Strongly regular decompositions and symmetric association schemes of a power of two

0.14 MB·

by  Hadi Kharaghani

Strongly regular decompositions and symmetric association
schemes of a power of two
Hadi Kharaghani∗ Sara Sasani† Sho Suda‡
7 January 23, 2017
1
0
2

n
Abstract
a
J Foranypositiveintegerm,thecompletegraphon22m(2m+2)verticesisdecomposed

9 into2m+1commutingstronglyregulargraphs,whichgiverisetoasymmetricassociation
1 scheme of class 2m+2−2. Furthermore, the eigenmatrices of the symmetric association

schemesaredeterminedexplicitly. Asanapplication,theeigenmatrixofthecommutative
]
O strongly regular decomposition obtained from the strongly regular graphs is derived.
C
. 1 Introduction
h
t
a
m A strongly regular graph with parameters (v,k,λ,µ) is a regular graph with v vertices of
degree k such that every two adjacent vertices have exactly λ common neighbors and
[
every two non-adjacent vertices have exactly µ common neighbors. A strongly regular

1 decomposition is a decomposition of the edge set of the complete graph with vertex set
v
V into strongly regular graphs with vertex set V. A strongly regular decomposition is
0
commutative if the adjacency matrices of strongly regular graphs are commutative. The
3
6 concept of strongly regular decompositions was introduced by van Dam [2] in order to
5 study more general situation of amorphous association schemes.
0 Inthispaper,weshowthatforanypositiveintegerm,thereisacommutativestrongly
1. regular decomposition of the complete graph of 22m(2m +2) vertices, into 2m strongly
0 regular graphs with parameters (22m(2m +2),22m +2m,2m,2m) and 2m +2 cliques of
7 size22m. Notethat2m+2cliquesofsize22m isastronglyregulargraphwithparameters
1 (22m(2m+2),22m−1,22m−2,0). In fact, the constructed strongly regular graphs with
:
v parameters (22m(2m+2),22m+2m,2m,2m) are symmetric (22m(2m+2),22m+2m,2m)-
i designs with symmetric incidence matrices with very large number of symmetries. Our
X
constructionmethod is basedon the generalizedHadamardmatrices obtainedfromfinite
r
a fields of characteristic two and symmetric Latin squares with constant diagonal of even
order.
One might wonder if the decomposition yields a symmetric association scheme, but
unfortunately this does not hold. We had to further decompose the edge sets of the
strongly regular graphs to obtain a symmetric association scheme, and determine the
eigenamtrices of the symmetric association scheme explicitly. As a corollary, we obtain
the eigenmatrix of the commutative strongly regular decomposition.
∗Department of Mathematics and Computer Science, University of Lethbridge, Lethbridge, Alberta, T1K
3M4, Canada. [email protected]
†Department of Mathematics and Computer Science, University of Lethbridge, Lethbridge, Alberta, T1K
3M4, Canada. [email protected]
‡Department of Mathematics Education, Aichi University of Education, 1 Hirosawa, Igaya-cho, Kariya,
Aichi 448-8542, Japan. [email protected]
1
2 Preliminaries
Let n be a positive integer. Let V be a finite set of size v and R (i ∈ {0,1,...,n}) be
i
a non-empty subset of V ×V. The adjacency matrix A of the graph with vertex set V
i
and edge set R is a v×v (0,1)-matrix with rows and columns indexed by the elements
i
of V such that (A ) =1 if (x,y)∈R and (A ) =0 otherwise. The pair (V,{R }n )
i xy i i xy i i=0
is said to be a commutative decomposition of the complete graph if the following hold:
(i) A =I , the identity matrix of order v.
0 v
(ii) n A =J , the all-ones matrix of order v.
i=0 i v
(iii) A is symmetric for i∈{1,...,n}.
Pi
(iv) For any i,j, A A =A A .
i j j i
Wealsorefertothesetofnon-zerov×v(0,1)-matricessatisfying(i)-(iv)asacommutative
decomposition. Note that the corresponding graph of each A is regular, because A and
i i
n
J = A commute. Let k denote the valency of the corresponding graph of A . A
v i=0 i i i
commutativedecomposition{A ,A ,...,A }isasymmetricassociationschemeofclassn
0 1 n
if therPe exist non-negativeintegers pk suchthat A A = n pk A . The non-negative
i,j i j k=0 i,j k
integers pk are said to be the intersection numbers. A commutative decomposition is a
i,j P
strongly regular decomposition if the corresponding graph of each A is strongly regular.
i
Note that a v×v (0,1)-matrix A is the adjacency matrix of a strongly regular graph if
and only if {I ,A,J −A−I } is a symmetric association scheme of class 2.
v v v
Wedefinetheeigenmatrixforcommutativedecompositions. Sincetheadjacencymatri-
ces are symmetric and commuting, there are maximal common eigenspaces V ,V ,...,V
0 1 t
of A ’s. Let E be the orthogonal projection Rv onto V for j ∈ {0,1,...,t}. Note that
i j j
the corresponding graph of each A is regular and one of them is connected, so we may
i
assume that V is spanned by the all-ones vector and thus E = (1/v)J . The matrices
0 0 v
E ,E ,...,E satisfyE E =δ E foranyi,j whereδ isthe Kroneckerdelta. Thenwe
0 1 t i j ij i ij
can express A as a linear combination of E :
j i
t
A = p E .
j ij i
i=0
X
We call a (t+1)×(n+1) matrix P = (p ) the eigenmatrix of the commutative
ij 0≤i≤t
0≤j≤n
decomposition. The following lemma was proved in [2].
Lemma 2.1. Let (V,{R }n ) be a commutative decomposition with orthogonal projec-
i i=0
tions E , j ∈ {0,1,...,t}. Then t ≥ n with equality if and only if (V,{R }n ) is a
j i i=0
symmetric association scheme.
For the case of symmetric associationschemes, we have t=n so that the eigenmatrix
P becomes a squarematrix ofordern+1. The vectorspace Aspanned by the adjacency
matrices over R is closed under the matrix multiplication and is said to be the Bose-
Mesner algebra. TheneachE is anelement inA, soit is writtenas alinear combination
j
of A as follows: for some real numbers q ,
i ij
n
1
E = q A .
j ij i
v
i=0
X
The matrices E ,E ,...,E are said to be the primitive idempotents of the symmetric
0 1 n
association scheme. The matrix P is also said to be the first eigenmatrix and the matrix
Q = (q )n is said to be the second eigenmatrix of the symmetric association scheme.
ij i,j=0
The valencies k (j ∈ {0,1,...,n}) satisfy k = p , and the multiplicities m (j ∈
j j 0j j
{0,1,...,n}) are defined by m = rank(E ) and satisfy m = q . The first and second
j j j 0j
eigenmatrices are related as the following.
2
Lemma 2.2. [1, Theorem 3.5(i)] Let (V,{R }n ) be a symmetric association scheme
i i=0
with first and second eigenmatrices P and Q. Let ∆ and ∆ be (n +1) × (n + 1)
k m
diagonal matrices with diagonal entries k ,k ,...,k and m ,m ,...,m , respectively.
0 1 n 0 1 n
Then it holds that ∆ P =Q⊤∆ where Q⊤ is the transpose of Q.
m k
Proof. Calculate the trace of A E in two ways.
i j
Finally we define the Krein numbers of the symmetric association scheme. Since the
Bose-Mesneralgebrahasa basis{A ,A ,...,A } consistingofdisjoint(0,1)-matrices,A
0 1 n
isclosedunderthe entrywiseproductdenoted◦. Thenforanyi,j,k∈{0,1,...,n},there
exist real numbers qk suchthat E ◦E = 1 n qk E . The realnumbers qk are said
i,j i j v k=0 i,j k i,j
to be the Krein numbers, and it is known that the Krein numbers are non-negative real
P
numbers [1, Theorem 3.8].
3 Association schemes of a power of two
From now let q = 2m be a power of two. We denote by F the finite field of q elements.
q
Let H be the multiplicative table of F , i.e., H is a q×q matrix with rowsandcolumns
q q q
indexed by the elements of F with (α,β)-entry equal to α·β. Then the matrix H is a
q q
generalized Hadamard matrixwithparameters(q,1)overtheadditivegroupofF . Letting
q
Gbeanadditivelywrittenfiniteabeliangroupoforderg,asquarematrixH =(h )gλ of
ij i,j=1
ordergλwithentriesfromGiscalledageneralized Hadamard matrix with the parameters
(g,λ) over G if for all distinct i,k ∈{1,2,...,gλ}, the multiset {h −h : 1≤ j ≤ gλ}
ij kj
contains exactly λ times of each element of G.
Let φ be a permutation representation of the additive group of F defined as follows.
q
Since q =2m, we view the additive group of F as Fm. Define R=J −I , and a group
q 2 2 2
homomorphism φ:Fq →GLq(R) as φ((xi)mi=1)=⊗mi=1Rxi.
Fromthe generalizedHadamardmatrix H andthe permutationrepresentationφ, we
q
constructq2 auxiliarymatrices;foreachα,α′ ∈Fq, define a q2×q2 (0,1)-matrixCα,α′ to
be
Cα,α′ =(φ(α(−β +β′)+α′))β,β′∈Fq.
Further, let x,y be indeterminates, we define Cx,α,Cy,α by Cx,α = Oq2 and Cy,α =
φ(α)⊗Jq for α∈Fq, where Oq2 denotes the zero matrix of order q2.
ItisknownthatasymmetricLatinsquareoforderv withconstantdiagonalexistsfor
any positive even integer v, see [3]. Let L=(L(a,a′))a,a′∈S be a symmetric Latin square
of order q+2 on the symbol set S = F ∪{x,y} with constant diagonal x. Write L as
q
L= a·P , where P is a symmetric permutation matrix of order q+2. Note that
a∈S a a
P =I .
x q+2
P
From the (0,1)-matrices Cα,α′’s and the Latin square L, we construct symmetric
designs as follows. For α∈F , we define a (q+2)q2×(q+2)q2 (0,1)-matrix N to be
q α
Nα =(CL(a,a′),α)a,a′∈S = Pa⊗Ca,α.
a∈FXq∪{y}
In order to show that each N is a symmetric design and study more properties, we
α
prepare a lemma on Cα,α′ and Pa.
Lemma 3.1. (i) For α∈F , C =qI ⊗φ(α)+(J +φ(α)−I )⊗J .
q a∈Fq∪{y} a,α q q q q
(ii) For a∈Fq∪{y} and α,α′P∈Fq, Ca,αCa,α′ =qCa,α+α′.
(iii) For distinct a,a′ ∈Fq∪{y} and α,α′ ∈Fq, Ca,αCa′,α′ =Jq2.
(iv) For α,α′,α′′ ∈Fq, (Iq ⊗φ(α′′))Cα,α′ =Cα,α′+α′′.
3
(v) For α,α′ ∈Fq, (Iq ⊗φ(α))Cy,α′ =Cy,α′.
(vi) P P =q(J −I ).
a,b∈Fq∪{y},a6=b a b q+2 q+2
Proof.P(i): For α,β,β′ ∈F , the (β,β′)-entry of C is
q γ∈Fq γ,α
P
φ(α) if β =β′
φ(γ(−β+β′)+α)= γ∈Fq
γX∈Fq (Pγ′∈Fqφ(γ′+α) if β 6=β′
Pqφ(α) if β =β′,
=
(Jq if β 6=β′,
which yields C =qI ⊗φ(α)+(J −I )⊗J . Therefore
γ∈Fq γ,α q q q q
P
C = C +C =qI ⊗φ(α)+(J +φ(α)−I )⊗J .
a,α γ,α y,α q q q q
a∈FXq∪{y} γX∈Fq
(ii): For a = y, Cy,αCy,α′ = (φ(α)⊗Jq)(φ(α′)⊗Jq) = qφ(α+α′)⊗Jq = qCy,α+α′.
For a,β,β′ ∈Fq, the (β,β′)-entry of Ca,αCa,α′ is
φ(a(−β+γ)+α)φ(a(−γ +β′)+α′)= φ(a(−β+β′)+α+α′)
γX∈Fq γX∈Fq
=qφ(a(−β+β′)+α+α′).
Thus we have Ca,αCa,α′ =qCa,α+α′.
(iii): The case of a ∈ F and a′ = y follows from the fact that C is a block matrix
q a,α
whose q×q sub-blockis a permutationmatrix. The caseofa,a′ ∈F ,a6=a′ followsfrom
q
a similar calculation to (ii) with the fact that {(a−a′)γ |γ ∈F }=F .
q q
(iv) and (v) are routine, and (vi) follows from the equations below. Recall that S =
F ∪{x,y}.
q
P P = P ( P )
a b a b
a,b∈FqX∪{y},a6=b a∈FXq∪{y} b∈SX\{x,a}
= P (J −I −P )
a q+2 q+2 a
a∈FXq∪{y}
= (J −P −I )
q+2 a q+2
a∈FXq∪{y}
=(q+1)(J −I )− P
q+2 q+2 a
a∈FXq∪{y}
=q(J −I ).
q+2 q+2
Now we are ready to prove results for N ’s. Note that the result for N being a
α 0
symmetric ((q+2)q2,q2+q,q)-design is well-known, see for example [4, Exercise 5.7].
Theorem 3.2. (i) For any α∈F , N is symmetric.
q α
(ii) For any α,β ∈F ,
q
N N =q2I ⊗φ(α+β)+qI ⊗φ(α+β)⊗J +q(J −I )⊗J .
α β q(q+2) q+2 q q(q+2) q(q+2) q
In particular, Nα2 = q2I(q+2)q2 +qJ(q+2)q2, that is, Nα is the incidence matrix of a
symmetric ((q+2)q2,q2+q,q)-design.
4
Proof. (i): It follows from the properties that the matrices P and C are symmetric
a a,α
for a∈F ∪{y} and α∈F .
q q
(ii): We use Lemma 3.1 to obtain:
N N = P P ⊗C C
α β a b a,α b,β
a,b∈XFq∪{y}
= P2⊗C C + P P ⊗C C
a a,α a,β a b a,α b,β
a∈FXq∪{y} a,b∈FqX∪{y},a6=b
= Iq+2⊗qCa,α+β + PaPb⊗Jq2
a∈FXq∪{y} a,b∈FqX∪{y},a6=b
=qIq+2⊗(qIq⊗φ(α+β)+(Jq +φ(α+β)−Iq)⊗Jq)+q(Jq+2−Iq+2)⊗Jq2
=q2I ⊗φ(α+β)+qI ⊗φ(α+β)⊗J +q(J −I )⊗J .
q(q+2) q+2 q q(q+2) q(q+2) q
The case of α=β follows from φ(0)=I .
q
Now we obtain our main result in this section.
Theorem 3.3. The set of matrices {Nα,Iq+2⊗(Jq2 −Iq2),I(q+2)q2 | α ∈ Fq} is a com-
mutative strongly regular decomposition of the complete graph on (q+2)q2 vertices.
Proof. It is easy to see that α∈FqNα +Iq+2 ⊗(Jq2 −Iq2) = J(q+2)q2 −I(q+2)q2. By
Theorem 3.2 (i), (ii) for α = β, the matrices N , α ∈ F , are the adjacency matrices of
P α q
stronglyregulargraphswithparameters((q+2)q2,q2+q,q,q). AndIq+2⊗(Jq2−Iq2)isthe
adjacencymatrix ofa stronglyregulargraphwithparameters((q+2)q2,q2−1,q2−2,0).
Thus they form a strongly regular decomposition. Furthermore N , α ∈ F , commute
α q
each other by Theorem 3.2 (ii) for α 6= β. Since each N has constant row and column
α
sums inthe off-diagonalblocks,eachNα commuteswithIq+2⊗(Jq2−Iq2). Thereforethe
set of matrices is a commutative strongly regular decomposition.
Unfortunately, as Theorem 3.2 (ii) showed, the matrices Nα, α ∈ Fq, Iq+2 ⊗(Jq2 −
Iq2),I(q+2)q2 donotformasymmetricassociationscheme. Nowwearegoingtorefinethe
adjacency matrices in order to obtain a symmetric association scheme.
For α∈F ,i∈{0,1,2,3},we define (0,1)-matrices A as
q α,i
A =I ⊗φ(α),
α,0 q(q+2)
A =I ⊗C ,
α,1 q+2 y,α
A =P ⊗C ,
α,2 y y,α
A =N −P ⊗C .
α,3 α y y,α
Note that A0,0 =I(q+2)q2 and α∈FqAα,0 =A0,1. Let F∗q =Fq\{0}.
Theorem 3.4. The set of maPtrices {A ,A ,A ,A | α ∈ F ,β ∈ F∗} is a sym-
α,0 β,1 α,2 α,3 q q
metric association scheme.
Proof. By the definitionofN , A arenon-zero(0,1)-matricessuchthat (A +
α α,i α∈Fq α,0
Aα,2+Aα,3)+ α∈F∗qAα,1 =J(q+2)q2. Since each Pa and Ca,α are symmePtric, each Aα,i
is symmetric. BPy α∈FqAα,0 = A0,1, it is enough to show that A := spanR{Aα,i | α ∈
F ,i∈{0,1,2,3}}is closed under the matrix multiplication.
q P
From Lemma 3.1, it follows that A′ = spanR{Aα,i | α ∈ Fq,i ∈ {0,1,2}} is closed
under matrix multiplication. Since A = A′ + spanR{Nα | α ∈ Fq}, we then need to
5
calculate A N ,N A and N N . The case N N follows from Theorem3.2 (ii). For
α,i β β α,i α β α β
the case of A N , we use Lemma 3.1 to obtain
α,i β
A N =N ,
α,0 β α+β
Aα,1Nβ =(Jq+2−Iq+2−Py)⊗Jq2 +qPy ⊗Cy,α+β,
Aα,2Nβ =(Jq+2−Iq+2−Py)⊗Jq2 +qIq+2⊗Cy,α+β.
FinallyN A followsfromtakingtransposeofA N becausealltheadjacencymatrices
β α,i α,i β
are symmetric.
FromtheproofofTheorem3.4,theintersectionnumberscanbedeterminedasfollows.
Precise calculations will be given in Appendix. Remark that the matrix A , which is
0,1
not an adjacency matrix, is included in the following proposition.
Proposition 3.5. Let α,β ∈F . The following hold.
q
(i) For l ∈{0,3}, A A =qA , and for l∈{1,2} A A =qA .
α,0 β,l α+β,l α,0 β,l β,l
(ii) A A =A A =qA and A A =qA .
α,1 β,1 α,2 β,2 α+β,1 α,1 β,2 α+β,2
(iii) A A = A A = A and A A = q2A + (q(−A +
α,1 β,3 α,2 β,3 γ∈Fq γ,3 α,3 β,3 α+β,0 γ∈Fq γ,0
A +A )+(q−2)A ).
γ,1 γ,2 γP,3 P
4 Eigenmatrices of symmetric association schemes and
strongly regular decompositions
We view F = Zm as the additive group. For α = (α ,...,α ),β = (β ,...,β ) ∈ Zm,
q 2 1 m 1 m 2
the inner product is defined by hα,βi = α β + ··· +α β . Then for β ∈ Zm, the
1 1 m m 2
irreduciblecharacterdenotedχ isχ (α)=(−1)hα,βi whereα∈Zm. Thecharacter table
β β 2
K of the abelian group Zm is a 2m×2m matrix with rows and columns indexed by the
2
elements of Zm with (α,β)-entry equal to χ (α). Note that χ (α) = χ (β). Then the
2 β β α
Schur orthogonality relation shows KK⊤ = 2mI2m, namely K is a Hadamard matrix of
order 2m. Moreover K is transformed by permuting rows and columns to the Sylvester-
type Hadamard matrix of order 2m. Here the Sylvester-type Hadamard matrix of order
2m is defined as the m tensor product of 1 1 .
1−1
To describe the primitive idempotents, let F , α∈F ,i∈{0,1,2,3},be
α,i q
(cid:0) (cid:1)
F = χ (γ)A .
α,i α γ,i
γX∈Fq
Lemma 4.1. Let α,β ∈F . The following hold.
q
(i) For l ∈{0,1,2,3}, F F =qδ F .
α,0 β,l α,β α,l
(ii) F F =F F =q2δ F and F F =q2δ F .
α,1 β,1 α,2 β,2 α,β α,1 α,1 β,2 α,β α,2
(iii) F F =F F =q2δ δ F andF F =q3δ F +q2δ δ (q(−F +
α,1 β,3 α,2 β,3 α,0 β,0 0,3 α,3 β,3 α,β α,0 α,0 β,0 0,0
F +F )+(q−2)F ).
0,1 0,2 0,3
Proof. (i) for l = 0 and (ii) for F F follow from the fact that for i ∈ {0,1}, the
α,1 β,1
group{A |α∈F } under the matrix multiplicationis isomorphicto Zm andthe Schur
α,i q 2
orthogonality. We prove only the case of F F . The other cases can be proved in a
α,3 β,3
6
similar manner.
Fα,3Fβ,3 = χα(γ)χβ(γ′)Aγ,3Aγ′,3
γ,Xγ′∈Fq
= χα(γ)χβ(γ′)(q2Aγ+γ′,0+q (−Aγ′′,0+Aγ′′,1+Aγ′′,2)+(q−2) Aγ′′,3)
γ,Xγ′∈Fq γX′′∈Fq γX′′∈Fq
=q2F F + χ (γ)χ (γ′)(q(−F +F +F )+(q−2)F )
α,0 β,0 α β 0,0 0,1 0,2 0,3
γ,Xγ′∈Fq
=q3δ F +q2δ δ (q(−F +F +F )+(q−2)F ).
α,β α,0 α,0 β,0 0,0 0,1 0,2 0,3
Let E ,E be
i α,j
1
E = (F +F +F ),
0 (q+2)q2 0,1 0,2 0,3
1 q
E = ( (F +F )−F ),
1 (q+2)q2 2 0,1 0,2 0,3
1
E = (F +F ), α∈F ,
α,1 2q2 α,1 α,2 q
1
E = (F −F ), α∈F ,
α,2 2q2 α,1 α,2 q
1
E = (qF +F ), α∈F ,
α,3 2q2 α,0 α,3 q
1
E = (qF −F ), α∈F .
α,4 2q2 α,0 α,3 q
NotethatE =E +E ,E =(q+1)E + E andE =(−q+1)E +2E +
0,1 0 1 0,3 2 0 γ∈F∗q γ,1 0,4 2 0 1
E . From Lemma 4.1, the following is readily obtained.
γ∈F∗ γ,1 P
q
PTheorem 4.2. The matrices E0,E1,Eβ,1,Eα,2,Eβ,3,Eβ,4, α∈Fq,β ∈F∗q, are the prim-
itive idempotents of the symmetric association scheme.
As a direct consequence of Theorem 4.2 and Lemma 2.2, the eigenmatrices of the
symmetric associationscheme are determined as follows.
Theorem 4.3. The first and second eigenmatrices P,Q of the symmetric association
scheme are given as follows:
A A A A
α,0 β,1 α,2 α,3
E 1 q q q2
0
E1  1 q q −2q 
P = Eβ′,1 1 qχβ′(β) qχβ′(α) 0 ,
 
Eα′,2 1 qχα′(β) −qχα′(α) 0 
 
Eβ′,3χβ′(α) 0 0 qχβ′(α) 
 
Eβ′,4χβ′(α) 0 0 −qχβ′(α)
 
 
E0 E1 Eβ′,1 Eα′,2 Eβ′,3 Eβ′,4
Aα,0 1 2q 2q +1 q2 +1 (q2 +1)qχβ′(α) (q2 +1)qχβ′(α)
Q= Aβ,1 1 2q (2q +1)χβ′(β) (q2 +1)χα′(β) 0 0 ,
Aα,2 1 2q (2q +1)χβ′(α) −(q2 +1)χα′(α) 0 0 
Aα,3 1 −1 0 0 (2q +1)χβ′(α) −(2q +1)χβ′(α)
 
where α,α′ ∈F ,β,β′ ∈F∗.
q q
7
Proof. The formula for the second eigenmatrix Q is derived from Theorem 4.2. Since
the valencies of A ,A ,A ,A are respectively 1,q,q,q2, the formula for the first
α,0 β,1 α,2 α,3
eigenmatrix P follows from Lemma 2.2.
Asacorollaryoftheformulaforthe firsteigenmatrixP,weobtainseveralcommuting
strongly regular graphs. In order to describe the eigenamtrix, we need the notion of the
permutation automorphism group of Sylvester-type Hadamard matrices. Define the per-
mutationautomorphism groupoftheSylvester-typeHadamardmatrixK =(χβ(α))α,β∈Fq
by
PAut(K)={(σ,τ)∈S(F )×S(F )|χ (σ(α))=χ (α) for any α,β ∈F },
q q τ(β) β q
where S(F ) denotes the set of all permutations on the set F . Note that for (σ,τ) ∈
q q
PAut(K), σ(0)=τ(0)=0.
Corollary 4.4. For anyσ ∈S(F ), theset ofmatrices {A , (A +A ),A +
q 0,0 α∈F∗ α,0 α,1 α,2
q
A |α∈F } is a commutative strongly regular decomposition.
σ(α),3 q P
If there exists τ ∈S(F ) such that (σ,τ)∈PAut(K), then the eigenmatrix is given by
q
A (A +A ) A +A
0,0 α∈F∗ α,0 α,1 α,2 σ(α),3
q
E 1 P q2−1 q2+q
0
E +E  1 q2−1 −q 
P = 1 0,2 .
Eβ′,1+Eτ(β′),3 1 −1 qχβ′(α) 
 
Eβ′,2+Eτ(β′),4 1 −1 −qχβ′(α) 
 
If there does not exist τ ∈ S(F ) such that (σ,τ) ∈ PAut(K), then the eigenmatrix is
q
given by
A (A +A ) A +A
0,0 α∈F∗ α,0 α,1 α,2 σ(α),3
q
E 1 P q2−1 q2+q
0
E +E  1 q2−1 −q 
1 0,2
P = Eβ′,1  1 −1 qχβ′(α) .
 
Eβ′,2  1 −1 −qχβ′(α) 
 
Eβ′,3  1 −1 qχβ′(σ(α)) 
 
Eβ′,4  1 −1 −qχβ′(σ(α)) 
 
In both cases above, α,α′ ∈F and β,β′ ∈F∗.
q q
Acknowledgments. HadiKharaghaniissupportedbyanNSERCDiscoveryGrant. Sho
Suda is supported by JSPS KAKENHI Grant Number 15K21075.
References
[1] E. Bannai and T. Ito, Algebraic Combinatorics I: Association Schemes, Ben-
jamin/Cummings, Menlo Park,CA, 1984.
[2] E. van Dam, Strongly regular decompositions of the complete graph, J. Algebraic
Combin. 17 (2003), 181–201.
[3] H. Kharaghani,New class of weighing matrices, Ars. Combin. 19 (1985), 69-72.
[4] D. R. Stinson, Combinatorial Designs: Constructions and Analysis, New York,
Springer, 2004.
8
A Intersection numbers
We calculate the intersection numbers of the symmetric association schemes. Let α,β ∈
F .
q
Proposition 3.5(i):
A A =I ⊗φ(α+β)=A .
α,0 β,0 q(q+2) α+β,0
A A =I ⊗((I ⊗φ(α))C )=I ⊗C =A .
α,0 β,1 q+2 q y,β q+2 y,β β,1
A A =P ⊗((I ⊗φ(α))C )=P ⊗C =A .
α,0 β,2 y q y,β y y,β β,2
A A =(I ⊗(I ⊗φ(α))(N −P ⊗C )
α,0 β,3 q+2 q β y y,β
=(Iq+2⊗(Iq ⊗φ(α))((CL(a,a′),β)a,a′∈S −Py ⊗Cy,β)
=((CL(a,a′),α+β)a,a′∈S −Py ⊗Cy,α+β
=N −P ⊗C
α+β y y,α+β
=A .
α+β,3
Proposition 3.5(ii):
A A =(I ⊗C )(I ⊗C )=qI ⊗C =qA .
α,2 β,2 q+2 y,α q+2 y,β q+2 y,α+β α+β,1
A A =qI ⊗C =qA .
α,1 β,2 q+2 y,α+β α+β,2
A A =(P ⊗C )(P ⊗C )=qI ⊗C =qA .
α,2 β,2 y y,α y y,β q+2 y,α+β α+β,1
Proposition 3.5(iii):
A A =(I ⊗C )(N −P ⊗C )
α,1 β,3 q+2 y,α β y y,β
=(I ⊗C )N −P ⊗(C C )
q+2 y,α β y y,α y,β
=((Jq+2−Iq+2−Py)⊗Jq2 +qIq+2⊗Cy,α+β)−qPy⊗Cy,α+β
=(Jq+2−Iq+2−Py)⊗Jq2
= A .
γ,3
γX∈Fq
A A =(P ⊗C )(N −P ⊗C )
α,2 β,3 y y,α β y y,β
=(P ⊗C )N −(P ⊗C )⊗(P ⊗C )
y y,α β y y,α y y,β
=(Jq+2−Iq+2−Py)⊗Jq2 +qIq+2⊗Cy,α+β)−qIq+2⊗Cy,α+β
=(Jq+2−Iq+2−Py)⊗Jq2
= A .
γ,3
γX∈Fq
We use the following formula for the permutation matrices.
P P = P ( P )
α β α β
α,β∈XFq,α6=β αX∈Fq β∈FXq,β6=α
= P (J −I −P −P )
α q+2 q+2 α y
αX∈Fq
= (J −P −I −P P )
q+2 α q+2 α y
αX∈Fq
=q(J −I )−( P )(I +P )
q+2 q+2 α q+2 y
αX∈Fq
=q(J −I )−(J −I −P )(I +P )
q+2 q+2 q+2 q+2 y q+2 y
=(q−2)(J −I )+2P .
q+2 q+2 y
9
This yields
Aα,3Aβ,3 =( Pγ ⊗Cγ,α)( Pγ′ ⊗Cγ′,β)
γX∈Fq γX′∈Fq
= PγPγ′ ⊗Cγ,αCγ′,β
γ,Xγ′∈Fq
=q Iq+2⊗Cγ,α+β + PγPγ′ ⊗Jq2
γX∈Fq γ,γ∈XFq,γ6=γ′
=qIq+2⊗(qIq ⊗φ(α+β)+(Jq−Iq)⊗Jq)+((q−2)(Jq+2−Iq+2)+2Py)⊗Jq2
=q2Iq(q+2)⊗φ(α+β)+qIq+2⊗(Jq −Iq)⊗Jq +((q−2)(Jq+2−Iq+2)⊗Jq2 +2Py⊗Jq2
=q2A + (q(−A +A +A )+(q−2)A ).
α+β,0 γ,0 γ,1 γ,2 γ,3
γX∈Fq
B Krein numbers
We calculate the entrywise product of the primitive idempotents. From the equations
below,wemayfindtheKreinnumbersforthesymmetricassociationschemebymakinguse
ofE =E +E ,E =(q+1)E + E ,E =(−q+1)E +2E + E .
0,1 0 1 0,3 2 0 γ∈F∗q γ,1 0,4 2 0 1 γ∈F∗q γ,1
For α,α′ ∈F ,β,β′ ∈F∗, the following are readily obtained by Theorem 4.3:
q q P P
1 q q−2
E ◦E = ( E + E ),
1 1 (q+2)q2 2 0 2 1
1 q
E ◦E = E ,
1 β,1 (q+2)q22 β,1
1 q
E ◦E = E ,
1 α,2 (q+2)q22 α,2
1 q+2 q−2
E ◦E = ( E + E ),
1 β,3 (q+2)q2 4 β,3 4 β,4
1 q−2 q+2
E ◦E = ( E + E ),
1 β,4 (q+2)q2 4 β,3 4 β,4
1 q+2
Eβ,1◦Eβ′,1 = (q+2)q2 2 Eβ+β′,1,
1 q+2
E ◦E = E ,
β,1 α,2 (q+2)q2 2 α+β,2
1 q+2
Eα,2◦Eα′,2 = (q+2)q2 2 Eα+α′,1,
1 q q+1 q−1
Eβ,3◦Eβ′,3 =Eβ,4◦Eβ′,4 = (q+2)q2(2 +1)( 2 Eβ+β′,3+ 2 Eβ+β′,4),
1 q q−1 q+1
Eβ,3◦Eβ′,4 = (q+2)q2(2 +1)( 2 Eβ+β′,3+ 2 Eβ+β′,4),
1 q+2 q+2
Eβ,1◦Eβ′,3 =Eβ,1◦Eβ′,4 = (q+2)q2( 4 Eβ′,3+ 4 Eβ′,4),
1 q+2 q+2
E ◦E =E ◦E = ( E + E ).
α,2 β,3 α,2 β,4 (q+2)q2 4 β,3 4 β,4
10