3.2 Application to Paley graphs
For a prime p≡ 1 (mod 4), the Paley graph Pp is the graph with vertex setZp
and an edge between x and y if and only if x−y = a2 for some non-zero a ∈ Zp. More generally, Paley graphs can also be defined in the same manner for any finite field Fq,q ≡1(mod 4), but we will only be concerned with the prime case.
Paley graphs are self-complementary, vertex and edge transitive, and
(p,(p−1)/2,(p−5)/4,(p−1)/4)-strongly regular (see [16] for these and other basic properties of Pp). Paley graphs have received considerable attention over the past decades because they exhibit many properties of random graphs G(p,1/2) where each edge is present with probability1/2. Indeed,Pp form a family ofquasi-random graphs, as shown in [26].
In this note we will be concerned with the independence number of Pp, i.e. the maximal cardinalitys(p)of a setB ⊂Zp such that the difference setB−B contains only quadratic non-residues (and zero). It is clear by self-complementarity that the independence number of Pp is equal to its clique-number. The general lower bound s(p) ≥ (12 +o(1)) log2p is established in [28], while it is proved in [47] that s(p) ≥ clogplog log logp for infinitely many primes p. The "trivial" upper bound s(p)≤ √p has been re-discovered several times (see [34, Theorem 3.9], [87, Problem 13.13], [25, Proposition 4.7], [16, Chapter XIII, Theorem 14], [86, Theorem 31.3], [78, Proposition 4.5], [33, Section 2.8] for various proofs). This bound is notoriously difficult to improve, and it is mentioned explicitly in the selected list of problems [33]. The only improvement we are aware of concerns the special case p = n2 + 1 for which it is proved in [90] that s(p) ≤ n − 1 (the same result was proved independently by T. Sanders – unpublished, personal communication). It is more likely, heuristically, that the lower bound is closer to the truth than the upper bound.
Numerical data [137,138] up top < 10000suggest (very tentatively) that the correct order of magnitude for the clique number ofPp isclog2p(see the discussion and the plot of the function s(p)at [139]).
In this note we prove the slightly improved upper bound s(p) ≤ √p−1 for the majorityof the primes p= 4k+ 1 (we will often suppress the dependence on p, and just write sinstead of s(p)). The proof has two cornerstones. The first is Delsarte’s bound as described in Theorem 3.1.4. The second is a "subclique trick" introduced in [106], which can be incorporated to the linear programming bound to yield an improvement. This will be described in Lemma 3.2.1 below.
We will denote the set of nonzero quadratic residues by Q, and that of nonzero non-residues by N Q. Note that 0∈/ Qand 0∈/ N Q.
3.2.1 The improved upper bound
We will first formulate the "subclique trick" introduced in [106], in the general setting. We will describe it in finite groups for simplicity.
Lemma 3.2.1. ( [106])AssumeB ={b1, . . . bm} ⊂ G is such thatbj−bk∈Ac∪{0}. Let h(x) = |B1|∑
y∈G1B(y)1B(x+y). Assume D⊂ G is such that any selection of k distinct elements of D contains two such that their difference falls in A. Then
∑
x∈D
h(x)≤k−1. (3.53)
Proof. Let us evaluate the sum in question:
∑
x∈D
h(x) = 1
|B|
∑
x,y∈G
1D(x)1B(y)1B(x+y) = (3.54) 1
|B|
∑
y∈G
1B(y)∑
x∈G
1D(x)1B(x+y).
The point is that the inner sum is ≤ k −1 for each y ∈ G. Indeed, if it were ≥ k for some y then there would exist distinct elements d1, . . . dk ∈ D such that the elements b1 =d1+y, . . . , bk =dk+y are all in B. By assumption, however, there would exist two of them, saydi anddj such that di−dj ∈A which contradicts that bi−bj ∈Ac.
In what way is this an improvement to Delsarte’s bound? The function h(x) trivially belongs to the set S+(G \ A), and satisfies ˆh ≥ 0. The point is that inequality (3.53) might introduce new linear constraints on h(x) if appropriate sets D⊂ G exist.
After this preparation we are in position to state the slightly improved upper bound on the independence number of Paley-graphs.
Theorem 3.2.2. ( [4]∗) Let p = 4k + 1 be a prime, and B ⊂ Zp, |B| = s, be a maximal set such that B−B ⊂N Q∪ {0}. The following hold:
(i) if n= [√p] is even then s2+s−1≤p (ii) if n = [√
p] is odd then s2 + 2s−2≤p.
Proof. Consider the function f(x) = 1B∗1−B(x) = ∑
y∈Zp1B(y)1B(x+y), which gives the number of representationsx=b−b′ with b, b′ ∈B. This function has the following properties:
f(0) =s (3.55)
fˆ(1) = ∑
x∈Zp
f(x) = s2 (3.56)
f(x)≥0 if x∈N Q, f(x) = 0 if x∈Q (3.57) f(γ) =ˆ ∑
x∈Zp
f(x)γ(x)≥0for all γ ∈Zˆp. (3.58) Using the notations of Section 3.1, properties (3.55), (3.56), (3.57) mean that f ∈ S+(N Q), and fˆ≥ 0 by (3.58). It is quite easy to determine the quantities λ+(Q) and λ−(Q), both of which turn out to be √1p. This implies the trivial bound
|B| ≤ √p, but we will impose further restrictions on f to get an improvement.
In order to use Lemma 3.2.1 we would need to identify a ’large’ setD⊂Zp such thatD−D⊂Q∪{0}. At first glance this seems to be impossible, as such setsDare cliques themselves, and hence are necessarily ’small’. However, we can circumvent this problem by considering the translated copies of the hypothetical set B. The details are as follows.
Letχ denote the quadratic multiplicative character, i.e. χ(t) = ±1according to whether t∈Q or t∈N Q (andχ(0) = 0). Let
φ(t) = ∑
b∈B
χ(t+b), (3.59)
giving the number of quadratic residues minus the number of non-residues in the shifted set t + B. If t ∈ −B then by assumption φ(t) = −s + 1 < 0 (for the last inequality note that s ≥ 2 for every p = 4k + 1). Also, ∑
t∈Zpφ(t) =
∑
b∈B
∑
t∈Zpχ(t+b) = 0. Therefore, φ(t) must also assume some positive values.
Let t0 be the place where φ assumes its maximum, φ(t0) >0 (note that t0 ∈ −/ B).
Let Bt0 = (t0 +B)∩Q denote the set of quadratic residues contained in t0 +B, and let r = |Bt0|. Then r > s2, and Bt0 is a set of quadratic residues such that Bt0 −Bt0 ⊂N Q∪ {0}. We claim that
s≤1 + p−1
2r . (3.60)
Let z ∈N Q be arbitrary, and consider the set Cz =zBt0. Then Cz ⊂N Q and Cz −Cz ⊂Q∪ {0}. By Lemma 3.2.1 we have
∑
x∈Cz
f(x)≤s. (3.61)
Summing up (3.61) for all z ∈N Q we obtain
Finally, putting together (3.55), (3.56), (3.57), (3.62) we conclude s2 = ∑ However, in the latter case we will need the stronger inequality r≥ s+32 .
Lemma 3.2.3. ( [4]∗) If s is odd, then r≥ s+32 . We also know the sum of the squares:
p−s By our assumption, all the positive values of φ(t) are equal to 1. Also, by the definition of φ we have φ(t) ≥ −s for all t. Notice, however, that the less trivial
inequality
−s+ 2≤φ(t)≤1 (3.66)
also holds. Indeed, φ(t) = −s is impossible because in that case the set B could be further extended by the element−t. Also, φ(t) = −s+ 1is impossible due to parity reasons.
The end of the argument is that the constraints (3.64), (3.65), (3.66) contradict each other. Indeed, from (3.64) and (3.65), we compute
p−s
Therefore, we conclude that there exists a t such that φ(t) > 1, and hence φ(t)≥3 and r≥ s+32 .
Now we can conclude the proof of the theorem. Indeed, if s≤ n−1, then both parts of the claim follow immediately. We assume therefore thats=n. If s is even we have seen thatr≥ s2+ 1, hence s2+s−1≤pfollows directly from (3.60). Ifs is odd, the trivial estimater ≥ s+12 combined with (3.60) just leads to the well-known s ≤ √p. But we have proved that r ≥ s+32 and hence s2+ 2s−2≤ p follows from (3.60).
Remark 3.2.4. It is clear from (3.60) that any improved lower bound on r will lead to an improved upper bound on s. If one thinks of elements of Zp as being quadratic residues randomly with probability1/2, then we expect thatr ≥ s2+c√
s.
This would lead to an estimate s ≤ √p−cp1/4. This seems to be the limit of this method. In order to get an improved lower bound on r one can try to prove non-trivial upper bounds on the third moment ∑
t∈Zpφ3(t). To do this, we would need that the distribution of numbers bb1−b2
1−b3 is approximately uniform on Q as b1, b2, b3 ranges over B. This is plausible because if s ≈ √p then the distribution of B−B must be close to uniform onN Q. However, we could not prove anything rigorous in this direction.
Remark 3.2.5. An alternative proof of Lemma 3.2.3 is as follows. The multiplica-tive character χ is a polynomial of degree p−21, χ(x) = xp−21, and therefore so is the function φ(t). The sum of the positive values of φ(t) is ≥ s(s−1), by (3.64).
However, the value +1 is assumed at most p−21 < s(s−1) times due to the degree of φ. Therefore, there must be other positive values, i.e. values where φ(t) ≥3. It is instructive to see how both proofs break down in Fp2 where the example C =Fp
shows that there exists a clique of sizep. In that case the functionφ(t)assumes the values −p+ 1 and +1 only (it becomes a constant polynomial overFp).
Remark 3.2.6. Theorem 3.2.2 gives the bounds≤[√
p]−1for about three quarters of the primes p = 4k+ 1. Indeed, part (ii) gives this bound for almost all p such that n= [√
p] is odd, with the only exception whenp= (n+ 1)2−3. Part(i)gives the improved bound s ≤ n−1 if n2+n−1 > p. This happens for about half of the primes p such that n is even. To make these statements rigorous we note that
√p/2 is uniformly distributed modulo one, when p ranges over primes of the form p= 4k+ 1: this is a special case of a result of Balog, [6, Theorem 1].