Combining Association Schemes

Transcription

1 Chapter 4 Combining Association Schemes 4.1 Tensor products In this chapter we take an association scheme on Ω 1 and an association scheme on Ω, and combine them to obtain an association scheme on Ω 1 Ω. We need some preliminary notions about the space R Ω 1 Ω of all functions from Ω 1 Ω to R. For f in R Ω 1 and g in R Ω, define the tensor product f g in R Ω 1 Ω by For example, f g:(ω 1,ω ) f (ω 1 )g(ω ) for (ω 1,ω ) in Ω 1 Ω. ( f χ Ω )(ω 1,ω ) = f (ω 1 ). For subspaces U of R Ω 1 and V of R Ω define the tensor product U V to be the subspace of R Ω 1 Ω spanned by { f g : f U and g V }. If {u 1,...,u m } is a basis for U and {v 1,...,v r } is a basis for V, then { ui v j : 1 i m, 1 j r } is a basis for U V, so dim(u V ) = dim(u)dim(v ). The natural bases for R Ω 1 and R Ω are {χ α : α Ω 1 } and { χ β : β Ω }. But χ α χ β = χ (α,β) and { χ (α,β) : (α,β) Ω 1 Ω } is a basis for R Ω 1 Ω, so R Ω 1 R Ω = R Ω 1 Ω. Lemma 4.1 If U is a subspace of R Ω 1 and V is a subspace of R Ω then the orthogonal complement of U V in R Ω 1 Ω is (U R Ω ) + (R Ω 1 V ). 55

2 56 CHAPTER 4. COMBINING ASSOCIATION SCHEMES Proof If u and f are in R Ω 1 and v and g are in R Ω then u v, f g = (α,β) Ω 1 Ω (u v)(α,β) ( f g)(α,β) = u(α)v(β) f (α)g(β) (α,β) Ω 1 Ω ( ) ( ) = u(α) f (α) α Ω 1 v(β)g(β) β Ω = u, f v,g. This is zero for all u in U and all v in V if either f U or g V. So (U V ) contains (U R Ω ) + (R Ω 1 V ). We shall complete the proof by showing that these two subspaces have the same dimension. To calculate dim ( (U R Ω ) + (R Ω 1 V ) ) we need to know dim ( (U R Ω ) (R Ω 1 V ) ). In general it is not straightforward to calculate the intersection of two tensor product spaces. However, every element of R Ω 1 Ω can be expressed in the form i x i y i with all of the x i different and all of the y i different. Such an element is in U R Ω if and only if each x i is in U ; similarly, such an element is in R Ω 1 V if and only if each y i is in V. Therefore (U R Ω ) (R Ω 1 V ) = U V. Now let Ω 1 = n 1 and Ω = n. Then ( ) dim (U R Ω ) + (R Ω 1 V ) ( ) ( = dim U R Ω + dim R Ω 1 V ) ( ) dim (U R Ω ) (R Ω 1 V ) ( ) ( = dim U R Ω + dim R Ω 1 V ) dim (U V ) = (n 1 dimu)n + n 1 (n dimv ) (n 1 dimu)(n dimv ) = n 1 n dimu dimv ) = dim (R Ω 1 Ω dim(u V ) ( = dim (U V ) ). Thus the two spaces must be equal. Tensor products of matrices are also defined. Let M be a matrix in R Ω 1 Ω 1 and N be a matrix in R Ω Ω. Since matrices are just functions on special sets, the preceding discussion shows that M N should be a function in R (Ω 1 Ω 1 ) (Ω Ω ).

3 4.1. TENSOR PRODUCTS 57 But it is convenient if M N is square when M and N are square, so we use the natural bijection between R (Ω 1 Ω 1 ) (Ω Ω ) and R (Ω 1 Ω ) (Ω 1 Ω ) and define M N to be the matrix in R (Ω 1 Ω ) (Ω 1 Ω ) given by M N ((α 1,α ),(β 1,β )) = M(α 1,β 1 )N(α,β ) for α 1, β 1 in Ω 1 and α, β in Ω. For example, I Ω1 I Ω = I Ω1 Ω ; J Ω1 J Ω = J Ω1 Ω ; M O Ω = O Ω1 N = O Ω1 Ω. A concrete way of constructing M N is to take each entry M(α,β) of M and replace it by the whole matrix N, all multiplied by M(α, β). Alternatively, one can start with the matrix N and replace each entry N(α,β) by the whole matrix M multiplied by N(α,β). So long as the rows and colus of M N are labelled, these two constructions give the same thing. Some standard facts about tensor products are gathered into the following proposition, whose proof is straightforward. Proposition 4. Let M, M 1 and M be matrices in R Ω 1 Ω 1 and N, N 1 and N be matrices in R Ω Ω. Then (i) if f R Ω 1 and g R Ω then in the sense that (M N)( f g) = (M f ) (Ng) ((M N)( f g))(ω 1,ω ) = (M f )(ω 1 ) (Ng)(ω ); (ii) if u is an eigenvector of M with eigenvalue λ and v is an eigenvector of N with eigenvalue µ then u v is an eigenvector of M N with eigenvalue λµ; (iii) (M N) = M N ; (iv) if M and N are symmetric then so is M N; (v) for scalars λ and µ, M (λn 1 + µn ) = λm N 1 + µm N and (λm 1 + µm ) N = λm 1 N + µm N;

4 58 CHAPTER 4. COMBINING ASSOCIATION SCHEMES (vi) (M 1 N 1 )(M N ) = M 1 M N 1 N. Corollary 4.3 If U is a subspace of R Ω 1 with orthogonal projector P and V is a subspace of R Ω with orthogonal projector Q then the orthogonal projector onto U V is P Q. Proof We need to prove that P Q is the identity on U V and is zero on (U V ). Proposition 4.(i) shows that, if u U and v V, then (P Q)(u v) = (Pu) (Qv) = u v. If x (U V ) then x = f y + z g, with f U, y R Ω, z R Ω 1 and g V, by Lemma 4.1. Then (P Q)x = (P Q)( f y) + (P Q)(z g) = P f Qy + Pz Qg = 0 because P f = 0 in R Ω 1 and Qg = 0 in R Ω. Note: tensor products are sometimes called Kronecker products. 4. Crossing For the rest of this chapter we need notation for two different association schemes which is as close as possible as to that used heretofore without being cumbersome. For symbols which are not already suffixed the obvious solution is to affix suffices 1 and. Otherwise the only clean solution is to use two different but related symbols, as shown in Table 4.1. We also need an abuse of notation similar to the one we made when defining the tensor product of matrices. If C Ω 1 Ω 1 and D Ω Ω then strictly speaking C D is a subset of (Ω 1 Ω 1 ) (Ω Ω ). We need to regard it as a subset of (Ω 1 Ω ) (Ω 1 Ω ). So we make the convention that, for such sets C and D, C D = {((α 1,α ),(β 1,β )) : (α 1,β 1 ) C and (α,β ) D}. This convention has the happy consequence that A C D = A C A D. (4.1) Definition For t = 1,, let Q t be a set of subsets of Ω t Ω t. The direct product Q 1 Q of Q 1 and Q is the set {C D : C Q 1 and D Q } of subsets of (Ω 1 Ω ) (Ω 1 Ω ). The operation of forming this direct product is called crossing Q 1 and Q.

5 4.. CROSSING 59 first scheme second scheme set Ω 1 Ω size of set n 1 n associate classes... C D... indexed by K 1 K number of classes s 1 s parameters p k i j q z xy valencies a i b x adjacency matrices A i B x strata... U e V f... indexed by E 1 E stratum projectors S e T f dimensions d e not used character table... C 1 C... and inverse D 1 D Table 4.1: Notation for two association schemes

6 60 CHAPTER 4. COMBINING ASSOCIATION SCHEMES Any sets of subsets of Ω 1 Ω 1 and Ω Ω can be crossed. It is clear that if Q t is a partition of Ω t Ω t for t = 1, then Q 1 Q is a partition of (Ω 1 Ω ) (Ω 1 Ω ). The result of crossing is most interesting when the two components have some nice structure, such as being association schemes. Theorem 4.4 Let Q 1 be an association scheme on Ω 1 with s 1 associate classes, valencies a i for i in K 1 and adjacency matrices A i for i in K 1. Let Q be an association scheme on Ω with s associate classes, valencies b x for x in K and adjacency matrices B x for x in K. Then Q 1 Q is an association scheme on Ω 1 Ω with s 1 s + s 1 + s associate classes, valencies a i b x for (i,x) in K 1 K and adjacency matrices A i B x for (i,x) in K 1 K. Proof Equation (4.1) shows that the adjacency matrices have the required form. For (i,x) in K 1 K, the set of (i,x)-th associates of (α 1,α ) is {(β 1,β ) Ω 1 Ω : β 1 C i (α 1 ) and β D x (α )}, which has size a i b x. The first two conditions for an association scheme are easy to check, for Diag(Ω 1 Ω ) = Diag(Ω 1 ) Diag(Ω ) and symmetry follows from Proposition 4.(iv). For the third condition we check the product of adjacency matrices: (A i B x ) ( ) A j B y = Ai A j B x B y ( ) ( ) = p k i ja k q z xyb z k K 1 z K = (k,z) K 1 K p k i jq z xya k B z, which is a linear combination of the adjacency matrices. So Q 1 Q is indeed an association scheme. The number of associate classes is one less than the number of sets in the partition Q 1 Q, which is equal to (s 1 + 1)(s + 1). So there are s 1 s + s 1 + s associate classes. Example 4.1 The direct product of the trivial association schemes n and m is indeed the rectangular association scheme R(n,m), with adjacency matrices I, I n (J m I m ), (J n I n ) I m and (J n I n ) (J m I m ). Theorem 4.5 Let the character tables of the association schemes Q 1, Q and Q 1 Q be C 1, C and C with inverses D 1, D and D. Let the strata for Q 1 be U e for e in E 1 and the strata for Q be V f for f in E. Then

7 4.. CROSSING 61 (i) the strata for Q 1 Q are U e V f for (e, f ) in E 1 E ; (ii) C = C 1 C ; (iii) D = D 1 D. Proof (i) The spaces U e V f for (e, f ) in E 1 E are pairwise orthogonal and sum to R Ω 1 Ω. Proposition 4.(ii) shows that every subspace of R Ω 1 Ω of the form U e V f is a sub-eigenspace of every adjacency matrix of Q 1 Q, so every stratum is a sum of one or more of these spaces. But the number of strata is equal to the number of adjacency matrices, by Theorem.6, which is (s 1 + 1)(s + 1). The same theorem shows that E 1 = s and E = s + 1. Hence the number of spaces of the form U e V f is the same as the number of strata, and so these spaces are exactly the strata. (ii) The eigenvalue of A i on U e is C 1 (i,e) and the eigenvalue of B x on V f is C (x, f ), so the eigenvalue of A i B x on U e V f is C 1 (i,e)c (x, f ), by Proposition 4.(ii). Hence C((i,x)(e, f )) = C 1 (i,e)c (x, f ) and so C = C 1 C. (iii) We know that D 1 C 1 = I Ω1 and D C = I Ω, so (D 1 D )(C 1 C ) = (D 1 C 1 ) (D C ) = I Ω1 I Ω = I Ω1 Ω. Thus D 1 D = (C 1 C ) 1 = C 1 = D. Example 4.1 revisited The trivial association scheme n has strata U 0 and U 0, and character table C n = same (1) different (n 1) U 0 U0 [ (1) (n 1) 1 1 n 1 1 ] with inverse D n = U 0 (1) U 0 (n 1) same different (1) (n 1) 1 1 n n. n 1 n 1 n (Why is it not correct to write nd n = C n?)

8 6 CHAPTER 4. COMBINING ASSOCIATION SCHEMES So the strata for n m are U 0 V 0, U 0 V 0, U 0 V 0 and U 0 V 0. The character table C n m is same (1) same row same colu other (m 1) (n 1) ((m 1)(n 1)) U 0 V 0 U 0 V0 U0 V 0 U0 V 0 (1) (m 1) (n 1) ((m 1)(n 1)) m 1 1 m 1 1 n 1 n (m 1)(n 1) (n 1) (m 1) 1 and the inverse D n m is same same row same colu other U 0 V 0 (1) U 0 V0 U0 V 0 U0 V 0 (m 1) (n 1) ((m 1)(n 1)) (1) (m 1) (n 1) ((m 1)(n 1)) 1 m 1 n 1 (m 1)(n 1) 1 1 n 1 (n 1) 1 m 1 1 (m 1) Example 4. The underlying set for the association scheme 3 5 consists of five colus of three points each, with the colus arranged in a pentagon as shown in Figure 4.1. The (names of the) associate classes in 3 are same and different, while those in 5 are same, edge and non-edge. Each ordered pair of these gives an associate class in 3 5. One point in Figure 4.1 is labelled 0. The remaining points are labelled i if they are i-th associates of the point 0, according to the following table. (same, same) 0 (same, edge) 1 (same, non-edge) (different, same) 3 (different, edge) 4 (different, non-edge) 5 The character tables of the component association schemes are C 3 = same (1) different () U 0 U0 [ (1) () ] 1 1 1

9 4.. CROSSING 63 and V 0 V 1 V (1) () () C 5 = same (1) edge () non-edge () from Example.4. Therefore C 3 5 is U 0 V 0 U 0 V 1 U 0 V U 0 V 0 U 0 V 1 U 0 V (1) () () () (4) (4) 0 (1) 1 () () 3 () 4 (4) 5 (4) Figure 4.1: The association scheme 3 5

10 64 CHAPTER 4. COMBINING ASSOCIATION SCHEMES 4.3 Isomorphism Definition Let Q 1 be an association scheme on Ω 1 with classes C i for i in K 1, and let Q be an association scheme on Ω with classes D j for j in K. Then Q 1 is isomorphic to Q if there are bijections φ:ω 1 Ω and π:k 1 K such that (α,β) C i (φ(α),φ(β)) D π(i). The pair (φ,π) is an isomorphism between association schemes. We write Q 1 = Q. If K 1 = K then an isomorphism (φ,π) is a strong isomorphism if π is the identity. In this situation an isomorphism in which π is not necessarily so constrained is a weak isomorphism. If Q 1 = Q then an isomorphism (φ,π) is an automorphism of the association scheme. Example 4.3 Let Q 1 be GD(,), which gives the strongly regular graph on the left of Figure 4., with classes C 0, C 1 and C, where C 1 = {(a,b),(b,a),(c,d),(d,c)}. Let Q be 4, which gives the strongly regular graph on the right of Figure 4., with classes D 0, D 1 and D, where Let and D 1 = {(0,1),(1,0),(1,),(,1),(,3),(3,),(3,0),(0,3)}. φ(a) = 0 φ(b) = φ(c) = 1 φ(d) = 3 π(0) = 0 π(1) = π() = 1. Then (φ,π) is an isomorphism from Q 1 to Q, so that GD(,) is isomorphic to 4. If we consider that the labels of the classes C i are the same as the labels of the classes D i then (φ,π) is a weak isomorphism but not a strong one, because it carries edges to non-edges. The two strongly regular graphs in Figure 4. are not isomorphic as graphs even though they define isomorphic association schemes. Proposition 4.6 Let 0, 1,..., s be a blueprint for Z n. Suppose that m is coprime to n. Let φ be the permutation of Z n defined by φ(ω) = mω for ω in Z n. If there is a permutation π of {0,...,s} such that {φ(ω) : ω i } = π(i) for i = 0,..., s then (φ,π) is a weak automorphism of the cyclic association scheme defined by this blueprint.

11 4.3. ISOMORPHISM 65 a c 0 1 b d 3 Q 1 Q Figure 4.: Two association schemes in Example 4.3 Example 4.4 (Example 1.8 continued) We know that 0, 1, is a blueprint for Z 13, where 0 = {0}, 1 = {1,3,4, 4, 3, 1} and = {,5,6, 6, 5, }. Let φ(ω) = ω for ω in Z 13. Then φ( 0 ) = 0, φ( 1 ) = and φ( ) = 1 so φ induces a weak automorphism of the cyclic association scheme defined by 0, 1,. It is evident that isomorphic association schemes have the same parameters, but the converse is not true. Example 4.5 Let Q 1 and Q be the association schemes of L(3,4) type defined by the Latin squares Π 1 and Π. Π 1 = A B C D D A B C C D A B B C D A Ω = Π = A B C D B A D C C D A B D C B A Draw edges between first associates. In both schemes p 1 11 = 4 so every edge is contained in four triangles. In Π every pair of points in the same row or colu or letter are contained in a Latin subsquare, so the edge is contained in two complete graphs of size 4: l r r c r r r l r c r (the letters r, c and l denote same row, same colu and same letter respectively). However, there are edges in Q 1 which are not contained in two complete graphs of size 4. For example, the triangles through {1,} are as follows.

12 66 CHAPTER 4. COMBINING ASSOCIATION SCHEMES So Q 1 cannot be isomorphic to Q. 1 l r c r 6 13 r r l r 3 4 c r Theorem 4.7 Crossing is commutative in the sense that Q 1 Q is isomorphic to Q Q 1. Proof Take φ((ω 1,ω )) = (ω,ω 1 ) and π((i,x)) = (x,i). Theorem 4.8 Crossing is associative in the sense that Q 1 (Q Q 3 ) = (Q 1 Q ) Q Nesting When we cross Q 1 with Q, the new underlying set can be thought of as the rectangle Ω 1 Ω. When we nest Q within Q 1 we replace each element of Ω 1 by a copy of Ω. Then the old adjacencies within Ω 1 apply to whole copies of Ω, while the old adjacencies within Ω apply only within each separate copy. Definition For t = 1,, let Q t be a set of subsets of Ω t Ω t. Suppose that Q 1 contains Diag(Ω 1 ). The wreath product Q 1 /Q of Q 1 and Q is the set of subsets {C (Ω Ω ) : C Q 1, C Diag(Ω 1 )} {Diag(Ω 1 ) D : D Q } of (Ω 1 Ω ) (Ω 1 Ω ). The operation of forming this wreath product is called nesting Q within Q 1. It is clear that Q 1 /Q is a partition of (Ω 1 Ω ) (Ω 1 Ω ) if Q 1 is a partition of Ω 1 Ω 1 containing Diag(Ω 1 ) and Q is a partition of Ω Ω. Theorem 4.9 Let Q 1 be an association scheme on a set Ω 1 of size n 1 with s 1 associate classes, valencies a i for i in K 1 and adjacency matrices A i for i in K 1. Let Q be an association scheme on a set Ω of size n with s associate classes, valencies b x for x in K and adjacency matrices B x for x in K. Then Q 1 /Q is an association scheme on Ω 1 Ω with s 1 + s associate classes, valencies a i n for i in K 1 \ {0} and b x for x in K, and adjacency matrices A i J Ω for i in K 1 \ {0} and I Ω1 B x for x in K.

13 4.4. NESTING 67 Proof Everything follows immediately from the definition except the fact that the product of two adjacency matrices is a linear combination of adjacency matrices. But (A i J Ω ) ( ( ) ) A j J Ω = Ai A j n J Ω = n p k i ja k J Ω k K 1 and = n k K 1 \{0} (A i J Ω )(I Ω1 B x ) = A i b x J Ω, p k i ja k J Ω + n p 0 i j x K I Ω1 B x, (I Ω1 B x )(I Ω1 B y ) = I Ω1 B x B y = k K q z xyi Ω1 B z. Example 4.6 The wreath product of the trivial association schemes b and k is indeed the group-divisible association scheme GD(b, k), with adjacency matrices (J b I b ) J k, I b I k and I b (J k I k ). This example shows immediately that nesting is not commutative. For example, 10/3 has valencies 1, and 7 while 3/10 has valencies 1, 9 and 0. Theorem 4.10 Let the strata for Q 1 be U e, for e in E 1, and the strata for Q be V f, for f in E. Then the strata for Q 1 /Q are U e V 0, for e in E 1, and R Ω 1 V f, for f in E \ {0}. Moreover, let C be the character table of Q 1 /Q, and D its inverse, so that the rows of C and colus of D are indexed by (K 1 \ {0}) K, while the colus of C and the rows of D are indexed by E 1 (E \{0}). If Q 1 and Q have character tables C 1 and C with inverses D 1 and D then and C(i,e) = n C 1 (i,e) for i K 1 \ {0} and e E 1, C(i, f ) = 0 for i K 1 \ {0} and f E \ {0}, C(x,e) = b x for x K and e E 1, C(x, f ) = C (x, f ) for x K and f E \ {0}, D(e,i) = 1 n D 1 (e,i) for e E 1 and i K 1 \ {0}, D(e, x) = d e n 1 n for e E 1 and x K, D( f,i) = 0 for f E \ {0} and i K 1 \ {0}, D( f,x) = D ( f,x) for f E \ {0} and x K.

14 68 CHAPTER 4. COMBINING ASSOCIATION SCHEMES Proof The spaces U e V 0, for e in E 1, and R Ω 1 V f, for f in E \ {0}, are pairwise orthogonal. Also, e E 1 (U e V 0 ) = R Ω 1 V 0 and (R Ω 1 V f ) = R Ω 1 V0, f E \{0} so the sum of these spaces is R Ω 1 Ω. Since E 1 (E \ {0}) = E 1 + E 1 = K 1 + K 1 = (K 1 \ {0}) K, in order to prove that the named subspaces are strata it suffices to show that they are sub-eigenspaces of every adjacency matrix. The entries in C come as part of this demonstration. Let i K 1 \ {0}, x K, e E 1, f E \ {0}, u U e, w R Ω 1 and v V f. Then (A i J Ω )(u χ Ω ) = A i u J Ω χ Ω = C 1 (i,e)u n χ Ω ; (A i J Ω )(w v) = A i w J Ω v = A i w 0 Ω = 0 Ω1 Ω ; (I Ω1 B x )(u χ Ω ) = I Ω1 u B x χ Ω = u b x χ Ω ; (I Ω1 B x )(w v) = I Ω1 w B x v = w C (x, f )v. Finally, we find the entries in D by expressing the stratum projectors as linear combinations of the adjacency matrices. If S e is the projector onto U e and T f is the projector onto V f then the projectors onto U e V 0 and R Ω 1 V f are S e T 0 and I Ω1 T f. But S e T 0 = ( ) D 1 (e,i)a i 1 J Ω i K 1 n = 1 n D 1 (e,i)a i J Ω + D 1(e,0) i K n 1 \{0} I Ω1 B x x K = 1 n D 1 (e,i)a i J Ω + d e i K n 1 \{0} 1 n I Ω1 B x, x K while ( ) I Ω1 T f = I Ω1 D ( f,x)b x. x K

15 4.4. NESTING 69 Using an obvious extension to the notation J and O, we have shown that C is E 1 E \ {0} K 1 \ {0} n [C 1 omitting 0-th row] O K1 \{0},E \{0} K diag(b)j K,E 1 [C omitting 0-th colu] while D is K 1 \ {0} K E 1 1 n [D 1 omitting 0-th colu] 1 n 1 n diag(d)j E1,K E \ {0} O E \{0},K 1 \{0} [D omitting 0-th row] Example 4.6 revisited For b/k we have n 1 = b, n = k, [ C 1 = [ C = 1 1 b k 1 1 ], D 1 = 1 [ b ], and D = 1 [ k 1 1 b k 1 1 ], ]. So C = (b 1)k k k 1 k 1 1 and D = 1 bk b 1 b 1 0 b(k 1) b. This agrees with Example., except for the order in which the rows and the colus are written. We see that it is not possible to simultaneously (i) label both the rows and the colus of C and D by (some of) the indices for Q 1 followed by (some of) the

16 70 CHAPTER 4. COMBINING ASSOCIATION SCHEMES indices for Q (ii) show both the diagonal associate class and the stratum W 0 as the first row and colu of these matrices. This is because the diagonal class of Q 1 /Q is labelled by the diagonal class of Q while the 0-th stratum of Q 1 /Q is labelled by the 0-th stratum of Q 1. As explained before, where there is a natural bijection, or even partial bijection, between K and E, it may not carry the 0-th associate class to the 0-th stratum. Theorem 4.11 Nesting is associative in the sense that Q 1 /(Q /Q 3 ) is isomorphic to (Q 1 /Q )/Q 3.