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. 2025 Sep 5;12(4):65. doi: 10.1007/s40687-025-00552-4

Approximate incidence geometry in the plane

Tuomas Orponen 1,
PMCID: PMC12413431  PMID: 40918593

Abstract

These are lecture notes for a mini-course given in Banff in June 2024. They discuss the problem of bounding the number of δ-incidences Iδ(P,L):={(p,)P×L:p[]δ} under various hypotheses on PR2 and LA(2,1). The main focus will be on hypotheses relevant for the Furstenberg set problem.

Introduction

These lectures are about approximate incidence geometry, but they are far from a complete introduction to this topic. A more accurate name for the contents would be introduction to upper bounds in linear approximate incidence geometry in the plane. For further reading, I mention a few important topics not covered in these notes, and some of the most recent references (as of summer 2025). I will not discuss the high-low method pioneered in [28] and applied, e.g. in [6, 8, 9, 23, 24, 53]. I will not discuss Falconer’s distance set problem, or other “curvilinear” incidence problems [20, 25, 26, 36, 55, 64]. I will not discuss incidence lower bound problems [8, 9, 14]. Finally, I will not discuss the Kakeya problem [32, 5961].

A prototypical problem in approximate incidence geometry is the following. Let PR2 be a finite set, and let L be a finite set of lines in the plane. Fix a (small) scale parameter δ>0, and consider the δ-incidences

Iδ(P,L):={(p,)P×L:p[]δ}.

Here and below, [A]δ stands for the closed Euclidean δ-neighbourhood of ARd.

Question 1

(Incidences) How large can |Iδ(P,L)| be in terms of P and L?

The following formulation is equivalent, but often gives a useful, slightly different perspective to the problem:

Question 2

(Rich lines) Given a finite set of points PR2, and a parameter r2, how large can be the set Lr,δ(P)={R2:|P[]δ|r} of lines in R2 whose δ-neighbourhood contains at least r points from P?

The answers to Questions 12 heavily depend on various hypotheses we place on P and L. If we make no hypotheses, the best possible answer is |Iδ(P,L)||P||L|. This is attained if P is contained in a single (δ/2)-disc BR2, and all the lines L also intersect B. To avoid this situation, one typically asks—at least—that both P and L are δ-separated; in the formulation of Question 2, one only counts δ-separated elements in Lr(P). The δ-separation of L is defined with respect to some natural metric on A(2,1), the space of all (affine) lines in R2. A common choice (also used in these notes) is

dA(2,1)(1,2):=πL1-πL2+|a1-a2|,

whenever j=Lj+aj, LjR2 is a 1-dimensional subspace, and ajLj. The notation πL refers to orthogonal projection to L, and · is the operator norm.

If both P and L are δ-separated, the sharp estimate for |Iδ(P,L)| is

|Iδ(P,L)|δ-1/3|P|2/3|L|2/3, 1.1

see [19] (or [11, Exercise 7.5] or [21, Theorem 1.5] for alternative arguments which yield (1.1) with a δ-ϵ-loss). This bound is (e.g.) attained whenever the points form a δ-net inside a Δ-disc BR2 for some δΔ1, and L a δ-net of lines intersecting B.

Despite its sharpness, (1.1) is not very useful for solving “interesting” problems. These problems tend to involve more hypotheses on P and L, which make the sharp bound for |Iδ(P,L)| more difficult to prove. We will return to this in later sections.

Bounds for δ-incidences under δ-separation

When δ=0, we abbreviate

I(P,L):=I0(P,L)={(p,)P×L:p},

and Lr(P):=Lr,0(P). The “exact incidence” p makes the problem of estimating |I(P,L)| easier than the problem in Question 1, although still non-trivial. Under no additional hypotheses on P and L, Szemerédi and Trotter [57] in 1983 proved that

|I(P,L)||P|2/3|L|2/3+|P|+|L|. 2.1

Equivalently, if PR2 is finite, then

|Lr(P)||P|2r3+|P|r,r2.

The Szemerédi-Trotter bound is quite sharp. Perhaps, the simplest example is where L={}, and P: then |I(P,L)|=|P|. This shows that the implicit constant in front of the |P| (similarly |L|) term cannot be lower than 1. This leaves open the sharpness of the term |P|2/3|L|2/3. The exponent 2/3 is sharp, indeed the simple Example 2.2 below shows that the implicit constant in front of the term |P|2/3|L|2/3 cannot be lower than 2-2/3:

Example 2.2

Consider P:={1,,n}×{1,,2n2} and

L:={(x,ax+b):a{1,,n}andb{1,,n2}}.

Note that every line =a,bL is n-rich, because if a{1,,n} and b{1,,n2}, then ax+b{1,,2n2} for all x{1,,n}. Thus, |I(P,L)|n|L|n4. On the other hand, if (2.1) holds with constant C>0, then

n4|I(P,L)|C|P|2/3|L|2/3+C|P|+C|L|=C22/3n4+2Cn3+Cn3.

This can only hold for all nN if C2-2/30.63.

Remark 2.3

The sharpest currently known form of (2.1) is |I(P,L)|C|P|2/3|L|2/3+|P|+|L| with C=2.44, see [1]. It is also known [52, Remark 4.2] that this inequality cannot hold with constant C<33/(4π2)31.27 (see also [3, Sect. 1.3]). In that example, P={1,,n}×{1,,n}, and L consists of (ϵn)-rich lines for any ϵ(0,1).

Crossing number proof

We start with a proof of (2.1) using the crossing number inequality discovered by Székely [56]. This proof also yields some information about δ-incidences, provided that both the points and lines are δ-separated:

Theorem 2.4

Let δ(0,1], and let PB(1)R2 be a δ-separated set. Let r2, and let L be a family of δ-separated lines such that |P[]δ|r for all L. Then,

|L||P|2r3+|P|r. 2.5

Equivalently, |Iδ(P,L)||P|2/3|L|2/3+|P|+|L|.

Here is the crossing number inequality (proved independently in [37] and [2]):

Lemma 2.6

(Crossing number inequality) Let G=(V,E) be a simple graph with |E|4|V|. Then, the crossing number cr(G) of G has the lower bound

cr(G)164|E|3|V|2.

The crossing number cr(G) is the minimal number of edge crossings in any planar drawing of G. In particular, cr(G)=0 if and only if G is a planar graph.

Proof of Theorem 2.4

We may assume that δ1100, since otherwise |P|1, and Theorem 2.4 is trivial. Fix L, so |P[]δ|r2. Since the set P[]δ is δ-separated, and δ is far larger than the width of the tube []δ, the points in P[]δ lie on a δ-Lipschitz graph over . In particular, they have a natural ordering (say, given by their projections to ). We form a set of unordered edges E(), in a graph with vertex set V=P, by placing an edge between any pair of consecutive points pq in this ordering. Then, we define

E:=LE().

The (Euclidean) length of any edge [p,q]E is δ. It may happen that [pq] lies in multiple families E(1),,E(k), but then the associated lines 1,,k all lie at distance δ from each other in A(2,1),1 Since L was assumed δ-separated, it follows that k1, i.e. the families E() have bounded overlap. Therefore,

|E|L|E()||L|(r-1)|L|r,

since r2. Now, if |E|<4|P|, we have r|L||P|, and therefore, |L||P|/r. This bound corresponds to the second term in (2.5).

Assume then that |E|4|P|. In this case, the crossing number inequality is applicable in the graph G=(P,E) and yields

(r|L|)3cr(G)|P|2.

We claim that cr(G)|L|2, which will complete the proof of (2.5). The idea is roughly: every crossing pair of edges (e1,e2) determines (hopefully) a unique pair of lines (1,2) such that e1E(1) and e2E(2). Therefore, cr(G)|L|2. The uniqueness is not entirely trivial, however. We know that the edges in E(j) lie on an O(δ)-Lipschitz graph over j, but this alone is not helpful (unless δ=0): It is well possible that two O(δ)-Lipschitz graphs intersect each other multiple times, see Fig. 1.

Fig. 1.

Fig. 1

The danger of many crossings

What saves the day is the fact that if 12, then all the crossings between E(1),E(2) must occur in a disc of radius O(δ). Let us do this carefully. We write

cr(G)(e1,e2)(e1,e2)iscrossing|{(1,2):ejE(j),j{1,2}}|(1,2)L×L|{(e1,e2)E:ejE(j),j{1,2}and(e1,e2)iscrossing}|.

It suffices to show that individually |{(e1,e2)E:}|1.

If 1==2, there are no crossing pairs (e1,e2) with e1,e2E(). So, consider 12. Let (e11,e21),,(e1m,e2m) be pairs of crossing edges with (e1i,e2i)E(1)×E(2) for all 1im. Then, every edge e1i and e2i meets the intersection [1]δ[2]δ, which is contained in a disc BR2 of radius δ. So, in fact ejiE(j,B):={eE(j):eB} for all 1im. Thus,

(e1i,e2i)E(1,B)×E(2,B),1im.

But |E(j,B)|1 for j{1,2} by the δ-separation of P, so m1, as claimed.

Cell decomposition proof

In this section, we give a second proof for (2.1), based on cell decompositions. This method was pioneered by Szemerédi and Trotter, but a simpler implementation can be achieved by the use of polynomials (see Lemma 2.11). The cell decomposition proof of (2.1) could also be modified to yield the δ-separated version stated in Theorem 2.4. We leave this as a (reasonably challenging) exercise for the reader.

We repeat below exactly what we are planning to prove:

Theorem 2.7

Let PR2 be a finite set. Then,

|Lr(P)||P|2r3+|P|r,r2. 2.8

Besides the idea of cell decompositions, a second ingredient is the following simple incidence estimate:

Proposition 2.9

For finite PR2 and LA(2,1),

|I(P,L)||P|1/2|L|+|P|. 2.10

Proof

For pP, write L(p):={L:p}. Using the definition of I(P,L) and Cauchy–Schwarz:

|I(P,L)|=pP|L(p)||P|1/2(p|L(p)|2)1/2=|P|1/2(,|P|)1/2.

Splitting the sum into diagonal and off-diagonal parts (where = and , respectively), we obtain |I(P,L)||P|1/2(|I(P,L)|+|L|2)1/2. This yields (2.10).

Here is the lemma from [27] on polynomial cell decompositions, we will need:

Lemma 2.11

Let dN be a degree. If PRn is a finite set, there exists a polynomial surface Z={poly=0} with poly0 of degree degZ:=deg(poly)=d such that every component ORd\Z contains |P|/dn points of P.

Remark 2.12

The components OR2\Z will be the “cells” in the “cell decomposition” proof of Theorem 2.7. Szemerédi and Trotter did not have access to Lemma 2.11, so they crafted their “cells” using straight lines and line segments. In many interesting, non-trivial cases (for example when P is a product set to begin with) crafting the cells by hand (using vertical and horizontal line) is easy.

Remark 2.13

The cases 2r1 of Theorem 2.7 are easy and can be obtained from the following 2-ends argument. Let L be a set of lines such that |P|r2 for all L. In particular, every L contains r2 pairs of distinct points (p,q)(P)2. As L varies, these pairs are distinct, so we obtain

r2|L|L|(P)2||P|2.

Rearranging yields |P|r|L|1/2. For 2r1, this coincides with the bound from Theorem 2.7. A key idea underlying the cell decomposition proof of Theorem 2.7 is to decompose P into smaller pieces PO, where we “expect” 2|PO|1 (although this inequality does not literally appear in the proof).

We are then equipped to prove Theorem 2.7.

Proof

We may assume that r4, since the opposite case follows from Remark 2.13. Write L:=Lr(P). We may also assume that |L|r. Namely when |L|r, we may use the elementary bound (2.10) to deduce that

r|L||I(P,L)||P|1/2|L|+|P|.

If the second term dominates, then |L||P|/r, and if the first term dominates, then |P|2r4, and therefore |L|r|P|2/r3.

Assume |L|r, and apply Lemma 2.11 with degree d:=(r/2) to obtain a non-trivial polynomial surface ZR2 with deg(Z)d such that |PO||P|/r2 for all connected components OR2\Z. These components will be denoted O.

Let

LZ:={L:Z}andLO:=L\LZ.

Since Z can contain d=r/2 distinct lines,2 We infer that |LZ|r/2, and therefore |LZ|12|L|, using the assumption |L|r. Thus |L|2|LO|, and it suffices to verify the bound (2.8) for LO in place of L.

Fix LO. By assumption |P|r4. In particular, contains r-13r/4 segments of the form [xy] with x,yP distinct and consecutive (i.e. there are no points of P between x and y). Let I be the collection of all such segments. A segment I=[x,y]I is cellular if x,yO for a common component OR2\Z, otherwise I is non-cellular. Note that every line LO contains d=r/2 non-cellular segments. Indeed, intersects Z somewhere on each non-cellular segment. So, if the number of non-cellular segments were >d, then |Z|>d, and hence Z (contrary to the definition of LO).

The cellular segments are denoted IO, and their number is

|IO|=OO|{IIO:IO}|OO|PO|2|P|r2OO|PO||P|2r2.

Every line LO contains 3r/4 segments, but only r/2 non-cellular segments. Therefore, contains r/4 cellular segments, denoted IO(). When LO varies, the collections IO() are clearly disjoint. Therefore,

r|L|r4·|LO|LO|IO()||IO||P|2/r2.

Consequently |P|r3/2|L|, as claimed.

Bounds for δ-incidences under s-dimensional separation

In Theorem 2.4, we saw that the Szemerédi-Trotter bound for δ-incidences is valid under the hypothesis that both P and L are δ-separated. For applications in “continuum” incidence problems in fractal geometry, this hypothesis on L,P is not so natural, see Remark 3.3. To keep the discussion concrete, we now introduce one distinguished “continuum” incidence question, which will follow us for the rest of these notes.

Problem 1

(Furstenberg set problem) Let s(0,1] and t[0,2]. Let FR2 be an (st)-Furstenberg set. This means that there exists a t-dimensional family LA(2,1) such that dimH(F)s for all L. What is the best lower bound for dimHF?

The Furstenberg set problem (for t=1) was proposed by Wolff [62, 63] in the late 90 s. After plenty of partial progress [4, 13, 15, 22, 2830, 34, 38, 48, 49], it was finally solved in 2023 by Ren and Wang [53]. The sharp lower bound for dimHF is

dimHFf(s,t):=min{s+t,(3s+t)/2,s+1}. 3.1

Here is a simple but illustrative example of Furstenberg sets:

Example 3.2

For s[0,1], let F=C×R, where C is an s-dimensional Cantor set. Then dimHF=s+1, and F is an (s, 2)-Furstenberg set: Every non-vertical line R2 satisfies dimH(F)=dimHC=s. In particular, the term “s+1” in (3.1) cannot be omitted.

Remark 3.3

The estimate (3.1) can be viewed as a continuum analogue of the Szemerédi-Trotter bound (2.1). We clarify this somewhat informally. Instead of aiming for (3.1), let us attempt to prove (the box dimension estimate) |F|δδ-f(s,t) for all δ>0 small enough. Here |·|δ is the δ-covering number.

Pick a δ-separated subset LδL with |Lδ|δ-t, and for each Lδ a δ-separated set Pδ()F with |Pδ()|δ-s. Let Pδ be a maximal δ-separated set in LδPδ()F. Then,

|Iδ(Pδ,Lδ)||Lδ|δ-sδ-(s+t). 3.4

(We used here that every line in Lδ is δ-incident to δ-s points in Pδ, although these points may not be the ones in Pδ().) On the other hand, if the Szemerédi-Trotter bound was valid under the δ-separation hypotheses on Pδ and Lδ, we could estimate

|Iδ(Pδ,Lδ)||Pδ|2/3|Lδ|2/3+|Pδ|+|Lδ||Pδ|2/3δ-2t/3+|Pδ|+δ-t.

Combining these estimates would lead to |F|δ|Pδ|min{δ-(s+t),δ-(3s+t)/2}, and eventually dim¯BFmin{s+t,(3s+t)/2}.

First, this is too good to be true: the term “s+1” in (3.1) is necessary, as we saw in Example 3.2. Second, the Szemerédi-Trotter bound is not valid under δ-separation alone; the sharp bound under this hypothesis was mentioned in (1.1). The argument above using (1.1) would produce the unsharp estimate dim¯BF(3s+t-1)/2.

One might try to use Theorem 2.4, namely the version of Szemerédi-Trotter valid under δ1/2-separation. This approach can—at best—yield an unsharp estimate. The argument above also runs into a difficulty: after extracting suitable δ1/2-separated sets Lδ1/2 and Pδ1/2, it seems hard to maintain any lower bound on the δ-incidences Iδ(Pδ1/2,Lδ1/2).

Discretising incidence problems involving Hausdorff dimension

We have seen that sharp upper bounds on |Iδ(P,L)| under δ-separation or δ-separation hypotheses on P and L fail to produce sharp estimates in the (st)-Furstenberg set problem. This is not surprising: The hypotheses dimHLt and dimH(F)s involve Hausdorff dimension, and that information is lost when passing to (merely) δ-separated or δ-separated subsets. We need to consider a separation condition that keeps the “Hausdorff dimension information”. Such a condition was introduced by Katz and Tao [34] in 2000:

Definition 3.5

((δ,s)-set) Let (Xd) be a metric space, and C,δ,s>0. A δ-separated set PX is called a (δ,s,C)-set if

|PB(x,r)|Crδs,xX,rδ.

If the value of the constant C>0 is irrelevant, a (δ,s,C)-set may be called a (δ,s)-set.

We also extend the definition of (δ,s)-sets to families of dyadic cubes. For δ2-N, a family of dyadic cubes PDδ is a (δ,s,C)-set if

|PQ|Crδs,QDr,rδ,

where PQ={pP:pQ}.

Remark 3.6

The (δ,s)-sets of these notes are sometimes called Katz-Tao (δ,s)-sets in the literature, while “(δ,s)-set” refers to sets PX satisfying |PB(x,r)|δrs|P|.

Example 3.7

The property of being a (δ,s)-set gets weaker as s increases. In fact, every finite δ-separated set in Rd is a (δ,d)-set.

Example 3.8

Every δ-separated set PB(1)R2 is a (δ,1)-set, because |PB(x,r)|1 for δrδ. On the other hand, for δr1, any r-disc can contain at most (r/δ)2r/δ discs of radius δ. In particular |PB(x,r)|r/δ for δr1.

Jump ahead to Fig. 2 for a rather “opposite” example of a (δ,1)-set.

Fig. 2.

Fig. 2

The train tracks example

We now record two standard tools for discretising continuum incidence problems involving Hausdorff dimension. Proposition 3.9 allows to find large (δ,s)-subsets, while Proposition 3.12 allows to find covers by (δ,s)-sets. We first recall that, for s>0, the s-dimensional Hausdorff content of KRd is defined by

Hs(K):=inf{jNdiam(Ej)s},

where the inf runs over all countable set families {Ej}jN such that KjNEj.

Proposition 3.9

(Large (δ,s)-subsets) Let KRd be a bounded set with Hs(K)=:τ>0. Then, for every δ(0,1], there exists a (δ,s,Cd)-set PK, and |P|dτδ-s.

The proof is taken from [18, Proposition A.1]. The idea is the same as in the proof of Frostman’s lemma (see, e.g. [40, Theorem 8.8]).

Proof

Without loss of generality, assume that δ=2-k for some kN (if δ(0,1] is arbitrary, first prove the proposition for the largest element in 2-N(0,δ]). We may also assume that K is contained in some dyadic cube Q0Dk0 with k0k.

Let P0 be a finite set containing exactly one point from KQ for all QD2-k(K). Then, |P0|=|K|δτδ-s as desired, but P0 may fail the (δ,s)-set condition. We modify P0 as follows. Fix Qk-1D2k-1(P0). If

|P0Qk-1|>(Qk-1)δs,

define P1Qk-1P0Qk-1 to be a maximal set with |P1Qk-1|((Qk-1)/δ)s. Then automatically

12(Qk-1)δs|P1Qk-1|(Qk-1)δs.

Repeat this for all cubes Qk-1D2k-1 to obtain P1. Then, repeat the procedure at all dyadic scales between δ=2-k and 2-k0: if Pj has already been defined, and there is a cube Qk-(j+1)D2k-(j+1) such that

|PjQk-(j+1)|>(Qk-(j+1))δs,

choose a subset Pj+1Qk-(j+1)PjQk-(j+1) satisfying

12(Qk-(j+1))δs|Pj+1Qk-(j+1)|(Qk-(j+1))δs. 3.10

The process stops in finitely many steps, because P0Q0.

We claim that for every xP0 (and not only xP!) there exists a dyadic cube QxQ0 such that (Qx)δ and

|PQx|12(Qx)δs. 3.11

If xP, then (3.11) holds for QxD2-k({x}). If xP0\P, the point x was deleted from P0 at some stage. Let Qx be the largest cube containing x, where point deletion occurred. If this happened while defining Pj+1, then (3.10) holds with Qk-(j+1)=Qx. But since Qx was the largest cube containing x where point deletion occurred, Pj+1Qx=PQx. This gives (3.11).

Let Q be the maximal elements in {Qx:xP0}. Then, Q is a disjoint cover of K, because every element of Dδ(P0)=Dδ(K) is contained in some element of Q, as we just showed. This combined with (3.11) yields

|P|=QQ|PQ|δ-sQQ(Q)sdκδ-s.

We claim that P is a (δ,s,Cd)-set. For QDl with k0lk, (3.10) yields |PQ|((Q)/δ)s. This implies |PB(x,r)|d(r/δ)s for all xR2 and rδ.

Proposition 3.12

(Covering by (δ,s)-sets) Let 0<sd, and let KRd be a compact set with Hs(K)=0. Then, for every k0N, there exists a sequence of (2-k,s,1)-sets {Pk}kk0 such that PkD2-k, and

Kkk0Pk.

A slightly different version of this proposition was first proved by Katz and Tao in [34], but for more recent incarnations, see [23, 45].

Proof of Proposition 3.12

Let KRd be a compact set with Hs(K)=0. Consider the following dyadic variant of the Hausdorff content:

Hδ,s(K):=minQ{QQ(Q)s:KQQQ},

where the “min” only runs over families Q of dyadic cubes with side-length δ. It will be crucial that the “min” exists thanks to the restriction to “large” cubes.

Fix k0N. Using the compactness of K, and Hs(K)=0, one may find δ>0 such that Hδ,s(K)2-k0s.3 This means that there exists a (finite) cover of K by dyadic cubes Q such that (Q)δ for all QQ, and

QQ(Q)s=Hδ,s(K)2-k0s. 3.13

It turns out that the sub-families Pk:={QQ:(Q)=2-k}, kN, are automatically (2-k,s,1)-sets, that is,

|{QPk:QQ}|r2-ks,QDr,r2-k.

If this inequality failed for some QDr, a new competitor for Hδ,s(K) is obtained by

Q:=(Q{Q})\{QPk:QQ}.

Moreover, the analogue of the sum (3.13) for Q is strictly smaller than the one for Q, because

(Q)s=rs<|{QPk:QQ}|·2-ks=QQ(Q)s.

This violates the minimality of the sum (3.13).

To summarise, we have now found a cover of K by (2-k,s,1)-sets, and it follows from (3.13) that kk0 for all (non-empty) families Pk.

Incidences between (δ,s)-sets of points and lines

We return to the problem of estimating |Iδ(P,L)|. After the introduction of (δ,s)-sets, a natural question is the following:

Problem 2

Let 0α,β2. Assume that PB(1)R2 is a (δ,α)-set, and LA(2,1) is a (δ,β)-set. How large can |Iδ(P,L)| be in terms of

  1. α and β?

  2. |P| and |L|?

We use here the notation “α” and “β” to avoid confusion with the parameters “s” and “t” appearing in the Furstenberg set problem.

A sharp answer to part (a) was found by Fu and Ren in [22], for all values of α,β[0,2]. They provide an explicit continuous function f:[0,2]2[0,) such that

|Iδ(P,L)|ϵδ-f(α,β)-ϵ,ϵ>0, 3.14

whenever P,L are as in Problem 2. In particular f(α,β)(1+α+β)/2 for α+β3. They also give examples showing that f(α,β) is the minimal function satisfying (3.14). The full formula for the (piecewise linear) function f is a bit complicated, see [22, Theorem 1.4]. The method of [22] also gives an answer to part (b), but I’m not sure if that is sharp for all cardinalities |P| and |L|. We will prove the bound f(α,β)(1+α+β)/2 in Theorem 3.22. For now, let us compare Problem 2 with the Furstenberg set problem.

First some bad news: Even a sharp answer to Problem 2 does not solve all the cases of the (st)-Furstenberg set problem. The reason is, roughly, that the hypothesis dimH(F)s available for every (st)-Furstenberg set is very strong, and the extremal pairs (P,L) for Problem 2.7 do not satisfy (a δ-discretised version) of this lower bound, e.g. in the form that P[]δ, L, would contain large (δ,σ)-sets for some σ>0.

Then, some good news: A sharp answer to Problem 2.7 still solves the cases t[2-s,2] of the Furstenberg set problem. In fact, we have the following proposition:

Proposition 3.15

Assume that f:[0,2]2[0,) is a continuous function with the following property. If PB(1) is a (δ,α)-set and LA(2,1) is a (δ,β)-set, then

|Iδ(P,L)|δ-f(α,β)

for all δ>0 small enough. Then, every compact (st)-Furstenberg set FR2 satisfies

f(dimHF,t)s+t. 3.16

Remark 3.17

The compactness hypothesis in Proposition 3.15 could be relaxed to boundedness, but verify this, we would need a version of Proposition 3.12 for bounded sets. This is true, see [23, Lemma 2].

By (3.14), the explicit Fu-Ren function f satisfies the hypothesis (3.16) up to the harmless “ϵ”, so one can derive lower bounds for dimHF from the inequality (3.16).

The bound f(α,β)(1+α+β)/2 for α+β3, proven in Theorem 3.22, leads to the lower bound dimHFs+1 for (st)-Furstenberg sets with with s+t2. Here’s why. Assume that s+t=2 (without loss of generality), and assume that there exists an (st)-Furstenberg set FR2 with dimHF<s+1. Then dimHF+t3, so

2=s+t(3.16)f(dimHF,t)(1+dimHF+t)/2,

and this can be rearranged to dimHF3-t=s+1. This is the sharp bound (predicted by (3.1)) in the range s+t2. The reader is encouraged to check that a similar application of Proposition 3.15 in the range s+t<2 would only yield unsharp results.

Proof of Proposition 3.15

By definition, there exists a line set LA(2,1) with dimHLt such that dimH(F)s for all L. It is easy to reduce matters to the case where our hypotheses are slightly strengthened as follows:

  1. Ht(L)1.

  2. Hs(F)1 for all L.

Fix α>dimHF. According to the Katz-Tao covering lemma, Proposition 3.12, we may find a sequence {Pk}kk0 of families of dyadic cubes such that PkD2-k is a (2-k,α)-set, and

Fkk0Pk.

In particular, for L fixed, the intersection F is covered by the sets Pk for every L. By our assumption (b), and the sub-additivity of Hausdorff content, there exists an index k()k0 such that Hs(Pk)k()-2.

For kk0, let Lk:={L:k()=k}. By assumption (a), there exists kk0 such that Ht(Lk)k-2. We now set δ:=2-k for this index kk0, and we abbreviate P:=Pk. Then, P is a (δ,α)-set. Define PP by selecting one point in each square in P.

Since Ht(Lk)k-2=(log1/δ)-2, by the subset finding lemma, Proposition 3.9, there exists a (δ,t)-set L¯Lk of cardinality |L¯|δ-t. Here, the “” notation hides factors of order log(1/δ). Moreover, since k()=k for all L¯, we know that

Hs(P)1,L¯. 3.18

This implies that |P[]2δ|δ-s for L¯, and consequently,

|I2δ(P,L¯)||L¯|δ-sδ-(s+t). 3.19

On the other hand, since P is a (δ,α)-set and L¯ is a (δ,t)-set, we have by the definition of the function f:

|I2δ(P,L¯)|δ-f(α,t).

These inequalities are only compatible (for all small δ>0) if f(α,t)s+t. Since α>dimHF was arbitrary, we infer f(dimHF,t)s+t by the continuity of f.

The train tracks example

We will shortly prove the bound f(α,β)(1+α+β)/2 in Problem 2, in particular f(1,1)3/2. Let us first see why this is sharp.

Example 3.20

Consider the points P and lines L drawn in the train tracks example in Fig. 2. There are |P|=δ-1 points organised in 1/δ “columns”. Both the separation and height of the columns are δ. There are also |L|δ-1 lines, with 1/δ lines incident to each column. In fact, every point pP is contained in []δ for 1/δ lines L. Therefore,

|Iδ(P,L)|=pP|{L:p[]δ}|δ-1|P|=δ-3/2.

Both P and L are (δ,1)-sets. Therefore, under the hypothesis that both P and L are (δ,1)-sets, the sharp upper bound for |Iδ(P,L)| cannot beat δ-3/2. This is weaker than the Szemerédi-Trotter bound in Theorem 2.4, which would predict |Iδ(P,L)|(|P||L|)2/3+|P|+|L|δ-4/3.

Remark 3.21

Recall from Remark 3.8 that δ-separated sets are (δ,1)-sets. The train tracks example demonstrates that the δ-separation of P and L in Theorem 2.4 cannot be relaxed to the hypothesis that both P and L are (δ,1)-sets.

One can next ask (at least) the following two questions:

  • (i)

    Would it suffice to assume that only one of the sets P or L is δ-separated—say P for concreteness—while L is merely δ-separated, or perhaps a (δ,1)-set?

  • (ii)

    Are the train tracks the only sharpness example if both P, L are (δ,1)-sets?

The answer to (i) is negative, as far as I know, but turns positive if one additionally assumes that |P| is “maximal”, that is |P|δ-1. Guth, Solomon, and Wang [28, Theorem 1.1] proved that if PB(1) is δ-separated with |P|δ-1, and LLr,δ(P) is δ-separated, then

|L|δ-ϵ|P|2r3,rδ-ϵ,ϵ>0.

In fact, this is only a special case of their result. The article [28] pioneered the powerful high-low method for studying δ-incidences. We will not cover the high-low method in these notes, but the only reason is the lack of space and time. The high-low method is a central tool in δ-discretised incidence geometry and was also used in the resolution of the (st)-Furstenberg set conjecture.

Regarding question (ii), the answer is roughly positive, except that the individual “tracks” (the columns of red balls in Fig. 2) may be rotated and translated arbitrarily, as long as the (δ,1)-set conditions are preserved. The precise statement is that if (P,L) is a pair of (δ,1)-sets with |Iδ(P,L)|δ-3/2, then P×L contains δ-1/2 “cliques” P×L such that |Iδ(P,L)||P||L|δ-1. This is a recent result of the author and Yi [51]. In Fig. 2, the “cliques” are the pairs P×L corresponding to each “track”.

Cases t[2-s,2] of the Furstenberg set problem

In this section, we prove a special case of Fu and Ren’s theorem [22, Theorem 1.4], which yields f(α,β)(1+α+β)/2 (in the notation of (3.14)), for α+β3. As discussed in Remark 3.17, this solves the (st)-Furstenberg set problem for t[2-s,2].

Theorem 3.22

Let s+t<3, and A,B1. Let PB(1)R2 be a (δ,s,A)-set, and let LA(2,1) be a (δ,t,B)-set. Then,

|Iδ(P,L)|s,tABδ-1|P||L|. 3.23

Remark 3.24

The bound (3.23) with an additional δ-ϵ-factor follows from [22, Theorem 1.5] and was originally proved using the high-low method. A finite field precedent is due to Vinh [58]. The δ-ϵ-free proof below is from [46], and it is based on classical Sobolev smoothing properties of the X-ray transform. Let us note that the idea of using X-ray transforms and related operators to study incidences has been applied several times prior to [46], for example in [17, 33].

We then gather some preliminaries for the proof of Theorem 3.22. The arguments are a little sketchy, for full details see [46].

Definition 3.25

(X-ray transform) For fS(R2), we define the X-ray transform XfC(A(2,1)) by the formula

(Xf)():=fdH1.

The X-ray transform is L2-smoothing of order 12, see [42, Theorem 5.3]. The following (special case) is [46, Theorem 2.16]:

Proposition 3.26

For every χCc(R2), there exists a constant Cχ>0 such that

X(fχ)H˙s+1/2CχfH˙s,fS(R2),-12s12.

The norm on the right hand side is the standard homogeneous Sobolev norm

fH˙s2:=|f^(ξ)|2|ξ|2sdξ.

On the left hand side, the Sobolev norm H˙s+1/2 is defined for functions in A(2,1): This can be accomplished rigorously by identifying A(2,1) with [0,1]×R, and then defining the Sobolev norm on [0,1]×R instead of A(2,1). We do not need the details here, except for the inequality

fgfH˙rgH˙-r. 3.27

The proof of this inequality can be summarised as Plancherel + Cauchy–Schwarz.

Sobolev norms of negative order can be expressed in terms of the Riesz energy as follows: If μ is a Radon measure on Rd, then

μH˙(s-d)/22=|μ^(ξ)|2|ξ|s-ddξd,sdμ(x)dμ(y)|x-y|s=:Is(μ),s(0,d). 3.28

For a proof, see [40, Lemma 12.12]. Regarding Riesz energies, we need the following elementary computation concerning measures supported on (δ,s)-sets:

Lemma 3.29

Let PB(1)Rd be a (δ,s,C)-set with s(0,d) and C1. Consider the measure μ:=δsH0|P, and let μδ:=μφδ be a standard mollification of μ. Then,

Iσ(μδ):=dμδ(x)dμδ(y)|x-y|σσC|P|δ2s-σ,σ(s,d).

Proof sketch

The mollification causes small technicalities omitted here. Essentially the proof boils down to the computation

δ2spq1|p-q|σδ2spPδr1r-σ|PB(p,r)|Aδs|P|δr1rs-σσA|P|δ2s-σ,

using s-σ<0.

We are then ready to prove Theorem 3.22.

Proof of Theorem 3.22

Recall that s+t<3. Choose σ>s and τ>t such that σ+τ=3. Note that either σ>1 or τ>1, and we assume with no loss of generality that τ>1.

Let PB(1) be a (δ,s,A)-set, and let LA(2,1) be a (δ,t,B)-set. We associate the following measures μ and ν to P and L, respectively:

μ:=δsH0|Pandν:=δtH0|L.

One may now check that

|Iδ(P,L)|δ-(s+t)1[]δ(p)dμ(p)dνδ().

Here νδ=νφδ is a mollification of ν (to make sense of this, identify A(2,1) with [0,1]×R). The inner integral is closely related to the X-ray transform of μ evaluated at : in fact

1[]δ(p)dμ(p)δ·(Xμδ)(),A(2,1),

where μδ=μφδ is a mollification of μ. Therefore,

|Iδ(P,L)|δ1-(s+t)(Xμδ)dνδ. 3.30

Next, we use the “duality” inequality (3.27) with r=τ/2-1, and eventually the boundedness of X between homogeneous Sobolev spaces (Proposition 3.26):

(Xμδ)dνδXμδH˙1-τ/2νδH˙τ/2-1μδH˙(1-τ)/2νδH˙τ/2-1.

Since τ(1,2) and σ+τ=3, we may use Lemma 3.29 to deduce

μδH˙(1-τ)/22τ(3.28)I3-τ(μδ)=Iσ(μδ)σA|P|δ2s-σ,

and similarly νδH˙τ/2-12Iτ(νδ)B|L|δ2t-τ. Plugging these estimates back into (3.30),

|Iδ(P,L)|δ1-(s+t)A|P|δ2s-σB|L|δ2t-τ=ABδ2-σ-τ|P||L|=ABδ-1|P||L|,

recalling that σ+τ=3.

Cases t[0,s] of the Furstenberg set problem

Every (st)-Furstenberg set with 0ts satisfies dimHFs+t. This follows from [38] or [31, Theorem A.1] (the special case s=t was contained in [30]). In contrast with the cases t[2-s,2], the sharp result in any of the cases t[0,2-s) cannot be derived by combining Proposition 3.15 with Fu and Ren’s (sharp) bound for |Iδ(P,L)| for (δ,s)-sets P and (δ,t)-sets L.

Exercise 1

What is the best lower bound you can obtain on the Hausdorff dimension of (st)-Furstenberg sets via Proposition 3.15, when t[0,2-s)?

Remark 3.31

It may appear puzzling that Fu and Ren’s sharp upper bound on δ-incidences between does not always yield a sharp lower bound on the dimension of (st)-Furstenberg sets. To illustrate the reason, fix α,β[0,2], s[0,1], and consider the following two questions concerning a (δ,α)-set PB(1) and a (δ,β)-set or lines LA(2,1):

  1. Is it possible that |P[]δ|δ-s for all L?

  2. Is it possible that P[]δ contains a (δ,s)-set with cardinality δ-s for all L?

The Fu-Ren bound always gives the correct answer to (Q1), but fails to see the difference between (Q1) and (Q2)—because the non-concentration conditions on P,L, and the number of δ-incidences are the same in both questions. So, whenever (α,β,s) is a triple relevant for the (st)-Furstenberg set problem such that (Q1)-(Q2) have opposite answers, the Fu-Ren bound fails to yield a sharp estimate on Furstenberg sets.

For example, take α=1=β and s=12. Now the answer to (Q1) is positive thanks to the “train tracks” Example 3.20. The answer to (Q2) is negative: A positive answer would essentially show that P is a (12,1)-Furstenberg set with “dimension” 1. However, it has been known since the early 2000 s [4, 34] that this is not possible. Indeed, now we know that (12,1)-Furstenberg sets have dimension 43 by (3.1).

Remark 3.31 gives hint for how to proceed dealing with the cases t[0,2-s): We need variants of incidence bounds which take into account the extra information present in (Q2). Consider the following generalised notion of δ-incidences:

Definition 3.32

Let PR2, and let F be an arbitrary collection of subsets of R2. We define Iδ(P,F):={(p,F)P×F:p[F]δ}.

Our familiar Iδ(P,L) is a special case of this definition. We will apply the definition in a case where each FF is a (δ,s)-subset of a line. In this context have the following incidence bound, which can be viewed as a δ-incidence counterpart of Proposition 2.9:

Proposition 3.33

Let PB(1) be δ-separated. Let 0ts1. Let F={F():L} be a family of sets, where L is a (δ,s,A)-set, and each F() is individually a (δ,s,B)-set. Then,

|Iδ(P,F)|ABδ-s|P||F|+|P|.

Remark 3.34

This proposition is “folklore” in the sense that the proof strategy has appeared in many places; I’m not sure where first, but it is certainly implicit, e.g. in the proof of [31, Theorem A.1]. A more explicit appearance is [48, Proposition 2.13].

Proof of Proposition 3.33

Write

|Iδ(P,F)|=pP|{L:p[F()]δ}||P|1/2(pP|{(,):p[F()]δ[F()]δ}|)1/2=|P|1/2(,|P[F()]δ[F()]δ|)1/2.

The value of the “diagonal” sum is precisely |Iδ(P,F)|. If the “diagonal” sum dominates, we obtain after rearranging |Iδ(P,F)||P|.

For the “off-diagonal” sum with , observe that [F()]δ[F()]δ[]δ[]δ is contained in a ball B,R2 of radius δ/dA(2,1)(,). Using the δ-separation of P, and the (δ,s,B)-set property of the sets F(), we deduce

|P[F()]δ[F()]δ||F()B,|BdA(2,1)(,)-s.

Therefore, using the (δ,s,A)-set property of L (which is implied by the (δ,t,A)-set property since ts),

|P[F()]δ[F()]δ|A1dA(2,1)(,)sABδ-s|L|.

Since |L|=|F|, this completes the proof.

Corollary 3.35

Every (st)-Furstenberg set FR2 with 0ts satisfies dimHFs+t.

Remark 3.36

Corollary 3.35 is evidently sharp: for example any product set F=A×B with dimHA=s and dimHB=t is an (st)-Furstenberg set, and it is often (if not always) the case that dimH(A×B)=s+t.

Proof of Corollary 3.35

The proof is similar to the proof of Proposition 3.15, we only point out the differences. We make a counter assumption: dimHF<σ+t for some σ<s.

Extract the (δ,σ+t)-set PDδ and the (δ,t)-set L¯L exactly the same way as in the proof of Proposition 3.15. Recall from (3.18) that

Hs(P)1,L¯.

In the proof of Proposition 3.15, this information was used in the (wasteful) way to deduce that |P[]2δ|δ-s. We now do something more sophisticated: applying the subset finding lemma, Proposition 3.9, we locate, for each L¯, a (δ,s)-set F() such that |P[F()]2δ|δ-s. Writing F:={F():L¯}, we then have

|I2δ(P,F)|δ-s|L¯|δ-s-t.

Compare this lower bound with (3.19)!

The family F satisfies the hypotheses of Proposition 3.33, so

δ-s-t|I2δ(P,F)|δ-s|P||L¯|+|P|

Recalling that |P|δ-σ-t with σ<s, and |L¯|δ-t, we arrive at a contradiction. It is worth noting that the (δ,σ+t)-property of P was not used here in any other ways than to ensure that P is δ-separated, and |P|δ-σ-t (this actually means that the Katz–Tao covering lemma, Proposition 3.12, is not really needed in the proof).

Formal δ-discretisation of the Furstenberg set problem

We discussed above (Remark 3.31) that the full solution of the (st)-Furstenberg set problem cannot be obtained from Problem 2 (incidences between (δ,α)-sets of points and (δ,β)-sets of lines). There nonetheless exists a δ-discretised incidence problem which implies lower bounds for the (st)-Furstenberg set problem. This is formalised by the next proposition:

Proposition 3.37

Fix s[0,1], t[0,2], and f[0,2]. Consider the following statement (P):

  • (P)
    For every η>0, there exists ϵ>0 such that the following holds. Let LA(2,1) be a (δ,t,δ-ϵ)-set with |L|δ-t+ϵ. For every L, let F()B(1) be a (δ,s,δ-ϵ)-set with |F()|δ-s+ϵ. Then, the union F:=LF() satisfies
    |F|δδ-f+η.

If (P) holds for some s,t,f, then every (st)-Furstenberg set FR2 satisfies dimHFf.

In fact, Ren and Wang [53] showed precisely that (P) in the previous proposition holds with the choice f=f(s,t)=min{s+t,(3s+t)/2,s+1} as in (3.1).

Proposition 3.37 is proven by a pigeonholing argument similar to the ones we have seen in Proposition 3.15 and Corollary 3.35, see [31, Lemma 3.3] for the details.

Cases (s, 1) under maximal separation

We have previously covered the cases t[0,s][2-s,2] of the (st)-Furstenberg set problem. The remaining cases t(s,2-s) are more complicated, and beyond the scope of these notes.

In this section, we establish property (P) in Proposition 3.15 with f=(3s+1)/2 (the sharp value) for all pairs (s, 1) with s(0,1], but under the additional hypothesis that the (δ,s)-sets F() are maximally separated. This is a special case of [21, Theorem 4], but the proof we present here is different, and avoids using the crossing number lemma.

For technical convenience, we work under the hypothesis that F() contains a δs-net in B(1). A modification of the argument would, however, work under the hypotheses that |F()|δ-s and F() is δs-separated.

As a lemma we need the following fact about (generalised) Kakeya sets:

Lemma 3.38

Let t[0,1], and let LA(2,1) be a (δ,1)-set such that diam(B(1))1 for all L. Then, the set U:=B(1)L[]δ satisfies Leb(U)δ|L|.

Proof

For every L, let F()B(1) be a maximal δ-separated set (in particular F() is a (δ,1)-set). Let PUB(1) be a maximal δ-separated set. Write F:={F():L}. Then, |I2δ(P,F)|δ-1|L|. Combining this with Proposition 3.33 with ts:=1, we obtain

δ-1|L||I2δ(P,L)|δ-1|P||L|+|P|,

which can be rearranged to |P|δ-1|L|. Thus Leb(U)δ2|P|δ|L|.

Remark 3.39

In the previous proof, we could have also used Theorem 3.22.

Theorem 3.40

Let s(0,1], and let LA(2,1) be such that diam(B(1))1 for all L. Set U:=B(1)L[]δ. Let F()B(1) be a δs-dense set for all L. Then, the set F:=LF() satisfies

|F|δδ-(3s+1)/2Leb(U).

In particular, if L is a (δ,1)-set with |L|δ-1, then |F|δδ-(3s+1)/2 (by Lemma 3.38).

Proof

Write T:={[]δB(1):L}, so U:=B(1)TTT. For QDδs(U)={QDδs:QU} fixed, we say that two tubes T,TT with TQTQ are Q-distinct if

TQ[T]2δ,

see Fig. 3 for intuition. Let MQ be the maximal cardinality of a Q-distinct subset of T.

Fig. 3.

Fig. 3

Here MQ=3, even though 9 tubes in total intersect Q

Claim 3.41

We have

MQLeb(QU)·δ-s-1,QDδs(U).

Proof

Let T1,,TMQT be a maximal Q-distinct collection. Then, every tube TT satisfies TQ[Tj]2δQ for some 1jMQ. In particular, QU is contained in the union of the sets [Tj]2δQ with 1jMQ. The Lebesgue measure of each of these sets is δs+1, so Leb(QU)MQδs+1, as claimed.

The proof of the next claim is a variant of the 2-ends argument in Remark 2.13.

Claim 3.42

We have

|F2Q|δMQ1/2,QDδs(U).

Proof

Let FF2Q be a maximal δ-separated set. Recall that the points in F() form a δs-net in B(1). Therefore, if T=[]δ, and TQ, the set T2Q contains two distinct points p,qF2Q with separation |p-q|δs, see Fig. 3.

Therefore, we obtain a map T(F2Q)2 which associates to every tube TT with TQ a pair (p,q)(F2Q)2 with separation |p-q|δs. It now remains to observe that this map is C-to-1 on every collection of Q-distinct tubes. Therefore |F2Q|2MQ, as claimed.

Finally, observe that Leb(QU)1/2δ-sLeb(QU) for QDδs(U), so

|F|δQDδs(U)|F2Q|δQDδs(U)MQ1/2δ-s+12QDδs(U)Leb(QU)1/2δ-3s+12QDδs(U)Leb(QU)=δ-3s+12Leb(U).

This completes the proof.

Applications and connections of Furstenberg sets

In 2000, Wolff [62, 63] had recently posed the (s, 1)-Furstenberg set problem, and it was, e.g. known that every (12,1)-Furstenberg set FR2 has dimHF1. The conjecture (now a theorem) states that dimHF4/3. In their influential 2001 paper, Katz and Tao [34] showed that an ϵ-improvement to the bound dimHF1 is logically equivalent (at a δ-discretised level) to an ϵ-improvement in a version of Falconer’s distance set conjecture, and also an ϵ-improvement in the δ-discretised version of the Erdős-Szemerédi sum-product problem. All of these equivalent ϵ-improvements were shortly afterwards obtained by Bourgain [4].

During the past 20 years, more connections have been discovered, for example to orthogonal and radial projections. The connection to orthogonal projections is straightforward: Non-trivial results on the δ-discretised Furstenberg set problem formulated in Proposition 3.37 imply non-trivial bounds on the dimension of exceptional sets of orthogonal projections. The mechanism is explained in [48, Sect. 3.2]. Notably, the full solution, due to Ren and Wang [53], implies the following sharp bound stated in [53, Theorem 1.2]:

Theorem 4.1

Let KR2 be analytic. Then, writing πe(x):=x·e for eS1 and xR2,

dimH{eS1:dimHπe(K)<σ}min{2σ-dimHK,0},0σmin{dimHK,1}.

A weaker estimate was earlier obtained by Kaufman [35].

The following sections contain brief accounts on the implications of Furstenberg set estimates to the (continuous) sum-product problem, to radial projections, and to arithmetic sums of fractal sets on the parabola.

The sum-product problem

The Erdős-Szemerédi sum-product conjecture [7] asks to prove that if AZ (original formulation) or AR (plausible extension) is a finite set, then max{|A+A|,|A·A|}ϵ|A|2-ϵ for all ϵ>0. This problem remains open and very actively studied. Elekes in the late 90 s [16] connected the problem to the Szemerédi-Trotter incidence bound and allowed him to establish the partial result max{|A+A|,A·A|}|A|5/4. This is nowadays far below the state-of-the-art in the discrete version of Erdős and Szemerédi’s problem (see [54]), but his argument has the benefit of extending easily to a “continuum” setting. The main idea is the following observation:

Lemma 4.2

Let A,B,CR be sets. Let F=(A+B)×(A·C). Then, the family of lines

L(B,C):={b,c={(x,cx-bc):xR}:bB,cC}

has the property that Fb,c contains the affine copy of A given by {(a+b,a·c):aA}.

Proof

Note that (a+b,a·c)b,c for all a,b,cR, since c(a+b)-bc=a·c. Therefore {(a+b,a·c):aA}Fb,c for all (b,c)B×C.

To use Lemma 4.2, one checks that the line set L(B,C)A(2,1) is diffeomorphic to B×C, and in particular dimHL(B,C)=dimH(B×C). Therefore, Lemma 4.2 tells us that F=(A+B)×(A·C) is an (st)-Furstenberg set with

s=dimHAandt=dimH(B×C).

One can now deduce various inequalities about dimHF using the Furstenberg set theorem. We only consider the case A=B=C. Write s=dimHA, and note that dimH(A×A)2s. Thus, F=(A+A)×(A·A) is an (s,2s)-Furstenberg set, and

dimH((A+A)×(A·A))=dimHF(3.1)min5s2,s+1.

Since dimH(A×B)dimHA+dimPB (where dimP is the packing dimension), it follows that either

dimH(A+A)min5s4,s+12ordimP(A·A)min5s4,s+12.

Here “54” is Elekes’ exponent. This argument also works well at a δ-discretised level, but we leave the details to the reader (or see [49, Corollary 6.6]). For further recent work on the continuum and δ-discretised sum-product problems, see [39, 43].

Radial projections and the dimension of quotients

In 2020, the best general lower bounds for the dimension of (st)-Furstenberg sets were the following:

dimHFmax{2s,s+t2},s(0,1],t[0,2]. 4.3

The 2s-bound is a special case of Corollary 3.35, whereas the (s+t2)-bound is due to Héra [29] (see also [41] for a partial result). In 2021, Shmerkin and myself [48] proved the following “ϵ-improvement” over the 2s-bound:

Theorem 4.4

For s(0,1) and t(s,2], the Hausdorff dimension of every (st)-Furstenberg set FR2 satisfies dimHF2s+ϵ(s,t) for some ϵ(s,t)>0.

Theorem 4.4 was applied in [50] to prove a new radial projection theorem, which, in turn, played a role in the solution of the full Furstenberg set problem (see [48, 53]). I will now explain (very informally) the connection between Theorem 4.4 and radial projections.

For xR2, the radial projection to x is the map πx:R2\{x}S1 defined by πx(y)=(x-y)/|x-y|. A good way to think about radial projections is the following: if FR2 and xR2\F, then dimHπx(F)=dimHLx, where

Lx={A(2,1):xandF}.

In [50], we proved the following:

Theorem 4.5

Let E,FR2 be disjoint Borel sets such that E is not contained on any line. Then, supxEπx(F)=min{dimHE,dimHF,1}.

The following argument is a very rough indication of how the ϵ-improvement in the Furstenberg set problem, Theorem 4.4, appears in the proof of Theorem 4.5.

“Proof” of Theorem 4.5

Write s:=min{dimHE,dimHF,1}. If necessary replacing F by an s-dimensional subset, we may assume that dimHF=s. To reach a contradiction, assume that dimHπx(F)=s-η for all xE, where η>0. In other words dimHLx=s-η for all xE.

The set L=xELxA(2,1) is a “dual” (s-η,s)-Furstenberg set: It contains an (s-η)-dimensional line family incident to every point in the s-dimensional set E. Since s-η<s, Theorem 4.4 implies

dimHL2(s-η)+ϵ

for some ϵ:=ϵ(s-η,s)>0. Or does it? A potential problem is that ϵ(0,s)=0 in Theorem 4.4, so if η=s (i.e. dimHπx(F)0 for xE), we are in trouble. However, we may assume here that ηs2. This part of the argument uses the hypothesis that E is not contained on a line, and is based on a much earlier result [44].

Now we pose a rather unrealistic simplifying assumption: since dimHLx=s-η, but dimHF=s, it is somewhat reasonable to expect that dimH(F)η for Lx. If this were the case, as we now assume, then F is an (η,t)-Furstenberg set with t=dimHL2(s-η)+ϵ. Consequently, by the second “classical” bound in (4.3),

s=dimHFη+2(s-η)+ϵ2=s+ϵ2.

This contradiction completes the “proof” of Theorem 4.5.

As a corollary, we obtain the following sum-quotient estimate:

Corollary 4.6

Let A,BR be Borel sets. Then

dimHA-BA-Bmin{dimHA+dimHB,1}.

Proof

We may assume that both AB contain at least two points. We apply Theorem 4.5 to the sets E=-A×B and F=-B×A. Since E is a Borel set not contained on a line, for every ϵ>0, there exists a point x=(-a,b)E such that

dimHπx(F\{x})min{dimHE,dimHF,1}-ϵmin{dimHA+dimHB,1}-ϵ.

Now, it remains to observe that dimHπx(F\{x}) agrees with the dimension of “slopes” spanned between the point x=(-a,b) and the set F\{x}, namely

a-b-b-(-a):(-b,a)(-B×A)\{(-a,b)}=a-ba-b:(-b,a)(-B×A)\{(-a,b)}.

Since the quotient set (A-B)/(A-B) contains all such slopes, the corollary follows.

Arithmetic sums of subsets of the parabola

The Furstenberg set problem has some hidden “curvature”, which can be brought to light by the following observation. The map Ψ(x,y):=(x,y2-x) sends every non-vertical line R2 to some translate of the standard up-ward pointing parabola P={(x,x2):xR}. More precisely, if (x0,y0):={(x,y):y=y0+2x0(x-x0)} is the line with slope 2x0 passing through (x0,y0), then

Ψ((x0,y0))=Ψ(x0,y0)+P.

The map Ψ is locally bi-Lipschitz, so s-dimensional subsets of lines are carried to s-dimensional subsets of the corresponding parabolas (for an alternative argument, observe that Ψ leaves the first coordinate fixed). The map Ψ also sends every t-dimensional set of lines to a family of parabolas of the form {x+P}xX, where dimHX=t.

Using the map Ψ, and its properties described above, one obtains (with the argument of [47, Sect. 1.1]) the following corollary of the Furstenberg set theorem:

Corollary 4.7

Let FP be a set with dimHF=s(0,1], and let ER2 be a set with dimHE=t[0,2]. Then,

dimH(E+F)mins+t,3s+t2,s+1. 4.8

In particular, dimH(F+F+F)min{5s2,s+1}.

Proof

The set Ψ-1(E+F) is an (st)-Furstenberg set, so (4.8) is implied by the Furstenberg set theorem. The bound dimH(F+F+F)min{5s2,s+1} follows by applying (4.8) to E=F+F and F (and noting first that dimHE2s).

Remark 4.9

The bound dimH(F+F+F)min{5s2,s+1} is a continuum analogue of an observation of Bourgain and Demeter [5, Proposition 2.15]. They showed that if ΛP is a finite set, then |Λ+Λ+Λ||Λ|5/2. They also ask in [5, Question 2.13] whether |Λ+Λ+Λ|ϵ|Λ|3-ϵ, and deduce a positive answer from the 2-decoupling theorem if Λ is |Λ|-C-separated. Analogously, it seems unlikely that the bound for dimH(F+F+F) in Corollary 4.7 is sharp. For further literature on this topic, see [10, 12, 47].

Acknowledgements

I would like to thank the reviewers for making numerous helpful suggestions to improve the presentation.

Funding

Open Access funding provided by University of Jyväskylä (JYU).

Data availability

No datasets were generated or analysed during this study.

Declarations

Conflict of interest

The author is supported by the Research Council of Finland via the project Approximate incidence geometry, Grant No. 355453, and by the European Research Council (ERC) under the European Union’s Horizon Europe research and innovation programme (Grant Agreement No. 101087499). There are no other conflicts of interest.

Footnotes

1

If [p,q]E(1)E(2), then the segment [pq] is contained in [2]δ[2]δ, which forces (1,2)δ.

2

If Z contained more lines than its degree, then almost every lines of R2 would be contained in Z. This would lead to Z=R2, and therefore poly0 contrary to Lemma 2.11.

3

First find a sequence {Qj} of arbitrary dyadic cubes covering K with (Qj)s122-k0s. Enlarge the cubes slightly to make them open, but still retaining (Qj)s2-k0s. Finally, pick a finite sub-cover, and let δ>0 be the smallest side-length in that sub-cover.

T.O. is supported by the Research Council of Finland via the project Approximate incidence geometry, Grant No. 355453, and by the European Research Council (ERC) under the European Union’s Horizon Europe research and innovation programme (Grant Agreement No. 101087499).

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