Association Schemes Association Schemes: Definition an Association
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Association schemes: definition Association schemes An association scheme of rank r on a finite set 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. 1/42 2/42 An association scheme with 15 points and 3 associate An association scheme defined by a Latin square 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 matrix 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 5/42 6/42 One adjacency matrix for the cube That hard condition 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 matrix multiplication. 7/42 8/42 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. 9/42 10/42 The trivial association scheme on 10 points The group 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 experiments The Hamming association scheme Hamm(4,2). Suppose that we are going to do an experiment 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 13/42 14/42 The cube association scheme is Hamm(3,2) Association schemes in coding theory 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. 15/42 16/42 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 19/42 20/42 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 ............. ............. ............. ............. ............. ............