An explicit incidence theorem in F_p

0.16 MB·

by  Harald Andres Helfgott

AN EXPLICIT INCIDENCE THEOREM IN F
p
HARALDANDRE´SHELFGOTT ANDMISHA RUDNEV
0
1
20 Aelebmsternatsc,tn.<Lept.PSe=ePA×asAa⊂setFpof×pFopin,tpsainptrhimeep.laAnsesuovmeertFhpa.tWPe=shAow×tAhahtatshne
n pairs of points in P determine cn1+2167 lines, where c is an absolute constant.
a ≥
We derive from this an incidence theorem: the number of incidences between
J
a set of n points and a set of n lines in the projective plane over Fp (n<√p) is
19 bounded by Cn32−106178, where C is an absolute constant.

]
O
C
1. Introduction
.
h
t In 1983 J. Beck proved the following incidence theorem in R2. Let P be a
a
m set of points in R2. Then either P contains cP points on a straight line, or the
| |
pairs of points of P determine at least cP 2 distinct lines, where c is an absolute
[
constant. (We write S for the number o|f |elements of a finite set S; c and C will
2 | |
henceforth denote some absolute constants, changing from one place to another.)
v
0 Beck’s original paper ([1]) was followed – in the same issue of the same journal –
8 by a result of Szemer´edi and Trotter stating that the number of incidences between
19 m straight lines and n points in R2 is O m+n+(mn)32 . The Szemer´edi-Trotter
1. theorem implies Beck’s theorem as a coro(cid:16)llary. (cid:17)
0 We prove a finite field version of Beck’s theorem.
0
1 Theorem 1. Let A F , where p is larger than an absolute constant. Let P =
p
v: A A F2. Let L =⊂L(P) be the set of all lines determined by pairs of points of
Xi ele×ment⊂s ofpP. If A < √p, then
| |
ar L(P) cP 1+2167,
| | ≥ | |
where c > 0 is an absolute constant.
All constants here and later are independent of p.
The statement L(P) cP 1+2167 cannot possibly hold in F2 in full generality,
| | ≥ | | p
as P = F2 itself generates only O(P ) lines. A non-trivial theorem of Beck type is
p | |
implicit in the well-known paper of Bourgain, Katz, and Tao ([4]) – namely, their
results imply that, if P F2 is a Cartesian product and P < p2−∆ (∆ > 0), then
⊂ p | |
L(P) = O P 1+δ(∆) for some δ(∆) > 0. However, there was no attempt to make
| | | |
explicit or optimise δ(∆).
(cid:0) (cid:1)
2000 Mathematics Subject Classification. 68R05,11B75.
1
2 HARALDANDRE´SHELFGOTTANDMISHARUDNEV
The problem of studying L(P) is an incidence problem, and [4] proves that for
n< p2−∆, the number of incidences between n lines and n points is O n 32−δ1(∆) .
| |
This statement happens to be on about the same level of generality(cid:16)as claimin(cid:17)g
that a set of n points that is a Cartesian product A A (with n = A2 < p2−∆)
× | |
determines Ω n1+δ2(∆) distinct lines. These results in [4] stem from a non-trivial
sum-product estimate in F proven in [4]. The sum-product estimate of [4] (the
p
(cid:0) (cid:1)
following formulation also includes Konyagin’s contribution [7]) says that, if A+A
and A A denote, respectively, the set of all sums and products of pairs of elements
of A ·F , then, as long as A < p1−∆, one has
p
⊂ | |
(1.1) max(A+A, A A) cA1+δ3(∆),
| | | · | ≥ | |
for some absolute c. The quantitative relation between the deltas was not estab-
lished.
(At this point one traditionally mentions the Erd¨os-Szemer´edi conjecture, which
states that, if A is a subset of integers, then max(A+A, A A) cA1+δ holds
| | | · | ≥ | |
for any 0 < δ < 1, where c is allowed to depend only on δ, but not on A.)
We return to F for the rest of the paper. On the level of the existenc|e|of positive
p
exponents,anontrivialsum-productestimateimpliesanon-trivialincidenceorBeck-
typetheorem,andconversely. Garaev [6]succeededinobtainingaquantitative sum-
product estimate in Fp: for a small enough (say A < √p) subset A of Fp, either
| |
A+AorA Ahascardinality atleast A1+114,uptoamultipleofapowerofclog A.
· | | | |
Katz and Shen ([8]) elaborated on a particular application of the Plu¨nnecke-Ruzsa
inequality in Garaev’s proof and improved the result to A1+113, up to a multiple of
| |
a power of clog A. Bourgain and Garaev ([3]) incorporated a covering argument
(whose variant w|e |cite as Lemma 3) and improved the estimate to A1+112, up to a
| |
multiple of a power of clog A. Li ([9]) showed that a multiple of a power of log A
| | | |
can be done away with; thus the best result known states that, for A < √p,
| |
max(A+A, A A) cA1+112,
| | | · | ≥ | |
where c is an absolute constant.
Our construction uses the techniques laid out in the above-mentioned papers and
locally follows rather closely the exposition from some of those papers. The main
result, Theorem 1, an explicit Beck type incidence theorem, implies the following
incidence theorem for n points and n straight lines in the projective plane P2(F ).
p
Theorem 2. If P and L are sets of points and lines in P2(F ) with P , L = n < p,
p
| | | |
then the number of incidences
I(P,L) = (p,l) P L : p l Cn23−106178
|{ ∈ × ∈ }| ≤
for some absolute C.
We call (p,l) P L : p l the set of incidences.
{ ∈ × ∈ }
Theproofof this theorem repeats thepigeonholing argument of [4] untilitmerges
with the proof of our Theorem 1. It is given at the end of this note.
AN EXPLICIT INCIDENCE THEOREM IN Fp 3
Remark 3. In both Theorems 1 and 2 one can easily extend the estimate to larger
sets, by a straightforward adaptation of Case (i) of the estimate (3.19) in the end
of our proof. We haven’t done so aiming at an estimate which does not contain p
explicitly; we leave the case of “larger” sets to the interested reader.
1.1. Acknowledgments. H. A.Helfgott is supportedin partby EPSRC grant EP-
E054919/1. The authors would like to thank M. Garaev, T. Jones and O. Roche-
Newton for their helpful comments on the draft of this paper.
2. Background in arithmetic combinatorics
We use the following largely standard arithmetic combinatorics lemmata. In
the sequel, in order to suppress constants, we will use the , , notations in
≪ ≫ ≈
estimates: X Y means X cY for some c, X Y means X cY
| | ≫ | | | | ≥ | | | | ≪ | | | | ≤ | |
for some c, X Y means X C Y and X c Y hold for some C , c .
1 2 1 2
| | ≈ | | | | ≤ | | | | ≥ | |
We abuse the English language in accordance with these notations by saying “at
least”, “at most”, or approximately in the sense conveyed by the symbols ,
≫ ≪
, , respectively. To avoid confusion, we will enclose “at least” and “at most” in
≈
quotation marks when we use them in this way.
Thus, for example, saying that X is “at least” Y means X Y .
| | | | | | ≫ | |
We adopt the following formulation of the Balog-Szemer´edi-Gowers theorem.
Lemma 1 (Balog-Szemer´edi-Gowers theorem). Let X,Y be additive sets of n el-
ements, and α (0,1). Suppose that there is a set of αn2 pairs of elements
∈
(x,y) X Y on which the sum x+y takes at most n distinct values.
∈ ×
Then there exist subsets X′ X, Y′ Y, with X′ , Y′ αn, such that
⊆ ⊆ | | | | ≫
Range of x+y on X′ Y′ α−5n.
| × |≪
The modern graph-theoretical proof of the Balog-Szemer´edi-Gowers theorem can
befoundin[13,Thm.2.29]. Theproofasappearsinthatstandardreferenceappears
to have a typographical error, however, which leads to an exponent of 4, rather
−
than to the correct (and somewhat weaker) exponent of 5 that we have above. For
−
a proof yielding the exponent 5, see [5].
−
We adopt the following form for the Plu¨nnecke-Ruzsa inequality, due to Ruzsa
([10]).
Lemma 2. Let Y; X ,...X be additive sets. Then there exists a non-empty subset
1 k
Y′ Y, such that
⊆
k Y +X
(2.1) Y′+X +...+X i=1| i| Y′ .
| 1 k| ≤ Y k−1 | |
Q | |
Ruzsa’s inequality immediately implies that
k Y +X
(2.2) X +...+X i=1| i|,
| 1 k|≤ Y k−1
Q | |
4 HARALDANDRE´SHELFGOTTANDMISHARUDNEV
for any “dummy set” Y. To alow for signs as well + signs, this inequality is often
−
used in conjunction with the Ruzsa distance inequality
X X X X
1 3 3 2
(2.3) X X | − || − |.
1 2
| − | ≤ X
3
| |
(See, e.g., [13, Lemma 2.6] for a brief proof of (2.3).)
Remark 4. Katz and Shen ([8]) showed that, at the expense of acquiring a constant
in the left-hand side of (2.1), one can make Y′ contain an arbitrarily large propor-
tion of Y. This trick was used to gain an improvement in ([8]) and subsequent
above-mentioned papers on the sum-product problem. We have not found a way to
take advantage of it here, as we shall be dealing with set families, controlling their
intersections, refinements therefore being apparently forbidden. I.e., we essentially
use Plu¨nnecke-Ruzsa only in the “crude” form (2.2).
Finally, we need the following covering lemma (see e.g. [3], [11], [9]; the proof is
a Cauchy-Schwartz type averaging argument).
Lemma 3. Let X and X be additive sets. Then for any ε (0,1) and some
1 2
∈
C(ε)
constant C(ε), there exist min(X +X , X X ) translates of X whose
1 2 1 2 2
X | | | − |
2
| |
union contains not less than (1 ε)X elements of X .
1 1
− | |
3. Proof of Theorem 1
WewillprovetheresultmoregenerallyforP = A A F F , A = A = n,
1 2 p p 1 2
× ⊂ × | | | |
n< √p.
Let L(P) bethe set of straight lines generated by pairs of elements of P. Suppose
that
(3.1) L(P) n2+2δ.
| | ≈
where δ < 1 . We will show how to reach a contradiction.
267
The approximately n4 pairs of distinct points of P are distributed between ap-
proximately n2+2δ lines. This implies that a positive proportion of approximately
n4 pairs of those points are supported on rich lines, meaning lines with “at least”
n1−δ points on each. These rich lines thus contain “at least” n5−δ collinear triples
of distinct points of P.
Hence, the equation
1 1 1
(3.2) x x x = 0, x ,x ,x A , y ,y ,y A
1 2 3 1 2 3 1 1 2 3 2
(cid:12) (cid:12) ∈ ∈
(cid:12) y y y (cid:12)
1 2 3
(cid:12) (cid:12)
(cid:12) (cid:12)
has “at least” (cid:12)n5−δ solutions(cid:12). Then, for some fixed and non-equal y ,y A , the
(cid:12) (cid:12) 1 2 ∈ 2
equation
1 1 1
(3.3) x x x = 0, x ,x ,x A , y A
1 2 3 1 2 3 1 3 2
(cid:12) (cid:12) ∈ ∈
(cid:12) y y y (cid:12)
1 2 3
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
AN EXPLICIT INCIDENCE THEOREM IN Fp 5
has “at least” n3−δ solutions. Since we can translate and dilate A , we can assume
2
y = 0, y = 1 withoutloss of generality. Thenthe condition (3.3) turnsinto a claim
1 2
that
(3.4) x (1 y )+x y A x ,x A , y A
1 3 2 3 1 1 2 1 3 2
− ∈ ∈ ∈
happens for “at least” n3−δ triples (x ,x ,y ). Assuming y = 0,1 does not change
1 2 3 3
6
the situation, as long as δ < 1. Let us define B as the set of elements of the form
b = y3 , y A 0,1 . Clearly B A = n and B does not contain zero.
1−y3 3 ∈ 2\{ } | | ≈ | 2|
(The set-up here is similar to that of Theorem C in Bourgain ([2]), which would
at this point yield the existence of δ, with a possibility of its quantitative estimate.
However, if one chases through Bourgain’s arguments, the value of δ appears to be
considerably smaller than what we obtain.)
Let B be the set of popular elements of B, meaning the set of all b B such
1
∈
the equation (3.4) has “at least” n2−δ solutions (x ,x ,y ) A A A with
1 2 3 1 1 2
y3 = b. By the pigeonhole principle, ∈ × ×
1−y3
B n1−δ.
1
| |≫
Besides, fixing b fixes y , so for each b B the condition
3 1
∈
1
x +bx A
1 2 1
∈ 1 y (b)
3
−
holds for “at least” n2−δ pairs (x ,x ) A A .
1 2 1 1
∈ ×
Applying the Balog-Szemer´edi-Gowers theorem for each b, yields the existence of
the subsets A1 and A2 of A , with Ai A 1−δ = n1−δ, i = 1,2, such that
b b 1 | b|≫ | 1|
(3.5) A1+bA2 n1+5δ.
| b b|≪
Let us restate that
(3.6) A1 , A2 , B n1−δ.
| b| | b| | 1| ≫
Let A = A1 A2. In view of (3.6) and by Cauchy-Schwartz, since each A is a
b b × b b
subset of A A,
×
1
2
B1 n2−2δ Ab n Ab Ab′ ,
| | ≪ | | ≤  | ∩ |
bX∈B1 b,bX′∈B1
 
so for some b B (which can be assumed non-zero), denoting A = A , we have
∗ ∈ 1 ∗ b∗
(3.7) A A B n2−4δ.
b ∗ 1
| ∩ |≫ | |
bX∈B1
Let B be a popular subset of B , namely a set such that for all b B we have
2 1 2
∈
(3.8) A A A n2−4δ;
b∧∗ b ∗
≡ | ∩ | ≫
clearly,
(3.9) B n1−5δ.
2
| | ≫
6 HARALDANDRE´SHELFGOTTANDMISHARUDNEV
Besides, each A is a cartesian product:
b∧∗
A = A1 A2 ,
b∧∗ b∧∗× b∧∗
and therefore, for i= 1,2
(3.10) Ai Ai Ai n1−4δ.
b∧∗ ≡ b∗ ∩ b ≫
We now apply the Plu¨nnecke-Ruzsa inequality (2.2) with k = 2, as well as (3.5),
and the set cardinality estimates (3.6) and (3.10) to draw the following conclusions.
For each b B we have
1
∈
A1+bA2 2
A1+A1 | b b| n1+11δ,
| b b|≤ A2 ≪
(3.11) | b|
A1+bA2 2
A2+A2 | b b| n1+11δ.
| b b|≤ A1 ≪
| b|
Furthermore, for each b B :
2
∈
b A2+b A2 b A2 +bA2
b A2+bA2 | ∗ b ∗ b∧∗|| ∗ b∧∗ b| n15δ b A2 +bA2
| ∗ b b|≤ A2 ≪ | ∗ b∧∗ b|
| b∧∗|
A1 +b A2 A1 +bA2
n15δ| b∧∗ ∗ b∧∗|| b∧∗ b| n1+29δ,
≪ A1 ≪
| b∧∗|
by (2.2), (3.10), (3.11), and (3.5).
Then
b A2+b A2 b A2 +bA2
b A2+bA2 | ∗ ∗ ∗ b∧∗|| ∗ b∧∗ b| n1+44δ.
| ∗ ∗ b| ≤ A2 ≪
| b∧∗|
and
b A2+bA2 bA2 +bA2
(3.12) b A2+bA2 | ∗ ∗ b∧∗|| b∧∗ b| n1+59δ.
| ∗ ∗ ∗|≤ A2 ≪
| b∧∗|
Throughout the rest of the proof, to save on the number of indices used, let us
refer to A2 as X and to b−1B as Y.
∗ ∗ 2
Let us use the symbol
(3.13) K = max X +yX , so K n1+59δ.
y∈Y | | ≪
Consider the equation
(3.14) x +yx˜ = x˜ +yx , x ,x ,x˜ ,x˜ X, y Y.
2 1 2 1 1 2 1 2
∈ ∈
|Y||X|4
This equation has at least solutions, as follows by applying Cauchy-Schwartz
K
for each individual Y and then summing over y Y. Equation (3.14) is equivalent
∈
to
x x˜ = y(x x˜ ), x ,x ,x˜ ,x˜ X, y Y.
2 2 1 1 1 2 1 2
− − ∈ ∈
AN EXPLICIT INCIDENCE THEOREM IN Fp 7
|Y||X|2
Hence, for some fixed (x˜ ,x˜ ) X X, the above equation has at least
1 2 ∈ × K
solutions. Let X = X x˜ , X = X x˜ be translates of X by x˜ and x˜ ,
1 1 2 2 1 2
− −
respectively. The equation
(3.15) v = yu, u X , v X , y Y
1 2
∈ ∈ ∈
has at least |Y||X|2 solutions. Consider the set X X in F2. The bound we have
K 1× 2 p
just given as to the number of solutions of equation (3.15) can be rephrased as
saying that the set of straight lines through the origin with slopes in Y makes at
|Y||X|2
least incidences with X X .
K 1× 2
|X|2
In the remainder of the proof will assume that 1, i.e., that the lines in
K ≫
question contain “at least” one point each on average. This follows immediately
from (3.6) and (3.13) once we assume δ < 1 .
61
Not less than 50% of the incidences specified by (3.15) are contributed by rich
|X|2
lines with “at least” points thereon. The number of rich lines is not greater
K
|Y||X|
than Y , and “at least” .
| | K
|Y||X|2
Those“at least” pointsofX X lyingonrichlinescan have X different
K 1× 2 | |
abscissae. Hence,thereisaverticalsetu X forsomenon-zerou X intersected
∗ 2 ∗ 1
× ∈
|Y||X|
by “at least” rich lines.
K
Thus, we have a subset Y (X u Y) of cardinality
1 2 ∗
⊂ ∩
Y X
(3.16) Y | || |.
1
| |≫ K
In the original notations, Y lies in the intersection of u b−1B and some translate
1 ∗ ∗ 2
of A2; besides,
∗
(3.17) Y n1−65δ,
1
| | ≫
by the bounds (3.6), (3.9), (3.12).
Let R betheset of all elements expressed via r = p−q, wherep,q,s,t areelements
s−t
of Y and s= t. Let us consider two cases: (i) R Y 2 and (ii) R < Y 2. Since
1 1 1
n< √p, R =6 Fp is a possibility only in case (i)|. |≥ | | | | | |
Let us consider Case (ii) first. For any ξ R and any non-equal y ,y Y , the
1 2 1
6∈ ∈
sum y +ξy has a single realisation, as
1 2
(3.18) y +ξy = y′ +ξy′, y ,y ,y′,y′ Y
1 2 1 2 1 2 1 2 ∈ 1
would imply ξ = y1−y1′. Thus, for every subset Y′ Y and any nonzero ξ R,
y2′−y2 1 ⊂ 6∈
(3.19) Y′+ξY′ = Y′ 2.
| 1 1| | 1|
For any y = 0 we have some p,q,s,t Y , such that ξ = p−q + y lies in the
6 ∈ 1 s−t
complement of R, for otherwise R+ y = R, which is possible only if R = F . In
p
{ }
particular, this holds for y = 1. Recall that Y is a subset of u b−1B , and so we
1 ∗ ∗ 2
8 HARALDANDRE´SHELFGOTTANDMISHARUDNEV
may regard p,q,s,t as elements of B . We have, then, some fixed p,q,s,t B such
2 2
∈
that
p q
(3.20) Y 2 Y′+ξY′ Y′+Y′+ − Y′ ,
| 1| ≪ | 1 1| ≤ | 1 1 s t 1|
−
for any Y′ Y , that constitutes a positive proportion of Y , to be chosen next.
1 ⊆ 1 1
We now use Lemma 3. Let us first show that for any b B we can cover
2
∈
99% of the elements of the sets bY (a subset of a translation of bA2) or bY (a
1 ∗ − 1
subset of a translation of bA2) by “at most” n24δ translates of the set A1. Indeed,
− ∗ ∗
A1 = A1 A1 is a subset of A1, and by Lemma 3 and (2.2), we can cover 99% of
b∧∗ b ∩ ∗ ∗
the elements of either bY or bY by “at most”
1 1
−
A1 +bY A1 +bA2 A1 +bA2 bA2 +bA2
(3.21) | b∧∗ 1| | b∧∗ ∗| | b∧∗ b∧∗|| b∧∗ ∗| n24δ
A1 ≤ A1 ≤ A1 A2 ≪
| b∧∗| | b∧∗| | b∧∗|| b∧∗|
translates of A1 , and hence of A1. In the last estimate we’ve used (3.5), (3.10),
b∧∗ ∗
and (3.11).
This altogether enables us to choose Y′ as a subset containing at least 50% of
1
Y , and such that (p q)Y′ gets covered by “at most” n48δ translates of A1 +A1.
1 − 1 ∗ ∗
Let us now, in the same vein, A˜2 be a a subset containing at least 50% of A2, such
∗ ∗
that (s t)A˜2 gets covered by “at most” n48δ translates of A1+A1. Then we apply
− ∗ ∗ ∗
Plu¨nnecke-Ruzsa to (3.20) as follows:
p q A˜2+Y′+Y′ A˜2 + p−qY′
Y′+Y′+ − Y′ | ∗ 1 1|| ∗ s−t 1|
| 1 1 s t 1| ≪ A˜2
− | ∗|
(3.22) |A2∗+A2∗+A2∗| A˜2+ p−qY′
≪ n1−δ | ∗ s t 1|
−
p q
n18δ A˜2+ − Y′ ,
≪ | ∗ s t 1|
−
afterapplyingPlu¨nnecke-Ruzsawithk = 3andthe“dummyset”b−1A1,using(3.6).
∗ ∗
The covering argument above implies that
p q
A˜2+ − Y′ n96δ A1+A1+A1+A1 n1+119δ,
| ∗ s t 1| ≪ | ∗ ∗ ∗ ∗|≪
−
by applying Pln¨necke-Ruzsa with the “dummy set” b A2 and k = 4, using (3.5) and
∗ ∗
(3.6).
Therefore returning to (3.20), we have
(3.23) Y′ 2 n1+137δ.
| 1| ≪
Comparing this with (3.17), and recalling that Y′ contains at least half of the el-
1
ements of Y , we conclude that 267δ 1, so δ 1 . This ends Case (ii) and
1 ≥ ≥ 267
essentially ends the proof of Theorem 1.
Indeed, to analyse Case (i) we observe that if R Y 2, then, summing the
1
| | ≥ | |
number of ordered quadruples (y ,y ,y′,y′) of elements of Y satisfying equation
1 2 1 2 1
(3.18)over all ξ R weget nomorethan2R Y′ 2 forthetotal numberof solutions,
∈ | || 1|
pentuples (y ,y ,y′,y′,ξ). Indeed, given ξ, the solutions can be either trivial, with
1 2 1 2
AN EXPLICIT INCIDENCE THEOREM IN Fp 9
(y ,y ) = (y′,y′) or not. The total number of non-trivial solutions is Y 4, since
1 2 1 2 | 1|
such a solution determines ξ; the total number of trivial ones is R Y 2. Under
1
| || |
the assumption of Case (i) the number of trivial solutions dominates the number
of non-trivial ones. Hence, by the pigeonhole principle and Cauchy-Schwartz, there
exists ξ = p−q R such that
s−t ∈
Y′ 4
Y′+ξY′ | 1| Y′ 2.
| 1 1|≫ R Y 2/R ≫ | 1|
1
| || | | |
for every subset Y′ Y with Y′ Y .
1 ⊂ 1 | 1| ≫ | 1|
This amounts to a simplification of (3.20), with Y˜′ replacing Y˜′+Y˜′ in its right-
hand side. The ensuing estimates, done in exactly the same way as in Case (ii) are
therefore better – by a factor of n18δ, as follows by inspection of (3.21) and (3.22).
This ends the proof of Theorem 1. (cid:3)
4. Proof of Theorem 2
We follow [4], Section 6. Suppose, for contradiction, that for some (P,L) one has
(4.1) I(P,L) n32−ǫ,
≈
for some ǫ > 0. We shall give the lower bound for such ǫ. Note that the role played
by the parameter n in this theorem is different from the proof of Theorem 1.
Let us first off erase the points in P incident to more than Cn21+ǫ lines of L,
without changing the notations (P,L). This can be done, as the maximum number
of incidences that can come from the set P of such points is
+
2
1
I(P ,L) = δ δ
+ pl ≤ Cn21+ǫ pl!
pX∈P+Xl∈L pX∈P+ Xl∈L
1 n2
= Cn21+ǫ l,Xl′∈LpX∈P+δplδpl′ ≪ Cn21+ǫ,
since any two distinct lines of L meet at at most a single point of P . (Here we use
+
the notation δ = 1 if the point p is incident to the line l, δ = 0 otherwise.)
pl pl
This having been done, let P be the set of popular points of P, each incident to
1
“at least” (recall that saying “at least” implies a suitable constant c) n12−ǫ lines of
L. We have I(P ,L) I(P,L) by the following argument:
1
≈
I(P,L) I(P1,L) = δp,ℓ < cn12−ǫ cn23−ǫ < 1I(P,L)
− ≤ 2
p∈P ℓ∈L p∈P
X X X
p∈/P1
for c small enough.
We can now refine L to the subset L of popular lines, each incident to “at
1
least” n21−ǫ points of P1. By the pigeonhole principle, it still contributes a positive
proportion of incidences. Let P P with respect to L . (That is, p P is an
2 1 1 1
element of P2 if and only if it is in⊂cident to “at least” n21−ǫ lines of L1.) ∈
10 HARALDANDRE´SHELFGOTTANDMISHARUDNEV
Once again, I(P ,L ) satisfies (4.1). (The refinement process could be iterated
2 1
any finite number of times, with the constants obviously getting worse.)
For p P , let P be a subset of all points of P , which are connected to p by
2 p 1
∈
some line from L . By definitions of P ,L we have P n1−2ǫ, for each p P .
1 2 1 p 2
| | ≫ ∈
Thus, by Cauchy-Schwartz
P n1−2ǫ P P P P ,
2 p 1 p¯ p˜
| | ≪ | |≤ | | | ∩ |
pX∈P2 p sp¯,Xp˜∈P2
and so
P 2
P P = P P P | 2| n2−4ǫ O(n2).
p¯ p˜ p¯ p˜ p¯
| ∩ | | ∩ |− | | ≫ P −
1
p¯,Xp˜∈P2 p¯,Xp˜∈P2 p¯X∈P2 | |
p¯6=p˜
Now since I(P2,L1) n23−ǫ and each point of P2 is incident to at most n12+ǫ
≈ ≪
lines of L, we have P n1−2ǫ, and thus
2
| |≫
P 2
P P | 2| n2−4ǫ.
p¯ p˜
| ∩ | ≫ P
1
p¯,Xp˜∈P2 | |
p¯6=p˜
Therefore one can fix some (p¯,p˜) (P P ), p¯= p˜, such that
2 2
∈ × 6
n2−4ǫ
P P P n1−4ǫ.
3 p¯ p˜
≡ | ∩ | ≫ P ≫
1
| |
Each point of P3 P1 is incident to “at least” n12−ǫ lines of the original set
⊆
of lines L. Without loss of generality, after a projective transformation we can
place the points p¯ and p˜ on the line at infinity, so that the “at most” n21+ǫ lines
of L emanating from these points can be viewed as being parallel to the x and y
1
coordinate axes. In other words, for some A,B of cardinality “at most” n21+ǫ each,
we have a subset P of A B of cardinality “at least” n1−4ǫ, such that the number
3
×
of incidences of P with L is
3
(4.2) I(P3,L) n32−5ǫ.
≫
Now, the number of triples of points of A B, which are collinear on some line
×
from L is then (by H¨older’s inequality or simply noticing that the smallest number
of triples is achieved with (4.2) as equality, with n lines, each supporting the same
number of points) “at least” n25−15ǫ A5−14+02ǫǫ. We can now merge with the proof
≫ | |
of Theorem 1 at its claim (3.2), with ǫ = δ = 1 . (cid:3)
40−2δ 10678
References
[1] J. Beck. On the lattice property of the plane and some problems of Dirac, Motzkin, and Erdo¨s
in combinatorial geometry. Combinatorica 3 (1983), 281–297.
[2] J. Bourgain. Multilinear Exponential Sums in Prime Fields Under Optimal. Entropy Condition
on the Source. Geom. Func.Anal. 18 (2009), 1477–1502.
[3] J. Bourgain, M.Z. Garaev. On a variant of sum-product estimates and explicit exponential sum
bounds in prime fields. Math. Proc. Cambridge Philos. Soc. 146 (2009), no. 1, 1–21.