flfl'flf’fii sfi, V $77 .1"??? l'.{ 3’.» 93%“ «a 1, ’1'! i n u 1’» I-A“ _vj~7.ff:¢o,t1‘.. 4 L“ ”2’“. 11 {arkhflfla rumm- “1.qu _'_,~ 11 a». fl} $171 13:“ :1», . 117. .6 '.: .u “r" vavég $9. "‘7‘“ uyfl'fl) .-,_‘ .5- ' u the”): J‘As'rav . 4.2..“le .- 1,5,; 260K: 1.- «‘Vn 1‘ 1 ~>.. U 1,‘ _ ‘1 .7 . T113319 for the Degree 0f PhD f 1..) A, 3. , 101111111113; -' ‘ - :; * :‘ 71;: «A -2. .L"';:‘ b1 o'.ru—.~...',.',.. <'v-o~'Q-‘~.--nn " 4- -; ' .o»:. V r—' "-0-, - . l'ra .n... .3112». -. 1‘” ~- a .1 a. 1 ' .{';-"‘ - - COD“ 1-". p»: or. r ('1; 'JM- r-nooosr w “trot." ,p 4 u...p....r.1, ,, Iva't‘w .11.... v.1 ' 3..» 1"" LIBRARY Ih65:3 This is to certify that the thesis entitled "Generalised Sylvester—Gallai Configurations" presented by Sonde Ndubeze Nwankpa has been accepted towards fulfillment of the requirements for PhoDo degree in Mathematics 56741791211,“ Major professog Date November 20, 1970 0—169 ABSTRACT GENERALISED SYLVESTER GALLAI CONFIGURATIONS BY Sonde Ndubeze Nwankpa In linear spaces it is of interest to study cutting hyperplanes of a family of, say, disjoint compact sets as opposed to studying supporting and separating hyperplanes. Kelly and Edelstein have shown that "if {Si} is a a finite collection of disjoint compact sets in a Hausdorff linear topological space 2, spanning a space of dimension d > 1 and if IUSiI = m then there is a line in Z, inter- secting precisely two of the sets of {Si]." Furthermore, they have shown that if d > 3 the condition that IUSiI = m can be removed. For d = 2 in real linear space there exists a large number of examples of what we now call general- ised Sylvester-Gallai configurations (i.e. a finite collection of disjoint finite sets spanning a space of dimension 2 such that no line cuts precisely two of the sets). For d = 3 only one example is known, the classically studied desmic configuration consisting of 3 sets of 4 points each. This thesis represents an effort to understand these generalised Sylvester-Gallai configurations better, both Sonde Ndubeze Nwankpa in the ordered linear setting, and in some general projective spaces. It is easily shown that a necessary condition for the existence of a GSG configuration of 3 sets in dimension d 2 3 is the existence of a completely self perspective arrangement (CSPA) in dimension d-l. Accordingly a considerable portion of the work is devoted to the study of CSPA particularly in the ordered projective plane. Our conjecture was (and is) that such an arrangement fails to exist in ordered projective space for sets of more than four points. We were only able to verify this for sets with fewer than nine points. This is the main result of Chapter 3. Generalised S.G. Configurations in which each Si consists of a single point are called simply S.G. configurations and have received considerable attention from several prominent sources. For example, J. P. Serre [AMM 73, p. 89 1966] asked if an S.G. configuration spanning complex projective 3-space exists. In Chapter 2 we show that such a set must contain at least 40 points. We also continue the program initiated by T. Motzkin LTAMS 70 (1951) 451-464] of character- izing abstract S.G. configurations of low orders or with other restrictions. For example, we completely analyze those which are subsets of 3 lines. Typical theorems proved in the thesis Theorem "If G is a GSG whose point sets {Si} span an ordered projective 3-space then no line intersects more than three of the sets." Sonde Ndubeze Nwankpa Theorem "If S is an S.G. configuration spanning complex projective 3-space then [S] 2 40." Theorem "If {Sl.Sz,F} is a multiply perspective arrangement of class [k,k.d] then k 2 2d_l." GENERALISED SYLVESTER GALLAI CONFIGURATIONS BY Sonde Ndubeze Nwankpa A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1970 GW/QO TO Hanna Neumann my mentor, Onyehuruchi my wife, and the children of Biafra. ii ACKNOWLEDGEMENTS The author wishes to express his gratitude to Professor L. M. Kelly for his helpful suggestions and guidance during the research. He would also like to thank M. Edelstein for the use of his unpublished results, and the guidance committee for their suggestions in the final preparation of the manu- script. He would like to thank the U. S. Government for their support through the Fulbright-Hayes Act Scholarship program administered by the Institute for International Education (I.I.E.). iii TABLE OF CONTENTS List of Figures. List of Symbols. CHAPTER 1. Introduction CHAPTER 2. Sylvester—Gallai Configurations. l. Configurations. 2. Complete characterizations of S.G. configura- tions for low orders not exceeding 12 3. S. G. Configurations as subsets of 3-1ines. 4. On the existence of S. G. configurations in complex 3-space CHAPTER 3. The Structure of a GSG Spanning Ordered 3-Space. 1. Introduction. 2. Completely self perspective sets of low orders. 3. Completely self perspective sets of order 8 CHAPTER 4. Desmic Triads in General Spaces. 1. Introduction. 2. Multiply perspective arrangements 3. Multiply perspective arrangements of class [k,k,d], k s 7. Bibliography Page iv vii 16 36 41 49 49 59 65 83 83 85 93 118 LIST OF FIGURES Figure Page 1.1, 1.2. . . . . . . . . . . . . . . . . . . . . . 3 1.3 . . . . . . . . . . . . . . . . . . . . . . . . 4 2.4, 2.5. . . . . . . . . . . . . . . . . . . . . .20 2.6 . . . . . . . . . . . . . . . . . . . . . . . .21 2.7 . . . . . . . . . . . . . . . . . . . . . . . .24 2.10. . . . . . . . . . . . . . . . . . . . . . . .26 2.11. . . . . . . . . . . . . . . . . . . . . . . .27 2.12. . . . . . . . . . . . . . . . . . . . . . . .28 2.13. . . . . . . . . . . . . . . . . . . . . . . .29 2.14. . . . . . . . . . . . . . . . . . . . . . . .30 2.15. . . . . . . . . . . . . . . . . . . . . . . .31 2.16. . . . . . . . . . . . . . . . . . . . . . . .32 2.17. . . . . . . . . . . . . . . . . . . . . . . .33 2.18. . . . . . . . . . . . . . . . . . . . . . . .34 2.19. . . . . . . . . . . . . . . . . . . . . . . .35 2.20. . . . . . . . . . . . . . . . . . . . . . . .37 2.22. . . . . . . . . . . . . . . . . . . . . . . .40 iv Figure 3.7, 3.8. 3.9, 3.10, 3.11 3.12, 3.13. 3.14. 3.15, 3.16, 3.17. For [SI = 8, con'd diagrams: Case 2. Case 3, 4 Case 5a,b Case 5c Case 6a Case 6b,c Case 7b Case 7a Case 8a,b Case 8c Case 9a Case 9b, 10 Case 11 Case 1. Page .47 .51 .53 .54 .59 .58 .60 .61 .62 .63 .64 .65 .66, .67 .68 .69 .70 .72 .73 .74 .75 .76 .77 .78 .79 .81 67 Figures Page Case 12 . . . . . . . . . . . . . . . . . . . . . .82 4.1 . . . . . . . . . . . . . . . . . . . . . . . .92 4.2 . . . . . . . . . . . . . . . . . . . . . . . .96 4.3 . . . . . . . . . . . . . . . . . . . . . . . 100 4.4, 4.5. . . . . . . . . . . . . . . . . . . . . 101 4.6 . . . . . . . . . . . . . . . . . . . . . . . 102 4.7 . . . . . . . . . . . . . . . . . . . . . . . 103 4.8 . . . . . . . . . . . . . . . . . . . . . . . 104 4.9 . . . . . . . . . . . . . . . . . . . . . . . 106 4.10. . . . . . . . . . . . . . . . . . . . . . . 108 4.11. . . . . . . . . . . . . . . . . . . . . . . 110 4.12, 4.13. . . . . . . . . . . . . . . . . . . . 111 4.144 . . . . . . . . . . . . . . . . . . . . . . 112 4.15, 4.16. . . . . . . .. . . . . . . . . . . . . 114 vi LIST OF SYMBOLS Symbol k-secant GSG. . . . . . . . GDC. SG [1x,23] points 1 and x separate points 2 and 3 (aa,b ) A configuration with 'a' points and 'b' 5 lines such that there are a points on one line and a lines on one point. (nl'nz'..°'nk) [P] , 5* . eij' 1 {A'B,F} , [k,p,d], {A,r} A Y B1 (46) . [k,p,d] SPA. (k,k, d) SPA. [k k d] MPA. vii Page 10 11 12 15 60 61 70 85 86 91 94 95 I NTRODUCTI 0N In the study of convex sets in linear spaces the con- cepts of supporting and separating hyperplanes is of central importance. While not as strong a case can be made for the study of cutting hyperplanes of a family of, say, disjoint compact sets in a linear space it does seem natural and possibly useful to prObe such matters. For example, we may ask if corresponding to any finite family of bounded, closed disjoint sets in a Banach space there is at least one hyperplane cutting exactly one set of the family. The answer is no. It is possible to con- struct two disjoint bounded closed sets in cO such that any hyperplane cutting one of the sets intersects the other set. However, if the sets, instead of being merely bounded and closed are also compact, then the corresponding question can be answered in the affirmative Definition 1.1: k-secant A line intersecting precisely k—members of a family of point sets is a k-secant of that family. Theorem 1.1: (Kelly and Edelstein) A finite family of disjoint compact sets in a topo- logical linear space has a l-secant. How about 2-secants? The answer has turned out to be rather surprising and leads us out of the linear topologi- cal setting into the linear combinatorial setting with.which this thesis will be concerned. The question about 2-secants was originally posed by B. Grunbaum [4] who proved that a finite family of disjoint continua in E not all subsets of the same line, must 2. have a 2-secant. He thoughthfthis as a generalization of the Sylvester phenomenon, namely that a finite non-linear set of points of En' or more generally in any ordered pro- jective space, is cut by some line in exactly two points. Herzog and Kelly (9) extended the Grunbaum result to a finite family of disjoint compact sets in En at least one of which is infinite. Edelstein, Herzog and Kelly (2) strengthened the above result to the following: Theorem 1.2: A finite family of disjoint compact sets in a topo— logical linear space at least one of which is infinite and not all of which are subsets of the same line is cut by some hyperplane in exactly two sets. That the condition of infinite number of points in one set cannot be dropped follows from the following example. Consider the Pappus configuration in E2: A, B, C on a line L, A’B’C’ on 1’ + z with AB’ flA’BzC A’C n AC’ I! w BC’ nB’C=A The sets {A,A’,A”} {B,B’,B”], {c.c’,c”] constitute a finite family of Figure 1.1 disjoint compact sets in E2 having no 2-secant. The hope was that this and possibly a few other con- figurations would be the only one standing in the way of improved version of theorem 1.2. Unfortunately, this is not the case. Consider as a further example, the vertices of a regular hexagon alternately carrying the number 1 and 2, together with the three points at infinity on the side lines carrying the number 3. The points carrying the same number con- stitute a set and the three such sets ,/ \ have no 2-secants. The interesting 1 2 3 thing about this example is that it is . 1., extendable to any regular 2n-gon. So Figure A in E2 we have an infinite class of examples of what we now call generalized Sylvester-Gallai configurations. Definition 1.2: A generalized Sylvester-Gallai configuration (GSG) is a finite family of disjoint finite sets not all on one line which has no 2-secant. Definition 1.3: If in the above each set of the family consists of a single point, then the configuration is called simply a Sylvester-Gallai configuration. At this stage Kelly and Edelstein [3] anticipated a flood of further counterexamples in E3 and En' For example they found that numbering the 8 vertices of a cube in E3 alternately with the integers l and 2, the center and three ideal points on the edges with the integer 3, yielded a counterexample. This is the classically studied desmic configuration discovered by Stephanos in 1890. g 3 However, they could find no other :f/fl example spanning E3 and this ’m-HLP //2&‘” _73 example failed to generalize to 27~‘%+_~.31' higher dimensions. Fig. 1'3 /’Efl‘w-m(i?/l Note incidentally that the l 5/ topology has completely vanished at this stage and one is concerned only with incidence structure and order. That the flood, alluded to above, would not materialize became evident from the discovery that in ordered projective spaces there are no generalized Sylvester-Gallai configurations spanning a space of more than three dimensions, specifically: Theorem 1.3: If {Si} is a finite collection of two or more non- empty disjoint finite sets in an ordered projective space such that Usi spans a subspace of dimension at least 4, then there exists a line (and therefore also a hyperplane) cutting precisely two of the sets. This theorem is proved in Chapter 3. In his Transactions'paper, Motzkin [8] initiated this study, showing among other thinge,that Sylvester-Gallai configurations exist in a wide variety of non-ordered projective spaces. It is, of course, clear that any finite projective space is a Sylvester-Gallai configuration but it is not so obvious that such configurations exist in projective spaces of infinite cardinality. Motzkin showed that such configurations always exist in projective spaces over a field which contains roots of unity other than 11. We continue this program in Chapter 2, characterizing such configurations of low orders as well as those which are subsets of three or four lines. The latter part of the chapter is concerned specifically with complex projective space. Specifically we try to cast some light on a ques- tion raised by J. P.Serre [AM 73 P. 89, 1966] concerning the existence of non-planar Sylvester-Gallai configurations in complex 3-space. We succeed only in establishing a rather high lower bound on the cardinality of such sets if, in fact, they do exist. In Chapter 3 we attack the problem of characterizing generalized Sylvester Gallai configurations in ordered projective three space. It has been conjectured that the classical desmic configuration is the lone element of this class but we were unable to prove or disprove this. However, we state and prove a number of unpublished theorems of Edelstein throwing considerable light on the structure of G S G's in ordered projective 3-space and go on from there to show that there are no analogues of the classical configuration in which each of the three sets has 5, 6, 7 or 8 points. The 8 point analysis involved a lengthy case-analysis which seems to preclude a generalization by these methods to higher orders. Chapter 4 is devoted to a study of the existence and characterization of generalized desmic configurations (GDC) in a variety of projective spaces. There is consider- able emphasis on the problem of obtaining configurations which span spaces of various dimensions. For example we show that a desmic triad in which each set consists of pre- cisely 7 points does not exist in any projective space, while one in which each set consists of exactly 6 points has a very special structure [see theorem 4.3] and exists only in a projective 3-space over a field of characteristic 3. Detailed analysis of the desmic triads resulted in the concept of multiply perspective and self perspective sets. As a final example of the variety of results in this chapter we cite the following theorems: Theorem 1.4; . . . d-l d-2 There eXists desmic triads of order q —- q for any prime p, where q = pk, d is dimension of the space of order q spanned by the configuration. Theorem 1.5: There is no desmic triad of order 10 spanning a projective space of 4 dimensions. CHAPTER 2 Sylvester - Gallai Configurations 1. Configurations: In our introductory remarks it did not seem necessary to abstract the notion of configurations. However, we cannot conveniently proceed much further without a more precise and a more general definition. Definition_gél: A structure consisting of two sets, P (points) and L (lines) together with a symmetric binary relation (incidence) between points and lines such that two points are incident with at most one line is a linear configuration. It is usual in such studies to identify a line in L with the set of points of P incident with that line. Definition 2.2: Two linear configurations are abstractly (combinatori- ally) isomorphic if there is a 1-1 correspondence between their points and lines respectively which preserves the rela- tion of incidence. Definition 2.3: A configuration is embeddable (or realizable) in a projective space if there is a 1-1 mapping of the points of 8 the configuration onto a set of points of the space and a 1-1 mapping of the set of lines of the configuration onto a set of lines of the space such that incidence is preserved. Definition 2.4: A configuration in which each two points are incident with a line is line complete. A point complete configuration is defined dually. In this new terminology we might recast the Sylvester- Gallai theorem thus: Thggg'gm 2.1: A necessary condition that a finite line complete con- figuration be embeddable in an ordered projective space is that some line of the configuration be incident with two and only two points. Since this theorem is so basic to this study we repro- duce a proof here adapted from that of Kelly and Moser [6] which employs the useful concept of residence of a point. This concept will recur in subsequent proofs. Definition 2.5: A line complete configuration of n points, n-l of which are on a line is a near pencil. Definition 2.6: The set of lines incident with a point of a configura- tion is called a pencil of the configuration. The point is 10 the vertex of the pencil and each line a ray. If the number of rays is m, the pencil is an m-pencil. If the numbers of points on the rays, arranged in descending order of magni- tude, are n1,n2,°'°nm, the pencil is said to be 9: type (n1.n2.---nm). Definition 2.7: A finite line complete configuration in which no line is incident with exactly two points is a Sylvester-Gallai configuration or an S.G. configuration. Definitionggggz If C is a finite line complete configuration in an ordered projective plane, with point set T, then the set of lines of C not through PET either consists of a single element or partition the plane into regions. In the latter case P is in one of these regions which is called the residence of P relative to g. A sideline of a residence is called a neighbor of P. We now proceed to prove theorem 2.1 stated earlier. There is no SG configuration in an ordered projective space. Case (1): The lines of the configuration form a partition of the plane which is a near pencil. Let P be the point not on i-th line of the pencil. Then the join of P to any point on L is a 2-secant. 11 Figure 2.1 The lines of the configuration, not through P, form a proper partition of the plane such that P has a residence with neighbor L. Figure 2.2 Suppose g has three points on it 1,2,3. Let x be a point of the boundary of the residence of P such that [1x, 23]. Claim that x * i , i = 1,2,3. Suppose not: i.e. x = 3 (say). Let 1’ be another neighbor of P. 1’ has at least two points z,4 on it. NOw either '53 intersects the residence of P or 43 intersects the resi- dence of P and z,3,P are collinear or z,3,P are 12 collinear and 1,2,3,4 are collinear. In the last case 2’ contains a third point m such that either .35 inter— sects the residence or 2’ is a 2-secant as required. We now suppose x # i, i = 1,2,3. The line IP has a third point Q on it otherwise we have a 2-secant as required. Thus (IE‘EE, 36 Ga], i.e. 16 =IIP and E5 intersect the residence .2 35 intersects the residence of P or .25 intersects the residence of P according as [10, PR] or [1P, QR] where l, P, Q, R are collinear as indicated. Therefore in every case we have a contradiction of the definition of residence of P, which implies z is a 2-secant as required. I Remark 2.2 (a) There is no S.G. configuration in En for all n. 2.2 (b): There is no S.G. configuration in any real Banach space. Every finite projective space is an S.G. configuration. Definition of a Scheme 2.9: Following the usual practice of using rectangular schemes to represent configurations of type (aa) in pro- jective geometry, we describe the following variant of it to represent any S.G. configuration of order n. 13 In constructing an S.G. configuration of order n the following method will be found convenient: We label the n points with the numbers 1 through n and label the n pen- cils at each point, similarly with the numbers (1) through (n). Then we set up a scheme of lines in which the lines incident with any given pencil are arranged in a column and the points incident with any given line are arranged in a row of a column. There will be n columns corresponding to the n pencils, but the number of rows per column will vary according to the number of lines in a pencil. The scheme must satisfy the following conditions (1) The numbers written in the rows of any one column must contain the number representing the pencil corresponding to the column to ensure that all lines of a pencil contain its vertex. (2) The numbers of a row of a column must be more than three in cardinality to avoid 2-secants. (3) Two different rows of a column cannot have two numbers in common, as this would make the straight lines corresponding to the rows coincide. Remark 2.3: We observe that a row of length k is repeated in the columns k times. An Illustration: An S.G. configuration of order 7 corresponds to the following scheme: 14 (l) (2) (3) (4) (5) (6) (7) 124 214 317 412 516 615 713 137 235 325 436 523 627 726 156 267 346 457 547 634 745 where each row in the columns after the vertical bar are re- peats. The resulting configuration is seen easily to be iso— morphic to the Fano plane. Definition 2.10: Two S.G. configurations are equivalent iff there is a 1-1 correspondence between the points and lines of the con- figurations which preserves incidence. Definition 2.11: Two schemes are equivalent if one can be obtained from the other by means of one or all of the following operations: (a) Interchange of any two numbers in each row of any column, corresponds to relabelling the points of a line. (b) Interchange of any two columns. This corresponds to the relabelling of the pencils hence of the points. (c) Interchange of any two rows of a column. This corresponds to relabelling the lines of a pencil. Definition 2.1;: Two S.G. configurations are said to be schematically equivalent if they have equivalent schemes. Theorem 2.5: Two S.G. configurations are equivalentzrfifthey are schematically equivalent. 15 2:29.12: Suppose the two configurations are denoted by C1 and C2 respectively. Furthermore suppose Cl and C2 are equivalent, i.e., there is a 1-1 correspondence between their points which preserves incidence. The 1-1 correspon- dence between points is an operation of type (b) i.e relabel- ling the points. The preservation of incidences are operations of types (a) and (c); i.e. Cl and C2 are schematically equivalent. Conversely if Cl and C2 are schematically equivalent all the operations involved are 1-1 correspond- ences and they preserve incidence; i.e. Cl and C2 are equivalent . I We now derive necessary numerical conditions for an S.G. to exist. Lemma 2.6: If a k-pencil F has the sequence (n1,n2,---,nk) associated with it. Then 1+F1W ni = n+k-l, where n is order of S.G. 16 21:29:: We have, counting all points but the vertex of the pencil (nl-l) + (nZ-l) +---+ (nk-l) = n-l k = Z n. - k = n-l 1 1 k i.e. 2‘. n. = n+k-l I 1 1 2. Complete characterizations of S. G. configurations for low orders not exceeding 12. Our procedure is as follows: For any given n as the order of an S.G. configuration we first of all seek the solution of the following system of equations and inequal- ities. then for each solution investigate the construction of an S. G. configuration. Lemma 2.7: An S. G. configuration of order 7 is not a subset of a S-pencil. 17 Proof: This follows from the fact that there are no integer solutions to the following system of equations and inequali- ties 5 Z n. = 11 l i ni 2 3 i = 1,2,3,4,5. Remark 2.8: Similarly, an S.G. configuration of order 7 is not a subset of a 4-penci1. Remark 229: For a 3-pencil the corresponding system of equations and inequalities has the solution (3,3,3). An S.G. configura— tion realizing this solution has already been constructed in the illustration above. Any two line complete configura- tion, each vertex of which has order 3, we easily see to be equivalent. Lemma 2.10: An S.G. configuration of order 9 is not a subset Of Ia k-pencil for k 2 5. Proof: This follows from the fact that there are, for k = 5 (say). no integer solutions to the system 18 ll [.4 0) Remark 2.11: There are solutions (3,3,3,3) and (5,3,3) (4,4,3) corresponding to a 4-pencil and two 3-pencils respectively. Remark 2.2: (a) (b) (C) We observe that if an S.G. configuration is a subset of a 3-pencil then we have It is also clear that an S.G. configuration is not a subset of a 2-pencil. Later we shall prove a theorem which determines completely all S.G. configur- ations which are subsets of a 3-penci1. Solutions (5,3,3) and (4,4,3) are impossible by (a). Theorem 2.13: There exists an S. G. configuration of order 9. 19 _Proof: This contains the 4-pencil corresponding to the solution (3,3,3,3) (l) (2) (3) (4) (5) (6) (7) (8) (9) 123 213 312 415 514 617 716 819 918 145 249 346 429 527 628 725 826 924 167 257 358 436 538 634 739 835 937 189 268 379 478 569 659 748 847 956 The above table satisfies all three conditions (a) Each column contains the integer representing the pencil in each of its rows. (b) Each row has length at least 3. (c) No two rows of a column have two integers in common. The scheme represents an S.G. configuration of order 9. I Remark 2.14: We observe that the above S.G. configuration is equi— valent to the configuration of type (9 12 4' 3)' It is always possible to represent a configuration by points in a projective plane in which the lines of two pencils of the configuration are faithfully represented by subsets of points on projective lines. If the projective line joining these two points is taken as an ideal line the pencils in the resulting affine plane are two independent sets of parallel lines and the resulting planar set of points and lines is called a grid diagram of the configuration. 2(3 Further linearities are sometimes conveniently suggested by means of simple arcs passing through a set of points of the grid diagram which correspond to sets of points which are linear in the configuration. Thus we might represent the seven point projective plane by a grid diagram in the real affine plane as follows: Slh 1K 7 I’m 4 6 } Fig. 2.4 1.1.1 > 1 2 3 As a second example consider a grid diagram representing the configuration (94, 123) F San IR A 9 F 18 ’ . 4.L 4/ ‘\ A F14}. 23.5 \ 4' \ In this second example the grid diagram is only a partial representation of the configuration while in the first example we have a complete representation of the seven point plane. Proceeding from the linearities of the grid diagram in example 2'we can construct a tabular scheme as follows: 21 312 593 123 634 716 856 213 415 918 346 514 145 658 725 819 257 428 926 378 527 167 617 738 837 269 436 935 359 568 189 629 749 824 284 479 947 It should be clear that up to the equivalence previously described the tabular scheme is unique and hence we claim 12 that the configuration (94, 3) is combinatorially unique. Definition 2.13: We now ask whether it is possible in some affine plane to complete the grid diagram so that we will have a com- plete representation of the (94, 12 configuration with 3) all lines of the configuration represented by linear subsets of the affine plane. Such a representation we refer to as an embedding of the configuration in the particular affine plane. 5A T» .T«\‘ (Lb) 40:11 A Flg. 2.6 (0.0) 1 2 5 7 8 I -% Fig. 2.7 l 2 3 4 The scheme below easily follows and is unique 4123 615 156 7458 836 9,10,64 4578 627 1234 726 8457 917 49106 638 179 719 829 928 69104 18,10 73,10 8110 935 2134 3124 5478 710964 267 359 516 1025 289 368 539 1037 25,10 37,10 52,10 1018 I We claim that Theorem 2.19: There is no 10 point S.G. configuration containing only 4-pencils Proof: Suppose S were such a pencil. Its grid diagram is shown below 25 53* We, 4 Fig. 2.8 6 7 _A. 1 2 '3 The pencil with vertex 1 must contain a 4-1ine and this 4-line will certainly contain either point 9 or 10. Thus the pencil with either vertex 9 or vertex 10 is not a 4-pencil. The unique S.G. configuration of order 10 has been combina- torially characterized. We now consider its possible embeddings. Theorem 2.23: A 10 point S.G. is embeddable in a Desarguesian plane, iff the coordinate ring has characteristic 3. 64A W\ 1!\ l\ 9 10 5 0.1) .. 9.1) B /r 10 2.17A / / \ '1' 6/ / I A Table for A 34 312 514 123 213 415 857 1095 634 346 526 145 256 436 839 1073 625 3710 578 1679 24810 48102 82410 10248 6179 389 5910 1811 2911 4711 8111 10611 6812 351112 531112 11012 2712 4912 8126 10112 61011 7310 9167 1118 12110 758 9211 1129 1227 7212 938 113512 123511 7169 9412 1147 1249 7411 9510 11610 1268 Table B 312 514 123 213 415 857 346 526 145 256 436 839 3710 578 1679 24810 48102 82410 389 5910 1811 2117 4119 8111 31112 511123 11012 2129 4127 8126 1095 643 7310 9167 1118 12110 1073 625 758 9212 1127 1229 10248 6179 7211 938 113512 123511 10116 6812 7169 9411 1149 1247 10121 61110 7412 9510 11610 1268 It is easy to check that 1A = and XB = 2. (see following diagram) Ax ll 5w 5 11 11 /[ 12 # ,/I/ 2 . , 8 / / / ’7 / / /' 10 1%;3—1... 44 1L. 4f; /’ 6 1 2 ‘5 1* .A: 1A = 6 B: XB = 2 i.e.; the S.G. configurations of order 12 represented by tables A and B are combinatorially distinct. 35 Embedding of A 11 '12 Eb) (bb) 8 90 , 1_ Fig. 2.19 / (amfiga) 5 (01) 6 4- / (ll) 1/ - (00) (10) Line 42: x + y = l = a + b = 1 - (1) For linearity l, 8, 11 x = yb_1a 4, 7, 11 x = (y-l)(a-1)_la - -1 =yb1a = (y-1)12 -a = 1 - 2a + a2 i.e. a2 - 2a + 1 = 0 = a = lgl- a b =_l§l- Thus we observe that S.G. configuration represented by table A embeds in a Desarguesian planeztfifthe ring is of charac- teristic 2, while that represented by table B embeds in a Desarguesian plane iff the ring contains 4th roots of unity. I Corollary 2%;;: There is one and only one S.G. configuration of order 12 in the complex plane whose structure is determined as a subset of a S-pencil coordinatized by 4th roots of unity. This completes the analysis of S.G. configuration of order 12. 3. S.G. configurations as subsets of 3 lines Theorem 2.33: A subset, S, of 3 non-concurrent lines in a projective 37 plane coordinatized over a field is a finite S.G. configura- tion if and only if S is projectively equivalent to the set of points {(1, ai, 0), (0, 1, -aj), (1, 0, ak)} i,j,k = l,2,---,n where {aa], a = 1,2,3,'°°,n are nth roots of unity. Proof: Suppose S a subset of 3 non- I concurrent lines L1, 22, L3 1n a Pappian plane,is a finite S.G. configuration. Now if 21 2 0 £1 = 01 and L3 0 21 = 02 are the vertices of the triangle of L reference with 01(1,0,0),02(010) and 03(001) then the coordinates . O 100 of Ai'Bj'Ck may respectively be 1( 1 taken to be A.(0,1,-a.), B.(1,0,b.), 3 1 1 J J k Ck(1,ck,0) With a1 = 1, cl = 1. It is easily seen that [[ai}| = |{bj]| = [(ck]| Now suppose Ai' Bj' Ck are collinear Then 0 1 -ai = 0 = (-l)(-bj) - aick = 0 l O bj :: D = aickl * 1 ck 0 1 J V With the assumption that C1 = 1 we see that the following points are collinear: (110), (10b) and (01 -a) i.e. 3 a, b e[bj} 9: for each a E {ai] b = a from the above equation (*) I: [bj}= {ai} 38 Similarly {bj}= {ck} i.e. {ai}= {bj}= {ck} , {ck} 3a3 E {ai} 9: for al, a2 6 {ai} Since {ai]= {bj} a3 = a1&2 i.e. {ai} is a subset of non-zero element of a field F which is closed under multiplication, therefore it is a multiplicative subgroup of F*. Therefore [ai} is a cyclic group (see Artin [12] page 49). i.e. H a positive integer n such that an = 1 for some a 6 {ai} i.e. the set {ai}= {bi}: {ck} are nth roots of unity. For the 2nd part of the proof, suppose [ai} is a cyclic subgroup of F* of some field F. Consider the following set of points [(1,ai,0), (0,1,-aj), (1,0,ak)}. We observe that the indicated coordinates satisfy a3. = aiak whenever any point Ai(1,ai,0) is joined by a line to a point Ak(l,0,a the line passes through a third point k) Aj(0,1,-aj) i.e. the indicated set form a finite S.G. configura— tion on three non-concurrent lines. I Theorem 2.34: A subset, S, of 3 concurrent lines in a projective plane coordinatized over afield F is a finite S.G. configura- tion iff F has finite characteristic and S is projectively equivalent to the set of points {(1,0,ai), (1,1,aj), (0,1,ak)} 39 i,j,k = 1,2,---,n where {am} a = 1,2,°°°,n is an additive subgroup of the coordinate field. Proof: Suppose S a subset of 3 concurrent lines 21,12.£ 3 in a Pappian plane is a finite S.G. configuration. If 03 = £1 n 22 0 23 and 01 6 £1. 02 6 £2 are the vertices of the triangle of 2 21 reference with 01(100) 02(010) and 03(001) then k may respectively be taken to the coordinates of Ai'Bj'C / 01(100 be Ai(01ai)' Bj(10bj) and Ck(llck)' With a = 0, b = 0 it is easily seen that [{ai}‘ = I[bj}‘ IfckH Now suppose Ai' B., Ck are collinear Then 0 1 a. = O = -[Ck-bj] + a. = O l , a. + b.l 1 J 1 0 bj = ck 1 1 c With the assumption that a1 = 0 we see that the following points are collinear: (010), (10b), and (11b) 3 c 6 {ck} such that for each b E {bj} 2b: :1). c {9k} 1 3} Similarly {q(}= {31}. 4O [ail = {133'} = {ck} - Since {a1} = {bj}= {ck} 3 a3 6 {a9}: for al, a2 6 {a1} 2 i.e. [ai} is a subset of the coordinate field F closed under addition, therefore it is an additive subgroup of the coordinate field. Conversely, suppose [ai} is an additive subgroup of F, then whenever we join the point we get a Ai(01ai) on to the p01nt Bj(10bj) on ‘1 Also if we join Bj(10bj) on ‘2 point Ck(11ck) on 11 we get a point Ai(01ai) on L3. to a point Ck(llck) on 13 22 since ai = ck - bj is an element of the additive group {a1} i.e. Ai' Bj' Ck are point of an S.G. and a subset of 3 concurrent lines. For the remaining part of the proof ‘we have to show that the field has finite characteristic. We have that C(11a), 13(100) “1 and A(0,1,a) are collinear. Also (11a), (010) and (10a) are collinear and (10a), (112a) and (01a) are collinear. Similarly (100), (112a), (012a) are collinear; (010), (112a), (102a) are collinear and (1023), (114a), (012a) are 1 ' 110 collinear and so on, for some a E F. Fig. 2.2 For the configuration to close we need the last point generated on L3 to coincide with some earlier point on L k . . . 23. Thus we need 2 a = 2 a for some p031t1ve integers k and z where L < k (say) 41 k-z = 2£(2 -1)a = 0 = 23(2m-l)a = 0 for some positive integer m. Since a # 0 we need either 2 = 0 i.e. coordinate field has characteristic 2 or p = 0 where p is a prime divisor of 2m-1 (since 2m-l containsaiprime factor + 2) i.e. characteristic p Thus in all cases the coordinate field has finite characteristic. This completes the proof of the theorem. I Remark 2.35: Theorem 2.36 corrects the mistake in the theorem on page 460 of Motzkin [8]. 4. 0n the existence of S.G. configurations in complex projective 3-space Motivated by the question raised by J. P. Serre (AMM731966 we, in this section,show that there is no non- planar Sylvester-Gallai configuration in complex projective 3-space with fewer than 40 points. It is convenient to confine the argument to the C3 setting but the reasoning for the most part is applicable to more general spaces and permits us eventually to characterize all non-planar Sylvester-Gallai configurations of fewer than 40 points in any projective 3-space. Lemma 2,36: If a Sylvester-Gallai configuration, S, is a subset of a 4-pencil having lines £1, £2, 23, £4 with vertex P 42 and if the associated sequence (n1,n2,n3,n4) has the properties nl-l = x 2 n2-1 = y 2 n3-l = z 2 n4-1 = w then '8' 2 x + 9. ££22£= Observe that x = [£1 0 (S-P)|: y = [22 n (S-P)|: z = I23 0 (S-P)‘: w = |z4 n (S—P)|. If Q is a point of 22 n (S-P), then the x lines joining Q to points of £1 n (S-P) intersect (L3 U L4) 0 (S-P) and hence z+w 2 x. Similarly y+z 2 x and y+w 2 x. Thus y+z+w 2 %x and ISI 2 x + gx. Now if x 2 6, |S| 2 x + 9. If x = 5 and w 2 3 then y + z 2 6, = |S| 2 x + 9. If x = 5 and w = 2, again y + z + 2w 2 2x, y + z 2 6. Keeping in mind that P must be an element of S we again have IS] 2 x + 9. If x = 4, it follows from theorem 2.28 that |S| 2 13 = 4 + 9. Finally if x = 3 then y, z, w are all 3 and |s| > x + 9. 1| Lemma 2.37: If in complex projective 3-space a Sylvester-Gallai configuration, S, is a subset of a pencil of four planes having axis 1, and if z’ is a line in one of these planes o with la n 5| 2 13, lz’ n SI 2 k 2 3, and IL n 2’ n s| = 0 then [s] 2 4o. 43 EEEEED Let B be any plane containing 1' and a point of S not on a. S n 8 is a planar Sylvester-Gallai configuration and is a subset of 4-penci1 with vertex P = L n L’. It follows from lemma 2.48 that I8 0 SI 2 k + 9. Now there are at least three distinct planes through 2’ intersecting S in points not on 1! Hence |s|23(9)+|ans|227+13=4o. I Theorem 2,383 If S is a Sylvester-Gallai configuration spanning complex projective 3-space, then ISI 2 40. ££29£= Let L be a line of C3. Case 1: |z n sI = 7. There is clearly a line 2’ skew to 1 such that IL' n SI = k 2 3. There are at least 7 distinct planes through 1’ intersecting points of S not on 2’. If a is any one of these planes then Ia n SI 2 k + 6. Thus Is] 2 7(6) + k 2 45. Case 2: IL 0 SI = 4. If there are 5 planes containing 2 and intersecting S-L then ISI 2 5(8) + 4 = 44. Since no Sylvester-Gallai configuration spanning C is a subset of a pencil of 3 44 3 planes (vide lemma) we may assume that S is a subset of precisely four planes containing 2. Suppose a and B are two of these planes with Ia fl SI = I8 n SI = 12. From lemma 2.32 detailing the structure of Sylvester- Gallai configuration of order 12 it is clear that there are then lines ‘1 in o. and z in B such that 2 IS 0 LII = IS 0 ‘2' = 4 and (z n 8) 0 S = ¢. There are 1 thus five distinct planes through intersecting S-L ‘2 2 and hence Is| 2 5(8) + 4 = 44. Suppose then, that one of the four planes through 2 contains 12 points of S. The only possible distribution of points of S on these four planes that does not immediately lead to the conclusions that ISI 2 40 is that in which three of the planes contains precisely 13 points and the remaining plane contains precisely 12 points. Suppose a is one of the planes with 13 points of S and B the one with 12 points. If P is a point of S n (Q-E) and Q is a point on S n (a-a) not on any of the four lines joining P to points of S n 1, then it follows from lemma 2.49 that ISI 2 40. If no such Q exists, then for at least one of the lines, say 2’, joining P to a point of S n L IS 0 z’I 2 4. There are now at least 5 planes containing 2’ and points if B n S and ISI 2 40. Case 3: I2 0 SI = 5 As in the previous cases we can proceed at once to assume that there are precisely four planes containing L and points of S - L. If each of these contain more than 45 13 points of S then ISI 2 4(9) + 5 = 4). We may thus assume that for one of the planes a, Id 0 SI = 12. It now easily follows that S n a is a subset of a 4-pencil two of which contain 5 points of S and the other two of which contain 3 points each. In such a configuration there is a line 2’ with either IL’ 0 SI = 4 or I1’ 0 SI = 3 and 2’ 0 z n S = 0. In the first instance the theorem follows from case 2 and in the latter from lemma 2.40. Case 4: IL 8 SI = 6. First observe that there is a line 2’ shew to 2 such that I2’ 0 SI 2 3. Now through 2’ there are at least six planes intersecting S-L’. Hence ISI 2 6(6) + 3 = 39. Now suppose as usual that S is a subset of a pencil of four planes through 2. For at least one of these planes, o. Ia n SI > 13. Let P e S n (o'z). If Q is a point of S n (a-L) not on any of the six lines joining P to the points of S n 2 then ISI 2 40 by lemma 2.49. If no such Q exists one of the six lines joining P to the points of S n L may be assumed to contain precisely 6 points of S. Thus Ia n SI 2 16. Since 3(7) + 16 = 37 < 39 it follows that either Id 0 SI 2 19 or for some second plane 5, I8 n SI 2 16. In either case ISI 2 40. Case 5: For all 1, IL 0 SI s 3. If for any plane, a, Id 0 SI > 13, then for any point P E S n a there are at least 7 lines containing P and 46 points of (S n o) - P. If 2’ is a line joining P to a point of S - a then L’ is contained in at least 7 distinct planes intersecting S - 2’. Hence ISI 2 7(6) + 3 = 45. We are thus led to assume that for all planes, a, contain- ing at least 3 non-linear points of S, lg 0 SI = 13 or Id 0 SI = 9. If Ia 0 SI = 13, then any point P of S n a is on 6 distinct lines containing points of (S 0 a) - P. If 2’, as above, is a line joining P to a point of S-a, then 1’ is on six planes containing points of s—z’. Thus Isl 2 6(6) + 3 = 39. Suppose, in fact, that |s| = 39, L n s c a, [S n oI = 13. There must be m+n planes each containing and intersecting S-L, with 39 = 10m + 6n + 3, m 2 1. This means m = 3, n = 1. That is S is a subset of a pencil of four planes through 2 three of which contain 13 points of S and one of which contains 6 points. If Q is a point of L n S and 2* a line joining Q to a point of S n (B-L), where B is one of the planes containing 13 points of S, then since Ia n SI = 13, there are 6 lines through Q in a each intersecting S. Hence, there are 6 planes containing 2* and points of S-£*. One of these planes is 8. Thus ISI 2 5(6) + 10 + 3 = 43. It remains to consider the case in which for all planes a intersecting S in three non collinear points Ia n SI = 9. We will show that such configurations do not exist. 47 {\ /' /\ A E 9I‘ G ’7IL/ .._——— --«~——'l ---"~““‘—'—""""_‘""‘_> i ,/ H’ I ~ ' /r/ / / I 1 '- Z _' I . 1, I F 2.-...2 “_ -_..._.._..-._... K I {.- ’ T I Xx/ C // 4f // Fig. 2.3 Suppose, in fact, that A,B,C are three collinear points of such a configuration S, and a and 8 two planes containing A,B,C and intersecting S - {A,B,C}. Let A,D,E be collinear points of S - [A,B} in a and A,D’,E’, be collinear points in S - {A,B} in B. We now introduce affine coordinates,taking the plane E,E’,C as ideal plane/in such a way that A(0,0,0), D(l,0,0), B(0,1,0), D’(0,0,l). From the known structure of’a 9 point planar S.G. configuration (see page 20 ) it follows that the remaining four points of a n S are H(a,l,0) F(1,%,0), G(a;%,0) and the ideal point I = AG 0 BF n DH 0 CE. Similarly in plane a point H’(0,1,a), F'(o,%,1), G’(O;%,a), and the ideal point I’ = AG’ n D’H’ n BF’ n CE’ are all points of B n S. 48 Now F,G,E F’,G’,E’ are in a plane Y. An easy computation shows that J(l,%3a) and K(a;§,l) are also in Y 0 S. Finally we observe that B, H’, E’, F, J, L are in a plane S, and that they are the six vertices of a complete quadrilateral. But a planar S.G. configuration of 9 points does not contain a complete quadrilateral as a subconfiguration so it follows that I6 n SI + 9. This contradiction shows that no S.G. configuration exists in C3 all of whose plane sections are 9 point S.G. configurations and completes the proof of the theorem. l Chapter 3 The structure of a GSG spanning ordered 3-space. Introduction: Definition 1.2 of the introduction described generalized Sylvester~Ga11ai configurations in the context of projective or linear spaces. We proceed now to define an abstract GSG. Definition 3.0: If the point set, S, of a finite linear configura- tion, C, is partitioned into sets [Si], i > 2 such that each pair of points from two different sets Si and Sj is on a line of C and if no line of C intersects exactly two of the sets {Si}, then C is a generalized Sylvester~ Gallai configuration or simply a GSG. The lines of the configuration are its secants and the sets Si are called the (constituent)sets of C. We first reproduce the Kelly-Edelstein proof that a GSG cannot span ordered projective k—space for k > 3. The remainder of the chapter is designed to throw light on their conjecture that the only GSG spanning ordered projec~ tive k-space, k > 2, is the 12 point desmic configuration (see introduction, page 4) discovered by Stephanos in 1890. 49 50 Mai-.1: There exists no GSG spanning ordered projective 4- space. Proof: Suppose, in fact, that S is a GSG spanning an ordered space, S4, of 4-dimensions. Let [Ai] be the constituent point sets of the configuration. Consider the line L 3 and p2 6 A2 and let S1 i together with pl spans S4. through p1 E A be a 3—space 1 not containing L. Then S The line L and each point of S-L determines a plane. All such planes may be described in a picturesque 'way as forming a book of planes with back L. Now L n Si = P and the planes of the book intersect Si in a bundle of lines with vertex P. Motzkin observed in his 1951 paper .[8] that for such a bundle in 3-space there exists a plane containing precisely two lines of the bundle. This follows, at once, if we con- sider a section of the bundle by a plane and apply theorem 2.1. The 3-space, $3, spanned by these two lines and L contains precisely two pages of our book. It is now an easy matter to check that if a collection of two or more finite non-empty and disjoint sets in 83 not on one, then there is a line intersecting precisely two lie on two planes and of the sets. This contradiction establishes the theorem. I 51 .EROOF: Fig. 3.1 Theorem 3.1 implies that GSG's in ordered projective spaces are confined to two and three dimensions. The next two theorems, due to M. Edelstein are the only ones known to us which throw much light on the structure of such sets. Definition 3.1: A k-secant of a collection of sets is simple if it intersects no set of the collection in more than one point. 52 _Theorem 3.2: If a finite family, {Si}, of disjoint finite sets spans an ordered projective space of dimension 3 and has no 2-secants, then all k-secants, k > 2, are simple. That is to say all k-secants of the sets of a GSG k > 1 are simple. .B£QQ£= Suppose, to the contrary, that 11 is a k-secant with k > 2 which is not simple. We may assume the labelling so that points 1 and 1’ are in S1 0 £1,226S20 9,33653001. Let V be a plane such that w 0 Si = 0 for i = 1,2,'°°,n. Now centrally project all points of 3 Si from 1’ onto w and let T be the set of these projections in w, and P the projection of 2. Let Q 6 T and X E 8 Si which projects into Q. Since X f 81 the line 1X 2contains a point Y E 8 Si other than X. The projection of Y onto w is a poiit of T on PQ. Hence PQ is not a 2-secant of T. It is now clear that T is not the point set of a near pencils and hence that the set of lines defined by the pairs of points of T which do not pass through P partition W into polygonal regions. It is shown in [G that if there are no 2-secants in T through P then the side lines of the residence of P are all 2-secants of T. Now if QR is a side line of the residence of P, with n Q, R E T, there is a point Z E U S 1 whose projection is 2 53 R, where X and Z are in different sets. I Hence the line joining X and Y must 1 1 contain a point of S1 since if it Fig. 3,2 2 X contains a point from a Si' RQ would not be a 2-secant of T? Thus we P G can assume a point 1” 6 S1 whose projection S* is on RQ. Now consider T U S*. It is easy to verify that the. lines defined by pairs of points of this set are precisely those defined by pairs of points of T. Hence the residence of P relative to T U S* is the same as the residence of P relative to T. Every line of T U S* through P contains at least three points of T U S*, but one of the side lines of the residence of P, namely QR, is not a 2- secant of T U S*. This is a contradiction and proves the theorem . I Theorem 3.3: If G is a GSG whose point sets {Si} span an ordered projective 3-space then no line intersects more than three of the sets. 2.299;: Suppose G has a k-secant z with k > 3. From theorem 3.2 we know that 1 must be simple and the nota— tion may be chosen so that z n Si = i, i = l,2,3,4,"',k, with points 1,2,'°',k in cyclic order. 54 As in the proof of theorem 3.2 we centrally project 8 Si from 1 onto a plane v with P the projection if 2 and T the set of projection of point if 9 Si. If Q + P is any point of T we again wish to show that the line PQ intersects T in yet another point. The argument then will proceed precisely as in theorem 3.2. To show that PQ contains a third point of T we consider a point X 6 2’ n 9 Si' where 2’ = 1Q, such that l and X are not separated by any points of 2’0 9 Si' Suppose X ¢ S3. The line 3X must then contain a point 0 3X. If 1 X’2 n 2’2 — n ’ ’— n K - o E 3 Si' X 4 0 £ — B E 3 Si then 1 2 3 4 l n of U Si. If not there is a point X’ = S a X B and and X are separated by two points of 3 1 n o 9 I U Si contrary to the definition of’ X. If X E S but 2 k > 4 the same argument shows that the line 4X has a n third point of U S on it. Thus in these cases there 2 i is a third point of T on line PQ. It remains to consider the case in which X 6 S3 and k = 4. n If the line 2X does not contain a point of U S 2 it must contain a point X* E 51' n Let X* 3 fl 2' = a’E U S. 2 i n X* 4 n 2’ = 5’6 3 Si' Again line 38’ contains a point w of S1 or the proof is complete. Since w i 83, XW contains a third point of U Si 0 z. This 55 point is clearly not 1, 2, or 3 and hence must be 4. Let a w n z = u 6 USi' Now 1 2 3 4 K 1 a’a’ A 1 4 u 3. Thus u + 2 and we have a contradiction to the assumption that k = 4. Thus in all cases the lines PQ intersect T in at least three points. The proof now proceeds in a completely analogous fashion to that of theorem 3.2 producing a contradiction to the assumption that In 0 USiI 2 3. I We have previously noted that the only known example of a three dimensional GSG in an ordered projective space is the 12 point desmic configuration consisting of three sets of four points each i.e. the configuration of type (12 16 4! 3). Since our efforts to characterize generalized Sylvester- Gallai configurations were not too fruitful we turned to the special case in which the number of sets in the configura- tion is three. It is easy to see that in this case ISlI ISZI = ISBI' We refer to such configurations as desmic triads or generalized desmic configurations (GDC). Definition 3.2: A GSG having exactly three sets all of whose 3-secants are simple is a desmic triad. In a desmic triad each point of set Si is a center of perspectivity for Sj and S (i,j,k) a permutation k of the integers 1, 2, 3. Thus sets Sj and Sk are 56 multiply perspective under ISiI different perspectivities. Thus the problem of constructing desmic triads in various spaces may be looked at as that of constructing two sets of n points which are perspective from n different centers. We will now show that a necessary condition that this be possible in a given space is that it be possible to find an n-point set in a plane of that space which is completely.§elf,perspective. The theorem showing the techniques used in the subsequent analysis will now be proved. Theorem 3:4: Suppose A, B, C are three sets of k points each, in a projective 3-space S which have no 2-secants. Let v be a plane of s with A n w = B n w = C n v = ¢. Let all secants be simple. Let m be a central projection of S onto w with center al 6 A. Define w(bi) = m(ci) = p e n m(B) = m(C) = P and m(ai) = q., i = 2,3,‘°',n. Then P l is self-perspective from each of the points qi. Proof: 57 Fig. 3.4 “*"P‘T‘ 2.1-2-2 .. q . Define qi: P 4 P as follows: “H Let Ckai 0 B = bj’ then 91(pk) = w(bj) (where kai is the line jOining ok to ai) Indeed Qi is uniquely defined, otherwise Ckai will be non-simple. C Also 9i is 1-1 Since 9i(pk) = 9i(p£) = o ai n B = k czai 0 B e Ckai = czai = ck = cg = pk = p2. Finally Qi is a self-perspectivity of P with center qi. Clearly pk, pj, qi are collinear since they are images of the collinear points bj’ ai under m. I ck, Remark 3;§: If no three points of P are collinear then the 91's are involutary. 58 Indeed; let ei(pk) = pj. Now for ei(pj) we consider cjai n B, Since cjai is coplanar with pkqui and no other points of P are in this plane we must have . _ . 2 _ . cjai n B bk' i.e. 9i(pj) — pk i.e. 9i — 1,Vi I Lemma 3.6: For each i, ei(p) I p V p e P Proof: Suppose the statement is false, then the points al'ai'bj’ and ok are collinear with pk. i.e. alai is a non-simple secant in A -«- I Lemma 3.7: ei(p) + ej(p) if i + j Proof: Suppose the statement is false; consider the accompany- ing figure. We observe that alp is a non-simple secant k cutting C at ck and ck’. *I Fig. 3.5 59 Theorem 3,8: A necessary condition for a GDC to exist in a project- ive 3-space is that in w the set P be self perspective from n-l centers. Proof: In defining the self-perspectivities Bi of P we obtain a 1—1 correspondence between the centers qi and the points af i = 2,"',n of A. I Remark 3.9: The qi's must be distinct from the pk's otherwise we have a non-Simple secant in A, namely since c b.a k j i is mapped by 8 onto pkqui if p3. = qi (say) then ai lies on the line 'aibjcj . The qi's need not be distinct points. The above theorem 3.12 holds true in any space, not only in ordered projective spaces since no order is involved in theorem 3.8 - 3.11 leading to it. Thus we shall make use of it in the more general analysis to be undertaken in chapter 4. 3.2 Completely self perspective sets of low orders. Suppose S a completely self perspective set in an ordered projective space and P, Q two distinct centers of perspectivity. 60 S is then a subset of two pencils [P] and [Q] formed by joining P and Q respectively to points of S. Let p and q be the number of rays in these pencils. If S* = [P] 0 [Q] then S C S* and IS*I= pq. It is convenient to consider the line PQ as the ideal line and to introduce affine coordinates. S may be then viewed as a subset of the "grid" 8* = [(ai.bj)lo al > (a1b1)62'bl)... It is, of course, clear that any quadruple of points no three collinear is a completely self perspective set, while no set of five points is of this type. We now proceed to show that there are no completely self perspective sets, S, in an ordered projective plane with ISI = 6,7,8. More general conclusions can be drawn about completely self perspective sets of orders 6 and 7 in more general spaces. This will be done in chapter 4. However the analysis of the case ISI = 8 by strictly algebraic arguments seems to be very involved. Theorem 3.10: There are no completely self—perspective sets, S, in an ordered projective plane with ISI = 6,7,8. 61 3J£ll Completely self perspective sets of order 6. ISI = 6. In the associated grid diagram either IS*I = 6 or IS*I = 9. In the first instance we may assume p = 2, q = 3 and S* = S {(allbl)l(a21b1)1(ajlbl)l(allb2)l(a2Ib2I(a-3Ib2)]° Q I; 5 6 . L‘ Fig. 3.7 > 1 2 3 P Let (albl) = l, (a2bl’ = 2, (a3,bl) = 3, (alb2) = 4, (a2b2) (a3,b2) = 6. If 915 denotes a perspectivity such that 915(1) = 5 then 915(2) = 6 or 915(2) = 4. The latter is immediately ruled out since (2) = 4 s 915(3) = 6 which violates 615 the assumption that the perspectivities are disjoint. If 915(2) = 6 then 915(3) = 4. But obvious order relations Show that lines 15, 26, 34 are not concurrent and 915 could not have a center. Hence ISI= IS*I = 6 is impossible Now suppose IS*I = 9, p = 3, q = 3. Qt »~ ,4 I I > } Fig. 3.8 79 Let T = [(albl),(alb3),(a3bl),(a3b3)], i.e. T is the set of "corners” of the grid. 62 It is immediately clear that S must contain at least two points of T and at least two opposite corners. Qa§e_l: S n T = [(al,bl),(a3,b3)). .5 6 = S = {(albl).(a2bl).(a3.b2).(a2b3).(a3.b3)) v Heavy dots indicate points of S, crosses 1 points of S*\S. 2 Fig. 3.9 923(1) = 4. or 923(1) = 5. or 923 (l) = 6. The following three grid diagrams Show that none of these is possible. 5 5 6 5 6 .-I..3$ ? , /? I V 3 4 3 :14 3t.“ 4 Fig. 3.10 k ‘ , ,7 "I? I ‘ \ .. . \ \ 1 2 1 2 " 1 In the first two respectively 923 and 946' 923 and 956 are not disjoint while in the third diagram obvious between- A to have a ness relations make it impossible for 23 center. Ease 2: S n T = ((319131)! (a3lb3)l (a3obl)} 5 6 _ 3 4 . (az'b3)I IK’ :f:>?, Fig. 3.11 1 2 916(4) = 2 916(3) = 5 But lines 16, 35, 24 do not concur and Q16 has no center. All other cases are isomorphic to either case 1 or case 2. Thus there are no completely self perspective sets of order 6 in an ordered projective space. 63 Again we remark that for ISI = 6 a more algebraic analysis can and will be made. But the presentation given here paves the way for the much more complicated treatment of ISI = 8. 3JLLZ. Completely self perspective sets of order 7. ISI 2 7. The situations in which IS*I = 8 or IS*I = 12 are easily seen to be combinatorially impossible. For IS*I = 9, p = q = 3 there are 3 non—isomorphic case depicted by the diagrams below T ..___.__r___...-..-.-i r Fig. 3.12 Associated with the A diagram there are two possibilities for S A A 6 l 6 2 7 7 4 4 5 Fig. 3.13 l 2 3 1 2 3 Since in Al no triples other than [1,2,3] and [1,4,6] are linear, S has at most two self perspectivities. For similar reasons there can be at most two self perspectivities of S in the A2 case. 64 Diagrams B and C give rise to essentially one possibility for S in each case B C 6 7 7 4 5 3 4 5 Fig. 3.14 k l 2 l 2 The analysis of linear triples shows that S in B has at most four possible self perspectivities while S in C can have at most two. Thus if S is a subset of an ordered projective space and ISI = 7 then S is not a completely self perspective set. 65 3.103. Completely self perspective sets of order 8. ISI = 8. If IS*I = 8 we have the essentially unique diagram shown A 5 "6 ’17 "8 \ 7 Fig. 3.15 1 2 3 4} The following diagrams dipict the possibilities for 916 4 I 15 A6 A7 [8 A N / Fig. 3.16 4 7 1 A , , '5 1 2 3 F 1 2 3 4 7 In neither case can 916 have a center, so IS*I = 8 is impossible. The case IS*I = 12 is very easily handled by methods discussed in chapter four while its analysis by ordered grid diagrams is a little awkward. Accordingly we refer the reader to chapter four for a discussion of this case. The discussion there shows that in an ordered projective space IS*I = 12, ISI = 8 is impossible. We turn to the final and most difficult case in which IS*I = 16. Let S* = [(ai,bj)} i,j = 1,2,3,4. (allb4) «---w- T: Fig. 3.17 (albl’ (a4,bl) 66 Our analysis is organized into a number of cases which can be described informally as follows: 1. 1 corner, 2 adjacent points 2. 1 corner, 1 adjacent, 1 non adjacent point 3. 1 corner, no adjacent points. 4. 2 corners adjacent, 2 non adjacent points. 5a,b,c. 2 corners adjacent, 1 adjacent point. 6a,b,c. 2 corners opposite, 2 adjacent points. 7a,b. 2 corners opposite, non adjacent points. 8a,b,c. 2 corners opposite, 1 adjacent point. 9a,b. 3 corners, 1 adjacent point. 10. 3 corners, non adjacent point. 11. 4 corners. 12. No corners. Case 1: S [‘1 T = [(allbl)t (azlbl)r (a11b2)}' It follows that S = [(al,bl), (a2,bl),(al,b2),(a2,b4,(a3,b3), (a3.b4).(a4.b2).(a4.b3)). 5 —-—-4 -—-~—-‘V6 7 8 Let gij represent the perspectivity in a complete set of perspectivities taking 1 into j. This notation is, of course, far from unique. 67 We propose to show, in this case, that there are no possibilities for 967' (D H U1 CD 0) \l (I) I M 967(8) = 3 967(8) = 4 l 7 “N H n) U1 N 0 U1 2__Lw_-._ii 0‘ UT I (7) U1 I 0‘ w p w 2 w p w 2 \ \ A; T L In each case the only possible center of the perspec— tivity 967 is seen to be inconsistent with the ordered arrangements of the remaining points. The impossibility is strongly suggested by the line with arrows. .Qe§§.;= s n T = [(al,bl),(a2,b1),(al,b3)}. S = [(al,bl),(a2,bl),(a4,b2),(a3,b2),(a4b3),(al,b3) (a2.b4).(a3.b4)} l 2 iiI 7 967(2) = 3 967(2) = 4 1 2 2 I 3 4 3 4 967 5 5 r/ A I A? 68 Hence is impossible. 967 s n T = {}. = S = {(a1.b1).(a3.bl).(al.b3).(a2.b2).(a4.b2).(a4.b3). (a3.b4).(a2.b4)}. 2 1 2 1 \ ’//l 422‘ 4 2-22_.>4 )3 -'-\<\d \, ’- / .2_ ”5‘224229 5 ,/ 6 5 , ”A //2/ ’2 ,/ I r r' 7 ./ ,. . 1 723/ 8 ’ k” 967(2) = 4 967(2) = 5 967(2) = 3 Hence 8 is impossible. . 67 Case 4: S n T = {(alrbl)o(a4tbl)t(allb3)} = s = {(a1.bl).(a4.b1)..(a4.b1)..(a2.b3>.(a3.b2).(a3.b4). (al.b3)..(a2.b4).('\ " / / 8 ,7 1 3 14\\ 4 ’ 5_ 4 ‘ 3 6 j 6 / [xi] ' r .111..- K 1 .1 7 //8 7 8 (52)(l3)(68) (51)(---) ‘—%//t7;/ C~_§F“Z F 2 3 .2” .._. y, 4 3 3 5 /6 / 5 61 5 6 / / /11- ’ / / 7 8 7 8\ 7 8 A1 n 81 = (86)(13) I Case 7a: B ‘374 G) 75 27 -4»~—~—;—4 |u—‘ ( w/ fie ‘ {I X l \‘ 1‘1 \ f1- $88 1 \ 4 L... 777’M'“%T"" (45)(18 1, )(36) 1 ~54,” 1 { ‘/ I 2’ 1 %;__17Aj4_4 ' .' I (7' r- .t-‘ ()~ 6 (24)(36 )(58) Grid diagrams 75 -/ 21 // / , ’/ 3.»; """ ’ j /4 / _ / , 5 *“‘*6 / /' 7 (45)(13)(68) __ 1 2 1 \ .1 \ / . T\ -1 \1 8 l ! 3 2. /§ ' i 1“" 11 (@ L11111J111 7 8 (23) (18) (46) 1 Z A .vf ) (28) (35) (14) 7- 8 (23) (46) (58) 7 p..- --- -141111 (24)(13)(68) 1 7 (28)(---) T‘fi K -1 76 8--) (4l)(--*) . [111 "8 1‘ (23)(---) 2 7£““'“ 1f?“ 7 / A31 1 / 1‘ '\ 41.6 (24)(18)(63) -,__1_- 17 1;--- 5 1,8 ' ,/ -/ -1--- 8 (25)(—‘-) 76 Case 8a: 5 H T = ((a1.bl).(a2.bl).(a1.b3)). (a): S (a2.b2).(a3.b3)1 (b): (D II ((al.b1).(a2.bl).(al.b2).(a2.b4).(a3.b2).(a3.b3). (a4.b2).(a4.b4)} (a4Ib2) I (a4Ib4) - __1__1-112 1.1.1__2 : { 12 3 / ' 4 3‘ /A A ¢ 1’ / 1 Fig. 3.31 A 6 , 5 - 6 971 ‘ '7 \ 7’78 7 8 (68)(--—) (64)(’-‘) (63)(—‘-) Thus 971 does not exist and configuration 8a is not completely self perspective. Case 8b: 1.11.1..- 2, L 2 A ;%r /: 4 J 3 ‘ ./ i , 971 5’ L6 \\\ (/ E//j _7 J11-L4::.11J J 8' 8 (64)(---) (68)(--‘) (63)(‘-‘) 871 is 1mposs1ble. I 77 Case 8c: 1__ 1 2 'WT . 2 .2 ””%:”§>;§_ 1 /r\3 11 V1 7 .1 TV (68)(---) (64)(---) (63)(---) 871 is impOSSible. I Case 9: s n T = {(al.bl).(a4.b1).(a1.b4)) (a): S ((al.b1).(a4.b1).(al.b4).(a2.b3).(a2.b4).(a3.b2) (a3.b3).(a4.b2). (b): s = {- 1 L ||I 'a‘: ( 1-- ._ _ a 1 -'i AM: Away 33 Amuv $3 33 82 Case 12: S = {(a1.b2).(al.b3).(a2.bl).(a2.b4).(a3.b1).(a4.b4). (a4.b2).(a4.b3)]. ' '. .-.-- f” '— ' '5 " ‘ " " """" T" 1-1 """"| 3 .. l J 9 T l 76 1 L \\1 i 1 2 5” i \V 1/ \ ‘/ /V , L i JL"‘ 1L ‘ (84)(---) (81)(---) (83)(---) (85)(---) 976 is 1mp0551b1e. I This completes the analysis of the situation in which IS*I = 16. All other cases are isomorphic to one of those analyzed above. Subject to completing the consideration of case in which |s*| = 12 in chapter 4 this theorem shows that there is no desmic triad of 24 points spanning an ordered projective k-space, k > 2. CHAPTER 4 Desmic triads in general spaces 4.1 Introduction In this chapter we continue the study of desmic triads but now in general spaces and not necessarily spanning three dimensions. If Z$_l, 22-1, Z§_l a finite projective space of dimension 11 then quite clearly are three hyperplanes in the three sets Zn-l - (Zn-l U Zn-l), Zn-l - (Zn-1 U Zn-l). 2 3 2 l 3 $3-1 - (XE-1 U 23-1) constitute a desmic triad. How much more general such triads can be is still not clear and a large segment of this chapter is concerned with this question. Many of the techniques of Chapter 3 are applicable to this more general setting with obvious modifications. Thus theorem 3.8 which asserts that a necessary condition for the existence of a desmic triad spanning a 3-space is the existence of a completely self-perspective set in the 2-spaces of that 3-space clearly extends to n-spaces. In Chapter 3 we found it occasionally convenient to express selffperspectivities as permutations in "factored" form. In this chapter much more extensive use is made of this technique in completely characterizing desmic triads of order 7 or less in all dimensions. The use of fixed sets in analyzing the equivalence of self-perspective arrangements 83 84 seems to be very closely related to some work of L. D. Cummings DJJ concerning steiner triple systems. We do not yet completely understand these connections but propose to look into the matter further. We have already observed in Chapter 3 that if A, B, C are three sets of a desmic triad then sets A and B are perspectively related by [C] disjoint perspectivities. Thus for the rest of this chapter we shall concentrate on the language of multiply perspective arrangements (to be defined shortly). The concept of multiply perspective arrangements is the same as that of multiply perspective sets but the new terminology is more convenient for the general treatment in this chapter. 4.2 Multiplvgperspective arrangements Definition 4.1: Let {A,B,F} denotes two disjoint sets A and B in a projective space 2 and a set F of disjoint perspective mappings of A onto B. Let ci be the center of Yi 6 T and C = {c1,c2,--- If |F| = p, |A| = |B| = k we 'Cp-l}' say that {A,B,F} is a multiply perspective arrangement of class [k,p,d], d > 1 where A U B spans a subspace of Z of dimension d and A n B = B n C = A n C = ¢. Notation: [k,p,d] MPA. Definition 4.2: Suppose {A,r] denotes a set A in a projective space 2 and a set P of disjoint self perspective mappings of A onto A including the identity. If |F| = p |A| = k we say that [A,F} is a self perspective arrangement of class [k,p,d] where A spans a subspace of Z of dimension d 2 2, A n C = D and C = [c1,c --,cp_l] is the set of centers. 2" It should be clear that if Y be expressed as a factored permutation then the points of A corresponding to each cycle must be linear. Keeping this in mind the condition A n C = ¢ can be described more explicitly in terms of permutation cycles of P as follows: (1) If a point in one cycle is on the line of a second cycle of the same Yi E F then the lines associated with 85 86 the two cycles are identical. (2) If two cycles have a pair of points in common then their associated lines are identical. Notation: [k,p,d] SPA _Definition 4.3: Two multiply perspective arrangements {81,SZ,P} and {S1,S2,F} are said to be equivalent if there exist l-l mappings 01: S1 4 82. 02: S l as g:1‘-.f, with 2' g(Yi) = ;i’ Yi 6 P such that the following diagram commutes $1 .1 l S1 Definition 4.4: Y 1 ——")52 _ l 02 Y 1 _ 4%52 Two self perspective arrangements (5,?) and (5,?) are equivalent if there exist g: T 4 f with g(Yi) = Yiel‘. l-l maps 0: S 4 S and such that ;i = o ly.o. Since o, y., §. may be regarded as permutations, on |S| = l 1 symbols this means that F and f are conjugate sets of permutations in the symmetric group on k symbols. f = 0 P O' -l Definition 4.4: If in the self-perspective arrangement {S.F}. ISI = I?! then the arrangement is called completely self perspective. 87 Remark 4.2: If (8,?) is a completely self perspective arrangement then the set of permutations is simply transitive on S. F, in general, is not a group. In chapter 3 we showed that it was impossible to realize completely self perspective arrangements of class [k,k,2] k = S, 6, 7, 8 in ordered projective spaces. We now propose to conduct a similar analysis for k s 7 in general Desarguesian spaces. In Chapter 3 our technique involved making a few combinatorial observations and then showing that these were incompatible with the order requirements of the embedding space. NOW’We are going to make similar combinatorial observations and then complete the analysis by making use of the conditions contained in the following theorem. Theorem 4.3: Let S = {l,2,°°-,n} denote the points of a completely self perspective arrangement {S.F}. Let f = {the permuta- tions of S representing the self-perspectivities P]. Let C = [C1,c2,°°°,cn] denote the centers of the perspectivi- ties 1‘. Then {if} satisfies the following conditions: (1) f is simply transitive on S. (2) Elements of f are pairwise disjoint. (3) 1 E f where l is the identity permutation of S. 88 2320;; Condition (1) follows from lemma 3.6 of Chapter 3. Condition (2) follows from the definition of completely self- perspective arrangement. Condition (3) is a consequence of IF] = ISI and the fact that the elements of T are disjoint. Remark 4.4: From now on we shall use F to represent both the self- perspectivities and their representing permutations. An illustration: If (8,?) is a self perspective arrangement of class [4,4,d] in a projective space 2, and S = [1,2,3,4] then d = 2 and T = (I, (12)(34), (l4)(32, (l3)(42)]. It is clear that no cycle of any Y can be of length three since Y would then have a fixed element and Y and I would fail to be disjoint. Theorem 4.6: If (8,?) is a completely self perspective arrangement then no cycle C, of y E F has length exceeding ‘3 where ISI = n. Proof: 2. 2. on a line L. Let S n L = P. Since F is simply transi- Suppose C has length R > The points of C are tive there are at least k-l distinct elements Yl'Y2'°"'Yk-l of F taking a point t E P into points of P. 89 The center of each Yi is on 2. If yi(p) = q, p 6 P, q 6 2—2 then the center of Yi is on line pq. Hence the center of Yi is L n pq = p 6 8. But this contradicts the definition of self-perspective arrangements. Thus yi(P) = P and yi(S-P) = S-P. But 0 < lS-P| < {21. Hence the k >-2r1 mappings Yi cannot be disjoint. I The next theorem establishes the result stated at the beginning of the introduction to this chapter. Theorem 4.7: If 21, 22, 23 are three distinct hyperplanes in a finite projective space 2d, d 2 3, then 21 - (22 U 23), 22 — (23 U 21) 23 - (22 U 21) form a desmic triad. Proof: 2i — (Zj U 2k) is not empty Since each line in ii contains at least 3 points. Let p E 21 - (22 U 23) g e 22 - (23 U 21). pq n 23 = X and k ¢ 22 U 21. Q.E.D. Corollary 4.7.1: There exists multiply perspective arrangement {A,B,F} of order [A] = [3' = IPI = qd"l - qd_2 where q = pk, p any prime and k a positive integer. Theorem 4.8: If (Sl,SZ,F) is a multiply perspective arrangement of class [k,k,d] then k 2 2d-1. 9O ££22£= (Induction on d) For d = 2 the theorem is trivially true. Let c be a center of one of the perspectivities of P. There exists a set P of d-l points in 81 such that c U P spans a space 2d_l. Let 81 n Zd-l = 81*, 82 n 26.1 = 82*, if F* is the set of restrictions to S1 of the perspectivities of P which have centers in Zd-l then it is easily seen that (81*, 82*, F*) is a multiply perspective arrangement of class (L. z. d-l). Hence by the induction hypothesis L 2 2d-2. Since the yi are disjoint and .. __ ° * . 'Sll — '82] — |P| if x 6 81 Fig. 4.1 and y 6 82-82* then there is a vi 6 P such that yi(x) = y. * We claim that y.(S *) c S -S *. For suppose z E S and 1 l 2 2 1 yi(z) = w E 82* Then the center of Yi is one lines xy and zw. Thus the only possibility for the center is the point x 6 81. But this is also impossible. Hence = 81* = 26-2 :: |sll = Zd-l which was to be proved. I One of the difficulties in analyzing self-perspective arrangements (8,?) for ISI = 8 comes in trying to list all conjugate classes of admissible sets of permutations. For example after some effort we discovered the following three sets of admissible perspectivities for n = 8. 91 I I (12) (34) (56) (78) (12) (34) (S6) (78) (13) (24) (57) (68) (13) (25) (47) (78) 1 (14) (23) (58) (67) (14) (26) (38) (57) 1‘ (15) (26) (37) (48) 1“2 (15) (28) (37) (46) (16) (25) (38) (47) (16) (27) (35) (48) (17) (28) (35) (46) (17) (24) (36) (58) (18)(27)(36)(45) (18)(23)(45)(67) I (12) (34) (56) (78) (13) (24) (57) (68) (14) (25) (38) (67) 1‘ (15) (26) (37) (48) (16) (23) (47) (58) (17) (28) (35) (46) (18) (27) (36) (45). To decide whether these are conjugate sets we could seek a permutation exhibiting the conjugacy. Another common scheme is to study the action of the sets F1 on the set 8 = (l,2,---,n). If for example T1 is simply transitive on the elements of 8 while T2 is not then F1 and F2 are not conjugate sets. However, in the case at hand all our admissible sets are simply transitive by construction. Another technique is to study fixed sets of F or sub- sets of T. Note for example the set {1,2,3,4} is fixed under the first three mappings of F1 while F2 contains no three mappings which fix a four element subset of 8. Thus F1 and F2 are not conjugate sets. Similarly, the first two maps in F3 fix the set {1,2,3,4} while no two trans- formations of F2 fix any four element set. In this way 92 we see that Fl. P2, F3 are in different conjugate classes. Thus if completely self perspective arrangements (Si'ri) exists i = 1,2,3 they will be pairwise inequivalent. Our investigations of completely self perspective arrangements of class (8,8,2) are still incomplete and we proceed now to an analysis of those of class (k,k,2) for k S 7. 3. Multiply perspective arrangements of class [k,k,d], k s 7 Our general procedure is as follows: If (S,r) is an SPA of class (k,p,2) then the sum of the lengths of the cycles in a particular y E T is k. Thus we first seek possible integral solutions of Z ki = k, ki 2 2. For small k this gives relatively few possible cycle lengths. The construction of the arrangement proceeds initially as in Chapters 2 and 3, making use of a grid diagram and then making extensive use of the conditions contained in theorem 4.3. If such a combinatorial representation exists then we try to find the nature of the embedding space by coordina- zation. Finally we investigate the possibility of "lifting" the various completely self perspective arrangements of class [k,k,d] to multiply perspective arrangements of class [k,k,d+l]. That is to say, we attempt to create desmk: triads of order k. Theorem 4.9: In any projective space 2, (8,?) is an SPA of class [4,4,d] iff d = 2,S=(L2£14} is a quadruple of points no three of which are linear and F = (I, (12)(34), (l3)(24),(l4)(23)). 93 94 £3353? Suppose (8,?) is an SPA of class [4,4,d] in a project- ive space 2. Since each of the 4 points has to be joined to the remaining 3, we have that d = 2. No three of the points are collinear since d > 1 by definition, and using disjointness and transitivity we have P = (I,(12(34), (13)(24, (l4)(23)]. The sufficiency is clear. In an affine 3~space over a field let s1 = (A(0,0,0), B = (1,1,0), 0(011), 0(1,0,1)) 52 — (Al(l,0,0), B1 = (0,1,0), 01(1,1,1), 01(0,0,1)}. Y1(A) = A1, Y1(B) = Bl. Y1(C) = c1. v1(D) = 01. Y2(A) = Bl. Y2(B) = A1. (2(0) = 01, Y2(D) = c1 F: Y3(A) = C1 Y3(B) = D1 Y3(C) = A1 V3(D) = B1 1 1 1 _ A1 Y4(A) =D Y4(B) =c (4(C) =B Y4(D) {s r} is an MPA of type [4,4,3]. A A 1'82' It should be clear that a configuration isomorphic to (81,82,T} is realizable in any affine space. Now let (81,82,f} be an MPA of type (4,4,3) in a Desarguesian affine space. _ - — - - — _ —1 —1 —1 —1 _ - - - — s1 — [A,B,C,D} 52 — (A ,B .C ,D }, F — (Yl.y2.v3.y4}- 95 l 1 We may assume that §1(A,B,C,D) = (Al,B ,C ,Dl). As a consequence of theorem 3.8 and 4.10 we can conclude that no three of the lines AAl, 881, 661, 551 are coplanar. Thus if y2(A) = 81 then (2(3) = A1, y2(D) = 61, y2(E) and if y3(A) = 51 then y3(fi)= A1, y3(E) = Bl, y3(§) = -1 - -1 . - - -1 - FlnallYt Y4 (A) I Y4 (C) = A I Y4 (D) = B Now there is an affine transformation T such that T(A) = A Tail) = A1, T031) = D1, T(Bl) = Bl, Tail) = A Thus T(81,82,f] = (81,82,F). This means that an MPA in Desarguesian affine space is affinely unique. Theorem 4.10: Any two multiply perspective arrangements of class (4,4,3) are combinatorially isomorphic. Jiheorem 4.11: Any two multiply perspective arrangements in a Desarguesian projective (affine) space are projectively (affinely) equivalent. -1 l 96 l // 2 Fig. 4.2: A geometric representation of the multiply perspective arrangement of class [4,4,3]. 97 In the case of (81.82,?) of class (k,k,2] it has been conjectured by M. Edelstein that in the real project- ive plane the figure must be rigid in the sense that the set of centers must lie on a line. We now show for k = 4 that this is not so by the following theorem. Theorem 4.12: An MPA of type [4,4,2] exists and is combinatorially unique. 1 . Case 1: ‘5; R,ald TR b'd 1,5 01 ’b,f s1 E O, .1 i P l ' 0,0 1,0 S]. = {PI Q! RI 8} 82 = {Pl.Ql.Rl.S}- l l l l . . Y1: P4P , QaQ ,RqR ,848 . Center: ideal pt on x ax1s l l l l . . y2: p45 , Q—p ,RaQ ,qu Center: ideal pt on y ax1s . l l l l . a d Y3° PaR , QaS ,RaP ,SaQ Center. a-b+l a-b+l f = b+d-1 . 1 1 1 1 . b f. a Y4- P40 , 04R ,RaS ,SaP Center. f-d+l f-d+l Select a, b, d, f such that af = b+d-1, a-b+l + O f-d+l + 0, d + f, a + b a, b, f, d + 0, 1 For example a = 2, b = 4, d = 7, f = 5 produces a multiply perspective arrangement of type (4,4,2) in the real affine plane such that no three of the four centers of perspectivity are on a line. For other properly chosen values of a,b,d A .‘ l:- _ . ... I ‘II. ... ll l I I III: I I l I I. I I . . and f the four centers will be linear. 98 two combinatorially equivalent MPA's of type [4,4,2] in real affine space which are not affinely equivalent. Case 2: Thus there exists To complete the picture of the multiply perspective arrangements of type [4,4,3] we examine the remaining combinatorially distinct arrangement and show that is not realizable in any Pappian plane. | ) l 1 l acéLxfic 01).1 . aflfifi C 1 It .’ I 5. L f Ajk %1' b C3' Y “ Ex ' 7 If ab-cd+d + 0 ad ab ab-cd+d )1 1 l a a l s ad-ab = d(a-c) ad ab ab-cd+d l O 1 = a b l 1 Y1- 1 Y2. A4B Y3: A-oCl B-oD1 C-oAl DaB l l BAAI c-.D1 l A4A B4B C4C D4D C - 1 Dec C = l Y4: A-oDl B-oCl C-oBl Dqu _ ab-cd+d Y _ ab-cd+d ad-ab ab d-cdi ad-ab = O 0 1 (D = O O a-c c l-c a-c = ab=cd. = x = a y = %§ ad ab d ad a d 0 l O l = O O l a b l a l l 99 == 0 O 1 =0 =0 d(a-l) =a(a-l) =a= d. But this is impossible Suppose then that ab-cd+d = 0. Then §-=‘§E% = :%i a b = c-l and d = a-l s ab - d(c-l) = 0 = ab - db = 0 = a = d and this is impossible. In this case if an SPA of class [5,5,2] exists the possible cycle length combination is (3,2). But this is impossible by Theorem 4.6. Thus there is no (8,?) of class [5,5,d] for all d, hence there is no MPA (81.82,?) of class [5,5,d] d 2 3. However MPA's of class [5,5,2] do exist. For example, the following figure exhibits an (81,82,F] of class [5,5,2] Fig. 4.3 J 2 _fl __ ._);3 2 /'/l The complete analysis of this class is not attempted in this thesis. Theorem 4.17: There are two distinct [8,F]'s as SPA's of class [6,6,2]. Proof: If (8,?) is an SPA of class [6,6,2] then the numerical conditions gives the possible cycle lengths combinations (3,3) and (2,2,2). 101 Case 1: Suppose (8,?) contains permutations having cycles of lengths (3,3). Its grid diagram is shown below Fig. 4.4 ._L_ l 5 3 V/ A representation for T follow easily from the grid diagram as I‘ = (I, (135) (246),(153) (264), (14) (25) (36), (16) (45) (23) , (12') (34) (56)). and it is clear that for this diagram this (8,?) of class [6,6,2] is unique up to an equivalence. Case 2: Suppose (8,?) contains permutations having all cycles of lengths (2,2,2). The grid diagram in this case is D 14 6 .11 given by Fig. 4.5 102 We thus derive a representation for ? as 1‘ = (I. (12)(34)(56). (13)(25) (46).(14) (26) (35).(15) (24) (36). (16)(23)(45)} which again is unique up to equivalence. Furthermore, the two SPA's of cases 1 and 2 are inequivalent by virtue of their cycle structure. I Remark 4.18: From the grid diagram of case 1, it is clear that we cannot have a ? in which all cycles of the permutations have lengths (3,3). Theorem 4.19: A Desarguesian affine plane 2 contains a unique self perspective arrangement (8,?) of class [6,6,2] with 3 points of 8 linear iff the associated coordinate ring contains V/:3. Two such arrangements in 2 are affinely equivalent. Proof: 4 (01) 2 L11) 6 (al) Fig. 4.6 1 (OO) 5 (10) 3 (a0) Line 16: ay - x = O ay = l-y Line 45: y + x - l = O y = (a+l)-l l n o x n DJ ‘8 + H I Line 23: (l-a)y - (x-a) ] 103 Using lines .12,'34, 56 ‘we Observe that the above equations are consistent. Thus we can choose a = fil4§~L:;- and this is possible in any division ring containing ~/:3. That this condition is also sufficient is easily checked. I Theorem 4 .29 : A Desarguesian affine plane 2 contains a self perspective arrangement (S,f) of class [6,6,2] with no three points of 8 linear iff the associated coordinate ring has characteristic 2. Two such arrangements in 2 are affinely equivalent. Proof: 4 (1b) 6 (ab) 2) Fig. 4.7 [(01) (11) 5 (a1) 1[(00) 3 (10) (a0) From lines 18; 24, 36' we have ab2 + 1 - a - b = 0] (1) From lines _6, 23, 45' we have abL-b+l - a+];] = O (2) From lines 12; 26, 35' we have a2b + l-b-a = 0 (3) (1) and (3) give a2b = ab2 e a = b Since ab + O, (2) gives -a2[2-2a] = O = a = 1 or 2 = 0 But a # 1, hence 2 = 0. That this condition is also suffi- cient is easily checked. I 104 Lifting: Theorem 4.21: An MPA (81.82,?) of class [6,6,3] exists in any projective 3-space whose coordinate ring is of characteristic 3. Proof: 4110111 51111) 6(211) z I l 1 l 171001) I 2’(101)’ 3J 3, from theorem 4.9. 115 § 4. Some general results In this last short section we give some general results, with particular reference to k = 8. Theorem 4.29: In a finite projective 4-space X4 , there exists. an MPA of class [p3,p3,4] where p is a prime and is also the order of the underlying field. Proof: This follows from theorem 4.9 since p3 2 8 V p I Corollary 4.30: There exists an MPA of class [8,8,4]. Corollagy 4.31: There exists an SPA of class [8,8,3]. This follows by projecting the MPA of class [8,8,4] from cum: of its centers onto 3-subspace of its embedding space. Remark 4.3;: The possible solutions to our numerical relations in case of k = 8 are (2,2,2,2), (4,2,2), (3,2,2) and (4,4). We have already made some remarks about the solution (2,2,2,2). For solution (4,2,2) we have Theorem 4.33: If an SPA (8,?) of class ]8,8,2] containing permu- tations with cycle lengths (4,2,2) exists then the coordi- nating field must have characteristic 2. 116 Proof: Consider the following representatives for the y's. Y56 = (1234)(56)(78) v57‘= (1432)(57)(68) Y58 = (13)(24)(58)(67) Let their corresponding centers be denoted by q56,q57, and q58° These centers are the diagonal points of the complete quadrangle with vertices 5,6,7,8 and furthermore they lie on the line 1234. Hence the result. I Theorem 4.34: There is no SPA of class [8,8,2] containing permuta- tions with cycle lengths (3,3,2). Proof: This follows from theorem 4.6 since when we compare one cycle of length3 against the rest of symbols we have an unequal splitting which is impossible by Theorem 4.6 I 117 CONCLUDING REMARKS Abstract Sylvester-Gallai configurations are very weak and inhomeogeneous structures. The class includes finite projective and affine spaces among others. Hence any effective classification or characterization of S.G. configura- tion is too much to expect. However, their complete absence in ordered projective spaces and their seeming scarcity in complex projective space suggests further efforts to under— stand their relation to the concept of order and to Whatever it is that inhibits their occurence in complex spaces. Certainly the curious behavior of generalized Sylvester- Gallai configurations in ordered projective spaces warrants more investigation. 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