Association schemes with given stratum dimensions: on a paper of Peter M. Neumann

In January 1969, Peter M. Neumann wrote a paper entitled"Primitive permutation groups of degree 3p". The main theorem placed restrictions on the parameters of a primitive but not 2-transitive permutation group of degree three times a prime. The paper was never published, and the results have been superseded by stronger theorems depending on the classification of the finite simple groups, for example a classification of primitive groups of odd degree. However, there are further reasons for being interested in this paper. First, it was written at a time when combinatorial techniques were being introduced into the theory of finite permutation groups, and the paper gives a very good summary and application of these techniques. Second, like its predecessor by Helmut Wielandt on primitive groups of degree 2p, it can be re-interpreted as a combinatorial result concerning association schemes whose common eigenspaces have dimensions of a rather limited form. This result uses neither the primality of p nor the existence of a permutation group related to the combinatorial structure. We extract these results and give details of the related combinatorics.

The proof of this theorem is also given in Chapter 5 of his book [19].It illustrates an extension of the methods of Schur rings using representation theory.He mentioned that, for a = 1, we have two examples: the groups S 5 and A 5 , acting on the set of 2-element subsets of {1, . . ., 5}.Now it is possible to show that there are no others.For example, using the Classification of Finite Simple Groups, all the finite primitive rank 3 permutation groups have been determined [8,10,12], and the observation can be verified by checking the list.
However, there is more to be said.Wielandt's proof falls into two parts.The first involves showing that the permutation character of G decomposes as 1 G + χ 1 + χ 2 , where 1 G is the principal character of G and χ 1 , χ 2 are irreducibles with degrees p − 1 and p.It follows from this that G has rank 3 and is contained in the automorphism group of a strongly regular graph, having the property that the eigenvalues of its adjacency matrix have multiplicities 1, p − 1, and p.Now the argument shows something much more general.Neither the existence of a rank 3 group of automorpisms nor the primality of p are needed.
First, a definition: a graph Γ is strongly regular with parameters (n, k, λ, µ) if it has n vertices, every vertex has k neighbours, and two vertices have λ or µ common neighbours according as they are joined by an edge or not.Every rank 3 group of even order is an automorphism group of a strongly regular graph, but not conversely; many strongly regular graphs have no non-trivial automorphisms.Any regular graph has the all-1 vector as an eigenvector; a regular graph is strongly regular if and only if its adjacency matrix, acting on the space orthogonal to the all-1 vector, has just two eigenvalues.
Theorem 1.2.Let Γ be a strongly regular graph on 2n vertices, with the property that the eigenvalues of the adjacency matrix, on the space of vectors orthogonal to the all-1 vector, have multiplicities n − 1 and n.Then either (a) Γ is a disjoint union of n complete graphs of size 2, or the complement of this; or (b) for some positive integer a, we have n = 2a 2 + 2a + 1, and up to complementation the parameters of the graph Γ are given by 2n = (2a + 1) 2 + 1, k = a(2a + 1), λ = a 2 − 1, µ = a 2 .
We are not aware of who first pointed this out.The result is given, for example, as [1,Theorem 2.20].
In the case a = 1, the complementary strongly regular graphs are the line graph of the complete graph K 5 and the Petersen graph.But, unlike in Wielandt's case, there are many others.For example, suppose that there exists a Steiner system S(2, a + 1, 2a 2 + 2a + 1).Then the strongly regular graph whose vertices are the blocks, two vertices adjacent if the corresponding blocks intersect, has the parameters given in the theorem.For example, when a = 2, the two Steiner triple systems on 13 points give non-isomorphic strongly regular graphs on 26 vertices.(We discuss examples further in the last section.) Now to the subject of this paper.In 1969, Peter Neumann wrote a long paper [13] extending Wielandt's result from 2p to 3p, where p is prime.His conclusion is that, if such a group is not 2-transitive, then p is given by one of three quadratic expressions in a positive integer a, or one of three sporadic values; the rank is at most 4, and the subdegrees are given in each case.Like Wielandt's, Neumann's proof falls into two parts: first find the decomposition of the permutation character, and then in each case find the combinatorial implications for the structure acted on by the group.In contrast to Wielandt, the first part is much easier, since in the intervening time, Feit [3] had given a characterisation of groups with order divisible by p having a faithful irreducible representation of degree less than p − 1.On the other hand, the second part is much harder; rather than just one possible decomposition of the permutation character, he finds eight potential decompositions, some of which require many pages of argument.
Again like Wielandt's, Neumann's conclusions have been superseded by results obtained using the classification of finite simple groups.For example, all the primitive permutation groups of odd degree have been classified [7,11].
The paper was never published.It happened that both Leonard Scott and Olaf Tamaschke had produced similar results.There was a plan for Neumann and Scott to collaborate on a joint paper, but for unknown reasons this never happened.The authors are grateful to Leonard Scott [17] for providing a scan of Peter Neumann's original typescript together with some historical material about the proposed collaboration.The second author has re-typed the paper and posted it on the arXiv [14].
Our task is to produce a combinatorial version of this, as we have seen for Wielandt's theorem.We give some historical background to the theorem with some comments on the place of Neumann's paper in the introduction of combinatorial methods into the study of permutation groups, and to check in detail that his arguments give combinatorial results which do not depend on either the existence of a primitive group or the primality of p. Indeed we find some families of parameters which do not occur in Neumann's case since the number of vertices is even.

History
The 1960s saw a unification of combinatorial ideas which had been developed independently in three different areas of mathematics.In statistics, R C. Bose and his colleagues and students developed the concept of an association scheme.Extracting information from experimental results requires inversion of a large matrix, and Bose realised that the task would be much simpler if the matrix belonged to a lowdimensional subalgebra of the matrix algebra; requiring entries to be constant on the classes of an association scheme achieves this.In the former Soviet Union, Boris Weisfeiler and his colleagues were studying the graph isomorphism problem, and developed the concept of a cellular algebra, an isomorphism invariant of graphs, to simplify the problem, and an algorithm, the Weisfeiler-Leman algorithm, to construct it.In Germany, Helmut Wielandt was extending the method of Schur rings to study permutation groups with a regular subgroup; by using methods from representation theory he was able to dispense with the need for the regular subgroup.These techniques were further developed by Donald Higman in the USA, under the name coherent configuration.
The three concepts are very closely related.We begin with Higman's definition.A coherent configuration consists of a set Ω together with a set {R 1 , R 2 , . . ., R r } of binary relations on Ω with the properties The number r is the rank of the configuration.Combinatorially, a coherent configuration is a partition of the edge set of the complete directed graph with loops.
If G is a permutation group on Ω, and we take the relations R i to be the orbits of G on Ω 2 , we obtain a coherent configuration.This was Higman's motivating example, which he called the group case.Not every coherent configuration falls into the group case; indeed, our task is to extend Neumann's results from the group case to the general case.
A coherent configuration is homogeneous if the diagonal is a single relation.In the group case, this means that the group is transitive.All the configurations in this paper will be homogeneous.
The notion of a cellular algebra is the same apart from an inessential small difference (the diagonal is replaced by some equivalence relation).Association schemes form a special case, where all the relations R i are symmetric.It follows that, in an association scheme, the diagonal is a single relation.(Statisticians deal with symmetric matrices, for example covariance matrices.) A coherent configuration with rank 2 is trivial: one relation is the diagonal, the other is everything else.For rank 3, we can suppose without loss that R 1 is the diagonal.There are then two possibilities: • R 3 is the converse of R 2 .Then R 2 is a tournament (an orientation of the edges of the complete graph on Ω); condition (d) shows that it is a doubly regular tournament [16].• R 2 and R 3 are symmetric.Then each is the edge set of a graph, and these graphs are strongly regular [1,Chapter 2].The definition of coherent configuration has an algebraic interpretation.Let A i be the adjacency matrix of the relation R i , the Ω × Ω matrix with (α, β) entry 1 if (α, β) ∈ R i .Then A 1 , . . ., A r are zero-one matrices satisfying the following conditions: (a) A 1 + • • • + A r = J, the all-1 matrix; (b) there is a subset of these matrices whose sum is the identity I; (c) for any i there is a j such that Condition (d) says that the linear span over C of A 1 , . . ., A r is an algebra (closed under multiplication), and condition (c) implies that this algebra is semi-simple.In the group case, it is the centraliser algebra of the permutation group, consisting of matrices which commute with every permutation matrix in the group.In the case of association schemes, it is known as the Bose-Mesner algebra of the scheme.In this case, all the matrices are symmetric, the algebra is commutative, and we can work over R. In the group case, the centraliser algebra is commutative if and only if the permutation character is multiplicity-free.
If the algebra is commutative, then the matrices are simultaneously diagonalisable; the common eigenspaces are called the strata of the configuration.In the rank 3 case where we have a strongly regular graph and its complement, the stratum dimensions are simply the multiplicities of the eigenvalues.We occasionally extend the use of the word "stratum" to the non-commutative case, where it means a submodule for the algebra spanned by the matrices which is maximal with respect to being a sum of isomorphic submodules.
In all cases which arise in Peter Neumann's paper, the algebra turns out to be commutative, although there are two potential cases where the permutation character is not multiplicity-free; both of these are eliminated.
It seems clear to the authors that, had the paper been published in 1969, it would have been very influential: it provides both a clear account of the theory and how it can be used to study permutation groups, and also a non-trivial example of such an application.The second author of the present paper read it at the start of his DPhil studies in Oxford under Peter Neumann's supervision, and considers himself fortunate to have been given such a good grounding in this area; he has worked on the interface of group theory and combinatorics ever since.

The results
The main theorems in this paper are the following.They are numbered to correspond to the eight cases in Neumann's paper.
If the eigenvalues of A 1 have multiplicities 1, n−1 2 , n−1 2 then one of the two following cases must hold: • n ≡ 1 (mod 4) and A 1 and A 2 are the adjacency matrices of conference graphs; • n ≡ 3 (mod 4) and A 1 and A 2 are the adjacency matrices of doubly regular tournaments.
Theorem 3.2.Let G be a strongly regular graph on 3n vertices.If the multiplicities of the eigenvalues of G are 1, n, 2n − 1 then G or its complement have the following parameters in terms of a non-negative integer a: Theorem 3.3.Let G be a strongly regular graph on 3n vertices.If the multiplicities of the eigenvalues of G are 1, 2n, n − 1 then either G or its complement is a disjoint union of n copies of K 3 or G or its complement have the following parameters for some non-negative integer a:

then one of the following hold:
• A 2 = A T 3 and the row sums of A 1 , A 2 ,, and A 3 are n − 2a − 1, n + a, and n + a respectively for some even integer a; 3 and the row sums of A 1 , A 2 , and A 3 are n + 2a + 1, n − a − 1, and n − a − 1 respectively for some odd integer a; • All matrices are symmetric and the row sums of A 1 , A 2 , A 3 are n + 2a + 1, n − a − 1, and n − a − 1 respectively for some non-negative integer a; • all matrices are symmetric and the row sums of A 2 , A 3 and A 4 are 2, n − 1 and 2(n − 1); 2 and the row sums of A 2 , A 3 and A 4 are 1, 1 and 3(n − 1).

Theorem 3.5. There exists no coherent configuration
Theorem 3.6.There is no strongly regular graph on 3n vertices with eigenvalue multiplicities

an association scheme and one of the following hold:
• n = 7 and the row sums of A 1 , A 2 , A 3 are 4, 8, and 8; • n = 19 and the row sums of A 1 , A 2 , A 3 are 6, 20, and 30; • n = 31 and the row sums of A 1 , A 2 , A 3 are 32, 40, and 20.

Theorem 3.8. There exists no coherent configuration
where the multiplicities of the eigenvalues of A 1 , . . ., A 5 are Algebraic Combinatorics, Vol. 6 #5 (2023) 4. The proofs 4.1.A lemma.We start with a lemma that will be used throughout the paper.
Lemma 4.1.Let A be a homogeneous coherent configuration on n points.Suppose that the dimension of a non-trivial stratum for A is at least n/3 − 1.Then one of the following happens: (a) One of the relations in A has at least n/3 connected components.
(b) Any matrix in A has the property that any eigenvalue λ apart from the row sum r satisfies |λ| < r.
Proof.We use the Perron-Frobenius Theorem, see [4].For any non-negative matrix A, one of the following holds: • Under simultaneous row and column permutations, A is equivalent to a matrix of the form B O O C .In our case the constancy of the row sum r means that r has multiplicity equal to the number of connected components; so there are at least n/3 connected components, and (a) holds.• A is decomposable, that is, under simultaneous row and column permutations it is equivalent to a matrix of the form , where X ̸ = O.But this contradicts the fact that the row sum is constant.• A is imprimitive, that is, equivalent under simultaneous row and column permutations to a matrix of the form But then re 2πik/t is a simple eigenvalue for k = 0, 1, . . ., t − 1, contrary to assumption.• A is primitive.Then the Perron-Frobenius Theorem asserts that there is a single eigenvalue with largest absolute value, as required.□ 4.2.Proof of Theorem 3.1.We first prove a lemma about strongly regular graphs that will be used in the proof of Theorem 3.1.
Firstly we define a special type of strongly regular graphs.A conference graph is a strongly regular graph on v vertices with parameters (v Let G be a strongly regular graph with parameters (n, k, λ, µ) and let k, r, s be the eigenvalues of the adjacency matrix of G.If r and s have equal multiplicities then G is a conference graph.
Proof.It is known for a strongly regular graphs that the multiplicities of r and s are respectively.Hence, if f = g then it follows that and thus G is a conference graph, as required.Moreover, Proof of Theorem 3.1.Since A is a coherent configuration, A 0 + A 1 + A 2 = J n and moreover A T i = A j for i, j ∈ {1, 2}.Hence, there are two possibilities.Either A i = A T i for i ∈ {1, 2} or A i = A T j for i, j ∈ {1, 2} and i ̸ = j.In the first case, the graphs with adjacency matrices A 1 and A 2 are undirected.Moreover, since A 1 and A 2 are symmetric, A is an association scheme and hence those graphs are strongly regular and one is the complement of the other.It follows by Lemma 4.2 that A 1 and A 2 are the adjacency matrices of conference graphs and in fact two copies of the same conference graph.Moreover, for a conference graph to exist, it is known that n ≡ 1 (mod 4).
In the second case, since A is a coherent configuration, it follows that A 1 and A 2 must have constant row and column sums and hence their digraphs are regular.Let G 1 , G 2 be the digraphs with adjacency matrices A 1 and A 2 respectively and V be the vertex set of those digraphs.For u, v ∈ V , we write u and vice versa and also that either (A k ) ij = 1 or (A k ) ji = 1 for k ∈ {1, 2}.Hence, G 1 and G 2 are regular tournaments.Also, notice that since A is a coherent configuration, it follows that for m, n ∈ {1, 2}, m ̸ = n, there exists a constant p m mn such that for any i, j ∈ V , such that mn .Hence, both G 1 and G 2 are doubly regular, and it is known that n ≡ 3 (mod 4) for doubly regular tournaments.□ 4.3.Proof of Theorem 3.2.
Proof.Let A 1 be the adjacency matrix of G and A 2 be the adjacency matrix of its complement.Since G is strongly regular, the eigenvalues of A 1 and A 2 have the same multiplicities.Moreover, if Reducing modulo n gives that k i ≡ s i (mod n).Therefore, since k i > s i by Lemma 4.1, it follows that k i − s i = ϵ i n for ϵ i ∈ {1, 2}.Therefore, Assume without loss of generality that ϵ 1 = 1 and ϵ 2 = 2.Then, k 1 = n + s 1 and also Also, we have that Since G is strongly regular and its eigenvalues have different multiplicities, it is not a conference graph, and hence its eigenvalues are integers.Hence, 16n − 5 = 3b 2 for some non-negative integer b.This gives us that 3b 2 + 5 ≡ 0 (mod 16).It follows that b = 3, 5, 11 or 13 (mod 16).We therefore need to examine the following four cases: In this case we get: 16n = 3(16a + 3) 2 + 5 ⇒ n = 48a 2 + 18a + 2. and s 1 = −4a − 1.Notice that only the negative solution works, since 16a + 2 is not divisible by 4. Consequently k 1 = 48a 2 + 14a + 1.We also get r 1 = 8a + 1 Now, using the formulae for the eigenvalues of strongly regular graphs, namely we get Solving this system we obtain λ = 16a 2 + 6a and µ = 16a 2 + 2a.
In this case we get: and s 1 = 4a + 1.Hence, k 1 = 48a 2 + 34a + 6.We also get r 1 = −8a − 3 As above, knowing r 1 , s 1 we can obtain λ and µ which in this case are equal to 16a 2 + 10a + 1 and 16a 2 + 14a + 3 respectively.Proof.Let A 1 be the adjacency matrix of G and A 2 be the adjacency matrix of its complement.Since G is strongly regular we know that the eigenvalues of A 1 and A 2 have the same multiplicities.Also, if Reducing modulo n gives that k i ≡ s i (mod n), and since by Lemma 4.1 either one of A 1 , A 2 is the disjoint union of n copies of K 3 or k i > |s i |.In the second case, it follows that k i − s i = ϵ i n for ϵ i ∈ {1, 2}.Also, as before, ϵ 1 + ϵ 2 = 3 and hence we may suppose without loss of generality that ϵ 1 = 1 and ϵ 2 = 2.Then, k 1 = n + s 1 and r 1 = −s1−1
Proof.Suppose for a contradiction that there exists such a strongly regular graph, and let A 1 be its adjacency matrix and A 2 be the adjacency matrix of its complement and suppose that k 1 , r 1 , s 1 and k 2 , r 2 , s 2 are the eigenvalues of A 1 and A 2 respectively.Then, for i ∈ {1, 2} we get Hence, (2s i − r i ) 2 ≡ 2s 2 i − r 2 i (mod n).By routine calculation, it follows that s i ≡ r i (mod n) and consequently k i ≡ r i (mod n).Therefore, k i = ϵ i n + r i and s i = η i n + r i for some ϵ i , η i ∈ {1, 2}.
Substituting into the trace equations and reducing modulo n 2 gives ).We now collect terms and divide by n and we get 0 it cannot be the case that both r 1 and r 2 are divisible by n.Hence, interchanging A 1 and A 2 if necessary we may assume that r 1 ̸ ≡ 0 (mod n).Then, However, by looking at the formulae for r 1 and s 1 for a strongly regular graph, we deduce that r 1 ̸ = s 1 , a contradiction.Similarly, if k 1 = 2n − 1, then k 2 = n which forces r 2 = s 2 = 0, again a contradiction.Hence, there is no strongly regular graph with those eigenvalue multiplicities.□ 4.6.Proof of Theorem 3.4.
Proof.Suppose first that at least some A i corresponds to a disconnected graph.There are two possibilities: either A 2 (say) is an undirected graph which consists of n disjoint triangles; or A 3 = A ⊤ 2 , corresponding to a converse pair of directed triangles.In the second case, I, A 2 + A 3 , A 4 satisfy the hypotheses of Theorem 3.3, and the last statement holds.So suppose that the first possibility occurs.
Let T 1 , . . ., T n be the connected components of A 2 (each a triangle), and let T i = {v i1 , v i2 , v i3 }.Now the common eigenspaces of I and A 2 have dimensions n and 2n.The first of these has the form V 0 ⊕ V 1 , where V 0 is the space of constant vectors and where n 1 c i = 0. Then V 1 is an eigenspace for A 2 as well; suppose that the eigenvalue is λ.Take a vector of the above form with c i = 1, c j = −1, and c k = 0 for k ̸ = i, j.The fact that it is an eigenvector of A 2 shows that v i1 is joined to −λ of the vertices of T j .So this number is independent of the choice of i and j and the particular vertex v i1 ∈ T i chosen; that is, in the graph A 2 , each vertex of T i is joined to −λ vertices of T j for j ̸ = i.We can suppose that λ = −1, so each vertex of T i is joined to one vertex of T j by an A 2 edge, and to two by A 3 edges.Hence A 2 has valency n − 1 and eigenvalue −1 on W ; for A 3 these numbers are 2(n − 1) and −2.Thus we have the penultimate case in the Theorem.So we may suppose that all the orbital graphs are connected.
Let k i , r i , s i , t i be the eigenvalues of A i , i ∈ {1, 2, 3}, with multiplicities 1, n, n, n−1 respectively.Firstly notice that t i must be a rational integer and r i and s i must either both be rational integers or algebraically conjugate algebraic integers.Then, we get Hence, n must divide k i −t i , and since by Lemma 4.1 and the connectivity assumption k i > t i , it follows that k i = ϵ i n + t i for some ϵ i > 0.Moreover, by [13, Equation (6.9)], ϵ 1 + ϵ 1 + ϵ 3 = 3 and hence ϵ i = 1 for all i ∈ {1, 2, 3}.Thus, k i = n + t i .
There are now two cases to consider.Either all matrices are symmetric or two of them, say A 2 and A 3 without loss of generality are such that A T 2 = A 3 .We first consider the second case.In this case the eigenvalues of A 2 and A 3 are the same.Hence, t 2 = t 3 and either r 2 = r 3 and s 2 = s 3 or r 2 = s 3 and r 3 = s 2 .Notice that the algebra spanned by the matrices of this coherent configuration is commutative and therefore A 2 and A 3 can be simultaneously diagonalised.Let U be the matrix that simultaneously reduces A 2 and A 3 .If r 2 = r 3 and s 2 = s 3 then U −1 A 2 U = U −1 A 3 U , which implies that A 2 = A 3 , a contradiction.Hence, r 2 = s 3 and r 3 = s 2 .Now adding A 2 and A 3 together produces an association scheme of the type arising in Theorem 3.3.Hence, n = 3a 2 +3a+1 and either We now show that if k 1 = n − 2a − 1 then a is even and if k 1 = n + 2a + 1 then a is odd.In the first case, the remaining eigenvalues of A 1 , A 2 , and A 3 are as shown below: and [13, Equation (6.8)] gives 3n(n + a)p 3 22 = (n + a) 3 + nrs(r + s) + (n − 1)a 3 .where p 3  22 is as defined on page 3. Eliminating rs and simplifying gives p 3 22 = a 3 + 3a 2 and since p 3 22 ∈ Z, a must be even.In the second case, the eigenvalues of A 1 , A 2 , and A 3 are the ones given below: where r +s = 1 by [13,Equation (6.6)].And [13, Equation (6.7)] gives rs = 1  2 a(5a+3) and from [13, Equation (6.8)] we get 3n(n − a − 1)p 3 22 = (n − a − 1) 3 + nrs(r + s) − (n − 1)(a + 1) 3 .Simplifying gives p 3 22 = a 2 + a−1 2 , and since p 3 22 ∈ Z, it follows that a is odd, as claimed.
We now consider the symmetric case.We get the following equations . From this we get 2r i s i = 1 + t i + 2t 2 i − 2n and hence we deduce that t i is odd.Also, we can calculate r i and s i and we find that r i , Without loss of generality we set Since A i is symmetric for all i ∈ {1, 2, 3}, it has real eigenvalues, and therefore (1) 3t 2 i ⩽ 4n − 1.Now, from [13, Equation (6.9)] we get (2) Since , not all of them can be negative.Let b be one of them such that b ⩾ 0.Then, it follows by ( 3) and ( 4 Setting n + b = x and refactorising we get the following quadratic in terms of x: By definition x is real and hence, the discriminant of this quadratic must be nonnegative.Therefore, By (5) we have that w < 2b + 1.Now since w ⩾ 0 it follows that 2b + 1 < 4b + 2 − w ⩽ 4b+ 2. Now, by (6), we get that w ⩽ 0 and hence w = 0, as claimed.Therefore, by ( 5), 4n − 1 = 3b 2 and hence b must be odd.We therefore set b = 2a + 1 for a ⩾ 0 and it follows that n = 3a 2 + 3a + 1.Now suppose without loss of generality that t 1 was b.Then from (2) and therefore t 2 = ±t 3 .
But we know that t 2 + t 3 = −1 − t 1 ̸ = 0 and hence Hence, Moreover, since we have shown that t i is odd, a must be even and Proof.As in 3.4, assume first that at least one of the graphs , then symmetrising as before gives an example with multiplicities 1, n − 1, 2n, as in Theorem 3.4; but before this symmetrisation there are six linearly independent matrices, which cannot have just four eigenspaces.So suppose that A 2 is symmetric, and corresponds to n disjoint triangles as in the case of Theorem 3.4.Then again the space V 1 of that proof is an eigenspace for all the matrices, and so for i ̸ = j a vertex in T i is joined to a constant number of vertices in T j .Then there can be at most three such matrices, or five altogether in the configuration (including I and A 2 ), contrary to assumption.
Let k i , r i , s i , t i be the eigenvalues of A i for i ∈ {1, . . ., 5} with multiplicities 1, n, n, n − 1 respectively.If the matrices Θ i,1 are as in [13], then they must be 2 × 2 matrices with eigenvalues r i , s i , where r i and s i are the eigenvalues of A i with multiplicity n.We know that r i , s i must necessarily be rational integers.Now from the linear trace equation we deduce that n must divide k i − t i and since by Lemma 4. In this section we deal with the cases arising in Theorems 3.7 and 3.8 together.We prove both statements through a series of lemmas that eliminate the case arising in Theorem 3.8 and force the parameters stated in Theorem 3.7.
Proof.Suppose for a contradiction that this is not the case.Then, since A is a homogeneous coherent configuration, one of the matrices say A 1 must be symmetric and A 2 , A 3 are such that A T 2 = A 3 .Then, A 2 and A 3 would have the same eigenvalues.Let k i , r i , s i , t i for i ∈ {1, 2, 3} be the eigenvalues of A 1 , A 2 , A 3 respectively with multiplicities 1, n + 1, n − 1, n − 1 respectively.Then, since A 2 = A T 3 , A 2 and A 3 have the same eigenvalues with the same multiplicities.Hence, s 2 + t 2 = s 3 + t 3 .But then, since by [13, Equation (6.9)] it follows that s 1 = t 1 and thus A 1 has three eigenvalues with multiplicities 1, n + 1, and 2(n − 1).However, Theorem 3.6 such a matrix cannot exist, a contradiction.Therefore, all matrices of A must be symmetric.□ For the remainder of the section, given a coherent configuration B we consider the association scheme A arising by adding every non-symmetric matrix and its transpose together to make a symmetric matrix.In this case notice that if B i has eigenvalues Proof.By the linear trace relation for A i we get Hence, k i ≡ −2r i (mod n − 1) and we can write Since by [13, Equation (6.9)] i r i = −1, it follows that i ϵ i = 3.Now suppose for a contradiction that ϵ i < 0. Since k i ⩾ 0 it follows that r i < 0. In particular, since by Lemma 4.1 |r i | < k i we have that −r i < (n − 1)ϵ i − 2r i and hence |r i | > n − 1 and thus |r i | ⩾ n.By the quadratic trace relation we get 2 .Hence, Dividing through by n and applying sqare roots gives us 2 < n, a contradiction.Hence ϵ i ⩾ 0 for all i.At this point, notice also that this bound on r i holds independently of the assumption on ϵ i and this claim will also be used later.□ Now considering the quadratic trace equation again and reducing modulo n − 1 we get

Association schemes with given stratum dimensions
We now show that in fact n − 1 divides 3r i (r i + 1).Lemma 4.5.If r i is as defined above, then n − 1 divides 3r i (r i + 1).
Proof.The trace equations give Now s i t i is a rational integer by assumption and also i − ϵ i is a product of consecutive integers it is even and hence 2(n − 1) must divide 6r i (r i + 1) and hence n − 1 divides 3r i (r i + 1), as claimed. □ We now prove another inequality that we will use later.
Consider the quadratic equation whose roots are s i and t i .Since s i and t i are real, it follows that the discriminant of this equation, namely ( 2 and hence using the trace equations we get 6n(ϵ i (n − 1) − 2r i ) − 2(ϵ i (n − 1) − 2r i ) 2 − 2(n + 1)r 2 i − (n − 1)(ϵ i + r i ) 2 ⩾ 0. This can be rearranged to give the required statement.□ From Lemma 4.4 we know that either one of the ϵ i 's is zero say ϵ 1 without loss of generality, or there are just three non-identity matrices and ϵ 1 = ϵ 2 = ϵ 3 = 1.We first consider the former case.In the case that ϵ 2 = ϵ 3 = ϵ 4 = 1 the inequality from Lemma 4.6 gives us 13 − 12r i − 9r 2 i ⩾ 0 for i ∈ {2, 3, 4} and since r i is integer, −2 ⩽ r i ⩽ 0. Since by [13, Equation (6.9)] r 1 , r 2 , r 3 , r 4 must sum up to −1, it follows that r 2 , r 3 , r 4 must sum up to 1, but this cannot hold since none of them can be positive.Now we examine the case where we have four matrices and ϵ 2 = 1 and ϵ 3 = 2.In this case Lemma 4.6 gives us The only way r 2 and r 3 could sum up to 1 is r 2 = 0 and r 3 = 1.In this case we get k 1 = 4, k 2 = 2, k 3 = 2 and checking for such coherent configurations in [6] we find that there is a unique coherent configuration with such row and column sums, but checking the rational eigenvalues using GAP [5] shows that the r i 's are not equal to −2, 0, 1 as we wish and hence there is no such association scheme.
First we look at the case where ϵ 2 = ϵ 3 = ϵ 4 = 1.By Lemma 4.6 we get −7r 2 i − 10r i + 17 ⩾ 0 Since r i is integer for i ∈ {2, 3, 4} this gives Again in this case we want the r i 's for i ∈ {2, 3, 4} to sum up to 1 but none of them is positive, so this case cannot hold.Now let ϵ 2 = 1 and ϵ 3 = 2.In this case Lemma 4.6 gives The only combination that could work is r 2 = 0 and r 3 = 1.In this case we would get k 1 = 4, k 2 = 3, k 3 = 4. Checking in [6] we do not find any coherent configurations with such row and column sums and appropriate eigenvalues and hence n = 4 cannot hold either.
For ϵ 2 = 1, ϵ 3 = 2, as shown in [13] we need k 1 = 4, k 2 = 8, k 3 = 8 and looking at [6] we deduce that there is a unique coherent configuration with such matrix row and column sums and hence, it is the one arising in [13].The corresponding s i 's and t i 's can be calculated to be Case 4: r 1 = −3, n = 10.
In this case it suffices to check the subcase ϵ 2 = 1, ϵ 3 = 2, since r 1 is odd and hence it cannot be the case that A 1 is the sum of a matrix and its transpose.Therefore, all the matrices in the initial coherent configuration must be symmetric and we must have four of them.In this case by Lemma 4.6 we get The (r 2 , r 3 ) pairs consistent with [13, Equation (6.9)] are (2, 0), (1, 1), (0, 2), (−1, 3) and all of those give row and column sums for which an association scheme does not exist.In this case, as shown in [13] k 1 = 6, k 2 = 20, k 3 = 30 and the corresponding s i 's and t i 's are .

□
We now deal with the case where ϵ 1 = ϵ 2 = ϵ 3 = 1.Notice that in this case, since the ϵ i 's are all odd, B = A and by Lemma 4.3 all matrices are symmetric.
Proof.Suppose for a contradiction that this is not the case and without loss of generality, let r 1 = r 2 .Then, since ϵ 1 = ϵ 2 , it follows that k 1 = k 2 .Thus, either s 1 = t 1 and s 2 = t 2 or s 1 = t 2 and s 2 = t 1 , and since our coherent configuration has rank 4, the matrices are simultaneously diagonalisable and it follows that But this means that s 3 = t 3 and thus A 3 is a matrix of the kind that Theorem 3.6 forbids, a contradiction.□ Lemma 4.9.Let a i = 3ri(ri+1) n−1 . Then, a i ⩽ 4 and if r i ⩾ 0, then a i ⩽ 3. Proof.Suppose for a contradiction that without loss of generality, a 1 = a 2 .Then, both r 1 and r 2 are roots of the equation 3r(r + 1) − a 1 (n − 1) = 0.
Since by Lemma 4.8 r 1 ̸ = r 2 , we must have r 1 +r 2 = −1.But from [13, Equation (6.9)], r 1 + r 2 + r 3 = −1 and hence r 3 = 0.But then, a 3 = 0, a contradiction.□ Lemma 4.11.If a > 0 and r is a root of the equation Proof.Notice that r + , it follows that r i is a root of the equation By Lemma 4.11 we get that where |η i | <  By interchanging A 2 and A 3 if necessary we may assume that r ⩾ 0. Then since by Lemma 4.8, r 1 , r 2 , r 3 are all different, it follows that r ̸ = 0 and r ̸ = 1.Hence, r ⩾ 2.Moreover, from Lemma 4.9 we know that 6r n − 1 • r + 1 2 ⩽ 3.
It follows that r + 1 ⩽ 6 and if 6r ̸ = n − 1 then since n − 1 divides 6r, r + 1 ⩽ 3. Now considering that 6r n−1 • r+1 2 must be integer and that the above inequality must hold for our choices of n and r we can check all cases and find that the only possibilities are: For n = 7 we see that k 1 , k 2 , k 3 are equal to 8, 2, 10 respectively and checking in [6], we see that there is no association scheme with such row and column sums.
If n = 19, then the trace equations give

Examples
In this section we provide examples with the parameters found in Theorems 3.1 to 3.8, in cases where they are known to exist.5.2.Theorem 3.2.For the second set of parameters arising in Theorem 3.2, a known example (with a = 0) is the triangular graph T (6) and its complement; no further examples are known.For the other sets of parameters, no known example with fewer than 512 vertices is known.Moreover, due to the large number of vertices that the given parameters force, it would be very hard to construct one.(2,3,19) is 11, 084, 874, 829 (see [9]); these give pairwise non-isomorphic graphs.There is no known example of graphs with the second set of parameters, and the nonexistence in the case a = 2 has been shown by Wilbrink and Brouwer [20].and PSL (2,19) respectively.These can be found in the GAP [5] database of primitive permutation groups as PrimitiveGroup(21,1) and PrimitiveGroup(57,1) respectively.
The database [6] gives the basis matrices for the first of these, and certifies its uniqueness.In the second case, the association scheme is also known to be unique [2]; the graph of valency 6 is the distance-transitive Perkel graph [15].Existence in the final case with 93 points is undecided, as far as we know.

Therefore, s 1 Case 1 : r 1 =
and t 1 are equal to 2 and −1 respectively.However, r 1 = −1 and k 1 = 2 but by Lemma 4.1 |s 1 | < k 1 , a contradiction.Hence, k 1 = 2 cannot hold.It now follows by Lemma 4.5 that n − 1 divides 18 or n − 1 divides 6.Using the inequality from Lemma 4.6 we deduce that either r 1 = −3 and n = 10 or n = 19, or r 1 = −2 and n = 3, 4, or 7. Now define A = {A i | ϵ i = 0}.Then, A must be a symmetric matrix of row sum k = k i and eigenvalue r = r i .What we have said above for matrices A i with ϵ i = 0 applies to A as well and therefore A must consist of only one summand, A 1 without loss of generality.Now since by Lemma 4.4 ϵ i = 3 there are two possibilities.There are either 5 matrices and ϵ 2 = ϵ 3 = ϵ 4 = 1 or there are 4 matrices and ϵ 2 = 2 and ϵ 3 = 1.Now we check this case individually to see which of those can hold.−2, n = 3.

5. 1 .
Theorem 3.1.The classic examples of symmetric conference graphs are the Paley graphs.The vertex set of such a graph is the set of elements of a finite field whose order is congruent to 1 (mod 4), and two vertices are connected by an edge if and only if their difference is a square in the field.Similarly, the classic examples of doubly regular tournaments are the Paley tournaments; the vertex set is the set of elements of a finite field of order congruent to 3 (mod 4), with an arc from a to b if b − a is a square.

5. 3 .
Theorem 3.3.For the first set of parameters arising in Theorem 3.3 and for a ⩾ 2, the graphs arising from Steiner systems of the type S(2, a + 1, n) with a ∈ {1, 2, 3} are known examples.The number of non-isomorphic Steiner systems

5. 4 . 5 . 5 .
Theorem 3.4.We do not have any examples for the first three cases of this theorem.For the last two (imprimitive) cases, there are examples.Neumann gives examples for the fourth case based on the group PSL(2, p − 1) where p is a Fermat prime.For the fifth case, the wreath product of the cyclic group of order 3 with any 2-transitive group of degree n gives an example.Theorem 3.7.The cases n = 21 and n = 57 are realised by the groups PGL(2, 7) Reducing modulo 2(n + t i ), we have 2n 2 − 6n ≡ 2t i (t i + 3), and after simplifying we deduce that 2 + t 2 ) − 2n(n − 1) = 0. i .Substitution for k i , r i , s i in terms of t i and algebraic manipulation gives2n 2 − 6n + 6t 2 i + 2t 3 i + (1 + t i )(4t 2 i − t i + 1) ≡ 0 (mod 6(n + t i )).i + 1)(3t i + 1) ≡ 0 (mod 2(n + t i )).
Now, if r i ⩾ 0, we get a i < 4, and hence a i ⩽ 3.If r i < 0 and n ⩾ 19, using the inequality from Lemma 4.4 stating that r i < 3 < 5 and so a i ⩽ 4. Now, if n < 19 and r i ⩽ 0 checking gives that a i ⩽ 3. □ Lemma 4.10.If none of a 1 , a 2 , a 3 are zero, then a 1 , 2 , a 3 are all different.
Suppose that this is not the case.Then, by Lemma 4.10, a 1 , a 2 , a 3 are all different.Since a i = 3ri(ri+1) a , as claimed.□Lemma 4.12.One of a 1 , a 2 , a 3 must be zero.Proof.