Sets with many pairs of orthogonal vectors overfinite fields

Omran Ahmadi Ali Mohammadian

School of Mathematics,Institute for Research in Fundamental Sciences (IPM),

Tehran, Iran

oahmadid@ipm.ir ali_m@ipm.ir

Abstract

Let n be a positive integer and B be a non-degenerate symmetricbilinear form over Fn

q , where q is an odd prime power and Fq is thefinite field with q elements. We determine the largest possible size ofa subset S of Fn

q such that |{B(x,y) |x,y ∈ S and x 6= y}| = 1. Wealso pose some conjectures concerning nearly orthogonal subsets of Fn

q

where a nearly orthogonal subset T of Fnq is a set of vectors in which

among any three distinct vectors there are two vectors x,y so thatB(x,y) = 0.

Keywords. Finite fields; Bilinear forms; Orthogonal sets; Unit distancesproblem

1 Introduction

Erdős in [6] posed his two famous distinct distances and unit distances prob-lems for the plane and later in [7] he considered the extension of these prob-lems to the d-dimensional Euclidean space. The distinct distances problemasks for fd(n), the minimum number of distinct distances among n points inthe d-dimensional Euclidean space, and the unit distances problem asks for

1

gd(n), the maximum number of the unit distances that can occur among npoints in the d-dimensional Euclidean space. He conjectured that there aretwo positive constants c1 and c2 such that

f2(n) > c1n√log n

and fd(n) > c2n2/d,

for any large enough n and any d > 3. Recently, Katz and Guth in [12] cameclose to settling this conjecture for the plane by proving

f2(n) > cn

log n,

for a positive constant c and any large enough n. Considering gd(n), the cased > 4 is easier to handle than the cases d = 2 and d = 3. In some cases it ispossible to give exact value of gd(n). For example, Erdős [7] showed that ifd > 4 is even and n ≡ 0 (mod 2d), then

gd(n) =n2(d− 2)

8+ n,

for any n large enough dependent on d.Recently, there has been a growing interest in the q-analogues of the

above problems, see [4,13–17] for example. By q-analogue problems we meanthat, instead of considering the points in the Euclidean spaces, one can con-sider points in the n-dimensional vector space over Fq and ask appropriateand similar questions, where Fq is the finite field with q elements. For in-stance, Iosevich and Rudnev in [16] defined the distance between two pointsx = (x1, . . . , xn) and y = (y1, . . . , yn) in Fn

q to be (x1−y1)2+ · · ·+(xn−yn)2and proved that if q is an odd prime power and S is a subset of Fn

q with|S| > cq(n+1)/2 for a sufficiently large constant c, then the set of distancesdetermined by pairs of points in S contains Fq \ {0}. Of course, one canchange the definition of distance used in [16] using an arbitrary quadraticform over Fn

q and ask questions analogous to the distinct distances and unitdistances problems.

In this article, we are interested in a q-analogue variant of the unit dis-tances problem as follows. Consider t ∈ Fq and assume that B(., .) is anon-degenerate symmetric bilinear form over Fn

q . We determine the largestpossible cardinality of a subset S of Fn

q so that B(x,y) = t, for every distinctvectors x,y ∈ S.

There are some results in the literature related to the special case oft = 0. Zame in [20] found the largest possible cardinality of a subset S ⊆ Fn

p ,

2

for a prime number p, so that the standard coordinate-wise inner product ofevery two distinct vectors in S is 0. In [19], the general case of non-degeneratesymmetric bilinear forms with t = 0 has been treated. Unfortunately, mainresult of [19] contains an error and although it is claimed that upper boundsobtained in [19] are tight, they are actually not. Our Theorem 3, correctsthe mistake of [19] by establishing sharp upper bounds.

There is a result related to the case q = 2 which has been obtained witha completely different motivation. Fix t ∈ F2. Erdős asked for the maximumnumber of subsets S1, . . . , Sm of an n-element set so that

|Si ∩ Sj | ≡ t (mod 2),

for every i, j with 1 6 i < j 6 m. This was independently answered byBerlekamp [2] for the case t = 0 and Graver [10] for the both cases t = 0and t = 1. The characteristic vector of a subset T of an n-element orderedset X = {x1, . . . , xn} is the vector υT = (υ1, . . . , υn) with

υi =

{1, if xi ∈ T ;0, otherwise.

For subsets T1, T2 ⊆ X, it is clear that |T1 ∩ T2| modulo 2 is equal to theusual inner product of vectors υT1 and υT2 as vectors over F2. Therefore,in order to answer the Erdős question, considering the characteristic vectorsof S1, . . . , Sm, it suffices to find the largest possible cardinality of a subsetV ⊆ Fn

2 such that the standard inner product of every two distinct vectorsin V is t. This is exactly what Berlekamp [2] and Graver [10] did.

This article is organized as follows. In Section 2, we review some basicfacts about bilinear forms over finite fields which will be needed in the subse-quent sections. In Section 3, we present our results and their proofs. Finally,in Section 4, we discuss the problem of nearly orthogonal sets in vector spacesover finite fields and present some conjectures about these sets.

2 Bilinear forms over finite fields

Throughout the paper, we let n be a positive integer and F be a field. Abilinear form over Fn is a map B : Fn × Fn → F such that

B(x,y) = x>Ay, for all x,y ∈ Fn,

where A is an n × n matrix over F . A bilinear form B is called symmetric(respectively, degenerate) if A is a symmetric matrix (respectively, det A =

3

0). If A is the identity matrix, then B is said to be the inner product over Fn.We say that two bilinear forms B1,B2 over Fn with corresponding matricesA1, A2 are equivalent if there exists an invertible matrix P over F such thatA2 = P>A1P . When F is a finite field of odd characteristic, we define η(B)to be equal to η(det A), where η is the quadratic character of F , that is thefunction η : F → C defined as follows:

η(a) =

0, if a = 0;

1, if a is a non-zero square;

−1, if a is not a square,

for any a ∈ F . The next theorem provides a classification of non-degeneratesymmetric bilinear forms over finite fields of odd characteristic.

Theorem 1. [11, p. 79] Let q be an odd prime power and ε be a fixed non-square in Fq. For every non-degenerate symmetric bilinear form B over Fn

q ,one of the following holds:

(i) If n is odd, then B is equivalent to the form

(x,y) 7−→ x1y1 − x2y2 + · · ·+ xn−2yn−2 − xn−1yn−1 + εxnyn;

(ii) If n is even, then B is equivalent to the form

(x,y) 7−→ x1y1−x2y2+· · ·+xn−3yn−3−xn−2yn−2+xn−1yn−1−εxnyn,

where x = (x1, . . . , xn), y = (y1, . . . , yn) and ε ∈ {1, ε} is such that η(B) =η((−1)bn/2cε).

In the rest of the article, we suppose that F is equipped with a non-degenerate symmetric bilinear form B. For simplicity, we write B(x,y) =x·y and B(x,x) = ||x|| if the reference to B is clear. A bijective linear mapσ : Fn → Fn is said to be isometry if σ(x)·σ(y) = x·y, for all x,y ∈ Fn.We will need the following theorem which is a special case of Witt’s theoremon extension of isometries [9, Theorem 2.44].

Theorem 2. Let q be an odd prime power, and let a, b be two vectors in Fnq

with ||a|| = ||b||. Then there exists an isometry σ over Fnq such that σ(a) = b.

4

3 Main results

We first introduce some notations which will be used in the sequel. Let S bea subset of Fn. We denote by 〈S〉 the subspace of Fn generated by S andwe let

S⊥ = {a ∈ Fn |a·v = 0, for all v ∈ S}.For convenience, we will write v⊥ instead of {v}⊥ and 〈v1, . . . ,vk〉 instead of〈{v1, . . . ,vk}〉. By matrix theoretic tools, it is easily verified that dimW +dimW⊥ = n, for any subspace W ⊆ Fn. We say S is orthogonal if s1·s2 =0, for every pair of distinct vectors s1, s2 ∈ S. As usual, ei denotes the i-thelement of the standard basis of Fn in which the i-th coordinate is equal to1 and all the other coordinates are zero. In what follows, we determine themaximum size of an orthogonal subset of Fn

q for each odd prime power q,correcting the mistakes of Theorem1.1 of [19] and Lemma5.1 of [14].

Theorem 3. Let q be an odd prime power and let S ⊆ Fnq \ {0} be an

orthogonal set. Then

|S| 6

q

n−12 , n is odd;

qn2 − 1, n is even and η(ε) = 1;

qn2−1 + 1, n is even and η(ε) = −1,

where ε is as in Theorem 1.

Proof. Suppose that B = {v1, . . . ,vk,vk+1, . . . ,vd} ⊆ S is a basis for 〈S〉such that ||vi|| = 0, for i = 1, . . . , k, and ||vi|| 6= 0, for i = k + 1, . . . , d. Notethat k can be equal to 0 or d. Fix v ∈ 〈S〉 \ {vk+1, . . . ,vd}. Since B is abasis for 〈S〉, there are scalars λ1, . . . , λd ∈ Fq so that

v =

d∑i=1

λivi. (1)

If we take v` with ` ∈ {k + 1, . . . , d} and multiply the both sides of (1) byv`, then from the orthogonality of S we derive that

0 = v·v` =d∑

i=1

λivi·v` = λ`||v`||.

Since ||v`|| 6= 0, it follows that λ` = 0. Therefore, λk+1 = · · · = λd = 0.This and the fact that S does not contain 0 imply that v ∈W \ {0}, whereW = 〈v1, . . . ,vk〉. Thus S ⊆ (W \ {0}) ∪ {vk+1, . . . ,vd} and hence

|S| 6 qk − 1 + d− k. (2)

5

Now from the orthogonality of S and the fact that ||v1|| = · · · = ||vk|| = 0,we deduce that v1, . . . ,vk ∈ S⊥. This means W ⊆ S⊥ and thus

k = dim W 6 dim 〈S〉⊥ = n− dim 〈S〉 = n− d, (3)

which implies that k + d 6 n. From k 6 d it follows that k 6 bn/2c.Moreover, by (3), we have d 6 n− k and using (2), we get

|S| 6 qk + n− 2k − 1. (4)

From the fact that q > 3, we conclude that the right hand side of (4) isthe largest possible when k takes its maximum value. Since k 6 bn/2c,to complete the proof, it suffices to show that k 6 n

2 − 1 when n is evenand η(ε) = −1. We prove this by induction on n. Since η(ε) = −1, usingTheorem 1 there is no vector x ∈ F2

q \ {0} with ||x|| = 0 and so the claimis valid for n = 2. So, assume that n > 4. Applying Theorem 2, it isnot hard to see that there exists a bijective linear map σ : Fn

q → Fnq which

preserves the orthogonality of vectors and σ(vk) = e1 + e2. Notice thatusing Theorem 1, ||e1 + e2|| = 0. This clearly implies that the first and thesecond components of each of the vectors σ(v1), . . . , σ(vk−1) are the same.For i = 1, . . . , k − 1, we consider the vector wi ∈ Fn−2

q obtained by deletingthe first and the second components of σ(vi). Obviously, {w1, . . . ,wk−1} isan orthogonal set in Fn−2

q and therefore k − 1 6 n−22 − 1, by the induction

hypothesis. This completes the proof of the theorem.

The following examples show that the bounds proved in Theorem 3 aresharp.

Example 4. Let n = 2m + 1 and B be the bilinear form given in Theorem1 (i). If we let

S = {(x1, x1, . . . , xm, xm, 0) |x1, . . . , xm ∈ Fq} ∪ {en},

then it is straightforward to see that S \ {0} is an orthogonal set of size qm.

Example 5. Let n = 2m and B be the bilinear form given in Theorem 1 (ii)with η(ε) = 1. Then the set

S = {(x1, x1, . . . , xm, xm) |x1, . . . , xm ∈ Fq} \ {0}

is orthogonal and achieves the bound of Theorem 3.

6

Example 6. Let n = 2m and B be the bilinear form given in Theorem 1 (ii)with η(ε) = −1. If we let

S = {(x1, x1, . . . , xm−1, xm−1, 0, 0) |x1, . . . , xm−1 ∈ Fq} ∪ {en−1, en},

then S \ {0} is clearly an orthogonal set of size qm−1 + 1.

Theorem 3 and the next theorem together determine the largest possiblesize of a subset S ⊆ Fn

q so that |{x·y |x,y ∈ S and x 6= y}| = 1, for eachodd prime power q.

Theorem 7. Let q be an odd prime power and t ∈ Fq \ {0}. Suppose thatS ⊆ Fn

q has the property that s1·s2 = t, for each pair of distinct vectorss1, s2 ∈ S. Then

|S| 6

q

n−12 , n is odd and η(εt) = 1;

qn−32 + 2, n is odd and η(εt) = −1;

qn2−1 + 1, n is even,

where ε is as in Theorem 1.

Proof. Suppose that B = {v1, . . . ,vk,vk+1, . . . ,vd} ⊆ S is a basis for 〈S〉such that ||vi|| = t, for i = 1, . . . , k, and ||vi|| 6= t, for i = k + 1, . . . , d. Notethat k can be equal to 0 or d. Let u,u0 be two distinct vectors in S \B.Since B is a basis for 〈S〉, there are scalars λ1, . . . , λd ∈ Fq so that

u =d∑

i=1

λivi. (5)

Multiplying the both sides of (5) by u0, we get

t = u·u0 =d∑

i=1

λivi·u0 =

(d∑

i=1

λi

)t.

Since t 6= 0, we getd∑

i=1

λi = 1. (6)

Further, for any ` ∈ {k + 1, . . . , d}, by multiplying the both sides of (5) byv` and using (6), we derive

t = u·v` = λ`||v`||+∑i 6=`

λivi·v` = λ`||v`||+ (1− λ`)t. (7)

7

Since ||v`|| 6= t, (7) yields that λ` = 0. Hence, λk+1 = · · · = λd = 0. Thismeans that S ⊆ A ∪ {vk+1, . . . ,vd}, where

A =

{λ1v1 + · · ·+ λk−1vk−1 +

(1−

k−1∑i=1

λi

)vk

∣∣∣∣∣ λ1, . . . , λk−1 ∈ Fq

}.

Therefore,|S| 6 qk−1 + d− k. (8)

For i = 2, . . . , k, we let ui = vi − v1. From the property of S given in thestatement of the theorem and since ||v1|| = · · · = ||vk|| = t, one can deducethat u2, . . . ,uk ∈ S⊥. We have

k − 1 = dim 〈u2, . . . ,uk〉 6 dim 〈S〉⊥ = n− dim 〈S〉 = n− d. (9)

So k+ d− 1 6 n. From k 6 d it follows that k 6 b(n+ 1)/2c. Moreover, by(9), we have d 6 n− k − 1 and using (8), we get

|S| 6 qk−1 + n− 2k + 1. (10)

From q > 3, one can conclude that the right hand side of (10) is the largestpossible when k takes its maximum value. Since k = b(n+1)/2c, to completethe proof, it is enough to show that k 6 n−1

2 when n is odd and η(εt) = −1.We prove this by induction on n. Since η(εt) = −1, there is no element x ∈ Fq

with εx2 = t and so the claim is trivial whenever n = 1. Now, assume thatn > 3. Applying Theorem 2, it is not hard to see that there exists a bijectivelinear map σ : Fn

q → Fnq which preserves the orthogonality of vectors and

σ(uk) = e1+e2. Since {v1}∪{u2, . . . ,uk−1} ⊆ u⊥k , the first and the secondcomponents of the vector σ(v1) and each of the vectors σ(u2), . . . , σ(uk−1)are the same. Thus the first and the second components of σ(vi) are thesame, for i = 1, . . . , k− 1. Let wi ∈ Fn−2

q be the vector obtained by deletingthe first and the second components of σ(vi), for i = 1, . . . , k − 1. Now, iteasily checked that wi·wj = t, for every i, j. So, the induction hypothesesimplies that k − 1 6 n−3

2 . This completes the proof of the theorem.

The following examples show that the bounds proved in Theorem 7 aresharp.

Example 8. Let n = 2m+1 and let B be the bilinear form given in Theorem1 (i). Fix t ∈ Fq with η(εt) = 1. Suppose

S = {(x1, x1, . . . , xm, xm, λ) |x1, . . . , xm ∈ Fq},

where λ ∈ Fq is such that t = ελ2. It is easy to check that |S| = qm ands1·s2 = t, for every pair of distinct vectors s1, s2 ∈ S.

8

In the rest of the examples we need the following well-known lemmawhose proof we bring for the sake of completeness.

Lemma 9. Let a, b, c ∈ Fq \{0}. Then there exist elements x, y ∈ Fq so thatax2 + by2 = c.

Proof. Let S = {as2 | s ∈ Fq} and T = {c − bt2 | t ∈ Fq}. We have |S| =|T | = (q + 1)/2. So S ∩ T 6= ∅. This means that there exist x, y ∈ Fq sothat ax2 = c− by2.

Example 10. Let n = 2m + 1, and let B be the bilinear form given inTheorem 1 (i). Fix t ∈ Fq with η(εt) = −1. By Lemma 9 there are α, β ∈ Fq

such that −α2 + εβ2 = t. It follows from η(εt) = −1 that α 6= 0. Let

S1 = {(x1, x1, . . . , xm−1, xm−1, 0, α, β) |x1, . . . , xm−1 ∈ Fq}

andS2 = {−tα−1en−1 + δen, en−2 + αen−1 + βen},

where

δ =

{1, β = 0;

0, β 6= 0.

If we set S = S1 ∪ S2, then it is easy to verify that |S| = qm−1 + 2 ands1·s2 = t, for every pair of distinct vectors s1, s2 ∈ S.Example 11. Let n = 2m, and let B be the bilinear form given in Theorem1 (ii). Fix t ∈ Fq \ {0}. Suppose

S = {(x1, x1, . . . , xm−1, xm−1, α, β) |x1, . . . , xm−1 ∈ Fq} ∪ {λen−1 + µen},

where α, β ∈ Fq are solutions to the equation α2 − εβ2 = t whose existenceis guaranteed by Lemma 9. Furthermore, suppose that

(λ, µ) =

{(α+ εα−1β, β + 1), α 6= 0;

(α+ 1, β + ε−1αβ−1), β 6= 0.

It is straightforward to check that S is of size qm−1+1 which has the propertythat s1·s2 = t, for every pair of distinct vectors s1, s2 ∈ S.Remark 12. Let q be a power of 2 and consider the inner product over Fn

q .By repeating the proofs of Theorems 3 and 7, it is easily seen that the upperbounds given in these theorems still hold, except for q = 2. Moreover, onecan check that for q = 2, the upper bounds given in Theorem 3 and Theorem7 hold for n > 6 and n > 7, respectively. So, our results in this section aregeneralizations of the corresponding results on F2 which are independentlypresented in the set theoretic language by Berlekamp [2] and Graver [10].

9

4 Nearly orthogonal sets

Let n > 2. A subset S of Fn \ {0} is said to be (k, l)-orthogonal if amongevery k distinct vectors in S, at least l of them are mutually orthogonal.Erdős posed the question of determining the largest possible cardinalityαn(k, l) of a (k, l)-orthogonal set in the Euclidean space Rn. This problemhas attracted the attention of many researchers, see [1, 8, 18] for example.Trivially, αn(2, 2) = n. A simple geometric observation in [8] shows thatα2(k, 2) = 2k − 2. In [18], Rosenfeld using an elegant algebraic argumentproved that αn(3, 2) = 2n. Another nice proof of Rosenfeld’s theorem wasgiven in [5]. Some variations of the Erdős problem have been considered bysome researchers. For instance, it has been shown in [3] that the maximumpossible size of a set S of vectors in Rn which has the property that amongevery three distinct vectors in S, two of them have a fixed angle greater thanπ/2 is the same as αn(3, 2) = 2n.

Continuing with the theme of this article, one can consider the q-analogueof (k, l)-orthogonal sets and ask a question similar to the question posed byErdős. More specifically, we give the following conjecture about the largestpossible cardinality of a (3, 2)-orthogonal set in Fn

q whenever q is odd.

Conjecture 13. Let q be an odd prime power and let S ⊆ Fnq \ {0} be a

(3, 2)-orthogonal set. Then

|S| 6

2qn−12 , n is odd and η(ε) = 1;

2qn−12 − q + 1, n is odd and η(ε) = −1;

2qn2 − q − 1, n is even and η(ε) = 1;

2qn2−1 + 2, n is even and η(ε) = −1,

where ε is as in Theorem 1.

The upper bounds in Conjecture 13 are tight as we have the followingexamples.

Example 14. Let n = 2m+1 and B be the bilinear form given in Theorem1 (i) with η(ε) = 1. If we put

S1 = {(x1, x1, . . . , xm, xm, 0) |x1, . . . , xm ∈ Fq} \ {0}

andS2 = {(0, x1, x1, . . . , xm, xm) |x1, . . . , xm ∈ Fq} \ {0},

then S1 ∪ S2 ∪ {e1, en} is a (3, 2)-orthogonal set of size 2qm.

10

Example 15. Let n = 2m+1 and B be the bilinear form given in Theorem1 (i) with η(ε) = −1. If we set

S1 = {(x1, x1, . . . , xm, xm, 0) |x1, . . . , xm ∈ Fq} \ {0}

and

S2 = {(xm, x1, x1, . . . , xm−1, xm−1, xm, 0) |x1, . . . , xm ∈ Fq} \ {0},

then S1 ∪ S2 ∪ {en, 2en} is a (3, 2)-orthogonal set of size 2qm − q + 1.

Example 16. Let n = 2m and B be the bilinear form given in Theorem1 (ii) with η(ε) = 1. If we let

S1 = {(x1, x1, . . . , xm, xm) |x1, . . . , xm ∈ Fq}

andS2 = {(xm, x1, x1, . . . , xm−1, xm−1, xm) |x1, . . . , xm ∈ Fq},

then (S1 ∪ S2) \ {0} is a (3, 2)-orthogonal set of size 2qm − q − 1.

Example 17. Let n = 2m and B be the bilinear form given in Theorem1 (ii) with η(ε) = −1. If we set

S1 = {(x1, x1, . . . , xm−1, xm−1, 0, 0) |x1, . . . , xm−1 ∈ Fq} \ {0}

and

S2 = {(0, x1, x1, . . . , xm−1, xm−1, 0) |x1, . . . , xm−1 ∈ Fq} \ {0},

then S1∪S2∪{e1, en−1, en, 2en} is a (3, 2)-orthogonal set of size 2qm−1+2.

For the inner product over Fn2 , by considering the characteristic vectors,

Conjecture 13 turns to the following interesting conjecture in terms of thesubsets of an n-element set. Note that the set theoretic analogues of Exam-ples 14 and 16 show that the upper bounds in the following conjecture aresharp.

Conjecture 18. Let n be a large enough integer, and let F be a family ofnon-empty subsets of an n-element set such that among every three distinctelements of F , there is a pair of sets whose intersection is of even cardinality.Then

|F | 6

{2

n+12 , n is odd;

2n2+1 − 3, n is even.

11

Although it seems that proving Conjecture 13 not to be very easy, as itis shown in the following, it is not difficult to establish an upper bound forthe size of a (3, 2)-orthogonal set which is a constant multiple of the boundpredicted by Conjecture 13.

Theorem 19. Let q be an odd prime power and let S ⊆ Fnq \ {0} be a

(3, 2)-orthogonal set. Then

|S| 6 3q

⌊n2

⌋.

Proof. First assume that every two linearly independent vectors in S areorthogonal to one another. Let B = {u1, . . . ,ud} ⊆ S be a basis for 〈S〉.By our assumption, B is an orthogonal set. We claim that S \B is also anorthogonal set. Consider two vectors x =

∑di=1 λiui and y =

∑di=1 µiui in

S \B, where λ1, . . . , λd, µ1, . . . , µd ∈ Fq. If x·y 6= 0, then it follows fromx·y =

∑di=1 λiµi||ui|| that there exists i0 for which λi0µi0 ||ui0 || 6= 0. This

means that there is not an orthogonal pair among x,y,ui0 , a contradiction.This establishes the claim. Now, by Theorem 3, we have |S| = |B|+|S\B| 62qbn/2c, and we are done.

Next assume that there are two linearly independent vectors v,w ∈ Swith v·w 6= 0. We prove the assertion by induction on n. Obviously,one may write S = Sv ∪ Sw ∪ Svw, where Sv = S \ (v⊥ ∪ {v}), Sw =S\(w⊥∪{w}) and Svw = S∩{v,w}⊥. Since S is a (3, 2)-orthogonal set, Svand Sw are two orthogonal sets. So, Theorem 3 implies that |Sv| 6 qbn/2c

and |Sw| 6 qbn/2c. From Svw ⊆ 〈v,w〉⊥, we deduce that (i) |Svw| 6 qfor n ∈ {2, 3}, and (ii) |Svw| 6 3qbn/2c−1 for any n > 4 by the inductionhypothesis. Now the assertion follows from |S| = |Sv| + |Sw| + |Svw| andq > 3.

Acknowledgements

The research of the second author was in part supported by a grant fromIPM(No. 92050405).

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