New York J. Math.7(2001)149–187.
Some Connections between Falconer’s Distance Set Conjecture and Sets of Furstenburg Type
Nets Hawk Katz and Terence Tao
Abstract. In this paper we investigate three unsolved conjectures in geomet- ric combinatorics, namely Falconer’s distance set conjecture, the dimension of Furstenburg sets, and Erd¨os’s ring conjecture. We formulate natural δ- discretized versions of these conjectures and show that in a certain sense that these discretized versions are equivalent.
Contents
1. Introduction 149
1.1. Notation 150
1.3. The Falconer distance problem 150
1.7. Dimension of sets of Furstenburg type 153
1.12. The Erd¨os ring problem 153
1.15. The main result 154
2. Basic tools 155
3. Arithmetic combinatorics 157
4. Bilinear Distance Conjecture implies Ring Conjecture 158 5. Ring Conjecture implies Discretized Furstenburg Conjecture 163 6. Discretized Furstenburg Conjecture implies Bilinear Distance Conjecture171
7. Discretization of fractals 176
8. Discretized Furstenburg Conjecture implies Furstenburg problem 178 9. Bilinear Distance Conjecture implies Falconer Distance Conjecture 180
References 186
1. Introduction
In this paper we studyFalconer’s distance problem, the dimension of sets of Furstenburg type, and Erd¨os’s ring problem. Although we have no direct progress
Received January 22, 2001.
Mathematics Subject Classification. 05B99, 28A78, 28A75.
Key words and phrases. Falconer distance set conjecture, Furstenberg sets, Hausdorff dimen- sion, Erd¨os ring conjecture, combinatorialgeometry.
ISSN 1076-9803/01
149
on anyof these problems, we are able to reduce the geometric problems to δ- discretized variants and show that these variants are all equivalent.
In order to state the main results we first must develop a certain amount of notation.
1.1. Notation. 0 < ε 1, 0 < δ 1 are small parameters. We use A B to denote the estimateA≤Cεδ−CεBfor some constantsCε,C, andA≈B to denote ABA.
We useB(x, r) =Bn(x, r) to denote the open ball of radiusr centered at xin Rn, andA=An to denote anyannulus in Rn of the formA:={x:|x| ≈1}.
IfAis a finite set, we use #Ato denote the cardinalityofA. For finite sets A, B, we saythatA is a refinement ofB ifA⊂B and #A≈#B.
IfEis contained in a subspace ofRn and has positive measure in that subspace, we use |E| for the induced Lebesgue measure of E. The subspace will always be clear from context.
For setsE, F of finite measure, we saythatE is a refinement ofF ifE⊂F and
|E| ≈ |F|. We saythatE isδ-discretizedifE is the union of balls of radius≈δ.
Definition 1.2. For any0 < α ≤n, we saythat a set E is a (δ, α)n-set if it is contained in a ballBn(0, C), isδ-discretized and one has
|E∩B(x, r)|δn(r/δ)α (1)
for allδ≤r≤1 andx∈Rn.
Roughlyspeaking, a (δ, α)n-set behaves like theδ-neighbourhood of anα-dimen- sional set in Rn. The condition (1) is necessaryto ensure that E does not con- centrate in a small ball, which would lead to some trivial counterexamples to the conjectures in this paper (cf. the “two ends” condition in [17], [18]).
If X, Y are subsets ofRn, we use X +Y to denote the set X+Y :={x+y : x∈X, y∈Y}. SimilarlydefineX−Y, and (whenn= 1)X·Y, X/Y,X2, √
X, etc. Note that X2 X ·X in general. Note thatX ×Y denotes the Cartesian product X ×Y := {(x, y) : x ∈ X, y ∈ Y} as opposed to the pointwise product X·Y :={xy:x∈X, y∈Y}. Unfortunatelythere is a conflict of notation between X2:={x2:x∈X}and X2 :={(x, y) :x, y∈X}; to separate these two we shall occasionallywrite the latter asX⊕2.
If a rectangleRhas sides of lengtha, bfor somea > b, we call thedirectionofR the directionω∈S1that the sides of lengthaare oriented on. This is onlydefined up to sign±.
1.3. The Falconer distance problem. For anycompact subsetK of the plane R2, define thedistance setdist(K)⊂RofKby
dist(K) :=|K−K|={|x−y|:x, y∈K}.
In [8] Falconer conjectured that if dim(K) ≥ 1, then dim(dist(K)) = 1, where dim(K) denotes the Hausdorff dimension ofK. As progress towards this conjecture, it was shown in [8] that dim(dist(K)) = 1 obtained whenever dim(K)≥3/2. This was improved to dim(K) ≥ 13/9 byBourgain [2] and then to dim(K) ≥ 4/3 by Wolff [21]. These arguments are based around estimates forL2 circular means of Fourier transforms of Frostman measures. However, it is unlikelythat a purely
Fourier-analytic approach will be able to improve upon the 4/3 exponent; for a discussion, see [21].
Now suppose that one onlyassumes that dim(K)≥1. An argument of Mattila [12] shows that dim(dist(K))≥ 12. One mayask whether there is anyimprovement to this result, in the following sense:
Distance Conjecture 1.4. There exists an absolute constant c0 > 0 such that dim(dist(K))≥ 12+c0 wheneverK is compact and satisfiesdim(K)≥1.
This is of course weaker than Falconer’s conjecture, but remains open.
One mayhope to prove this conjecture byfirst showing aδ-discretized analogue.
As a naive first approximation, we mayask the informal question of whether (for 0 < δ, ε 1) the distance set of a (δ,1)2 set of measure ≈ δ can be (mostly) contained in a (δ,1/2)1set.
Unfortunately, this problem has an essentially negative answer, as the counterex- ample
(x1, x2) :x1=k√
δ+O(δ), x2=O(√
δ) for somek∈Z, k=O
δ−1/2 (2)
shows1 . A substantial portion of the distance set of (2) is contained in the δ- neighbourhood of an arithmetic progression of spacing δ1/2, and this is a (δ,12)1
set.
Figure 1. An example to remember. Few blurred distances but manyblurred points.
This obstruction to solving Conjecture 1.4 can be eliminated byreplacing the above informal problem with a “bilinear” variant in which an angular separation condition is assumed:2
Bilinear Distance Conjecture 1.5. LetQ0,Q1,Q2be three cubes inB(0, C)of radius≈1satisfying the separation condition
|(x1−x0)∧(x2−x0)| ≈1 for allx0∈Q0, x1∈Q1, x2∈Q2. (3)
For eachj= 0,1,2, letEj be a(δ,1)2 subset of Qj, and letD be a(δ,1/2)1 subset of R. Then
|{(x0, x1, x2)∈E0×E1×E2:|x0−x1|,|x0−x2| ∈D}|δ3−c1 (4)
wherec1>0 is an absolute constant.
1This counterexample also appears in Fourier-based approaches to the distance problem. See [21].
2This idea is frequently used in related problems, see, e.g., [15], [1], [20]. Other discretizations are certainly possible, providing of course that (2) is neutralized.
The estimate (4) is triviallytrue whenc1= 0. Also, if it were not for condition (3) one could easilydisprove (4) for any c1 >0 bymodifying (2). Conjecture 1.5 is also heuristicallyplausible from analogywith results on the discrete distance problem such as Chung, Szemer´edi and Trotter [4]. We remark that the arguments in that paper require the construction of three cubes satisfying (3), and involve the Szemer´edi-Trotter Theorem (which maybe considered as a result concerning the discrete analogue of the Furstenburg problem).
Q0 Q1
Q2
Figure 2. In the bilinear distance conjecture, the points are split into three camps.
In Section9we prove:
Theorem 1.6. A positive answer to the Bilinear Distance Conjecture 1.5 implies a positive answer to the Distance Conjecture 1.4.
Although this implication looks plausible from discretization heuristics, there are technical difficulties due to the presence of the counter-example (2), and also by
the fact that several scales maybe in playwhen studying the Hausdorff dimension of a set.
1.7. Dimension of sets of Furstenburg type. We now turn to a problem arising from the work of Furstenburg, as formulated in work of Wolff [19], [21].
Definition 1.8. Let 0 < β ≤1. We define a β-set to be a compact setK ⊂R2 such that for everydirection ω ∈S1 there exists a line segment lω with direction ω which intersectsKin a set with Hausdorff dimension at leastβ. We letγ(β) be the infimum of the Hausdorff dimensions ofβ-sets.
In [19] the problem of determining γ(β) is formulated. At present the best bounds known are
max(β+1
2,2β)≤γ(β)≤3 2β+1
2;
see [19]. This problem is clearlyconnected with the Kakeya problem (which is essentiallyconcerned with the higher-dimensional analogue ofγ(1)). Connections to the Falconer distance set problem have also been made; see [21].
The most interesting value of β appears to be β = 1/2. In this case the two lower bounds onγ(β) coincide to becomeγ(12)≥1. We ask:
Furstenburg Problem 1.9. Is it true that γ(12)≥1 +c2 for some absolute con- stantc2>0? In other words, is it true that 12-sets must have Hausdorff dimension at least1 +c2?
One canδ-discretize this problem as:
Discretized Furstenburg Conjecture 1.10. Let 0 < δ 1, and let Ω be aδ- separated set of directions, and for eachω ∈ΩletRω be a(δ,12)2 set contained in a rectangle of dimensions ≈1×δ oriented in the direction ω. Let E be a (δ,1)2 set. Then
|{(x0, x1)∈E×E :x1, x0∈Rω for someω∈Ω}|δ2+c3 (5)
for some absolute constantc3>0.
As before, this conjecture is heuristicallyplausible from analogywith discrete incidence combinatorics, in particular the Szemer´edi-Trotter Theorem [14]. Unlike the case with the distance problem, the set (2) does not provide a serious threat, and so one does not need to go to a bilinear framework.
The Discretized Furstenburg Conjecture1.10is related to the Furstenburg Con- jecture 1.9in much the same waythat the Kakeya maximal function conjecture is related to the Kakeya set conjecture. In Section8we show:
Theorem 1.11. A positive answer to the Discretized Furstenburg Conjecture1.10 implies a positive answer to the Furstenburg Problem 1.9.
1.12. The Erd¨os ring problem. We consider a problem of Erd¨os, namely:
Ring Problem 1.13. Does there exist a subringR ofRwhich is a Borel set and has Hausdorff dimension strictly between0 and1?
This problem is connected to Falconer’s distance problem; for instance, Falconer [8] used results on the distance problem to show that Borel subringsRofRcould
not have Hausdorff dimension strictlybetween 1/2 and 1. Essentially, the idea is to use the fact that dist(R×R)⊆√
We concentrate on the specific problem of whether a subring can have dimensionR.
exactly1/2; it seems reasonable to conjecture that such rings do not exist. A positive answer to Conjecture1.4would essentiallyimplythis conjecture.
IfRis a ring of dimension 1/2, then of courseR+RandRRalso have dimension 1/2. This leads us to the followingδ-discretization of the above conjecture.
Ring Conjecture 1.14. Let0< δ 1, and letA⊂Abe a(δ,12)1 set of measure
≈δ1/2. Then at least one ofA+AandAA has measure δ12−c4, wherec4>0is an absolute constant.
The dimension condition (1) is crucial, as the trivial counterexampleA:= [1,1+
δ1/2] demonstrates. In principle the discretized ring conjecture gives a negative answer to the Erd¨os ring problem, but we have not been able to make this rigorous.
For the discrete version of this problem, when measure is replaced bycardinality, there is a result of Elekes [6] that when A has finite cardinality#A, at least one of A+Aand AA has cardinality#A5/4 . The proof of this result exploits the Szemer´edi-Trotter Theorem. This is heuristic evidence for Ring Conjecture 1.14 if one accepts the (somewhat questionable) analogybetween discrete models and δ-discretized models.
It mayappear that the ring hypothesis is being under-exploited when reducing to Ring Conjecture1.14, since one is onlyusing the fact thatR+RandRRare small.
However, we shall see in Proposition 4.2 that control on A+A and AA actually implies quite good control on other arithmetic expressions such as AA−AA or (A−A)2+ (A−A)2 (after passing to a refinement), so the ring hypothesis is not being wasted.
1.15. The main result.
One Ring to rule them all, One Ring to find them, One Ring to bring them all, and in the darkness bind them. [16]
As one can see from the previous discussion, there have been manypartial con- nections drawn between the Falconer, Furstenburg, and Erd¨os problems. The main result of this paper is to consolidate these connections into:
Main Theorem 1.16. Conjectures 1.5,1.10, and1.14are logically equivalent.
We shall prove this theorem in Sections3-6.
In particular, in order to make progress on the Falconer and Furstenburg prob- lems it suffices to prove the Ring Conjecture1.14. This appears to be the easiest of all the above problems to attack. It seems likelythat one needs to exploit some sort of “curvature” between addition and multiplication to prove this conjecture, although a naive Fourier-analytic pursuit of this idea seems to run into difficul- ties. This mayindicate that a combinatorial approach will be more fruitful than a Fourier approach. The fact thatRis a totallyordered field mayalso be relevant, since the analogue of Erd¨os’s ring problem is false for non-ordered fields such as the complex numbers C or the finite field Fp2. (Unsurprisingly, the analogues of
Falconer’s distance problem and the conjectures for Furstenburg sets also fail for these fields; see, e.g., [19].)
These problems are also related to the Kakeya problem in three dimensions, although the connection here is more tenuous. A proof of Conjecture 1.14would probablylead (eventually!) to an alternate proof of the main result in [10], namely that Besicovitch sets3 in R3have Minkowski dimension strictlygreater than 5/2, and would not relyas heavilyon the assumption that the line segments all point in different directions. Veryinformally, the point is that the arguments in [10] can be pushed a bit further to conclude that a Besicovitch set of dimension exactly 5/2 must essentiallybe a “Heisenberg group” over a ring of dimension 1/2. We shall not pursue this connection in detail as it is somewhat lengthyand would not directlyyield anynew progress on the Kakeya problem.
In conclusion, these results indicate that the possibilityof 1/2-dimensional rings is a fundamental obstruction to further progress on the Falconer and Furstenburg problems, and mayalso be obstructing progress on the Kakeya conjecture and related problems (restriction, Bochner-Riesz, Stein’s conjecture, local smoothing, etc.). It also appears that substantiallynew techniques are needed to tackle this obstruction, possiblyexploiting the ordering of the reals.
2. Basic tools
In this section 0< ε1 is fixed, butδis allowed to vary. As in other sections, the implicit constants here are not allowed to depend onδ.
To clarifymanyof the arguments in this paper, it mayhelp to know that almost all estimates of the form AB which occur in this paper are sharp in the sense that the converse boundABis usuallytrivial to prove. It is this sharpness which allows us to pass from one expression to another without losing verymuch in the estimates (if one does not mind the implicit constants in thenotation increasing veryquickly).
A typical application of this philosophy is:
Cauchy-Schwarz 2.1. Let A,B be sets of finite measure, and let∼be a relation between elements of Aand elements ofB. If
|{(a, b)∈A×B:a∼b}| ≥λ|A||B|
for some0< λ≤1 then
|{(a, b, b)∈A×B×B:a∼b, a∼b}| ≥λ2|A||B|2. Proof. We can rewrite the hypothesis as
A|{b∈B:a∼b}|da≥λ|A||B|
and the conclusion as
A|{b∈B :a∼b}|2 da≥λ2|A||B|2.
The claim then follows from Cauchy-Schwarz.
3A Besicovitch set is a set which contains a unit line segment in every direction.
The next lemma deals with the issue of how to refine aδ-discretized set to become a (δ, α)n set for suitableα.
Refinement 2.2. Let 0 < δ 1 be a dyadic number, 0 < α < n, K 1 be a constant, and let E be a δ-discretized set in Bn(0, C) such that |E| δn−α. Then one can find a set Eδ for all dyadic δ < δ ≤ 1 which can be covered by δKεδ−α balls of radiusδ, and a set (δ, α)n set E∗ (with the implicit constants in the definition of a (δ, α)n set depending onK)such that
E⊆E∗∪
δ<δ≤1
Eδ. Proof. Define the setsEδ by
Eδ :=
x∈Rn:|E∩B(x, δ)| ≥δ−Kεδn(δ/δ)α andE∗ by
E∗:=
E\
δ<δ≤1
Eδ
+B(0, δ).
The required properties onEδ andE∗ are then easilyverified.
Separation 2.3. Let X be a (δ, α)n set in Rn for some 0 < α < n such that
|X| ≈δn−α. Then there exist refinements X1, X2 of X which respectively live in cubesQ1,Q2 of size and separation≈1with |Q1|=|Q2|, and|X1|,|X2| ≈δn−α. Proof. By(1) we see that
|X∩Q| ≤10−n|X|
for all cubesQof side-lengthδC1ε, ifC1is a sufficientlylarge constant. The claim then follows bycovering B(0, C) with such cubes, extracting the top 5n cubes in that collection which maximize |X∩Q|, picking two of those cubesQ1,Q2 which are not adjacent, and settingXi :=X∩Qi fori = 1,2. We leave the verification
of the desired properties to the reader.
For anyfunctionf inR2, define the Kakeya maximal functionfδ∗(ω) forω∈S1 by
fδ∗(ω) := sup
R
1
|R|
R|f|,
whereR ranges over all 1×δrectangles oriented in the directionω.
The following estimate can be found in [5] (see also Lemma 6.2):
Kakeya 2.4 (C´ordoba’s estimate). We have fδ∗2f2.
Dually, if we setRωbe a collection ofδ×1 rectangles oriented in aδ-separated set of directions, then
ω
χRω 21.
3. Arithmetic combinatorics
We shall prove Theorem1.16byshowing that
Bilinear Distance +3Discretized Ring
u}tttttttttttt
tttttttttttt Discretized Furstenburg aiKKKKK
KKKKKKKK KKKKK
KKKKKKKK
We shall need a number of standard results concerning the cardinalityof sum-sets A+B and difference sets A−B, and partial sum-sets {a+b: (a, b)∈G}, where Gis a large subset ofA×B.
We first give the results in a discrete setting.
Lemma 3.1. [13] SupposeA1, A2 are finite subsets ofR such that
#(A1+A2)≈#A1≈#A2. Then we have
#(Ai1± · · · ±AiN)≈#A1
for all choices of signs ± and i1, . . . , iN ∈ {1,2}, where the implicit constants depend on N. Also, we can find a refinement A1 of A1 and a real number xsuch that x+A1 is a refinement ofA2.
Proof. Most of these results are in [13]. For the last result, observe that the discrete function χ−A1∗χA2 has an l1 norm ≈(#A1)2 and is supported in a set of cardinality ≈ #A1 bythe results in [13]. Thus one can find an x such that χ−A1∗χA2(x)#A1, and the claim follows bysettingA1=A1∩(A2−x).
We also need Bourgain’s variant of the Balog-Szemer´edi Theorem [3] (as used in Gowers [9]), namely:
Lemma 3.2. [3]LetN 1be an integer, and letA,B be finite subsets ofRsuch that
#A,#B≈N.
Suppose there exists a refinementGof A×B such that
#{a+b: (a, b)∈G}N.
Then we can find refinementsA,BofAandBrespectively such thatG∩(A×B) is a refinement of A×B, and for all (a, b)∈A×B we have
#{(a1, a2, a3, b1, b2, b3)∈A×A×A×B×B×B:
a−b= (a1−b1)−(a2−b2) + (a3−b3)} ≈N5. In particular, we have
#(A−B)≈N.
We can easilyreplace these discrete lemmata with δ-discretized variants as fol- lows.
Corollary 3.3. SupposeA, B are finite unions of intervals of length≈δsuch that
|A+B| ≈ |A| ≈ |B|.
Then we have
|A± · · · ±A| ≈ |A|
for all choices of signs ±, with the implicit constants depending on the number of signs. Also, we can find a refinementA ofAand a real numberxsuch thatx+A is a refinement ofB.
Perfection 3.4. Letrδ, and letA,B be finite unions of intervals of length≈δ such that
|A|,|B| ≈r.
Suppose there exists a refinementGof A×B such that
|{a+b: (a, b)∈G}|r.
Then we can findδ-discretized refinementsA,B ofAandB respectively such that G∩(A×B)is a refinement of A×B, and for all(a, b)∈A×B we have
|{(a1, a2, a3, b1, b2, b3)∈A×A×A×B×B×B:
a−b= (a1−b1)−(a2−b2) + (a3−b3)}| ≈r5. In particular, we have
|A−B| ≈r.
To obtain these corollaries, we first observe that anyδ-discretized setAcontains the ≈δ-neighbourhood of a discrete set A∗ of cardinality#A∗ ≈ |A|/δ which is contained in an arithmetic progression of spacing≈δ. The claims then follow by applying the previous lemmata toA∗,B∗. (See also the proof of [3], Lemma 2.83).
We also observe the trivial estimate
|A+B||A|,|B|
(6)
for all setsA,B.
If we also assume that the setsA,Bare contained in the annulusAthen one can also obtain analogues of (6) and the above two Corollaries in which addition and subtraction are replaced bymultiplication and division respectively. This simply follows byapplying a logarithmic change of variables. In the next section we shall use the fact that multiplication distributes over addition, to obtain hybrid versions of the above results.
4. The Bilinear Distance Conjecture 1.5 implies the Ring Conjecture 1.14.
Assume that the Bilinear Distance Conjecture1.5is true for some absolute con- stantc1>0. In this section we show how the Ring Conjecture1.14follows.
Let 0< ε1 be fixed. We mayassume that δ is sufficientlysmall depending onε, since the Ring Conjecture is trivial otherwise. We mayalso assume thatδis dyadic. Assume for contradiction that one can find a (δ,12)1-setA⊂Aof measure
|A| ≈δ1/2such that
|A+A|,|A·A|δ1/2 (7)
Figure 3. A set which contradicts the distance conjecture if a half-dimensional ring exists and constitutes its vertical and hori- zontal sets of projections.
We will obtain a contradiction from this, and it will be clear from the nature of the argument that one can in fact show that at least one ofA+A, A·Ahas measure δ12−c4 for some absolute constantc4>0 depending onc1.
From Separation 2.3one can find refinementsA1, A2 ofA which are contained in intervals of size and separation ≈1 and have measure|A1|,|A2| ≈ δ1/2. From the additive and multiplicative versions of (6) we thus have
|A1|,|A2|,|A1+A2|,|A1A2| ≈δ1/2. (8)
Heuristically, the idea is to apply the Bilinear Distance Conjecture1.5withE0, E1,E2equal toA1×A1,A1×A2,A2×A1respectively. The difficulty with this is that we cannot quite control the distance set
(A1−A1)2+ (A1−A2)2accurately using (8). However, this difficultycan be avoided if we pass to various refinements ofA.
We turn to the details. From (8) and Perfection3.4withA, B, G set toA1,A2, A1×A2 respectively, and some re-labeling, we can find δ-discretized refinements C,D ofA1,A2 respectivelysuch that
|{(a1, a2, a3, a4, a5, a6)∈A⊕6:d−c= (a1−a4)−(a2−a5) + (a3−a6)}| ≈δ5/2 (9)
for all (c, d)∈C×D. From construction we have
|c−d| ≈1 for allc∈C, d∈D.
(10)
Lemma 4.1. We have
|A·A·A·(C−D)/(A·A)|=
a1a2a3
a4a5 (c−d) :a1, a2, a3, a4, a5∈A, c∈C, d∈D ≈δ1/2. Proof. The lower bound is clear from (10) and the multiplicative version of (6), so it suffices to show the upper bound.
Fixa1, a2, a3, a4, a5,c,d. Bymultiplying (9) bya1a2a3/a4a5, which is≈1, we see that
|{(e1, e2, e3, e4, e5, e6)∈(A·A·A·A/(A·A))⊕6: a1a2a3
a4a5 (d−c) = (e1−e4)−(e2−e5) + (e3−e6)}|δ5/2. Integrating this over all possible values of a1aa42aa53(d−c) and using Fubini’s theorem we obtain
|A·A·A·A/(A·A)|6δ5/2|A·(C−D)|.
On the other hand, from (7) and the multiplicative form of Corollary3.3we have
|A·A·A·A/(A·A)| ≈δ1/2.
The claim follows bycombining the above two estimates.
From (8) and the multiplicative version of (6) we have
|C|,|D|,|CD| ≈δ1/2.
From the multiplicative form of Perfection3.4withA:=CandB := 1/D, we may thus find refinementsC,D ofC,D respectivelysuch that ifc∈C,d∈D, then (11) |X| ≈δ5/2, where
X ={(c1, c2, c3, d1, d2, d3)∈C×C×C×D×D×D:cd= (c1d1)(c2d2)−1(c3d3)}.
Lemma 4.2. We have
|CD−CD|=|{cd−cd :c, c ∈C, d, d∈D}| ≈δ1/2.
Proof. As before, the lower bound is immediate from the additive and multiplica- tive versions of (6), so it suffices to show the upper bound.
Fix c, c, d, d, and let (c1, c2, c3, d1, d2, d3)∈ X with X in (11). Then we have the telescoping identity
cd−cd=x1−x2+x3−x4
where
x1:= (c1−d)d1c3d3
c2d2 x2:= d(c−d1)c3d3 c2d2
x3:= dc(c3−d2)d3
c2d2 x4:= dcd2(c2−d3) c2d2 .
Indeed, we have the identities c1d1c3d3
c2d2 =cd dd1c3d3
c2d2 =cd−x1
dcc3d3
c2d2 =cd−x1+x2 dcd2d3
c2d2 =cd−x1+x2−x3
cd= dcd2c2
c2d2 =cd−x1+x2−x3+x4.
As a consequence of these identities, (10) and some algebra we see the map (c1, c2, c3, d1, d2, d3)→(x1, x2, x3, x4, c2, d2)
is a diffeomorphism onX (recall thatc,d,c,dare fixed). From (11) we thus have
|{(x1, x2, x3, x4, c2, d2)∈(A·A·A·(C−D)/(A·A))⊕4×C×D:
cd−cd =x1−x2+x3−x4}|δ5/2. Integrating this over all values ofcd−cd and using Fubini’s theorem we obtain
|CD−CD|δ5/2|A·A·A·(C−D)/(A·A)|4|C||D|.
The claim then follows from Lemma4.1.
From the above lemma and the multiplicative form of (6) we have
|C|,|D|,|CD| ≈δ1/2.
From the multiplicative version of Corollary3.3we can therefore find a refinement F ofCand a real numberx≈1 such thatxF is a refinement ofD. In particular, sinceF F−F F is a subset ofx−1(CD−CD), we thus see that
|F F−F F| ≈ |F F| ≈ |F| ≈δ1/2. From Corollary3.3we thus have
|F F−F F −F F +F F +F F −F F−F F +F F| ≈δ1/2. Since (F−F)2⊂F F −F F−F F +F F, we thus have
|(F−F)2+ (F−F)2|δ1/2.
The set F is a (δ,12)1 set with measure≈δ1/2. From Separation2.3 we mayfind refinementsF1, F2ofFwhich are contained in intervalsI1,I2of size and separation
≈1 such that|I1|=|I2|and|F1|,|F2| ≈δ1/2. Define
E0:=F1×F1, E1:=F1×F2, E2:=F2×F1, Q0:=I1×I1, Q1:=I1×I2, Q2:=I2×I1.
It is clear thatQ0, Q1, Q2obey(3) and thatE0, E1, E2 are (δ,1)2 sets of measure
≈δcontained inQ0,Q1,Q2 respectively.
LetDdenote the set
D=
(F2−F1)2+ (F1−F1)2.
Clearly D is aδ-discretized set of measure |D| δ1/2 which lives in A. In fact, from the size and separation ofF1 andF2we have
|D| ≈δ1/2. (12)
Also, we have
|x1−x0|,|x2−x0| ∈D for allx0∈E0,x1∈E1,x2∈E2. In particular, we have
|{(x0, x1, x2)∈E0×E1×E2:|x0−x1|,|x0−x2| ∈D}|=|E0||E1||E2| ≈δ3. (13)
We are almost readyto applythe hypothesis (4), however the one thing which is missing is thatDneed not satisfy(1). To rectifythis we shall remove some portions fromD.
ApplyRefinement2.2 to obtain a covering D⊂D∗∪
δ<δ1
Dδ
with the properties asserted in Refinement2.2, andKequal to a large constant to be chosen shortly.
Proposition 4.3. For allδ> δ, we have
|{(x0, x1)∈E0×E1:|x0−x1| ∈Dδ}|δ2δKε/100 and
|{(x0, x2)∈E0×E2:|x0−x2| ∈Dδ}|δ2δKε/100.
Proof. Fixδ. We mayassume thatεis sufficientlysmall depending onK, andδ is sufficientlysmall depending onKandε, since the claim is trivial otherwise.
Byreflection symmetryit suffices to prove the first estimate. Suppose for con- tradiction that
|{(x0, x1)∈E0×E1:|x0−x1| ∈Dδ}|δ2δKε/100. From Cauchy-Scwartz2.1we thus have
|{(x0, x1, x1)∈E0×E1×E1:|x0−x1| ∈Dδ,|x0−x1| ∈Dδ}|δ3δKε/50. Writex1= (x1,y1),x1= (x1,y1). Observe that
|{(x0, x1, x1)∈E0×E1×E1:|x1−x1|δKε/10}|δ3δKε/20.
This is because for fixed x1, x1 can onlyrange in a set of measure δ1/2δKε/20 thanks to (1) and the fact thatF1 is a (δ,12)1set. Subtracting the two inequalities we obtain (ifδis sufficientlysmall)
|{(x0, x1, x1)∈E0×E1×E1:
|x0−x1| ∈Dδ,|x0−x1| ∈Dδ,|x1−x1|δKε/10}|δ3δKε/50. Since|E1| ≈δ, we maythus findx1, x1∈E1 such that
|x1−x1|δKε/10 (14)
and
|{x0∈E0:|x0−x1| ∈Dk,|x0−x1| ∈Dδ}|δδKε/50. (15)
From Refinement2.2Dδ can be covered byδKεδ−1/2intervals inAof length δ. From this fact, (14), and the geometryof annuli which intersect non-tangentially, we see that the set in (15) can be covered byδ−1δ2Kε balls of radiusδ−Kε/5δ. SinceE0 is a (δ,1)2 set, we see from (1) that
LHS of (15)δ−1δ2Kεδδ−Kε/5.
But this contradicts (15) ifδis sufficientlysmall. This concludes the proof of the
proposition.
From (13) and the above proposition we see that (ifKis a large enough absolute constant, andδ is sufficientlysmall depending onε,K)
|{(x0, x1, x2)∈E0×E1×E2:|x0−x1|,|x0−x2| ∈D∗}|δ3. (16)
From (12) we have |D∗|δ1/2. From elementarygeometryand a change of vari- ables we have
|{x0∈E0:|x0−x1|,|x0−x2| ∈D∗}||D∗|2
for all x1 ∈ E1, x2 ∈ E2. Integrating this over x1 and x2 and comparing with the previous we thus see that|D∗| ≈δ1/2. But then (16) contradicts (4) (withD replaced by D∗), ifε is sufficientlysmall depending onc1 and δsufficientlysmall depending onε. The full claim of the proposition follows bya modification of this argument, providing thatc4 is sufficientlysmall depending onc1.
5. Ring Conjecture 1.14 implies Discretized Furstenburg Conjecture 1.10
Assume that the Ring Conjecture1.14is true for some absolute constantc4>0.
In this section we show how the Discretized Furstenburg Conjecture1.10follows.
The main idea is that R is a half-dimensional ring then R×R contains a one dimensional set of lines each of which contain half dimensional sets. That manyof these lines are parallel seems hardlyconsequential and we will deal with it byan appropriatelychosen projective transformation.
Let 0< ε1 be fixed. We mayassume that δ is sufficientlysmall depending onε, since (5) is trivial otherwise, and mayassumeδis dyadic as before. LetE, Ω, Rωbe as in the Discretized Furstenburg Conjecture1.10. Assume for contradiction that
|{(x0, x1)∈E×E:x1, x0∈Rωfor some ω∈Ω}|δ2 (17)
We will obtain a contradiction from this, and it will be clear from the nature of the argument that (5) in fact holds for some absolute constantc3>0 depending onc4. It will be convenient to define the non-transitive relation ∼ bydefining x∼y if and onlyif x, y∈ Rω for some ω ∈ Ω. We also writex1, . . . , xn ∼y1, . . . ym if xi∼yj for all 1≤i≤nand all 1≤j≤m.
From (17) we then have
|{(x0, x1)∈E×E:x0∼x1}|δ2. (18)
Roughlyspeaking, the idea will be to findx1, x1∈E and a refinementE ofE such thatx0∼x1, x0∼x1for allx0∈E, and such that there are manyrelations between pairs of points in E. Then after a projective transformation sending x1, x1to the cardinal points at infinitywe can transformEto a Cartesian product of
two (δ,12)1 sets of measure≈δ1/2, at which point the ring structure of these sets can be easilyextracted.
We turn to the details. From (18) and the fact that |E| ≈δ, we see that
|{(x0, x1)∈E×E:x0∼x1}|δ2 (19)
where
E ={x0:|{x1∈E:x0∼x1}| ≈δ}
provided the constants are chosen appropriately.
LetC2 be a large constant to be chosen later, and letE1 be the set E1=
x1∈E:
ω∈Ω
χRω(x1)≤δ−C2εδ−1/2
. From Kakeya2.4and Chebyshev we have
|E\E1|δ2C2εδ and thus
|{(x0, x1)∈E×(E\E1) :x0∼x1}|δ2C2εδ2.
If we then chooseC2 large enough, and δis small enough depending onC2 and ε, we thus see from (19) that
|{(x0, x1)∈E×E1:x0∼x1}|δ2. (20)
In particular, we have |E1| ≈ δ as before. Henceforth C2 is fixed so that (20) applies.
From (20) and Cauchy-Schwarz2.1we have
|{(x0, x1, x1)∈E×E1×E1:x0∼x1, x1}|δ3. (21)
LetC3 be a large constant to be chosen later.
Lemma 5.1. IfC3 is large enough, andδis small enough depending onC3 andε, we have
(x0, x1, x1)∈E×E1×E1:x0∼x1, x1;|(x1−x0)∧(x1−x0)| ≥δC3ε≈δ3. Proof. From (20) it suffices to show that
(22) (x0, x1, x1)∈E×E1×E1:
x0∼x1, x1;|(x1−x0)∧(x1−x0)| ≤δC3εδC3ε/8δ3. (The constant 8 is non-optimal, but this is irrelevant for our purposes). In order to have |(x1−x0)∧(x1−x0)| ≤δC3ε
one must either have|x1−x1|δC3ε/2, or that|x1−x1|δC3ε/2andx0is within δC3ε/2 of the line joiningx1−x1.
Let us consider the contribution of the former case. SinceE1is a (δ,1)2 set, we see that each pair (x0, x1) contributes a set of measure δC3ε/2δ to (22). From Fubini’s Theorem we thus see that the contribution of this case to (22) is acceptable.
Now let us consider the contribution of the latter case. ByFubini’s Theorem again it suffices to show that
|{x0∈E:x0∈S}|δC3ε/8δ for anystripS of widthδC3ε/2.
Figure 4. A Furstenburg set when viewed fromx1 andx1. Note how this resembles a projective transformation of Figure3.
FixS. From the definition ofE and Fubini’s Theorem it suffices to show that
|{(x0, x2)∈E×E:x0∈S, x2∼x0}|δC3ε/8δ2. From the definition of∼, we can estimate the left-hand side by
ω∈Ω
|S∩Rω||Rω|.
