Subconstituent algebras of Latin squares

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1 University of South Florida Scholar Commons Graduate Theses and Dissertations Graduate School 2008 Subconstituent algebras of Latin squares Ibtisam Daqqa University of South Florida Follow this and additional works at: Part of the American Studies Commons Scholar Commons Citation Daqqa, Ibtisam, "Subconstituent algebras of Latin squares" (2008). Graduate Theses and Dissertations. This Dissertation is brought to you for free and open access by the Graduate School at Scholar Commons. It has been accepted for inclusion in Graduate Theses and Dissertations by an authorized administrator of Scholar Commons. For more information, please contact

2 Subconstituent Algebras of Latin Squares by Ibtisam Daqqa A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy Department of Mathematics College of Arts and Sciences University of South Florida Major Professor: Brian Curtin, Ph.D. Masahiko Saito, Ph.D. Xiang-dong Hou, Ph.D. Brendan T. Nagle, Ph.D. Date of Approval: November 29, 2007 Keywords: Terwilliger algebra, Bose-Mesner algebra, Association scheme, Strongly regular graph, Fusions c Copyright 2008, Ibtisam Daqqa

3 Dedication To my loving parents, my supportive husband, and my three lovely daughters, Leen, Dana, and Yasmeen.

4 Acknowledgements I would like to gratefully and sincerely thank Dr. Brian Curtin for his guidance, understanding, patience, and kindness. His mentorship was paramount in providing a well-rounded experience consistent with my long-term career goals. For everything you ve done for me, Dr. Brian, I thank you. I would also like to thank the Department of Mathematics at University of South Florida, especially those members of my doctoral committee for their input, valuable discussions and accessibility. Many thanks go to the former and current administrative staff in the department, Beverly, Nancy, and Mary Ann for their kindness and assistance. I would like to thank many friends who have helped me stay sane through these difficult years, especially Dr. Kheira, and Dr. Ibrahimou and his wife. I greatly value their friendship and I deeply appreciate their help. I m also grateful to Dr. Hassan Alnajjar and his wife who helped me adjust to a new country. Finally, and most importantly, I would like to thank my husband Nour Aldeen. His support, encouragement, quiet patience and unwavering love were undeniably the bedrock upon which the past ten years of my life have been built. I thank my parents, for their faith in me and allowing me to be as ambitious as I wanted. I also thank my brothers and my sisters for their endless help and support. Above all I must thank Allah who encouraged and comforted me in the whole process of this dissertation.

5 Table of Contents List of Tables List of Figures Abstract iii iv v 1 Introduction 1 2 Algebraic Preliminaries The Bose-Mesner Algebra The Dual Bose-Mesner Algebra The Subconstituent Algebra and its Modules The Subconstituent Algebra of a Latin Square Latin Squares and Bose-Mesner Algebras Some Permutations Cycle Modules Decomposition into Irreducible Modules Some Intermediate Modules Collecting Cycle Modules The Fourth Subconstituent Other Results Cayley Tables of Finite Groups Small Latin Squares i

6 4 Strongly Regular Graphs from a Latin Square Strongly Regular Graphs Strongly Regular Graphs from a Latin Square Fusions G G G Isomorphisms Isomorphisms of Bose-Mesner Algebra Isomorphisms of Subconstituent Algebras Equivalences of Latin Squares Isomorphisms and Latin Squares References 72 Appendices 80 Appendix A: Permitted Roots of Unity Appendix B: Computer Code About the Author End Page ii

7 List of Tables 5.1 The numbers of Latin squares of various sizes Equivalence classes of Latin squares Number of possible structural isomorphism classes Number of possible Bose-Mesner isomorphism classes Number of possible abstract isomorphism classes Irreducible T -modules for n = Irreducible T -modules for n = Irreducible T -modules for n = Irreducible T -modules for n = Irreducible T -modules for n = 9, pt Irreducible T -modules for n = 9, pt Irreducible T -modules for n = 10, pt Irreducible T -modules for n = 10, pt Irreducible T -modules for n = 10, pt iii

8 List of Figures 1.1 Two Cayley tables Two point sequences with respect to (1, 1, 1) The interleaving of cycles (Lemma 3.2.5, Definition 3.2.7) The action on u 1,i, u 2,i, u 3,i W (C 1, C 2, C 3 ) The action on v 1,i W (C 1, C 2, C 3 ) The action on v 2,i W (C 1, C 2, C 3 ) The action on v 3,i W (C 1, C 2, C 3 ) The action on u ɛ i W ɛ (C 1, C 2, C 3 ) The action on v1 ɛ W ɛ (C 1, C 2, C 3 ) The action on v2 ɛ W ɛ (C 1, C 2, C 3 ) The action on v3 ɛ W ɛ (C 1, C 2, C 3 ) iv

9 Subconstituent algebra of Latin Squares Ibtisam Daqqa Abstract Let n be a positive integer. A Latin square of order n is an n n array L such that each element of some n-set occurs in each row and in each column of L exactly once. It is wellknown that one may construct a 4-class association scheme on the positions of a Latin square, where the relations are the identity, being in the same row, being in the same column, having the same entry, and everything else. We describe the subconstituent (Terwilliger) algebras of such an association scheme. One also may construct several strongly regular graphs on the positions of a Latin square, where adjacency corresponds to any subset of the nonidentity relations described above. We describe the local spectrum and subconstituent algebras of such strongly regular graphs. Finally, we study various notions of isomorphism for subconstituent algebras using Latin squares as examples. v

10 1 Introduction Let n be a positive integer. A Latin square of order n is an n n array L such that each element of some n-set occurs in each row and in each column of L exactly once. The simplest examples are the Cayley table of a group, as in Figure 1.1. Sudoku puzzles, when completed, form a Latin square Z 5 Z 2 Z 2 Figure 1.1: Two Cayley tables Latin squares were introduced by Euler in 1779 in the context of a puzzle in recreational mathematics. Since that time Latin squares have found applications in a variety of mathematical areas. Such fields include group theory, graph theory, finite geometries, coding theory, design theory, cryptography, and statistics. These connections are developed in the general references [42, 43, 69]. Let L denote a Latin square of order n 3, and encode L with the set X = {(i, j, L(i, j)) 1 i, j n}. Define five relations on X: For all x = (i, j, L(i, j)), x = (i, j, L(i, j )) X, R 0 (identity): xr 0 x if x = x, R 1 (same row): xr 1 x if i = i and x x, 1

11 R 2 (same column): xr 2 x if j = j and x x, R 3 (same entry): xr 3 x if L(i, j) = L(i, j ) and x x, R 4 (everything else): xr 4 x if i i, j j, and L(i, j) L(i, j ). It is well-known that (X, {R i } 4 i=0) is a symmetric association scheme [5, 41], which we refer to as the association scheme of L. Hence the characteristic matrices of the five relations comprise the basis of Hadamard idempotents of a Bose-Mesner algebra M, which we refer to as the Bose-Mesner algebra of L (see [5]). Bose-Mesner algebras first arose in statistical designs [15, 16], in centralizer algebras of permutation groups [90], and in connection with distance-transitive graphs [12]. A period of growth in the subject occurred in the 1980 s after Delsarte demonstrated applications to codes and designs [41] and as the classification of finite simple groups motivated and aided the study of distance-transitive graphs. Details can be found in [5, 10, 13, 17, 51]. Bose- Mesner algebras have been generalized in several directions, such as coherent algebras [53, 54], table algebras [1, 2, 3, 4], group-like association schemes [91]. More recently, connections have been developed to link invariants [22, 61, 62], quantum groups [38, 60], fusion algebras [8], maximal abelian subalgebras [52, 75], commuting squares [6], and subfactors [31, 63, 64]. In this work we study the subconsitutent (or Terwilliger) algebras of the association scheme of a Latin square. The subconstituent algebra refines the Bose-Mesner algebra by encoding additional combinatorial information concerning the relation between each point and the fixed base point. It is known that the (algebraic) isomorphism class of the Bose-Mesner algebra of a Latin square depends only upon its order and no other property of the Latin square. One of our motivations is to better distinguish Latin squares using subconstituent algebras. Subconstitutent algebras have been studied in several papers. Themes of these papers include complete description of the subconstituent algebra of some class of association scheme [7, 11, 19, 20, 21, 30, 44, 49, 83, 88], partial description of the subconstituent algebra of some class of association scheme [23, 24, 29, 36, 45, 55, 58, 59, 84, 86, 87], descriptions of the association schemes whose subconstituent algebra satisfies some condition [25, 39], correspondences between certain combinatorial (often local) and algebraic 2

12 conditions [27, 32, 39, 50, 72, 73], algebraic connections [33, 34, 38, 89, 84], and generalizations [46, 47, 60, 81]. Fix a base point p = (r p, c p, e p ) X. Let T denote the subconstituent algebra of the association scheme of L with respect to p. We describe the action on the irreducible T -modules in terms of a substructure of the Latin square which we now describe. By the definition of a Latin square any two coordinates of an element of X uniquely determine the third: We refer to this as the Latin square property. Form a sequence of points as follows. Pick x 1 X such that x 1 R 1 p write x 1 = (r p, c 1, e 1 ). Once x 1 is chosen, all subsequent points are uniquely determined by the Latin square property. Given x i = (r p, c i, e i ), let y i X be the unique point such that y i R 2 p and y i R 3 x i write y i = (r i, c p, e i ). Let z i be the unique z i X such that z i R 3 p and z i R 1 y i write z i = (r i, c i+1, e p ). Finally, let x i+1 X be the unique point such that x i+1 R 1 p and x i+1 R 2 z i write x i+1 = (r p, c i+1, e i+1 ). Repeat this process until x k+1 = x 1. View x 1, x 2,...,x k as a cycle in a permutation of the n 1 points in relation R 1 with p, where the other cycles constructed similarly. The permutation constructed depends upon the base point p. Figure 1.2 interprets two sequences of points in a Latin square with respect to (1, 1, 1): x 1 = (1, 2, 2), y 1 = (2, 1, 2), z 1 = (2, 3, 1), x 2 = (1, 3, 3), y 2 = (3, 1, 3), z 2 = (3, 2, 1) and x 1 = (1, 4, 4), y 1 = (4, 1, 4), z 1 = (4, 4, 1). This gives rise to a permutation with one 1-cycle and one 2-cycle, namely (x 1 x 2 )(x 1) Figure 1.2: Two point sequences with respect to (1, 1, 1)... In Chapter 3, we describe the irreducible T -modules. We summarize the results now. There is always a unique 5-dimensional irreducible T -module, and there are n 2 6n + 7 many mutually isomorphic 1-dimensional T -modules in the standard module. The 3

13 remaining T -modules are related to the cycles of the permutation constructed above. If there is just one cycle (with k = n 1), then there is a 6-dimensional irreducible T -module associated with each k th root of unity except 1 itself. If there is more than one cycle, then for each cycle of length k there is a 6-dimensional irreducible T -module associated with each k th root of unity. However, in this case one of the irreducible T - modules associated with 1 must be dropped to obtain linear independence. The T -action on each of these (n 2)-many 6-dimensional irreducible T -modules is entirely determined by the associated root of unity: Two such modules are isomorphic if and only if they are associated with the same root of unity. One may define a strongly regular graph with vertex set X for each subset C of coordinates 1, 2, 3 by declaring distinct x = (i, j, L(i, j)) and x = (i, j, L(i, j )) to be adjacent if for some coordinate c C, x(c) = x (c). In Chapter 4 we describe the local spectrum and subconstituent algebra of these strongly regular graphs using the cycles described above. The key idea is that the irreducible T -modules remain modules (no longer irreducible) for the subconstituent algebras of these strongly regular graphs. Thus we show how to decompose these already small irreducible T -modules into irreducible modules for the strongly regular graphs subconstituent algebras. Then the local spectrum is determined by acting on these modules by an appropriate matrix. In Chapter 5, we discuss isomorphisms of subconstituent algebras. The subconstituent algebra of a Latin square with respect to some base point is isomorphic to M 5 M l 6 M 1, where l is the number of mutually nonisomorphic irreducible T -modules of dimension 6. Many distinct cycle structures may give rise to the same value of l. However, the action of the subconstituent algebra on each irreducible T -module is uniquely determined by the cycle structure of the permutation constructed above. In Appendix A we gives tables relating cycle structure in small examples to possible irreducible modules, and in Appendix B we give Mathematica code which produces the irreducible modules of the subconstituent algebra of a Latin square and the related strongly regular graphs. 4

14 2 Algebraic Preliminaries 2.1 The Bose-Mesner Algebra In this section we recall some background material. We first recall Bose-Mesner algebras and some of their basic properties. General references for the subject include [10, 17, 51]. Let X denote a finite, nonempty set, and let M X denote the complex algebra of matrices with complex entries whose rows and columns are indexed by X. For A M X and for x, y X, let A(x, y) denote the (x, y)-entry of A. For A, B M X, let A B denote the Hadamard product of A and B: (A B)(x, y) = A(x, y)b(x, y). The ordinary matrix product of A and B will be denoted by juxtaposition: AB. For A M X, let t A denote the transpose of A. A Bose-Mesner algebra on X is a commutative subalgebra of M X which is closed under Hadamard product, which is closed under transposition, and which contains the identity matrix I and the all-ones matrix J. Let M denote a (d+1)-dimensional Bose-Mesner algebra on X. The basis of Hadamard idempotents of M is the unique basis {A i } d i=0 such that A 0 = I, (2.1.1) A i A j = δ ij A i (0 i, j d), (2.1.2) d A i = J, (2.1.3) i=0 where δ ij denotes the Kronecker symbol. Let A 0, A 1,..., A d be an ordering of the Hadamard idempotents of M. Then relative to this ordering, the intersection numbers 5

15 p h ij of M are defined by A i A j = d h=0 ph ija h (0 i, j d). (2.1.4) The basis of primitive idempotents of M is the unique basis {E i } d i=0 such that E 0 = n 1 J, (2.1.5) E i E j = δ ij E i, (2.1.6) d E i = I. (2.1.7) i=0 Let E 0, E 1,..., E d be an ordering of the primitive idempotents of M. Then relative to this ordering, the Krein parameters qij h of M are defined by E i E j = n 1 d h=0 qh ije h (0 i, j d). (2.1.8) Relative to the orderings of the Hadamard and primitive idempotents, the eigenvalues P (j, i) and the dual eigenvalues Q(j, i) of M are defined respectively by A i = d P (j, i)e j (0 i d), (2.1.9) j=0 E i = n 1 d Q(j, i)a j (0 i d). (2.1.10) j=0 The (d + 1) (d + 1) matrices P and Q with (j, i)-entries P (j, i) and Q(j, i) are called the eigenmatrix and dual eigenmatrix of M, respectively. It is known that the each of these sets of parameters (the intersection numbers, the Krein parameters, the eigenmatrix, and the dual eigenmatrix) determines the others. See, for example, [10] for precise details. 2.2 The Dual Bose-Mesner Algebra We now recall from [84] the dual Bose-Mesner algebra of a Bose-Mesner algebra M on X. Fix p X. For each A M, let ρ(a) M X denote the diagonal matrix with (x, x)-entry ρ(a)(x, x) = A(p, x) (x X). Let M = ρ(m). We refer to M as the 6

16 dual Bose-Mesner algebra of M with respect to p. Observe that ρ : M M is a linear bijection and ρ(a B) = ρ(a)ρ(b). Set Ei = ρ(a i ). Then {Ei } d i=0 is a basis of M. We refer to {Ei } d i=0 as the basis of dual idempotents. Applying ρ to (2.1.2) and (2.1.3) gives Ei Ej = δ ij Ei (0 i, j d), (2.2.11) d Ei = I. (2.2.12) i=0 Set A i = ρ(ne i ) (0 i d). Then {A i } d i=0 is a basis of M. We refer to {A i } d i=0 as the basis of dual Hadamard idempotents of M. Applying ρ to (2.1.5), (2.1.7), and (2.1.8) gives A 0 = I, (2.2.13) d A i = ne0, (2.2.14) i=0 A i A j = d h=0 qh ija h, (0 i, j d). (2.2.15) Applying ρ to (2.1.9) and (2.1.10) gives A i = d Q(j, j=0 i)e j (0 i d), (2.2.16) Ei = n 1 d P (j, j=0 i)a j (0 i d). (2.2.17) 2.3 The Subconstituent Algebra and its Modules We recall from [84] some facts concerning the subconstituent algebra of a Bose-Mesner algebra M on X. Fix p X. The subconstituent (or Terwilliger) algebra of M with respect to p is the subalgebra of M X generated by M M. By (2.1.1), (2.1.4), (2.2.11), and (2.2.12), T is also generated by {Ei A j Ek 0 i, j, k d}. Lemma [84] For all h, i, j (0 h, i, j d), E i A h E j = 0 if and only if p h ij = 0. (2.3.18) 7

17 Theorem [84] Subconstituent algebras are semisimple. We may appeal to Wedderburn theory [40] to describe subconstituent algebras by their irreducible modules. Let V = C X denote the column vector space with entries indexed by X. Endow V with the Hermitian inner product defined by u, v = t u v. Observe that M X acts on V by left-multiplication. We refer to V as the standard module for T. By a T -module we mean a linear subspace U of V which is closed under the action of T : Au U for all A T and for all u U. Let Λ be an index set for the isomorphism classes of irreducible T -modules. Let V λ be the sum of all irreducible T -modules in the isomorphism class of irreducible T -modules indexed by λ Λ. Each V λ is an orthogonal direct sum of mutually isomorphic irreducible T -modules. While the direct summands of V λ are not necessarily unique, their number is. Write mult(λ) to denote this number, and for any irreducible T -module W contained in V λ, set mult(w ) = mult(λ). We refer to mult(w ) as the multiplicity of W. Note that V = λ Λ V λ (orthogonal direct sum). We note the following consequence of this discussion. Lemma Suppose W and W are non-isomorphic irreducible T -modules. Then W and W are orthogonal to one another. The primitive central idempotents of T are also indexed by Λ as they are in bijective correspondence with the V λ. For each subspace V λ there is a unique primitive central idempotent ϕ λ such that V λ = ϕ λ V, and conversely for each primitive central idempotent ϕ, ϕv = V λ for some λ Λ. Let W V λ be an irreducible T -module. Then the map taking L ϕ λ T to the endomorphism w Lw (w W ) is an isomorphism. Thus ϕ λ T = End C (W ). However, End C (W ) is isomorphic to the k k complex matrix algebra, where k is the dimension of W. Now T = λ Λ ϕ λ T (direct sum). Thus T is isomorphic to a direct sum of complex matrix algebras. For all elements x of X, define [x] V to be the characteristic vector of x, ie, the vector with a one in the row indexed by x and zeros everywhere else. Observe that the set {[x] x X} is the standard basis of V. If S is a multi-set of elements of X, define 8

18 [S ] = x S [x], where each summand occurs once for each occurrence in S. For all x X, define Γ i (x) = {y X xr i y}. Lemma [84] For all x X, E i [x] = A i [x] = [Γ i (x)]. [x] if pr i x, 0 otherwise, In particular, [Γ Ei A j Ek i (p) Γ j (x)] if xr k p, [x] = 0 otherwise. Proof. Now since E i From the definition of A i, 0 i d, we have A i [x] = yr i x [y ] = [Γ i (x)]. is diagonal where E i (y, y) = 1 if yr i p and zero any where else, we have Ei [x] = δ hi [x], where h such that pr h x. For Ei A j Ek, we have E i A j E k [x] = = Ei A j [x] if xr k p, 0 otherwise, [Γ j (x) Γ i (p)] if xr k p, 0 otherwise. The support of a vector v V is the set supp(v) = {x X v(x) 0}. Lemma Let U V be a subset of nonzero vectors. Suppose that for all subsets S U and all u U\S, the symmetric difference of s S supp(s) and supp(u) is nonempty. Then U is linearly independent. Proof. Suppose that k i=1 α iu i = 0, where u i U. Let S = {u 2, u 3,..., u k } and u = u 1. Then by assumption the symmetric difference of s S supp s and supp(u) is nonempty. In particular, the sum is nonzero unless α 1 = 0. Proceeding by induction, we find that α i = 0 for all i, so U is linearly independent. 9

19 We recall some facts about a special irreducible T -module. Lemma [84] For all i, j, k (0 i, j, k d) E j A k E i [Γ i (p)] = p k ij [Γ j (p)]. Lemma [84] For all i, j, k (0 i, j, k d) and for all x X [Γ Ei JEk i (p)] if xr k p, [x] = 0 otherwise. Proof. By Lemma 2.3.4, E j A k E i [Γ i (p)] = xr i p E j A k E i [x] = xr i p [Γ k (x) Γ j (p)] = xr i p yr k x yr j p p k ij [y ] = p k ij [Γ j (p)]. Theorem [84] There is an irreducible T -module with basis {[Γ i (p)] 0 i d}. We refer to as the primary T -module and denote by P. Proof. Observe that {[Γ i (p)] 0 i d} spans a T -module by Lemma This lemma also implies that this module is irreducible. These vectors are linearly independent by Lemma 2.3.5, so the result follows. 10

20 3 The Subconstituent Algebra of a Latin Square 3.1 Latin Squares and Bose-Mesner Algebras Let L denote a Latin square of order n 3. Encode L with the set X = {(i, j, L(i, j)) 1 i, j n}. (This encoding is sometimes referred to as the orthogonal array representation of L). To describe the associated Bose-Mesner algebra we first define five relations on X as follows: For all x = (i, j, L(i, j)), x = (i, j, L(i, j )) X, R 0 (identity): xr 0 x if x = x, (3.1.1) R 1 (same row): xr 1 x if i = i and x x, (3.1.2) R 2 (same column): xr 2 x if j = j and x x, (3.1.3) R 3 (same entry): xr 3 x if L(i, j) = L(i, j ) and x x, (3.1.4) R 4 (everything else): xr 4 x if i i, j j, and L(i, j) L(i, j ). (3.1.5) Now (X, {R i } 4 i=0) is a commutative association scheme, and hence the characteristic matrices of the five relations comprise the basis of Hadamard idempotents of a Bose- Mesner algebra M, which we refer to as the Bose-Mesner algebra of L (see [5]). The equivalence of Bose-Mesner algebras and commutative association schemes is wellknown [10, 17]. Given a Latin square L and its association scheme (X, {R i } d i=0), define for i = 0, 1, 2, 3, 4, matrices A i M X by A i (x, y) = 1 if xr i y and 0 otherwise (x, y X). Then {A i } 4 i=0 is the basis of Hadamard idempotents of a Bose-Mesner algebra, which we refer to as the Bose-Mesner algebra of L. We now recall some facts concerning the Bose-Mesner algebra of a Latin square. For completeness we provide brief proofs of these facts. Their intersection numbers are well-known. 11

21 Theorem [5] Let L denote a Latin square of order n, and let M denote the Bose- Mesner algebra of M. (i) The intersection numbers of M are given by (p 0 ij) = (p 1 ij) = (p 2 ij) = (p 3 ij) = n n n n 2 3n n n n n 2 n 2 n 2 5n n n n 2 0 n 2 0 n 2 n 2 5n n n n n 2 n 2 0 n 2 5n + 6,,,, 12

22 (p 4 ij) = n n n 3 1 n 3 n 3 n 3 n 2 6n Proof. Let A r = A 1 +I, A c = A 2 +I, and A e = A 3 +I where A 1, A 2, A 3 are the respective adjacency matrices of the relations R 1, R 2, R 3 on X. Then A r is the matrix of the union of relations R 0 and R 1, etc. To compute p k ij we will compute A i A j for 0 i, j 3. To do so, let R r = R 0 R 1, R c = R 0 R 2, R e = R 0 R 3. Note that (A r A r )(x, y) = γ X A r (x, γ)a r (γ, y) = {γ : γr r x, γr r y} = na r (x, y). i.e A r A r = na r. Similarly A 2 c = na c and A 2 e = na e. Also note that (A r A c )(x, y) = {γ : γr r x, γr c y} = 1. Thus A r A c = J. Similarly A c A r, A r A e, A e A r, A c A e, and A e A c are equal to J. Now using the above computations we compute p h ij for i j, 0 i, j 3 and 0 h 4: A 1 A 1 = (A r I)(A r I) = A 2 r 2A r + I = (n 2)(A r I) + (n 1)I = (n 2)A 1 + (n 1)A 0. Thus p 0 11 = n 1, p 1 11 = n 2, and p 2 11 = p 3 11 = p 4 11 = 0. Similarly we compute all other intersection numbers p h ij, for 0 i, j 3, i j, and 0 h 4. Using the symmetry we deduce p h ij, for 0 i, j 3, i < j. Now for the p h ij with at least one of i, j equal to 4, note that for 0 h d, d i=0 ph ij = k j where k j is the sum of each row in A j (the valency). In the Latin square case k 0 = 1,k 1 = k 2 = k 3 = n 1, and k 4 = n 2 3n

23 3.2 Some Permutations We now begin our study of the subconstituent algebra of the Bose-Mesner algebra of a Latin square. In this section we show that certain elements of the subcontsituent algebra induce permutations on Γ 1 (p), on Γ 2 (p), and on Γ 3 (p). Notation Let L denote a Latin square of order n 3 and with symbol set {1, 2,..., n}. Let X denote the set {(i, j, L(i, j)) 1 i, j n}. Let M denote the Bose-Mesner algebra of L. Fix p = (r p, c p, e p ) X, and let T denote the subconstitutent algebra of M with respect to p. Lemma With Notation 3.2.1, fix a permutation i, j, k of 1, 2, 3. For each x Γ i (p), the row of E i A j E k indexed by x has a unique entry equal to one and all other entries are equal to zero. For each y Γ k (p), the column of E i A j E k indexed by y has a unique entry equal to one and all other entries are equal to zero. All other entries of E i A j E k are zero. Proof. Immediate from the definitions of Ei, A j, and Ek, and the fact that pj ik = 1 by Theorem We note the action of E i A j E k Latin square case. on the standard basis of the standard module in the Lemma With Notation 3.2.1, fix a permutation i, j, k of 1, 2, 3. Let x X. Then E i A j E k [x] = δ x(k),p(k) [y ], where y(i) = p(i), y(j) = x(j), and y(k) is uniquely determined by the Latin square property. Proof. The sum in Lemma runs over all y such that y(i) = p(i) and y(j) = x(j) by the definitions of the relations. There is exactly one such y by the Latin square property. Lemma With Notation 3.2.1, for all permutations i, j, k of 1, 2, 3, the principal minor of Ei A j Ek A iej A k Ei indexed by Γ i (p) is a permutation matrix and every other entry of Ei A j Ek A iej A k Ei is zero. 14

24 Proof. In light of Lemma 5.3.2, it is enough to treat just E 1A 2 E 3A 1 E 2A 3 E 1 acting on an element of Γ 1 (p). By Lemma 3.2.3, for all (r p, c, e) Γ 1 (p) E 1A 2 E 3A 1 E 2A 3 E 1 [r p, c, e] = E 1A 2 E 3A 1 E 2 [r, c p, e] = E 1A 2 E 3 [r, c, e p ] = [r p, c, e ], where c and e are uniquely determined by the Latin square property. Note that (r p, c, e ) Γ 1 (p). Thus the principal minor of E1A 2 E3A 1 E2A 3 E1 indexed by Γ 1 (p) is a permutation matrix. All other entries are zero by Lemma Lemma With reference to Lemma 3.2.4, the following are equivalent. (i) E1A 2 E3A 1 E2A 3 E1 induces a k-cycle on Γ 1 (p) of the form (r p, c 1, e 1 ), (r p, c 2, e 2 ),..., (r p, c k, e k ). (ii) E2A 3 E1A 2 E3A 1 E2 induces a k-cycle on Γ 2 (p) of the form (r 1, c p, e 1 ), (r 2, c p, e 2 ),..., (r k, c p, e k ). (iii) E3A 1 E2A 3 E1A 2 E3 induces a k-cycle on Γ 3 (p) of the form (r 1, c 2, e p ), (r 2, c 3, e p ),..., (r k, c 1, e p ). Proof. (i) (ii): By Lemma 3.2.3, E 2A 3 E 1 [r p, c i, e i ] = [r i, c p, e i ], E 2A 3 E 1 [r p, c i+1, e i+1 ] = [r i+1, c p, e i+1 ]. By (i), E 1A 2 E 3A 1 E 2A 3 E 1 [r p, c i, e i ] = [r p, c i+1, e i+1 ]. Hence E 2A 3 E 1A 2 E 3A 1 E 2 [r i, c p, e i ] = E 2A 3 E 1 E 1A 2 E 3A 1 E 2A 3 E 1 [r p, c i, e i ] = E 2A 3 E 1 [r p, c i+1, e i+1 ] = [r i+1, c p, e i+1 ]. 15

25 [r i 1, c i, e p ] [rp., c i, e i ] [r i, c p, e i ] [r i, c i+1, e p ].. [rp, c i+1, e i+1 ] 321 Figure 3.1: The interleaving of cycles (Lemma 3.2.5, Definition 3.2.7) Thus (i) implies (ii). The other consequences are proven in a similar manner. See Figure 3.1, where we abbreviate ijk = E i A j E k. Corollary With reference to Lemma 3.2.4, the cycle structure of the permutation on Γ i (p) induced by Ei A j Ek A iej A k Ei is independent of i, j, k for all permutations i, j, k of 1, 2, 3. We refer to the common cycle structure as the cycle structure of L with respect to p. Proof. Clear from Lemma Definition With reference to Lemma 3.2.5, we refer to the triple of k-cycles C 1 = ((r p, c 1, e 1 ),(r p, c 2, e 2 ),..., (r p, c k, e k )), C 2 = ((r 1, c p, e 1 ), (r 2, c p, e 2 ),..., (r k, c p, e k )), C 3 = ((r 1, c 2, e p ), (r 2, c 3, e p ),..., (r k, c 1, e p )) as an interleaved triple of k-cycles. 3.3 Cycle Modules We produce a T -module for each interleaved triple of cycles. We will need to sum over elements of X with one of the three coordinates fixed. Since no triples other than those in X are considered here, we shall not explicitly write this condition. We shall place dots over the two coordinates which vary in the summation, e.g. (r i, ċ, ė), to remind ourselves that the triple must be an element of X. Thus although the two coordinates vary, they do not do so independently. Theorem With Notation 3.2.1, fix an interleaved triple C 1, C 2, C 3 of k-cycles as in Definition

26 (i) For 1 h k and 0 j 4, the vectors u 1,h := [r p, c h, e h ], u 2,h := [r h, c p, e h ], u 3,h := [r h, c h+1, e p ], v 1,h := [r h, ċ, ė], v 2,h := v 3,h := [Γ j (p)] (r h,ċ,ė) Γ 4 (p) (ṙ,c h,ė) Γ 4 (p) (ṙ,ċ,e h ) Γ 4 (p) [ṙ, c h, ė], [ṙ, ċ, e h ], span a T -module. We refer to this T -module as the cycle module of C 1, C 2, C 3 and denote it W (C 1, C 2, C 3 ). (ii) The action of the generators E i A j E k of T is as shown in Figures , where the subscripts are taken modulo k. In Figures we abbreviate ijk = Ei A j Ek and ijk = [Γ i ] Ei A j Ek. The action of E i A 4 Ek and E i are omitted as they can be deduced from (2.1.3) and (2.2.11). All other omitted actions are zero. Proof. In light of Lemma 5.3.2, it is enough to prove the theorem for one of each type of vector. By Lemma 2.3.1, Theorem 3.1.1, and (2.2.11), the generators Ei A j Ek of T that act on u 1,h in a nonzero manner have ijk {011, 101, 111, 23, 321, 421, 431, 341, 241, 441}. We do not derive formulae for ijk {341, 241, 441} since their action is deduced from (2.1.3), and we do not do so for ijk = 101 since E1 = E1A 0 E1 acts as the identity on u 1,h by (2.2.11). By Lemma 3.2.3, E2A 3 E1u 1,h = u 2,h and E3A 2 E1u 1,h = u 3,h. The remaining actions are deduced using Lemma Since Γ 0 (p) = {p} and (r p, c h, e h )R 1 p, we find E0A 1 E1u 1,h = [r p, c p, e p ]. Also, E 1A 1 E 1u 1,h = (r p,ċ,ė) p,(r p,c h,e h ) [r p, ċ, ė] = [Γ 1 (p)] u 1,h, 17

27 E 4A 2 E 1u 1,h = E 4A 3 E 1u 1,h = [ṙ, c h, ė] = v 2,h, (ṙ,c h,ė) Γ 4 (p) (ṙ,ċ,e h ) Γ 4 (p) [ṙ, ċ, e h ] = v 3,h. By Lemma 2.3.1, Theorem 3.1.1, and (2.2.11), the generators Ei A j Ek of T which act on v 1,h in a nonzero manner have ijk {044, 404, 124, 134, 144, 214, 234, 244, 314, 324, 344, 414, 424, 434, 444}. As above, we needn t derive formulae for ijk {144, 244, 344, 444, 044, 404}. By Lemma 2.3.4, E 3A 1 E 4v 1,h = = E3A 1 E4 [r h, ċ, ė] (r h,ċ,ė) Γ 4 (p) [r h, c h+1, e p ] = (r h,ċ,ė) Γ 4 (p) (r h,ċ,ė) u 3,h = (n 2)u 3,h, ċ c h,c p E2A 1 E4v 1,h = (n 2)u 2,h, E2A 3 E4v 1,h = E2A 3 E4 [r h, ċ, ė] = (r h,ċ,ė) Γ 4 (p) (r h,ċ,ė) (r h,c p,e h ),(r h,c h+1,e p) = [Γ 2 (p)] u 2,h, E1A 3 E4v 1,h = [Γ 1 (p)] u 1,h, E3A 2 E4v 1,h = [Γ 3 (p)] u 3,h, E1A 2 E4v 1,h = [Γ 1 (p)] u 1,h+1, E4A 1 E4v 1,h = E4A 1 E4 [r h, ċ, ė] = (r h,ċ,ė) Γ 4 (p) [r h, ċ, ė] [r h, ċ, ė ] (r h,ċ,ė) Γ 4 (p) (r h,ċ,ė ) (r h,ċ,ė),(r h,c h+1,e p),(r h,c p,e h ) = (n 3)v 1,h, 18

28 E4A 2 E4v 1,h = E4A 2 E4 [r h, ċ, ė] = (r h,ċ,ė) Γ 4 (p) [ṙ, ċ, ė ] (r h,ċ,ė) Γ 4 (p) (ṙ,ċ,ė ) (r h,ċ,ė),(r p,ċ, ),(,ċ,e p) = [Γ 4 (p)] v 2,h+1 v 1,h, E4A 3 E4v 1,h = [Γ 4 (p)] v 3,h v 1,h... v 2,i v 3,i v 1,i v 2,i u 3,i u 1,i u 2,i.. u. 3,i [r p, c p, e p ] P u 1,i+1 Figure 3.2: The action on u 1,i, u 2,i, u 3,i W (C 1, C 2, C 3 ) v 3,i. v 1,i. v 2,i I n n n I 124 u 1,i u 2,i u 3,i u 1,i+1 Figure 3.3: The action on v 1,i W (C 1, C 2, C 3 ) I v 1,i 1. v 2,i.. v 3,i n n n I 234. u 2,i 1 u 3,i 1 u 1,i u 2,i Figure 3.4: The action on v 2,i W (C 1, C 2, C 3 ).. 19

29 .. v 2,i. v 3,i.. v 1,i I n n n I 314 u 3,i 1 u 1,i u 2,i u 3,i Figure 3.5: The action on v 3,i W (C 1, C 2, C 3 ) Decomposition into Irreducible Modules In this section we describe the decomposition of each cycle module into irreducible T - modules. Lemma The primary module P is an irreducible T -submodule of each cycle module. Proof. Clear from Theorems and Lemma With Notation 3.2.1, fix an interleaved triple of k-cycles C 1, C 2, C 3 as in Definition Assume 1 k < n 1. (i) There is an irreducible T -submodule of W (C 1, C 2, C 3 ) spanned by u 1 1 := u 1 2 := u 1 3 := v1 1 := v2 1 := v3 1 := k u 1,j k n 1 [Γ 1 (p)] = j=1 k u 2,j k n 1 [Γ 2 (p)] = j=1 k u 3,j k n 1 [Γ 3 (p)] = j=1 k v 1,j k n 1 [Γ 4 (p)] = j=1 k v 2,j k n 1 [Γ 4 (p)] = j=1 k v 3,j k n 1 [Γ 4 (p)] = j=1 k j=1 k j=1 k j=1 k [r p, c j, e j ] k n 1 [Γ 1 (p)], [r j, c p, e j ] k n 1 [Γ 2 (p)], [r j, c j+1, e p ] k n 1 [Γ 3 (p)], j=1 (r j,ċ,ė) Γ 4 (p) k j=1 (ṙ,c j,ė) Γ 4 (p) k j=1 (ṙ,ċ,e j ) Γ 4 (p) [r j, ċ, ė] k n 1 [Γ 4 (p)], [ṙ, c j, ė] [ṙ, ċ, e j ] k n 1 [Γ 4 (p)], k n 1 [Γ 4 (p)]. 20

30 We denote this T -module W 1 (C 1, C 2, C 3 ). (ii) The action of the generators Ei A j Ek on u1 1, u 1 2, u 1 3, v1, 1 v2, 1 and v3 1 is as shown in Figures with ɛ = 1, where the action of Ei A 4 Ek and E i are omitted as they can be deduced from (2.1.3) and (2.2.11). All other omitted actions are zero. (iii) If n 5, then u 1 1, u 1 2, u 1 3, v1, 1 v2, 1 and v3 1 are linearly independent. Proof. (ii): In light of Lemma 5.3.2, it suffices to show that the generators Ei A j Ek of T act on u 1 1 and v1 1 as claimed. As in the proof of Theorem 3.3.1, we need only consider the action of Ei A j Ek with ijk {011, 111, 231, 321, 421, 431} on u1 1. By Lemma and Theorem 3.3.1, E 0A 1 E 1u 1 1 = k [r p, c p, e p ] E 1A 1 E 1u 1 1 = E 2A 3 E 1u 1 1 = E 4A 3 E 1u 1 1 = k j=1 (r p,ċ,ė) p,(r p,c j,e j ) = k [Γ 1 (p)] = j=1 k (n 1) (n 1)[r p, c p, e p ] = 0, [r p, ċ, ė] k [r p, c j, e j ] j=1 k k [r j, c p, e j ] (n 1) [Γ 2 (p)] = u 1 2, k j=1 (ṙ,ċ,e j ) Γ 4 (p) k E 3A 2 E 1u 1 1 = u 1 3, E4A 2 E1u 1 1 = v2, 1 k E0A 4 E4v 1 1 = j=1 (ṙ,ċ,e j ) Γ 4 (p) j=1 (r j,ċ,ė) Γ 4 (p) [ṙ, ċ, e j ] [ṙ, ċ, e j ] [r p, c p, e p ] = k(n 2)[r p, c p, e p ] k (n 1) E 1A 1 E 1 [Γ 1 (p)] k (n 1) (n 2)[Γ 1 (p)] = u 1 1, k (n 1) E 4A 3 E 1 [Γ 1 (p)] k (n 1) [Γ 4 (p)] = v 1 3, k (n 1) E 0A 4 E 4 [Γ 4 (p)] k (n 1) (n 1)(n 2)[r p, c p, e p ] = 0, 21

31 E 4A 2 E 4v 1 1 = k [Γ 4 (p)] j=1 = k [Γ 4 (p)] k = v1 1 v2, 1 ( k E1A 2 E4v 1 1 = [Γ 1 (p)] j=1 = j=1 = u 1 1. (ṙ,c j,ė) [[Γ 4 (p)]] j=1 (ṙ,c j,ė) Γ 4 (p) [ṙ, c j, ė] [ṙ, c j, ė] ) k [r p, c j+1, e j+1 ] j=1 k k [r p, c j+1, e j+1 ] + (n 1) [Γ 1 (p)] k(n 3) (n 1) [Γ 4 (p)] k(n 3) (n 1) [Γ 4 (p)] k(n 2) (n 1) [Γ 1 (p)] The other actions are similarly verified. (i): By (ii), W 1 (C 1, C 2, C 3 ) is a T -module. To show that W 1 (C 1, C 2, C 3 ) is irreducible, we show that W 1 (C 1, C 2, C 3 ) T u for any nonzero u W 1 (C 1, C 2, C 3 ). First suppose Ei u 0 for some i with 1 i 3: Say i = 3. Then E3u = αu 1 3 E3W 1 (C 1, C 2, C 3 ), and E1A 2 E3u = αu 1 1, are nonzero elements of E1W 1 (C 1, C 2, C 3 ). Now W 1 (C 1, C 2, C 3 ) T u by Theorem Now suppose Ei u = 0 for 1 i 3, so u = α 1 v1 1 + α 2 v2 1 + α 3 v3 1 for some scalars α i (i = 1, 2, 3) which are not all zero. Applying E2A 1 E4, E1A 2 E4, and E1A 3 E4 to u we get (3.4.6) (3.4.8). At least one of these coefficients is nonzero since (3.4.9) has no nonzero solution. Now W 1 (C 1, C 2, C 3 ) T u. (iii): We now show that u 1 1, u 1 2, u 1 3, v1, 1 v2, 1 and v3 1 are linearly independent whenever n 5. We note that if n 5, the supports are not distinct so the following argument wouldn t work. Suppose u = β 1 u β 2 u β 3 u α 1 v1 1 + α 2 v2 1 + α 3 v3 1 = 0. Then β 1 = β 2 = β 3 = 0 by Lemma Now the following (3.4.6) (3.4.8) are zero: E2A 1 E4u = (n 2)α 1 u 1 2 α 2 u 1 2 α 3 u 1 2 = ((n 2)α 1 α 2 α 3 )u 1 2, (3.4.6) E1A 2 E4u = ((n 2)α 2 α 1 α 3 )u 1 1, (3.4.7) E1A 3 E4u = ((n 2)α 3 α 1 α 2 )u 1 1. (3.4.8) 22

32 Since u 1 1 and u 1 2 are linearly independent, their coefficients in (3.4.6) (3.4.8) are zero. Thus (n 2)α 1 α 2 α 3 = (n 2)α 2 α 1 α 3 = (n 2)α 3 α 1 α 2 = 0. (3.4.9) Equations (3.4.9) have no non-zero solutions. Hence u 1 1, u 1 2, u 1 3, v1, 1 v2, 1 and v3 1 are linearly independent. The case k = n 1 which was excluded from Lemma behaves differently. Lemma With reference to Theorem 3.3.1, suppose k = n 1. Then k u j,i = [Γ j (p)] (j = 1, 2, 3), i=1 k v j,i = [Γ 4 (p)] (j = 1, 2, 3). i=1 Proof. Clear. Lemma With Notation 3.2.1, fix an interleaved triple of k-cycles C 1, C 2, C 3 as in Definition Assume 1 < k n 1. Let ɛ 1 be a k th root of unity. (i) There is an irreducible T -submodule of W (C 1, C 2, C 3 ) spanned by u ɛ 1 := u ɛ 2 := u ɛ 3 := v1 ɛ := v2 ɛ := k k ɛ j u 1,j = ɛ j [r p, c j, e j ], j=1 k ɛ j u 2,j = j=1 k ɛ j u 3,j = j=1 k ɛ j v 1,j = j=1 k ɛ j [r j, c p, e j ], j=1 k ɛ j [r j, c j+1, e p ], j=1 k j=1 j=1 (r j,ċ,ė) Γ 4 (p) k ɛ j v 2,j = k j=1 j=1 (ṙ,c j,ė) Γ 4 (p) ɛ j [r j, ċ, ė], ɛ j [ṙ, c j, ė], 23

33 v ɛ 3 := k ɛ j v 3,j = j=1 k ɛ j [ṙ, ċ, e j ]. j=1 (ṙ,ċ,e j ) Γ 4 (p) We denote this T -module W ɛ (C 1, C 2, C 3 ). (ii) The action of the generators Ei A j Ek on these vectors is as shown in Figures , where the action of Ei A 4 Ek and E i are omitted as they can be deduced from (2.1.3) and (2.2.11). All other omitted actions are zero. (iii) If n 5, then u ɛ 1, u ɛ 2, u ɛ 3, v1, ɛ v2, ɛ and v3 ɛ are linearly independent. Proof. (ii): The action follows from Theorem Note that k > 1, so the sum of all k th roots of unity is zero. Thus for example, E1A 2 E4v 1 ɛ = = = ( k ɛ j E1A 2 E4v 1,j j=1 k ɛ j ([Γ 1 (p)] u 1,j+1 ) j=1 k ɛ j )[Γ 1 (p)] ɛ 1 j=1 j=1 = ɛ 1 u ɛ 1, k E4A 3 E4v 1 ɛ = ɛ j E2A 3 E4v 1,j = j=1 k ɛ j ([Γ 4 (p)] v 3,j v 1,j ) j=1 = v3 ɛ v1 ɛ k E4A 2 E1u ɛ 1 = ɛ j E4A 2 E1u 1,j = j=1 k ɛ j v 2,j = v2. ɛ j=1 k ɛ j+1 u 1,j+1 The remaining actions are computed similarly. (i): Arguing as in Lemma gives that span(u ɛ 1, u ɛ 2, u ɛ 3v1, ɛ v2, ɛ v3) ɛ is closed under the action of the generators Ei A j Ek of T and that W ɛ (C 1, C 2, C 3 ) is irreducible. 24

34 (iii): To show linearly independence, let v = β 1 u ɛ 1+β 2 u ɛ 2+β 3 u ɛ 3+α 1 v ɛ 1+α 2 v ɛ 2+α 3 v ɛ 3 = 0. By Lemma we have β 1 = β 2 = β 3 = 0. To show that α 1 = α 2 = α 3 = 0, apply E 1A 2 E 4, E 1A 3 E 4, and E 2A 1 E 4 to v to find ( ɛα 1 + (n 2)α 2 α 3 )u ɛ 1 = 0, ( α 1 ɛα 2 + (n 2)α 3 )u ɛ 1 = 0, ((n 2)α 1 α 2 α 3 )u ɛ 2 = 0. Thus α 1 ɛ + α 2 (n 2) α 3 = α 1 α 2 ɛ + (n 2)α 3 = (n 2)α 1 α 2 α 3 = 0. Solving these three equations gives α 1 = α 2 = α 3 = 0. Hence u ɛ 1, u ɛ 2, u ɛ 3, v ɛ 1, v ɛ 2, and v ɛ 3 are linearly independent. 25

35 .... v ɛ 2 v ɛ 3 v ɛ ɛ u ɛ ɛ123 u ɛ u ɛ 3.. ɛ Figure 3.6: The action on u ɛ i W ɛ (C 1, C 2, C 3 ) v ɛ I n 3 v ɛ 1 ɛ( 424 I).. v ɛ n ɛ n u ɛ 3 u ɛ 1 u ɛ 2 Figure 3.7: The action on v ɛ 1 W ɛ (C 1, C 2, C 3 ) v ɛ 1. ɛ 1 ( 414 I) n 3 v ɛ I.. v ɛ n 2 ɛ ɛ ɛ(n 2).... u ɛ 1 u ɛ 2 u ɛ 3 Figure 3.8: The action on v ɛ 2 W ɛ (C 1, C 2, C 3 ) v ɛ I n 3 v ɛ I.. v ɛ n ɛ n u ɛ 2 u ɛ 3 u ɛ 1 Figure 3.9: The action on v ɛ 3 W ɛ (C 1, C 2, C 3 ) 26

36 Lemma With Notation 3.2.1, fix an interleaved triple of k-cycles C 1, C 2, C 3 as in Definition Let ɛ and δ be distinct k th roots of unity. Then W ɛ (C 1, C 2, C 3 ) and W δ (C 1, C 2, C 3 ) are non-isomorphic T -modules. Moreover, W ɛ (C 1, C 2, C 3 ) and W δ (C 1, C 2, C 3 ) are orthogonal. Proof. Suppose ɛ 1. Observe that k E1A 2 E3A 1 E2A 3 E1u ɛ 1 = E1A 2 E3A 1 E2A 3 E1 ɛ j [r p, c j, e j ] k = ɛ j [r p, c j+1, e j+1 ] j=1 j=1 k = ɛ 1 ɛ j+1 [r p, c j+, e j+1 ] j=1 = ɛ 1 u ɛ 1. Also E 1A 2 E 3A 1 E 2A 3 E 1 acts as zero on u ɛ 2, u ɛ 3, v ɛ 1, v ɛ 2, v ɛ 3. Similarly, if δ 1, then E 1A 2 E 3A 1 E 2A 3 E 1u δ 1 = δ 1 u δ 1, and E 1A 2 E 3A 1 E 2A 3 E 1 acts as zero on u δ 2, u δ 3, v δ 1, v δ 2, v δ 3. Thus the result holds in this case. Suppose δ = 1. Now by Lemma and Theorem 3.1.1, E 1A 2 E 3A 1 E 2A 3 E 1u 1 1 = E 1A 2 E 3A 1 E 2A 3 E 1( = k j=1 = u 1 1. k j=1 [r p, c j+1, e j+1 ] k n 1 [Γ 1 (p)] [r p, c j, e j ] k n 1 [Γ 1 (p)]) Also E 1A 2 E 3A 1 E 2A 3 E 1 acts as zero on u 1 2, u 1 3, v 1 1, v 1 2, v 1 3. It follows that W ɛ (C 1, C 2, C 3 ) and W δ (C 1, C 2, C 3 ) are non-isomorphic. The orthogonality of these two modules follows from Lemma Theorem With Notation 3.2.1, fix an interleaved triple of k-cycles C 1, C 2, C 3. (i) If k n 1, then W (C 1, C 2, C 3 ) has orthogonal direct decomposition into irreducible 27

37 T -modules W (C 1, C 2, C 3 ) = P ɛ C W ɛ (C 1, C 2, C 3 ). ɛ k =1 (ii) If k = n 1, then W (C 1, C 2, C 3 ) has orthogonal direct decomposition into irreducible T -modules W (C 1, C 2, C 3 ) = P W ɛ (C 1, C 2, C 3 ). ɛ C ɛ k =1,ɛ 1 Proof. It is clear from Lemmas and that the sums in (i) and (ii) are orthogonal (and hence direct). First suppose k < n 1. Then by orthogonality, the sum in (i) spans a subspace of dimension 6k+5. But by construction, dim W (C 1, C 2, C 3 ) 6k+5. Thus (i) holds. Next suppose k = n 1. Then by orthogonality, the sum in (ii) spans a subspace of dimension 6k 1. Also by Lemma dim W (C 1, C 2, C 3 ) 6k 1. Thus (ii) holds. Note that Theorem holds with no restriction on n, but for n 2, the only irreducible T -module is the primary module. For n = 3, there is just one main class of Latin squares, that of the Cayley table of Z 3, which has cycle structure 2 1 with respect to all points (see Section 3.9). 3.5 Some Intermediate Modules We give a common generalization of Theorem and Lemmas and which allows us to produce some nice submodules of a cycle module. Lemma With Notation 3.2.1, fix an interleaved triple of k-cycles C 1, C 2, C 3 as in Definition Pick nonnegative integers l and m such that k = lm. Let ɛ be any 28

38 complex m th root of 1. For i = 1,..., l set u ɛ,l 1,i = u ɛ,l 2,i = u ɛ,l 3,i = v ɛ,l 1,i = v ɛ,l 2,i = v ɛ,l 3,i = m ɛ j [r p, c jl+i, e jl+i ], j=1 m ɛ j [r jl+i, c p, e jl+i ], j=1 m ɛ j [r jl+i 1, c jl+i, e p ], j=1 m ɛ j [r jl+i, ċ, ė], j=1 (r jl+i,ċ,ė) Γ 4 (p) m ɛ j [ṙ, c jl+i, ė], j=1 (ṙ,c jl+i,ė) Γ 4 (p) m ɛ j [ṙ, ċ, e jl+i ]. j=1 (ṙ,ċ,e jl+i ) Γ 4 (p) For t = 1, 2, 3, let U t = {u ɛ,l t,i }l i=1 and V t = {v ɛ,l t,i }l i=1. Then ( 3 t=1u t ) ( 3 t=1v t ) {[Γ t ]} 4 t=0 spans a T -module. Moreover if n 5, then the following hold. (i) If k < n 1 or ɛ 1, then the 6l + 5 vectors in ( 3 t=1u t ) ( 3 t=1v t ) {[Γ t ]} 4 t=0 are linearly independent. (ii) If k = n 1 and ɛ = 1, then ( 3 t=1u t ) ( 3 t=1v t ) {[Γ t ]} 4 j=0 span a (6l 1)-dimensional T -module. Proof. It is easy to see from Lemma and Theorem that span(( 3 t=1u t ) ( 3 t=1v t ) {[Γ t ]} 4 t=0) is closed under the action of the generators Ei A j Ek of T, and hence is T -module. (i): The dimension of this module is at most 6l + 5 by construction, and it is at least this large since it contains the primary module and l-many 6-dimensional orthogonal irreducible submodules by Lemmas and Hence equality holds. It follows that the given vectors are linearly independent. (ii): Suppose ɛ = 1 and k = n 1. The dimension is at most 6l 1 by construction and Lemma It is at least this large by Lemma Hence equality holds. 29

39 Parts (i) and (ii) of Lemma fail if n 5 since the dimension of E4V is too small. We note that the cycle module W (C 1, C 2, C 3 ) appears in Lemma in the case m = 1, l = k (in which case ɛ = 1). The irreducible submodule W ɛ (C 1, C 2, C 3 ) appears in Lemma in the case m = k, l = Collecting Cycle Modules We begin by extending Notation Latin squares with order at least 5. To avoid degenerate situations, only consider Notation Let L denote a Latin square of order n 5 and with symbol set {1, 2,..., n}. Let X denote the set {(i, j, L(i, j)) 1 i, j n}. Let M denote the Bose-Mesner algebra of L. Fix p = (r p, c p, e p ) X, and let T denote the subconstitutent algebra of M with respect to (r p, c p, e p ). Let I 1, I 2,..., I m denote the interleaved cycles of L with respect to p. Denote the elements of I j as C1, j C2, j C3. j Use X(j) to refer to object X associated with W (C1, j C2, j C3); j for example u ɛ 1(j). Lemma With Notation Fix two distinct interleaved triples of cycles C 1, C 2, C 3 and C 1, C 2, C 3. Suppose the C i are k-cycles and the C i are k -cycles. Let l be a positive integer such that l k and l k, and let ɛ be an l th root of unity. Then W ɛ (C 1, C 2, C 3 ) and W ɛ (C 1, C 2, C 3) are isomorphic T -modules. Proof. Note that neither k nor k is n 1 since there are two distinct interleaved triples of cycles. Thus W 1 (C 1, C 2, C 3 ) and W 1 (C 1, C 2, C 3) are both defined. However, the case ɛ = 1 need not be treated separately. Suppose the modules W ɛ (C 1, C 2, C 3 ) and W ɛ (C 1, C 2, C 3) have respective bases {u ɛ 1, u ɛ 2, u ɛ 3, v1, ɛ v2, ɛ v3} ɛ and {s ɛ 1, s ɛ 2, s ɛ 3, t ɛ 1, t ɛ 2, t ɛ 3}. Define a linear map φ : W ɛ (C 1, C 2, C 3 ) W ɛ (C 1, C 2, C 3) by φ(u ɛ i) = s ɛ i and φ(vi) ɛ = t ɛ i (1 i 3). It is clear that φ is a bijection. To show that for all A T and v V, φ(av) = Aφ(v) it is enough to treat the case where A is of the form Ei A j Ek and v is one of the basis elements uɛ i or vi. ɛ For example, φ(e 3A 2 E 1u ɛ 1) = φ(ɛv ɛ 2) = ɛφ(u ɛ 3) = ɛv ɛ 3 = E 3A 2 E 1s ɛ 1 = E 3A 2 E 1φ(u ɛ 1). 30

40 The result follows from Theorem (Figures ). Lemma With Notation If ɛ and δ are both primitive l th roots of unity, then mult(w ɛ (C 1, C 2, C 3 )) = mult(w δ (C 1, C 2, C 3 )). Proof. Clear since the multiplicity of each is the number of interleaved cycles with length divisible by l. In contrast to the situation for non-isomorphic irreducible T -modules, the isomorphic irreducible T -modules contained in distinct cycle modules are generally not orthogonal to one another. It turns out that with the exception of the dependencies noted in the following lemma, isomorphic irreducible T -modules constructed so far are linearly independent. Lemma With Notation 3.6.1, suppose m 1 for i = 1, 2, 3, m u 1 i (j) = 0 j=1 and m vi 1 (j) = 0. j=1 Proof. Suppose that C j 1 has length k(j) (1 h m). Recall that u 1 1(j) = (r p,ċ,ė) C j 1 [r p, ċ, ė] k(j) (n 1) [Γ 1 (p)], so m u 1 1(j) = j=1 m j=1 (r p,ċ,ė) C j 1 [r p, ċ, ė] = [Γ 1 (p)] [Γ 1 (p)] = 0. m j=1 k(j) (n 1) [Γ 1 (p)] Similarly, m j=1 u1 2(j) = m j=1 u1 3(j) = 0. Note that v 1 1(j) = k(j) h=1 (r h (j),ċ,ė) [[Γ 4 (p)]] [r h (j), ė, ċ] k(j) (n 1) [Γ 4 (p)]. 31

41 Thus k(j) m m v1(j) 1 = [r h (j), ċ, ė] [Γ 4 (p)] j=1 j=1 h=1 (r h (j),ċ,ė) [[Γ 4 (p)]] = [Γ 4 (p)] [Γ 4 (p)] = 0. Similarly, m j=1 v1 2(j) = m j=1 v1 3(j) = 0. When m = 1, Lemma restates Lemma Lemma With Notation 3.6.1, the following set is linearly independent: m 1 j=1 {u1 1(j), u 1 2(j), u 1 3(j), v 1 1(j), v 1 2(j), v 1 3(j)}. Proof. For 1 j m, let w1(j) 1 = u 1 1(j) + k(j) k(j) n 1 [Γ 1 (p)] = h=1 [r p, c h (j), e h (j)]. Since the first coordinate of the support of each w 1 1(j) are disjoint, the set {w 1 1(j)} m j=1 is linearly independent by Lemma It follows that {u 1 1(j)} m 1 j=1 Similarly {u 1 2(j)} m 1 j=1 and {u1 3(j)} m 1 j=1 are linearly independent sets. Let u = 3 h=1 m 1 j=1 α (j) h u1 h(j) + 3 h=1 m 1 j=1 β (j) h v1 h(j), is linearly independent. 32

42 and suppose u = 0. Then α (j) 1 = α (j) 2 = α (j) 3 = 0 by Lemma and the above. Applying E 1A 2 E 4, E 1A 3 E 4, and E 2A 1 E 4 to u gives 0 = ( β (1) 1 + (n 2)β (1) 2 β (1) 3 )u 1 1(1) + + ( β (m 1) 1 + (n 2)β (m 1) 2 β (m 1) 3 )u 1 1(m 1), 0 = ( β (1) 1 β (1) 2 + (n 2)β (1) 3 )u 1 1(1) + + ( β (m 1) 1 β (m 1) 2 + (n 2)β (m 1) 3 )u 1 1(m 1), 0 = ((n 2)β (1) 1 β (1) 2 β (1) 3 )u 1 2(1) + + ((n 2)β (m 1) 1 β (m 1) 2 β (m 1) 3 )u 1 2(m 1). Hence for 1 j m, β (j) 1 + (n 2)β (j) 2 β (j) 3 = 0, β (j) 1 β (j) 2 + (n 2)β (j) 3 = 0, (n 2)β (j) 1 β (j) 2 β (j) 3 = 0. As in the proof of Lemma 3.4.2, it follows that β (j) 1 = β (j) 2 = β (j) 3 = 0 for all j (1 j m 1). The choice of omission in Lemma is was arbitrary. Lemma With Notation 3.6.1, let ɛ be a primitive l th root of unity other than 1. Let I i 1, I i 2,..., I ip be all interleaved cycles with order divisible by l. Then the following set is linearly independent: p j=1 {uɛ 1(i j ), u ɛ 2(i j ), u ɛ 3(i j ), v ɛ 1(i j ), v ɛ 2(i j ), v ɛ 3(i j )}. Proof. Let u = 3 p h=1 j=1 α (j) h uɛ h(i j ) + 3 p h=1 j=1 β (j) h vɛ h(i j ), 33