A note on [D. G. Higman 1988 : Strongly regular designs and coherent configurations of type [{3\atop {\;}}\;\;{2\atop 3}]]
aa r X i v : . [ m a t h . C O ] F e b A note on [D. G. Higman 1988 : Strongly regular designs andcoherent configurations of type [ ]] Akihide HanakiDepartment of Mathematics, Faculty of Science,Shinshu University, Matsumoto 390-8621, Japane-mail : [email protected]
Abstract
In 1988, D. G. Higman gave properties of parameters of strongly regular designs without proof. Itincluded an error. We will give proofs and correct the error.
In 1988, D. G. Higman gave properties of parameters of strongly regular designs without proof in [2].He wrote “The details, which are routine, will be omitted here”. In 2014 at Villanova, Mikhail Klin saidto the author that Higman’s paper included an error. We will give proofs and correct the error.This short note was first written in 2016. The author would like to submit this by a request of MikhailKlin and Alyssa Sankey.We will consider the following fifteen equations in [2, page 414].(1) f = f ,(2) n S = n S ,(3) P ( n − S ) = ( k − N ) S ,(4) a k = N S ,(5) b ℓ = ( S − N − S ,(6) N + P ( k − N ) = k + λ N + µ ( S − N − N P + P ( k − P ) = λ P + µ ( S − P ),(8) N a + P ( S − a ) = S + a λ + b ( k − λ − N b + P ( S − b ) = a µ + b ( k − µ ),(10) S + a N + b ( S − N −
1) = S + a N + b ( S − N − a P + b ( S − P ) = a P + b ( S − P ),(12) P ( k − N ) = P ( k − N ),(13) S + a k + b ℓ = S + a k + b ℓ = S S ,(14) S + a r − b ( r + 1) = S + a r − b ( r + 1),(15) S + a s − b ( s + 1) = S + a s − b ( s + 1) = 0. We use the same notation in [2] or [1]. However, the notation f i was used redundantly, relations andmultiplicities, in [2]. We use f i for multiplicities and σ i for relations. We also use σ i for the adjacencymatrices of the relations. Lemma 1.
Equations (1) and (2) hold.Proof.
Since | I | = 2, we may assume that f = f . Counting 1’s in σ , we have (2). In [2], (15) is “ S + a S − b ( s + 1) = S + a S − b ( s + 1) = 0” and this is incorrect. Table of multiplications
It is not difficult to make a table of multiplications. σ σ σ σ σ σ σ σ σ k σ + λ σ + µ σ ( k − λ − σ + ( k − µ ) σ σ σ ( k − λ − σ + ( k − µ ) σ ( n − k − σ + ( n − k + λ ) σ + ( n − k + µ − σ σ σ σ σ σ σ N σ + P σ ( k − N ) σ + ( k − P ) σ σ ( S − N − σ + ( S − P ) σ ( n − S − k + N ) σ + ( n − S − k + P − σ σ σ σ σ σ σ σ σ σ k σ + λ σ + µ σ ( k − λ − σ + ( k − µ ) σ σ σ ( k − λ − σ + ( k − µ ) σ ( n − k − σ + ( n − k + λ ) σ + ( n − k + µ − σ σ σ σ σ σ σ N σ + P σ ( k − N ) σ + ( k − P ) σ σ ( S − N − σ + ( S − P ) σ ( n − k − S + N ) σ + ( n − k − S + P − σ σ σ σ σ σ N σ + P σ ( S − N − σ + ( S − P ) σ σ σ ( k − N ) σ + ( k − P ) σ ( n − k − S + N ) σ + ( n − k − S + P − σ σ σ σ S σ + a σ + b σ ( S − a ) σ + ( S − b ) σ σ ( S − a ) σ + ( S − b ) σ ( n − S ) σ + ( n + a ) σ + ( n + b ) σ σ σ σ σ σ N σ + P σ ( S − N − σ + ( S − P ) σ σ σ ( k − N ) σ + ( k − P ) σ ( n − k − S + N ) σ + ( n − k − S + P − σ σ σ σ S σ + a σ + b σ ( S − a ) σ + ( S − b ) σ σ ( S − a ) σ + ( S − b ) σ ( n − S ) σ + ( n + a ) σ + ( n + b ) σ Other products are zero. We can read all intersection numbers p kij by the table.Valencies are as follows. v = 1 v = S v = k v = n − S v = ℓ = n − k − v = S v = 1 v = n − S v = k v = ℓ = n − k − By [2, 3.6 (vi)], we have p kij v k = p ikj ∗ v i . Lemma 2.
Equations (3), (4), and (5) hold.Proof.
The equation (3) holds by p v = p , v , (4) by p v = p v , and (5) by p v = p v .2 By regular representations
Let M be a right regular representation of the adjacency algebra, namely M : σ i ( p tsi ) st . We can readintersection numbers p ust from the table of multiplications in § Lemma 3.
Equations (6), (7), (8), (9), (10), (11), and (12) hold.Proof.
We consider the equation M ( σ i ) M ( σ j ) = P k =1 p kij M ( σ k ). The equation (6) is from ( M ( σ ) M ( σ )) ,(7) from ( M ( σ ) M ( σ )) , , (8) from ( M ( σ ) M ( σ )) , (9) from ( M ( σ ) M ( σ )) , (10) from ( M ( σ ) M ( σ )) ,(11) from ( M ( σ ) M ( σ )) , , and (12) is from ( M ( σ ) M ( σ )) . By definition, { σ , σ , σ } forms an ( n , k ; λ , µ )-SRG, and { σ , σ , σ } forms an ( n , k ; λ , µ )-SRG.The character tables are χ i k i ℓ i ϕ i r i − − r i f i ψ i s i − − s i g i for i = 1 ,
2. We are assuming that f = f in (1). Then the character table of the coherent configurationis σ σ σ σ σ s χ = χ + χ k ℓ k ℓ ϕ = ϕ + ϕ r − − r r − − r f = f ψ s − − s g ψ s − − s g Lemma 4.
Equations (13), (14), and (15) hold.Proof.
The equation (13) is by χ ( σ σ ) = χ ( σ σ ), (14) by ϕ ( σ σ ) = ϕ ( σ σ ), and (15) is by ψ ( σ σ ) = ψ ( σ σ ) = 0 and ψ ( σ σ ) = ψ ( σ σ ) = 0. Remark.
In [2, page 414, line 18], p must be p . Remark.
In [2, page 416, Table 1], we must exchange µ i and λ i . References [1] D. G. Higman,
Coherent algebras , Linear Algebra Appl. (1987), 209–239.[2] , Strongly regular designs and coherent configurations of type [ ], European J. Combin. (1988), no. 4, 411–422. In [2], M ai is considered. It is a part of the regular representation omitting zero parts.is considered. It is a part of the regular representation omitting zero parts.