# New estimates on the size of $(\alpha,2\alpha)$-Furstenberg sets

Daniel Di Benedetto Joshua Zahl

###### Abstract

We use recent advances on the discretized sum-product problem to obtain new
bounds on the Hausdorff dimension of planar $(\alpha,2\alpha)$-Fursterberg
sets. This provides a quantitative improvement to the $2\alpha+\epsilon$ bound
of Héra-Shmerkin-Yavicoli. In particular, we show that every $1/2$-Furstenberg
set has dimension at least $1+1/4536$.

##  1 Introduction

In [24], Wolff defined a class of Besicovitch-type sets, inspired by the work
of Furstenberg [7], which he called Furstenberg sets.

######  Definition 1.1.

For $0<\alpha\leq 1$, we say that a compact set $E\subset\mathbb{R}^{2}$ is a
(planar) _$\alpha$ -Furstenberg set_ if for every direction $\omega\in S^{1}$,
there is a line $\ell\subset\mathbb{R}^{2}$ with direction $\omega$ and
$\dim_{H}(E\cap\ell)\geq\alpha$.

Here and throughout, $\dim_{H}$ denotes Hausdorff dimension. In particular,
every Besicovitch set in the plane is also a $1$-Furstenberg set. Wolff
considered the problem of estimating

$\gamma(\alpha)=\inf\\{\dim_{H}(E):E\text{ is an }\alpha\text{-Furstenberg
set}\\},$

and showed that

$\max\Big{\\{}2\alpha,\alpha+\frac{1}{2}\Big{\\}}\leq\gamma(\alpha)\leq\frac{3\alpha}{2}+\frac{1}{2}.$
(1)

Wolff conjectured that the upper bound is sharp. When $\alpha=1/2$, the two
lower bound estimates coincide, and we have $1\leq\gamma(1/2)\leq 5/4$.

In the influential paper [14], Katz and Tao connected the problem of
estimating $\gamma(1/2)$ to the Falconer distance conjecture, and also to a
discretized variant of the Erdős-Szemerédi sum-product conjecture, which they
called the discretized ring conjecture. This latter conjecture involves
objects called $(\delta,\alpha)_{n}$-sets, which are discretized analogues of
Borel subsets of $\mathbb{R}^{n}$ that have Hausdorff dimension $\alpha$. To
define these sets precisely, we introduce the parameters
$n,\alpha,\varepsilon,C,\delta$, which are chosen in that order.

######  Definition 1.2.

We say a set $E\subset\mathbb{R}^{n}$ is a _$(\delta,\alpha)_{n}$ -set_ if $E$
is a union of balls of radius $\delta$, and for each ball $B$ of radius $r$,
we have $|E\cap B|\leq C\delta^{n-\varepsilon}(r/\delta)^{\alpha}$.

The discretized ring conjecture says that a $(\delta,1/2)_{1}$-set must expand
substantially under addition and/or multiplication.

######  Conjecture 1.3 (Katz-Tao).

There exists an absolute constant $c_{1}$, so that if $\varepsilon>0$ is
chosen sufficiently small and $C$ sufficiently large, the following holds for
all sufficiently small $\delta>0$. Let $E\subset[1,2]$ be a
$(\delta,1/2)_{1}$-set, with $|E|\geq\delta^{1/2+\varepsilon}$. Then

$|E+E|+|E.E|\geq\delta^{-c_{1}}|E|.$

Katz and Tao proved that a positive answer to Conjecture 1.3 would imply that
$\gamma(1/2)\geq 1+c_{2}$, for some (small) $c_{2}>0.$ Shortly thereafter,
Bourgain [1] established Conjecture 1.3.

######  Theorem 1.4 (Bourgain).

Conjecture 1.3 holds for some $c_{1}>0$.

As a consequence of [14] and [1], we have that $\gamma(1/2)\geq 1+c_{2}$.
While the precise values of $c_{1}$ and $c_{2}$ were not computed explicitly
in [1, 14], they are extremely small. Moreover, even if Conjecture 1.3 was
proved for the conjectured optimal value $c_{1}=1/2$, the reduction in [14]
would still yield a bound for $\gamma(1/2)$ that is much smaller than the
conjectured value $\gamma(1/2)=5/4$.

In [8], Guth, Katz, and the second author found a simple new proof of Theorem
1.4 that also gives explicit quantitative estimates on the size of $c_{1}$.

######  Theorem 1.5 (Guth-Katz-Zahl).

Conjecture 1.3 holds for all $c_{1}<1/136$.111One of the arguments in [8] is
inefficient, and Victor Lie observed that with a small modification the
exponent can be improved to $1/128$.

Armed with Theorem 1.5 (or more precisely, a new, more technical variant), we
prove new quantitative estimates on $\gamma(\alpha)$. First, we will introduce
a class of Furstenberg-type sets, which have generated recent interest due to
their connections with projection theory. In the definition below, we identify
a set $L$ of lines in $\mathbb{R}^{2}$ with the corresponding points in the
affine Grassmannian $A(2,1)$. Since the latter is a manifold, this allows us
to define $\dim_{H}(L)$.

######  Definition 1.6 ($(\alpha,\beta)$-Furstenberg set).

For $0<\alpha\leq 1,\ 0<\beta\leq 2$, we say that a set
$E\subset\mathbb{R}^{2}$ is a (planar) _$(\alpha,\beta)$ -Furstenberg set_ if
there exists a set $L$ of lines, with $\dim_{H}(L)\geq\beta$, so that
$\dim_{H}(E\cap l)\geq\alpha$ for each $l\in L$.

We define

$\gamma(\alpha,\beta)=\inf\\{\dim_{H}(E):E\text{ is an
}(\alpha,\beta)\text{-Furstenberg set}\\}.$

Numerous authors [5, 10, 11, 12, 16, 17, 18] have obtained lower bounds for
$\gamma(\alpha,\beta)$ in different regimes. The most relevant for our
discussion is the following result of Héra, Shmerkin, and Yavicoli, which used
a variant of Theorem 1.4 due to Bourgain [2] to obtain bounds on
$\gamma(\alpha,2\alpha)$.

######  Theorem 1.7 (Héra, Shmerkin, and Yavicoli).

For each $0<\alpha<1$, there exists $c(\alpha)>0$, which depends continuously
on $\alpha$, so that $\gamma(\alpha,2\alpha)\geq 2\alpha+c(\alpha)$.

Note that when $\alpha=1/2$, this recovers the previous bound $\gamma(1/2)\geq
1+c_{2}$ (possibly for a different value of $c_{2}$) discussed above.

While the authors in [12] do not explicitly compute $c(\alpha)$, the value
obtained by their arguments would be extremely small. Using Theorem 1.5 (or
more precisely, a technical variant of the theorem), we obtain explicit
estimates on the size of $c(\alpha)$. This estimate is stronger than what
would be obtained by merely computing the value of $c(\alpha)$ from the
arguments in [12].

######  Theorem 1.8.

$\gamma(\alpha,2\alpha)\geq 2\alpha+\frac{\alpha(1-\alpha)}{6(155+68\alpha)}$.
In particular, $\gamma(1/2)\geq\gamma(1/2,1)\geq 1+1/4536$.

Our bound is novel provided $\alpha$ is not much larger than $1/2$. For larger
$\alpha$, the bound $\gamma(\alpha,2\alpha)\geq 2\alpha+(1-\alpha)(2\alpha-1)$
recently obtained by Dąbrowski, Orponen, and Villa [5] is superior.

###  1.1 Thanks

The authors would like to thank Alexia Yavicoli and Victor Lie for helpful
comments and suggestions on a previous version of this manuscript. The first
author was supported in part by a Four Year Doctoral Fellowship from the
University of British Columbia. The second author was supported by a NSERC
Discovery Grant.

##  2 Discretized $(\alpha,\beta)$-Furstenberg sets

To prove Theorem 1.7, Héra, Shmerkin, and Yavicoli introduced an object they
called a discretized $(\alpha,\beta)$-Furstenberg set. Before defining this
object, we will introduce some additional notation. If
$X\subset\mathbb{R}^{n}$, we define the $\delta$-covering number
$\mathcal{E}_{\delta}(X)$ to be the minimum number of balls of radius $\delta$
required to cover $X$. If $X$ is itself a union of balls of radius $\delta$,
then $\mathcal{E}_{\delta}(X)\sim\delta^{-n}|X|$, where the implicit constant
depends on $n$. Abusing notation, we say a set $X\subset\mathbb{R}^{n}$ (which
need not be a union of balls of radius $\delta$) is a
$(\delta,\alpha)_{n}$-set if for each ball $B$ of radius $r\geq\delta$, we
have $\mathcal{E}_{\delta}(E\cap B)\leq
C\delta^{-\varepsilon}(r/\delta)^{\alpha}$. In particular, if $X$ is a union
of balls of radius $\delta$, then it is a $(\delta,\alpha)_{n}$-set in the
sense given here if and only if it is a $(\delta,\alpha)_{n}$-set using the
previous definition, though the associated constant $C$ will differ in these
two definitions by a multiplicative factor depending on $n$. Since
$n\in\\{1,2\\}$ throughout this paper, this distinction will not be important.

Next, if $L$ is a set of lines in $\mathbb{R}^{2}$, we can identify $L$ with a
subset of the affine Grassmannian $A(2,1)$. For concreteness, we will
parameterize a coordinate patch of $A(2,1)$ as follows: if $l$ is the line
$y=mx+b$ with $(m,b)\in[0,1]^{2},$ we define $\iota(l)=(m,b)$ and
$\operatorname{dir}(l)=(1,m)/\sqrt{1+m^{2}}$. In the discussions that follow,
we can assume that every line is of the above form. We define the distance
between two lines $l,l^{\prime}$ to be the Euclidean distance between the
points $\iota(l)$ and $\iota(l^{\prime}),$ and we define
$\angle(l,l^{\prime})=\angle(\operatorname{dir}(l),\operatorname{dir}(l^{\prime}))$.

We say a set $L$ of lines is $\delta$-separated if
$\operatorname{dist}(\iota(l),\iota(l^{\prime}))\geq\delta$ for all distinct
$l,l^{\prime}\in L$. A set $L$ of lines is said to be a
$(\delta,\beta)_{2}$-set if $\iota(L)\subset[0,1]^{2}$ is a
$(\delta,\beta)_{2}$-set. With these definitions, we recall the following
definition from [12].

######  Definition 2.1.

For $0<\alpha\leq 1,$ $0<\beta\leq 2$, $\varepsilon>0$, $C>0$, and $\delta>0$,
we say that a set $E\subset\mathbb{R}^{2}$ is a (planar) _discretized
$(\alpha,\beta)$-Furstenberg set_ if $E\supset\bigcup_{l\in L}P_{l}$, where:

  1. (i)

$L$ is a $\delta$-separated, $(\delta,\beta)_{2}$-set of lines (with implicit
parameters $\varepsilon$ and $C$), and $\\#L\geq
C^{-1}\delta^{\varepsilon-\beta}$.

  2. (ii)

For each $l\in L,$ $P_{l}\subset l$ is a $(\delta,\alpha)_{1}$-set (with
implicit parameters $\varepsilon$ and $C$), and $\\#\mathcal{P}_{l}\geq
C^{-1}\delta^{\varepsilon-\alpha}$.

The authors in [12] proved Theorem 1.7 by showing that every discretized
$(\alpha,2\alpha)$-Furstenberg set must have large $\delta$ covering number.
The relationship between these two statements is as follows.

######  Proposition 2.2 ([12], Lemma 3.3).

Let $0<\alpha<1,$ $0<\beta<2$, and $s\geq 0$. Suppose that there exists
$\varepsilon>0$ and $C>0$ so that for all sufficiently small $\delta>0$, every
discretized $(\alpha,\beta)$-Furstenberg set $E$ satisfies
$\mathcal{E}_{\delta}(E)\geq\delta^{-s}.$ Then every
$(\alpha,\beta)$-Furstenberg set has Hausdorff dimension at least $s$.

Thus, in order to prove Theorem 1.8, it suffices to establish the
corresponding statement for discretized Furstenberg sets. Our result in this
direction is the following.

######  Proposition 2.3.

Define $c(\alpha)=\frac{\alpha(1-\alpha)}{6(155+68\alpha)}$. Then for every
$0<\alpha<1$ and every $c<c(\alpha)$, there exists $\varepsilon>0$ and $C>0$
so that the following holds for all $\delta>0$ sufficiently small. Let $E$ be
a discretized $(\alpha,2\alpha)$-Furstenberg set (with implicit parameters
$\varepsilon$ and $C$). Then $\mathcal{E}_{\delta}(E)\geq\delta^{-2\alpha-c}.$

At a qualitative level, the connection between sum-product estimates, point-
line incidences, and Furstenberg sets is well known. In [4], Bourgain, Katz,
and Tao proved a sum-product estimate over $\mathbb{F}_{p}$, and used this to
obtain a point-line incidence bound in $\mathbb{F}_{p}^{2}$. The argument used
by Héra, Shmerkin, and Yavicoli [12] is also based on these ideas, but uses
Bourgain’s discretized projection theorem [2] instead of the discretized sum-
product theorem. Our main contribution is that our arguments are carefully
structured to obtain an effective quantitative relationship between the
discretized sum-product theorem and the size of $(\alpha,2\alpha)$-Furstenberg
sets. One major difficulty is that while both Theorem 1.5 and Proposition 2.3
involve sets that satisfy non-concentration conditions, not all paths between
these two non-concentration conditions are equally efficient. To overcome this
difficulty, we carefully construct a sequence of point-line configurations
that allow us to translate between the two notions of non-concentration. See
Figure 1 for an overview of this translation step.

##  3 Tools from combinatorics

We will begin by introducing several standard tools from additive
combinatorics. See [23] for an introduction to the topic. We will also state
and prove a more technical version of Theorem 1.5.

We begin with some definitions. If $A,B$ are finite subsets of an Abelian
group, and $E\subset A\times B$, we define the partial sum and partial
difference sets

$A\overset{E}{+}B=\\{a+b:(a,b)\in E\\},\quad A\overset{E}{-}B=\\{a-b:(a,b)\in
E\\}.$

If $E=A\times B$, then we will abbreviate this as $A+B$ (resp. $A-B$); this is
the usual Minkowski sum and Minkowski difference.

We recall the following version of the Balog-Szemerédi-Gowers theorem. The
version stated here follows by making a small modification to Lemma 2.2 from
[3]. The exact formulation given below can be found in [13, Lemma 8].

######  Lemma 3.1 (Balog-Szemerédi-Gowers).

Let $A,B$ be finite subsets of an Abelian group, and let $E\subset A\times B$.
Then there is a set $A^{\prime}\subset A$ such that

$\\#A^{\prime}\gtrsim\frac{\\#E}{\\#B},\quad\textrm{and}\quad\\#(A^{\prime}-A^{\prime})\lesssim\frac{(\\#A)^{4}(\\#B)^{3}(A\overset{E}{-}B)^{4}}{(\\#E)^{5}}.$

Next, we will introduce some notation to describe the additive and
multiplicative energy of discretized sets. Recall that if $A,B$ are subsets of
an Abelian group (with group operation $+$), we define the additive energy

$E(A,B)=\\#\\{(a,a^{\prime},b,b^{\prime})\in A^{2}\times B^{2}\colon
a+b=a^{\prime}+b^{\prime}\\}=\sum_{a\in A}\sum_{b\in
B}\\#\big{(}(a+B)\cap(b+A)\big{)}.$

The following result is a simple consequence of the Cauchy-Schwarz inequality;
a proof can be found in [23, Corollary 2.10].

######  Lemma 3.2.

Let $A,B$ be finite subsets of an Abelian group. Then

$E(A,B)\leq E(A,A)^{1/2}E(B,B)^{1/2}.$

If $A\subset\mathbb{R}$ and $\delta>0$, we define
$A_{\delta^{+}}=N_{\delta}(A)\cap(\delta\mathbb{Z})$ and
$A_{\delta^{\times}}=N_{2\delta}(A)\cap 2^{\delta\mathbb{Z}}$; note that with
this definition, for each $x\in A$ there is a point $y\in A_{\delta^{+}}$ with
$|x-y|\leq\delta$. Similarly, since $\frac{d}{dx}2^{x}<4$ for $x\in[1,2]$,
every point $x\in[1,2]$ has distance less than $2\delta$ from
$2^{\delta\mathbb{Z}}$; thus if $A\subset[1,2]$, then for each $x\in A$ there
is a point $y\in A_{\delta^{\times}}$ with $|x-y|\leq 2\delta.$ In practice,
we will always have $A,B\subset[1,2]$.

$A_{\delta^{+}}$ (resp. $A_{\delta^{\times}}$) is a subset of the Abelian
group $\delta\mathbb{Z}$ (resp. $2^{\delta\mathbb{Z}}$). We define

$E_{+}^{\delta}(A,B)=E(A_{\delta^{+}},B_{\delta^{+}}),\quad
E_{\times}^{\delta}(A,B)=E(A_{\delta^{\times}},B_{\delta^{\times}}).$

We conclude this section by stating the precise variant of Theorem 1.5 that we
will use to prove Proposition 4.1.

######  Proposition 3.3.

For each $0<\alpha<1$ and $\varepsilon>0$, the there exists $\delta_{0}>0$ so
that the following holds for all $0<\delta\leq\delta_{0}$. Let
$K_{1},K_{2},K_{3},K_{4}\geq 1$, and let $A\subset[1,2]$ be a
$\delta$-separated set satisfying the following properties

  1. 1.

$\\#(A)\geq K_{1}^{-1}\delta^{-\alpha}$.

  2. 2.

If $J$ is an interval of length at least $\delta$, then $\\#(A\cap J)\leq
K_{2}|J|^{\alpha}\delta^{-\alpha}$.

  3. 3.

$\mathcal{E}_{\rho}(A-A)\leq K_{3}\mathcal{E}_{\rho}(A)$ for all
$\delta\leq\rho\leq 1$.

  4. 4.

$E_{\times}^{\delta}(A,A)\geq K_{4}^{-1}(\mathcal{E}_{\delta}(A))^{3}$.

Then

$K_{1}^{6}K_{2}^{6+\alpha}K_{3}^{2(9+4\alpha)}K_{4}^{4(2+\alpha)}\geq\delta^{-\alpha(1-\alpha)+\varepsilon}.$
(2)

The proof of Proposition 3.3 is nearly identical to the proof of Theorem 1.1
from [8]. We will briefly remark on the differences between the two theorems.
First, in [8], the authors set $K_{3}=K_{4},$ and they did not track the
explicit dependence on $K_{1}$ and $K_{2}$. Note that when $\alpha=1/2$,
$K_{1}\sim K_{2}\sim 1$ and $K_{3}=K_{4}=K$, then (2) asserts that
$K\gtrsim\delta^{-1/128+\varepsilon}$; this recovers the result from [8], with
the improvement observed by Victor Lie.

Second, the statement of Theorem 1.1 from [8] replaces Item 3 with the weaker
requirement that $\mathcal{E}_{\delta}(A-A)\leq K_{3}\mathcal{E}_{\delta}(A),$
and replaces Item 4 with the stronger requirement that
$\mathcal{E}_{\delta}(A.A)\leq K_{4}\mathcal{E}_{\delta}(A)$. The authors in
[8] begin by finding a large subset $A^{\prime}\subset A$ that satisfies Item
3 (in the form stated above), and then use their upper bound on
$\mathcal{E}_{\delta}(A.A)$ (which also holds with $A^{\prime}$ in place of
$A$) to obtain Item 4, with $A^{\prime}$ in place of $A$. In our setting, we
are not allowed to replace $A$ with a subset $A^{\prime}\subset A$, since Item
4 might fail for $A^{\prime}$. This is why we impose the stronger condition in
Item 3 at the outset. we will sketch a proof of Proposition 3.3 in Appendix A.

###  3.1 Moran-regular trees

In this section, $k$ and $T$ will be large integers; later we will set
$k=\lfloor 1/\varepsilon\rfloor,$ and we will select $T$ to be an integer so
that $2^{-kT}\leq\delta<2^{-k(T+1)}.$ With this choice of $k$ and $T$, we have
$2^{k}\leq 2^{1/\varepsilon}\sim 1$ and $2^{T}\leq\delta^{-2\varepsilon}$.

######  Definition 3.4.

A $2^{T}$-adic interval is a subset of $[0,1)$ of the form
$[s2^{-jT},(s+1)2^{-jT})$ for some $j\geq 0$ and $0\leq s\leq 2^{T}$. A
$2^{T}$-adic square is a Cartesian product of $2^{T}$-adic intervals. In what
follows, we will simply use “square” to refer to a $2^{T}$-adic square.

Note that two distinct squares are either disjoint, or one contains the other.
If $S$ is a square of side-length $2^{-jT}$ that contains a square
$S^{\prime}$ of side-length $2^{-(j+1)T}$, then we say $S^{\prime}$ is a child
of $S$.

######  Definition 3.5.

If $A\subset[0,1)^{d}$ and if $S$ is a square, we say $S$ meets $A$ if $S\cap
A\neq\emptyset$. We say $A$ is discretized at scale $2^{-kT}$ if every square
of side-length $2^{-kT}$ intersects at most point point from $A$.

######  Definition 3.6.

If $A\subset[0,1)^{d}$ is discretized at scale $2^{-kT}$, and if
$N_{1},\ldots,N_{k}$ are integers, we say that $A$ is Moran-regular with
branching $(N_{0},\ldots,N_{k-1})$ if for each $j=0,\ldots,k-1$, every square
$S$ of side-length $2^{-jT}$ that meets $A$ has exactly $N_{j}$ children
$S^{\prime}\subset S$ that meet $A$.

The next lemma says that every discretized set has a large subset that is
Moran-regular. See [15, Lemma 3.4] or [19, Lemma 2.2] for a detailed proof.

######  Lemma 3.7.

Let $A\subset[0,1)^{d}$ be discretized at scale $2^{-kT}$. Then there exists a
set $A^{\prime}\subset A$ and integers $N_{0},\ldots,N_{k-1}$ so that
$\\#A^{\prime}\geq(4T)^{-k}(\\#A)$, and $A^{\prime}$ is Moran-regular with
branching $(N_{0},\ldots,N_{k-1})$. Furthermore, we can ensure that each of
the integers $N_{0},\ldots,N_{k-1}$ are of the form $2^{z}$ for some non-
negative integer $z$.

We can also refine a collection of discretized sets, so that they are all
Moran regular with the same branching.

######  Corollary 3.8.

Let $\mathcal{A}$ be a collection of subsets of $[0,1)^{d}$, each of which is
discretized at scale $2^{-kT}$. Then there exist integers
$N_{0},\ldots,N_{k-1}$; a set $\mathcal{A}^{\prime}\subset\mathcal{A}$; and
for each $A\in\mathcal{A}^{\prime}$, a set $A^{\prime}\subset A$ so that

$\sum_{A\in\mathcal{A}^{\prime}}\\#A^{\prime}\geq(4T)^{-2k}\sum_{A\in\mathcal{A}}\\#A,$

and for each $A\in\mathcal{A}^{\prime}$, $A^{\prime}$ is Moran-regular with
branching $(N_{0},\ldots,N_{k-1})$.

###### Proof.

Apply Lemma 3.7 to each $A\in\mathcal{A}$, and let $A^{\prime}\subset A$ and
$(N^{A}_{0},\ldots,N^{A}_{k-1})$ be the output of that lemma. Since each
$N^{A}_{j}$ is a power of two between $2^{0}$ and $2^{T}$, there are
$(T+1)^{k}$ possible values for $(N^{A}_{0},\ldots,N^{A}_{k-1})$. The result
now follows from pigeonholing. ∎

Finally, we observe that if $A\subset[0,1)^{d}$ is discretized at scale
$2^{-kT}$ and Moran-regular with branching $(N_{0},\ldots,N_{k-1})$, then for
each square $S$ of side-length $2^{-jT}$, either $S\cap A=\emptyset,$ or
$\\#(S\cap A)=\prod_{i=j}^{k-1}N_{i}.$

##  4 A discretized incidence theorem

In this section, we will prove a discretized incidence theorem for collections
of points and lines satisfying certain non-concentration assumptions. In
Section 5 we will show that the incidence theorem proved in this section
implies Proposition 2.3.

Throughout this section, we fix $0<\alpha<1$ and a small number
$\varepsilon>0$. We will let $\delta>0$ be a small parameter. If $A$ and $B$
are quantities that depend on $\delta$, we write $A\lessapprox B$ if there are
constants $C=C(\alpha)$ and $\delta_{0}=\delta_{0}(\alpha,\varepsilon)>0$ so
that $A\leq\delta^{-C\varepsilon}B$ for all $0<\delta\leq\delta_{0}$. If
$A\lessapprox B$ and $B\lessapprox A$, we write $A\approx B$.

Let $P\subset[0,1]^{2}$ be a set of points and let $L$ be a set of lines in
$\mathbb{R}^{2}$ (as before, all lines will be of the form $y=mx+b$, with
$(m,b)\in[0,1]^{2}$). We say that a point $p$ is _incident_ (or
$\delta$-incident, if we wish to emphasize the role of $\delta$) to a line $l$
if $p\in N_{\delta}(l).$ We write $I_{\delta}(P,L)$ to denote the set of pairs
$(p,l)\in P\times L$ of incident points. If $I\subset P\times L$, define
$I(p)=\\{l\in L\colon(p,l)\in I\\}$, and define $I(l)=\\{p\in P\colon(p,l)\in
I\\}.$

Our task is to bound the number of incidences between $P$ and $L$, under the
following non-concentration conditions. Let $V\geq 1,$ and let $I\subset
I_{\delta}(P,L)$ be a set of incidences with the following properties.

  1. (A)

For each line $l\in L$,

$\\#I(l)\geq\delta^{-\alpha+\varepsilon}V.$ (3)

  2. (B)

For each $l\in L$ and each ball $B$ of radius $r\geq\delta$,

$\\#(I(l)\cap B)\leq\delta^{-\varepsilon}(r/\delta)^{\alpha}V.$ (4)

  3. (C)

For each ball $B$ of radius $r\geq\delta$,

$\\#(B\cap\iota(L))\leq\delta^{-\varepsilon}(r/\delta)^{2\alpha}.$ (5)

  4. (D)

Each point $p\in P$ is incident to approximately an average number of lines.
More precisely,

$\delta^{\varepsilon}\frac{(\\#I)}{(\\#P)}\leq\\#I(p)\leq\delta^{-\varepsilon}\frac{(\\#I)}{(\\#P)}.$
(6)

  5. (E)

For each point $p\in P$, the lines incident to $p$ point in a set of
directions that satisfies a “robust transversality” condition. Specifically,
for each unit vector $v\in S^{1}$, we have

$\\#\\{l\in
I(p)\colon\angle(l,v)\leq\delta^{\varepsilon}\\}\leq\frac{1}{2}\\#I(p).$ (7)

  6. (F)

For each point $p\in P$, the lines incident to $p$ point in a set of
directions that forms a $(\delta,\alpha)_{1}$ set. Specifically, for each unit
vector $v\in S^{1}$ and each $r\geq\delta$, we have

$\\#\\{l\in I(p)\colon\angle(l,v)\leq
r\\}\leq\delta^{-\varepsilon}(r/\delta)^{\alpha}.$ (8)

######  Proposition 4.1.

Let $P\subset B(0,1)\subset\mathbb{R}^{2}$ be a set of points, let $L$ be a
set of lines, let $V\geq 1$, and let $I\subset I_{\delta}(P,L)$. Suppose that
$P,L,$ and $I$ satisfy Properties (A)-(F). Then

$\\#P\gtrapprox\delta^{-2\alpha-c(\alpha)}(\delta^{2\alpha}\\#L)V^{c(\alpha)/\alpha},\qquad
c(\alpha)=\frac{\alpha(1-\alpha)}{6(155+68\alpha)}.$ (9)

######  Remark 4.2.

Note that Proposition 4.1 as stated above formally implies the (superficially)
weaker statement, where the inequalities “$A\leq\delta^{\varepsilon}B$” in
(3), (4), (5), (6), and (8) are replaced by “$A\lessapprox B$.”

######  Remark 4.3.

The most interesting case is when $V=1$ and $\\#L\approx\delta^{-2\alpha}$.
However, in order to use Proposition 4.1 to obtain bounds on the size of
$(\alpha,2\alpha)$-Furstenberg sets, we will need to apply a two-ends
reduction to ensure that (7) holds. Because of this, we must also consider
other values of $V\geq 1$, and sets of lines with $\\#L<\delta^{-2\alpha}$.

The remainder of this section will be devoted to the proof of Proposition 4.1.
In the arguments that follow, we will make statements of the following type:
If $l\in L$, then

$\\#\\{(p,q)\in I(l)^{2}\colon\operatorname{dist}(p,q)\approx
1\\}\gtrapprox\delta^{-2\alpha}V^{2}.$ (10)

What this means is that if the constant $C_{1}$ is chosen appropriately
(possibly depending on $\alpha$, but not depending on $\varepsilon$ or
$\delta$), then there exists a constant $C_{2}$ (again, possibly depending on
$\alpha$, but not depending on $\varepsilon$ or $\delta$) so that for all
$\delta$ sufficiently small we have

$\\#\\{(p,q)\in I(l)^{2},\
\operatorname{dist}(p,q)\geq\delta^{C_{1}\varepsilon}\\}\geq\delta^{C_{2}\varepsilon}\delta^{-2\alpha}V^{2}.$
(11)

For this example, (11) (and hence (10)) is true with $C_{1}=3/\alpha$ and
$C_{2}=3$. Indeed, by (3), $\\#I(l)\geq\delta^{\varepsilon-\alpha}V.$ Once
$p\in I(l)$ has been chosen, there are $\geq\delta^{\varepsilon-\alpha}V$
elements $q\in I(l)$. By (4) with $r=\delta^{C_{1}\varepsilon}$ and with
$C_{1}$ as above, at most
$\delta^{-\varepsilon}(r/\delta)^{\alpha}V\leq\delta^{2\varepsilon-\alpha}V$
such $q$ satisfy $\operatorname{dist}(p,q)\leq\delta^{C_{1}\varepsilon}$. Thus
if $\delta>0$ is sufficiently small (in this case $\delta<2^{-1/\varepsilon}$
will suffice), then there are at least
$\frac{1}{2}\delta^{2\varepsilon-2\alpha}V^{2}\geq\delta^{C_{2}\varepsilon}\delta^{-2\alpha}V^{2}$
pairs $(p,q)\in I(l)^{2}$ with
$\operatorname{dist}(p,q)\geq\delta^{C_{1}\varepsilon}$. Arguments of this
type will be used frequently in our proof without further comment.

In the arguments that follow, we will construct a sequence of increasingly
complicated configurations of points and lines. The main steps of this
construction are illustrated in Figure 1. An important tool for constructing
these configurations is Hölder’s inequality. For example, suppose
$I^{\prime}\subset I$. Define $m(l)=\\#I^{\prime}(l)$, let $k\geq 2$, and let
$k^{\prime}$ be the conjugate exponent to $k$. Then by Hölder’s inequality we
have

$(\\#I^{\prime})^{k}=\Big{(}\sum_{l\in L}1\cdot
m(l)\Big{)}^{k}\leq(\\#L)^{k/k^{\prime}}\Big{(}\sum_{l\in
L}(m(l))^{k}\Big{)},$

and thus

$\\#\\{(p_{1},\ldots,p_{k},l)\in P^{k}\times L\colon(p_{i},l)\in I^{\prime}\
\textrm{for}\ i=1,\ldots,k\\}\geq(\\#I^{\prime})^{k}(\\#L)^{-k/k^{\prime}}.$

Arguments of this type will be used in our proof without further comment.

$p$ $q$ (a) Step 1

$p_{1}$ $q$ $p_{2}$ $p_{3}$ $p_{4}$ (b) Step 2

$p_{1}$ $q$ $p_{2}$ $p_{4}$ $p_{5}$ (c) Step 3

$p_{5}$ $q$ (d) Step 4

Figure 1: The main steps in translating point-line incidences to a sum-product
statement

###### Proof of Proposition 4.1.

#### Counting triples relative to a fixed line.

Let $l_{0}\in L$. We claim that

$\begin{split}\\#&\\{(p,l,q)\in I(l_{0})\times L\times P:(p,l),(q,l)\in
I;\operatorname{dist}(p,q)\gtrapprox
1;\angle(l_{0},l)\geq\delta^{\varepsilon};\\\ &\textrm{there are}\
\gtrapprox\delta^{-\alpha}V\ \textrm{points incident to}\ l\ \textrm{between}\
p\ \textrm{and}\ q\\}\\\
&\gtrapprox\Big{(}\delta^{-\alpha}V\Big{)}\Big{(}\delta^{-\alpha}(\\#L)(\\#P)^{-1}V\Big{)}\Big{(}\delta^{-\alpha}V\Big{)}\\\
&=\delta^{-3\alpha}(\\#L)(\\#P)^{-1}V^{2}.\end{split}$ (12)

The set in (12) is constructed as follows: first we select a point $p\in
I(l_{0})$; by (3) there are $\gtrapprox\delta^{-\alpha}V$ choices for $p$.
Then we select a line $l\in I(p)$ with
$\angle(l_{0},l)\geq\delta^{\varepsilon}$; by (6) and (7), there are
$\gtrapprox\frac{(\\#I)}{(\\#P)}\approx\delta^{-\alpha}(\\#L)(\\#P)^{-1}V$
choices for $l$. Finally we select a point $q\in I(l)$ with
$\operatorname{dist}(p,q)\gtrapprox 1$ and for which there are
$\gtrapprox\delta^{-\alpha}V\ \textrm{points incident to}\ l\
\textrm{between}\ p\ \textrm{and}\ q$. By (3) and (4), there are
$\gtrapprox\delta^{-\alpha}V$ choices for $q$.

For each $q\in P$, let $L(q)\subset L$ be the set of lines $l$ so that there
exists $p\in P$ such that $(p,l,q)\in\eqref{triplesplq}$. For each $l\in
L(q)$, there are $\lessapprox 1$ points $p\in P$ so that
$(p,l,q)\in\eqref{triplesplq}$ (indeed, such a point $p$ must be contained in
$N_{\delta}(l)\cap N_{\delta}(l_{0})$, and these two lines intersect at angle
$\approx 1$).

#### Applying a two-ends reduction.

Next we will apply a two-ends reduction; see [21] for an introduction to this
topic. Fix $q\in P$, and for each $r\in[\delta,\pi]$ and $v\in S^{1}$, define

$f(r,v)=r^{-\varepsilon}\\#\\{l\in L(q)\colon\angle(l,v)\leq r\\}.$

This function is upper semi-continuous (indeed, for every
$(r,v)\in[\delta,\pi]\times S^{1}$, there exists $\tau>0$ so that
$f(r^{\prime},v^{\prime})\leq f(r,v)$ for all $|r-r^{\prime}|<\tau,\
|v-v^{\prime}|<\tau$), and thus achieves its maximum on the compact set
$[\delta,\pi]\times S^{1}$. Let $(r_{q},v_{q})$ be a pair that achieves this
maximum.

Define

$L^{\prime}(q)=\\{l\in L(q)\colon\angle(l_{i},v_{q})<r\\}.$

Since $f(\pi,v)=\pi^{-\varepsilon}\\#L(q)$ for any $v\in S^{1}$, we have
$f(r_{q},v_{q})\geq f(\pi,v_{q})\geq\pi^{-\varepsilon}\\#L(q)$, and hence
$\\#L^{\prime}(q)\geq(r_{q}/\pi)^{\varepsilon}\\#L(q)\geq(\delta/\pi)^{\varepsilon}\\#L(q)$.

We claim that for each $\delta\leq r\leq r_{q}$, there at most
$(r/r_{q})^{\varepsilon}(\\#L^{\prime}(q))^{4}$ quadruples
$(l_{1},\ldots,l_{4})\in(L^{\prime}(q))^{4}$ so that $\angle(l_{3},l_{4})<r$.
Indeed; once $l_{1},l_{2},l_{3}$ have been selected, $l_{4}$ must satisfy
$\angle(l_{3},l_{4})\leq r$, and hence the number of such $l_{4}$ is
$r^{\varepsilon}f(r,\operatorname{dir}(l_{3}))$. Since
$f(r,\operatorname{dir}(l_{3}))\leq f(r_{q},v_{q})$, we have
$r^{\varepsilon}f(r,\operatorname{dir}(l_{3}))\leq
r^{\varepsilon}f(r_{q},v_{q})=(r/r_{q})^{\varepsilon}\\#L^{\prime}(q)$.

An identical argument for each pair of distinct indices $i\neq j$ shows there
there are at most $\binom{4}{2}(r/r_{q})^{\varepsilon}(\\#L^{\prime}(q))^{4}$
quadruples $(l_{1},\ldots,l_{4})\in(L^{\prime}(q))^{4}$ so that
$\angle(l_{i},l_{j})<r$ for some $i\neq j$. Selecting $r\sim r_{q}$
appropriately, we have

$\\#\\{(l_{1},\ldots,l_{4})\in(L^{\prime}(q))^{4}\colon\angle(l_{i},l_{j})\gtrsim
r_{q}\ \textrm{for each}\ i\neq
j\\}\gtrsim(\\#L^{\prime}(q))^{4}\approx(\\#L(q))^{4}.$ (13)

By dyadic pigeonholing, we can refine the set on the LHS of (12) by a factor
of $1/|\log\delta|$, and find a number $d$ such that $r_{q}\sim d$ for every
point $q\in P$ that contributes to at least one triple from (12). Note that
Inequality (12) remains true for this (slightly) smaller set of triples.

Finally, recall that at the beginning of this section we fixed a choice of
$l_{0}\in L$. We write $d=d_{l_{0}}$ to emphasize that the value of $d$
determined above depends on $l_{0}$.

#### Applying Hölder’s inequality.

Define

$\begin{split}\mathcal{N}_{l_{0}}=&\\{(p_{1},\ldots,p_{4},l_{1},\ldots,l_{4},q)\in
I(l_{0})^{4}\times L^{4}\times P:\\\ &(p_{i},l_{i}),(q,l_{i})\in I;\
\operatorname{dist}(p_{i},q)\approx 1;\
\angle(l_{0},l_{i})\geq\delta^{\varepsilon},\ i=1,\ldots,4;\\\
&\qquad\operatorname{dist}(p_{1},p_{2})\sim\operatorname{dist}(p_{2},p_{3})\sim\operatorname{dist}(p_{3},p_{4})\sim
d_{l_{0}}\\},\end{split}$

where in the above set we chose the labeling so that $p_{1},\ldots,p_{4}$
appear in that order along $l_{0}$. Applying Hölder’s inequality to the
refinement of (12) described above, we have

$\mathcal{N}_{l_{0}}\gtrapprox\delta^{-12\alpha}(\\#L)^{4}(\\#P)^{-7}V^{8}.$
(14)

Note that if
$(p_{1},\ldots,p_{4},l_{1},\ldots,l_{4},q)\in\mathcal{N}_{l_{0}}$, then
$\operatorname{dist}(l_{0},q)\approx 1$. By (12), we also have that there are
$\gtrapprox\delta^{-\alpha}V$ points $p_{5}\in P$ with $(p_{5},l_{3})\in I$ so
that $p_{5}$ is between $p_{3}$ and $q$ on the line $l_{3}$. Thus by (4) and
the fact that $\angle(l_{0},l_{3})\approx 1$, we have that there are
$\gtrapprox\delta^{-\alpha}V$ points $p_{5}\in P$ so that (i)
$(p_{5},l_{3})\in I$, (ii) $\operatorname{dist}(l_{0},p_{5})\approx 1$, (iii)
$\operatorname{dist}(p_{5},q)\approx 1$, and (iv) $p_{5}$ is contained in the
line segment between $p_{3}$ and $q$ (i.e. $p_{5}$ is contained inside the
triangle $\Delta_{p_{1},p_{4},q}$ spanned by $p_{1},p_{4},q$). Items (i)–(iv)
also imply that

$\begin{split}\operatorname{dist}(q,\overline{p_{1}p_{5}})\gtrapprox
d_{l_{0}},\quad\textrm{and}\quad\operatorname{dist}(q,\overline{p_{4}p_{5}})\gtrapprox
d_{l_{0}},\end{split}$ (15)

where $\overline{p_{1}p_{5}}$ denotes the line spanned by $p_{1}$ and $p_{5}$,
and similarly for $\overline{p_{4}p_{5}}$. Thus,

$\begin{split}&\\#\\{(p_{1},\ldots,p_{4},l_{1},\ldots,l_{4},q,p_{5})\in\mathcal{N}_{l_{0}}\times
P:(p_{5},l_{3})\in I;\ \operatorname{dist}(l_{0},p_{5})\approx 1;\\\
&\qquad\operatorname{dist}(p_{5},q)\approx 1;\
p_{5}\in\Delta_{p_{1},p_{4},q};\
\operatorname{dist}(p,\overline{p_{1}p_{5}})\gtrapprox d_{l_{0}};\
\operatorname{dist}(p,\overline{p_{4}p_{5}})\gtrapprox d_{l_{0}}\\}\\\
&\gtrapprox\delta^{-13\alpha}(\\#L)^{4}(\\#P)^{-7}V^{9}.\end{split}$ (16)

#### Summing over the lines.

Recall that the set described in (16) depends on the choice of line $l_{0}$.
Let $\mathcal{D}$ be the disjoint union of such sets over all lines $l_{0}\in
L$, i.e.

$\begin{split}\mathcal{D}=\\{&(l_{0},p_{1},\ldots,p_{4},l_{1},\ldots,l_{4},q,p_{5})\colon(p_{1},\ldots,p_{4},l_{1},\ldots,l_{4},q,p_{5})\in\mathcal{N}_{l_{0}}\times
P;\ (p_{5},l_{3})\in I;\\\ &\operatorname{dist}(l_{0},p_{5})\approx 1;\
\operatorname{dist}(p_{5},q)\approx 1;\ p_{5}\in\Delta_{p_{1},p_{4},q};\
\operatorname{dist}(p,\overline{p_{1}p_{5}})\gtrapprox d_{l_{0}};\
\operatorname{dist}(p,\overline{p_{4}p_{5}})\gtrapprox
d_{l_{0}}\\}.\end{split}$ (17)

By (16), we have

$\\#\mathcal{D}\gtrapprox\delta^{-13\alpha}(\\#L)^{5}(\\#P)^{-7}V^{9}.$

After dyadic pigeonholing, we can suppose there is a number $d$ so that so
that $d_{l_{0}}\sim d$ for a $|\log\delta|^{-1}$ fraction of the tuples in
$\mathcal{D}$. Abusing notation, we will continue to refer to this refined set
as $\mathcal{D}$.

#### Fixing four special points.

Next, we will use pigeonholing to select a choice of
$(l_{0},p_{1},p_{2},p_{4},p_{5})$ that occur in many tuples from
$\mathcal{D}$. There are $\\#L$ choices for $l_{0}$. Once $l_{0}$ has been
specified, there are at most $\delta^{-\alpha-\varepsilon}V$ choices for
$p_{1}$. Next, there are $\lessapprox d^{\alpha}\delta^{-\alpha-\varepsilon}V$
choices for each of $p_{2}$ and $p_{4}$. Finally, there are $\leq\\#P$ choices
for $p_{5}$. Thus there is a choice of $(l_{0},p_{1},p_{2},p_{4},p_{5})$ that
appears in

$\gtrapprox\frac{\delta^{-13\alpha}(\\#L)^{5}(\\#P)^{-7}V^{9}}{(\\#L)(\delta^{-\alpha}V)(d^{\alpha}\delta^{-\alpha}V)^{2}(\\#P)}=d^{-2\alpha}\delta^{-10\alpha}(\\#L)^{4}(\\#P)^{-8}V^{6}$

tuples from $\\#\mathcal{D}$. Fix this choice of
$l_{0},p_{1},p_{2},p_{4},p_{5}$. Note that the point $p_{3}$ and line $l_{3}$
are determined up to multiplicity $\approx 1$.

Write $p\sim q$ if there is a line $l\in L$ with $(p,l),(q,l)\in I$. Since
$\operatorname{dist}(p_{i},q)\approx 1,\ i=1,2,4,5$, if $p_{i}\sim q$ then
there are $\approx 1$ lines $l_{i}\in L$ incident to both $p_{i}$ and $q$.
Thus if we define

$\begin{split}Q=\\{q\in P\colon p_{1}\sim q;\ p_{2}\sim&q;\ p_{4}\sim q;\
p_{5}\sim q;\ \operatorname{dist}(l_{0},q)\approx 1;\
\operatorname{dist}(p_{5},q)\approx 1;\\\ &p_{5}\in\Delta_{p_{1},p_{4},q};\
\operatorname{dist}(q,\overline{p_{1}p_{5}})\gtrapprox d;\
\operatorname{dist}(q,\overline{p_{4}p_{5}})\gtrapprox d\\},\end{split}$

then

$\\#Q\gtrapprox d^{-2\alpha}\delta^{-10\alpha}(\\#L)^{4}(\\#P)^{-8}V^{6}.$
(18)

#### Rescaling.

If $R$ is a rectangle, we define $\operatorname{dir}(R)\in S^{1}$ to be the
direction of the major axis of $R$. In practice, we will always consider
rectangles with eccentricity $>1$, so this is well defined. The quantity
$\operatorname{dir}(R)$ is meaningful up to uncertainty $\pm e^{-1}$, where
$e$ is the eccentricity of $R$.

Let $\\{R\\}$ be a maximal set of rectangles of dimensions $4\times 100d$,
which are centered at the midpoint of the line segment $p_{1}$—$p_{4}$, and
point in $d$-separated directions. Note that the union of these rectangles
covers $[0,1]^{2}$ (and thus covers $P$), and these rectangles are boundedly
overlapping in the following sense: if $x\in[0,1]^{2}$ has distance $\approx
1$ from the midpoint of $p_{1}$—$p_{4}$, then $\lessapprox 1$ rectangles from
this collection contain $x$. In particular, each point $q\in Q$ is contained
in $\approx 1$ of these rectangles.

Next, observe that if $q\in Q$ is contained in a rectangle $R$, and if $l\in
L$ with $(p_{1},l),(q,l)\in I$, then since
$\operatorname{dist}(p_{1},q)\approx 1$ and since $p_{1}$ and $q$ are both
contained in $R$, there is a constant $C_{0}=C_{0}(\alpha)$ (which depends on
the implicit constants in the above quasi-inequalities) so that if $\delta>0$
is selected sufficiently small, then

$\angle(l,\operatorname{dir}(R))\leq\delta^{-C_{0}\varepsilon}d.$ (19)

The same holds true with $p_{2}$ and $p_{4}$ in place of $p_{1}$. Crucially,
the same also holds true with $p_{5}$ in place of $p_{1}$—since
$p_{5}\in\Delta_{p_{1},p_{4},q}$, we must have $p_{5}\in R$.

By dyadic pigeonholing, we can select a set $\mathcal{R}$ of rectangles from
the collection described above, and an integer $N$, so that the following
holds:

  * •

$\\#\Big{(}Q\cap\bigcup_{R\in\mathcal{R}}R\Big{)}\gtrapprox\\#Q$

  * •

$N\leq\\#(R\cap Q)\leq 2N$ for each $R\in\mathcal{R}$.

  * •

The quantity $\\#\\{l\in
I(p_{1})\colon\angle(l,\operatorname{dir}(R))\leq\delta^{-C_{0}\varepsilon}d\\}$
is approximately the same for each $R\in\mathcal{R}$.

  * •

The above item holds for each of $p_{2},p_{4}$ and $p_{5}$ in place of
$p_{1}$.

Define $M=\\#\mathcal{R}$. By (18), for each $R\in\mathcal{R}$ we have

$\\#(Q\cap R)\gtrapprox
d^{-2\alpha}\delta^{-10\alpha}(\\#L)^{4}(\\#P)^{-8}V^{6}M^{-1},$ (20)

while by the third item above and (3), (6), for each $R\in\mathcal{R}$ we have

$\\#\\{l\in
I(p_{1})\colon\angle(l,\operatorname{dir}(R))\leq\delta^{-C_{0}\varepsilon}d\\}\lessapprox\\#I(p_{1})M^{-1}\lessapprox\delta^{-\alpha}(\\#L)(\\#P)^{-1}VM^{-1},$
(21)

and similarly for each of $p_{2},p_{4}$ and $p_{5}$ in place of $p_{1}$.

For the remainder of the argument, we will fix a rectangle $R\in\mathcal{R}$.
Define $Q_{1}=Q\cap R$. Define

$L(p_{1})=\\{l\in
I(p_{1})\colon\angle(l,\operatorname{dir}(R))\leq\delta^{-C_{0}\varepsilon}d\\}.$

Recall that $l\in I(p_{1})$ implies that $p_{1}\in N_{\delta}(l)$. Abusing
notation (but hopefully not the sensibilities of our readers), we shall
slightly perturb each line $l\in L(p_{1})$ so that $p_{1}\in l$. If $q\in Q$
was a point with $q\in N_{\delta}(l)$ prior to this perturbation, then $q\in
N_{2\delta}(l)$ after the perturbation; this is harmless for our arguments.
Define $L(p_{2}),L(p_{4})$, and $L(p_{5})$ similarly. The cardinality of these
sets is controlled by (21).

Observe that for each $q\in Q_{1}$, there are $\approx 1$ lines $l_{1}\in
L(p_{1})$ (resp. $l_{2}\in L(p_{2})$ and $l_{4}\in L(p_{4})$) that are
incident to $q$. The converse, however is not true; three lines
$l_{1},l_{2},l_{4}$ intersect in a rectangle of dimensions roughly
$\delta\times\delta/d$; the long axis of this rectangle makes an angle
$\lesssim d^{-1}$ with the long axis of $R$ (since both of these rectangles
have eccentricity about $d^{-1}$, the two rectangles are as “parallel” as
possible). By (4), there could be as many as $\lessapprox d^{-\alpha}V$ points
$q\in Q_{1}$ in this rectangle that are incident to $l_{1},l_{2}$ and $l_{4}$.
Let $Q_{2}\subset Q_{1}$, with

$\\#Q_{2}\gtrapprox d^{\alpha}V^{-1}(\\#Q_{1})\gtrapprox
d^{-\alpha}\delta^{-10\alpha}(\\#L)^{4}(\\#P)^{-8}V^{5}M^{-1},$ (22)

so that each rectangle dimensions $\delta\times\delta/d$ that is parallel to
$R$ contains at most one point from $Q_{2}$.

Let $\tilde{\delta}=\delta/d$. Let
$\tilde{T}\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ be the composition of a
translation, rotation, and dilation that first sends $p_{1}$ to $(0,0)$, then
sends $p_{4}$ to a point on the $x$-axis, and finally stretches
$\mathbb{R}^{2}$ in the $x$-direction, so that the image of $p_{4}$ becomes
$(1,0)$. The image of $R$ under $\tilde{T}$ is a rectangle of dimensions
$\approx 1\times 1$. Let $\tilde{p}_{i}$ and $\tilde{L}(p_{i})$, $i=1,2,3,5$
be the image of $p_{i}$ and $L(p_{i})$, respectively, under $\tilde{T}$. Let
$\tilde{Q}_{2}$ be the image of $Q_{2}$ under $T$

We have that $\tilde{p}_{1}=(0,0)$, $\tilde{p}_{4}=(1,0)$, and $\tilde{p}_{2}$
is contained in the $\delta$-neighborhood of the $x$-axis and has distance
$\approx 1$ from $\tilde{p}_{1}$ and $\tilde{p}_{4}$. Abusing notation, we
will replace $\tilde{p}_{2}$ with a point on the $x$-axis, and modify the set
$\tilde{L}(p_{2})$ accordingly.

The purpose of this transformation is that now the points in $\tilde{Q}_{2}$
are $\tilde{\delta}$-separated, and each $q\in\tilde{Q}_{2}$ satisfies

$\operatorname{dist}(q,x\textrm{-axis})\approx
1,\quad\operatorname{dist}(q,\overline{\tilde{p}_{1}\tilde{p}_{5}})\approx
1,\quad\operatorname{dist}(q,\overline{\tilde{p}_{4}\tilde{p}_{5}})\approx 1.$
(23)

Finally, the lines in $\tilde{L}(p_{1})$ satisfy a re-scaled version of the
non-concentration estimate (8). For each unit vector $v\in S^{1}$ and each
$r\geq\delta$, we have

$\\#\\{l\in\tilde{L}(p_{1}),\angle(l,v)\leq
r\\}\lessapprox(dr/\delta)^{\alpha}=(r/\tilde{\delta})^{\alpha},$ (24)

and similarly for $\tilde{L}(p_{2}),$ $\tilde{L}(p_{4}),$ $\tilde{L}(p_{5}).$

#### Applying a projective transformation.

Next, we apply a projective transformation $T^{\dagger}$ that sends the
$x$-axis to the line at infinity. We choose this transformation so that lines
containing $(0,0)$ (and in particular, lines in $\tilde{L}(p_{1})$) are mapped
to vertical lines; lines containing $(1,0)$ (and in particular, lines in
$\tilde{L}(p_{4})$) are mapped to horizontal lines; and lines containing
$\tilde{p}_{3}$ (and in particular, lines in $\tilde{L}(p_{3})$) are mapped to
lines with slope 1.

Recall that $\tilde{p}_{1},$ $\tilde{p}_{2}$, and $\tilde{p}_{3}$ have
separation $\approx 1$; $\tilde{Q}_{2}$ is contained in a ball centered at the
origin of radius $\approx 1$; and by (23),
$\operatorname{dist}(\tilde{Q}_{2},\ x\textrm{-axis})\approx 1$. Thus
$T^{\dagger}$ does minimal damage to $\tilde{Q}_{2}$—the set remains contained
in a ball centered at the origin of radius $\approx 1$; it is
$\approx\tilde{\delta}$ separated; and if $l\in\tilde{L}(p_{1})$ was
$\tilde{\delta}$-incident to some $q\in\tilde{Q}_{2}$, then the image of $l$
remains $\approx\tilde{\delta}$-incident to the image of $q$.

Recall from (17) that $\operatorname{dist}(p_{5},\overline{p_{1}p_{4}})\approx
1$, and thus $\operatorname{dist}(\tilde{p}_{5},x\textrm{-axis})\approx 1$.
This means that the image of $\tilde{p}_{3}$ under $T^{\dagger}$ has distance
$\lessapprox 1$ from the origin, so we can apply a translation and suppose
that the image of $\tilde{p}_{3}$ is $(0,0)$. After this translation we still
have that the image of $\tilde{Q}_{2}$ is contained in a ball centered at the
origin of radius $\approx 1$. Abusing notation, we will continue to refer to
this transformation as $T^{\dagger}$

Let $L(p_{1})^{\dagger}$, $L(p_{2})^{\dagger}$, $L(p_{4})^{\dagger}$,
$L(p_{5})^{\dagger}$ denote the images of $\tilde{L}(p_{1})$,
$\tilde{L}(p_{2})$, $\tilde{L}(p_{4})$, $\tilde{L}(p_{5})$ under
$T^{\dagger}$. $T^{\dagger}$ maps the line
$\overline{\tilde{p}_{1},\tilde{p}_{5}}$ to the $y$-axis, and the line
$\overline{\tilde{p}_{4},\tilde{p}_{5}}$ to the $x$-axis. Thus by (23), each
$q^{\dagger}\in T^{\dagger}(\tilde{Q}_{2})$ has distance $\approx 1$ from the
$x$ and $y$ axes. For convenience, we will suppose that each $q^{\dagger}\in
T^{\dagger}(\tilde{Q}_{2})$ is contained in the first quadrant; this can
always be achieved by applying a suitable reflection and refining
$\tilde{Q}_{2}$ by a factor of at most 4. Next, by dyadic pigeonholing, we can
suppose there are numbers $M_{x},M_{y}\approx 1$ so that after refining
$\tilde{Q}_{2}$ by at most $|\log\delta|^{-1}$, we have that
$T^{\dagger}(\tilde{Q}_{2})$ is contained in
$[M_{x},2M_{x}]\times[M_{y},2M_{y}]$. Thus if we compose $T^{\dagger}$ with
the anisotropic scaling $S(x,y)=(M_{x}^{-1}x,M_{y}^{-1}y)$ and define
$Q^{\dagger}=S\circ
T^{\dagger}(\tilde{Q}\cap[M_{x},2M_{x}]\times[M_{y},2M_{y}])$, then
$Q^{\dagger}\subset[1,2]^{2}$. Abusing notation slightly, we will replace
$L(p_{1})^{\dagger}$, $L(p_{2})^{\dagger}$, $L(p_{4})^{\dagger}$,
$L(p_{5})^{\dagger}$ with their image under $S$.

Finally, we will choose a value of $\delta^{\dagger}\approx\tilde{\delta}$ so
that after further refining $Q^{\dagger}$ by a $\approx 1$ factor we, we can
ensure that $Q^{\dagger}$ is $\delta^{\dagger}$-separated, and if
$l\in\tilde{L}(p_{1})$ was $\tilde{\delta}$-incident to some
$q\in\mathcal{Q}_{2}$, then the image of $l$ will be
$\delta^{\dagger}$-incident to the image of $q.$

Recall that the lines in $L(p_{1})^{\dagger}$ are parallel to the $y$-axis;
let $X\subset\mathbb{R}$ be the set of $x$-intercepts of these lines. The
lines in $L(p_{4})^{\dagger}$ are parallel to the $x$-axis; let
$Y\subset\mathbb{R}$ be the set of $y$-intercepts of these lines. The lines in
$L(p_{2})^{\dagger}$ are of the form $x+y=t$; let $Z\subset\mathbb{R}^{2}$ be
the set of $t$-values for such lines. The lines in $L(p_{5}){\dagger}$ are of
the form $y=mx$; let $W$ be the set of slopes (i.e. $m$ values) of these
lines.

The sets $X,Y,Z,W$ are $\approx\delta^{\dagger}$ separated. Since each
$q^{\dagger}\in Q_{2}^{\dagger}$ is contained in $[1,2]^{2}$, each line $l\in
L(p_{1})^{\dagger}$ that is incident to at least one point of
$Q_{2}^{\dagger}$ must have $x$-intercept in $[1,2]$. Similarly, each line
$l\in L(p_{4})^{\dagger}$ that is incident to at least one point of
$Q_{2}^{\dagger}$ must have $y$-intercept in $[1,2]$. Thus we can suppose that
$X,Y\subset[1,2]$ (i.e. we can discard any elements of $X$ or $Y$ for which
this condition fails, since such lines cannot be incident to any points in
$Q_{2}^{\dagger}$).

The angular non-concentration (24) becomes a spatial non-concentration
condition on the sets $X$ and $Y$. Specifically, for each interval
$J\subset\mathbb{R}$ we have

$\begin{split}&\\#(X\cap J)\lessapprox(|J|/\delta^{\dagger})^{\alpha},\\\
&\\#(Y\cap J)\lessapprox(|J|/\delta^{\dagger})^{\alpha}.\end{split}$ (25)

Note that the “$\lessapprox$” in the above inequalities represents a
multiplicative factor of the form $\delta^{-C\varepsilon}$, not
$(\delta^{\dagger})^{-C\varepsilon}$. (21) becomes

$\begin{split}&\\#X\lessapprox\delta^{-\alpha}(\\#L)(\\#P)^{-1}VM^{-1},\\\
&\\#Y\lessapprox\delta^{-\alpha}(\\#L)(\\#P)^{-1}VM^{-1},\\\
&\\#Z\lessapprox\delta^{-\alpha}(\\#L)(\\#P)^{-1}VM^{-1},\\\
&\\#W\lessapprox\delta^{-\alpha}(\\#L)(\\#P)^{-1}VM^{-1}.\end{split}$ (26)

Recall that
$X_{\delta^{\dagger}}=N_{\delta^{\dagger}}(X)\cap(\delta^{\dagger}\mathbb{Z})$.
Let $E\subset X_{\delta^{\dagger}}\times Y_{\delta^{\dagger}}$, with $(x,y)\in
E$ if there are points $x^{\prime}\in X$, $y^{\prime}\in Y$ with
$|x-x^{\prime}|\leq\delta^{\dagger},$ $|y-y^{\prime}|\leq\delta^{\dagger},$ so
that the corresponding lines $l_{1}\in L(p_{1})^{\dagger}$, $l_{4}\in
L(p_{4})^{\dagger}$ with intercepts $x^{\prime}$ and $y^{\prime}$ are
$\delta^{\dagger}$-incident to a common point of $Q_{2}^{\dagger}$. By (22),

$\\#E\gtrapprox
d^{-\alpha}\delta^{-10\alpha}(\\#L)^{4}(\\#P)^{-8}V^{5}M^{-1}.$

Furthermore, if $(x,y)\in E$, then the corresponding point
$q\in\tilde{Q}_{2}^{\dagger}$ is incident to a line of the form $x+y=t$, where
$|t-(x-y)|\leq 4\delta^{\dagger}$. This means that $x-y$ is an element of
$\delta^{\dagger}\mathbb{Z}$ that is contained in $N_{4\delta^{\dagger}}(Z)$.
But since $Z$ is $\approx\delta^{\dagger}$ separated, by (26) we have that

$\\#(X_{\delta^{\dagger}}\overset{E}{-}Y_{\delta^{\dagger}})\leq
4\mathcal{E}_{\delta^{\dagger}}(Z)\lessapprox\\#Z\lessapprox\delta^{-\alpha}(\\#L)(\\#P)^{-1}VM^{-1}.$

#### Bounding the sum-set.

Next, we will use the Balog-Szemerédi-Gowers theorem to find a large set
$A_{0}\subset X_{\delta^{\dagger}}$ whose sum-set is small; use Lemma 3.1 to
extract a set $A_{0}\subset X_{\delta^{\dagger}}$ with

$\begin{split}\\#A_{0}&\gtrapprox\frac{d^{-\alpha}\delta^{-10\alpha}(\\#L)^{4}(\\#P)^{-8}V^{5}M^{-1}}{\delta^{-\alpha}(\\#L)(\\#P)^{-1}VM^{-1}}\\\
&=\Big{(}d^{-2\alpha}\delta^{-8\alpha}(\\#L)^{3}(\\#P)^{-7}V^{4}\Big{)}(\delta^{\dagger})^{-\alpha},\end{split}$
(27)

and

$\begin{split}\\#(A_{0}-A_{0})&\lessapprox\frac{\big{(}\delta^{-\alpha}(\\#L)(\\#P)^{-1}VM^{-1}\big{)}^{12}}{\big{(}d^{-\alpha}\delta^{-10\alpha}(\\#L)^{4}(\\#P)^{-8}V^{5}M^{-1}\big{)}^{6}}(\\#A)\\\
&=\Big{(}d^{6\alpha}\delta^{48\alpha}(\\#L)^{-12}(\\#P)^{36}V^{-18}M^{-6}\Big{)}(\\#A).\end{split}$
(28)

Next, we will find a large set $A\subset A_{0}$, so that $A+A$ is has small
covering number at every scale (c.f. Item 3 from Proposition 3.3). Our
arguments are similar to the proof of Lemma 2.1 from [8]. Let
$k=1/\varepsilon$ and select $T$ so that $2^{kT}\leq\delta^{\dagger}\leq
2^{k(T+1)}$. Apply Lemma 3.7 to $A_{0}$; let $A\subset A_{0}$ and
$(N_{0},\ldots,N_{k-1})$ be the output from that lemma. $A$ continues to
satisfy (27). We will show that for each $\delta^{\dagger}\leq\rho\leq 1,$ we
have

$\mathcal{E}_{\rho}(A-A)\lessapprox\Big{(}d^{6\alpha}\delta^{48\alpha}(\\#L)^{-12}(\\#P)^{36}V^{-18}M^{-6}\Big{)}\mathcal{E}_{\rho}(A).$
(29)

First, we will verify (29) for $\rho$ of the form $2^{-jT},\ 1\leq j\leq k$.
Since each $2^{T}$-adic interval of length $2^{-jT}$ contains
$\prod_{i=j}^{k-1}N_{i}$ points from $A$, we have

$\\#(A-A)=\mathcal{E}_{\delta}(A-A)\geq\prod_{i=j}^{k-1}N_{i}\mathcal{E}_{\rho}(A-A).$

Thus

$\mathcal{E}_{\rho}(A-A)\leq\Big{(}\prod_{i=j}^{k-1}N_{i}\Big{)}^{-1}\\#(A-A)\lessapprox\Big{(}\prod_{i=j}^{k-1}N_{i}\Big{)}^{-1}\Big{(}d^{6\alpha}\delta^{48\alpha}(\\#L)^{-12}(\\#P)^{36}V^{-18}M^{-6}\Big{)}(\\#A_{0}).$

(29) now follows from the inequality $\\#A\approx\\#A_{0}$, and the
observation that
$\mathcal{E}_{\rho}(A)=\big{(}\prod_{i=j}^{k-1}N_{i}\big{)}^{-1}(\\#A)$.

Next, to verify (29) for arbitrary $\rho,$ select $1\leq j\leq k$ with
$2^{-jT}\leq\rho\leq 2^{-(j-1)T}$ and let $\rho^{\prime}=2^{-jT}$. Then (29)
follows from the observation that
$\mathcal{E}_{\rho}(A-A)\leq\mathcal{E}_{\rho^{\prime}}(A-A),$ and
$\mathcal{E}_{\rho^{\prime}}(A)\approx\mathcal{E}_{\rho}(A)$.

#### Bounding the multiplicative energy.

We still have that every point in $E\cap(A\times Y_{\delta^{\dagger}})$ is
$\delta^{\dagger}$-incident to a line of the form $y=mx$, with $m\in W$. Thus
by Cauchy-Schwarz,

$\\#\\{(p,p^{\prime})\in E\cap(A\times Y)\colon p,p^{\prime}\ \textrm{are
$\delta^{\dagger}$-incident to a common line}\ y=mx,\ m\in
W\\}\gtrapprox\frac{\big{(}\\#(E\cap(A\times Y))\big{)}^{2}}{\\#W}.$

Since $A,Y\subset[1,2]$, each pair $(p,p^{\prime})$ in the above set
contributes to the multiplicative energy $E^{\delta^{\dagger}}_{\times}(A,Y)$,
and thus we have

$E^{\delta^{\dagger}}_{\times}(A,Y)\gtrapprox\frac{\big{(}\\#(E\cap(A\times
Y))\big{)}^{2}}{\\#W}.$

Since $Y\subset[1,2]$ is $\approx\delta^{\dagger}$-separated, we have
$E^{\delta^{\dagger}}_{\times}(Y,Y)\lessapprox(\\#Y)^{3}$. By Lemma 3.2, we
have

$E^{\delta^{\dagger}}_{\times}(A,A)\geq\frac{\big{(}E^{\delta^{\dagger}}_{\times}(A,Y)\big{)}^{2}}{E^{\delta^{\dagger}}_{\times}(Y,Y)}\gtrapprox\frac{\big{(}\\#(E\cap(A\times
Y))\big{)}^{4}}{(\\#W)^{2}(\\#Y)^{3}}.$

But

$\\#\big{(}E\cap(A\times Y)\big{)}\gtrapprox(\\#E)\frac{\\#A}{\\#X},$

so we get

$\begin{split}E^{\delta^{\dagger}}_{\times}(A,A)&\gtrapprox\frac{(\\#E)^{4}(\\#A)^{4}}{(\\#W)^{2}(\\#X)^{4}(\\#Y)^{3}}\\\
&\gtrapprox\frac{(\\#E)^{5}}{(\\#W)^{2}(\\#X)^{4}(\\#Y)^{4}}(\\#A)^{3}\\\
&\gtrapprox\big{(}d^{-5\alpha}\delta^{-40\alpha}(\\#L)^{10}(\\#P)^{-30}V^{15}M^{5}\big{)}(\\#A)^{3}.\end{split}$
(30)

#### Applying discretized sum-product.

To summarize, we have a $A\subset[1,2]\cap(\delta^{\dagger}\mathbb{Z})$, with

$\\#A\gtrapprox\Big{(}d^{2\alpha}\delta^{8\alpha}(\\#L)^{-3}(\\#P)^{7}V^{-4}\Big{)}^{-1}(\delta^{\dagger})^{-\alpha},$
(31)

which satisfies the non-concentration condition

$\\#(A\cap J)\lessapprox|J|^{\alpha}(\delta^{\dagger})^{-\alpha},$ (32)

and

$\begin{split}\mathcal{E}_{\rho}(A-A)&\lessapprox\Big{(}d^{6\alpha}\delta^{48\alpha}(\\#L)^{-12}(\\#P)^{36}V^{-18}M^{-6}\Big{)}\mathcal{E}_{\rho}(A),\quad\delta^{\dagger}\leq\rho\leq
1,\\\
E^{\delta^{\dagger}}_{\times}(A,A)&\gtrapprox\Big{(}d^{5\alpha}\delta^{40\alpha}(\\#L)^{-10}(\\#P)^{30}V^{-15}M^{-5}\Big{)}^{-1}(\\#A)^{3}.\end{split}$
(33)

Comparing this with Proposition 3.3 and using the bounds (31), (32), and (33),
we conclude that

$\begin{split}&\Big{(}d^{2\alpha}\delta^{8\alpha}(\\#L)^{-3}(\\#P)^{7}V^{-4}\Big{)}^{6}\\\
&\cdot\Big{(}d^{6\alpha}\delta^{48\alpha}(\\#L)^{-12}(\\#P)^{36}V^{-18}M^{-6}\Big{)}^{18+8\alpha}\Big{(}d^{5\alpha}\delta^{40\alpha}(\\#L)^{-10}(\\#P)^{30}V^{-15}M^{-5}\Big{)}^{8+4\alpha}\\\
&=d^{4\alpha(40+17\alpha)}\delta^{16\alpha(77+34\alpha)}(\\#L)^{-2(157+68\alpha)}(\\#P)^{6(155+68\alpha)}V^{-12(39+17\alpha)}M^{-4(37+17\alpha)}\\\
&=(\delta^{2\alpha}\\#P)^{6(155+68\alpha)}(\delta^{2\alpha}\\#L)^{-2(157+68\alpha)}d^{4\alpha(40+17\alpha)}V^{-12(39+17\alpha)}M^{-4(37+17\alpha)}\\\
&\hskip
284.52756pt\gtrapprox(\delta^{\dagger})^{-\alpha(1-\alpha)},\end{split}$

i.e.

$(\delta^{2\alpha}\\#P)^{6(155+68\alpha)}\gtrapprox
d^{-3\alpha(53+23\alpha)}M^{4(37+17\alpha)}\delta^{-\alpha(1-\alpha)}(\delta^{2\alpha}\\#L)^{2(157+68\alpha)}V^{12(39+17\alpha)}.$

It is now clear that the worst case occurs when $d$ is as large as possible
(i.e. $d\sim 1$) and $M$ is as small as possible (i.e. $M\sim 1$). We conclude
that

$\\#P\gtrapprox\delta^{-2\alpha-c(\alpha)}(\delta^{2\alpha}\\#L)^{\frac{157+68\alpha}{3(155+68\alpha)}}V^{\frac{2(39+17\alpha)}{155+68\alpha}},\quad
c(\alpha)=\frac{\alpha(1-\alpha)}{6(155+68\alpha)}.$

The final step is to verify that for all $\alpha\in(0,1)$,
$\frac{157+68\alpha}{3(155+68\alpha)}\leq 1$ and
$\frac{2(39+17\alpha)}{155+68\alpha}\leq c(\alpha)/\alpha$. ∎

######  Remark 4.4.

When $c=1/2$, the discretized sum-product theorem (Proposition 3.3) gives a
gain of roughly $\delta^{-1/100}$. The final gain Proposition 4.1 is roughly
50 times, or two orders of magnitude, worse than this. Roughly speaking, one
order of magnitude comes from repeated pigeonholing and use of Cauchy-Schwarz,
while the other order of magnitude comes from the use of Balog-Szemerédi-
Gowers. While some further optimizations are likely possible, it appears that
such optimizations would yield rather modest gains. In particular, Proposition
3.3 requires that the difference set $A-A$ be small (rather than merely
requiring that $E_{+}(A,A)$ be large), so it appears difficult to avoid using
the Balog-Szemerédi-Gowers theorem). While a more efficient sequence of
pigeonholing and Cauchy-Schwarz arguments might improve the final bound, since
the current argument only loses (roughly) one order of magnitude at this step,
any improvement here would at best yield a modest improvement in the final
bound.

##  5 From Proposition 4.1 to discretized Furstenberg

Our goal in this section is to prove that Proposition 4.1 implies Proposition
2.3, and hence Theorem 1.8. First, we will briefly describe the differences
between Propositions 4.1 and 2.3. Both are incidence theorems about
collections of points and lines, but Proposition 4.1 requires two additional
non-concentration conditions.

Property E is sometimes called “robust transversality,” and it asserts that
for each $p\in P$, a typical pair of lines incident to $p$ point in well-
separated directions. With this condition, Proposition 4.1 can also be viewed
as a bilinear incidence theorem. Thus our reduction from Proposition 4.1 to
2.3 will employ “bilinear to linear” type arguments, which were first used by
Tao, Vargas, and Vega in the context of the Kakeya and Restriction problems
[22]. A prototype for the specific arguments used here can be found in [9,
§2.3]. See also [21] for a similar argument in a slightly different setting.

Property (F) is also an assertion about the distribution of directions
$\\{\operatorname{dir}(l)\colon(p,l)\in I\\}$. This property imposes stronger
non-concentration conditions at small scales (specifically at length scales
between $\delta$ and $(\delta^{-2\alpha}/\\#P)^{1/\alpha}$), but it says
nothing about non-concentration at larger scales. To the best of our
knowledge, the arguments used to obtain the assertion in Property (F) are
novel, and we hope that similar arguments might be helpful for related
problems.

###### Proof of Proposition 2.3.

Fix $0<\alpha<1$ and $c<c(\alpha)=\frac{\alpha(1-\alpha)}{6(155+68\alpha)}$.
We will show that if the parameters $\varepsilon>0$ and $C$ are chosen
appropriately, then for all $\delta>0$ sufficiently small, if
$E\supset\bigcup_{l\in L_{0}}P_{l}$ is a discretized
$(\alpha,2\alpha)$-Furstenberg set (with implicit parameters $\varepsilon$ and
$C$), then

$\mathcal{E}_{\delta}(E)\geq\delta^{-2\alpha-c}.$ (34)

In particular, (34) will follow from the estimate

$\mathcal{E}_{\delta}\Big{(}\bigcup_{l\in
L_{0}}P_{l}\Big{)}\gtrapprox\delta^{-2\alpha-c(\alpha)}.$ (35)

#### Constructing the incidence arrangement.

Our goal is to construct a $\delta$-separated set $P\subset\bigcup_{l\in
L_{0}}P_{l}$, and an incidence relation $I\subset P\times L$ that satisfies
the hypotheses of Proposition 4.1. While it is tempting to simply choose $P$
to be a maximal $\delta$-separated subset of $\bigcup_{l\in L_{0}}P_{l}$, it
is possible that a set constructed in this fashion will have the following
undesirable property: let $R$ be a rectangle of dimensions $r\times\delta$, so
that many lines $l\in L_{0}$ intersect $R$ in a line-segment of length roughly
$r$; while Item (ii) from Definition 2.1 guarantees that each set $R\cap
P_{l}$ has cardinality $\lessapprox(r/\delta)^{\alpha}$, the union
$R\cap\bigcup P_{l}$ might be much larger. For technical reasons that will
become apparent later in the proof, we would like to avoid this type of
incidence arrangement.

To avoid this problem, we will use a stopping-time type argument to construct
an incidence arrangement $(P,L,I)$ so that for each rectangle $R$ of this
type, only a few lines from $L_{0}$ that have large intersection with $R$
contribute points to $R$.

Let $k=\lfloor 1/\varepsilon\rfloor$ and select $T$ so that
$2^{-kT}\leq\delta/6<2^{-k(T-1)}$. Refining each set $\\{P_{l}\\}_{l\in
L_{0}}$ by a factor of at most $(2\delta)^{-\varepsilon}$, we can suppose that
each set $P_{l}$ is discretized at scale $2^{-kT}$. Apply Corollary 3.8 to the
collection of sets $\\{P_{l}\\}_{l\in L_{0}}$, and let $L_{1}\subset L_{0}$;
$\\{P_{l}^{\prime}\\}_{l\in L_{1}};$ $(N_{0},\ldots,N_{k-1})$ be the output
from the corollary. We have

$\sum_{l\in L_{1}}\\#P_{l}^{\prime}\gtrapprox\sum_{l\in
L}\\#P_{l}\gtrapprox\delta^{-3\alpha}.$ (36)

Enumerate the elements of $L_{1}$ as $l_{1},\ldots,l_{M}$,
$M\approx\delta^{-2\alpha}$. Will will construct a point set
$P_{1}\subset\bigcup_{l\in L}P_{l}$ and an incidence relation $I_{1}\subset
P_{1}\times L_{1}$ as follows.

We begin by setting $P_{1}=\emptyset$ and $I_{1}=\emptyset$. Let $m=1$, and
add all of the points in $P_{l_{1}}^{\prime}$ to $P_{1}$; add all of the
incidences $\\{(p,l_{1})\colon p\in P_{l_{1}}^{\prime}\\}$ to $I_{1}$. For
each $2^{T}$-adic square $S$ for which $P_{l_{1}}^{\prime}\cap
S\neq\emptyset$, we say that $l_{1}$ has _contributed points in square $S$_.
We have now “processed” the line $l_{1}$.

Suppose that we have processed the lines $l_{1},\ldots,l_{m-1}$. We shall
process the line $l_{m}$ as follows. To begin, we declare all $2^{T}$-adic
squares inside $[0,1]^{2}$ to be “undominated.” We perform the following
procedure for each $j=0,\ldots,k$: consider each $2^{T}$-adic square
$S\subset[0,1)^{2}$ of side-length $2^{-jT}$ that is not contained in a
(larger) dominated square.

  * •

If $S\cap{P^{\prime}_{l_{m}}}=\emptyset$, we ignore $S$, and it remains
undominated (squares of this type will not be relevant for successive steps of
the algorithm).

  * •

If instead $P^{\prime}_{l_{m}}\cap S\neq\emptyset$, and hence
$\\#(P^{\prime}_{l_{m}}\cap S)=\prod_{i=j}^{k-1}N_{i}$, then either (A) or (B)
must hold:

    * (A):

there is at least one line $l_{n},\ n<m$ so that:

      * (i):

$l_{n}$ has contributed points in $S$,

      * (ii):

$\angle(l_{m},l_{n})\leq(\frac{\pi}{2})2^{(j-k)T}$,

      * (iii):

$\operatorname{dist}(l_{m}\cap S,l_{n}\cap S)\leq(\sqrt{2})2^{-kT}.$

    * (B):

no such line exists.

If Option (A) occurs, select a line $l_{n}$ with this property; add the pairs
$\\{(p,l_{m})\colon p\in S\cap P^{\prime}_{l_{n}}\\}$ to $I_{1}$; and mark the
square $S$ as dominated. We will say that “$l_{m}$ is dominated by $l_{n}$ on
square $S$.” Note that the conditions in Option (A) imply that $l_{n}\cap
S\subset N_{\delta}(l_{m})$ (indeed, there is a point in $S$ where $l_{m}$ and
$l_{n}$ have separation at most $(\sqrt{2})2^{kT}\leq\delta/4$, and the
condition on $\angle(l_{m},l_{n})$ then ensures that $l_{n}\cap S\subset
N_{\delta}(l_{m})$. Thus each of the incidences $(p,l_{m})$ added in this step
satisfy $p\in N_{\delta}(l_{m})$. If Option (B) occurs, the square $S$ remains
undominated.

Repeat this process for $j=1,2,\ldots,k$. We say a $2^{T}$-adic square $S$ is
a _maximal purely undominated square_ if $S$ does not contain a dominated
square, but every $2^{T}$-adic square $S^{\prime}\supset S$ _does_ contain a
dominated square. Each undominated square $S$ is contained in a unique maximal
purely undominated square. For each maximal purely undominated square $S$ for
which $S\cap P_{l_{m}}^{\prime}\neq\emptyset$, add the points in $S\cap
P_{l_{m}}^{\prime}$ to $P_{1}$, and add the corresponding incidences
$(p,l_{m})$ to $I_{1}$. For each $2^{T}$-adic square $S$ (maximal or not) for
which $P_{l_{m}}^{\prime}\cap S$ is undominated, we say that $l_{m}$ has
contributed points in square $S$.

After the above steps have been completed, we have now processed the line
$l_{m}$. Observe that at this point, the incidence arrangement
$(P_{1},L_{1},I_{1})$ has the following properties:

  * •

If $(p,l_{m})\in I_{1}$, then $p\in N_{\delta}(l_{m})$

  * •

For each $2^{T}$-adic square $S$, we have

$\\#\\{p\in P_{1}\cap S\colon(p,l_{m})\in I_{1}\\}=\\#(P_{l_{m}}^{\prime}\cap
S).$ (37)

As a consequence of (37) when $S=[0,1)^{2}$, we have
$\\#I_{1}(l_{m})\gtrapprox\delta^{-\alpha}.$ Furthermore, if $\delta\leq r\leq
1$, then

$\\#(B(x,r)\cap I_{1}(l_{m}))\lessapprox(r/\delta)^{\alpha}$ (38)

for all balls $B(x,r)$. Indeed, choose $j$ so that $2^{-jT}\leq
r<2^{-(j-1)T}$. Then $B(x,r)$ intersects $\approx 1$ squares of side-length
$2^{-jT}$, so (38) follows from (37), plus the fact that $P_{l_{m}}^{\prime}$
is a $(\delta,\alpha)_{1}$-set.

As additional lines are processed, we will add more points to $P_{1}$ and more
incidences to $I_{1}$, but this will not disrupt the properties of $l_{m}$
described above.

After all of the lines in $L_{1}$ have been processed in the above fashion,
the incidence arrangement $(P_{1},L_{1},I_{1})$ satisfies Properties (B) (with
$V=1$) and (C) from Proposition 4.1, as well as the estimate
$\\#I_{3}\approx\delta^{-3\alpha}$. We claim that $P_{1}$ is discretized at
scale $2^{-kT}$, and thus $\lessapprox 1$ points $p\in P_{1}$ can be contained
in any ball of radius $\delta$. To prove this claim, note that if $S$ is a
square of side-length $2^{-kT}$, then at most one line $l\in L_{1}$ can
contribute points to $S$—if two such lines $l,l^{\prime}$ both contributed
points to $S$, then we must have that either $\angle(l,l^{\prime})>\pi/2$
(which is impossible), or $\operatorname{dist}(l\cap S,l^{\prime}\cap
S)>(\sqrt{2})2^{-kT}=\operatorname{diameter}(S)$ (which is impossible).

Since $P_{1}$ is discretized at scale $2^{-kT}$, in order to establish (35) it
suffices to prove that

$\\#P_{1}\gtrapprox\delta^{-2\alpha-c(\alpha)}.$ (39)

Finally, we observe that if $p\in P_{1}$, if $1\leq j\leq k-1$, and if $S$ is
a square of side-length $2^{-jT}$ that contains $p$, then there is exactly one
line $l\in I_{1}(p)$ that contributes points in square $S$.

#### Non-concentration of lines through a point.

Our next task is to extract a set of points $P_{2}\subset P_{1}$ and a set of
incidences $I_{2}\subset I_{1}$ so that for each $p\in P_{2}$, the
distribution of directions of lines incident to $p_{2}$ satisfies Property
(F).

For each $p\in P_{1}$, we will identify the set $I_{1}(p)$ with a subset of
$[0,1]$ by identifying the line $y=mx+b$ with the point $m$; call this set
$A_{p}$. Since $L_{0}$ is $\delta$-separated, at most 16 points in $A_{p}$ can
intersect any interval of length $2^{-kT}\leq\delta/6$ (indeed, the set of
such points in $A_{p}$ corresponds to lines in $L_{1}$ whose image under
$\iota$ is contained in a ball of radius $2\delta$). In particular, by
refining each set $A_{p}$ by a factor of 16, we can suppose that $A_{p}$ is
$2^{-kT}$-discretized. By removing on additional point if necessary, we can
also suppose that $A_{p}\subset[0,1)$.

Apply Corollary 3.8 to the collection of sets $\\{A_{p}\\}_{p\in P_{1}}$, and
let $P_{2}\subset P_{1}$; $\\{A_{p}^{\prime}\\}_{p\in P_{2}};$
$(M_{0},\ldots,M_{k-1})$ be the output from the corollary. Each set
$A_{p}^{\prime}$ has cardinality $\prod_{j=0}^{k-1}M_{j}$, and

$(\\#P_{2})\prod_{j=0}^{k-1}M_{j}=\sum_{p\in
P_{2}}\\#A_{p}^{\prime}\approx\sum_{p\in
P_{1}}\\#A_{p}^{\prime}\approx\\#I_{1}\approx\delta^{-3\alpha}.$ (40)

Define $I_{2}$ to consist of those incidences $(p,l)\in P_{2}\times L_{1}$ so
that the slope of $l$ is in $A_{p}^{\prime}.$

Recall that if $0\leq j\leq k$ and if $J\subset[0,1)$ is an $2^{T}$-adic
interval of side-length $2^{-jT}$ that meets $A_{p}$, then $\\#(J\cap
A_{p}^{\prime})=\prod_{i=j}^{k-1}M_{i}$. Our next task is to control the size
of this product. In theory, each $M_{i}$ could be as large as $2^{T}$. We will
show that on average, each $M_{i}$ has size at most $2^{\alpha T}$. More
precisely, we will prove that there exists a constant $C_{0}=C_{0}(\alpha)$ so
that

$\prod_{i=j}^{k-1}M_{i}\leq\delta^{-C_{0}\varepsilon}\Big{(}\frac{2^{(j-k)T}}{\delta}\Big{)}^{\alpha},\quad\textrm{for
each}\ j=0,\ldots,k-1.$ (41)

We will prove (41) by induction—the base case will be when $j$ is large (i.e.
$j$ is close to $k-1$, which corresponds to small length scales), and the
induction step will go from $j$ to $j-1$. Observe that since the points in
$A_{p}^{\prime}$ are $2^{-kT}$-separated, we have the trivial estimate
$M_{i}\leq 2^{T}\leq\delta^{-2\varepsilon}.$ Thus since
$2^{(j-k)T}\delta^{-1}\geq 2^{-kT}\delta^{-1}\geq\delta^{2\varepsilon},$ we
have

$\prod_{i=j}^{k-1}M_{i}\leq\delta^{-(2k-2j-2)\varepsilon}\Big{(}\frac{2^{(j-k)T}}{\delta}\Big{)}^{\alpha},$

and thus (41) holds for $j\geq k-(C_{0}-2)/2$.

Suppose now that (41) has been established for all $j^{\prime}>j$; our task is
to prove (41) for $j$. If $M_{j}\leq 2$ then this follows immediately from
(41) for $j+1$. Suppose now that $M_{j}\geq 3$.

Since $(\\#I_{2})(\\#L_{1})\gtrapprox\delta^{-\alpha}$, there exists a line
$l\in L$ with $\\#I_{2}(l)\gtrapprox\delta^{-\alpha}$. Fix this choice of $l$.
Define $\rho=2^{(j-k)T}$. By (38), each ball $B$ of radius
$\delta^{1-\varepsilon}\rho^{-1}$ satisfies $\\#(I_{2}(l)\cap
B)\lessapprox(\delta^{1-\varepsilon}\rho^{-1}/\delta)^{\alpha}\lessapprox\rho^{-\alpha},$
and hence we can select a set $P^{\prime}\subset I_{2}(l)$ that is
$(\delta^{1-\varepsilon}\rho^{-1})$-separated, with
$\\#P^{\prime}\gtrapprox(\delta/\rho)^{-\alpha}$.

Let $m$ be the slope of $l$. For each $p\in P^{\prime}$, we have that $m\in
A_{p}$, and hence the $2^{T}$-adic interval $J$ of length $\rho$ containing
$m$ intersects $A_{p}$. Thus for each $p\in P^{\prime},$ there are
$\geq\prod_{i=j}^{k-1}M_{i}$ lines $l^{\prime}\in I_{2}(p)$ with
$|m-m^{\prime}|\leq\rho$ (here $m^{\prime}$ is the slope of $l^{\prime}$). On
the other hand, the lines $l^{\prime}\in I_{2}(p)$ with $|m-m^{\prime}|\leq
2^{-T}\rho$ are contained in a union of at most two $2^{T}$-adic intervals of
length $2^{-(j-1)T}$; each of these intervals contains
$\prod_{i=j+1}^{k-1}M_{i}=M_{j}^{-1}\prod_{i=j}^{k-1}M_{i}\leq\frac{1}{3}\prod_{i=j}^{k-1}M_{i}$
points.

Thus for each $p\in P^{\prime}$ there are at most
$\frac{2}{3}\prod_{i=j}^{k-1}M_{i}$ lines $l^{\prime}\in I_{2}(p)$ with
$|m-m^{\prime}|\leq 2^{-T}\rho$. We conclude that

$\\#\\{l^{\prime}\in I_{2}(p)\colon\delta^{\varepsilon}\rho\leq
2^{-T}\rho\leq|m-m^{\prime}|\leq\rho\\}\geq\frac{1}{3}\prod_{i=j}^{k-1}M_{i}.$

Call the above set of lines $L(p)$. Since the points in $P^{\prime}$ are
$\delta^{1-\varepsilon}\rho^{-1}$ separated and contained in the
$\delta$-neighborhood of $l$, if $p,p^{\prime}\in P^{\prime}$ are distinct,
then the sets $L(p)$ and $L(p^{\prime})$ are disjoint (indeed, a line
$l^{\prime}\in L(p)\cap L(p^{\prime})$ must intersect $N_{\delta}(l)$ in an
interval of length $\geq\delta^{1-\varepsilon}\rho^{-1}$, but this is
impossible since the slopes of $l$ and $l^{\prime}$ differ by at least
$\delta^{-T}\rho\geq\delta^{-\varepsilon}\rho)$. Thus

$\\#\bigcup_{p\in
P^{\prime}}L(p)\gtrapprox(\\#P^{\prime})\prod_{i=j}^{k-1}M_{i}\gtrapprox\rho^{-\alpha}\prod_{i=j}^{k-1}M_{i}.$

The lines in the above set satisfy
$\operatorname{dist}(\iota(l),\iota(l^{\prime}))\leq\rho$, and thus

$\prod_{i=j}^{k-1}M_{i}\lessapprox(\\#P^{\prime})^{-1}\\#(B(x,\rho)\cap\iota(L_{1}))\lessapprox(\rho/\delta)^{-\alpha}(\rho/\delta)^{2\alpha}=(\rho/\delta)^{\alpha}.$
(42)

Selecting $C_{0}$ sufficiently large based on the implicit constants in the
above quasi-inequality (these constants in turn depend only on $\alpha$), we
conclude that (41) holds for $j$. This completes the induction step.

By (41), if $p\in P_{2}$ and $J\subset S^{1}$ is an interval whose image
$J^{\prime}\subset[0,1]$ is a $2^{T}$-adic interval, then

$\\#\\{l\in I_{2}(p)\colon\operatorname{slope}(l)\in
J\\}\lessapprox(|J|/\delta)^{\alpha}.$

Now, let $v\in S^{1}$ and let $\delta\leq r\leq 1$. Then there is an index $j$
so that $2^{-jT}\leq r<2^{-(j-1)T}\leq\delta^{-\varepsilon}2^{-jT}$, and thus
the set of unit vectors $\\{v^{\prime}\in S^{1}\colon\angle(v^{\prime},l)\leq
r\\}$ correspond to a set of slopes $m$ that can be covered by a union of
$\approx 1$ intervals of length $2^{-jT}$. We conclude that

$\\#\\{l\in I_{2}(p)\colon\angle(l,v)\leq r\\}\lessapprox(r/\delta)^{\alpha}.$
(43)

Our incidence arrangement $(P_{2},L_{1},I_{2})$ now satisfies Properties (B),
(C), (D), and (F) from Proposition 4.1 (with $\lessapprox$ in place of
$\leq\delta^{-\varepsilon}$ in some instances), as well as the estimate
$\\#I_{3}\approx\delta^{-3\alpha}$.

#### Robust transversality of lines through a point.

Our goal in this section is to find a set of incidences $I_{3}\subset I_{2}$
with $\\#I_{3}\approx\delta^{-3\alpha}$ that continues to satisfy Hypotheses
(B), (C), (D), and (F), but which also satisfies a re-scaled version of
Property (E).

Let $j_{0}\geq 0$ be an index so that $M_{j_{0}}>100$, and $M_{j}\leq 100$ for
each $j=0,\ldots,j_{0}-1$; we can assume that such an index must exist, since
otherwise we would have
$\\#I_{2}(p)=\\#A_{p}^{\prime}=\prod_{j=0}^{k-1}M_{j}\leq\prod_{j=0}^{k-1}100\leq
100^{1/\varepsilon}$ for each $p\in P_{2}$. But then
$\\#P_{2}\geq\\#I_{2}/100^{1/\varepsilon}\gtrapprox\delta^{-3}$, which would
establish (39).

For each $p\in P_{2}$, define $A_{p}^{\prime\prime}=A_{p}^{\prime}\cap J$,
where $J$ is a $2^{T}$-adic interval of length $2^{-j_{0}T}$ that intersects
$A_{p}^{\prime\prime}.$ By construction we have

$\\#A_{p}^{\prime\prime}=(N_{0}N_{1}\cdots
N_{j_{0}-1})^{-1}\\#A_{p}^{\prime}\gtrsim\\#A_{p}^{\prime},$

and for any $2^{T}$-adic interval $J^{\prime}\subset J$ of length
$2^{-(j_{0}+1)T}$, we have

$\\#(A_{p}^{\prime\prime}\cap
J^{\prime})\leq\frac{1}{100}\\#A_{p}^{\prime\prime}.$

Let $I_{3}\subset I_{2}$ consist of those pairs $(p,l)\in I_{2}$ for which
$\operatorname{slope}(l)\in A_{p}^{\prime\prime}$. Define $s=2^{-j_{0}T}$. We
record the following properties of this incidence arrangement.

  * •

Properties (B) (with $V=1$) and (C) continue to hold for
$(P_{2},L_{1},I_{3})$, and $\\#I_{3}\gtrapprox\delta^{-3}$.

  * •

(43) (and thus Property (F)) continues to hold with $I_{3}$ in place of
$I_{2}$

  * •

For each $p\in P_{2}$, $\\#I_{3}(p)=\\#I_{3}(\\#P_{2})^{-1},$ and thus
Property (D) continues to hold.

  * •

For each $p\in P_{2},$ the set $\\{\operatorname{slope}(l)\colon l\in
I_{3}(p)\\}\subset[0,1]$ is contained in an interval of length $s$. If
$J\subset[0,1]$ is an interval of length $\delta^{\varepsilon}s$, then

$\\#\\{l\in I_{3}(p)\colon\operatorname{slope}(l)\in
J\\}\leq\frac{1}{20}\\#I_{3}(p).$ (44)

(Indeed, the slopes of the lines in the above set correspond to points in
$A_{p}^{\prime\prime}$ that are contained in a union of at most two
$2^{T}$-adic intervals of length $2^{-(j_{0}+1)T}$).

For each $p\in P_{2}$, let $v_{p}\in S^{1}$ be a vector of the form
$(1,m)/\sqrt{1+m^{2}}$, where $m$ is the midpoint of the interval containing
the set of slopes $\\{\operatorname{slope}(l)\colon l\in I_{3}(p)\\}$. We can
suppose $v_{p}=(\cos\theta,\sin\theta)$, with $\theta\in[0,\pi/4]$.

#### Partitioning into rectangles.

Let $\mathcal{Z}_{0}=([0,1]\times[-1,1])\cap(\frac{s}{2}\mathbb{Z})^{2}$. For
each $z=(m,b)\in\mathcal{Z}_{0}$, let $l_{z}$ be the line $y=mx+b$, and define

$P_{2}^{z}=\\{p\in P_{2}\colon p\in N_{s}(l_{z}),\ \angle(v_{p},l_{z})\leq
s/2\\},\quad L_{1}^{z}=\bigcup_{p\in P_{2}^{z}}I_{3}(p).$

We claim that

$P_{2}=\bigcup_{z\in\mathcal{Z}_{0}}P_{2}^{z},$ (45)

and each $p\in P_{2}$ is contained in at most 9 sets in the above union.
Indeed, for each $p\in P_{2}$, there exists at least one, and at most three
$m\in[0,1]\cap(\frac{r}{2}\mathbb{Z})$ with the following property: if $v_{m}$
is the direction of the vector $(1,m)$, then $\angle(v_{p},v_{m})\leq s/2$.
Once this choice of $m$ has been fixed, there is at least one, and at most
three $b\in[-1,1]\cap(\frac{r}{2}\mathbb{Z})$ with the following property: if
$z=(m,b)$, then $p\in N_{s}(l_{z})$.

Observe as well that each $l\in L_{1}$ can be contained in at most 100 sets of
the form $L_{1}^{z}$: if $l\in L_{1}^{z}$ then $l\in I_{3}(p)$ for some $p\in
P_{2}^{z}$, which implies
$\angle(l,l_{z})\leq\angle(l,v_{p})+\angle(v_{p},l_{z})\leq s$. Next, since
$p\in N_{\delta}(l)\cap N_{s}(l_{z})\cap[0,1]^{2}$, we have that
$\operatorname{dist}(l\cap[0,1]^{2},l_{z})\leq s+\delta$. But this implies
that $\operatorname{dist}(\iota(l),\iota(l_{z}))\leq 3s$. The result now
follows from the fact that $\mathcal{Z}_{2}$ is $s/2$-separated.

Thus if we define $I_{3}^{z}=I\cap(P_{2}^{z}\times L_{1}^{z})$, then
$I_{3}=\bigcup_{z\in\mathcal{Z}_{0}}I_{3}^{z}$, and each $(p,l)\in I_{3}$ is
contained in at most 9 sets in the above union. By refining $\mathcal{Z}_{0}$
by a factor of at most 9, we can suppose that the sets $\\{P_{2}^{z}\\}$,
$\\{L_{1}^{z}\\}$, and thus $\\{\mathcal{I}_{3}^{z}\\}$ are disjoint. Using
dyadic pigeonholing, select $\mathcal{Z}\subset\mathcal{Z}_{0}$ so that the
sets $\\{P_{2}^{z}\\}_{z\in\mathcal{Z}}$, $\\{L_{1}^{z}\\}_{z\in\mathcal{Z}}$
(and thus $\\{\mathcal{I}_{3}^{z}\\}_{z\in\mathcal{Z}}$) are disjoint,
$\sum_{z\in\mathcal{Z}}\\#I_{3}^{z}\gtrsim\\#I_{3},$ and $\\#P_{2}^{z}$ (and
hence $\\#I_{3}^{z}$) is approximately the same (up to a multiplicative factor
of 2) for each $z\in\mathcal{Z}$. In particular,

$\\#P_{2}^{z}\approx(\\#P_{2})/(\\#\mathcal{Z}),\quad\textrm{and}\quad\\#I_{3}^{z}\approx(\\#I_{3})/(\\#\mathcal{Z})\quad\textrm{for
each}\ z\in\mathcal{Z}.$ (46)

Geometrically, we have partitioned (a large portion of) the arrangement
$(P_{2},L_{1},I_{3})$ into sub-arrangements of the form
$(P_{2}^{z},L_{1}^{z},I_{3}^{z})$, each of which is contained in the region
$[0,1]^{2}\cap N_{3s}(l_{z})$, and each of which contributes a roughly equal
number of incidences. Our next step is to analyze each of these sub-
arrangements.

#### Averaging over segments.

For the remainder of our arguments we will fix a choice of $z\in\mathcal{Z}$
with

$\\#L_{1}^{z}\gtrapprox\\#L_{1}(\\#\mathcal{Z})^{-1}\gtrapprox\delta^{-2\alpha}(\\#\mathcal{Z})^{-1}.$
(47)

Let $P=P_{2}^{z},$ $L=L_{1}^{z}$ and $I=I_{3}^{z}$. Let $\mathcal{S}$ be the
set of $2^{T}$-adic squares in $[0,1)^{2}$ of side-length $2^{(j_{0}-2-k)T}$
(note that $2^{(j_{0}-2-k)T}\approx\delta/s$). We claim that the arrangement
$(P,L,I)$ has the following properties: If $S\in\mathcal{S}$ and if $p\in
P\cap S$, then

  * (i):

There is exactly one line $l\in I(p)$ that contributes points to the square
$S$.

  * (ii):

Every other line $l^{\prime}\in I(p)$ is dominated by $l$ on some square
$S^{\prime}\supset S$.

Item (i) holds in general (see the discussion following (39)). For Item (ii),
since $l^{\prime}\in I(p)$ and $l^{\prime}\neq l$, we have that $l$
contributed points in the square $S$ containing $p$, and $l^{\prime}$ did not
contribute points in $S$. Let $S^{\prime}$ be the largest $2^{T}$-adic square
containing $p$ so that $l^{\prime}$ was dominated by $l$ on $S^{\prime}$ (the
square $S^{\prime}$ with this property is necessarily unique). We will show
that $S^{\prime}\supset S$. If $S^{\prime}\supsetneq S$ then we are done. If
not, then at the step in the construction of $(P_{1},L_{1},I_{1})$ when
$l^{\prime}$ was processed, the square $S$ would not be contained in a
(larger) dominated square. But we also have that

$\operatorname{dist}(l\cap S,l^{\prime}\cap S)\leq\operatorname{dist}(l\cap
S^{\prime},l^{\prime}\cap S^{\prime})\leq(\sqrt{2})2^{-kT},$

and

$\angle(l,l^{\prime})\leq\angle(l,l_{z})+\angle(l_{z},l^{\prime})\leq
2s=2\cdot 2^{-j_{0}T}\leq(\frac{\pi}{2})2^{([k-(j_{0}-2)]-k)T},$

which implies that $l$ is dominated by $l^{\prime}$ on square $S$.

Next, observe that if $l,l^{\prime}$ both contribute points to $S$, then since
$\angle(l,l^{\prime})\leq(\frac{\pi}{2})2^{((j_{0}+1)-k)T}$, we must have
$\operatorname{dist}(l\cap S,l^{\prime}\cap S)>(\sqrt{2})2^{-kT}.$

Let $L(S)\subset L$ be the set of lines that contribute points to the square
$S$. Then the above discussion shows that if $l,l^{\prime}\in L(s)$ are
distinct, then the sets $S\cap I(l)$ and $S\cap I(l^{\prime})$ are disjoint.
We have

$P=\bigsqcup_{S\in\mathcal{S}}\bigsqcup_{l\in L(S)}(S\cap I(l)).$

Since each set $P_{l}$ is a $(\delta,\alpha)_{1}$-set, each of the sets
$(S\cap I(l))$ have cardinality at most

$\delta^{-\varepsilon}\Big{(}\frac{2^{(j_{0}-2-k)T}}{\delta}\Big{)}^{\alpha}\lessapprox\Big{(}\frac{\delta/s}{\delta}\Big{)}^{\alpha}=s^{-\alpha}$

Thus by dyadic pigeonholing, there is a number $M$ with

$1\leq M\lessapprox s^{-\alpha}$ (48)

so that

$\\#P\approx\sum_{S\in\mathcal{S}}\sum_{\begin{subarray}{c}l\in L(S)\\\
M\leq\\#(S\cap I(l))<2M\end{subarray}}\\#(S\cap I(l)).$

Choose $Q\subset P$ by selecting one point from $S\cap I(l)$ for each square
$S\in\mathcal{S}$ and each $l\in L(S)$ in the above sum. Define
$I_{Q}=I\cap(Q\times L)$. In particular, (43) and (44) continue to hold with
$Q$ in place of $P$, and $I_{Q}$ in place of $I_{2}$ or $I_{3}$. We have
$\\#I_{Q}\gtrapprox M^{-1}\delta^{-3\alpha}$.

#### Rescaling

Define $\tilde{\delta}=\delta/s$. Let $T\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$
be a translation in the direction $\operatorname{dir}(l_{z})^{\perp}$ that
sends $l_{z}$ to a line through the origin, composed with an anisotropic
dilation by $s^{-1}$ in the direction $\operatorname{dir}(l_{z})^{\perp}$; the
image of $N_{s}(l)\cap[0,1]^{2}$ under this linear map is contained in
$B(0,10)$, and it has volume $\sim 1$. Let $\tilde{Q}$ (resp. $\tilde{L}$) be
the image of $Q$ (resp. $L$) under $T$. Let
$\tilde{I}_{Q}\subset\tilde{Q}\times\tilde{L}$ be the incidence relation
corresponding to $I_{Q}$. If $(\tilde{q},\tilde{l})\in\tilde{I}_{Q}$ then
$\tilde{q}\in N_{\tilde{\delta}}(\tilde{l})$.

If $S\in\mathcal{S}$ and $l,l^{\prime}\in L(S)$, then $T(l\cap S)$ and
$T(l^{\prime}\cap S)$ are $\gtrapprox\delta/s=\tilde{\delta}$ separated. Thus
for each such square $S$, $\tilde{Q}\cap T(S)=T(Q\cap S)$ is
$\gtrapprox\tilde{\delta}$-separated. But since the squares in $\mathcal{S}$
have side-length $\approx\tilde{\delta}$, each ball $B\subset\mathbb{R}^{2}$
of radius $\tilde{\delta}$ intersect at most $\approx 1$ sets of the form
$\\{T(S)\colon S\in\mathcal{S}\\}$. We conclude that $\\#(\tilde{Q}\cap
B)\lessapprox 1$ for each ball $B\subset\mathbb{R}^{2}$ of radius
$\tilde{\delta}$. Refining the set $\tilde{Q}$ by a $\approx 1$ factor, we can
suppose that $\tilde{Q}$ is $\tilde{\delta}$-separated; this refinement does
not affect the validity of any of our previous estimates.

Define $V=\min(1,(s^{\alpha}M)^{-1})$; by (48) we have
$V\approx(s^{\alpha}M)^{-1}$, and thus $M^{-1}\delta^{-\alpha}\approx
V\tilde{\delta}^{-\alpha}$. We will record the following properties of
$\tilde{Q},\ \tilde{L}$, and $\tilde{I}$.

$\displaystyle\\#P\gtrsim M\\#\tilde{Q},$ (49)
$\displaystyle\\#\tilde{I}_{Q}(q)=(\\#\tilde{I})(\\#\tilde{Q})^{-1}\quad\textrm{for
each}\ q\in\tilde{Q},$ (50) $\displaystyle\\#\tilde{I}_{Q}(l)\lessapprox
M^{-1}\delta^{-\alpha}\approx\tilde{\delta}^{-\alpha}V\quad\textrm{for each}\
l\in\tilde{L},$ (51) $\displaystyle\\#\tilde{I}\gtrapprox
M^{-1}(\delta^{-\alpha}\\#L^{\prime})\gtrapprox\tilde{\delta}^{-\alpha}(\\#\tilde{L})V.$
(52)

(50) shows that $(\tilde{Q},\tilde{L},\tilde{I})$ satisfies Property (D) from
Proposition 4.1. For each $l\in\tilde{L}$ and each ball $B(r,x)$, with
$r\geq\tilde{\delta}$,

$\\#\\{\tilde{q}\in\tilde{Q}\cap
B(t,x)\colon(\tilde{q},\tilde{l})\in\tilde{I}\\}\leq
M^{-1}\delta^{-\varepsilon}(r/\delta)^{\alpha}\approx(r/\tilde{\delta})^{\alpha}V,$
(53)

c.f. Property (B) from Proposition 4.1. For each ball $B(t,x)$, the set of
lines

$\\{l\in L\colon\iota(T(l))\in B\\}=\\{T^{-1}(l)\colon\iota(l)\in B\\}$

is contained in a ball $B^{\prime}$ of radius $\approx st$. Thus

$\\#(\iota(\tilde{L})\cap
B)\leq\delta^{-\varepsilon}(st/\delta)^{2\alpha}=\delta^{-\varepsilon}(t/\tilde{\delta})^{2\alpha}.$
(54)

c.f. Property (C) from Proposition 4.1.

Next, for each $q\in\mathcal{Q}$ we have the following analogue of (43): for
each $v\in S^{1}$ and each $r\geq\tilde{\delta}$,

$\\#\\{\tilde{l}\in\tilde{I}_{Q}(\tilde{q})\colon\angle(\tilde{l},v)\leq
r\\}\lessapprox(sr/\delta)^{\alpha}\leq(r/\delta)^{\alpha}.$ (55)

Indeed, (55) follows from (43), since if $q=T^{-1}(\tilde{q}),$ then
$\tilde{I}_{Q}(\tilde{q})=\\{T(l)\colon l\in I_{Q}(q)\\}$. Since the latter
set of lines obeys (43), the former obeys (55). c.f. Property (F) from
Proposition 4.1

Finally, for each $\tilde{q}\in\tilde{Q}$ we have the following analogue of
(44): If $v\in S^{1}$, then

$\\#\\{l\in\tilde{I}_{Q}(\tilde{q})\colon\angle(l,v)\leq\delta^{\varepsilon}\\}\leq\frac{1}{10}\\#I_{Q}(\tilde{q}).$
(56)

Again, (56) follows from (44) by the same reasoning as above. Thus
$(\tilde{Q},\tilde{L},\tilde{I})$ satisfies Property (E) from Proposition 4.1.

######  Remark 5.1.

Observe that arrangement $(\tilde{Q},\tilde{L},\tilde{I})$ satisfies Property
(E) with room to spare—if we select a smaller set of incidences
$\tilde{I}_{Q}^{\prime}\subset\tilde{I}_{Q}$ so that

$\\#\tilde{I}^{\prime}_{Q}(\tilde{q})\geq\frac{1}{4}\\#\tilde{I}_{Q}(\tilde{q})\quad\textrm{for
each}\ q\in\tilde{Q},$

then Property (E) will remain true with $\tilde{I}_{Q}^{\prime}$ in place of
$\tilde{I}_{Q}$; we will need this flexibility below.

We have shown that the arrangement $(\tilde{Q},\tilde{L},\tilde{I}_{Q})$
satisfies all of the hypotheses of Proposition 4.1 (or more precisely, an
equivalent version of these hypotheses, as discussed in Remark 4.2) except
Property (A), which is only satisfied on average. To fix this, we shall use
the following graph refinement lemma from [6].

######  Lemma 5.2 (Graph refinement).

Let $G=(A\sqcup B,E)$ be a bipartite graph. Then there is a sub-graph
$G^{\prime}=(A^{\prime}\sqcup B^{\prime},E^{\prime})$ so that
$\\#E^{\prime}\geq\\#E/2$; each vertex in $A^{\prime}$ has degree at least
$\frac{\\#E}{4\\#A}$; and each vertex in $B^{\prime}$ has degree at least
$\frac{\\#E}{4\\#B}$.

Apply Lemma 5.2 to $(\tilde{Q}\sqcup\tilde{L},\ \tilde{I}),$ and let
$(\accentset{\approx}{Q}\sqcup\accentset{\approx}{L},\
\accentset{\approx}{I})$ be the resulting refinement. We can verify that
$(\accentset{\approx}{Q}\sqcup\accentset{\approx}{L},\
\accentset{\approx}{I})$ now satisfies Properties (A) - (F) of Proposition
4.1. Indeed, by (50), each $\accentset{\approx}{q}\in\accentset{\approx}{Q}$
satisfies

$\frac{1}{4}(\\#\tilde{I})(\\#\tilde{Q})^{-1}\leq\\#\accentset{\approx}{I}(\accentset{\approx}{q})\leq(\\#\tilde{I})(\\#\tilde{Q})^{-1},$

and hence Property (E) remains true (see the discussion in Remark 5.1). (55),
and thus Property (F) remains true for similar reasons. Similarly for Property
(D).

Properties (B) and (C) remain true because
$(\accentset{\approx}{Q}\sqcup\accentset{\approx}{L},\
\accentset{\approx}{I})$ is a refinement of $(\tilde{Q}\sqcup\tilde{L},\
\tilde{I})$. Finally, Property (A) holds because by (51) and (52), for each
$\accentset{\approx}{l}\in\accentset{\approx}{L}$ we have

$\tilde{\delta}^{-\alpha}V\lessapprox\frac{1}{4}(\\#\tilde{I})(\\#\tilde{L})^{-1}\leq\\#\accentset{\approx}{I}(\accentset{\approx}{l})\lessapprox\tilde{\delta}^{-\alpha}V.$

Note as well that (47), (51) and (52) imply that

$\\#\accentset{\approx}{L}\gtrapprox\delta^{-2\alpha}(\\#\mathcal{Z})^{-1}.$
(57)

Applying Proposition 4.1, we conclude that

$\\#\tilde{Q}\geq\\#\accentset{\approx}{Q}\gtrapprox\tilde{\delta}^{-2\alpha-c(\alpha)}(\tilde{\delta}^{2\alpha}\\#\accentset{\approx}{L})V^{c(\alpha)/\alpha}\gtrapprox\tilde{\delta}^{-2\alpha-c(\alpha)}(\tilde{\delta}^{2\alpha}\delta^{-2\alpha}(\\#\mathcal{Z})^{-1})V^{c(\alpha)/\alpha},$
(58)

where the final inequality used (57). And hence by (46) and (49),

$\begin{split}\\#P_{2}&\gtrapprox(\\#\mathcal{Z})(\\#P)\\\
&\gtrapprox(\\#\mathcal{Z})M\tilde{\delta}^{-2\alpha-c(\alpha)}(\tilde{\delta}^{2\alpha}\delta^{-2\alpha}(\\#\mathcal{Z})^{-1})V^{c(\alpha)/\alpha}\\\
&\gtrapprox
M(\delta/s)^{-2\alpha-c(\alpha)}(s^{-2\alpha})V^{c(\alpha)/\alpha}\\\
&\gtrapprox
M\delta^{-2\alpha-c(\alpha)}s^{c(\alpha)}(s^{\alpha}M)^{-c(\alpha)/\alpha}\\\
&\gtrapprox\delta^{-2\alpha-c(\alpha)}M^{1-c(\alpha)/\alpha}\\\
&\geq\delta^{-2\alpha-c(\alpha)}.\end{split}$

Since $P_{2}\subset P_{1}$, we obtain (39). ∎

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##  Appendix A Proof of Proposition 3.3

In this section, we will sketch the proof of Proposition 3.3. As discussed in
Section 3, this proof does not contain any new ideas; it is simply the proof
of Theorem 1.1 from [8], except we keep track of the dependence on
$K_{0},\ldots,K_{4}$. We also preserve two distinct estimates that were merged
too early in [8], which allows for a slightly stronger final bound. The proof
sketch presented here is not self contained—the reader should compare with
Theorem 1.1 from [8].

To prove Proposition 3.3, it suffices to establish the estimate

$K_{1}^{6}K_{2}^{6+\alpha}K_{3}^{2(9+4\alpha)}K_{4}^{4(2+\alpha)}\gtrapprox\delta^{-\alpha(1-\alpha)}.$
(59)

To begin, we will record some facts about the set $A$ that were proved in [8].
For comparison, the set $A$ in the statement of Proposition 3.3 corresponds to
the set $A^{\prime}$ from [8] (In [8], $A^{\prime}\subset A$ was selected to
that $\mathcal{E}_{\rho}(A^{\prime}+A^{\prime})\leq
K_{3}\mathcal{E}_{\rho}(A^{\prime})$ for all $\delta\leq\rho\leq 1$. In
Proposition 3.3, this property already holds for $A$).

The authors in [8] use their assumed upper bound on $\mathcal{E}_{\delta}(A)$
to obtain the following lower bound on the multiplicative energy of $A$:

$\sum_{a,b\in
A}\\#\big{(}(a.A)_{\delta^{+}}\cap(b.A)_{\delta^{+}}\big{)}\gtrsim
K_{4}^{-1}(\\#A)^{3}.$ (60)

We claim that our hypothesis $\mathcal{E}_{\delta}^{\times}(A,A)\geq
K_{4}^{-1}(\\#A)^{3}$ implies the analogous statement, with $\delta$ replaced
by $2\delta$. Indeed, for each $a,b\in A$, select $a^{\prime},b^{\prime}\in
A_{\delta^{\times}}$ with $|a-a^{\prime}|<2\delta,\ |b-b^{\prime}|<2\delta$.
Then

$\\#\big{(}(a.A)_{2\delta^{+}}\cap(b.A)_{2\delta^{+}}\big{)}\gtrsim\\#\big{(}(a^{\prime}.A_{\delta^{\times}})\cap(b^{\prime}.A_{\delta^{\times}}).$

By (60) (with $2\delta$ in place of $\delta$) and pigeonholing, we conclude
that there exists $b\in A$ so that

$\sum_{a\in
A}\\#\big{(}(a.A)_{2\delta^{+}}\cap(b.A)_{2\delta^{+}}\big{)}\gtrsim
K_{4}^{-1}(\\#A)^{2}.$ (61)

(61) is the analogue of (8) from [8]. From here, the argument proceeds in a
almost identical fashion, except we track the dependence on
$K_{1},K_{2},K_{3},K_{4}$. We will briefly highlight a few of the key
estimates from [8], and give the analogous statements.

The authors in [8] select a number $K_{4}^{-1}\leq\rho\leq 1$ and a set
$\bar{A}\subset A$, with $\\#\bar{A}\gtrsim|\log\delta|^{-1}(\\#A)$, so that

$\\#\big{(}(aA)_{2\delta^{+}}\cap(bA)_{2\delta^{+}}\big{)}\sim(\\#A)K_{4}^{-1}\rho^{-1}\quad\textrm{for
each}\ a\in\bar{A}.$

They then select a set $A_{1}\subset\bar{A}$ with
$\\#A_{1}\gtrapprox\\#\bar{A}$, so that $A_{1}+A_{1}$ has small
$\rho$-covering number for all $\delta\leq\rho\leq 1$. The key estimates we
need are the following analogues of Lemmas 3.6 and 3.7 from [8]. Let
$a_{1},a_{2},b_{2},b_{2}\in A_{1}$ and let $d_{1}=a_{1}-b_{1},$
$d_{2}=a_{2}-b_{2}$. Then

$\mathcal{E}_{\delta}(d_{1}A+d_{2}A)\lessapprox
K_{1}K_{2}K_{3}^{8}K_{4}^{4}\rho^{4}\max(|d_{1}|,|d_{2}|)^{\alpha}\\#(A),$
(62)

and for each $k\geq 2$ there is a set $A_{2}\subset A_{1}$, with
$\\#A_{2}\geq\frac{1}{4}\\#A_{1}$, so that

$|d_{1}A_{2}+\underbrace{d_{2}A_{2}+\ldots+d_{2}A_{2}}_{\text{$k$
times}}|\lessapprox
K_{1}K_{2}K_{3}^{7+k}K_{4}^{4}\rho^{5-k}\max(|d_{1}|,|d_{2}|)^{\alpha}\\#(A).$
(63)

Next, let $s$ and $\gamma$ be parameters that we will specify below.
Proceeding as in [8], we define

$B=\Big{\\{}\frac{a_{1}-a_{2}}{a_{3}-a_{4}}\colon a_{i}\in A_{1},\
|a_{3}-a_{4}|>\delta^{\gamma}\Big{\\}}.$

Lemma 4.1 from [8] says that either there is a point $b\in B\cap[0,1]$ so that
both $b/2$ and $(b+1)/2$ have distance at least $s$ from $B$, or the
$s$-covering number of $B\cap[0,1]$ is comparable to $s^{-1}$. The former
situation is called the “gap case,” while the latter is called the “dense
case.”

Our arguments in both of these cases mirror those of [8], with (62) and (63)
in place of Lemmas 3.6 and 3.7. The arguments from [8] show that in the dense
case, we must have either

$K_{2}K_{3}^{8}K_{4}^{4}\gtrapprox\delta^{\alpha}s^{-1}\quad\textrm{and/or}\quad
K_{1}^{3}K_{2}K_{3}^{8}K_{4}^{4}\gtrapprox\delta^{-\gamma\alpha},$ (64)

while in the gap case we must have

$K_{1}^{3}K_{2}K_{3}^{10}K_{4}^{4}\gtrapprox\left(\delta^{\gamma-1}s\right)^{\alpha}.$
(65)

Until this point, the arguments above have proceeded in parallel to those in
[8]. The improvement over [8] comes from properly choosing $s$ and $\gamma$;
we choose $s$ and $\gamma$ so that

$s=\left(K_{1}^{3}K_{2}^{2}K_{3}\delta^{3\alpha/2}\right)^{2/(\alpha+2)},\quad\delta^{\gamma}=K_{3}^{1/\alpha}\delta^{1/2}s^{-1/2}.$
(66)

We can suppose that $0<s<1$, since otherwise
$K_{1}^{3}K_{2}^{2}K_{3}>\delta^{-3\alpha/2}$, which is a stronger estimate
than (59) (recall that by assumption, $K_{4}\geq 1$). Similarly, we can
suppose that $0<\delta^{\gamma}<1$ (and hence $\gamma>0$), since otherwise
$K_{3}^{4/\alpha}\delta^{2-2\alpha}\geq\left(K_{1}^{6}K_{2}^{4}\right)\geq 1,$
and hence $K_{3}\geq\delta^{-\frac{\alpha(2-\alpha)}{4}},$ which again is a
stronger estimate than (59) (recall that by assumption, $K_{1},K_{2}K_{4}\geq
1$ and $\alpha\in(0,1)$).

Since at least one of the three inequalities from (64) and (65) must hold, we
can verify that with this choice of $s$ and $\gamma$, (64) and (65) imply
(59).