Stevens, S. , & Zeeuw, F. D. (2017). An improved point‐line incidence bound over arbitrary fields. Bulletin of the London Mathematical Society, 49(5), 842-858. https://doi.org/10.1112/blms.12077
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This document is made available in accordance with publisher policies. Please cite only the published version using the reference above. Full terms of use are available: http://www.bristol.ac.uk/red/research-policy/pure/user-guides/ebr-terms/ arXiv:1609.06284v4 [math.CO] 18 Jul 2017 F 1.1. notation: x ob h ubrof number the be to (1) 1.2. x 4]adZh 4]wr ucsfli xedn h Szemer´edi-Tr the extending in successful were C [42] Zahl and [40] nteseilcase special the in hsbudhsbe mrvduigohrmtos(h urn be current (the 4 methods exponent other the using has improved been has bound This 2 max su (2) the Szemer´edi-Trotter th is the from application deduced famous set who One finite [10], 39]). 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Part of the reason for this may be that the bound (1) can fail for large sets when the field F has a finite subfield G; in particular, if we take = G2 and let be the set of all lines in G2, then ( , ) G 3 but m2/3n2/3 PG 8/3. L OverI P finiteL ≈ fields, | | the extremal situations≈ | | of ‘very small’ or ‘very large’ sets are relatively well understood (where size is relative to the cardinality of the field). For 4/3 very small sets, Grosu [14] achieved the optimal bound O(N ) over Fp, if m,n N and 5N < log log log p. For very large sets, Vinh [41] proved ( , ) mn/q≤ + 2 6 18 I P L ≤ q1/2√mn over any finite field F . When m = n = N q3/2, this bound also meets q ≈ the Szemer´edi-Trotter bound O(N 4/3). However, for ‘medium-size’ sets in between these extremes, these results (and their proofs) have little to say. Over any field F, a simple argument (see e.g. [8, Corollary 5.2] or [9, Claim 2.2]) using the Cauchy-Schwarz inequality gives the following combinatorial bound. Lemma 1. Let be a set of m points in F2 and a set of n lines in F2. Then P L ( , ) min m1/2n + m,mn1/2 + n . I P L ≤ n o Bourgain, Katz and Tao [8] were the first to establish a non-trivial incidence bound in F2. They proved ( , ) N 3/2−ǫ for m,n N = pα, with 0 <α< 2 p I P L ≪ ≤ and ǫ = ǫ(α) > 0. They achieved this by first proving a more general but weaker version of Elekes’s sum-product bound (2), and then, roughly speaking, running Elekes’s argument in reverse. An explicit value ǫ = 1/10678 for α = 1 was found by Helfgott and Rudnev [18], and further improvements to ǫ appeared in the work of Jones [20, 19], with the best bound summarised below. Theorem 2. (Jones [19]) Let be a set of m points in F2 and a set of n lines P p L in F2, with m,n N Range of n Best bound n Bourgain, Katz and Tao [8] first made the observation that if there are many incidences, a large grid-like structure must exist in the point set, and this idea was later refined by Jones [19]. Our proof of Theorem 3 is based on the same idea: We cover the point set by grid-like structures that are projectively equivalent to Cartesian products, and then we bound the incidences on each Cartesian product using the following new incidence bound. Theorem 4. Let A, B F be sets with A = a, B = b, a b and ab2 n3. Let be a finite set of lines⊂ of size n. If| F| has positive| | characteristic≤ p, assume≤ an L p2. Then ≪ (A B, ) a3/4b1/2n3/4 + n. I × L ≪ When a = b = m1/2, the bound in Theorem 4 becomes O(m5/8n3/4 + n), which improves a similar statement in Aksoy Yazici et al. [1] (for comparison, the bound in [1] is O(m3/4n2/3 +n)). Note that our bound has the somewhat unusual property that it is better for uneven products, i.e., for products A B of a fixed size ab, the bound gets better as a/b decreases. × We use Theorem 4 to prove Theorem 3, but it is also interesting in its own right. Indeed, in some applications of point-line incidence bounds, the point set is a Cartesian product; see Sections 5.1 and 5.2 for several examples. Theorem 4 can also be used to obtain a near-sharp estimate on the number of collinear quadruples in a planar point. We refer the reader to Petridis [29] for this estimate, and a subsequent work of Murphy et al. [25] for a wealth of applications. In general, Theorem 4 is quantitatively weaker than the Szemer´edi-Trotter bound (1). If, for instance, we consider Cartesian products = A B with A = B and P × | | | | = = A 2, then Theorem 4 gives the bound ( , ) A 11/4, whereas (1) |L| |P| | | I P L ≪ | | gives ( , ) A 8/3 (over R). Nevertheless, Theorem 4 is optimal for certain sets ofI pointsP L and≪ | lines,| as the following construction of Elekes [11] demonstrates. Example 5. Let a,c be integers. If F has positive characteric p, assume that 2ac Using the condition 2ac < p, we have = 2a2c and = ac2. Every line in contains a points of , so ( , )= a2|P|c2. Theorem 4 yields|L| the matching boundL P I P L ( , ) a3/4(2ac)1/2(ac2)3/4 a2c2 . I P L ≪ ≪ Note that Example 5 works for any choice of the sizes and ; roughly |P| |L| speaking, we can set a ( 2/ )1/3 and c ( 2/ )1/3. However, the choice of and may force≈ the|P| Cartesian|L| product≈ |L|= A|P|B to be rather uneven, in the|P| sense that|L| B is much larger than A . GivenP that the× Szemer´edi-Trotter bound (1) is better for| more| balanced products| | over R, it appears likely that Theorem 4 is not optimal for all choices of A , B and L . | | | | | | 1.4. Applications. Theorem 3 and Theorem 4 lead to several improved bounds in well-known problems. We summarize the applications here, and in Section 5 we will provide some background and full proofs for most of the results. First of all, we use Theorem 4 to deduce the sum-product bounds max A + A , A A A 6/5 and A (A + 1) A 6/5 {| | | · |} ≫ | | | · | ≫ | | for a finite set A F (assuming A p5/8 in positive characteristic p). The first bound was originally⊂ obtained by| Roche-Newton,| ≪ Rudnev and Shkredov [33], but here we recover it using Elekes’s original argument for (2). The second inequality improves slightly on a result from [35]. We also reprove (assuming A p2/3) the bounds | | ≪ A + AA A 3/2 and A(A + A) A 3/2, | | ≫ | | | | ≫ | | which were first proved in [33] and [1] respectively. Next, we use Theorem 4 to prove that the polynomial f(x, y)= x2 + xy satisfies f(A, B) N 5/4 | | ≫ for any finite sets A, B F with A = B = N (assuming N p2/3 in positive characteristic p). This improves⊂ the| | exponent| | for a result of Bourgain≪ [6]. We give two geometric applications of Theorem 3. For a set in F2, write 2 2 P ∆( ) = (px qx) + (py qy) : p, q for the set of squared ‘Euclidean distances’P { determined− by .− We prove that∈ P} P ∆( ) m8/15, | P | ≫ assuming m p15/11 if F has positive characteristic p, and assuming ∆( ) = 0 if 1 is a square≪ in F. This improves on a result of Bourgain, Katz andP Tao6 { [8].} Next,− we prove the following analogue of a theorem of Beck, improving on Jones [19]. For a finite set of m points in F2 (if F has positive characteristic p, assume P m p7/6), either has Ω(m) points on a line, or determines Ω(m8/7) lines. ≪Finally, we recordP in this note that our TheoremP 3 can be plugged into a result F3 of Lewko [24] to give an improved restriction estimate for the paraboloid in p. A contextual discussion of this problem as well as definitions of forthcoming notation are found in [24, 27]. Specifically, combining [24, Theorem 2] with Theorem 3 shows ∗ (assuming 1 is not a square in Fp) that the statement (2 68/19 + ǫ) 1 − 2 R 2 →F3 ≪ holds for all ǫ> 0 for the paraboloid S defined by z = x + y in p, which means that ∨ (gdσ) 68/19+ǫ g L2(F3 ,dσ). k kL (S,dσ) ≪k k p For comparison, the exponent 68/19 = 18/5 2/95 3.579 improves on Lewko’s exponent 745/207 = 18/5 1/1035 3.599, which− in turn≤ improved on the exponent − ≥ ANIMPROVEDPOINT-LINEINCIDENCEBOUND 5 18/5 obtained by Mockenhaupt and Tao [27]. We refer the interested reader to [24] for a thorough treatment of this problem. 1.5. Discussion. We will deduce Theorem 3 from Theorem 4, and we will derive Theorem 4 from the following point-plane incidence bound of Rudnev [34]. Theorem 6 (Rudnev). Let be a set of r points in F3 and let be a set of s planes in F3, with r s. If FR has positive characteristic p, supposeS that r p2. Let k be the maximum≤ number of collinear points in . Then ≪ R ( , ) r1/2s + ks. I R S ≪ This bound is tight if k r1/2; if k of the points are on a line, and all s planes contain that line, then there≥ are ks incidences. Over R, there are known to be better bounds for small values of k. Theorem 6 is based on a ground-breaking result of Guth and Katz [15], which states that a set of N lines in F3, with no N 1/2 lines on a quadric surface, determines O(N 3/2) intersection points. This result played a role in the resolution of the Erd˝os distinct distance problem in R2 in [15]. The proof in [15] uses interpolation to capture the lines in an algebraic surface of degree N 1/2, and then analyses the intersections between the lines in each irreducible component of that surface. The result in [15] was stated over R, but essentially the same proof works over any field, if in positive characteristic one adds the restriction N p2 (see Rudnev [34] and Koll´ar [22, Corollary 40]). A recent work by the second≪ author [43] shortened Rudnev’s proof, removing some of the technicalities. To summarise, a proof from scratch of Theorem 3 would proceed as follows: We cover by grids, for each grid we convert the point-line incidences to point-plane incidences,P these are then converted to line-line intersection points in space (as in [34]), and these are finally bounded using the algebraic techniques in [15]. This gives a nice picture of the connections between the different types of incidence bounds. Previous approaches to incidence problems over finite fields relied on techniques from additive combinatorics, and in particular the Balog-Szemer´edi-Gowers (BSG) theorem [13]. A number of recent papers, including [33, 1, 35], have successfully replaced the traditional application of BSG by more geometric arguments, leading to several quantitative improvements. Our proof of Theorem 3 is another example where BSG is replaced by the geometry inherent within the problem. An obvious question is how to further improve the bounds in Theorem 3 and Theorem 4. We note that, even if the main term in the bound of Theorem 4 were improved to O(m2/3n2/3), our proof would not lead to the same bound in Theorem 3 (the result would be O(m8/11n8/11)). Another interesting open problem, first posed by Bourgain [6], is whether similar bounds can be obtained for non-linear objects, like circles, conics, or other algebraic curves. Over R and C such bounds are known (see e.g. [9, 36]), and over Fp Bourgain [7] proved an incidence bound for hyperbolas. 1.6. Organisation. The rest of the paper is structured as follows. In Section 2 we prove Theorem 4, while in Section 3 and Section 4 we use Theorem 4 to prove Theorem 3. Finally, in Section 5, we work through some of the applications of these new incidence bounds. 6 SOPHIESTEVENSANDFRANKDEZEEUW 2. A point-line incidence bound on Cartesian products As a first step towards Theorem 3, we prove a stronger point-line incidence bound when the point set is a Cartesian product. The idea is to think of point- line incidences as solutions (x,y,s,t) of the equation xs + t = y, where (x, y) is a point and (s,t) represents a line. The number of such solutions can be related to the number of solutions of xs+t = x′s′ +t′. Such a bilinear equation in six variables can then be turned into a point-plane incidence problem, to which Theorem 6 applies. The fact that the point set is a Cartesian product is crucial, because it allows us to ‘split’ the variable x from the variable y. The key to relating the solutions of xs+t = y to the solutions of xs+t = x′s′ +t′ is the Cauchy-Schwarz inequality. Although it is fairly standard, let us explicitly formalise this connection, since we will use it again in Section 5. Lemma 7. Let X and Y be finite sets, and let ϕ : X Z be a function, where Y Z. Then → ⊂ (x, y) X Y : ϕ(x)= y Y 1/2 (x, x′) X X : ϕ(x)= ϕ(x′) 1/2. |{ ∈ × }| ≤ | | · |{ ∈ × }| Proof. Set Xz := x X : ϕ(x) = z for any z Z. By the Cauchy-Schwarz inequality, we have{ ∈ } ∈ (x, y) X Y : ϕ(x)= y = X Y 1/2 X 2. |{ ∈ × }| | y| ≤ | | | y| yX∈Y yX∈Y Combining this with X 2 X 2 (x, x′) X X : ϕ(x)= ϕ(x′) | y| ≤ | z| ≤ |{ ∈ × }| yX∈Y zX∈Z proves the lemma. Proof of Theorem 4. Recall that = n, A = a, B = b, a b and ab2 n3. First we show that, by modifying|L|, we can| assume| | the| following≤ three properties.≤ L There are no vertical lines in . • We remove all vertical lines from L ; they together contribute at most ab incidences, and the assumption ab2L n3 implies ab a3/4b1/2n3/4. We have b2 an. ≤ ≤ • Given that there≤ are no vertical lines, we have the bound (A B, ) an, since each line from intersects each of the a vertical linesI covering× L A≤ B at most once. If b2 >L an, then we get (A B, ) an a3/4b1/2n3/4×. I × L ≤ ≤ At most a1/2n1/2 lines of are concurrent or parallel. • We iteratively remove any pencilL (a pencil is a set of concurrent lines) of 1/2 1/2 more than a n concurrent or parallel lines. Let ni be the number of lines in the i-th pencil that we remove (not counting those that were removed earlier). Then the i-th pencil is involved in at most ab + ni in- cidences. We need at most n/(a1/2n1/2) = a−1/2n1/2 steps to remove all such pencils. In pruning the line set in this manner, we discount at most a−1/2n1/2 ab + n a3/4b1/2n3/4 + n incidences, where we used the · i ≤ assumption b2 Pan. ≤ ANIMPROVEDPOINT-LINEINCIDENCEBOUND 7 Since has no vertical lines, the affine dual ∗ := (c, d) F2 : y = cx + d of is well-defined.L Then we have1 L { ∈ ∈ L} L ( , )= (x,y,s,t) A B ∗ : xs + t = y . I P L |{ ∈ × × L }| If we set E := (x,s,t,x′,s′,t′) (A ∗)2 : xs + t = x′s′ + t′ , { ∈ × L } then Lemma 7 (with X = A ∗, Y = B, Z = F and ϕ(x,s,t)= xs + t) gives × L (3) (A B, )= b1/2 E 1/2. I × L | | We bound E using the point-plane incidence bound in Theorem 6. Define a point set and a| plane| set by := (x, s′,t′) A ∗ , := xs + t = x′s′ + t′ : (x′,s,t) A ∗ . R { ∈ × L } S { ∈ × L } We have = = an and E = ( , ). To apply|R| Theorem|S| 6 we need| | toI checkR S its conditions. The condition that the number of points is O(p2) follows from the assumption that an p2. The condition that there are at most as many points as planes clearly holds,≪ since = . Because of the product structure of = A ∗, the maximum number of|R| collinear|S| points in is bounded by the maximumR of ×a and L the maximum number of collinear points in R∗. The former is bounded by a1/2n1/2, using the fact that a n, which follows fromL a b and ab2 n3. The latter equals the maximum≤ number of concurrent lines≤ in , which by≤ our earlier assumption is also bounded by a1/2n1/2. Therefore, we canL apply Theorem 6 with k = a1/2n1/2 to obtain ( , ) 1/2 + k a3/2n3/2. I R S ≪ |R| |S| |S| ≪ Combining this with (3) and E = ( , ) gives | | I R S (A B, ) b1/2 E 1/2 a3/4b1/2n3/4 , I × L ≪ | | ≪ proving the theorem. Note that the term n in the bound of Theorem 4 comes only from the step in the proof where we ensured that at most a1/2n1/2 lines of are concurrent or parallel. Also observe that we could have stated the bound inL the slightly stronger form Ca3/4b1/2n3/4 + n for a constant C. 3. Finding a Cartesian product In order to apply Theorem 4 to an unstructured point set, we require a means to find large grids in a point set with many incidences. This approach was first taken in the original incidence bound over Fp in [8], where the authors showed that if a point set has many incidences, then a large subset of the points can be captured inside the intersection of two relatively small pencils. Then they used the fact that the set of intersection points of two pencils is projectively equivalent to a Cartesian product. This approach was quantitatively refined by Jones in [19], who showed that, after carefully ‘regularising’ the points, can be efficiently partitioned into a number of subsets, each of which is covered byP two relatively small pencils. Our approach is also based on the fact that if a set is ‘regular’ in the sense that each point lies on a similar number of lines, then there are two pencils whose intersection covers many points of . This fact is captured in Lemma 8 below. P 1We abuse notation by denoting an element in A × B ×L∗ by (a, b, c, d) instead of (a, b, (c, d)). 8 SOPHIESTEVENSANDFRANKDEZEEUW This lemma is a quantitative version of Proposition 4 of [19]. We avoid asymptotic notation in this section, because in the next section we will apply Lemma 8 inside an induction, where we have to be careful with the dependence of the constants. We denote by pq the line in F2 containing the points p, q F2. ∈ Lemma 8. The following holds for any constants c2 >c1 > 0. Let be a set of m points and a set of n lines, such that between c K and P L 1 c2K lines of pass through each point in . Assume K 4n/(c1m), K 8/c1 and K3 26nL2/(c3m). P ≥ ≥ ≥ 1 Then there are distinct points p1, q1 and a set G p1q1 of cardinality 4 4 9 2 ∈ P ⊆ P\ G c1K m/(2 n ), such that G is covered by at most c2K lines from through p| ,|≥ and by at most c K lines from through q . L 1 2 L 1 Proof. Let := ℓ : ℓ ( , )/(2n) . L1 { ∈ L | ∩P|≥I P L } Then we have ( , 1) ( , )/2, since the set of lines not contained in 1 contribute fewerI thanP L n ≥ I( P, L)/(2n) = ( , )/2 incidences to ( , ). LetL p be a point incident· I toP L at least ( I, P )L/(2m) lines in .I SuchP L a point 1 ∈ P I P L1 L1 exists since the set of points that are incident to fewer than ( , 1)/2m lines contribute fewer than m ( , )/(2m)= ( , )/2 incidencesI toP L( , ). · I P L1 I P L1 I P L1 Note that the assumptions of the lemma imply ( , ) c1Km, so we have ( , )/(2n) c Km/(2n) and ( , )/(2m) I(P( L, ≥)/2)/(2m) c K/4. I P L ≥ 1 I P L1 ≥ I P L ≥ 1 Thus the point p1 is incident to at least c1K/4 lines from 1, and each line in 1 is incident to at least (c Km/(2n)) 1 points in p . ItL follows that L 1 − P\{ 1} := q p : p q Q { ∈ P\{ 1} 1 ∈ L} satisfies c K c Km c2K2m (4) 1 1 1 1 , |Q| ≥ 4 2n − ≥ 24n where in the last inequality we used the assumption K 4n/(c m). ≥ 1 The points in still have the property that between c1K and c2K lines of pass through them,Q so we can repeat the argument above, with in the role of L, and the same line set . We let Q P L := ℓ : ℓ ( , )/(2n) . L2 { ∈ L | ∩Q|≥I Q L } As above, we have ( , 2) ( , )/2, and there is a point q1 that is incident to at least I(Q, L )/(2≥ I )Q Lc K/4 lines in . Thus q is∈ incident Q to I Q L2 |Q| ≥ 1 L2 1 at least (c1K/4) 1 lines from 2 other than the line p1q1, and each line in 2 is incident to at least− ( , )/(2nL) c K /(2n) points in . Thus the set L I Q L ≥ 1 |Q| P := q p q : q q R { ∈ Q\ 1 1 1 ∈ L2} satisfies c K c K c K c K c2K2 c4K4m 1 1 1 |Q| 1 1 1 |Q| = 1 |Q| 1 , |R| ≥ 4 − 2n − ≥ 8 · 4n 25n ≥ 29n2 where in the second inequality we used K 8/c1 in the first factor, and both (4) and K3 26n2/(c3m) in the second factor,≥ while in the last inequality we used (4). ≥ 1 As p1 is incident to at most c2K lines, is covered by at most c2K lines from that pass through p , and therefore so isQ . Similarly, is covered by at L 1 R⊂Q R ANIMPROVEDPOINT-LINEINCIDENCEBOUND 9 most c2K lines from that pass through q1. Therefore, we can choose the point set G as a subset of Lwith G c4K4m/(29n2). This concludes the proof. R | |≥ 1 4. Proof of Theorem 3 We will prove that there exists a constant C such that, for all and with n7/8 11/15 (E, ) < C 2−15m n11/15 < 2−11I. I L So E also contributes at most 2−11I incidences to I. Let A := (E D) be the remaining points. By definition of D and E, every P\ ∪ point in A is incident to at least c1K and at most c2K lines of . From the previous paragraph, we know that A contributes at least 1 2 2−11 LI incidences to I. −11 − · 15 We repeatedly use Lemma 8 with c1 = 2 and c2 = 2 , to get the following sequence of grid-like subsets. Let A := A. We iteratively choose G A as 1 i ⊂ i in Lemma 8, so there exist distinct points pi, qi such that Gi is covered by at most 215K lines from through p , and by at most 215K lines from through L i L qi. Then we set Ai+1 = Ai Gi and repeat. We terminate this process at the −15\ s-th step when As+1 2 m (allowing for the possibility that s = 0, which happens if A | 2−15|m ≤, and the process is empty). This results in a sequence A A | | ≤A , with 1 ⊇ 2 ⊇···⊇ s+1 c4K4 A (2−11)4K4(2−15m) K4m G 1 | i| | i|≥ 29n2 ≥ 29n2 ≥ 268n2 10 SOPHIESTEVENSANDFRANKDEZEEUW As the Gi are disjoint by construction, the process terminates after at most m 268n2 s ≤ min G ≤ K4 i{| i|} steps. It is a straightforward calculation to show that throughout the process, the 3 6 2 3 conditions K 4n/(c1 Ai ), K 8/c1 and K 2 n /(c1 Ai ) of Lemma 8 hold if C is chosen sufficiently≥ | large.| ≥ ≥ | | We may apply the inductive assumption to bound (A , ) < C(2−15m)11/15n11/15 =2−11I. I s+1 L Thus the subsets G ,...,G contribute at least 1 3 2−11 I I/2 incidences, 1 s − · ≥ and in particular we have s (5) I 2 (G , ). ≤ I i L Xi=1 We now show that each Gi is projectively equivalent to a Cartesian product. We refer to Richter-Gebert [31] for an introduction to the projective plane and projective transformations, and for the following facts. The affine plane F2 can be extended to a projective plane by adding a line λ at infinity. There are two points α, β on the line at infinity such that all lines through α (except for λ) are horizontal lines in the affine plane, and the lines through β (except for λ are vertical lines in the affine plane. Projective transformations are those bijections of the projective plane that preserve collinearities and point-line incidences, and we call two sets projectively equivalent if there is a projective transformation that maps one bijectively to the other. For any two points p, q there is a projective transformation that sends p and q to α and β (see for instance [31, Theorem 3.4]). For each i, we let τi be a projective transformation sending pi and qi to α and β. The preimage of the line at infinity is then the line piqi, and from Lemma 8 we have G p q = , so τ maps G into the affine plane. Also, if p q happens to be in i ∩ i i ∅ i i i i , then it has no incidences with Gi, so we can ignore it when bounding (Gi, ). L 2 15 15 I L The set Hi = τ(Gi) F is covered by 2 K horizontal lines and 2 K vertical ⊆ 15 lines, so it is contained in a Cartesian product Xi Yi with Xi = Yi 2 K. Since projective transformations preserve incidences,× we have | (H| , )=| | ≤(G , ). I i L I i L We apply Theorem 4 to bound the incidences on each product Xi Yi. In positive characteristic, the extra condition of Theorem 4 holds, since the assumption× m−2n13 p15 gives ≪ X Kn m−4/15n26/15 p2. | i||L| ≪ ≪ ≪ Therefore, we can apply Theorem 4 to obtain (letting c∗ denote the implicit constant in Theorem 4) (G , ) (X Y , ) c∗(215K)3/4(215K)1/2n3/4 5. Applications In this section we give a few corollaries of Theorems 3 and 4. These are meant to give an impression of the possible applications, and we are certain there are more. None of the proofs in this section are new, but we include them here to make it easy for the reader to verify the resulting exponents, as well as the extra condition in positive characteristic. We briefly introduce each problem, but refer to the relevant papers for a more detailed background. 5.1. Sum-product-type bounds. As a first application, we reproduce the best known sum-product bound, which was first proved by Roche-Newton, Rudnev and Shkredov [33], also using Theorem 6. Here we show that it follows from Theorem 4 using the same argument that Elekes [10] used to derive the sum-product bound (2) over R from the Szemer´edi-Trotter theorem. Corollary 9. Let A F be a finite set. If F has positive characteristic p, then assume A p5/8. Then⊂ | | ≪ max A + A , A A A 6/5. {| | | · |} ≫ | | Moreover, if one of A + A , A A is O( A ), then the other is Ω( A 3/2). | | | · | | | | | Proof. Set Mmax := max A + A , A A and Mmin := min A + A , A A . Define a point set and line{| set by | | · |} {| | | · |} := (A + A) (A A), := y = a′(x a) : (a,a′) A A . P × · L { − ∈ × } 2 2 If F has positive characteristic, we need to verify the condition Mmin A p 6/5 | | 6/≪5 of Theorem 4. Either Mmin A , and we are done, or Mmin A , so 2 2 ≫ | | 5/8 ≪ | | that Mmin A p follows from the assumption A p . The other condition | | ≪ 2 3 6| | ≪ of Theorem 4 is that MminMmax = A ; if this failed, it would imply 2 ≤ |L| | | Mmax A and we would be done. The≥ line | |y = a′(x a) contains the point (a′′ + a,a′a′′) for any choice of a′′ A, so each of the A 2 lines− gives at least A incidences. Applying Theorem 4 gives∈ | | | | A 3 I( , ) M 3/4 M 1/2 A 6/4, | | ≤ P L ≪ min max| | so (6) M 3 M 2 A 6, min max ≫ | | which implies the two statements in the corollary. The inequality A + A 2 A A 3 A 6 was obtained in [33] with the condition | | | · | ≫ | | A p5/8, and A + A 3 A A 2 A 6 was obtained in [1] with the condition | | ≪ | | | · | ≫ | | A p3/5. Equation (6) combines both these inequalities, and improves the |condition| ≪ for the second one. As a second application, we prove a lower bound on the size of the set A (A+1), a question raised by Bourgain [6]. Our argument is again in the style of Elekes,· and it was used over R by Garaev and Shen [12]. Our bound is a slight improvement on a result of Rudnev, Shkredov and Stevens [35] (also based on Theorem 6), who proved the same bound up to logarithms. Corollary 10. Let A F be a finite set. If F has positive characteristic p, then assume A p5/8. Then⊂ | | ≪ A (A + 1) A 6/5. | · | ≫ | | 12 SOPHIESTEVENSANDFRANKDEZEEUW Proof. Define a point set and line set by := (A (A + 1)) (A (A + 1)), := y = a (x/(a′ +1)+1):(a,a′) A A . P · × · L { · ∈ × } Applying Theorem 4 gives the same calculation as in the proof of Corollary 9, with M and M replaced by A (A + 1) . min max | · | As another application of this kind, we consider the sets A + BC and A(B + C) for finite sets A, B, C F. Barak, Impagliazzo and Wigderson [4] used [8] to ⊂ 1+ǫ prove that there is an ǫ > 0 such that A + AA A for every A Fp with A 5.2. An expanding polynomial. As another application of Theorem 4, we prove an explicit expansion bound for the polynomial f(x, y)= x2 +xy. This problem was first considered by Bourgain [6], who used the result of [8] to prove the following. For all 0 <ǫ< 1 there is δ > 0 such that if A, B F have size A = B = N pǫ, ⊂ p | | | | ≈ then f(A, B) N 1+δ, where f(A, B) := f(a,b) : (a,b) A B . In other words, f is a| two-variable| ≫ expander. Corollary 11{ shows that x+yz∈ and× x}(y +z) are three- variable expanders, but establishing expansion for two-variable polynomials appears to be harder. Note that, in spite of Corollary 10, g(x, y)= xy+x is not an expander in this sense, because for distinct sets A, B of size N we can have g(A, B) N. As far as we know, no explicit exponents have been published| for Bou|rgain’s ≪ problem over finite fields. Hegyv´ari and Hennecart [17] and Shen [37] generalised Bourgain’s bound to polynomials of a similar form. Over R, it is known that f(A, B) N 4/3 for A, B R with A = B = N, and there is a general theory |of which| polynomials ≫ are expanders⊂ (see| | Raz,| Sharir,| and Solymosi [30]). We prove the explicit expansion bound f(A, B) N 5/4 over any field, using a proof similar to that of [17, Theorem 4]. | | ≫ Corollary 12. Consider the polynomial f(x, y)= x2 +xy, and finite sets A, B F with A = 0 . If F has positive characteristic p, then assume A 2 B p2. Then⊂ 6 { } | | | | ≪ f(A, B) min A 1/2 B 3/4, B 2 . | | ≫ {| | | | | | } Proof. Define E := (a,b,a′,b′) A B A B : f(a,b)= f(a′,b′) . { ∈ × × × } By Lemma 7 (with X = A B, Y = f(A, B), Z = F and ϕ(a,b)= f(a,b)) we get × (7) A B f(A, B) 1/2 E 1/2. | || | ≤ | | | | On the other hand, we can bound E using Theorem 4, by viewing a solution of the equation a2 + ab = (a′)2 + a′b′ as| an| incidence between the point (b,b′) and the line a2 + ax = (a′)2 + a′y. Define a point set and line set by := B B, := ax a′y = (a′)2 a2 : (a,a′) A A, a = a′ . P × L − − ∈ × 6 ± The lines in are distinct (because of the restriction a = a′), and ( , ) E /2. The conditionL B p2 of Theorem 4 follows directly6 ± fromI theP L assumption≥ | | A 2 B p2. The| ||L| other ≪ condition of Theorem 4 is that B 3 3 = A 6. If | | | | ≪ | | ≤ |L| | | this fails, then B > A 2 gives f(A, B) B > A 1/2 B 3/4, where the first inequality is obtained| | by| considering| | the values| ≥ | of| f(x,| y|) with| |x any fixed nonzero element of A (using the assumption A = 0 ). So we can apply Theorem 4 to get 6 { } E 2 ( , ) B 3/4 B 1/2( A 2)3/4 + A 2 = A 3/2 B 5/4 + A 2. | |≤ I P L ≪ | | | | | | | | | | | | | | Together with (7) this gives the bound in the corollary. 5.3. Distinct distances. As mentioned in Section 1, Guth and Katz [15] solved the distinct distance problem in R2 (up to a logarithmic factor). Write d(q, r) = 2 2 (qx rx) + (qy ry) for the squared Euclidean distance between two points q and r, and− write ∆(− ) = d(q, r): q, r for the set of distances determined by . Guth and KatzP proved{ that for ∈R P}2 we have ∆( ) / log . P A related problem is the ‘pinnedP ⊂ distance’ problem,| P which| ≫ |P| asks for|P| the existence of a point from which many distinct distances occur. We write ∆q( )= d(q, r): r for the set of distances ‘pinned’ at q. The approach of [15] doesP not{ apply ∈ P} 14 SOPHIESTEVENSANDFRANKDEZEEUW to this variant, and the best known bound is due to Katz and Tardos [21], who 2 0.86 proved that for any R , there is a q such that ∆q( ) . The finite field versionP ⊂ of this problem∈P was first considered| P | by ≫Bourga |P| in, Katz F2 α and Tao [8], who proved that for p, with = p and 0 <α< 2, there is 1/P2+ ⊂ǫ |P| a q such that ∆q(P ) for some ǫ = ǫ(α) > 0. For large α, explicit versions∈P of this statement| | are ≫ |P| known, with the current best due to Hanson, Lund, and Roche-Newton [16], who proved that p4/3 implies that there is a q |P| ≥ ∈ P with ∆q( ) p. As far as we know, for α < 4/3 no explicit values have been published| P for| ≫ǫ in the statement of Bourgain, Katz and Tao. Here we prove that for α 15/11 we can take ǫ = 1/30 (but we note that for α 4/3 this is weaker than the≤ bound of [16]). Our proof is essentially that of [8], but≥ we take some more care to deal with the case where 1 is a square in F. To avoid degeneracies, [8] proved− their theorem only for finite fields in which 1 is not a square2. Indeed, if 1 is a square in F, then the plane F2 has isotropic− lines. The defining property− of an isotropic line is that the distance between any two points on the line is zero. Explicitly, for r = (r , r ) F2, there are two x y ∈ isotropic lines passing through r, defined via the equations (y ry)= ı (x ry), 2 − ± · − where ı = 1. We use λr and µr to denote the isotropic lines of r. A point set contained in− an isotropic line has ∆( )= 0 ; we exclude this case explicitly. P P { } Corollary 13. Let be a set of m points in F2. If F has positive characteristic, assume m p15/11.P If 1 is a square in F, assume that ∆( ) = 0 . Then there exists a point≪ q such− that P 6 { } ∈P ∆ ( ) m8/15. | q P | ≫ In particular, we have ∆( ) m8/15. | P | ≫ Proof. For two distinct points r, s F2, the set ∈ ℓ := q F2 : d(q, r)= d(q,s) rs { ∈ } is a line (over R, this is the perpendicular bisector of r and s). If 1 is not a square in F, then for a fixed point r, distinct s give distinct lines. However,− when 1 is a − square in F, then for any two points s,t on one of the isotropic lines λr,µr we have ℓrs = ℓrt, which would cause a problem in the counting argument below. To deal with these isotropic lines, first observe that if any line contains at least m/3 points of , but has a point q outside that line, then ∆q( ) m. By the assumption thatP ∆( )P= 0 , we know that is not contained inP an≫ isotropic line, so we can assume thatP 6 any{ } isotropic line containsP at most m/3 points of . For P any fixed r , this implies that there are at least m/3 points not on λr or µr. We will use∈ this P observation below. For a fixed r , define the set of lines ∈P := ℓ : s , d(r, s) =0 . Lr { rs ∈P 6 } We have m, so Theorem 3 gives |Lr|≤ ( , ) m22/15. I P Lr ≪ Distinct s with d(r, s) = 0 give distinct lines in . To see this, assume (by 6 Lr translating the point set) that r = (0, 0), so that the line ℓrs is given by 2sxx + 2This is not stated in the journal version of [8], but it is mentioned in Section 7 of the later version arXiv:math/0301343v3. ANIMPROVEDPOINT-LINEINCIDENCEBOUND 15 2 2 2 2 2syy = sx + sy. This implies that distinct s give distinct lines, unless sx + sy = 0, which is excluded in the definition of r. It follows that an incidence q ℓrs corresponds to exactly two triples (q,r,s)L, (q,s,r) 3 such that d(q, r)= d(q,s∈ ) = 0 (over R, these triples form isosceles triangles, at∈P least when the three points are6 not collinear). Thus we have ( , ) (q,r,s) 3 : d(q, r)= d(q,s) =0 I P Lr ≈ |{ ∈P 6 }| rX∈P = (r, s) 2 : d(q, r)= d(q,s) =0 . |{ ∈P 6 }| qX∈P For every q , Lemma 7 (with X = (λq µq), Y = ∆q( ) 0 , Z = F 0 and ϕ(r)= d(q,∈P r)) gives: P\ ∪ P \{ } \{ } (r, δ) ∆ ( ): d(q, r)= δ =0 2 (r, s) 2 : d(q, r)= d(q,s) =0 |{ ∈P× q P 6 }| . |{ ∈P 6 }| ≥ ∆ ( ) | q P | As we assumed that at least m/3 points of are not on an isotropic line through any given point in , we have, for every q P , P ∈P (r, δ) ∆ ( ): d(q, r)= δ =0 m/3. |{ ∈P× q P 6 }| ≥ Hence we may conclude that 2 3 37/15 m m m ( , r) . ≫ I P L ≫ ∆q( ) ≥ maxq ∆q( ) rX∈P qX∈P | P | | P | This gives max ∆ ( ) m8/15. q | q P | ≫ For A p2/3, using Theorem 4 in the proof above gives ∆ (A A) | | ≪ | q × | ≫ A 5/4, improving on [1, Corollary 13(a)]. Petridis [28] improved this to ∆ (A | | | q × A) A 3/2, but his proof does not seem to apply to unstructured point sets as in Corollary| ≫ | | 13. 5.4. Beck’s theorem. As a final application, we show that Theorem 3 leads to a sharper bound over general fields for a theorem of Beck [5], also known as ‘Beck’s theorem of two extremes’. We say that a line ℓ is determined by a point set if ℓ contains at least two points of . Beck proved that for any set of m pointsP in R2, either (i) has Ω(m) pointsP on a line, or (ii) determines Ω(P m2) distinct lines. P P F2 Naturally there are similar results in other settings. In p, Helfgott and Rudnev [18] established that if m q determines cm2 lines, with P ⊂ q c depending on m/q. Aksoy Yazici et al. [1] used Theorem 6 to show that = A A F2 determines P × ⊆ Ω(m3/2) lines over an arbitrary field F (assuming m p4/3 in positive character- istic). We present a more general, albeit weaker, result≪ for an unstructured point set, improving on [19]. We deduce it from the incidence bound in Theorem 3 us- ing a standard argument. Using Theorem 4 instead of Theorem 3 in our proof of 16 SOPHIESTEVENSANDFRANKDEZEEUW Corollary 14 would give the same result as in [1]. Our proof works for m p7/6, but we note that for m>p it is weaker than the result of Alon [2]. ≪ Corollary 14. Let be a set of m points in F2. If F has positive characteristic p, suppose that m p7P/6. Then one of the following is true: ≪ (i) has Ω(m) points on a line; P (ii) determines Ω(m8/7) lines. P Proof. Let be the set of lines determined by . Partition into log2 m sets suchL that P L ⌊ ⌋ Lj ⊆ L = ℓ :2j ℓ < 2j+1 . Lj { ∈ L ≤ | ∩ P| } 7/8 8/7 Consider j with j > m . We can assume that each j < m , since otherwise (iiL) holds.|L Thus,| in positive characteristic, we can use|L | the assumption 7/6 −2 13 15 m p to get m j p . This lets us apply Theorem 3 to and j . Since≪ every line in gives|L | at≪ least 2j incidences, Theorem 3 gives P L Lj 2j m11/15 11/15, |Lj | ≪ |Lj | which leads to m11/4 . |Lj | ≪ (2j )15/4 2j Each line in j contains Θ(2 ) pairs of points, so all the lines in j together contain L− L O(m11/4(2j ) 7/4) pairs of points. 1/2 7/8 j 1/2 j For j with m j m , Lemma 1 gives 2 j m j , so 2 1/2 L ≤ |L | ≤ |L | ≪ 15/|L8 | ≪ m , and then the number of pairs of points on lines of j is O(m ). Finally, if 1/2 j L −j j m , then Lemma 1 gives 2 j m, so j 2 m, and the number of |L |≤ j|L | ≪ |L | ≪ pairs of points on lines of j is O(2 m). L 3/7 j Let C be a large constant and let U be the union of all j for Cm 2 m/C. By the estimates above, there are then at most L ≤ ≤ m11/4 m m2 O + m15/8 log m + m = O (Cm3/7)7/4 C · C pairs of points on lines in U. 1 m For sufficiently large C, this quantity is less than 2 2 . Thus, the remaining Ω(m2) pairs of distinct points of must lie outside U. 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Sophie Stevens, Department of Mathematics, University of Bristol, Bristol BS8 1TW, United Kingdom E-mail address: [email protected] Frank de Zeeuw, Department of Mathematics, Ecole´ Polytechnique Fed´ erale´ de Lausanne, Lausanne CH-1015, Switzerland E-mail address: [email protected]