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On three variable expanders over finite fields

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Using additive character sum estimates, we study expander property of the function x1 + P x2, x3.. We give an alternative proof using spectra of sum–product graphs in the case Keywords:

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 World Scientific Publishing Company

DOI: 10.1142/S1793042113501182

ON THREE-VARIABLE EXPANDERS OVER FINITE FIELDS

LE ANH VINH

University of Education Vietnam National University Hanoi, Vietnam vinhla@vnu.edu.vn

Received 26 February 2013 Accepted 15 September 2013 Published 9 December 2013

Let Fqbe the finite field withq elements and P (x, y) be a polynomial in F q[x, y] Using

additive character sum estimates, we study expander property of the function x1 +

P (x2, x3 ) We give an alternative proof using spectra of sum–product graphs in the case

Keywords: Expanders; extractors; spectral graphs; finite fields.

Mathematics Subject Classification 2010: 11L40, 11T30, 11E39

1 Introduction

LetFq be the finite field with q elements, where q is an odd prime power, and let E

be a finite subset of Fd

q , the d-dimensional vector space over Fq Given a function

f :Fd

q → F q, define

f (E) = {f(x) : x ∈ E},

the image of f under the subset E We say that f is a d-variable expander with expansion index  if |f(E)| ≥ C  |E| 1/d+ for every subset E, possibly under some general density or structural assumptions on E The question of whether certain

polynomials have the expander property has been studied in various classical

prob-lems For example, given a finite subset E ⊂ R d, the Erd˝os distance problem [6] deals with the function ∆ :Rd ×R d → R where ∆(x, y) = x−y It is conjectured

that ∆ is a 2d-variable expander with expansion index 1/2d, i.e |∆(E, E)|  |E| 2/d.

This problem in the Euclidean plane has recently been solved by Guth and Katz [13]

They showed that a set of N points inR2has at least cN/ log N distinct distances. For the latest developments on the Erd˝os distance problem in higher dimensions, see [22, 29], and the references contained therein Here and throughout, X  Y means that X ≥ CY for some large constant C and X  Y means that Y = o(X).

689

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LetFq denote a finite field with q elements, where q, a power of an odd prime, is

viewed as an asymptotic parameter ForE ⊂ F l

q (l ≥ 2), the finite analogue of the

classical Erd˝os distance problem is to determine the smallest possible cardinality of the set

∆(E) = {x − y = (x1− y1)2+· · · + (x l − y l)2: x, y ∈ E} ⊂ F q

The first non-trivial result on the Erd˝os distance problem in vector spaces over finite fields is due to Bourgain, Katz, and Tao [3], who showed that if q is a prime,

q ≡ 3 (mod 4), then for every ε > 0 and E ⊂ F2

q with|E| ≤ C ε q2, there exists δ > 0 such that|∆(E)| ≥ C δ |E|1+δ for some constants C

ε , C δ The relationship between

ε and δ in their arguments, however, is difficult to determine In addition, it is

quite subtle to go up to higher-dimensional cases with these arguments Iosevich and Rudnev [20] used Fourier analytic methods to show that there are absolute

constants c1, c2 > 0 such that for any odd prime power q and any set E ⊂ F d

l of cardinality|E| ≥ c1q l/2, we have

In [33], Vu gave another proof of (1.1) using the graph-theoretic method (see also [30] for a similar proof) Iosevich and Rudnev reformulated the question in analogy with the Falconer distance problem: how large does E ⊂ F l

q , l ≥ 2, needed to be

ensure that ∆(E) contains a positive proportion of the elements of F q The above result implies that if |E| ≥ 2q l+12 then ∆(E) = F q directly in line with Falconer’s result in Euclidean setting that for a setE with Hausdorff dimension greater than

(l + 1)/2, the distance set is of positive measure At first, it seems reasonable that the exponent (l + 1)/2 may be improvable, in line with the Falconer distance conjecture

described above However, Hart, Iosevich, Koh and Rudnev discovered in [17] that

the arithmetic of the problem makes the exponent (l + 1)/2 best possible in odd

dimensions, at least in general fields In even dimensions, it is still possible that the

correct exponent is l/2, in analogy with the Euclidean case In [4], Chapman et al.

took a first step in this direction by showing that ifE ⊂ F2

q satisfies|E| ≥ q 4/3then

|∆(E)| ≥ cq This is in line with Wolff’s result for the Falconer conjecture in the

plane which says that the Lebesgue measure of the set of distances determined by

a subset of the plane of Hausdorff dimension greater than 4/3 is positive.

Another well-known problem, the sum–product estimate problem, also can be interpreted as a result about expanders This problem deals with the fact that for

a given set if one function is non-expanding then it may imply that the another

function is an expander For any subset A, B ⊂ Z, define

A + B = {a + b : a ∈ A, b ∈ B}, A · B = {ab : a ∈ A, b ∈ B}.

Erd˝os and Szemer´edi [7] conjectured that for any subset A⊂ Z, either |A + A| or

|A · A| is large, that is

max(|A + A|, |A · A|)  |A|2.

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To support this conjecture, Erd˝os and Szemer´edi only gave the bound

max(|A + A|, |A · A|)  |A| 1+

for a positive constant  > 0 Explicit bounds on  have been studied by many

researchers (see [5] and the references therein) The current best known bound

 ≥ 3/14 − δ where δ → 0 as |A| → ∞ is due to Solymosi [26] These bounds hold in the more general context of finite subsets ofR In this case, the best known bound, due to Solymosi [28], is ≥ 1/3 The sum–product problems have also been

explored in the context of the finite fieldsFq In this context, one may not rely on the topological structure of the real spaces The first non-trivial result is due to Bourgain, Katz and Tao [3] If A ⊂ F p , p is a prime, and if |A| ≤ p 1− for some

 > 0, then there exists δ > 0 such that then

max(|A + A|, |A · A|)  |A| 1+δ . The relationship between  and δ in their arguments, however, is difficult to

determine Quantitative versions of this estimate have been developed by various researchers, see, for example, [8,9,18,19,27,31,33] and the references therein

A related question that has recently received attention is the following Let

A ⊂ F q , how large does A need to be to make sure that

F

q ⊂ dA2= AA + · · · + AA(d times).

Bourgain [2] showed that ifA ⊂ F q of cardinality|A| ≥ Cq 3/4then

A · A + A · A + A · A = F q

Glibichuk [10] proved in the case of prime fields Fp that for d = 8, one can take

|A| > √p, and extended [11] this result to arbitrary finite fields under a weaker assumption Glibichuk and Konyagin [12] proved that if A is a subgroup ofF

p (p is

a large prime), and |A| > p δ , δ > 0, then kA2=Zp with k 41/δ.

In [16,17], a geometric approach to this problem has been developed In particu-lar, it was proved that if|A| > q 1/2+1/2d thenF

q ⊂ dA2, and if|A| > q 1/2+1/2(2d−1)

then |dA2| ≥ q/2 In the most studied case, d = 2, |2A2|  q whenever |A|  q 2/3 and 2A2 = F

q whenever |A|  q 3/4 S´ark¨ozy [23, 24] also studied the expanding

property of the set |2A2| and |A + A + A · A| Using additive character sum

esti-mates, he proved that|A+A+A·A|  q whenever |A|  q 2/3 and A+A+A ·A = F q

whenever|A|  q 3/4 These results imply that f1(x1, x2, x3, x4) = x1x2+ x3x4and

f2(x1, x2, x3, x4) = x1+ x2+ x3x4 are four-variable expanders for|A|  q 2/3

Fur-thermore, Garaev [9] considered these functions over some special sets A, B, C, D

to obtain new results on the sum–product problem in finite fields The author reproved these results using graph theory methods in [32] Using bounds of multi-plicative character sums, Shparlinski [25] extended the class of sets which satisfy this property Shparlinski also asked for the size of the set |A + B · C| for large

subsets A, B, C ⊂ F q More precisely, he proved that

q − |A + B · C|  |A||B||C| q3 .

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Note that, this result is only effective when |A||B||C|  q2 In prime fields, it follows from the result of Glibichuck and Konyagin [12] that for|A| < q 1/2one has

that |A + A · A|  |A| 7/6 This shows that under these constraints one has that

f (x1, x2, x3) = x1+ x2x3is a three-variable expander of expansion index 1/18 Our first result is that f (x1, x2, x3) = x1+ x2x3 is also a three-variable expander when

|A|  p 1/2 More precisely, we have the following theorem.

Theorem 1.1 For any subsets A, B, C ⊆ F q , we have

|A + B · C|  min



q, |A||B||C|

q



.

In [14, 15], Gyarmati and S´ark¨ozy studied solvability of the general equations

a + b = f (c, d) and ab = f (c, d), a ∈ A, b ∈ B, c ∈ C, d ∈ D where A, B, C, D ⊂ F q

and f (x, y) ∈ F q [x, y] We will use these results to study the expanding property

of the polynomial x1 + f (x2, x3) where f (x, y) ∈ F q [x, y] Our next result is the

following theorem

Theorem 1.2 Assume that f (x, y) ∈ F q [x, y] is not of the form g(x) + h(y) with

g(x), h(x) ∈ F q [x] Write f (x, y) in the form

f (x, y) =

n



k=0

with g k (y) ∈ F q [y], and let K denote the greatest k value with the property that

g k (y) is not identically constant : g K (y) ≡ c and either K = n or g K+1 , , g n (y)

are identically constant Denote the degree of the polynomial g K (y) by D and assume

that (K, q) = 1 (note that K, D > 0 by the assumption) For any subsets A, B, C ⊆

Fq , we have

|A + f(B, C)|  min



q, |A||B||C|

q(D + (K − 1)q 1/2)



.

Although Theorem 1.1 can be obtained directly from Theorem 1.2 by setting

f (x, y) = xy, we choose to present a graph-theoretic proof to show how different

techniques can be used to deal with problems of this kind Using the same

tech-niques, we extend this result in the setting of the finite cyclic ring Let m be a large

integer andZm=Z/mZ be the ring of residues mod m Let γ(m) be the smallest prime divisor of m and τ (m) be the number of divisors of m We identifyZmwith

{0, 1, , m − 1} Define the set of units and the set of non-units in Z mbyZ×

mand

Z0

m, respectively We have the following finite ring analogue of Theorem1.1

Theorem 1.3 For any subsets A, B, C ⊆ Z m , we have

|A + B · C|  min



m, |A||B||C|γ(m)

(τ (m)) 1/2 m2



.

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Note that Theorem 1.3 is most effective when m has only few prime divisors and γ(m)  m 1/ for some  > 0 In the next section, we will prove Theorem 1.2

using additive character sum estimates Finally, we give graph-theoretic proofs of Theorems1.1and1.3in Sec 3

2 General Three-Variable Expanders

We will prove Theorem1.2in this section Our first tool is the following character sum estimate proved by Gyarmati and S´ark¨ozy in [14]

Lemma 2.1 ([ 14, Theorem C]) Assume that α(x), β(x) are complex-valued

functions on Fq , ψ is a non-trivial additive character of Fq , f (x, y) ∈ F q [x, y],

and f (x, y) is not of the form g(x) + h(y) with g(x), h(x) ∈ F q [x] Write f (x, y) in

the form

f (x, y) =

n



k=0

with g k (y) ∈ F q [y], and let K denote the greatest k value with the property that

g k (y) is not identically constant : g K (y) ≡ c and either K = n or g K+1 , , g n (y)

are identically constant Denote the degree of the polynomial g K (y) by D and assume

that (K, q) = 1 (note that K, D > 0 by the assumption) Write

x∈F q



y∈F q

α(x)β(y)ψ(f (x, y)),

x∈F q

|α(x)|2 and Y = 

y∈F q

|β(y)|2.

Then we have

|S| ≤ (XY q(D + (K − 1)q 1/2))1/2 .

Using [14, Theorem C], Gyarmati and S´ark¨ozy [15, Theorem 3] studied the solv-ability of equation

a + b = f (c + d), a ∈ A, b ∈ B, c ∈ C, d ∈ D,

where A, B, C, D ⊆ F q We mimic their arguments to study the equation

a + f (b, c) − a1− f(b1, c1) = 0, a ∈ A, b ∈ B, c ∈ C, a1∈ A1,

b1∈ B1, c1∈ C1, (2.2)

where A, B, C, A1, B1, C1⊂ F q

Lemma 2.2 Assume that q is a prime power, f (x, y) ∈ F q [x, y], f (x, y) is not

of the form g(x) + h(y) Define K and D as in Lemma 2.1, and assume that

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(K, q) = 1 If A, B, C, A1, B1, C1 ⊂ F q , and the number of solutions of Eq (2.2) is

denoted by N, then we have



N − |A||B||C||A1||B1||C1|

q



 ≤ q(D +(K −1)q 1/2)

|A||B||C||A1||B1||C1|. (2.3)

Proof We first recall some character sum estimates over the finite fieldFq Let Ψ

be the set of all additive characters ofFq , let ψ0be the principal character, and let

Ψ∗ ⊂ Ψ be the set of all non-principal characters We have the following identities:



ψ∈Ψ ψ(z) =



q z = 0,

and



z∈F q

ψ(z) =



q ψ = ψ0,

Note that if the field is Zp , then the characters are just e 2πia p and the identity fol-lows by summing up the geometric series For more information about the additive characters, we refer the reader to [21, Sec 11.1]

Therefore, we have

N = 1 q



a∈A,b∈B,c∈C

a1∈A1,b1∈B1,c1∈C1



ψ∈Ψ ψ(a − a1+ f (b, c) − f(b1, c1)).

Separating the principle character term ψ = ψ0, we obtain

N = |A||B||C||A1||B1||C1|

q

+1

q



ψ∈Ψ ∗



a∈A,b∈B,c∈C

a1∈A1,b1∈B1,c1∈C1

ψ(a − a1+ f (b, c) − f(b1, c1)),

which implies that

N − |A||B||C||A1||B1||C1|

q

= 1

q



ψ∈Ψ ∗





a∈A ψ(a)

 



a1∈A1

¯

ψ(a1)

 

b∈B,c∈C

ψ(f (a, b))

×

b ∈B ,c ∈C

¯

ψ(f (a1, b1))

.

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N − |A||B||C||A1||B1||C1|

q





1 q



ψ∈Ψ ∗









a∈A ψ(a)















a1∈A1

¯

ψ(a1)











b∈B,c∈C ψ(f (a, b))











b1∈B1,c1∈C1ψ(f (a¯ 1, b1))





.

By using Lemma2.1and Cauchy’s inequality, it follows that



N − |A||B||C||A1||B1||C1|

q





1 q



ψ∈Ψ ∗









a∈A ψ(a)















a1∈A1

¯

ψ(a1)





(|B||C||B1||C1|) 1/2 q(D + (K − 1)q 1/2)

≤ (D + (K − 1)q 1/2)(|B||C||B1||C1|) 1/2



ψ∈Ψ









a∈A ψ(a)







2

1/2

×



ψ∈Ψ









a∈A

¯

ψ(a)







2

1/2

whence, by the identity



ψ∈Ψ









h∈F q

z h ψ(h)







2

= q 

h∈F q

|z h |2

(for any complex number z h ∈ C),



N − |A||B||C||A1||B1||C1|

q





≤ (D + (K − 1)q 1/2)(|B||C||B1||C1|) 1/2 (q |A|) 1/2 (q |A1|) 1/2

= q(D + (K − 1)q 1/2)(|A||B||C||A1||B1||C1|) 1/2 ,

which completes the proof of Lemma2.2

We are now ready to give a proof of Theorem 1.2 For any a ∈ F q and three

subsets A, B, C ⊆ F q , denote by N a (A, B, C) the number of solutions of x1 +

f (x2, x3) = a with x1∈ A, x2∈ B, x3∈ C Note that

|A + f(B, C)| = |{a : N a (A, B, C) > 0 }|, (2.6) and



a∈F

N a (A, B, C) = |A||B||C|. (2.7)

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On the other hand, let

T = |{(x1, x2, x3, y1, y2, y3)∈ A × B × C

× A × B × C : x1+ f (x2, x3) = y1+ f (y2, y3)}|,

then

a∈F q

N2

a (A, B, C).

It follows from Lemma2.2that



T − |A|2|B| q2|C|2 ≤ q(D + (K − 1)q 1/2)|A||B||C|.

This implies that

T ≤ |A|2|B|2|C|2

q + q(D + (K − 1)q 1/2)|A||B||C|. (2.8)

By Cauchy’s inequality,

|{a : N a (A, B, C) > 0 }|

a∈F q

N2

a (A, B, C) ≥





a∈Fq N a (A, B, C)







2

. (2.9)

Putting (2.6)–(2.9) together, we have

|A + f(B, C)| ≥ |A|2|B|2|C|2 |A|2|B|2|C|2

q + q(D + (K − 1)q 1/2)|A||B||C|

 min



q, |A||B||C|

q(D + (K − 1)q 1/2)



.

This concludes the proof of Theorem1.1

3 The Expander x1 + x2 x3

We first take a detour to recall some graph theory concepts For a graph G with

n vertices, let λ1 ≥ λ2 ≥ · · · ≥ λ n be the eigenvalues of its adjacency matrix

The quantity λ(G) = max {λ2, −λ n } is called the second eigenvalue of G A graph

G = (V, E) is called an (n, d, λ)-graph if it is d-regular, has n vertices, and the

second eigenvalue of G is at most λ It is well known (see [1, Chap 9] for more details) that if λ is much smaller than the degree d, then G has certain random-like

properties

For two (not necessarily) disjoint subsets of vertices U, W ⊂ V , let e(U, W ) be

the number of ordered pairs (u, w) such that u ∈ U, w ∈ W , and (u, w) is an

edge of G For a vertex v of G, let N (v) denote the set of vertices of G adjacent

to v and let d(v) denote its degree Similarly, for a subset U of the vertex set, let

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N U (v) = N (v) ∩ U and d U (v) = |N U (v) | We will need the following well-known

fact (see, for example, [1])

Lemma 3.1 ([ 1, Corollary 9.2.5]) Let G = (V, E) be an (n, d, λ)-graph For

any two sets U, W ⊂ V, we have



e(U,W) − d |U||W | n  ≤ λ|U||W |.

3.1. Sum–product graphs in finite fields

LetFq denote the finite field of q elements For any d ≥ 1, the sum–product graph

SPq,d is defined as follows The vertex set of the sum–product graph SPq,d is the

set V (SP q,d) = Fq × F d

q Two vertices U = (a, b) and V = (c, d) ∈ V (SP q,d)

are connected by an edge, (U, V ) ∈ E(SP q,d ), if and only if a + c = b · d Our

construction is similar to that of Solymosi in [27]

Lemma 3.2 For any d ≥ 1, the sum–product graph, SP q,d , is a (q d+1 , q d , q d/2

)-graph.

Proof It is easy to see that SPq,d is a regular graph of order q d+1 and of valency

q d We now compute the eigenvalues of this multigraph For any a, c ∈ F q and

b = d ∈ F d q, the system

a + u = b · v, c + u = d · v, u ∈ F q , v ∈ F d q

has q d−1 solutions (We can argue as follows There are q d−1 possibilities of v such that (b − d) · v = a − c For each choice of v, there exists a unique u satisfying the system.) If b = d and a = c, then the system has no solution Hence, for any two vertices U = (a, b) and V = (c, d) ∈ V (SP q,d ), if b = d then U and V have exactly

q d−1 common neighbors, and if b = d and a = c then U and V have no common

neighbors Let A be the adjacency matrix of SB q,d (λ) It follows that

A2= AA T = q d−1 J + (q d − q d−1 )I − q d−1 E, (3.1)

where J is the all-one matrix, I is the identity matrix, and E is the adjacency matrix

of the graph B E , where for any two vertices U = (a, b) and V = (c, d) ∈ V (SP q,d),

(U, V ) is an edge of B E if and only if a = c and b = d Since SP q,d is a q d-regular

graph, q d is an eigenvalue of A with the all-one eigenvector 1 The graph SP q,d

is connected so the eigenvalue q d has multiplicity one Besides, choose b, d ∈ F d q

such that b · d = 2a = 0, then SP q,dcontains a triangle with three vertices (−a, 0),

(a, b), and (a, d), which implies that the graph is not bipartite Hence, for any other

eigenvalue θ, |θ| < q d Let v θ denote the corresponding eigenvector of θ Note that

v θ ∈ 1, so J v θ= 0 It follows from (3.1) that

2− q d + q d−1 )v θ=−q d−1 Ev θ (3.2)

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Hence, v θ is also an eigenvector of E Since B E is a disjoint union of q d copies of

the complete graph K q , B E has eigenvalues q − 1 with multiplicity q d, and−1 with

multiplicity q d+1 − q d One corresponding eigenvector of the eigenvalue q − 1 is the

all-one eigenvector 1, and other corresponding eigenvectors are in the orthogonal space 1 Plug into Eq (3.2), A has eigenvalues qdwith multiplicity 1, and all other eigenvalues are±q d/2 and 0 The lemma follows

3.2. Sum–product graphs in finite rings

LetZm=Z/mZ m , be the finite cyclic rings of m elements Notice that we identify

Zm with {0, 1, , m − 1} Similarly to the above, the sum–product graph SP m,d

is defined as follows The vertex set of the sum–product graph SPm,d is the set

V (SP m,d) = Zm × Z d

m Two vertices U = (a, b) and V = (c, d) ∈ V (SP m,d) are

connected by an edge, (U, V ) ∈ E(SP m,d ), if and only if a + c = b · d.

Theorem 3.3 For any d ≥ 1, the sum–product graph SP m,d is a

m d+1 , m d ,

2τ (m) m

d γ(m) d/2



− graph.

Proof It is easy to see that SPm,d is a regular graph of order m d+1 and valency

m d We now compute the eigenvalues of this multigraph For any a, c ∈ Z m and

b = d ∈ Z d

m, we count the number of solutions of the following system:

a + u = b · v mod m, c + u = d · v mod m, u ∈ Z m , v ∈ Z d m (3.3)

For each solution v of

(b − d) · v = a − c mod m, (3.4)

there exists a unique u satisfying the system (3.3) Therefore, we only need to

count the number of solutions of (3.4) Let n be the largest divisor of m such that

all coordinates of b − d are also divisible by n If n  a − c then (3.4) has no solution

Suppose that n | a − c Let γ = (a − c)/n ∈ Z m/n and x = (b − d)/n ∈ Z d m/n We

first count the number of solutions v ∈ Z d

m/n of

Suppose that m/n = p r1

1 p r t t Let S i be the number of solutions v ∈ Z d p ri

i of

x · v = γ mod p r i

Since n is the largest divisor of m such that all coordinates of b − d are also divisible

by n, there exists an index x j which is not divisible by p i So for any choice of

v k ∈ Z p ri

i , k = j, we can always find a unique v j ∈ Z p ri

i , which satisfies (3.6)

This implies that S i = p r i (d−1)

i By the Chinese Remainder Theorem, the number

... q d One corresponding eigenvector of the eigenvalue q − is the

all-one eigenvector 1, and other corresponding eigenvectors are in the orthogonal space 1... q,d)

are connected by an edge, (U, V ) ∈ E(SP q,d ), if and only if a + c = b · d Our

construction is similar to that of Solymosi in... graph-theoretic proofs of Theorems1.1and1.3in Sec

2 General Three- Variable Expanders< /b>

We will prove Theorem1.2in this section Our first tool is the following character sum estimate proved

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