 3.4, 3.5, 3.7. The sets of seven nodal cubics corresponding to each of them are shown in Figures 3.8, 3.9, 3.10. We need to find a set with two cubics that each have an isolated node (or a loop passing through none of the seven points apart from the node). The five points that are not a node of one of the cubics must moreover lie in the same cyclic ordering on the two odd components. The first condition is achieved only for the configurations (G, 6), (H, 6), (K, 6) and R. But none of them satisfies the second condition. Lemma 13.8 Let C9 have three nests and let Oi be any one of the nonempty ovals. All of the lines through pairs of ovals interior to Oi cut either all of the principal segments [Aj Ak ] or all of the nonprincipal segments [Aj Ak ]0 , where Aj (Ak ) ranges over the set of ovals interior to Oj (Ok ). Proof: Assume there are at least three ovals E, F, G inside of Oi . By Bezout’s theorem with the conic through E, F, G, Aj , Ak , the lines EF , EG, F G must all cut the same segment [Aj Ak ] or [Aj Ak ]0 . Let A0k be another oval interior to Ok . If EF cuts the principal segment [Aj Ak ] and the nonprincipal segment [Aj A0k ]0 , then EF cuts the principal segment [Ak A0k ] which is interior to Ok . This contradicts Lemma 13.1. Definition 13.3 Let C9 have three nests. A nonempty oval Oi of C9 is separating if any line EF joining two ovals E, F interior to Oi cuts the principal segments [Aj Ak ], where Aj (Ak ) ranges over the set of ovals interior to Oj (Ok ). Otherwise, Oi is nonseparating. By Lemma 13.3, a nonempty oval with a jump is nonseparating. Lemma 13.9 Let A be an empty oval interior to Oi or Oj ; we denote by FA the pencil of lines based at A and sweeping out Ok from the first interior oval to the last one. All of the pencils obtained with the αi + αj choices of A give rise to the same Fiedler chain of empty ovals, the sequence of Ok . The sequence of O3 with a jump is formed of three jump sequences (int, ext, int), containing respectively l1 , l2 and l3 ovals. Proof: Let A be interior to Oi and A0 be interior to Oi ∪ Oj . If Ok has no jump, Lemma 13.9 follows immediately from Lemma 13.1 if A0 is in the same nest as A, and from Lemma 13.8 otherwise. Let O3 have a jump and B, C be the first and the last oval met by FA . Let A0 be an oval interior to
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
A
J
O1 O2
O3
G
F
B D
C E
O1
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FIGURE 13.5 Curve C9 with three nests and several jumps
O2
M curves of degree 9 with four or three nests
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O1 ∪O2 and consider FA0 sweeping out O3 from B to C. We need to prove that any three ovals met successively by FA are swept out in the same ordering 0 by FA . If at least two of the three ovals are interior, this follows from Lemmas 13.1, 13.2, 13.3, 13.8. Now let the exterior group contain l2 ≥ 2 ovals, and let D1 , D2 be two of them. Assume that FA meets successively B, D1 , D2 , C and FA0 meets successively B, D2 , D1 , C. Then, the conic ABA0 D2 D1 cuts C9 at 20 points, a contradiction. This finishes the proof. Let O3 have a jump; we name some of the ovals of C9 in such a way that the jump is determined by A, B, C, D for any choice of A interior to O1 or O2 , and A1 , A2 , C, D, B lie in convex position. The ovals B, C, D are distributed in the three jump sequences (int, ext, int), and the jump is odd if each of these sequences contains an odd number of ovals. The pencil of lines FD : B → O1 → O2 → C sweeps out a sector (DA1 , DA2 ) which is the union of two triangles determined by D, A1 and A2 , one of them being principal. By Bezout’s theorem between C9 and the lines (DA1 , DA2 ), the oval O3 intersects only one of these two triangles. Definition 13.4 Let C9 have a jump. If O3 intersects the principal triangle DA1 A2 the oval D is front, otherwise D is back, see Figure 13.6. If all of the ovals in the exterior jump sequence are front (resp. back), then O3 is crossing (resp. bending). Lemma 13.10 Let C9 be as in Figure 13.2, and choose a supplementary empty oval A0i interior to Oi . The lines A0i Aj , A0i Ak , Aj Ak , and J give rise to new triangles T00 , T10 , T20 , T30 . Let E be another empty oval. 1. If Oi is nonseparating, then: E ∈ Tl ⇐⇒ E ∈ Tl0 for l = 0, . . . 3. 2. If Oi is separating (say A0i ∈ Ti ), then: E ∈ Tl ⇐⇒ E ∈ Tl0 for l = j, k; and E ∈ T0 ∪Ti ⇐⇒ E ∈ T00 ∪Ti0 . In the latter case, one of the three possibilities hereafter is realized: E ∈ T0 ∩ T00 , E ∈ Ti ∩ Ti0 , or E ∈ Ti ∩ T00 is interior to Oi . Proof: It follows from Bezout’s theorem with the conic through Ai , A0i , Aj , Ak , E. Definition 13.5 The empty ovals in T0 ∪ T1 ∪ T2 ∪ T3 are triangular, the empty ovals in Q1 ∪ Q2 ∪ Q3 are quadrangular. It follows from Lemma 13.10 that these properties are independent from the choice of A1 , A2 , A3 . Lemma 13.11 Let C9 have a jump. 1. If D is front, there are no ovals in T3 . 2. If D is back, there are no ovals in T0 ∪ T1 ∪ T2 .
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane A
1
O3 D
J
A
1
A2
B
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O1 O2
C
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FIGURE 13.6 Jump in O3 , D front, D back Proof: Let C2 be the conic through A1 , A2 , B, C and another empty oval E. The triple (A1 , A2 , A3 = B) determines triangles Tl , l = 0, . . . 3, and the triple (A1 , A2 , A03 = C) determines triangles Tl0 , l = 0, . . . 3. The oval O3 is nonseparating. We apply Lemma 13.10 (1) with i = 3. 1. Let E ∈ T3 ; one has E ∈ T3 ∩T30 , thus C2 = A1 EA2 CB. By Bezout’s theorem with C9 , the arc CB of C2 lies inside of O3 . Thus, D > C2 , and E < A1 A2 CDB. The conic A1 A2 CDB cuts C9 at 20 points, a contradiction. 2. Let E ∈ T0 ; one has E ∈ T0 ∩ T00 . The conic C2 = A1 EA2 BC cuts C9 at 20 points, a contradiction. Let E ∈ T1 ∪ T2 . By symmetry, we may suppose that E ∈ T1 . As E ∈ T1 ∩ T10 , one has a priori C2 = A1 A2 ECB or A1 A2 BCE. By Bezout’s theorem with C9 , one must have C2 = A1 A2 ECB, and the arc CB of C2 lies inside of O3 . Thus, D < C2 and E < A1 A2 CDB. The conic A1 A2 CDB cuts C9 at 20 points, a contradiction. If C9 has both front and back ovals (O3 is neither crossing nor bending), then there are no triangular ovals. Let αi+ and αi− be the numbers of positive and negative ovals interior to Oi . Lemma 13.9 implies that αi+ − αi−  ≤ 2. The equality is achieved if and only if i = 3 and O3 has an odd jump. Let us call base ovals the empty principal ovals A1 , A2 , A3 . For i = 1, 2, 3, we denote by i the contributions of Oi to Λ+ − Λ− , and by ηi the contributions of Ai (i , ηi ∈ {±1}). Consider now the contributions to Λ+ − Λ− brought by the nonprincipal ovals in each of the seven zones. Let λ0 , λ1 , λ2 , λ3 be these contributions for the triangles T0 , T1 , T2 , T3 , and µ1 , µ2 , µ3 be those for the quadrangles Q1 , Q2 , Q3 , see Figure 13.2. Lemma 13.12 One has: λ0 + µ1 − λ1 = − 21 (3 + η3 + 2 + η2 ) λ0 + µ2 − λ2 = − 12 (3 + η3 + 1 + η1 )
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λ0 + µ3 − λ3 = − 12 (2 + η2 + 1 + η1 ) 3λ0 + µ1 + µ2 + µ3 − λ1 − λ2 − λ3 + 1 + 2 + 3 + η1 + η2 + η3 = 0 λ0 − λ1 − λ2 − λ3 = − 12 (Λ+ − Λ− ) = Π+ − Π− − 4 Proof: Apply Fiedler’s theorem to the pencils of lines FA1 : A3 → T0 ∪Q1 ∪ T1 → A2 , FA2 : A1 → T0 ∪ Q2 ∪ T2 → A3 and FAP → T0 ∪ Q3 ∪ T3 → A1 . 3 : A2 P Subtracting λ0 + µ1 + µ2 + µ3 + λ1 + λ2 + λ3 + i + ηi = Λ+ − Λ− from the fourth identity, and combining with RokhlinMishachev’s formula yields the last identity. Corollary 13.1 Let C9 be an M curve of degree 9 with real scheme hJ q 1hα1 i q 1hα2 i q 1hα3 i q βi. Then β > 0. Proof: Let β = 0. Then λ0 − λ1 − λ2 − λ3 = 0, and Π+ − Π− = 4. So C9 has an odd jump, and hence an exterior oval, a contradiction. See also [13], [47].
13.2
Inequalities
Let C9 be as previously an M curve of degree 9 with real scheme hJ q 1hα1 i q 1hα2 i q 1hα3 i q βi. Let us perform a Cremona transformation cr : (x0 : x1 : x2 ) → (x1 x2 : x0 x2 : x0 x1 ) with base points A1 , A2 , A3 , distributed in the corresponding base ovals. The respective images of the lines A1 A2 , A2 A3 , A3 A1 are the points A3 , A1 , A2 . One has cr(T0 ) = T0 . For any point E that does not lie on a base line, we shall call its image also E. The curve C9 is mapped onto a curve C18 of degree 18 with three singular points. The complex orientation of C9 induces via cr a complex orientation of C18 . We shall call the main part of C18 the piece formed by the images of J and the principal ovals, see Figure 13.7 where cr(Ai ) and cr(Oi ) stand for the images of the ovals Ai and Oi . In addition to its main part, C18 has 22 ovals, that are the images of the nonprincipal ovals of C9 . An oval A of C18 will be said to be interior, exterior, positive, negative, triangular, quadrangular if its preimage is. Let O = cr(J ). One has Int(O) = cr(T0 ∪ T1 ∪ T2 ∪ T3 ), Ext(O) = cr(Q1 ∪ Q2 ∪ Q3 ). The triangular ovals of C18 lie inside of O, the quadrangular ovals lie outside. A configuration of ovals of C18 lies in convex position if there exists a line L such that the ovals lie in convex position in the affine plane RP 2 \ L. Let π+ (π− ) be the numbers of positive (negative) injective pairs formed by the triangular ovals and O. One has π+ − π− = λ1 + λ2 + λ3 − λ0 . Definition 13.6 Let {i, j, k} = {1, 2, 3}. A base line Ai Aj and a conic C2 are mutually maximal if C2 cuts Ai Aj and C2 cuts each component cr(Ok ) and cr(Ak ) at four points.
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A1
cr(Q2)
O= cr(J )
cr(T2)
cr(T3)
cr(Q3 )
T0 A3
cr(O2 ) cr(A2 ) cr( O2 ) cr(A2 )
A2
cr(T1) cr(Q1)
cr(O1 )
cr(A1)
cr( A1) cr(O1 ) cr(O ) cr(A3 ) 3 cr(O3 ) cr(A3 )
FIGURE 13.7 Main part of C18 Lemma 13.13 If a base point Ai lies inside of a conic C2 , then the two base lines Ai Aj and Ai Ak are maximal with respect to C2 . Proof: Let Ai be interior to C2 and let Pk , Qk = C2 ∩ Ai Aj , Pj , Qj = C2 ∩ Ai Ak . The conic C2 meets Pk , Pj , Qk , Qj in this ordering. Each arc of C2 joining two consecutive points cuts cr(Ok ), cr(Ak ), cr(Oj ) and cr(Aj ). Lemma 13.14 Let C2 be a conic passing through five ovals E1 , . . . E5 of C18 . Let Ai , Aj be two of the base points, lying outside of C2 , such that the line Ai Aj cuts C2 . If any of the following conditions is verified, then Ai Aj is maximal with respect to C2 . 1. Each arc of C2 \ (C2 ∩ Ai Aj ) passes through an oval Em that is exterior to cr(Ok ), or cuts O. 2. Int(C2 ) ∪ Int(O) is orientable and Ai , Aj lie on different arcs of O \ (O ∩ C2 ). The proof is easy and left to the reader. Lemma 13.15 Let C2 be a conic passing through five ovals E1 , . . . E5 of C18 , and having at least four intersection points with O. Then one of the three base lines, say A1 A3 , is nonmaximal with respect to C2 , and the points A1 , A3 lie outside of C2 .
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Proof: If the three base lines are maximal with respect to C2 , then C2 cuts the images of the principal ovals at 24 points, O at four points, and the union ∪Ei , i = 1, . . . , 5 at 10 points, hence in total 38 intersections. This is a contradiction. A base line, say A1 A3 , is nonmaximal; the points A1 , A3 must lie outside of C2 by Lemma 13.13. Lemma 13.16 Let C2 = E1 E2 E3 E4 E5 be a conic satisfying: Int(O) ∪ Int(C2 ) is orientable, E1 , E3 , E5 are quadrangular, and E2 , E4 are triangular. We call principal arcs of C2 the five arcs determined by Ei , i = 1 . . . 5. Consider the set S of arcs of O \ (O ∩ C2 ), each arc of S is isotopic with fixed endpoints to an arc of C2 . 1. There exist four arcs s1 , . . . , s4 of S such that: The two endpoints of si , i = 1, 2, 3 lie respectively on the principal arcs Ei Ei+1 , Ei+1 Ei+2 of C2 , and the two endpoints of s4 lie respectively on the principal arcs E4 E5 , E1 E2 . The arcs s1 , s3 are exterior and the arcs s2 , s4 are interior to C2 . Any other arc s of S has its two endpoints on the same principal arc E1 E2 , E2 E3 , E3 E4 or E4 E5 of C2 . 2. One of the base lines, say A1 A3 , is nonmaximal for C2 and A1 , A3 lie both on the same arc s = s1 or s3 of S. Proof: Let s be an arc of S that is isotopic with fixed endpoints to an arc σ of C2 . If σ contains Ei and not Ej , then the line Ei Ej cuts s; if s is exterior to C2 and σ contains Ei , Ej , then the line Ei Ej cuts s twice. These obvious facts will be used several times in the proof. As only one of the ovals Ei , Ei+1 , i = 1, . . . 4 is triangular, each of the principal arcs Ei Ei+1 of C2 cuts O an odd number of times, hence C2 has at least four intersection points with O. Lemmas 13.15 and 13.14(2) imply that a base line, say A1 A3 , is nonmaximal for C2 and the points A1 , A3 lie on the same arc s of S, exterior to C2 . By Bezout’s theorem, each of the ten lines Ei Ej has at most two intersection points with O. As E2 , E4 are both triangular, the line E2 E4 cuts O at two points M, N , say with the cyclic ordering M, E2 , E4 , N . As moreover Int(C2 )∪Int(O is orientable, the arc s1 (s3 ) of S containing M (N ) is exterior to C2 and isotopic with fixed endpoints to an arc σ1 (σ3 ) of C2 containing E2 (E4 ). Suppose that σ1 contains also one of the points E1 , E3 , or σ3 contains one of the points E3 , E5 , say E3 ∈ σ1 . The line E2 E3 cuts s1 twice. As E2 is triangular and E3 quadrangular, there exists some other arc s01 ∈ S exterior to C2 , with one endpoint only on the principal arc E2 E3 of C2 . The arc s01 must cut either the line E2 E4 between E2 and M , or the line E2 E3 . This is a contradiction. The other three cases are similar, thus s1 and s3 are as required in 1). Let now s2 be an arc of S interior to C2 with one endpoint only on the principal arc E2 E3 of C2 ; s2 cannot cut the line E2 E4 , hence its other endpoint is on the principal arc E3 E4 of C2 . Let now s4 (s04 ) be an arc of S interior to C2 with one endpoint only on the principal arc E1 E2 (E4 E5 ). The arc s4 (s04 ) cannot cut the line E2 E4 . Moreover, if s4 has its endpoint on
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the principal arcs E1 E2 and E4 E5 , then the line E1 E3 cuts s2 and s4 . Hence, either s04 = s4 , or both arcs have their second endpoint on the principal arc E1 E5 of C2 . Let S 0 = {s1 , s2 , s3 , s4 , s04 }. All of the lines Ei Ej , (i, j) 6= (1, 5) have two intersections with S 0 . If s04 6= s4 , the line E1 E5 cuts each of the two arcs s4 and s04 . Hence, any arc s ∈ / S 0 has both endpoints on the same principal arc of C2 . By Lemma 13.15, one base line, say A1 A3 , is nonmaximal with respect to C2 , and A1 , A3 are exterior to C2 . We will prove that the arc s containing A1 , A3 must be s1 or s3 . Assume that s is another arc of S. The endpoints of s are on the same principal arc of C2 ; up to symmetry, we may assume that this arc is E1 E2 , E2 E3 or E5 E1 . Let us start with the case where the arc is E1 E2 . Consider the conics C2 (A1 ) = A1 E2 E3 E4 E5 and C2 (A3 ) = A3 E2 E3 E4 E5 , see the first picture of Figure 13.8. One of the two base points lies in the interior of the conic passing through the other, say A3 lies in the interior of C2 (A1 ). Hence, A2 A3 is maximal with respect to C2 (A1 ). The other two base lines, passing through A1 , are of course also maximal with respect to C2 (A1 ) (the other base point A2 may be inside or outside of this conic). Then, C2 (A1 ) cuts O at six points, each of cr(Oi ), cr(Ai ), i = 1, 2, 3 at four points, and each of the ovals E2 , . . . , E5 at two points. Hence, in total 38 intersection points with C18 , a contradiction. The cases where s has its endpoints on one of the principal arcs E2 E3 or E5 E1 are similar, letting C2 (Ai ), i = 1, 3 be respectively: Ai E3 E4 E5 E2 and Ai E1 E2 E3 E4 , see the second and third pictures of Figure 13.8. To finish the proof, one needs to show that the principal arc E5 E1 of C2 cannot cut O (otherwise stated s04 = s4 ). Assume the contrary. Up to symmetry, we may assume that s = s1 . We find again a contradiction, using conics C2 (Ai ) = Ai E3 E4 E5 E1 , see the last picture of Figure 13.8. Lemma 13.17 Let the curve C18 contain six ovals B1 , D3 , B2 , D1 , B3 , D2 lying in convex position in some affine plane RP 2 \ L, such that B1 , B2 , B3 are triangular, and the segments [Bi Bj ] interior to the convex hull are also interior to O. Then, one of the ovals D1 , D2 , D3 must also be triangular. Proof: Let Bi , Di , i = 1, 2, 3 satisfy the conditions of the lemma and assume that D1 , D2 , D3 are all quadrangular. Denote by H the hexagon bounding the convex hull. Each of the six edges [Bi Dj ] of H cuts O once. Let P1 , Q1 , P2 , Q2 , P3 , Q3 be the successive intersections of O with the edges [D2 B1 ], [B1 D3 ], . . . [B3 D2 ]. Connect pairwise these six points by arcs of O interior to H; these arcs must be Q1 P2 , Q2 P3 , Q3 P1 . Connect pairwise the six points with exterior arcs. By Bezout’s theorem with the lines Di Dj , these arcs must be Q1 P1 , Q2 P2 , Q3 P3 , see the first picture of Figure 13.9. Consider the three conics B1 D3 D1 B3 D2 , B3 D2 D3 B2 D1 and B2 D1 D2 B1 D3 . Bezout’s theorem with the six lines supporting H implies that Int(O) ∪ Int(C2 ) is orientable. By Lemma 13.16(1), the arc Dk Di of Bi Dk Di Bk Dj does not cut O. Each oval Bi lies inside of the conic C2 determined by the other five ovals. Hence, each exterior arc of O \ (O ∩ C2 ) is entirely contained in another conic.
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FIGURE 13.8 Contradiction with the conic C2 (A1 )
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Thus, any choice of base points leads to a contradiction with Lemma 13.16(2), see the second picture of Figure 13.9. The proofs of the next two statements involve conics passing through five empty ovals of C18 . (These conics are images by cr of rational quartics passing through the preimage of the five ovals, and having nodes at A1 , A2 , A3 .) Several times, we will find a conic that is maximal with respect to the three base lines. The maximality always follows from Lemma 13.14(1): each base line separates, on this conic, a pair of exterior ovals. Lemma 13.18 Let the base ovals A1 , A2 , A3 of C9 be such that T0 contains only exterior ovals of C9 . One λ0  ≤ 3. If λ0 = ±3, then the nonempty P has P ovals are all separating and i + ηi = ∓6. Moreover, one of the quadrangles Q1 , Q2 , Q3 is empty. Proof: Note that the choice of Ai is unique if Oi is separating, and arbitrary if Oi is nonseparating. Perform the Cremona transformation cr, and denote by Bi , i = 1, . . . , n the ovals of C18 lying in T0 . We shall say that two ovals E, F of C18 are consecutive for some pencil of lines FD : E → F if this pencil meets no other oval between E and F . Assume there exist Bi , Bj , Bk , Bl such that Bl lies in the triangle Bi Bj Bk that does not cut the base lines. Consider the pencil of conics FBi Bj Bk Bl ; the conics of this pencil are all maximal with respect to the three base lines and cut O at four points. They intersect C18 at 36 points, and the other empty ovals of C18 cannot be swept out. This is a contradiction. Thus, the ovals of T0 lie in convex position in this triangle; let ∆ be their convex hull. Let Bi , Bj be two adjacent vertices of ∆. Assume that Bi , Bj are consecutive for the pencil FBk sweeping out the edge [Bi Bj ] of ∆. Then, Bi , Bj are consecutive for the (maximal) pencil FD sweeping out [Bi Bj ] for any choice of another empty oval D of C18 . Indeed, assume that some oval D does not fulfill this condition. There exists an oval E such that the conic C2 through Bi , Bj , Bk , D, E is Bk Bi EDBj or Bk Bi DEBj . This conic is maximal with respect to the three base lines and cuts O at four points, a contradiction. Thus, the ovals of T0 , ordered by the convexity, split into a number N of successive Fiedler chains, whose base points need not be specified. Assume λ0  ≥ 3. There exist three chains bringing the same contribution +1 or −1 to λ0 . Choose three ovals B1 , B2 , B3 distributed in the three chains. Let [B1 B2 ], [B1 B3 ] and [B2 B3 ] be the segments contained in T0 determined by the three points. For {i, j, k} = {1, 2, 3}, the pencil of lines FBi sweeping out [Bj Bk ] must meet an oval Di outside of T0 . One gets thus a configuration of six ovals. Consider the conics determined by the three ovals in T0 and two of the other ovals. By Bezout’s theorem with C18 , these three conics are: B1 D3 B2 D1 B3 , B2 D1 B3 D2 B1 , B3 D2 B1 D3 B2 . (If some Di is interior to the conic passing through the other five ovals, then this is also the case for the other two ovals Dj , Dk .) The pencil FB1 sweeping out the edge [B2 B3 ] of ∆ sweeps out the arc B2 B3 of B3 D2 B1 D3 B2 and the arcs B2 D1 B3 of the other two conics. These three arcs are thus isotopic with fixed endpoints. The same argument applies with pencils based at B2 and B3 . Consider the interiors
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Q1 B1
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FIGURE 13.9 The six ovals in convex position, the three conics
D1
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FIGURE 13.10 The hexagon H and the six triangles Z1 , . . . Z6 of the three conics; their union is orientable. Let L be a line that does not cut this union, the points B1 , D3 , B2 , D1 , B3 , D2 lie in convex position in the affine plane RP 2 \L. The hexagon H = B1 D3 B2 D1 B3 D2 gives rise to a natural cyclic ordering of the six lines supporting its edges. Let Zl , l ∈ {1, . . . , 6} be the six triangles that are supported by triples of consecutive lines, such that the intersection of Zl and H is a common edge, see Figure 13.10 (the exact position of the triangles is not relevant, but their cyclic ordering is). Let L, L0 ˆ the sector (L, L0 ) be two lines and D be a point, we denote by (L, L0 , D) that does not contain D. The base lines are distributed in the three zones: ˆ1 ) ∪ (B3 B2 , B3 D1 , B ˆ1 ), (B3 B1 , B3 D2 , B ˆ2 ) ∪ (B1 B3 , B1 D2 , B ˆ2 ) (B2 B3 , B2 D1 , B ˆ ˆ and (B1 B2 , B1 D3 , B3 ) ∪ (B2 B1 , B2 D3 , B3 ). See Figure 13.11; the three zones are the M¨ obius bands bounded by the bold (plain or dotted) segments, and the three thin dotted lines correspond to admissible positions of the base lines. Each triangle Zl has a nonempty intersection with T0 and cuts only one base line, and each base line is cut by two triangles having Di , i = 1, 2 or 3 as a common vertex. Let for example: Z1 , Z2 cut A1 A2 ; Z3 , Z4 cut A2 A3 ; Z5 , Z6 cut A3 A1 . Then: Z1 ∪ Z2 ⊂ T0 ∪ cr(Q3 ∪ T3 ), Z3 ∪ Z4 ⊂ T0 ∪ cr(Q1 ∪ T1 ), Z5 ∪ Z6 ⊂ T0 ∪ cr(Q2 ∪ T2 ). The remaining 18 ovals lie in ∪Zl . Indeed, if there was some oval E in another zone, then one of the three conics through E, B1 , B2 , B3 , Di , i = 1, 2, 3 would cut C18 at 38 points, a contradiction. Let E, F be two ovals in the same triangle. If the line EF cuts one of the affine segments [B1 B2 ], [B2 B3 ],[B1 B3 ], then the conic C2 determined by E, F, B1 , B2 , B3 intersects C18 at 38 points, a contradiction. One proves similarly that all of the ovals in a triangle Zl are consecutive for each of the three pencils of lines FBi , i = 1, 2, 3. Thus, there is a natural cyclic ordering of the ovals of C18 given by the pencils of lines FBi , i = 1, 2, 3 sweeping out the six triangles (and the pencils FDi , i = 1, 2, 3 sweeping out the four triangles that do not have Di as vertex). The intersection of each line Ai Aj with ∪Zl is contained in the
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D2
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B3 D1
FIGURE 13.11 The three zones containing the base lines segment [Ai Aj ] interior to O. Hence, by Bezout’s theorem between C18 and the lines supporting H, the zones cr(Int(Oi ) ∩ (Qj ∪ Qk )), i = 1, 2, 3 do not cut ∪Zl . Thus, the interior ovals are all triangular. Otherwise stated, the nonempty ovals of C9 are all separating. Let Z1 be the triangle having Bi and Dj as vertices, and Z6 be the other triangle having Bi as vertex. Let E, F be two ovals in Z1 , such that Bi , E, F, Dj are successive in the cyclic ordering. If F is in T0 , then E is also. Indeed, replace Bi by F in the configuration of six ovals. The triangles Z1 , Z6 are replaced by new triangles Z10 , Z60 , such that Z10 ⊂ Z1 , Z6 ⊂ Z60 . One must have E ∈ Z60 ∩Z1 . But Z1 ⊂ T0 ∪cr(Q3 ∪T3 ) and Z60 ⊂ T0 ∪ cr(Q2 ∪ T2 ), hence E ∈ T0 . The 22 ovals in the affine zone ∪Zl form a cyclic Fiedler chain splitting into six consecutive subchains in the zones T0 , cr(T1 ∪Q1 ), T0 , cr(Q2 ∪T2 ), T0 , cr(Q3 ∪T3 ). In this cyclic chain, an oval is, say, positive if and only if it is the image by cr of a positive oval in T1 ∪T2 ∪T3 or of a negative oval in T0 ∪ Q1 ∪ Q2 ∪ Q3 . Thus, λ0 + µ1 + µ2 + µ3 − λ1 − λ2 − λ3 = 0. Combining P P this with the fourth identity in Lemma 13.12, one gets: 2λ0 = − i − ηi . Any six successive ovals distributed in the six subchains lie in convex position. By Lemma 13.17, one of the zones cr(Ti ∪ Qi ) contains only triangular ovals. Otherwise stated, there are no ovals of C9 in the quadrangle Qi . Proposition 13.2 Let Ti , i ∈ {1, 2, 3} contain only exterior ovals. One has λi  ≤ 3. If λi  = 3, then λi = +3 and λ0 − λ1 − λ2 − λ3 = −2. Proof: Perform the Cremona transformation cr, and denote by B1 , . . . , Bn
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the exterior ovals of C18 lying in cr(Ti ). Let Bq , Br , Bs be three such ovals; the triangle Bq Br Bs whose edges do not cut the base lines is entirely contained in cr(Ti ). Similar arguments as in the proof of Lemma 13.18 yield the following: The ovals of cr(Ti ) lie in convex position in cr(Ti ); let ∆ be their convex hull. Two adjacent vertices Bq , Br of ∆ that are consecutive for some pencil FBs sweeping out the edge [Bq Bs ] are also consecutive for any (maximal) pencil FD , based at another oval D and sweeping out [Bq Bs ]. Thus, one may speak of a Fiedler chain of ovals in cr(Ti ), without referring to a base point. Assume that λi  ≥ 3, so there exist three Fiedler chains of ovals in cr(Ti ), bringing each the same contribution +1 or −1 to λi . Choose B1 , B2 , B3 to be three ovals distributed in the three chains. Let [B1 B2 ], [B1 B3 ] and [B2 B3 ] be the segments contained in cr(Ti ) determined by the three points. For {q, r, s} = {1, 2, 3}, the pencil of lines FBq sweeping out [Br Bs ] must meet an oval Dq outside of cr(Ti ). One gets thus a configuration of six ovals. Consider the conics determined by the ovals B1 , B2 , B3 and two of the other ovals. By Bezout’s theorem with C18 , these three conics are: B1 D3 B2 D1 B3 , B2 D1 B3 D2 B1 , B3 D2 B1 D3 B2 ; the union of their interiors is orientable. Choose a line at infinity L that does not cut this union; the points B1 , D3 , B2 , D1 , B3 , D2 lie in convex position in the affine plane RP 2 \ L. The hexagon H0 = B1 D3 B2 D1 B3 D2 gives rise to a natural cyclic ordering of the six lines supporting its edges. Let Zl , l ∈ {1, . . . , 6} be the six triangles that are supported by triples of consecutive lines. Here we fix that Z1 ∩ H0 = [B1 D3 ], Z2 ∩ H0 = [D3 B2 ] . . . Z6 ∩ H0 = [D2 B1 ]. All of the remaining ovals of C18 lie in ∪Zl . There is a natural cyclic ordering of the ovals of C18 given by the pencils of lines FBq , q = 1, 2, 3 sweeping out the six triangles (and the pencils FDq , q = 1, 2, 3 sweeping out the four triangles that do not have Dq as vertex). The 22 ovals form thus a cyclic Fiedler chain. By Lemma 13.17, one of the ovals D1 , D2 , D3 is triangular. Say D3 is triangular. Let {i, j, k} = {1, 2, 3}; the ovals B1 , B2 are separated from D3 in Int(O) by [Aj Ak ]. Therefore, the line Aj Ak lies in the zone ˆ3 ) ∪ (B2 B1 , B2 D3 , B ˆ3 ). The zone cr(Qi ∪ Ti ) is a triangle T (B1 B2 , B1 D3 , B with vertices A1 , A2 , A3 . If Dq , q ∈ {1, 2} lies in T , then Dq lies in cr(Qi ). The boundary of cr(Ti ) is a topological circle [Aj Ak ]∪(O∩T ). The ovals B1 , B2 , B3 lie inside of this circle, and D1 , D2 , D3 lie outside. The edge [Dq B3 ], q ∈ {1, 2} does not cut [Aj Ak ], so it cuts O. The part O ∩ T cuts each of the lines D2 B3 , D1 B3 twice. Hence, the other part O \ (O ∩ T ) does not cut the segments [D2 B3 ], [D1 B3 ] (see Figure 13.12), D1 and D2 are quadrangular. We may assume that B1 and B3 are respectively the first and the last triangular oval met by the pencil of lines FD2 sweeping out O. By Lemma 13.17, Z1 ∪ Z2 contains only triangular ovals. The complete pencil FD2 starting at B1 and sweeping out successively Z1 , Z2 , Z3 , Z4 , Z5 ∪ Z6 meets four consecutive chains of ovals, alternatively triangular and quadrangular. Indeed, assume there exist more triangular and quadrangular subchains. Then, there exist two ovals E, F met successively by FD2 , such that: E, F ∈ Z3 , E is quadrangular, F is triangular; or E, F ∈ Z4 , F is quadrangular, E is triangular. Let E, F ∈ Z3 . If the line EF cuts the edge [D1 B3 ] of H0 , then Bezout’s theorem with the conic EF D1 B2 B3
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FIGURE 13.12 The hexagon H0 yields a contradiction. Otherwise, B2 , E, F, D1 , B3 , D2 lie in convex position and contradict Lemma 13.17. The case where E, F ∈ Z4 is similar. The chain of triangular ovals containing B3 lies in Z4 . As the line Aj Ak does not cut this triangle, all of the ovals of this chain must lie in cr(Ti ). The other chain of triangular ovals splits into three consecutive subchains containing respectively B1 , D3 and B2 , such that the first and the third subchain lie in cr(Ti ), and the second subchain lies in cr(T0 ∪ Tj ∪ Tk ). Indeed, assume there exist more subchains in each zone. There exist two ovals E, F met successively by FD2 such that: E, F ∈ Z1 , E ∈ cr(T0 ∪ Tj ∪ Tk ), F ∈ cr(Ti ); or E, F ∈ Z2 , F ∈ cr(T0 ∪ Tj ∪ Tk ), E ∈ cr(Ti ). Let E, F ∈ Z1 . The line EF cuts the edge [D3 B2 ] of H0 and the conic EF D3 B1 B2 intersects C18 at 38 points, which is a contradiction. The case where E, F ∈ Z2 is similar. We have proved that if Ti contains only exterior ovals, then λi  ≤ 3. If λi = ±3 then the triangular subchain containing D3 brings a contribution λ0 − λj − λk = ±1, hence λ0 − λ1 − λ2 − λ3 = ∓2. On the other hand, the last identity of Lemma 13.12 implies that λ0 − λ1 − λ2 − λ3 ≤ 0.
13.3
M curves with three nests and a jump
Let C9 be an M curve of degree 9 with three nests and a jump arising from B, C, D; see Figure 13.6 where one makes A3 = B. Consider the pencil of conics FA1 A2 A3 C . Bezout’s theorem implies that the 21 ovals of C9 \ {O1 , O2 , O3 , A1 , A2 , A3 , C} are swept out in the portion A1 A3 ∪ A2 C → A1 A2 ∪A3 C → A2 A3 ∪A1 C if O3 is crossing, and in the portion A1 A3 ∪A2 C →
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A1 A2 ∪ A3 C otherwise (there exists some back oval). In both cases, the 21 ovals are distributed in two Fiedler chains. Denote by P, Q the pair of starting points of the chains, where P is a point of tangency with O3 and Q is a point of tangency with J . Denote by P 0 , Q0 the endpoints of the chains, where P 0 is a point of tangency with O3 and Q0 is a point of tangency with J . The pair of Fiedler chains is then: (P → P 0 , Q → Q0 ) or (P → Q0 , Q → P 0 ). The complete pencil of conics FA1 A2 A3 C divided in three portions by the double lines A1 A3 ∪ A2 C, A1 A2 ∪ A3 C, A2 A3 ∪ A1 C is mapped by cr onto the pencil of lines FC , divided in three portions by the lines A2 C, A3 C, A1 C. For D front, the pencil FC : A2 → A3 → A1 meets successively Q, D, P, {Q0 , P 0 }. If there exists some triangular oval E, then E ∈ T0 ∪ T1 ∪ T2 and the ordering is Q, D, P, Q0 , E, P 0 . For D back, the pencil FC : A2 → A3 → A1 meets successively {P, Q}, P 0 , D, Q0 . If there exists some triangular oval F , then F ∈ T3 and the ordering is P, F, Q, P 0 , D, Q0 . Figure 13.13 shows the relevant part of C18 along with the pencil FC . The oval D in the upper (resp. lower) picture is front (resp. back). In both pictures, we have placed the tangency points P 0 , Q0 , P, Q in such a way that there may exist triangular ovals. By Lemma 13.11, if C9 has triangular ovals, then all of the ovals in the exterior jump sequence are front (O3 is crossing) or back (O3 is bending). Lemma 13.19 All of the ovals in T0 ∪ T1 ∪ T2 ∪ T3 are consecutive for the maximal portion of FA1 A2 A3 C formed by conics intersecting J . If O3 is crossing, they form a Fiedler chain: triangular ovals → P 0 . Thus, λ0 − λ1 − λ2 = 0 or −3 . If O3 is bending, they form a Fiedler chain: P → triangular ovals. Thus, λ3 = 0 or −3 . Proof Let G be a triangular oval of C9 (G = E ∈ T0 ∪ T1 ∪ T2 , or G = F ∈ T3 ). Consider the rational pencil of cubics based at A1 , A2 , A3 , C, D, G, with node at A1 . We shall use Bezout’s theorem with a cubic of the pencil passing through a supplementary oval H of C9 . Let us perform the Cremona transformation cr(A1 , A2 , A3 ); the rational pencil of cubics sweeping out C9 is mapped onto a pencil of conics FA1 GCD sweeping out C18 . The five reducible cubics correspond to five particular conics: the three double lines, and the two conics passing through A2 and A3 respectively. A cuspidal cubic corresponds to a conic that is tangent to the base line A2 A3 , and there is at most one such conic. So the nongeneric cubics of the pencil are the five reducible cubics, plus possibly one cuspidal cubic (see also Section 2.1). A given position of the three double lines with respect to the base lines A1 A2 , A1 A3 , A2 A3 gives rise to one or two admissible cyclic orderings of the five particular conics. There are two orderings if and only if the conics passing through A2 and A3 are consecutive. Let O3 be crossing, let E ∈ T0 ∪ T1 ∪ T2 and H be an oval met after E by the pencil of conics FA1 A2 A3 C : A1 A3 ∪ A2 C → A1 A2 ∪ A3 C → A1 C ∪ A2 A3 . We shall prove that H must be also in T0 ∪ T1 ∪ T2 . Assume first that E ∈ T0 ∪ T1 . Perform the Cremona transformation and consider then the pencil of conics FA1 ECD . The admissible positions of the double lines
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FIGURE 13.13 Curve C18 along with the pencil of lines FC , D front, D back
186
Pencils of Cubics and Algebraic Curves in the Real Projective Plane A1 A3
A1 A2
E
A3
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C
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C
FIGURE 13.14 E ∈ T0
A1
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FIGURE 13.15 E ∈ T1
A1 A3
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FIGURE 13.16 F ∈ T3
D
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M curves of degree 9 with four or three nests
E
187
A1
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missing portion
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E C A1 E A3 D C
FIGURE 13.17 E ∈ T0
A2
C A1 A2 CDA3 A1
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A1 A2 D CE
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
A1
A1 A3
E
A2
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C
C
A1
A1
E
A3 C
A2
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FIGURE 13.18 Missing portion, ED ∩ A1 C ∈ T1 , ED ∩ A1 C ∈ T0 with respect to the base lines are shown in Figures 13.1413.15. Figures 13.1713.22 show the pencils of conics along with the corresponding rational pencils of cubics. The case E ∈ T0 is shown in Figure 13.17. There exists a conic C2 in the piece A1 EA3 DC → A1 C ∪ ED of FA1 ECD , that is tangent to the base line A2 A3 . The preimage of C2 is a cuspidal cubic. The portion C2 → A1 C ∪ ED is not represented in Figure 13.17. This missing portion depends namely on whether the intersection of lines ED ∩ A1 C lies in T1 or in T0 , see Figure 13.14. Both situations are displayed in Figure 13.18. If E ∈ T1 , there are four possible sequences of singular cubics (all reducible) for FA1 A1 A2 A3 ECD , see Figures 13.1913.22. Figures 13.19 and 13.20 fit to the first picture of Figure 13.15, Figure 13.21 to the second one, and Figure 13.22 to the last one. Let H be one of the remaining ovals of C18 . If E ∈ T0 , then Bezout’s theorem yields that H must be swept out in the portion FA1 ECD : A1 A2 DCE → A1 E ∪ CD → A1 EA3 DC → A1 C ∪ ED. Moreover, if H is met after E by the pencil of lines FC : A2 → A3 → A1 , then H lies inside of O = cr(J ). If E ∈ T1 , then H is swept out by FA1 ECD in the portion: A2 A1 ECD → A1 E ∪ CD → A1 C ∪ ED (case 1) A1 DCEA3 → A1 E ∪ CD → A1 C ∪ ED (case 2) A2 A1 ECD → A1 E ∪ CD → A1 C ∪ ED (case 3) A2 A1 ECD → A1 E ∪ CD → A1 EA3 DC → A1 C ∪ ED (case 4) In all cases, if H is met after E by the pencil of lines FC : A2 → A3 → A1 , then H must lie inside of O = cr(J ).
M curves of degree 9 with four or three nests
189
A1
A1 A2
A3 E C
A1 E
A3 D
D
A1 E
CD
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A1 A2 CD A3
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EC
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A1 A2
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E C D
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FIGURE 13.19 E ∈ T1 case 1
D
A1 A2 EC A3
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A1 A3
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190
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
A1
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A3 E C A1 E
A2
A3 D
D
A1 E
CD
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A1 A2 CD A3
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E A1 C
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E A1 A2 ED A3 A1
A1 A3
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E
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C
D A1 D
EC
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E C
A2
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A3 D
A1 A2 DCE
A1 A3
A1 A3
A1 DCE A3
FIGURE 13.20 E ∈ T1 , case 2
A2
C
E
A1 A2 ECD
A1 A2
E C D
E
A1 A2 EC A3
A1 A3
A2
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C E A1 ECD A3
M curves of degree 9 with four or three nests
191
A1
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A3
A2 E C
A1 E
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A2 D
D
C E
A1 E
CD
A1 A2 CD A3 A1
A1 A2
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E A1C
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ED
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A1 D
A1 CE A3 D
D
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C
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EC
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C
D
A2 A1 CD E
FIGURE 13.21 E ∈ T1 , case 3
E A1 A2 A3 CE
A1
E
E
A1
A1
E
C
D
D
A1 D A3 EC
A3
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A1 A2 EDA3
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192
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
A1
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A2 D
E C D A1 E
A1 E
CD
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A2 C
A1 A2CDA3
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A1 E A3 DC
A1 A2
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D C
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A1 A2 A3 DE
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ED A1
A1 A2
E
C
A1 D
D
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EC
A3 D
A2 A1 ECD
FIGURE 13.22 E ∈ T1 , case 4
A1 A2
C
E A1 A2 A3 CE
A1
E
E A1CDA3 E
A1
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E
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A1
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E A2 A1 ECD
M curves of degree 9 with four or three nests
193
A1
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F
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C D A3 A2 FCD
A1 A3
C
A3 DCA2 F
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F A3
D FC
D C A3 D
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A1 F A2 CA3 A1
A3
A2 C
A3 C
F
F A3
D
D A3 C
FD
A1 A2
C
A3
A3 A1 FDC
A3 F
FIGURE 13.23 F ∈ T3
CD
C
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D
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F A2
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C A1 F A2 D A3
A1 A3
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A3 A1 FCD F A2
A3 D A3 F
C A3 A1 A2 CD
194
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
The case E ∈ T2 is deduced from the case E ∈ T1 by performing the axial symmetry switching (A1 , A3 ) with (A2 , C). Let now O3 be noncrossing and F ∈ T3 . Let H be an oval met before F by the pencil of conics FA1 A2 A3 C : A1 A3 ∪ A2 C → A1 A2 ∪ A3 C. We shall prove that H must also be in T3 . Perform the Cremona transformation cr, and consider then the pencil of conics FA3 F CD . The double lines of this pencil are shown in Figure 13.16. The pencil of conics along with the corresponding rational pencil of cubics is shown in Figure 13.23. If H is one of the remaining ovals of C18 , then H must be swept out by FA3 F CD in the portion A3 A1 F DC → A3 F ∪ CD → A3 A2 F CD. Moreover, if H is met before F by the pencil of lines FC : A2 → A3 , then H must lie inside of O. Lemma 13.20 Let C9 be an M curve of degree 9 with three nests and a jump. One of the three possibilities hereafter arises: 1. λ0 − λ1 − λ2 − λ3 = 0 and Π+ − Π− = 4, C9 is odd, odd, even with an odd jump. 2. O3 is crossing, λ0 − λ1 − λ2 = −1, 3 = 1, Π+ − Π− = 3, 3. O3 is bending, λ3 = 1, 3 = −1, Π+ − Π− = 3 In cases 2) and 3), the curve C9 is odd, even, even with odd jump or odd, odd, odd. Proof: By Lemma 13.12, λ0 − λ1 − λ2 − λ3 = Π+ − Π− − 4 ≤ 0. Combining with Lemma 13.19 allows to conclude.
13.4
End of the proof, using two Orevkov formulas
Let Cm be an M curve of degree m = 4d + 1, d ≥ 2. In this subsection, a nest N of depth n is a configuration of ovals (o1 , . . . , on ) such that oi lies in the interior of oj for all pairs i, j with j > i. A nest is maximal if it is not a subset of a bigger nest of Cm . We assume that there exist four maximal nests Ni , i ∈ {1, 2, 3, 4} of Cm verifying: if F is a pencil of conics based in the four innermost ovals of the nests, any conic of F intersects the union of the four nests and J at a minimum of 2m − 2 points. Let Vi be the outermost oval of the nest Ni . We shall call big ovals the ovals that belong to the union of the nests Ni , and small ovals the other ovals. For S, s ∈ {+, −}, let πsS (Ni ) be the numbers of pairs of ovals (O, o) with signs (S, s) such that O is an oval + + of Ni and o is an empty oval contained in Int(O). Let πi = (π− − π+ )(Ni ), − − πi0 = (π+ − π− )(Ni ). Let ΠSs (Ni ) be the number of pairs (O, o) with signs (S, s), where o is a small oval in Int(O). Denote by Pi the number of positive ovals in Ni , and by Ni the number of nonempty ovals among them. Denote by Qi the number of negative ovals in Ni , and by Mi the number of nonempty
M curves of degree 9 with four or three nests
195
ovals among them. Let p1 , . . . , p4 be four points distributed in the innermost ovals of the four nests. If Ni , Nj , Nk have all depth d, call principal triangle pi pj pk the triangle pi pj pk that does not intersect J . The formulas hereafter are proven in [62] (with slightly different notations): Formula 3 (Orevkov) Let Cm be such that the nests Ni , i ∈ {1, 2, 3, 4} have respective depths d, d, d, d−1, and pl , l ∈ {1, 2, 3, 4} lies in the principal triangle determined by the other three points pi , pj , pk , then: πi + πj + πk + πl0 = Ni2 + Nj2 + Nk2 + Ml2 Formula 4 (Orevkov) Let Cm be such that: The nests Nl , l ∈ {1, 2, 3, 4} all have depth d, some Vi , i ∈ {1, 2, 3} coincides with V4 , but the nests N 0 i , N 0 4 , Nj , Nk are pairwise disjoint, with Nl0 = Nl \ Vl . Let us denote by V the oval + Vi = V4 , and by T the principal triangle p1 p2 p3 . Let Πl = Π+ − (Nl ) − Π+ (Nl ), − − + − 0 Πl = Π+ (Nl ) − Π− (Nl ). Let Int (V ) = Int(V ) \ T , Int (V ) = Int(V ) ∩ T . For any big oval O 6= V ⊂ Nl , let Int± (O) = Int(O). For l ∈ {i, 4}, denote by ˜ S (Nl ) the numbers of pairs (O, o) with signs (S, s) where O ⊂ Nl is big and Π s ˜ − (Nl ) − Π ˜ − (Nl ). ˜l = Π ˜ + (Nl ) − Π ˜ + (Nl ), Π ˜0 = Π o ⊂ IntS (O) is small. Let Π + − − + l If pi , i ∈ {1, 2, 3} lies in the principal triangle pj pk p4 , then: ˜ 0 + Πj + Πk + Π ˜ 4 = Q2 − 2Qi + P 2 − Pj + P 2 − Pk + P 2 − P4 + ν(V ), Π 4 i i j k where ν(V ) = 0 if V is positive, and 1 if V is negative. Let C9 be an M curve of degree 9 with real scheme hJ q 1hα1 i q 1hα2 i q 1hα3 i q βi. The complex scheme of the nest Oi is encoded as follows: 1i hαi+ q αi− i, one has αi+ + αi− = αi and αi+ − αi−  ≤ 2 (the equality occurs if and only if Oi has an odd jump). Let νi ∈ ± be the sign of αi+ − αi− . The short complex scheme Si of the nest Oi is defined as follows: Si = i if αi+ − αi− = 0, Si = (i , νi ) if αi+ − αi− = ±1, Si = (i , νi , νi ) if αi+ − αi− = ±2. The short complex scheme of C9 is the triple (S1 , S2 , S3 ). Assume that Oi is separating and αi is even. Let Ai and A0i be the two extremal interior ovals, such that Ai ∈ Ti0 and A0i ∈ T0 . If A0i is negative, we say that Oi is up, otherwise Oi is down, see Figure 13.24. The short complex type S¯i of the nest Oi is defined as follows. Let Si = (i , νi ), then S¯i = (i , νi , s) if Oi is separating, and (i , νi , n) otherwise. Let Si = i , then S¯i = (i , u), (i , d) or (i , n) depending on whether Oi is up, down or nonseparating. Let Si = (i , νi , νi ), then Oi has a jump and is hence nonseparating. The short complex type coincides with the short complex scheme. Let S¯ = (S¯1 , S¯2 , S¯3 ), this triple is the short complex type of C9 . Choose base ovals Ai , i = 1, 2, 3 and let Ni = (Ai , Oi ), i = 1, 2, 3. If Oi is positive, then πi = αi− − αi+ , πi0 = 0, Ni = 1, Mi = 0. If Oi is negative, then πi = 0, πi0 = αi+ − αi− , Ni = 0, Mi = 1. Lemma 13.21 Let C9 have some exterior oval B ∈ Ti , i ∈ {0, 1, 2, 3}. Then
196
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
Ei = 0, where E0 = π1 + π2 + π3 − (N1 + N2 + N3 ), E1 = π10 + π2 + π3 − (M1 + N2 + N3 ), E2 = π1 + π20 + π3 − (N1 + M2 + N3 ), E3 = π1 + π2 + π30 − (N1 + N2 + M3 ). Proof: Formula 3 applies making Ni , i = 1, 2, 3 and N4 = B. Let Oi be separating. Again, choose base ovals A1 , A2 , A3 , and let A4 be a fourth oval, interior to Oi , lying in Ti . Let Nl = (Al , Ol ), l = 1, 2, 3 and ˜0 + Π ˜ 4 − (Q2 − 2Qi + P 2 − P4 + ν(Oi )), and N4 = (A4 , Oi ). Let: Fi = Π 4 i i 2 Gl = Pl − Pl − Πl . It is easily seen that Gl depends only on Si , and Fi depends only on S¯i . Lemma 13.22 Let C9 have three nests and separating Oi . Then, Fi = Gj + Gk Proof: Formula 4 applies with the nests Nl , l = 1, 2, 3, 4. In Tables 13.113.2, we computed the terms appearing in Lemmas 13.2113.22, for all possible complex schemes and types of nests. Lemma 13.23 Let C9 be an M curve of degree 9 with three nests. If the union T0 ∪ T1 ∪ T2 ∪ T3 is empty, then C9 verifies: S1 , S2 ∈ {(+, −), (−, +)}, S3 ∈ {(+, −, −), (−, +, +)}. Proof: One has λ0 −λ1 −λ2 −λ3 = 0, hence by Lemma 13.12, Λ+ −Λ− = 0, Π+ − Π− = 4. Proof of Theorem 13.2: Assume that α1 , α2 , α3 are even. By Lemma 13.20, C9 has no jump, so this curve realizes one of the four complex schemes of Table 13.3. The last column Z of this table contains the indices of the triangles Ti , i ∈ {0, 1, 2, 3} that may contain exterior ovals, according to Lemma 13.21. In Table 13.4, we assume that Oi is separating and compute the term Fi − Gj − Gk . In Table 13.5, we enumerate the admissible short complex types for C9 along with the data Z. The first two types contradict Lemma 13.23. For the other short complex types, choose each Ai , i ∈ {1, 2, 3} in such a way that (Ai , Oi ) is a positive pair. The last identity in Lemma 13.12 reads λ0 − λ1 − λ2 − λ3 = −4. Compute the values of λ0 , λ1 , λ2 , λ3 . For the last four types, one gets λ0 ≤ −4, which contradicts Lemma 13.18. For the remaining two types, one has λ3 = 4 or 5, which contradicts Proposition 13.2. The last two propositions in this chapter bring supplementary information about the complex orientation and rigid isotopy class of the curves with three nests. For a more advanced study, see [25]. Proposition 13.3 Let C9 be an M curve with three nests, and A1 , A2 , A3 be base ovals such that T0 contains only exterior ovals. Then λ0  ≤ 2.
M curves of degree 9 with four or three nests
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O1
O1
FIGURE 13.24 O1 up, O1 down
Sl πl − 0 + 0 (−, +) 0 (+, −) 1 (−, −) 0 (+, +) −1 (−, +, +) 0 (+, −, −) 2 (−, −, −) 0 (+, +, +) −2
πl0 0 0 1 0 −1 0 2 0 −2 0
Nl 0 1 0 1 0 1 0 1 0 1
Ml 1 0 1 0 1 0 1 0 1 0
Gl 0 1 0 0 0 2 0 −1 0 3
TABLE 13.1 Parameters that depend on Sl
S¯i Fi (−, d) 0 (−, u) 1 (+, d) 0 (+, u) 1 TABLE 13.2 Fi depends on S¯i
S¯i Fi (−, −, s) 1 (−, +, s) 0 (+, −, s) 0 (+, +, s) 1
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
S1 − − − +
S2 − − + +
S3 − + + +
E0 0 −1 −2 −3
E1 −1 −2 −3 −2
E2 −1 −2 −1 −2
E3 Z −1 (0) 0 (3) −1 ∅ −2 ∅
TABLE 13.3 Short complex schemes even, even, even, along with Ei , i = 0, . . . 3 and Z
S¯i Sj (−, d) − (−, d) − (−, d) + (−, u) − (−, u) − (−, u) + (+, d) − (+, d) − (+, d) + (+, u) − (+, u) − (+, u) +
Sk − + + − + + − + + − + +
Fi − Gj − Gk 0 −1 −2 −1 −2 −3 0 −1 −2 −1 −2 −3
TABLE 13.4 The cases with Fi − Gj − Gk 6= 0 are forbidden
S¯1 (+, n) (−, n) (−, n) (−, n) (−, n) (−, d) (−, d) (−, d)
S¯2 (+, n) (+, n) (−, n) (−, n) (−, n) (−, n) (−, d) (−, d)
S¯3 Z (+, n) ∅ (+, n) ∅ (+, n) (3) (+, d) (3) (−, n) (0) (−, n) (0) (−, n) (0) (−, d) (0)
TABLE 13.5 Admissible short complex types even, even, even, along with Z
M curves of degree 9 with four or three nests
199
S1 S2 S3 E0 − − (−, −) 0 − (−, −) (−, −) 0 (−, −) (−, −) (−, −) 0 + + (+, +) −2 + (+, +) (+, +) −5 (+, +) (+, +) (+, +) −6 TABLE 13.6 Admissible short complex schemes along with E0 S1 S2 S3 E1 E2 E3 − − (−, −) −1 −1 −2 − (−, −) (−, −) −1 −2 −2 (−, −) (−, −) (−, −) −2 −2 −2 TABLE 13.7 Short complex schemes left, along with E1 , E2 , E3 Proof: Let C9 satisfy the conditions of the proposition and assume that λ0  > 2. By Lemma 13.18, one has λ0 = ±3, i = ηi = ∓1 for i = 1, 2, 3, and the nonempty ovals are all three separating. At least one of the nests contains an odd number of ovals by Theorem 13.2. The admissible short complex schemes are listed in Table 13.6, along with the value of E0 . By Lemma 13.21, one must have E0 = 0, which excludes three cases. In Table 13.7, we compute the values of E1 , E2 , E3 for the remaining cases. By Lemma 13.21, the triangles T1 , T2 , T3 contain no exterior ovals. By Lemma 13.18, one of the quadrangles, say Q1 , is empty. Applying the first identity of Lemma 13.12, with λ0 = 3, µ1 = 0 yields λ1 = 1. Hence, the short complex scheme S1 is −. The short complex type S¯1 is (−, u). For both cases S2 , S3 = −, (−, −) or (−, −), (−, −), one has F1 − G2 − G3 = −1. This contradicts Lemma 13.22. Proposition 13.4 Let C9 be an M curve with three nests and a jump arising from B, C, D where B, C are the extreme ovals in the sequence of O3 , and A1 , A2 , C, D, B lie in convex position. Assume there exist front and back ovals. Then the exterior jump sequence splits into two consecutive chains (Gi ), (Hj ), where the Gi are front and the Hj are back. Proof: Consider two supplementary ovals E, F such that B, D, E, F, C are disposed in this ordering in the sequence of O3 (see 13.9). Assume that 1. D, F are front and E is back, or 2. E is front and D, F are back. Perform the Cremona transformation cr and consider the pencil of conics FA2 CDE . The double lines of this pencil for cases 1) and 2) are shown in
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Pencils of Cubics and Algebraic Curves in the Real Projective Plane
A1 A3 C
A1
A2 F
E case 1
A2
A3 D
E C
F
D
case 2
FIGURE 13.25 The double lines Figure 13.25; in either case, there are two possible positions of F . In all four subcases, F is swept out in the portions CE∪A2 D → CD∪A2 E → CA2 ∪DE. Note that the base points A1 and A3 are swept out outside of this portion. Perform the Cremona transformation back; the conic through C, A2 , D, E, F is mapped onto a rational cubic C3 passing through A1 , A2 , A3 , C, D, E, F , with node at A2 . Figure 13.26 shows the pair of possible cubics C3 for both cases. Any one of these cubics cuts C9 at 29 points, a contradiction.
M curves of degree 9 with four or three nests
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C
F
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case 1 A1
A1 A3
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FIGURE 13.26 The possible cubics C3
A2 F
C
14 More restrictions
CONTENTS 14.1 14.2
14.1
M curves of degree 9 or 11 with one nonempty oval . . . . . . . . . . . Curves of degree 11 with many nests . . . . . . . . . . . . . . . . . . . . . . . . . . . .
203 208
M curves of degree 9 or 11 with one nonempty oval
Let Cm be a curve of odd degree m with one unique nonempty oval; its real scheme is hJ q β q 1hαii. Consider the union of all principal triangles whose vertices are interior ovals. If this union is a disc bounded by a polygon, we say that it is the convex hull of the interior ovals, and call it ∆. It was first observed by S. Orevkov that for m ≥ 13, the interior ovals may have no convex hull. Lemma 14.1 If m ≤ 11, then the interior ovals have a convex hull ∆. Proof: Assume there exist four interior ovals A, B, C, D, such that D lies in the principal triangle ABC. One has either ABC = ABD ∪ BCD ∪ ACD, or J cuts an odd number of times each of the three segments joining D to A, B and C in ABC. Assume the latter case occurs. Up to the action of S3 on A, B, C, there exist a piece of J and a piece of the nonempty oval O in ABC realizing one of the six positions displayed in Figure 14.1. Note that as A, B are interior ovals, the segment [AB] has two supplementary intersection points with O. The segment [AB]0 cuts J at least once, and O at least twice. The line AB cuts Cm at 13 points or more, so m ≥ 13. For m ≤ 11, one constructs ∆ inductively, adding one oval after the other. Start with the principal triangle ∆3 determined by three ovals. Assume one has obtained with n ovals a disc ∆n bounded by a polygon. Add a new oval X, if X is in ∆n , then ∆n+1 = ∆n . Otherwise, consider the pencil of lines FX sweeping out ∆n , the extreme ovals Y, Z met by the pencil are vertices of ∆n . One has ∆n+1 = ∆n ∪ XY Z. Lemma 14.2 Let Cm be a dividing curve of odd degree m = 2k + 1 with real scheme hJ q β q 1hαii. If the interior ovals have a convex hull, then: 1 − k − β ≤ Π+ − Π− ≤ k − 1 + β − where = 0 if β = 0, = 1 otherwise. 203
204
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
Lemma 14.3 Let Cm be an M curve of odd degree m ≥ 7 with one unique nonempty oval and at least one exterior empty oval. 1. If O is negative, then: β+ − β− ≡ (k − 1)2 (mod 3). 2. Assume that the interior ovals have a convex hull. If O is positive, then: k 2 − 3k + 1 ≤ 2β+ ≤ 2β. If O is negative, then: k 2 − 5k + 7 ≤ 2β+ + 2β ≤ 4β. In both cases, one has: β ≥ (k 2 − 5k + 7)/4. Remark: In degree 7, there exist M curves realizing 14 different real schemes: hJ q β q 1hαii (1 ≤ β ≤ 13) and hJ q 15i, see Section 11.2. The M curves with a nonempty oval verify: Π+ − Π− ∈ {0, 1, −1, 2}. Proof of Lemmas 14.214.3: Consider the convex hull ∆ of the interior ovals. Let A be an oval placed at a vertex. The pencil FA sweeping out ∆ sweeps out all of the interior ovals and possibly some of the β exterior ovals, giving rise to k − 2 Fiedler chains. We consider here not the complete chains, but the longest subchains whose extremities are interior ovals. There are no J jumps between the interior ovals, and two consecutive interior ovals in a chain have thus alternating orientations with respect to J . A chain has j jumps if it is formed of 2j + 1 subchains with alternatively interior and exterior ovals. A chain with j jumps brings a contribution to Π+ − Π− whose absolute value is less than or equal to j + 1. The total number of jumps is at most β. Taking into account the contribution of the last oval A one gets the inequality Π+ −Π−  ≤ k−1+β. If the equality holds, then the ovals placed at the vertices of ∆ all have the same orientation. Assume there exists some exterior oval B lying outside of ∆, and consider the pencil of lines FB sweeping out ∆. There are no J jumps between the interior ovals, and these ovals are distributed in k − 2 chains with at most β − 1 jumps in total. Thus, Π+ − Π−  ≤ k + β − 3. If the equality Π+ − Π−  = k − 1 + β holds, then the exterior ovals all lie inside of ∆ and some edge of ∆ intersects the nonempty oval O. Let A be a vertex of ∆, and an extremity of such an edge. The pencil of lines based at A sweeping out ∆ meets all of the other empty ovals, and these ovals are distributed in k − 2 chains whose extremal ovals are all interior ovals having the same orientation as A. The last oval of one chain must be connected with a point on O; this oval forms therefore a negative pair with O, see Figure 14.2. Thus, Π+ − Π− = 1 − k − β. This finishes the proof of Lemma 14.2. The RokhlinMishachev formula may be rewritten for the M curves with one unique nonempty oval: O positive:
Π+ − Π− + 1 + β+ − β− = k 2 − 2k
More restrictions
205 3(Π+ − Π− ) − 1 + β+ − β− = k 2 − 2k
O negative:
Lemma 14.3 follows immediately from these formulas, combined with the inequality Π+ − Π− ≤ k − 2 + β. A
D
B
A
C A
D
B
D
A
C B
A
C
D
B
C B
D
C
A
D
B
C
FIGURE 14.1 The dotted arcs in the triangles are pieces of O, the plain arcs are pieces of J
A
∆
FIGURE 14.2 The last oval of one chain is connected with O Remark: Consider a curve Cm of arbitrary degree m, a real line L, and the line pencils with base points on L. The sets of complex conjugated arcs obtained fiber the curve Cm , and one may recalculate its Euler characteristic. The nonreal intersection points of Cm with L are meridianlike singularities (like the poles of the Earth fibered by the meridians), the real intersection points are parallellike singularities (like the poles of the Earth fibered by the parallels). Each of these singularities brings a contribution of +1. The real inflection points of Cm and the tangencies at non realpoints of Cm correspond to the bifurcations, and they bring each a contribution of −1. The tangent lines are actually bitangents. Denote by ir the number of real inflection points of Cm , and by t” the number of bitangents to pairs of complex conjugated
206
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
nonreal points. One has χ = 2 − 2g = m − ir − 2t”, hence ir + 2t” = m(m − 2), which is the famous Klein formula. The fact that one can recover this formula this way was first observed by T. Fiedler in his PhD thesis. Proposition 14.1 There do not exist M curves of degree m = 2k+1 (k = 3, 4 or 5), having one unique nonempty oval that contains all others in its interior. Otherwise stated, the real schemes: hJ q 1h14ii (k = 3), hJ q 1h27ii (k = 4), hJ q 1h44ii (k = 5) are not realizable. Proof: Assume there exists some curve Cm contradicting Proposition 14.1, and let O be its nonempty oval. Denote by α± the numbers of positive and negative interior ovals, and let λ = α+ − α− . For O positive, λ = 1 − k 2 + 2k; for O negative λ = (k − 1)2 /3. By Lemma 14.2, λ = Π+ − Π−  ≤ k − 1. By Lemma 14.3, the nonempty oval can be negative only if k ≡ 1 (mod 3), so k = 4. The only possibility is a curve of degree 9 with complex scheme hJ q1− h15+ q12− ii. If O is positive and k > 3, then λ = 1−k 2 +2k < 1−k < 0, a contradiction. The only possibility is a curve of degree 7 with complex scheme hJ q 1+ h6+ q 8− ii. The real scheme hJ q 1h14ii is not realizable, see Section 11.2, and the complex scheme hJ q 1− h15+ q 12− ii has been excluded more recently in [35]. We give here an alternative method to exclude both complex schemes. Let Cm , m = 7 or 9 realize one of these complex schemes. Let P be a point of the nonempty oval O such that O is locally convex at P : if one chooses a point P 0 close enough to P on O, the principal segment [P P 0 ] lies inside of O. Let L be the line tangent to O at P . We consider the complete pencils of lines Lt,Q based at a mobile point Q that follows the line L, with the starting and ending positions at P . Each of the pencils contains the line L, tangent to O at P . Near the start, Q is close to P on L, another line of the pencil is tangent to O, at a point P 0 close to P , and the two points P, P 0 are S connected by a pair of conjugated arcs forming a small circle in CA ∩ ( CLt,Q ). As Q moves away from P , one gets a set of growing circles, all tangent to one another at P . Near the end of the motion, when Q moves towards P from the other side of L, we have symmetrically a set of shrinking circles, that lie on the other side of P on CA as shown in Figure 14.3. As CA is not a sphere, there must have been at some moment a bifurcation in the set of conjugated imaginary arcs, after which the two tangency points P, P 0 with O are no longer connected one to the other. Each tangency point is then connected to an interior oval whose orientation (with respect to J ) coincides with that of O; this oval is the starting oval of one of the k − 1 interior chains determined by the pencil. But for any choice of base point outside of O, the interior ovals are distributed in k − 1 chains whose ovals have alternating orientations with respect to J , each chain bringing a contribution +1 to Π+ − Π− . Performing a bifurcation would lead to a contradiction with Fiedler’s theorem, see Figure 14.3. (See also [13] where a similar argument is used in a different proof.) Proposition 14.2 There do not exist M curves of degree 2k + 1 with the
More restrictions
207
∆ Q P L
∆ Q P L
P P’
FIGURE 14.3 M curves hJ q 1+ h6+ q 8− ii and hJ q 1− h15+ q 12− ii
208
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
following real schemes: hJ q1q1h26ii (k = 4); hJ q1q1h43ii, hJ q2q1h42ii (k = 5). Proof: For k = 5, one has by Lemma 14.3: β ≥ 2 if O is negative, and β ≥ 6 if O is positive. Assume O is negative and β = 2. As β+ − β− ≡ 1 (mod 3); one must have β+ = 0, β− = −2. Hence, 2β+ + 2β = 4 < k 2 − 5k + 7, a contradiction. For k = 4, one has by Lemma 14.2: β+ ≥ 3 or β+ − β− ≡ 0 (mod 3). Neither condition is realized if β = 1. Note that for m = 7, the real scheme hJ q 1 q 1h13ii is realizable, with the two admissible complex schemes: hJ q 1+ q 1+ h6+ q 7− ii and hJ q 1+ q 1− h7+ q 6− ii, see Table 11.3. For m = 9, the nonrealizability of the schemes hJ q 1h27ii, hJ q 1 q 1h26ii proved above had previously been announced in [13], but unfortunately the proofs were never published and went lost. Ninth degree M curves hJ q β q 1hαii with 1 ≤ α ≤ 19, α = 22, 23 have been constructed by A. Korchagin [52]. The cases α = 20, 21 have been realized by Orevkov [59]. By Lemma 14.3, the only admissible schemes for β = 2 and 3 are: hJ q 1+ q 1− q 1− h14+ q 11− ii, hJ q 3+ q 1+ h10+ q 14− ii, hJ q 3+ q 1− h13+ q 11− ii, hJ q 3− q 1− h14+ q 10− ii. Finally, note that the proofs and results presented here are valuable also for pseudoholomorphic curves.
14.2
Curves of degree 11 with many nests
Theorem 14.1 Let C11 be a real algebraic dividing curve of degree 11 in RP 2 . The curve C11 cannot contain any one of the following two configurations of ovals: 1. eight disjoint nests 1hαi i, i = 1, . . . 8 of depth one, or 2. seven disjoint nests 1hαi i, i = 1, . . . 7 of depth one, such that some of them are surrounded by an oval. This result was conjectured by Benoˆıt Chevallier after systematic constructions [9]. Proof: Let C11 be a dividing curve of degree 11; this curve has an odd number 2K + 1 of ovals (2K + 1 = 45 if C11 is an M curve). Assume that C11 has eight disjoint nests of depth one. For each nest with exterior oval Oi , choose one of the interior ovals Oi0 . Let X = 8 × 1h1i ∪ J be the union of the ovals Oi , i = 1, . . . 8, Oi0 , i = 1, . . . 8, along with the pseudoline. Take eight points 1, . . . 8 distributed on the eight interior ovals O10 , . . . O80 and consider the pencil of cubics they determine; this pencil has a ninth base point 9. By Bezout’s theorem, the 2K − 15 ovals of C11 that are not parts of X are cut by cubics that intersect X at less than 33 points. To achieve this condition, a cubic must have an oval O contained entirely in one of the eight nonempty ovals, say O1 , and passing through exactly two base points: 1 and 9. The
More restrictions
209
cubics that intersect X at less than 33 points form a connected portion P of the pencil, bounded by two cubics C3 (1), C3 (2) that are tangent from the inside to the oval O1 , see Figure 14.4 (where 9 has been placed arbitrarily outside of O10 , it could also be inside). Endow C11 with one of the two possible complex orientations. Let us apply Fiedler’s theorem using the piece of pencil P, formed by dividing cubics. The oval O1 induces compatible orientations on C3 (1) and C3 (2). The orientations of the cubics of P should therefore change an even number of times. But this portion of pencil sweeps out the 2K − 15 remaining ovals of C11 , and there are thus 2K − 15 changes of orientation, which is a contradiction. Assume now that C11 has seven disjoint nests of depth one. For each nest with exterior oval Oi , choose one of the interior ovals Oi0 . Let Y = 7 × 1h1i ∪ J be the union of the ovals Oi , i = 1, . . . 7, Oi0 , i = 1, . . . 7, along with the pseudoline. Take seven points 1, . . . 7 distributed on the seven ovals Oi0 and consider the seven nodal cubics passing through 1, . . . 7, with respective nodes at 1, . . . 7. We call a combinatorial cubic a topological type (cubic, seven points). Up to the action of the symmetric group S7 on the seven points, there are exactly 14 possible sets of seven combinatorial cubics (see Theorem 1.2 and Figures 3.83.10). In each set, there is a cubic with a loop passing through at least one point among 1, . . . 7 other than the node. Such a cubic with node at, say 1, intersects O1 ∪O10 at eight points (counted with multiplicities), each other pair Oi , Oi0 at four points, and the pseudoline J at one point. The cubic cuts C11 at 33 real points. A supplementary oval surrounding one of the nests would bring two more intersection points, contradicting Bezout’s theorem.
O1
1 9
C3 (1) FIGURE 14.4 The pencil of cubics sweeping out C11
C3 (2)
15 Totally real pencils of cubics
CONTENTS 15.1 15.2
15.1
Two real schemes of sextics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nodal pencil again . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
211 213
Two real schemes of sextics
The motivation for this chapter comes from a simple observation. Consider the hyperbolic curve of degree m = 2k or 2k + 1, consisting of k nested ovals plus one pseudoline if m is odd. The lines of a pencil based at a point in the innermost oval intersect all this curve at m real points. This fact may be used to prove that the hyperbolic curves are dividing. More generally, let us say that a real pencil of curves is totally real with respect to some real algebraic curve A if it has only real intersections with CA. Rokhlin proved in [70] that if an algebraic curve is swept out by a totally real pencil of lines, then this curve is dividing. The argument generalizes to pencils of curves of higher degrees. Can conversely any dividing curve be endowed with some totally real pencil? The answer to this question has been found recently to be affirmative, even when the base points of the pencil are not contained in the curve, see [53]. But the technique developed in that paper does not give an explicit description of the pencil. We study here the case of M − 2sextics with real scheme h2 q 1h6ii or h6 q 1h2ii. By a congruence due to Kharlamov [45], these two real schemes are of type I. Recall that the rigid isotopy class of a nonsingular sextic is determined by its real scheme, plus its type I or II (see Section 11.1). Thus, any two M − 2sextics realizing the same real scheme among our two are rigidly isotopic. We shall call here the interior ovals inner ovals, and the exterior empty ovals outer ovals. Theorem 15.1 The M − 2sextics with real schemes h2 q 1h6ii or h6 q 1h2ii satisfy the following properties. 1. Any M −2sextic with one of these real schemes may be endowed with a totally real pencil of cubics. It suffices to choose eight base points distributed on the eight empty ovals. 2. The empty ovals of an M − 2sextic with real scheme h2 q 1h6ii 211
212
Pencils of Cubics and Algebraic Curves in the Real Projective Plane 1
1
8
X
7 6
5
5 4
4 8
6
3
8 Y
Y
X
5
3
4
1
7 3
7
6
X
4
5
6
3
8
Y
7 6
5
X
8 1
X Y 4
5
3
1
7 Y
1
8
4
8 1 5
6 4 3
X
X 3 7 8
6
7 Y
6
7
5
Y
1
X Y 4
3
FIGURE 15.1 Pencil of cubics with eight base points distributed in the eight empty ovals of h2 q 1h6ii lie in convex position. Let 1, . . . 8 be eight generic points distributed in successive empty ovals, such that 8 and 1 lie in the two exterior ovals and the oval containing 7 forms a negative injective pair with the nonempty oval O. Then, the successive distinguished cubics of the pencil determined by 1, . . . 8 are as shown in Figure 15.1, where {X, Y } = {2, 9} (9 is the ninth base point), and O is represented with a dotted line. If 2 is chosen on the empty oval instead of inside, then the pencil is totally real. 3. Along a rigid isotopy between two sextics with real scheme h2q1h6ii, this totally real pencil of cubics may be deformed continuously, preserving total reality, with at most one type of degeneration, after which the positions of the two base points 2 and 9 on the distinguished cubics are swapped. Before stepping forward to the proof, let us give examples of dividing curves that may be endowed with totally real pencils. For an M quartic h4i, a pencil of conics with the four base points distributed in the four ovals will do. Similarly, the curve of degree 8 with real scheme h1h1i q 1h1i q 1h1i q 1h1ii formed by four nests of depth two lying outside one another admits a totally real pencil of conics, and it suffices to choose the base points inside of the innermost ovals. For an M quintic hJ q6i, one finds a suitable pencil of cubics with six base points distributed on the six ovals, and two further chosen on the pseudoline J . As this pseudoline must cut any cubic an odd number of times, the required 15 real intersections are granted. More generally, the M curves of degree m may always be endowed with suitable pencils of degree m − 2, see [34], page 348.
Totally real pencils of cubics
15.2
213
Nodal pencil again
Lemma 15.1 Consider a real sextic C6 with one nonempty oval O. There is a natural cyclic ordering of the empty ovals of C6 , consistent with the ordering given by the complete pencils of lines based at the inner ovals and the partial pencils of lines based at outer ovals and sweeping out O. For this cyclic ordering, the set of empty ovals splits in two consecutive chains (inner, outer). Moreover, the inner ovals lie in convex position inside of O. If C6 has at most two outer ovals, then all of the empty ovals lie in convex position. Proof of the lemma: Given an empty oval X of C6 we will often have to consider one point chosen in the interior of X. For simplicity, we call this point also X, and it will be clear from the context whether we speak of the oval or of the point. Let A, B be two exterior ovals, and let C, D be two interior ovals. By Bezout’s theorem with the line CD, one segment [CD] is entirely contained in O. Assume that the line AB cuts this segment, and consider the pencil of conics based at A, B, C, D. The conics of this pencil cut the set of ovals A, B, C, D, O at 12 points, so the other empty ovals cannot be swept out; this is a contradiction. So the line AB must cut the other segment of line [CD]0 . Let now C6 have more than two inner ovals, and let E be a third inner oval. The points C, D, E determine four triangles in RP 2 , one of them is entirely contained in O, let us call it principal triangle CDE. Let F be another empty oval. Using a pencil of conics based at C, D, E, F , we prove that F cannot lie inside of the principal triangle CDE. Assume that two outer ovals A and B lie in two different (nonprincipal) triangles CDE. Then the conic through A, B, C, D, E cuts the set of ovals A, B, C, D, E, O at 14 points; this is a contradiction. Note also that the cyclic ordering gives rise to a choice of a principal segment [XY ] connecting two consecutive ovals X, Y . If X, Y are inner ovals, this segment lies inside of O. Proof of 1): Let us consider the M − 2sextics h2 q 1h6ii and h6 q 1h2ii, and a pencil of cubics with eight base points distributed on the eight empty ovals; all of the cubics of this pencil cut the set of empty ovals at 16 points. Assume that the pencil is not totally real; then it has a cubic C3 that does not cut O. This cubic is maximal, its oval O passes through the inner ovals of C6 , and its pseudoline J passes through the outer ovals of C6 as shown in Figure 15.2. Note that the ninth base point lies on the pseudoline, outside of O. The cyclic orderings of the interior ovals on O and of the exterior ovals on J are consistent with the natural cyclic ordering defined above. This fact follows from Bezout’s theorem between C3 and the pencils of lines based at inner ovals. In an affine plane where the line at infinity cuts none of the principal segments [XY ] (as in Figure 15.2), the naturally ordered empty ovals have alternating orientations by Fiedler’s theorem. Let us move in the pencil of
214
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
FIGURE 15.2 The cubic C3 cubics, starting from C3 = C3 (0), in one of the two possible directions. In the upper part of Figure 15.3, the cubic C3 is drawn in bold, and the dotted oval is the oval O(t) of some cubics C3 (t) close to C3 . During the motion, two mobile arcs with fixed endpoints, one arc s on the oval O(t), the other s0 on the odd component J (t), move towards each other until they are glued with a nonisolated double point. Let us denote by C3 (1) the corresponding singular cubic. For the sextic with six inner ovals, let us denote the base points of the pencil by 1, . . . 8 as indicated in Figure 15.3. The mobile arcs 27, 56 and 34 of O(t) are outside of the oval O of the initial cubic C3 . Thus, s is one of these three arcs. Note that each mobile arc of O(t) is homotopic with fixed extremities to a segment of line contained in O. The line 68 cuts the segment of lines [27] interior to O, so this line cuts C3 (t) at 6, 8 and a point of the arc 27. During the motion, the mobile arc 56 must stay inside of a zone delimited by the lines 68, 54, and the arc 56 of the initial oval O. As J (t) may a priori intersect only the side 54, and this only once, this zone is not cut by J (t), see the upper part of Figure 15.3. Symmetrically, the mobile arc 34 is bound to stay inside of a zone that is not cut by J (t) either. Therefore, the singular cubic C3 (1) is obtained by attaching the arc 27 of O(t) to some arc of the pseudoline. The singular cubic C3 (1) cuts C6 at 18 points. If one provides both curves with a complex orientation, and perturbs all double points of their product according to these orientations, one gets a curve C9 of degree 9 with L − 1 = 10 ovals. This curve is dividing, with a complex orientation induced by that of the initial curves, see [70]. RokhlinMishachev’s formula reads: 2(Π+ − Π− ) + Λ+ − Λ− = L − 1 − k(k + 1) = −10. The n = 2 or 6 base points chosen on the exterior ovals divide the pseudoline J (t) in n arcs, and one of them contains s0 or is the whole of s0 . Clearly, the data Π± , and Λ± depend neither on the actual position of s0 nor on the position of the double point of C3 (1) inside or outside of O. Up to a symmetry with respect to a vertical axis passing through the middle of the figure, changing C3 (1), the orientation of O may be chosen at leisure. In the middle part of Figure 15.3, we have drawn possible cubics C3 (1) for either sextic with the nonempty oval O enhanced with some orientation. In the bottom part, we have represented
Totally real pencils of cubics 1
8
7 6
215
5
4
2 3
FIGURE 15.3 The next singular cubic in the pencil leads to a contradiction the two curves C9 obtained. One gets 2(Π+ − Π− ) + Λ+ − Λ− = −8 in both cases, which is a contradiction. Proof of 2) and 3): We use here the notations from Part 2. Let us determine the list L(1, . . . 8) realized by eight generic points distributed in the eight empty ovals of the sextic h2q1h6ii as indicated in the upper part of Figure 15.3. By Bezout’s theorem, one has ˆ8 = 1±, ˆ1 = 8±, and n ˆ ∈ {8±, 1±} for n = 2, . . . 7. On the other hand, if ˆ8 = 1+, then ˆ1 ∈ {8+, 6±, 4±, 2±}, see Table 9.3. So the list L(1, . . . 8) has either a pair ˆ8 = 1+, ˆ1 = 8+, or a pair ˆ8 = 1−, ˆ 1 = 8− (the second pair is deduced from the first by the action of the element (+1)(48) of D8 ). Table 5.4 shows the 32 lists with ˆ8 = 1+, ˆ1 = 8+. Only two of them satisfy the condition on the n ˆ , n = 2, . . . 7, namely the list L1 = max(ˆ 8 = 1+) and L32 = max(ˆ1 = 8+). Up to the action of (+1)(48), let us assume that L(1, . . . 8) is max(ˆ1 = 8+) or max(ˆ1 = 8−). These two lists are nodal, max(ˆ 1 = 8−) gives rise to nine combinatorial pencils (Figure 8.2 and Table 8.1), and max(ˆ 1 = 8+) is realizable with the first three combinatorial pencils among those nine. (The two lists differ by a nonessential elementary change.) Let us choose a cubic C3 of the pencil, provide it with an orientation, perturb the union C3 ∪ C6 and check RokhlinMishachev’s formula for the obtained curve C9 . Assume that one of the first seven pencils of Table 8.1 is realized and choose C3 to be (1−, 3). We find a contradiction for both
216
Pencils of Cubics and Algebraic Curves in the Real Projective Plane
orientations of the nonempty oval O of C6 . Thus, the pencil is one of the last two. Let us now choose C3 to be the cubic (1+, 12), the case where O and the oval surrounding 7 form a positive pair yields again a contradiction. One checks that the other orientation of O yields no contradiction. The pencil is nontotally real if 2 and 9 lie together inside of the oval 2 of C6 : a bad cubic is a cubic with an oval passing through 2 and 9, that is entirely contained in the empty oval 2. To avoid this problem, let us choose the point 2 on the corresponding empty oval 2. This finishes the proof of 2). The last point 3) is an immediate consequence of 2).
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Index
Lconfiguration ordered, 3 unordered, 3 Ldeformation, 3 Qconfiguration, 169 ordered, 3, 31 unordered, 3, 12, 31, 161 Qdeformation, 3 adjacency graph, 4, 20, 46, 53 adjacent triple, 19, 45, 46 base line, 165, 173 base oval, 173 base point, 86, 208 base points, 59, 123, 211 Bezout theorem, 131, 136, 157, 163, 168, 184, 209 bitangent, 205 chamber, 3, 40, 132 close pair, 86, 103 essential, 86 inessential, 86 code, 4, 18 combinatorial equivalence, 3 point by point, 3, 20, 43 complex conjugation, 132 complex orientation, 132, 214 complex scheme, 132, 155, 164, 195, 206 short, 195 configuration αconfiguration, 12, 18, 161 βconfiguration, 12, 18, 161, 168 δconfiguration, 12, 18, 161 γconfiguration, 12, 18, 161 almost generic, 3, 20
generic, 3 lineconic, 43 conicdiagram, 31 conicwall, 4, 18, 31 ordered, 19 unordered, 19 convex hull, 138, 157, 176, 203 convex position, 60, 65, 156, 165, 173, 212 Cremona transformation, 7, 173, 184 cubic combinatorial, 6, 60, 83, 85, 104, 209 completely reducible, 93 cuspidal, 6, 7, 184 distinguished, 60, 84, 85, 212 nodal, 35, 59, 83, 124 reducible, 6, 7, 35, 61, 91, 97, 159, 184 singular, 61, 214 curve M curve, 131, 134, 138, 145, 155, 163, 206 dividing, 132, 203, 208, 211 hyperbolic, 211 maximal, 131 of type I, 132 of type II, 132 real algebraic, 131 real pseudoholomorphic, 134, 143 deformation equivalence, 4 point by point, 4, 20, 43 diagram ndiagram, 17 marked, 12 223
224 unmarked, 12 discriminantal hypersurface, 59, 132 distance, 75 elementary arc, 61, 124 elementary change, 78, 105 essential, 87 inessential, 87, 105 elementary pair, 78, 91 essential, 87, 103 inessential, 87 Euler characteristic, 59, 205 Fiedler chain, 136, 158, 168, 184, 204 Fiedler theorem, 136, 168, 173, 209, 213 fold, 136 group dihedral, 66, 67, 159 icosahedral, 19 monodromy, 4, 19, 40 symmetric, 3 Hilbert’s sixteenth problem, 131 inflection point, 93, 205 injective pair, 131 negative, 132, 173, 212 positive, 132, 173 jump, 138, 163, 166, 183, 204 O1 jump, 155 J jump, 138, 159, 204 odd, 171, 172, 194 jump sequence, 169 Klein formula, 206 Korchagin conjectures, 145 letter triple, 50 linewall, 4, 18, 31, 40 ordered, 19 refined, 4, 43, 50 unordered, 19 list, 65, 168, 215 almost generic, 97
Index extremal, 71 generic, 65 maximal, 71, 84 minimal, 71 nodal, 83, 105, 215 nongeneric, 65 nonnodal, 85 principal, 69, 81, 91 symmetric, 97 loop, 7, 59, 67, 85, 124, 168, 209 main part, 173 maximal, 173 nest, 145, 163, 208 deep, 131, 155 maximal, 194 node, 6, 83, 209 isolated, 59, 124, 168 nonisolated, 59, 124 odd component, 59, 67, 84, 85, 169 orbit, 19, 40, 69, 83, 92, 168 orbit data, 35 orbits, 106 Orevkov formulas, 156, 195 oval, 59, 124, 131 back, 171, 184, 199 base, 172 bending, 171, 184, 194 big, 194 crossing, 171, 184, 194 down, 195 empty, 131, 165 even, 132 exterior, 131, 139, 173 front, 171, 184, 199 inner, 155, 211 interior, 139, 173 median, 155 negative, 132, 172, 173 nonempty, 163, 203 nonseparating, 169, 195 odd, 132 outer, 155, 211 positive, 132, 172, 173
Index principal, 165, 175 quadrangular, 171, 173 separating, 169, 178, 195 small, 194 triangular, 171, 173, 176, 184 up, 195 Pappus theorem, 95 Pascal theorem, 95 pencil combinatorial, 84 maximal, 136 nonnodal, 106 totally real, 211 pencil of conics, 7, 184, 212 pencil of cubics, 59, 208, 211 combinatorial, 6, 60 generic, 59 Hessian, 93 nodal, 90, 103 rational, 6, 7, 161, 168, 184 singular, 59, 90, 123 pencil of lines, 136, 143, 204, 213 maximal, 168 Petrovskii inequalities, 133 principal segment, 138, 156, 213 principal triangle, 134, 138, 203, 213 projective transformation, 43 pseudoline, 59, 124, 131, 165, 212 quadrangle, 178 quadruple, 35 quasipositive braid, 143, 144 quintic, 134, 212 real scheme, 131, 155, 164, 203, 211 indefinite, 132 of type I, 132 of type II, 132 RiemannHurwitz formula, 144 rigid isotopy, 3, 132, 212 Rokhlin congruence, 133 Rokhlin formula, 136 RokhlinMishachev formula, 136, 138, 158, 168, 173, 204, 214
225 sextic, 211 short complex type, 195 stratification, 3 subcode, 31 subconfiguration, 6, 31, 45 sublist, 67 SylvesterGallai theorem, 95 symplectic form, 140 wall, 3, 17, 40 Welschinger invariant, 60