Association schemes: definition Association schemes An association scheme of rank r on a finite Ω (sometimes called points) R. A. Bailey is a colouring of the elements of Ω Ω [email protected], [email protected] × (sometimes called edges) by r colours such that (i) one colour is exactly the main diagonal; G. C. Steward lecture, (ii) each colour is symmetric about the main diagonal; Gonville and Caius College, Cambridge (iii) if (α, β) is yellow yellow then there are exactly pred,blue points γ such that (α,γ) is red and (γ, β) is blue (for all values of yellow, red and blue).

10 May 2013 The non-diagonal classes are called associate classes.

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An association scheme with 15 points and 3 associate An association scheme defined by a classes

1 2 3 4 If (α, β) is blue 1 A B C D α β 2 B C D A white blue 1 3 C D A B blue white 1 4 D A B C blue blue 1 black pink 4 pink black 4 pink pink 4

different cells in same row red different cells in same column green different cells in same letter yellow 3/42 other different cells black 4/42

The association scheme defined by the cube Association schemes: alternative definition

ABCDEFGH

EF A The adjacency Ai for colour i is the Ω Ω matrix with @ × @ B 1 if (α, β) has colour i @ Ai(α, β) = @ C 0 otherwise.  A B D Colour 0 is the diagonal, so E D C (i) A0 = I (identity matrix); @ F @ (ii) every Ai is symmetric; @ G (iii) A A = pk A ; @ i j ∑ ij k H k H G (iv) ∑Ai = J (all-1s matrix). edge (distance 1) yellow i distance 2 blue opposite (distance 3) red

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Ablue = ABCDEFGH AiAj(α, β) = ∑Ai(α,γ)Aj(γ, β) A 0 0 1 0 0 1 0 1 γ 0 0 0 1 1 0 1 0 = γ : (α,γ) has colour i and (γ, β) has colour j B   |{ }| C  1 0 0 0 0 1 0 1    k   = p if (α, β) has colour k. D  0 1 0 0 1 0 1 0  ij    0 1 0 1 0 0 1 0  E   So   k F  1 0 1 0 0 0 0 1  AiAj = ∑pijAk.     k G  0 1 0 1 1 0 0 0     1 0 1 0 0 1 0 0  For an association scheme, H     we turn counting into .

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How many edges of colour i at each point? How many edges of colour i at each point in this association scheme?

AiAi(α,α) = ∑Ai(α,γ)Ai(γ,α) γ = γ : (α,γ) has colour i and (γ,α) has colour i |{ }| i ai = γ : (α,γ) has colour i by symmetry white 1 |{ }| purple 8 blue 4 0 = pii. black 2

0 Put ai = pii. Then there are precisely ai edges of colour i at each point.

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The trivial association scheme on 10 points The divisible association scheme GD(3,5).

A0 = I The set Ω is A = J I 1 − partitioned into 3 “groups” of size 5. This is the only association scheme Different points in with rank 2. the same “group” are 1st associates (green). Points in different “groups” are 2nd associates (red).

11/42 12/42 Association schemes in design of The Hamming association scheme Hamm(4,2).

Suppose that we are going to do an to compare 5 varieties of wheat. The points are We can use 15 plots of land in a single field. 4 vectors in Z2. If the plots are all alike, they form the trivial association scheme How many equal on 15 plots. coordinates? If the plots are not all alike, we may have to group them into 4 white three blocks: one block has the five plots near to the trees; 3 yellow another block has the five stony plots; and the third block has 2 black the five plots near to the stream. Now we have the group 1 green divisible association scheme GD(3,5). 0 red

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The cube association scheme is Hamm(3,2) Association schemes in

000 001 011 010 100 101 111 110 You receive the following top-secret message: 100 101 000 @ SHOOT BORDS @ 001 @ @ 011 Should you 000 001 010 SHOOT BORIS? 100 011010 @ 101 SHOOT BIRDS? @ @ 111 SHOOT LORDS? @ 110 111110 SHOOT BONDS? edge (distance 1) yellow 2 equal coordinates distance 2 blue 1 equal coordinate Coded versions of possible messages should have opposite (distance 3) red 0 equal coordinates large Hamming distance between them.

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The Johnson association scheme J(6,3). Association schemes in genetics

12 13 14 15 23 24 25 34 35 45 12 13 14 The points are 15 subsets of size 3 of a set of size 6. 23 What is the size of 24 their intersection? 25 34 3 white 2 green 35 1 black 45 0 red Types of offspring from cross-breeding five pure breeds of dog. Two different types may have one or zero parental lines in common. This is J(5,2). 17/42 18/42 How do we verify the hard condition? Where can I get to in two red steps?

12 13 14 15 23 24 25 34 35 45 We need to evaluate A A for 0 i r 1 and 0 j r 1. i j ≤ ≤ − ≤ ≤ − 12 A = I and IA = A I = A , so 0 i i i 13 we only need to evaluate AiAj for 1 i r 1 and 1 j r 1. ≤ ≤ − ≤ ≤ − 14 If Ai does not have a constant number of 1s in each row and column then we do not have an association scheme. 15 If Ai has exactly ai entries 1 in each row and column then 23 AiJ = aiJ. So 24 Ai(A0 + + Ar 1) = AiJ = aiJ = ai(A0 + + Ar 1). So ··· − ··· − 25 we only need to evaluate AiAj for 1 i r 2 and 1 j r 2. ≤ ≤ − ≤ ≤ − 34 If A A = ∑ pk A then A A is symmetric, so i j k ij k i j 35 k AiAj = (AiAj)> = A>A> = AjAi = ∑ p Ak, so j i k ij 45 we only need to evaluate A A for 1 i j r 2. i j ≤ ≤ ≤ − 2 A2 = 3I + 0A + 1A . When r = 3 we only need to evaluate A1. red red blue

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Strongly regular graphs Association schemes of rank 3

A graph is strongly regular if 0 I every point is in p11 edges; 1 I every edge is in p11 triangles; 2 I every non-edge is in p11 two-paths; I the graph is neither null nor complete.

...... An association scheme has rank 3 ...... 12...... if and only if ...... v ...... each of its non-identity classes forms the edges of a strongly ...... regular graph...... 35 ...... v...... 4534 ...... 0 ...... p = 3 v...... v 11 ...... 15 ...... 23 ...... v ...... v ...... 1 ...... p = 0 ...... 11 ...... 2 ...... p11 = 1 ...... 2414 ...... v v ...... 25 ...... 13 21/42 22/42 v v

Distance-regular graphs The icosahedral association scheme

Suppose that a graph G with v vertices is connected and simple (no multiple edges) and has diameter (maximum distance) d. Let A be the v v matrix whose (α, β)-entry is equal to i × 1 if the distance from α to β in G is i The 12 vertices and (0 otherwise 30 edges of the icosahedron form a distance-regular The graph G is distance-regular if A A is a linear combination 1 i graph with of Ai 1, Ai and Ai+1 for all i. − diameter 3. Theorem distance 1 turquoise If G is distance-regular with diameter d then its distance classes form distance 2 blue an association scheme of rank d + 1. distance 3 purple

If r 4 then not all association schemes of rank r arise in this ≥ I have a wonderful cardboard model of this . . . but it was too way. large to fit on the train. 23/42 24/42 Cyclic association schemes (circuits) Association schemes and permutation groups

The n-circuit is an association scheme with rank n 1 1 + − . 2   If G is a transitive permutation group on Ω, If the distance from α to β is d then (α, β) has colour d. it induces a permutation group on Ω Ω. × ...... Give (α, β) the same colour as (γ,δ) if and only if ...... g g ...... there is some g in G with (α , β ) = (γ,δ)...... v ...... The colour classes are the orbitals of ...... G ...... I ...... Transitivity = (i); ...... ⇒ ...... I v...... v These are also the orbitals are self-paired (ii); ...... ⇐⇒ ...... distance-regular ...... I (iii) is always satisfied...... graphs...... v v 25/42 26/42

The Bose–Mesner Real symmetric matrices

(i) A = I (identity matrix); 0 Theorem (ii) every Ai is symmetric; k If A is an n n real then the following hold. (iii) AiAj = ∑pijAk; × k (i) All eigenvalues of A are real. (iv) ∑Ai = J (all-1s matrix). (ii) If the distinct eigenvalues of A are λ1,..., λs, then the i minimal of A is (X λ )(X λ ) (X λ ). − 1 − 2 ··· − s (iii) Eigenvectors of A corresponding to different eigenvalues are Let be the set of all real linear combinations of the Ai. A orthogonal to each other (w. r. t. the usual inner product). This is a real vector space. (iv) A is diagonalizable, The matrices A0,..., Ar 1 are linearly independent n − which that R has a basis consisting of eigenvectors of A, (in position (α, β), only one of the A-matrices is non-zero), n which means that R = W1 Ws, so one basis for is A0,A1,...,Ar 1 , and the dimension of ⊕ · · · ⊕ A { − } A where W is the eigenspace for eigenvalue λ . is equal to the rank of the association scheme. i i (v) Let Qi be the matrix of orthogonal projection onto Wi Condition (iii) shows that is closed under multiplication, so A (this means that Qiv = v if v Wi and Qiv = 0 if v W⊥). it is an algebra. ∈ ∈ i Then Qi is a polynomial in A. A A = A A for all i and j, so is a . i j j i A 27/42 28/42

How do you find the eigenvalues of a real symmetric The easy fruit matrix A?

I Solve the equation det(A xI) = 0 − If u is the all-1 vector then Aiu = aiu (Cambridge Maths students). (because Ai has ai non-zero entries in each row and column). I Put it into Matlab So u is an eigenvector of all matrices in , so we can confine the (Lisbon students). A rest of our search to vectors which are orthogonal to u. I If the matrix is patterned, guess an eigenvector and try it (RAB). I Find the minimal polynomial.

29/42 30/42 A lucky guess for the cube Finding the minimal polynomial: I

12 13 14 15 23 24 25 34 35 45 12 A 0 0 1 0 0 1 0 1 1 1 − 13 B  0 0 0 1 1 0 1 0   1   1  − 14 C  1 0 0 0 0 1 0 1   1   1       −        15 D  0 1 0 0 1 0 1 0   1   1  Ablue =     =  −  23 E  0 1 0 1 0 0 1 0   1   1     −          24 F  1 0 1 0 0 0 0 1   1   1     −          25 G  0 1 0 1 1 0 0 0   1   1     −    34 H  1 0 1 0 0 1 0 0   1   1     −          35 We have found an eigenvector with eigenvalue 1. − 45 2 = + + = Ared 3I 0Ared 1Ablue and AredJ 3J.

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Finding the minimal polynomial: II Commuting projectors

Theorem A2 = 3I + 0A + 1A and A J = 3J. red red blue red Let P be the matrix of orthogonal projection onto the subspace U, and Put A = Ared. let Q be the matrix of orthogonal projection onto the subspace V. If PQ = QP then PQ is the matrix of orthogonal projection onto the A2 = 3I + (J A I) and AJ = 3J. subspace U V. − − ∩ Proof. 2 A = 2I A + J. 2 2 2 − (PQ) = PQPQ = P Q = PQ, so PQ is idempotent. Since PQ is also symmetric, it is the matrix of orthogonal projection onto its image. Call this W. A3 = 2A A2 + AJ = 2A (2I A + J) + 3J = 2I + 3A + 2J. − − − − v W PQv = v P2Qv = Pv PQv = Pv v = Pv v U. ∈ ⇒ ⇒ ⇒ ⇒ ⇒ ∈ A3 2A2 5A + 6I = 0. Similarly, v W v V, so W U V. − − ∈ ⇒ ∈ ⊆ ∩ If v U V then v = Pv = Qv so PQv = Pv = v so v W. The minimal polynomial is X3 2X2 5X + 6 = ∈ ∩ ∈ 2 − − Therefore U V W, and so U V = W. (X 3)(X + X 2) = (X 3)(X 1)(X + 2). ∩ ⊆ ∩ − − − − 33/42 34/42

Commuting symmetric matrices Mutual eigenspaces of the Bose–Mesner algebra

Theorem The Bose–Mesner algebra consists of symmetric matrices, A Let A and B be real symmetric n n matrices. If AB = BA then there is commutative, and has dimension r, where r is the rank of the × are mutually orthogonal subspaces W1,...,Ws such that association scheme. n So there are mutually orthogonal subspaces W ,..., W of Rn I R = W1 Ws; 1 s ⊕ · · · ⊕ such that if M then each eigenspace of M is either one of I each Wi is contained in an eigenspace of A and an eigenspace ∈ A the W or the direct sum of two or more of W ,..., W . of B; i 1 s I the matrix of orthogonal projection onto each subspace is a Let Pj be the orthogonal projector onto Wj. Then Pj is a polynomial in A0, A1,..., Ar 1, so Pj . polynomial in A and B. − ∈ A P1,..., Ps are linearly dependent because PjP = 0 if j = k. Proof. k 6 Let be the eigenvalue of on . Then Let the eigenprojectors of A be P1,..., Pm and the cij Ai Wj

eigenprojectors of B be Q1,..., Qt. The former are s in A, and the latter are polynomials in B, so they commute. A = c P for i = 0, . . . , r 1. i ∑ ij j − Apply the previous theorem to each nonzero product PiQj. j=1 2 I = I = (P1 + + Pm)(Q1 + + Qt) = ∑i ∑j PiQj, ··· ··· Hence P ,...,P is another basis for , and so s = r. so k Wk is the whole space. { 1 s} A L 35/42 36/42 Strata The character table

We have seen that if cij is the eigenvalue of Ai on Wj then

r The subspaces W1,..., Wr are called strata. Ai = ∑ cijPj for i = 0, . . . , r 1. j=1 − The projectors Pj are sometimes called minimal idempotents. The dimension d = dim(W ) is sometimes called j j Let C be the r r matrix with entries c . The columns of C are the (number of) degrees of freedom for W . × ij j called the characters of the association scheme, and C is called The 1-dimensional space spanned by the all-1 vector is always the character table of the association scheme. a stratum. (The conventions are slightly different from those in group theory.) 1 To express Pj as a linear combination of the Ai, we need C− .

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Two bases Possible generalizations

A0,...,Ar 1 and P1,...,Pr are both bases for . { − } { } A A0,...,Ar 1 is more useful for interpretation, and is easier for { − } doing addition. I Allow Ai and Ai> to be two different colours, P ,...,P is easier for doing multiplication, including finding but still insist on commutativity { 1 r} inverses (and generalized inverses). (P. Delsarte, coding theory, 1973, thesis) I Allow A and A to be two different colours, To transfer back and forth between the two bases, we need to i i> but do not insist on commutativity know both C and C 1. − (natural approach for permutation groups, Theorem I. Schur, 1933; H. Wielandt, 1964; D. G. Higman, 1964, 1967) (now called homogeneous coherent configurations)

1 1 1 1 I Allow the diagonal to be more than one class, C− = diag(d1,...,dr)C> diag ,..., . Ω a0 ar 1 which forces the loss of commutativity and symmetry | |  −  (called coherent configurations by D. G. Higman). In general, there is no easy way of finding C, and no natural labelling for the strata.

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Some history: I Some history: II

I Relevant idea for permutation groups (I. Schur, 1933). I Definition of association scheme, in (R. C. Bose and K. R. Nair, 1939). I Coherent configurations (Higman, 1975, 1976, after ten I A strongly regular graph on 100 points with valency 22 years of lecturing on them). (D. M. Mesner, 1956, thesis). I Breakdown of barriers with U.S.S.R., and realisation that I Bose–Mesner algebra (Bose and Mesner, 1959). cellular are the same as coherent configurations I Multidimensional association schemes (J. N. Srivastava, (1990). 1961, thesis; Bose and Srivastava, 1964; much more by I Death of Mesner, and realisation that he had Srivastava over the years). pre-discovered the Higman–Sims graph (2002). I Effectively homogeneous coherent configurations, but I Death of Higman (13/2/2006). unnamed (C. R. Nair, 1964) I A strongly regular graph on 100 points with valency 22 I Death of Srivastava (18/11/2010). and large automorphism group (D. G. Higman and I Realisation that multidimensional association schemes are C. C. Sims, 1968). the same as coherent configurations (2011). I Cellular algebras (B. Yu. Weisfeiler and A. A. Lehman, 1968, possibly earlier in Russian?; much more by others in the U.S.S.R. over the years). 41/42 42/42