On the Association Schemes of QuadraticForms

Mitsugu Hirasaka

Abstract

The association scheme of quadratic forms (or the quadratic formsscheme, for short) is one of the known P- and Q-polynomial schemesand its first eigenmatrix is represented by using the Askey-Wilsonpolynomials. We consider two fission schemes of the quadratic formsscheme in characteristic 2 and describe the first eigenmatrix of one ofthese fission schemes, and compute some intersection numbers of theother.

1 Introduction

Properties of quadratic forms defined on a vector space V over Fq arequite different depending on whether q is even or odd. If q is odd, there is aone-to-one correspondence between the set of quadratic forms and the set ofsymmetric bilinear forms. Hence the study of quadratic forms is reduced tothat of symmetric bilinear forms. But if q is even, this is no longer the case.Let f be a quadratic form on V and let Bf be the symmetric bilinear formassociated to f . Then we have

Bf (x, x) = 2f(x) = 0 for any x ∈ V ,

so that Bf is alternating. Given an alternating bilinear form B, there aremany quadratic forms associated to B.

This is reflected on the structure of quadratic forms scheme, which isone of the known P- and Q-polynomial schemes. Let X be the set ofquadratic forms on V . By defining some relations we can construct a P-

1

and Q-polynomial scheme, which is called the quadratic forms scheme. Thequadratic forms scheme has the same parameters as the alternating formsscheme with one dimension larger, but they are not isomorphic (see [4]).

In this paper we consider two fission schemes of the quadratic formsscheme in characteristic 2 and describe the first eigenmatrix of one of thesefission schemes, and compute some intersection numbers of the other.

2 Elementary properties of quadratic forms

Let V be a vector space over Fq with dimV = n, where q is a prime power.

Definition 2.1 A function f from V to Fq is called a quadratic form on Vif the following conditions hold:

(i)f(au) = a2f(u) for any u ∈ V, a ∈ Fq

(ii)f(u + v) = f(u) + f(v) + Bf (u, v) for any u, v ∈ V ,where Bf is a symmetric bilinear form on V .

Let B be a symmetric bilinear form on V . Then the radical of B is definedby the following:

Rad(B) := u ∈ V |B(u, v) = 0 for any v ∈ V .Next we define the radical of a quadratic form f to be

Radf := RadBf ∩ f−1(0)

and the rank of f to be

rank(f) := dimV − dim(Radf).

Definition 2.2 Let f be a quadratic form on V . f is called nondegenerateif Radf = 0.

Proposition 2.1 Let q be even. Let f be a nondegenerate quadratic formon V . Then there is a basis e1, · · · , en of V such that

f(u) =j∑

i=1

x2i−1x2i+

x2j+1x2j+2, if dimV = 2j + 2 · · · (I)x2j+1x2j+2 + x2

2j+1 + ξx22j+2, if dimV = 2j + 2 · · · (II)

x22j+1, if dimV = 2j + 1

where u =∑n

i=1 xiei, and ξ is an element of Fq for which t2+t+ξ is irreduciblein Fq[t].

2

(Proof)(see [3], p.59) 2

Let GL(V ) be the general linear group on V . Let X be the set ofquadratic forms on V . If we define the action of GL(V ) on X by

GL(V ) 3 T, X 3 f, Tf(v) = f T−1(v) for any v ∈ V ,

then X is partitioned into the following orbits by the above proposition:f ∈ V |rank(f) = 2l (l = 1, 2, · · ·) are divided into type (I) and type (II),which are respectively called hyperbolic and elliptic type, while

f ∈ V |rank(f) = 2l − 1 (l = 1, 2, · · ·)are themselves orbits.

3 Preliminaries on association schemes

Definition 3.1 (Association scheme) Let X be a finite set.Let Ri (i = 0, . . . , d) be subsets of X ×X. The configuration (X, Rii=0,...,d)is called an association scheme if it satisfies the following conditions:(i)R0 = (x, x)|x ∈ X(ii)R0 ∪R1 ∪ . . . ∪Rd = X ×X.

For any i, j (i 6= j), Ri ∩Rj = φ.

(iii)For any i, there is some i′ ∈ 0, 1, . . . , d such that

Rti := (y, x)|(x, y) ∈ Ri = Ri′

(iv)|z ∈ X|(x, z) ∈ Ri, (z, y) ∈ Rj| is constant whenever (x, y) ∈ Rk, whichis denoted by pk

ij, called the intersection number.

Definition 3.2 (Association scheme) Let X be a finite set.Let Ri(i = 0, . . . , d) be subsets of X ×X. We define the adjacency matrix ofRi as follows:

(Ai)xy :=

1 if (x, y) ∈ Ri

0 otherwise

The configuration (X, Rii=0,...,d) is called an association scheme if it satis-fies the following conditions:

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(i)A0 = I, where I is the identity matrix.(ii)A0 + . . . + Ad = J , where J is the all 1 matrix.(iii)For any i, there is some i′ ∈ 0, 1, . . . , d such that

tAi = Ai′ .

(iv)For any i, j,

AiAj =d∑

k=0

pkijAk.

Note that Definition 3.1 and Definition 3.2 are equivalent.Let (X, Rii=0,...,d) be an association scheme. (X, Rii=0,...,d) is called

commutative if AiAj = AjAi for any i, j, and symmetric if Ati = Ai for

any i. Suppose that (X, Rii=0,...,d) is a commutative association scheme.Then each Ai is diagonalizable by a unitary matrix and A0, A1, · · · , Ad arepairwise commutative. Hence A0, A1, · · · , Ad are simultaneously diagonal-izable by a unitary matrix. Let W be a vector space over the complex fieldwith dimW = |X|. By the above fact W can be written as the orthogonalsum of maximal common eigenspaces, say

W = W0 ⊥ W1 ⊥ · · · ⊥ Wr.

We may assume W0 :=< (1, · · · , 1) >C since < (1, · · · , 1) >C is a maximalcommon eigenspace. Also we see r = d. (see [1], p.58) Now we consider< A0, · · · , Ad >C , which is called the Bose-Mesner algebra and is a commu-tative ring. We know that there is a unique basis E0, · · · , Ed consisting ofprimitive idempotents. (see [1], p.58)

Definition 3.3 Let (X, Rii=0,...,d) be an association scheme.The first eigenmatrix of (X, Rii=0,...,d) is the transformation matrix of thebasis E0, · · · , Ed to A0, A1, · · · , Ad. i.e.

(A0, A1, · · · , Ad) = P (E0, · · · , Ed)

where P is in Md+1(C) and written as:

P =

p0(0) p1(0) · · · pd(0)p0(1)· · ·

p0(d) · · · pd(d)

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where pi(j) is the eigenvalue of Ai on Wj.

Definition 3.4 Let (X, Rii=0,...,d) be a symmetric association scheme. Theassociation scheme (X, Rii=0,...,d) is called a P-polynomial scheme with re-spect to the ordering R0, R1, . . . , Rd if

Ai = vi(A1) (i = 0, 1, · · · , d)

for some polynomials vi(x) of degree i.(X, Rii=0,...,d) is called a Q-polynomial scheme with respect to the orderingE0, E1, · · · , Ed if

Ei = v∗i (E1) (i = 0, 1, . . . , d)

for some polynomials v∗i (x) of degree i under the Hadamard product.The association scheme (X, Rii=0,...,d) is called a P- and Q-polynomialscheme if it is both P-polynomial and Q-polynomial scheme.

4 Quadratic forms schemes

Let V be a vector space over Fq with dimV = n, where q is a power of 2.

Lemma 4.1 Let X be the set of quadratic forms on V . Define the relationsRi on X as follows:

Ri 3 (f, g) ⇐⇒ rank(f − g) = i

Then X = (X, Rii=0,...,n) is a symmetric association scheme.

(Proof) Define σg, πT , κv : X → X for any g ∈ X, T ∈ GL(V ), v ∈ V bythe following:

σg(f) = f + g

πT (f) = f T−1

κv(f) = f + Bf (·, v)2

for any f ∈ X.Let G =< σg, πT , κv|g ∈ X, T ∈ GL(V ), v ∈ V > . Then G acts on Xtransitively. The orbits on X×X of GL(V ) are exactly Rii=0,...,n. Since we

5

know that if a permutation group of X acts transitively then the relations onX defined by the orbits of G on X ×X satisfy the definition of associationschemes (see [1], p.53), the above lemma holds. 2

Proposition 4.1 Let Y = (X, ∆ii=0,...,d), where ∆i = R2i−1 ∪ R2i andd = [n+1

2]. Then Y is a P- and Q-polynomial scheme. Moreover the first

eigenmatrix of Y is represented by Askey-Wilson polynomials.More precisely, let PY be the first eigenmatrix of Y.

PY =

1 θ0 · · · vd(0)1 θ1 · · ·· · · · · · vi(θj) · · ·1 θd · · · vd(d)

Then vi(θj) = kiui(θj),where

ui(θj) = 3Φ2

(q−2i, q−2j, 0, ; q2, q2

q−n−1, q−n,

)

with

θj = θ0 + (q−2j − 1)h, θ0 =(qn − 1)(qn+1 − 1)

q2 − 1, h =

q2n+1

q2 − 1

and ki := p0ii, Φ is defined by the following:

r+1Φr

(a1, · · · , ar+1, ; p, xb1, · · · , br+1,

)=

∞∑

t=0

(a1; p)t · · · (ar+1; p)txt

(b1; p)t · · · (br; p)t(p; p)t

where

(a; p)t =

(1− a) · · · (1− apt−1) (t = 1, 2, · · ·)

1 t = 0

(Proof)(see [1], p.308, [2], p.290) 2

6

5 Computation of PXLet Ai0≤i≤n be the adjacency matrices of X . Let Mi0≤i≤d be the adja-cency matrices of Y . Then the following lemmas hold.

Lemma 5.1 The following equality holds.

A1A2j = (q2j − 1)A2j + q2jA2j+1 (j = 1, 2, · · ·)(Proof) For a fixed (0, δ) ∈ Rk, we will compute

pk1 2j = |γ ∈ X|rank(γ) = 1, rank(γ + δ) = 2j|.

Note that for a quadratic form f , rank(Bf ) = 2j if and only if rank(f) = 2jor 2j + 1. In this caserank(γ + δ) = 2j ⇒ rank(Bδ+γ) = 2j

⇐⇒ rank(Bδ + Bγ) = 2j⇐⇒ rank(Bδ) = 2j⇐⇒ rank(δ) = 2j or 2j + 1

Hence we have pk1 2j = 0 for k 6= 2j or 2j + 1. Suppose rank(δ) = 2j. Then

γ|Radδ = 0 if and only if rank(γ + δ) = 2j. There are q2j − 1 such γ withrank(γ) = 1, so p2j

1 2j = q2j − 1. Similarly we see p2j+11 2j = q2j. 2

Lemma 5.2 The eigenvalues of A1 are −1, qn − 1.

(Proof) This is clear, since A21 = (qn − 1)A0 + (qn − 2)A1 by a similar calcu-

lation as Lemma 5.1. 2

Let W be a vector space over the complex field with dimW = |X|. Recallfrom section 3 that W is decomposed into an orthogonal sum of the maximalcommon eigenspaces of A0, · · · , An, say

W = W0 ⊥ · · · ⊥ Wn.

Similarly W is decomposed into an orthogonal sum of the maximal eigenspacesof M0, · · · ,Md, say

W = U0 ⊥ · · · ⊥ Ud.

We may assume that the eigenvalue of M1 on Ui is θi, where θi (i = 0, 1, · · · , d)is defined in Proposition 4.1.

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Proposition 5.1 If n is even, then for any i ∈ 1, · · · , d there is a unique2-subset j1(i), j2(i) of 1, · · · , n such that Ui = Wj1(i) ⊥ Wj2(i) where theeigenvalue of A1 on Wj1(i) is −1, that of A1 on Wj2(i) is qn − 1.

If n is odd, then for any i ∈ 1, · · · , d − 1 there is the unique 2-subsetj1(i), j2(i) of 1, · · · , n such that Ui = Wj1(i) ⊥ Wj2(i) where the eigenvalueof A1 on Wj1(i) is −1, that on Wj2(i) is qn − 1. Moreover Ud is a maximalcommon eigenspace of A0, · · · , An and the eigenvalue of A1 on Ud is −1.

(Proof) Since Ui (i = 1, · · · , d) is a common eigenspace of M0, · · · ,Md,for any i ∈ 1, · · · , d we can denote Ui = Wl1(i) ⊥ · · · ⊥ Wlr(i) for somel1(i), · · · , lr(i) ∈ 1, · · · , n.

First we show r ≤ 2. Suppose r ≥ 3. Since the eigenvalues of A1 are −1and qn−1, there exists a pair in Wl1(i), · · · ,Wlr(i) such that A1 has the sameeigenvalue on them. We may assume that they are Wl1(i),Wl2(i) without lossof generality. We can determine the eigenvalue of A2 on Wl1(i) ⊥ Wl2(i) sincethe eigenvalue of A1 + A2 on Wl1(i) ⊥ Wl2(i) is only v1(θi). By Lemma 5.1.we can determine the eigenvalue of A3 on Wl1(i) ⊥ Wl2(i). Inductively wecan determine the eigenvalue of Ak (j = 0, · · · , n) on Wl1(i) ⊥ Wl2(i). Thiscontradicts to the maximality of Wl1(i). Hence we have r ≤ 2.

If n is even then we have r = 2 for any i, since d = [n+12

]. Interchagingl1(i) and l2(i) if necessary, we may assume that the eigenvalue of A1 on Wl1(i)

is −1, that on Wl2(i) is qn − 1.If n is odd then there exists the only element of U1, · · · , Ud such that the

eigenvalue of A1 is either −1 or qn−1. Moreover for any other element there isthe unique 2-subset j1(i), j2(i) of 1, · · · , n such that Ui = Wj1(i) ⊥ Wj2(i)

where the eigenvalue of A1 on Wj1(i) is −1, that on Wj2(i) is qn − 1.Moreover we can compute PX explicitly as follows. Let αk be the eigen-

value of Ak on Wj1(i) Lemma5.1 we obtain the following equalities:

(−1)α2j = (q2j − 1)α2j + q2jα2j+1

α2j−1 + α2j = pj(i)

The solution is given by

−α2j = α2j+1 = −j∑

k=0

pk(i)

8

Let βk be the eigenvalue of Ak on Wj2(i). Similarly we obtain the followingequalities:

(qn − 1)β2j = (q2j − 1)β2j + q2jβ2j+1

β2j−1 + β2j = pj(i)

The solution is given by

β2j+1 = −j∑

k=0

pk(i)j−k∏

l=0

(1− qn−2j+2l)

β2j+1 = pj+1(i)− β2j+2

Now we show that the eigenvalue of A1 on Ud is −1 if n is odd. Suppose not.Since n = 2d− 1, the eigenvalue of An on Wd2 is

βn(d) = −d−1∑

k=0

pk(d)d−1−k∏

l=0

(1− q2d−1−2(d−1)+2l)

= −d−1∑

k=0

pk(d)(q; q2)d−k.

On the other hand since Md = An, we see βn(d) = pd(d). We know

θd = −d−1∑

k=0

pk(d).

Now we observe

−d−1∑

k=0

(1− (q : q2)d−k)pk(d).

Note that sign of pk(d) is (−1)k by the facts that θd is the minimum eigenvalueof the P-polynomial scheme Y . This can be seen from the interlacing propertyof orthogonal polynomials (see [1], p.203).Note also that sign of (1 − (q : q2)d−k) is (−1)k+d. So (1 − (q : q2)d−k)pk(d)is positive for all k or negative for all k, which is a contradiction.

The calculation in the proof of Proposition 5.1 gives the following theo-rem.

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Theorem 5.1 If n is even, the first eigenmatrix of X is given by

PX =

1 β1(0) · · · βn(0)1 β1(1) βn(1)1 α1(1) αn(1)· · · · · · βj(i) · · ·· · · · · · αj(i) · · ·1 α1(d) · · · αn(d)

,

and if n is odd,

PX =

1 β1(0) · · · βn(0)1 β1(1) βn(1)1 α1(1) αn(1)· · · · · · βj(i) · · ·· · · · · · αj(i) · · ·1 β1(d− 1) · · · · · ·1 α1(d− 1) · · · · · ·1 α1(d) · · · αn(d)

,

where

β2j+1(i) = −j∑

k=0

pk(i)j−k∏

l=0

(1− qn−2j+2l), β2j+2(i) = pj+1(i)− β2j+1,

−α2j(i) = α2j+1(i) = −j∑

k=0

pk(i).

6 On the parameter of ZWe recall from section 2 that GL(V ) acts on X and its orbits are

Λ2j+1 = f ∈ V |rank(f) = 2j + 1 (0 ≤ j ≤ d− 1)

Λ2j+ = f ∈ V |rank(f) = 2j, f is non-twisted (1 ≤ j ≤ bn− 1

2c),

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Λ2j− = f ∈ V |rank(f) = 2j, f is twisted (1 ≤ j ≤ bn− 1

2c).

We construct an association scheme Z by using GL(V )-orbits. Let X bethe set of quadratic forms on V . The relations R2j+1, R2j+ , R2j− are definedas follows:

(f, g) ∈ Rk ⇐⇒ f − g ∈ Λk,

wherek ∈ 0, 1, 2+, 2−, 3, 4+, 4−, · · · , .

Then Z = (X, Rii=0,1,2+,2−,···) is an association scheme. Suppose that qis two. We will compute some of the intersection numbers of Z. In par-ticular we want to know the coefficients in the expression of A2−A2j+1 =∑

k∈0,1,2+,2−,3,4+,4−,···, pk2 2j+1Ak. This equation can be thought a key to ob-

tain a recurssive equation with respect to the eigenvalues of A2j− ’s.Now we give some lemmas in order to calculate these coefficients. Fix a

basis e1, · · · , en of V .

Lemma 6.1 Let B, B′ be bilinear forms on V with rank(B) = b andrank(B) = c. Let W be a complement of Rad(B) in V and write U =Rad(B −B′) ∩W . Then b− c ≤ rank(B −B′) ≤ b + c. Moreover:(i)rank(B −B′) = b + c if and only if V = Rad(B) + Rad(B′)(and B|U is nondegenerate).(ii)rank(B −B′) = b− c if and only if Rad(B) ⊆ Rad(B′) and dimU = c(and B|U is nondegenerate). In this case B′ is uniquely determined byB′|U = B|U and Rad(B′) = v ∈ V |B(v, U) = 0.

(Proof) (see [2], p.283, Lemma9.5.5) 2

Lemma 6.2 Let f, g be quadratic forms on V with rank(f) = b and rank(g) =c. Then b− c ≤ rank(f − g) ≤ b + c.

(Proof) In the case that both b and c are even, since rank(f) = rank(Bf ), rank(g) =rank(Bg) it follows from Lemma 6.1.Otherwise we have

rank(f − g) ≤ rank(Bf−g) + 1≤ rank(Bf ) + rank(Bg) + 1≤ b + c

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Replacing f by f + g, we have b− c ≤ rank(f − g). 2

By Lemma 6.2 we see that pk2− 2j+1 = 0 for any k ∈ H

where H := 0, 1, 2+, 2−, · · · , (2j − 2)−, (2j + 4)+, (2j + 4)−, (2j + 5), · · ·.Lemma 6.3 Given any alternating form B on V with rankB = 2, there isa unique element f ∈ Λ−2 such that Bf = B.

(Proof) Since q is two, it is clear. 2

Remark 6.1 If we represent quadratic forms on V as homogeneous polyno-mials of degree 2 in x1, · · · , xn, any element of Λ−2 is represented as the formof ρ2 + ρφ + φ2 where

ρ =n∑

i=1

εixi, φ =n∑

i=1

µixi ( εi, µi ∈ F2, i = 1, · · · , n).

6.1 Computation of p2j+

2− 2j+1

For a fixed (0, δ) ∈ R2j+ , we want to compute |γ ∈ Λ2−|γ+δ ∈ Λ2j+1.| Since

we know that Z is an association scheme, we may assume δ =∑j

i=1 x2i−1x2i.Then we have

Rad(Bδ) =< e2j+1, · · · , en >C .

We divide into the following cases with respect to ρ and φ:

(i)ρ(Rad(Bδ)) = 0, φ(Rad(Bδ)) = 0

(ii)ρ(Rad(Bδ)) 6= 0, φ(Rad(Bδ)) 6= 0(iii)ρ(Rad(Bδ)) = 0, φ(Rad(Bδ)) 6= 0

Lemma 6.4 In the case (i) there is no element of γ ∈ Λ2−|γ + δ ∈ Λ2j+1,which holds if δ ∈ Λ2+ .

(Proof) In this case we have

Radδ = RadBδ ⊆ RadBγ = Radγ.

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So we have

Radδ ⊆ γ−1 ∩ δ−1 ∩RadBγ ∩RadBδ ⊆ (γ + δ)−1 ∩RadBδ+γ.

Hence we obtain the following formula:rank(γ + δ) = dimV − dimRad(γ + δ)

= dimV − dimRad(Bγ+δ) ∩ (γ + δ)−1(0)≤ dimV − dimRad(Bγ+δ) ∩Rad(Bδ)= dimV − dimRadBδ

= n− (n− 2j) = 2jThis proves the lemma. 2

Lemma 6.5 In the case (ii) we have φ(u) = ρ(u) for any u ∈ Rad(Bδ),which holds if δ ∈ Λ2+ .

(Proof) Otherwise there is u ∈ Rad(Bδ) such that φ(u) 6= ρ(u). We mayassume that φ(u) = 1 and ρ(u) = 0. By the hypothesis of (ii) there ex-ists v ∈ Rad(Bδ) such that ρ(v) = 1 and u 6= v. We consider the subspaceU =< u, v >. Since U ⊂ Rad(Bδ) and u, v is a hyperbolic pair with respectto Bδ+γ and RadBγ + RadBδ, we have rank(Bδ+γ) = 2j + 2 by Lemma 6.1.This is a contradiction. 2

Using the above lemmas and replacing ρ by ρ + φ we can reduce thecase (ii) to the case (iii). Hence it is enough to consider the only case (iii).By the above lemmas we can denote ρ by

∑2ji=1 εixi.

δ + γ =j∑

i=1

Ni

where Ni = x2i−1x2i + ρ2 + ρφ + φ2

= φ2 +∑j

i=1(ε2i−1x2i−1 + ε2ix2i)φ + (ε2i−1x2i−1)2 + (ε2ix2i)

2 + x2i−1x2i

Ni =

x2i−1x2i if(ε2i−1, ε2i) = (0, 0)x2i−1(x2i + φ) + x2

2i−1 if (ε2i−1, ε2i) = (1, 0)x2i(x2i−1 + φ) + x2

2i if (ε2i−1, ε2i) = (0, 1)(x2i + φ)(x2i−1 + φ) + x2

2i + x22i−1 + φ2 if (ε2i−1, ε2i) = (1, 1)

13

It follows from the above formula that the contribution of φ2 appears in theonly case of (ε2i−1, ε2i) = (1, 1). This contribution is associated to the valueof rank(δ+γ), i.e. it hold that rank(δ+γ) = 2j+1 if and only if the numberof contributions of φ2 is odd.We count the number the choices of (ε1, ε2, · · · , ε2j) for which |i|(ε2i−1, ε2i) =

(1, 1)| is even. This number is given by∑∞

t=0

(j2t

)(22 − 1)j−2t.

Lemma 6.6

∞∑

t=0

(j

2t

)(22 − 1)j−2t = 22j−1 + 22j

(Proof) Let Xj =∑∞

t=0

(j2t

)(22 − 1)j−2t and Yj =

∑∞t=0

(j

2t+1

)(22 − 1)j−2t−1.

We have

(22 − 1 + 1)j =∞∑

s=0

(j

s

)(22 − 1)j−2t = Xj + Yj

and

Yj =∞∑

t=0

(

(j − 1

2t

)+

(j − 1

2t + 1

))(22 − 1)j−2t−1 = Xj−1 + 3Yj−1.

Hence we can obtain the following equations.

Xj + Yj = 22j

Yj = Xj−1 + 3Yj−1

We solve the above equations to complete the proof of the lemma. 2

By Lemma 6.6 and ρ 6= 0 we see that the choice of ρ is 22j−1 + 22j − 1and the choice of φ is 2n − 22j. Note that γ = ρ2 + ρφ + φ2 = ρ(ρ + φ) + φ2.Since (ρ+φ)(Radδ) 6= 0, this implies that we number the choice of γ exactlytwice. Hence we obtain the following proposition.

Proposition 6.1

p2j+

2− 2j+1 = (22j−1 + 22j − 1)(2n−1 − 22j−1)

14

6.2 Computation of p2j−2− 2j+1

For a fixed (0, δ) ∈ R2j− , we compute |γ ∈ Λ2−|γ + δ ∈ Λ2j+1|. We mayassume

δ = x22j−1 + x2

2j + x2j−1x2j +j−1∑

i=1

x2i−1x2i.

By Lemma 6.4, 6.5, we may assume γ = ρ2+ρφ+φ2 where ρ =∑2j

i=1 εixi, φ =∑ni=1 µixi (εi, µi ∈ F2, i = 1, · · · , n).

We haveδ + γ = x2

2j−1 + x22j +

∑ji=1 x2i−1x2i + ρ2 + ρφ + φ2

= x22j−1+x2

2j+φ2+∑j

i=1(εix2i−1+εix2i)φ+(εix2i−1)2+(εix2i)

2+x2i−1x2i.For any i ∈ 1, 2, · · · , j − 1 the contribution of φ2 appear in the only caseof (ε2i−1, ε2i) = (1, 1).For i = j

x2j−1x2j + x22j−1 + x2

2j if(ε2j−1, ε2j) = (0, 0)x2j−1(x2j + φ) + x2

2j if(ε2j−1, ε2j) = (1, 0)x2j(x2j−1 + φ) + x2

2j−1 if(ε2j−1, ε2j) = (0, 1)(x2j + φ)(x2j−1 + φ) + φ2 if(ε2j−1, ε2j) = (1, 1)

¿From the above formula we see that the contribution of φ2 appear in thecase of (ε2j−1, ε2j) 6= (0, 0). Now we consider the choice of (ε1, · · · , ε2j−2)respectively. If (ε2j−1, ε2j) = (0, 0), we can take (Xj−1− 1)’s. If (ε2j−1, ε2j) =(1, 0), (0, 1), (1, 1), we can take Yj−1’s. Hence These are summed up to

3Yj−1 + Xj−1 − 1 = Yj − 1.

We obtain the following proposition.

Proposition 6.2

p2j−2− 2j+1 = (22j−1 − 22j − 1)(2n−1 − 22j−1)

6.3 Computation of p2j+2+

2− 2j+1

¿From Lemma 6.1 we see that if γ ∈ γ ∈ Λ2−|γ + δ ∈ Λ2j+1 then we find a2-space U ⊆ W which is not isotropic with respect to Bγ; and conversely anysuch U determines Bγ uniquely. Moreover Bγ determines γ ∈ Λ2− uniquelyby Lemma 6.3. Note that for given such U rank(γ + δ) is not necessarily2j + 1, which depends on the choice of U . i.e.

15

Lemma 6.7 Let δ ∈ Λ2− ∪ Λ2−, keeping notations above. If δ|U is elliptic,then (γ + δ)|U = 0, which follows that rank(γ + δ) = 2j. If δ|U is hyperbolic,then (γ + δ)|U 6= 0, which follows that rank(γ + δ) = 2j + 1.

(Proof) Suppose that δ|U is elliptic. Since dimU = 2 and q = 2, γ is alsoelliptic, which implies (γ + δ)|U = 0. So we have (δ + γ)−1 ⊆ U . Then wehave

(γ + δ)−1 ⊆ RadBδ+γ = RadBδ ⊕ U.

Hence we have rank(γ + δ) = 2j.Suppose that δ|U is hyperbolic. Similarly we have (γ + δ)|U 6= 0. In this

case we have RadBδ+γ = RadBδ ⊕ U . Since there exist an element u ∈ Usuch that (δ + γ)(u) = 1. Hence we have rank(δ + γ) = 2j + 1. Hence wemust find 2-space U ⊆ W which δ|U is hyperbolic. 2

The following lemmas hold.

Lemma 6.8 Let δ be a nondegenerate twisted quadratic form on W . LetΘ = U ⊆ W | δ|U is nontwisted and dimU = 2. Let Ω be the set ofunordered pairs u, v of W such that δ(u) = δ(v) = 0 and Bδ(u, v) = 1.Then there is a one-to-one correspondence between Θ and Ω.Moreover if dimW = 2j + 2 then |Ω| = 22j−1(22j+1 + 2j − 1).

(Proof) (see [3], p.60) 2

We obtain the intersection number by these lemmas.

Proposition 6.3 p2j+2+

2− 2j+1 = 22j−1(22j+1 + 2j − 1).

6.4 Computation of p2j+2−2− 2j+1

Note that δ is twisted.

Lemma 6.9 Let δ be a nondegenerate twisted quadratic form on W . LetΘ = U ⊆ W | δ|U is nontwisted and dimU = 2. Let Ω be the set ofunordered pairs u, v of W such that δ(u) = δ(v) = 0 and Bδ(u, v) = 1.Then there is a one-to-one correspondence between Θ and Ω.Moreover if dimW = 2j + 2 then |Ω| = 22j−1(22j+1 − 2j − 1).

16

(Proof) (see [3], p.60) 2

By Lemma 6.7, 6.9 we can compute the intersection number as follows:

Proposition 6.4

p2j+2+

2− 2j+1 = 22j−1(22j+1 − 2j − 1).

6.5 Computation of p2j+32− 2j+1

Use Lemma 6.1. In this case whenever δ|U is ellptic or hyperbolic, we haverank(γ + δ) = 2j + 1. So we must count the number of choices of U ⊆ Wsuch that Bδ|U is nonisotropic. i.e.

|U ⊆ W |dimU = 2| − |U ⊆ W |dimU = 2 Bδ|U is isotropic|

Lemma 6.10 Let B be an alternating form on W . If B is nondegenerateand dimW = 2l, then the number of k-subspace of W is

[2j

k

]:=

(22l − 1) · · · (22l−k+1 − 1)

(2k − 1) · · · (2− 1),

and the number of isotropic k-spaces in W is

[j

k

]Πk−1

i=0 (2j+1−i−1 + 1).

(Proof) (see [2], p.274) 2

We can calculate this number as follows.

Proposition 6.5

p2j+32− 2j+1 =

22j(22j+2 − 1)

3.

17

6.6 Computation of p2j+12− 2j+1

For a fixed (0, δ) ∈ R2j+1, we compute |γ ∈ Λ2−|γ + δ ∈ Λ2j+1|.Fix δ =

∑ji=1 x2i−1x2i + x2

2j+1. We may assume that γ has the form

φ2 + φρ + ρ2

where

ρ =n∑

i=1

εixi, φ =n∑

i=1

µixi (εi, µi ∈ F2, i = 1, · · · , n).

We consider the following cases:(i)ρ(Rad(Bδ)) = 0, φ(Rad(Bδ)) = 0(ii)ρ(Rad(Bδ)) 6= 0, φ(Rad(Bδ)) 6= 0(iii)ρ(Rad(Bδ)) 6= 0, φ(Rad(Bδ)) = 0

Now we consider the case (i).Let W be a complement of Rad(Bδ) in V and write U = RadBγ+δ ∩W . ByLemma 6.1 we have

2j − 2 ≤ rank(Bγ+δ) ≤ 2j + 2.

The assumption of (i) follows rank(Bγ+δ) ≤ 2j, dimU = 2, and RadBδ ⊆RadBγ. By Lemma 6.7 it hold that rank(Bγ+δ) 6= 2j − 2 if and only if Bδ|Uis not nondegenerate. Hence the choice of such U is as follows:

[j

2

]Π2−1

i=0 (2j+1−i−1 + 1) =(22j − 1)(22j−2 − 1)

3

Moreover U ⊆ W determines γ ∈ Λ2− uniquely by the condition RadBγ =v ∈ V |Bδ(v, U) = 0 and Lemma 6.1. Since rank(γ + δ) = 2j + 1 for anysuch γ, the number of choices of γ in the case (i) is

(22j − 1)(22j−2 − 1)

3.

Next we consider the case (ii). Recall Lemma 6.5. Replacing ρ by ρ we seethat the case (ii) is reduced to the case (iii).

Finally we consider the case (iii). In this case we may assume φ =∑2ji=1 µixi, ρ = ρ1 + ρ2 where ρ1 =

∑2ji=1 εixi, ρ2 =

∑ni=2j+1 εixi. We di-

vide the case (iii) into the two cases:

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(iii)-1 ρ2 6= x2j+1

(iii)-2 ρ2 = x2j

In the case (iii)-1 whenever we choose φ, ρ1, ρ2 we have rank(γ + δ) = 2j +1.The number of choices of φ is 22j−1 and that of ρ is 2n−22j+1. Note that wecount the number of choices of γ exactly twice. Hence we obtain the numberof choices of γ is

(22j − 1)(2n − 22j+1)

2.

Next we consider the case (iii)-2. Note that the value of rank(γ+δ) dependson the choice of (ε1, · · · , ε2j, µ1, · · · , µ2j). We research the contributions of ρ2

similar to section 6.1. Observe the following equation:

γ+δ =j∑

i=1

x2i−1x2i+(ε2i−1x2i−1+ε2ix2i)ρ+(ε22i−1+µ2

2i−1)x22i−1+(ε2

2i+µ22i)x

22i.

The number of choices of (ε2i−1, ε2i, µ2i−1, µ2i) is 16. The cases of (ε2i−1, ε2i, µ2i−1, µ2i)which appear the contribution of ρ2 are

(1, 1, 0, 0), (1, 0, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), (1, 0, 1, 1), (1, 1, 1, 1).

Hence It holds that rank(γ + δ) = 2j + 1 if and only if

|i|(ε2i−1, ε2i, µ2i−1, µ2i) ∈ S|

is odd where

S := (1, 1, 0, 0), (1, 0, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), (1, 0, 1, 1), (1, 1, 1, 1).

So Kj :=∑∞

t=0

(j

2t+1

)62t+110j−2t−1 is the number we need.

Lemma 6.11

Kj = 24j−1 − 22j−1

(Proof) Let Lj :=∑∞

t=0

(j2t

)62t10j−2t. Then we can obtain the following equa-

tions similar to Lemma 6.6:

Lj + Kj = 24j

Kj = 6Lj−1 + 10Kj−1

19

We solve the above equations to complete the proof of the lemma. 2

Since we count the number of choices of γ exactly twice in this case,

24j−1 − 22j−1

2= 24j−2 − 22j−2

is the number we need.We sum the number in the case (i), (iv)-1, (iv)-2 up to obtain the intersectionnumber.Proposition 6.6

p2j+12− 2j+1 =

(22j − 1)(3 · 2n − 22j+2 − 2)

6.

6.7 Computation of p2j−12− 2j+1

For a fixed (0, δ) ∈ R2j−1, we compute |γ ∈ Λ−2 |γ + δ ∈ Λ2j+1|. Fixδ = x2

2j−1 +∑j−1

i=1 x2i−1x2i. We may assume that γ has the form φ2 + φρ + ρ2

where ρ =∑n

i=1 εixi, φ =∑n

i=1 µixi (εi, µi ∈ F2, i = 1, · · · , n). Moreoverwe have that ρ(Rad(Bδ)) 6= 0, φ(Rad(Bδ)) 6= 0 and there exists i0 > 2jsuch that εi0 6= µi0 by Lemma 6.4.

We divide into the following cases:(i)ρ(Rad(δ)) = 0, φ(Rad(δ)) = 0(ii)ρ(Rad(δ)) 6= 0, φ(Rad(δ)) 6= 0(iii)ρ(Rad(δ)) = 0, φ(Rad(δ)) 6= 0.

In the case (i) there is no possibility.In the case (ii) the number of choices of φ is 2n − 22j−1, the number of

choices of ρ which satisfies the above properties is 2n− 22j. We note that wenumber γ exactly 6 times to see that the number we need is

(2n − 22j−1)(2n − 22j)

6.

In the case (iii) We can denote ρ by x22j−1 +

∑2j−2i=1 εixi. We have

δ + γ =j−1∑

i=1

(x2i−1x2i + ε22i−1x

22i−1 + ε2

2ix22i) + ρφ + φ2,

which implies rank(δ + γ) = 2j. Hence there is no possibity.We obtain the intersection number as follows:

20

Proposition 6.7

p2j−12− 2j+1 =

(2n − 22j−1)(2n − 22j)

6.

These propositions prove the following theorem.

Theorem 6.1 For any j ≥ 1 we have thatA2−A2j+1 = 1

6(2n − 22j−1)(2n + 22j)A2j−1 +

(22j−1 + 2j−1 − 1)(2n−1 − 22j−1)A2j+ +(22j−1 − 2j−1 − 1)(2n−1 − 22j−1)A2j− +

16(22j − 1)(3 · 2n − 22j+2 − 2)A2j+1 +

22j−1(22j+1 + 2j − 1)A2j+2+ +22j−1(22j+1 − 2j − 1)A2j+2− +

1322j(22j+2 − 1)A2j+3

Acknowledgement

I wish to express my deepest gratitude to E. Bannai and A. Munamasa fortheir advices and encouragement.

References

[1] E. Bannai and T. Ito, Algebraic Combinatorics I:Associationschemes, Benjamin/Cummings, Menlo Park, CA, 1984.

[2] A. E. Brouwer, A. M. Cohen and A. Neumaier, Distance RegularGraphs, Springer-Verlag,1989.

[3] M. Suzuki, A Finite Simple Group Theory, Kinokuniya Book Store,1987.

[4] Y. Egawa, Association schemes of quadratic forms, J. of Combinato-rial Theory (A) 38(1985)1-14

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