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A structural Szemerédi–Trotter Theorem for Cartesian Productsthanks: This research project was done as part of the 2020 NYC Discrete Math REU, supported by NSF awards DMS-1802059, DMS-1851420, DMS-1953141, and DMS-2028892.

Adam Sheffer Department of Mathematics, Baruch College, City University of New York, NY, USA. [email protected]. Supported by NSF award DMS-1802059 and by PSCCUNY award 63580.    Olivine Silier California Institute of Technology, Pasadena, CA 91125. [email protected]. Supported by the Lynn A. Booth and Kent Kresa SURF Fellowship
Abstract

We study configurations of nn points and nn lines that form Θ(n4/3)\Theta(n^{4/3}) incidences, when the point set is a Cartesian product. We prove structural properties of such configurations, such that there exist many families of parallel lines or many families of concurrent lines. We show that the line slopes have multiplicative structure or that many sets of yy-intercepts have additive structure. We introduce the first infinite family of configurations with Θ(n4/3)\Theta(n^{4/3}) incidences. We also derive a new variant of a different structural point–line result of Elekes.

Our techniques are based on the concept of line energy. Recently, Rudnev and Shkredov introduced this energy and showed how it is connected to point–line incidences. We also prove that their bound is tight up to sub-polynomial factors.

1 Introduction

The Szemerédi–Trotter theorem [24] is a central result in discrete geometry. The many variants and generalizations of this theorem are now considered as an entire subfield. The Szemerédi-Trotter theorem and its variants have a wide range of applications, in combinatorics, harmonic analysis, theoretical computer science, number theory, model theory, and more (for example, see [2, 3, 4, 14, 22]). It is thus awkward that hardly anything is known about the cases where this theorem is asymptotically tight.

Let 𝒫\mathcal{P} be a set of points and let \mathcal{L} be a set of lines, both in 2\mathbb{R}^{2}. An incidence is a pair (p,)𝒫×(p,\ell)\in\mathcal{P}\times\mathcal{L}, such that the point pp is on the line \ell. We denote the number of incidences in 𝒫×\mathcal{P}\times\mathcal{L} as I(𝒫,)I(\mathcal{P},\mathcal{L}). The Szemerédi–Trotter theorem provides an asymptotically tight upper bound for I(𝒫,)I(\mathcal{P},\mathcal{L}).

Theorem 1.1.

(Szemerédi and Trotter) Let 𝒫\mathcal{P} be a set of mm points and let \mathcal{L} be a set of nn lines, both in 2\mathbb{R}^{2}. Then

I(𝒫,)=O(m2/3n2/3+m+n).I(\mathcal{P},\mathcal{L})=O(m^{2/3}n^{2/3}+m+n).

The structural Szemerédi–Trotter problem asks to characterize the point–line configurations that have an asymptotically maximal number of incidences. We focus on sets of nn points and nn lines that form Θ(n4/3)\Theta(n^{4/3}) incidences. Most of our results can be extended to mm point and nn lines.

Around the middle of the 20th century, Erdős [9] discovered a configuration of nn points and nn lines that form Θ(n4/3)\Theta(n^{4/3}) incidences. In this construction, the point set is a n×n\sqrt{n}\times\sqrt{n} section of the integer lattice 2\mathbb{Z}^{2}. In the early 2000s, Elekes [8] discovered a simpler configuration of nn points and nn lines that form Θ(n4/3)\Theta(n^{4/3}) incidences. In Elekes’s configuration, the point set is an n1/3×n2/3n^{1/3}\times n^{2/3} section of the integer lattice 2\mathbb{Z}^{2}.

To obtain more point-line configurations, we can take one of the above configurations and perform transformations that preserve most of the incidences. We can apply projective transformations, point–line duality, remove some points and lines, and add additional points and lines. Up to such transformations, the only known configurations with Θ(n4/3)\Theta(n^{4/3}) incidences were Erdős’s and Elekes’s. This led to questions such as

  • Are these the only two configurations?

  • If there are more configurations, are there only sporadic configurations? Or is there an infinite family of configurations with continuous parameters?

  • Up to the above transformations, is the point set always a lattice?

  • What properties must the set of lines satisfy?

Hardly anything is known about such structural questions. The only non-trivial result that we are aware of is by Solymosi [23].

Theorem 1.2.

(Solymosi) For every constant integer kk, the following holds for every sufficiently large nn. Let 𝒫\mathcal{P} be a set of nn points and let \mathcal{L} be a set of nn lines, both in 2\mathbb{R}^{2}, such that I(𝒫,)=Θ(n4/3)I(\mathcal{P},\mathcal{L})=\Theta(n^{4/3}). Then there exists a set of kk of the points, no three on a line, such that there is a line of \mathcal{L} passing through each of the (k2)\binom{k}{2} point pairs.

A recent result of Mirzaei and Suk [15] can be seen as quantitative variant of Theorem 1.2, for specific types of subgraphs. A recent result of Hanson, Roche-Newton, and Zhelezov [12, Theorem 1.8] considers the case where the point set is A×AA\times A, where AA is a set of rational numbers with a very small product set. They show that, in this case, the number of incidences is significantly smaller than n4/3n^{4/3}.

Our structural results. In this work, we prove several results for the structural Szemerédi–Trotter problem. First, we provide the first infinite family of point–line configurations with Θ(n4/3)\Theta(n^{4/3}) incidences. The constructions of Erdős and Elekes are the two extreme cases of this family.

Theorem 1.3.

The following holds for every 1/3<α<1/21/3<\alpha<1/2. Let 𝒫\mathcal{P} be a section of the integer lattice 2\mathbb{Z}^{2} of size nα×n1αn^{\alpha}\times n^{1-\alpha}. Then there exists a set \mathcal{L} of nn lines such that

I(𝒫,)=Θ(n4/3).I(\mathcal{P},\mathcal{L})=\Theta(n^{4/3}).

For a proof of Theorem 1.3 and additional details, see Section 4. This family of constructions answers some structural questions and leads to new ones. For example, in all of the constructions of this family, there are Θ(n1/3)\Theta(n^{1/3}) families of Θ(n2/3)\Theta(n^{2/3}) parallel lines. We wonder whether there exist configurations with Θ(n4/3)\Theta(n^{4/3}) incidences that do not have this property, possibly after applying some transformations.

A follow-up work by Larry Guth and the second author [11] provides constructions where the point set is a Cartesian product of generalized arithmetic progressions. This prompts the question: Do all configurations consist of a Cartesian product of sets with a small sum set? For the definition of a sum set and more information, see Section 3.

We derive properties of the set of lines in the case where the point set is a Cartesian product. Part (b) of the this result relies on additive and multiplicative energies. For a definition of these energies, see Section 3.

Theorem 1.4.

      
(a) For 1/3<α<1/21/3<\alpha<1/2, let A,BA,B\subset\mathbb{R} satisfy |A|=nα|A|=n^{\alpha} and |B|=n1α|B|=n^{1-\alpha}. Let \mathcal{L} be a set of nn lines in 2\mathbb{R}^{2}, such that I(A×B,)=Θ(n4/3)I(A\times B,\mathcal{L})=\Theta(n^{4/3}). Then at least one of the following holds:

  • There exists 12αβ2/31-2\alpha\leq\beta\leq 2/3 such that \mathcal{L} contains Ω(n1β/logn)\Omega(n^{1-\beta}/\log n) families of Θ(nβ)\Theta(n^{\beta}) parallel lines, each with a different slope.

  • There exists 1αγ2/31-\alpha\leq\gamma\leq 2/3 such that \mathcal{L} contains Ω(n1γ/logn)\Omega(n^{1-\gamma}/\log n) disjoint families of Θ(nγ)\Theta(n^{\gamma}) concurrent lines.

(b) Assume that we are in the case of Ω(n1β/logn)\Omega(n^{1-\beta}/\log n) families of Θ(nβ)\Theta(n^{\beta}) parallel lines. There exists n2βtn3βn^{2\beta}\leq t\leq n^{3\beta} such that, for Ω(n1β/log2n)\Omega(n^{1-\beta}/\log^{2}n) of these families, the additive energy of the yy-intercepts is Θ(t)\Theta(t). Let SS be the set of slopes of these families. Then

E×(S)t=Ω(n3α/log12n).E^{\times}(S)\cdot t=\Omega(n^{3-\alpha}/\log^{12}n).

We can think of Theorem 1.4(b) as another split into two cases: Either the set of slopes has a large multiplicative energy or many sets of yy-intercepts have large additive energies. More intuitively and less accurately, either the set of slopes is somewhat similar to a geometric progression, or many sets of yy-intercepts are somewhat similar to arithmetic progressions. In all of the configurations that we are aware of, there are Θ(n1/3)\Theta(n^{1/3}) sets of Θ(n2/3)\Theta(n^{2/3}) parallel lines, the energy E×(S)E^{\times}(S) is small, and the sets of yy-intercepts have large additive energies. One possibility is that all of the constructions have these properties. For more information, see Section 4.

For the proof of Theorem 1.4, see Section 5. A follow-up work of the first author and Junxuan Shen [20] shows that, for lattices and several other cases, the concurrent case of Theorem 1.4(a) cannot happen. Thus, if the concentric case can happen, then it takes place with very different point sets.

The current work contains several additional related results, which are stated in the part titled Additional results below.

Line energy. Not knowing much about the structural Szemerédi–Trotter problem is part of a more general phenomena. We do not know much about the structural variants of most of the related problems. For example, not much is known about the structural distinct distances theorem (for more information, see [18]). The main exception is the characterization of sets with few ordinary lines by Green and Tao [10]. In that paper, Green and Tao find structure by relying on tools from additive combinatorics. In the current work, we derive other structural results using different tools from additive combinatorics.

Elekes [5, 6, 7] considered a line \ell that is defined by y=ax+by=ax+b as the linear function f(x)=ax+bf_{\ell}(x)=ax+b. He was then able to compose lines and to consider the inverse of a line. This allowed Elekes to derive new combinatorial properties of lines. Recently, Rudnev and Shkredov [17] further developed Elekes’s ideas. Given a set \mathcal{L} of non-axis-parallel lines, they consider the quantity

|{(1,2,3,4)4:f11f2=f31f4}|.\left|\left\{(\ell_{1},\ell_{2},\ell_{3},\ell_{4})\in\mathcal{L}^{4}\ :\ f_{\ell_{1}}^{-1}\circ f_{\ell_{2}}=f_{\ell_{3}}^{-1}\circ f_{\ell_{4}}\right\}\right|.

We refer to this quantity as the line energy of \mathcal{L} and denote it as E()E(\mathcal{L}). A detailed and rigorous discussion of line energy can be found in Section 3.

Rudnev and Shkredov derived an interesting connection between point-line incidences and line energy.

Theorem 1.5.

(Rudnev and Shkredov) Let A,BA,B\subset\mathbb{R} be finite sets and let \mathcal{L} be a set of non-axis-parallel lines in 2\mathbb{R}^{2}. Then

I(A×B,)=O(|B|1/2|A|2/3E()1/6||1/3+|B|1/2||).I(A\times B,\mathcal{L})=O\left(|B|^{1/2}|A|^{2/3}E(\mathcal{L})^{1/6}|\mathcal{L}|^{1/3}+|B|^{1/2}|\mathcal{L}|\right).

Rudnev and Shkredov also derived upper bounds for E()E(\mathcal{L}). The following upper bound was derived afterwards by Petridis, Roche-Newton, Rudnev, and Warren [16].

Theorem 1.6.

(Petridis, Roche-Newton, Rudnev, and Warren) Let \mathcal{L} be a set of non-axis-parallel lines in 2\mathbb{R}^{2}, such that at most mm lines have the same slope and every point is incident to at most MM lines. Then

E()=O(m1/2||5/2+M||2).E(\mathcal{L})=O\left(m^{1/2}|\mathcal{L}|^{5/2}+M|\mathcal{L}|^{2}\right).

The paper [16] contains a dual formulation of Theorem 1.6. When reading that dual formulation, it may seem as if we need to add the term m||2m|\mathcal{L}|^{2} to the bound of Theorem 1.6. However, since m||m\leq|\mathcal{L}|, we have that m||2m1/2||5/2m|\mathcal{L}|^{2}\leq m^{1/2}|\mathcal{L}|^{5/2}.

Additional results. In Section 4, we show that the main term of Theorem 1.5 is tight up to sub-polynomial factors.

Theorem 1.7.

For every ε>0{\varepsilon}>0, in the first term of the bound of Theorem 1.5, no exponent could be decreased by an ε{\varepsilon}.

Elekes [7] studies lines that contain many points of a Cartesian product.

Theorem 1.8.

(Elekes) There exists a constant c>0c>0 that satisfies the following for every nn. Let AA be a set of nn reals and let 0<α<10<\alpha<1. Let \mathcal{L} be a set of non-axis-parallel lines, each incident to at least αn\alpha n points of A×AA\times A. Then at least one of the following holds:

  • There exist αcn\alpha^{c}n lines of \mathcal{L} that are parallel.

  • There exist αcn\alpha^{c}n lines of \mathcal{L} that are concurrent.

Petridis, Roche-Newton, Rudnev, and Warren [16] derived a quantitative variant of Theorem 1.8.

Theorem 1.9.

(Petridis, Roche-Newton, Rudnev, and Warren) Let AA be a set of nn reals and let α<1\alpha<1 satisfy α=Ω(n1/2)\alpha=\Omega(n^{-1/2}). Let \mathcal{L} be a set of kk non-axis-parallel lines, each incident to at least αn\alpha n points of A×AA\times A. Then at least one of the following holds:

  • There exist α12n2k3\alpha^{12}n^{-2}k^{3} lines of \mathcal{L} that are parallel and E+(A)=Ω(α14k3)E^{+}(A)=\Omega(\alpha^{14}k^{3}).

  • There exist α6n1k2\alpha^{6}n^{-1}k^{2} lines of \mathcal{L} that are concurrent. Also, there exists ss such that E×(As)=Ω(α8nk2)E^{\times}(A-s)=\Omega(\alpha^{8}nk^{2}). (The set AsA-s is obtained by subtracting ss from every element of AA.)

While Theorems 1.8 and 1.9 seem similar to Theorem 1.4, there are several significant differences between these two structural problems. First, it is not difficult to find a configuration that satisfies the concurrent case of Theorem 1.8. On the other hand, it is plausible that the concurrent case of Theorem 1.4 does not exist.

To see another difference between the two above scenarios, we apply Theorem 1.9 to the structural problem of Theorem 1.4. When nn points and nn lines form Θ(n4/3)\Theta(n^{4/3}) incidences, those incidences originate from Θ(n)\Theta(n) lines that are incident to Θ(n1/3)\Theta(n^{1/3}) points (see the proof of Lemma 5.1 below). In our case there are |A|2=n2|A|^{2}=n^{2} points, so Θ(n2)\Theta(n^{2}) lines are incident to Θ(n2/3)\Theta(n^{2/3}) points. That is, k=n2k=n^{2} and α=Θ(n1/3)\alpha=\Theta(n^{-1/3}). Then, Theorem 1.9 leads to the trivial statement that there exist Ω(1)\Omega(1) parallel lines. While the two structural problems have a different behavior, the proofs of Theorem 1.9 and of Theorem 1.4(a) both rely on Theorem 1.5 and Theorem 1.6.

We introduce a variant of Theorem 1.9 that does lead to interesting results in the case of Θ(n4/3)\Theta(n^{4/3}) incidences. This variant is in the style of Theorem 1.4(b). By the pigeonhole principle, there exists 0β10\leq\beta\leq 1 such that \mathcal{L} contains Ω(k1β/logk)\Omega(k^{1-\beta}/\log k) families of Θ(kβ)\Theta(k^{\beta}) parallel lines, each with a different slope. We provide an upper bound on kk in terms of the multiplicative energy of these slopes and of the additive energy of their yy-intercepts.

Theorem 1.10.

Let AA be a set of nn reals and let α<1\alpha<1 satisfy α=Ω(n1/2log2k)\alpha=\Omega(n^{-1/2}\log^{2}k). Let \mathcal{L} be a set of kk non-axis-parallel lines, each incident to at least αn\alpha n points of A×AA\times A. Consider 0β10\leq\beta\leq 1 such that \mathcal{L} contains Ω(k1β/logk)\Omega(k^{1-\beta}/\log k) families of Θ(kβ)\Theta(k^{\beta}) parallel lines. There exists k2βtk3βk^{2\beta}\leq t\leq k^{3\beta} such that, for Ω(k1β/log2k)\Omega(k^{1-\beta}/\log^{2}k) of these families, the additive energy of the yy-intercepts is Θ(t)\Theta(t). Let SS be the set of slopes of these families. Then

k=O(n1/4t1/4E×(S)1/4log3nα3/2).k=O\left(\frac{n^{1/4}t^{1/4}E^{\times}(S)^{1/4}\log^{3}n}{\alpha^{3/2}}\right).

The statement of Theorem 1.10 is long and technical, but it also has a simple intuition: If many lines are incident to many points of A×AA\times A, then the slopes have a large multiplicative energy or the yy-intercepts of many parallel families have large additive energies. For a proof of Theorem 1.10, see Section 6.

To see that Theorem 1.10 provides a reasonable bound in the case of Θ(n4/3)\Theta(n^{4/3}) incidences, we consider Erdős’s construction with n2n^{2} points and lines (see for example [19]). Like all of the known constructions, the set of lines contains n2/3n^{2/3} families of n4/3n^{4/3} parallel lines. Each of those lines forms Θ(n2/3)\Theta(n^{2/3}) incidences, so α=n1/3\alpha=n^{-1/3}. A simple variant of our proof of Lemma 4.1 shows that E×(S)=O(n4/3+ε)E^{\times}(S)=O(n^{4/3+{\varepsilon}}), for every ε>0{\varepsilon}>0. The sets of yy-intercepts are arithmetic progressions, so t=Θ(n4)t=\Theta(n^{4}). Plugging this into Theorem 1.10 leads to the bound k=O(n25/12+ε)k=O(n^{25/12+{\varepsilon}}). This is not far from the correct bound k=Θ(n2)k=\Theta(n^{2}), so Theorem 1.10 cannot be significantly improved.

For the opposite case of Theorem 1.10, we set A={20,21,22,,2n1}A=\{2^{0},2^{1},2^{2},\ldots,2^{n-1}\} and consider the lines that are defined by y=2jxy=2^{j}\cdot x with j{0,1,2,,n/2}j\in\{0,1,2,...,n/2\}. In this case, there are kk families of a single parallel line, so t=1t=1. Every line contains Θ(n)\Theta(n) points of A×AA\times A, so α=Θ(1)\alpha=\Theta(1). The line slopes form a geometric progression, so E×(S)=Θ(n3)E^{\times}(S)=\Theta(n^{3}). Plugging the above into Theorem 1.10 leads to the bound k=O(nlog3n)k=O(n\log^{3}n), which is tight up to the polylogarithmic factor.

Many of the results in this paper can be extended to other fields, to mm points and nn lines, and to other problems.

Acknowledgments. The authors thank Misha Rudnev, Junxuan Shen, Ilya Shkredov, and Audie Warren for helpful conversations. We also thank Chi Hoi Yip and the anonymous referees for spotting issues and helping to improve this work.

2 Preliminaries

In this section we introduce a few tools that we use in our proofs. We suggest to quickly skim this section and return to it when necessary.

Euler’s totient function. For a positive integer bb, the Euler totient function of bb is

ϕ(b)=|{a: 1a<b and GCD(a,b)=1}|.\phi(b)=\left|\{\ a\ :\ 1\leq a<b\quad\text{ and }\quad\mathrm{GCD}(a,b)=1\}\right|.

The following lemma collects a few basic properties of the totient function.

Lemma 2.1.

Let rr be a positive integer. Then
(a) i=1rϕ(i)=3π2r2+O(rlogr)\displaystyle\sum_{i=1}^{r}\phi(i)=\frac{3}{\pi^{2}}r^{2}+O(r\log r),
(b) i=1riϕ(i)=Θ(r3)\displaystyle\sum_{i=1}^{r}i\cdot\phi(i)=\Theta(r^{3}),
(c) i=1rϕ(i)/i2=O(logr)\displaystyle\sum_{i=1}^{r}\phi(i)/i^{2}=O(\log r).

Proof.

Part (a) is a classic formula. For example, see [13, Theorem 330]. For part (b), we note that

i=1riϕ(i)\displaystyle\sum_{i=1}^{r}i\cdot\phi(i) ri=1rϕ(i)=O(r3), and that\displaystyle\leq r\cdot\sum_{i=1}^{r}\phi(i)=O(r^{3}),\quad\text{ and that }
i=1riϕ(i)\displaystyle\sum_{i=1}^{r}i\cdot\phi(i) >i=r/2riϕ(i)>(r/2)(i=1rϕ(i)i=1r/2ϕ(i))=Ω(r3).\displaystyle>\sum_{i=r/2}^{r}i\cdot\phi(i)>(r/2)\left(\sum_{i=1}^{r}\phi(i)-\sum_{i=1}^{r/2}\phi(i)\right)=\Omega(r^{3}).

Combining the above bounds implies that i=1riϕ(i)=Θ(r3)\sum_{i=1}^{r}i\cdot\phi(i)=\Theta(r^{3}).

By definition, ϕ(i)i\phi(i)\leq i holds for every positive integer ii. This implies that

i=1rϕ(i)/i2i=1r1/i=O(logn).\sum_{i=1}^{r}\phi(i)/i^{2}\leq\sum_{i=1}^{r}1/i=O(\log n).

Rich lines. Let 𝒫\mathcal{P} be a set of points in 2\mathbb{R}^{2}. For an integer r2r\geq 2, a line 2\ell\subset\mathbb{R}^{2} is rr-rich with respect to 𝒫\mathcal{P} if \ell is incident to at least rr points of 𝒫\mathcal{P}. The following result is often called a dual Szemerédi–Trotter theorem. This is because each of Theorems 1.1 and 2.2 can be derived from the other by only using elementary arguments.

Theorem 2.2.

Let 𝒫\mathcal{P} be a set of mm points in 2\mathbb{R}^{2} and let rr be an integer larger than one. Then, the number of rr-rich lines with respect to 𝒫\mathcal{P} is

O(m2r3+mr).O\left(\frac{m^{2}}{r^{3}}+\frac{m}{r}\right).

3 Energies

In this section we describe line energy in more detail. We begin with a brief description of additive and multiplicative energies. An expert reader may wish to skip this part.

Additive and multiplicative energies. Let AA\subset\mathbb{R} be a finite set. The sum set of AA is

A+A={a+a:a,aA}.A+A=\{a+a^{\prime}\ :\ a,a^{\prime}\in A\}.

It is not difficult to verify that |A+A|=Ω(|A|)|A+A|=\Omega(|A|) and |A+A|=O(|A|2)|A+A|=O(|A|^{2}). Intuitively, a small sum set implies that AA is similar to an arithmetic progression. The additive energy of AA is

E+(A)=|{(a,b,c,d)A4:a+b=c+d}|.E^{+}(A)=|\{(a,b,c,d)\in A^{4}\ :\ a+b=c+d\}|.

After fixing the values of a,ba,b, and cc, at most one dAd\in A satisfies a+b=c+da+b=c+d. This implies that E+(A)|A|3E^{+}(A)\leq|A|^{3}. By considering the case where a=ca=c and b=db=d, we get that E+(A)|A|2E^{+}(A)\geq|A|^{2}.

Additive energy is a central object in additive combinatorics. Intuitively, it provides information about the structure of AA under addition. We now consider one example of this. For xx\in\mathbb{R}, we define

rA+(x)\displaystyle r_{A}^{+}(x) =|{(a,a)A2:a+a=x}|,\displaystyle=|\{(a,a^{\prime})\in A^{2}\ :\ a+a^{\prime}=x\}|,
rA(x)\displaystyle r_{A}^{-}(x) =|{(a,a)A2:aa=x}|.\displaystyle=|\{(a,a^{\prime})\in A^{2}\ :\ a-a^{\prime}=x\}|.

Since every pair from A2A^{2} contributes to exactly one rA+(x)r_{A}^{+}(x), we have that xrA+(x)=|A|2\sum_{x\in\mathbb{R}}r_{A}^{+}(x)=|A|^{2}. For a fixed xx\in\mathbb{R}, the number of solutions from AA to a+b=c+d=xa+b=c+d=x is rA+(x)2r_{A}^{+}(x)^{2}. This implies that E+(A)=xrA+(x)2E^{+}(A)=\sum_{x}r_{A}^{+}(x)^{2}. Then, the Cauchy-Schwarz inequality leads to

E+(A)=xA+ArA+(x)2(xA+ArA+(x))2xA+A1=(|A|2)2|A+A|=|A|4|A+A|.E^{+}(A)=\sum_{x\in A+A}r_{A}^{+}(x)^{2}\geq\frac{\left(\sum_{x\in A+A}r_{A}^{+}(x)\right)^{2}}{\sum_{x\in A+A}1}=\frac{(|A|^{2})^{2}}{|A+A|}=\frac{|A|^{4}}{|A+A|}.

Thus, a small sum set implies a large additive energy. In particular, if |A+A|=Θ(|A|)|A+A|=\Theta(|A|) then E+(A)=Ω(|A|3)E^{+}(A)=\Omega(|A|^{3}).

For finite A,BA,B\subset\mathbb{R}, we define the additive energy of AA and BB as

E+(A,B)=|{(a,b,c,d)A×B×A×B:a+b=c+d}|.E^{+}(A,B)=|\{(a,b,c,d)\in A\times B\times A\times B\ :\ a+b=c+d\}|.

We can rearrange the above condition as ac=dba-c=d-b. Then, the Cauchy-Schwarz inequality implies that

E+(A,B)=xrA(x)rB(x)\displaystyle E^{+}(A,B)=\sum_{x\in\mathbb{R}}r_{A}^{-}(x)\cdot r_{B}^{-}(x) (xrA(x)2)1/2(xrB(x)2)1/2\displaystyle\leq\Big{(}\sum_{x\in\mathbb{R}}r_{A}^{-}(x)^{2}\Big{)}^{1/2}\cdot\Big{(}\sum_{x\in\mathbb{R}}r_{B}^{-}(x)^{2}\Big{)}^{1/2}
=E+(A)1/2E+(B)1/2.\displaystyle\hskip 85.35826pt=E^{+}(A)^{1/2}\cdot E^{+}(B)^{1/2}. (1)

The product set of AA is

AA={aa:a,aA}.AA=\{a\cdot a^{\prime}\ :\ a,a^{\prime}\in A\}.

The multiplicative energy of AA is

E×(A)=|{(a,b,c,d)A4:ab=cd}|.E^{\times}(A)=|\{(a,b,c,d)\in A^{4}\ :\ a\cdot b=c\cdot d\}|. (2)

Product sets and multiplicative energies are similar to sum sets and additive energies. For example, a small product set implies a large multiplicative energy. Intuitively, a set with a small product set is similar to a geometric progression.

Line energy. In the following, when considering lines, we refer only to non-axis-parallel lines in 2\mathbb{R}^{2}. We associate a line \ell with the function f(x)=cx+df_{\ell}(x)=cx+d. The composition of the lines \ell and \ell^{\prime} that are defined by y=cx+dy=cx+d and y=cx+dy=c^{\prime}x+d^{\prime} is

=ff=cf(x)+d=ccx+cd+d.\ell\circ\ell^{\prime}=f_{\ell}\circ f_{\ell^{\prime}}=c\cdot f_{\ell^{\prime}}(x)+d=cc^{\prime}x+cd^{\prime}+d.

The set of non-axis-parallel lines under the above composition form the affine group of \mathbb{R}. However, for our purposes, we prefer to think of the group elements as lines in 2\mathbb{R}^{2}. We also associate a line \ell that is defined by y=cx+dy=cx+d with the point (c,d)2(c,d)\in\mathbb{R}^{2}. We refer to the plane that contains \ell as the primal plane and to the plane that contains (c,d)(c,d) as the dual plane. With this dual notation, the group operation is

(c,d)(c,d)=(cc,cd+d).(c,d)\circ(c^{\prime},d^{\prime})=(cc^{\prime},cd^{\prime}+d). (3)

The identity element is (1,0)(1,0) and the inverse of (c,d)(c,d) is (1/c,d/c)(1/c,-d/c). This group is non-commutative. For example, we have that

(2,3)(2,4)=(4,11)(4,10)=(2,4)(2,3).(2,3)\circ(2,4)=(4,11)\neq(4,10)=(2,4)\circ(2,3).

We define the line energy of a set of lines \mathcal{L} as

E()=|{(1,2,3,4)4:f11f2=f31f4}|.E(\mathcal{L})=\left|\left\{(\ell_{1},\ell_{2},\ell_{3},\ell_{4})\in\mathcal{L}^{4}\ :\ f_{\ell_{1}}^{-1}\circ f_{\ell_{2}}=f_{\ell_{3}}^{-1}\circ f_{\ell_{4}}\right\}\right|.

When writing fj=cjx+djf_{\ell_{j}}=c_{j}x+d_{j}, we also have the dual formulation

E()=|{(1,2,3,4)4:(c1,d1)1(c2,d2)=(c3,d3)1(c4,d4)}|.E(\mathcal{L})=\left|\left\{(\ell_{1},\ell_{2},\ell_{3},\ell_{4})\in\mathcal{L}^{4}\ :\ (c_{1},d_{1})^{-1}\circ(c_{2},d_{2})=(c_{3},d_{3})^{-1}\circ(c_{4},d_{4})\right\}\right|.

Rudnev and Shkredov [17] refer to E()E(\mathcal{L}) as an energy, but do not give it an explicit name. The recent paper [16] refers to E()E(\mathcal{L}) as affine energy, because of its connection with the affine group. We use the name line energy, since think of E()E(\mathcal{L}) as a property of a set of lines.

We note that

(c,d)1(c,d)=(1c,dc)(c,d)=(cc,ddc).(c,d)^{-1}\circ(c^{\prime},d^{\prime})=\left(\frac{1}{c},\frac{-d}{c}\right)\circ(c^{\prime},d^{\prime})=\left(\frac{c^{\prime}}{c},\frac{d^{\prime}-d}{c}\right).

Since we only consider non-axis-parallel lines, c0c\neq 0 and the above is well-defined. A quadruple (1,2,3,4)4(\ell_{1},\ell_{2},\ell_{3},\ell_{4})\in\mathcal{L}^{4} contributes to E()E(\mathcal{L}) if and only if

(c2c1,d2d1c2)=(c4c3,d4d3c4).\left(\frac{c_{2}}{c_{1}},\frac{d_{2}-d_{1}}{c_{2}}\right)=\left(\frac{c_{4}}{c_{3}},\frac{d_{4}-d_{3}}{c_{4}}\right).

Simplifying leads to the system

c2c3\displaystyle c_{2}\cdot c_{3} =c1c4,\displaystyle=c_{1}\cdot c_{4}, (4)
c3(d2d1)\displaystyle c_{3}(d_{2}-d_{1}) =c1(d4d3).\displaystyle=c_{1}(d_{4}-d_{3}). (5)

3.1 Cartesian products of lines

In the dual plane, we can consider a set of lines that is a Cartesian product. That is, a Cartesian product C×DC\times D consists of the points that are dual to lines with a slope from CC and a yy-intercept from DD. The following bound for Cartesian products of lines is a warm-up towards similar arguments that we use in the following sections.

Theorem 3.1.

Consider finite sets C,D{0}C,D\subset\mathbb{R}\setminus\{0\}. Let \mathcal{L} be a set of lines in 2\mathbb{R}^{2} that is dual to C×DC\times D. Then

E(C×D)E×(C)E+(D).E(C\times D)\leq E^{\times}(C)\cdot E^{+}(D). (6)
Proof.

The number of solutions to (4) is E×(C)E^{\times}(C). We fix a solution to (4) and derive an upper bound on the number of corresponding solutions to (5). That is, we fix c1,c2,c3,c4c_{1},c_{2},c_{3},c_{4}, and consider the number of valid values for d1,d2,d3,d4d_{1},d_{2},d_{3},d_{4}.

We set s=c1/c3s=c_{1}/c_{3} and rephrase (5) as

d2d1=s(d4d3).d_{2}-d_{1}=s(d_{4}-d_{3}). (7)

By the Cauchy-Schwarz inequality, the number of solutions to (7) is

krD(k)rD(k/s)\displaystyle\sum_{k\in\mathbb{R}}r^{-}_{D}(k)r^{-}_{D}(k/s) (krD(k)2)1/2(krD(k/s)2)1/2\displaystyle\leq\left(\sum_{k\in\mathbb{R}}r^{-}_{D}(k)^{2}\right)^{1/2}\left(\sum_{k\in\mathbb{R}}r^{-}_{D}(k/s)^{2}\right)^{1/2}
=(E+(D))1/2(E+(D))1/2=E+(D).\displaystyle\hskip 142.26378pt=\left(E^{+}(D)\right)^{1/2}\left(E^{+}(D)\right)^{1/2}=E^{+}(D).

There are E×(C)E^{\times}(C) solutions to (4) and each corresponds to at most E+(D)E^{+}(D) solutions to (5). This implies that E(C×D)E×(C)E+(D)E(C\times D)\leq E^{\times}(C)\cdot E^{+}(D). ∎

To put Theorem 3.1 in context, we consider Elekes’s construction from [8]. In particular, we consider the point set

𝒫={(i,j): 1in1/3/2 and 1j2n2/3},\mathcal{P}=\left\{\ (i,j)\ :\ 1\leq i\leq n^{1/3}/2\quad\text{ and }\quad 1\leq j\leq 2n^{2/3}\ \right\},

and the set of lines

={y=ax+b: 1an1/3 and 1bn2/3}.\mathcal{L}=\left\{\ y=ax+b\ :\ 1\leq a\leq n^{1/3}\quad\text{ and }\quad 1\leq b\leq n^{2/3}\ \right\}. (8)

We note that |𝒫|=||=n|\mathcal{P}|=|\mathcal{L}|=n, that I(𝒫,)=Θ(n4/3)I(\mathcal{P},\mathcal{L})=\Theta(n^{4/3}), and that the set of lines is a Cartesian product.

Claim 3.2.

The set \mathcal{L} from (8) satisfies that E()=O(n8/3logn)E(\mathcal{L})=O(n^{8/3}\log n). This is tight, possibly up to the logarithmic factor.

Proof.

We write C={1,2,,n1/3}C=\{1,2,\ldots,n^{1/3}\} and D={1,2,,n2/3}D=\{1,2,\ldots,n^{2/3}\}. Then \mathcal{L} is dual to the Cartesian product C×DC\times D. We define rC÷(x)=|{(c,c)C2:c/c=x}|r_{C}^{\div}(x)=|\{(c,c^{\prime})\in C^{2}\ :\ c/c^{\prime}=x\}|. Since 0C0\notin C, the condition ab=cda\cdot b=c\cdot d from (2) is equivalent to a/c=d/ba/c=d/b. This implies that E×(C)=xrC÷(x)2E^{\times}(C)=\sum_{x}r_{C}^{\div}(x)^{2}. We note that rC÷(1)2=n2/3r_{C}^{\div}(1)^{2}=n^{2/3} and that rC÷(x)=rC÷(1/x)r_{C}^{\div}(x)=r_{C}^{\div}(1/x) for every xx. This implies that

E×(C)\displaystyle E^{\times}(C) =O(n2/3)+2s=1n1/31t<sGCD(s,t)=1rC÷(t/s)2=O(n2/3)+2s=1n1/31t<sGCD(s,t)=1(n1/3s)2.\displaystyle=O(n^{2/3})+2\sum_{s=1}^{n^{1/3}}\sum_{1\leq t<s\atop\mathrm{GCD}(s,t)=1}r_{C}^{\div}(t/s)^{2}=O(n^{2/3})+2\sum_{s=1}^{n^{1/3}}\sum_{1\leq t<s\atop\mathrm{GCD}(s,t)=1}\left(\frac{n^{1/3}}{s}\right)^{2}.

In the final transition above, we note that every (c,c)C(c,c^{\prime})\in C that satisfy c/c=t/sc/c^{\prime}=t/s can be written as c=atc=at and c=asc^{\prime}=as where 1an1/3/s1\leq a\leq n^{1/3}/s. Thus, the number of representations of t/st/s as c/cc/c^{\prime} is n1/3/s\lfloor n^{1/3}/s\rfloor. For a fixed ss, the sum over tt contains ϕ(s)\phi(s) identical elements. We may thus write

E×(C)O(n2/3)+2s=1n1/3ϕ(s)(n1/3s)2=O(n2/3)+2n2/3s=1n1/3ϕ(s)s2=O(n2/3logn).\displaystyle E^{\times}(C)\leq O(n^{2/3})+2\sum_{s=1}^{n^{1/3}}\phi(s)\left(\frac{n^{1/3}}{s}\right)^{2}=O(n^{2/3})+2n^{2/3}\sum_{s=1}^{n^{1/3}}\frac{\phi(s)}{s^{2}}=O(n^{2/3}\log n).

In the last transition, we applied Lemma 2.1(c).

Since DD is an arithmetic progression, we have that E+(D)=Θ(|D|3)=Θ(n2)E^{+}(D)=\Theta(|D|^{3})=\Theta(n^{2}). Theorem 3.1 implies that

E()E×(C)E+(D)=O(n2/3logn)Θ(n2)=O(n8/3logn).E(\mathcal{L})\leq E^{\times}(C)\cdot E^{+}(D)=O(n^{2/3}\log n)\cdot\Theta(n^{2})=O(n^{8/3}\log n).

To show that the above bound is close to tight, we claim that there are many solutions to (4) and (5). We rephrase (4) as c2/c4=c1/c3c_{2}/c_{4}=c_{1}/c_{3}. There are n2/3n^{2/3} solutions in CC to c2/c4=c1/c3=1c_{2}/c_{4}=c_{1}/c_{3}=1. For each of these solutions, (5) becomes d2d1=d4d3d_{2}-d_{1}=d_{4}-d_{3}. The number of solutions of this equation is E+(D)=Θ(n2)E^{+}(D)=\Theta(n^{2}). Thus, there are Ω(n8/3)\Omega(n^{8/3}) solutions to (4) and (5). In other words, E()=Ω(n8/3)E(\mathcal{L})=\Omega(n^{8/3}). ∎

Claim 3.2 shows that Theorem 3.1 is tight, possibly up to a logarithmic factor. Combining Claim 3.2 with Theorem 1.5 leads to

I(𝒫,)=O(n1/3n2/9(n8/3logn)1/6n1/3+n1/3n)=O(n4/3log1/6n).I(\mathcal{P},\mathcal{L})=O\left(n^{1/3}n^{2/9}(n^{8/3}\log n)^{1/6}n^{1/3}+n^{1/3}n\right)=O(n^{4/3}\log^{1/6}n).

This implies that the bound of Theorem 1.5 is tight, possibly up to a logarithmic factor. However, it is possible that this tightness is achieved by the less interesting term |B|1/2|||B|^{1/2}|\mathcal{L}|. In Theorem 1.7, we show that the main term of the bound of Theorem 1.5 is tight up to sub-polynomial factors.

4 New constructions

In this section we prove Theorem 1.3 and Theorem 1.7. We first recall the statement of each theorem.

Theorem 1.3. The following holds for every 1/3<α<1/21/3<\alpha<1/2. Let 𝒫\mathcal{P} be a section of the integer lattice 2\mathbb{Z}^{2} of size nα×n1αn^{\alpha}\times n^{1-\alpha}. Then there exists a set \mathcal{L} of nn lines such that

I(𝒫,)=Θ(n4/3).I(\mathcal{P},\mathcal{L})=\Theta(n^{4/3}).
Proof.

We may assume that

𝒫={(i,j): 0i<nα and n1α/2<jn1α/2}.\mathcal{P}=\bigg{\{}(i,j)\ :\ 0\leq i<n^{\alpha}\quad\text{ and }\quad-n^{1-\alpha}/2<j\leq n^{1-\alpha}/2\bigg{\}}. (9)

If 𝒫\mathcal{P} is a different section of 2\mathbb{Z}^{2} of size nα×n1αn^{\alpha}\times n^{1-\alpha}, then we obtain (9) after a translation of 2\mathbb{R}^{2}. Such a translation does not affect the number of incidences.

We consider the set of lines

={y=(a+bc\displaystyle\mathcal{L}=\bigg{\{}y=\bigg{(}a+\frac{b}{c} )(xi)+d:(a,b,c,d)4, 0a<n12α/4,\displaystyle\bigg{)}\cdot(x-i)+d\ :\ (a,b,c,d)\in\mathbb{N}^{4},\ 0\leq a<n^{1-2\alpha}/4,
1b<cnα1/3,GCD(b,c)=1, 0i<c, 0d<n1α/4}.\displaystyle 1\leq b<c\leq n^{\alpha-1/3},\ \mathrm{GCD}(b,c)=1,\ 0\leq i<c,\ 0\leq d<n^{1-\alpha}/4\bigg{\}}. (10)

In (10), after fixing 1<cnα1/31<c\leq n^{\alpha-1/3}, there are cc possible values for ii and ϕ(c)\phi(c) possible values for bb. Lemma 2.1(b) implies that

||=c=2nα1/3(n12α4n1α4cϕ(c))=n23α16c=2nα1/3(cϕ(c))=n23α16Θ((nα1/3)3)=Θ(n).|\mathcal{L}|=\sum_{c=2}^{n^{\alpha-1/3}}\Big{(}\frac{n^{1-2\alpha}}{4}\cdot\frac{n^{1-\alpha}}{4}\cdot c\cdot\phi(c)\Big{)}=\frac{n^{2-3\alpha}}{16}\cdot\sum_{c=2}^{n^{\alpha-1/3}}(c\cdot\phi(c))=\frac{n^{2-3\alpha}}{16}\cdot\Theta((n^{\alpha-1/3})^{3})=\Theta(n).

At the end of the proof, we revise \mathcal{L} to ensure that ||=n|\mathcal{L}|=n.

We consider a,b,c,d,ia,b,c,d,i that satisfy the restrictions in (10) and also 0x<nα0\leq x<n^{\alpha}. Then

y\displaystyle y =(a+bc)(xi)+d<n12α4nα+n1α4=n1α2,\displaystyle=\left(a+\frac{b}{c}\right)(x-i)+d<\frac{n^{1-2\alpha}}{4}\cdot n^{\alpha}+\frac{n^{1-\alpha}}{4}=\frac{n^{1-\alpha}}{2},
y\displaystyle y =(a+bc)(xi)+d>n12α4(nα1/3)=n2/3α4.\displaystyle=\left(a+\frac{b}{c}\right)(x-i)+d>\frac{n^{1-2\alpha}}{4}\cdot(-n^{\alpha-1/3})=\frac{-n^{2/3-\alpha}}{4}. (11)

Thus, yy is always in the range of the yy-coordinates of the points from (9), although yy may not be an integer.

Consider a line \ell\in\mathcal{L}. We note that (a+bc)(xi)+d\left(a+\frac{b}{c}\right)(x-i)+d is an integer for every cc’th value of x{0,1,,nα1}x\in\{0,1,\ldots,n^{\alpha}-1\}. This implies that \ell is incident to a point from every cc’th column of 𝒫\mathcal{P}. By combining this with (11), we get that \ell is incident to Θ(nα/c)\Theta(n^{\alpha}/c) points of 𝒫\mathcal{P}.

For 1<cnα1/31<c\leq n^{\alpha-1/3}, let c\mathcal{L}_{c} be the set of lines of \mathcal{L} that are defined with this cc. By inspecting (10), we note that

|c|=n12α4n1α4cϕ(c)=n23αcϕ(c)16.|\mathcal{L}_{c}|=\frac{n^{1-2\alpha}}{4}\cdot\frac{n^{1-\alpha}}{4}\cdot c\cdot\phi(c)=\frac{n^{2-3\alpha}\cdot c\cdot\phi(c)}{16}.

Lemma 2.1(a) implies that

I(𝒫,)=c=2nα1/3|c|Θ(nαc)\displaystyle I(\mathcal{P},\mathcal{L})=\sum_{c=2}^{n^{\alpha-1/3}}|\mathcal{L}_{c}|\cdot\Theta\left(\frac{n^{\alpha}}{c}\right) =c=2nα1/3Θ(n22αϕ(c))\displaystyle=\sum_{c=2}^{n^{\alpha-1/3}}\Theta\left(n^{2-2\alpha}\cdot\phi(c)\right)
=Θ(n22αc=2nα1/3ϕ(c))=Θ(n4/3).\displaystyle=\Theta\left(n^{2-2\alpha}\sum_{c=2}^{n^{\alpha-1/3}}\phi(c)\right)=\Theta(n^{4/3}).

We still need to change \mathcal{L} so that ||=n|\mathcal{L}|=n. We recall that ||=Θ(n)|\mathcal{L}|=\Theta(n). If ||<n|\mathcal{L}|<n then we add arbitrary lines to \mathcal{L} until ||=n|\mathcal{L}|=n. If ||>n|\mathcal{L}|>n then we repeatedly remove a line with the smallest number of incidences, until ||=n|\mathcal{L}|=n. In either case, the number of incidences is asymptotically unchanged. ∎

Repeating the above construction with α=1/2\alpha=1/2 leads to Erdős’s construction. Repeating it with α=1/3\alpha=1/3 leads to Elekes’s construction.

The following lemma provides additional properties of the set of lines that was introduced in Theorem 1.3.

Lemma 4.1.

Let SS be the set of slopes of the lines from (10). Then |S|=Θ(n1/3)|S|=\Theta(n^{1/3}) and, for every ε>0{\varepsilon}>0, we have that

E×(S)=O(n2/3+ε).E^{\times}(S)=O(n^{2/3+{\varepsilon}}).
Proof.

By inspecting (10), we get that

S={a+bc:(a,b,c)3, 0a<n12α4, 1b<cnα1/3,GCD(b,c)=1}.S=\bigg{\{}a+\frac{b}{c}\ :\ (a,b,c)\in\mathbb{N}^{3},\ 0\leq a<\frac{n^{1-2\alpha}}{4},\ 1\leq b<c\leq n^{\alpha-1/3},\ \mathrm{GCD}(b,c)=1\bigg{\}}. (12)

Lemma 2.1(a) implies that

|S|=n12α4c=2nα1/3ϕ(c)=Θ(n12αn2α2/3)=Θ(n1/3).|S|=\frac{n^{1-2\alpha}}{4}\cdot\sum_{c=2}^{n^{\alpha-1/3}}\phi(c)=\Theta(n^{1-2\alpha}\cdot n^{2\alpha-2/3})=\Theta(n^{1/3}).

Let qq be a positive integer. Let AA be a set of rational numbers, where all the denominators and numerators are at most qq. Theorem 3 of Shteinikov [21] states that

E×(A)|A|2qO(1/loglogq).E^{\times}(A)\leq|A|^{2}\cdot q^{O(1/\log\log q)}.

We note that every element of SS can be written as a rational number with the denominator being at most nα1/3n^{\alpha-1/3} and the numerator at most n2/3αn^{2/3-\alpha}. Theorem 3 of Shteinikov [21] states that, for every ε>0{\varepsilon}>0,

E×(S)|S|2nO(1/loglogn)=O(n2/3+ε).E^{\times}(S)\leq|S|^{2}\cdot n^{O(1/\log\log n)}=O(n^{2/3+{\varepsilon}}).

We are now ready to prove Theorem 1.7.

Theorem 1.7. For every ε>0{\varepsilon}>0, in the first term of the bound of Theorem 1.5, no exponent could be decreased by an ε{\varepsilon}.

Proof.

Let α=1/3+ε\alpha=1/3+{\varepsilon}^{*}, where the value of ε>0{\varepsilon}^{*}>0 is determined below. With this value of α\alpha, we consider the point set 𝒫\mathcal{P} from (9) and the line set \mathcal{L} from (10). We recall that I(𝒫,)=Θ(n4/3)I(\mathcal{P},\mathcal{L})=\Theta(n^{4/3}) and that 𝒫\mathcal{P} is a section of the integer lattice of size nα×n1αn^{\alpha}\times n^{1-\alpha}.

Assume for contraction that the bound of Theorem 1.5(a) holds after decreasing one of the exponents of the first term by ε{\varepsilon}. We apply this improved bound on 𝒫\mathcal{P} and \mathcal{L}, to obtain that

I(𝒫,)=O(n(1α)/2n2α/3E()1/6n1/3/nε/10+n(1α)/2n).I(\mathcal{P},\mathcal{L})=O\left(n^{(1-\alpha)/2}n^{2\alpha/3}E(\mathcal{L})^{1/6}n^{1/3}/n^{{\varepsilon}/10}+n^{(1-\alpha)/2}n\right).

Since α>1/3\alpha>1/3, the second term of the above bound is asymptotically smaller than n4/3n^{4/3}. Since I(𝒫,)=Θ(n4/3)I(\mathcal{P},\mathcal{L})=\Theta(n^{4/3}), this term cannot dominate the bound. This leads to

n4/3=O(n5+α6ε10E()1/6).n^{4/3}=O(n^{\frac{5+\alpha}{6}-\frac{{\varepsilon}}{10}}\cdot E(\mathcal{L})^{1/6}).

Rearranging and recalling that α=1/3+ε\alpha=1/3+{\varepsilon}^{*} gives that

E()=Ω(n3α+3ε/5)=Ω(n8/3ε+3ε/5).E(\mathcal{L})=\Omega(n^{3-\alpha+3{\varepsilon}/5})=\Omega(n^{8/3-{\varepsilon}^{*}+3{\varepsilon}/5}). (13)

To derive an upper bound on E()E(\mathcal{L}), we use a variant of the proof of Theorem 3.1. Let SS be the set of slopes of \mathcal{L}, as defined in (12). Lemma 4.1 implies that E×(S)=O(n2/3+ε)E^{\times}(S)=O(n^{2/3+{\varepsilon}^{*}}). In other words, there are O(n2/3+ε)O(n^{2/3+{\varepsilon}^{*}}) solutions to (4) when (c1,c2,c3,c4)S4(c_{1},c_{2},c_{3},c_{4})\in S^{4}. By inspecting (10), we note that O(n2/3)O(n^{2/3}) lines of \mathcal{L} have the same slope. Thus, for every solution of (4), there are O(n2)O(n^{2}) possible values for d1,d2,d3d_{1},d_{2},d_{3}. After fixing those, at most one value of d4d_{4} satisfies (5). This implies that

E()=O(n2/3+εn2)=O(n8/3+ε).E(\mathcal{L})=O(n^{2/3+{\varepsilon}^{*}}\cdot n^{2})=O(n^{8/3+{\varepsilon}^{*}}).

By setting ε<3ε/10{\varepsilon}^{*}<3{\varepsilon}/10, we get a contradiction to (13). Indeed, in this case 8/3+ε<8/3ε+3ε/58/3+{\varepsilon}^{*}<8/3-{\varepsilon}^{*}+3{\varepsilon}/5. ∎

5 The main structural result

In this section we prove Theorem 1.4. Our proof requires the following lemma.

Lemma 5.1.

Consider a set 𝒫\mathcal{P} of nn points and a set \mathcal{L} of nn lines, both in 2\mathbb{R}^{2}, such that I(𝒫,)=Θ(n4/3)I(\mathcal{P},\mathcal{L})=\Theta(n^{4/3}). Then there exists a subset \mathcal{L}^{\prime}\subseteq\mathcal{L} such that ||=Θ(n)|\mathcal{L}^{\prime}|=\Theta(n), I(𝒫,)=Θ(n4/3)I(\mathcal{P},\mathcal{L}^{\prime})=\Theta(n^{4/3}), and every line of \mathcal{L}^{\prime} is incident to Θ(n1/3)\Theta(n^{1/3}) points of 𝒫\mathcal{P}.

Proof.

Recall that o()o(\cdot) means “asymptotically strictly smaller than.” We write I(𝒫,)=c1n4/3+o(n4/3)I(\mathcal{P},\mathcal{L})=c_{1}n^{4/3}+o(n^{4/3}). Let 1\mathcal{L}_{1} be the set of lines of \mathcal{L} that are incident to at most c1n1/3/4c_{1}n^{1/3}/4 points of 𝒫\mathcal{P}. Then

I(𝒫,1)c1n1/34|1|c1n4/34.I(\mathcal{P},\mathcal{L}_{1})\leq\frac{c_{1}n^{1/3}}{4}\cdot|\mathcal{L}_{1}|\leq\frac{c_{1}n^{4/3}}{4}. (14)

Let c2c_{2} be the constant that is hidden by the O()O(\cdot)-notation in the bound of Theorem 2.2. Let 2\mathcal{L}_{2} be the set of lines of \mathcal{L} that are incident to at least (10c2n)1/3/c11/2(10c_{2}n)^{1/3}/c_{1}^{1/2} points of 𝒫\mathcal{P}. Theorem 2.2 implies that

|2|c2n210c2n/c13/2+c2n(10c2n)1/3/c11/2=c13/2n10+c11/2c22/3n2/3101/3.|\mathcal{L}_{2}|\leq\frac{c_{2}n^{2}}{10c_{2}n/c_{1}^{3/2}}+\frac{c_{2}n}{(10c_{2}n)^{1/3}/c_{1}^{1/2}}=\frac{c_{1}^{3/2}n}{10}+\frac{c_{1}^{1/2}c_{2}^{2/3}n^{2/3}}{10^{1/3}}.

When nn is sufficiently large, we may assume that

|2|<c13/2n8.|\mathcal{L}_{2}|<\frac{c_{1}^{3/2}n}{8}.

Ackerman [1] proved that the O()O(\cdot)-notation in the bound of Theorem 1.1 may be replaced with the constant 2.442.44. This leads to

I(𝒫,2)2.44(|𝒫|2/3|2|2/3+|𝒫|+|2|)<2c1n4/33+O(n).I(\mathcal{P},\mathcal{L}_{2})\leq 2.44\cdot(|\mathcal{P}|^{2/3}|\mathcal{L}_{2}|^{2/3}+|\mathcal{P}|+|\mathcal{L}_{2}|)<\frac{2c_{1}n^{4/3}}{3}+O(n). (15)

We set =(12)\mathcal{L}^{\prime}=\mathcal{L}\setminus(\mathcal{L}_{1}\cup\mathcal{L}_{2}) and note every line of \mathcal{L}^{\prime} is incident to Θ(n1/3)\Theta(n^{1/3}) points of 𝒫\mathcal{P}. Combining (14) and (15) leads to

I(𝒫,)=I(𝒫,)I(𝒫,1)I(𝒫,2)>c1n4/3/12o(n4/3).I(\mathcal{P},\mathcal{L}^{\prime})=I(\mathcal{P},\mathcal{L})-I(\mathcal{P},\mathcal{L}_{1})-I(\mathcal{P},\mathcal{L}_{2})>c_{1}n^{4/3}/12-o(n^{4/3}).

Since every line of \mathcal{L}^{\prime} is incident to Θ(n1/3)\Theta(n^{1/3}) points of 𝒫\mathcal{P}, we have that

||=Θ(I(𝒫,)/n1/3)=Θ(n).|\mathcal{L}^{\prime}|=\Theta(I(\mathcal{P},\mathcal{L}^{\prime})/n^{1/3})=\Theta(n).

We are now ready to prove Theorem 1.4. We first recall the statement of this result.

Theorem 1.4.       
(a) For 1/3<α<1/21/3<\alpha<1/2, let A,BA,B\subset\mathbb{R} satisfy |A|=nα|A|=n^{\alpha} and |B|=n1α|B|=n^{1-\alpha}. Let \mathcal{L} be a set of nn lines in 2\mathbb{R}^{2}, such that I(A×B,)=Θ(n4/3)I(A\times B,\mathcal{L})=\Theta(n^{4/3}). Then at least one of the following holds:

  • There exists 12αβ2/31-2\alpha\leq\beta\leq 2/3 such that \mathcal{L} contains Ω(n1β/logn)\Omega(n^{1-\beta}/\log n) families of Θ(nβ)\Theta(n^{\beta}) parallel lines, each with a different slope.

  • There exists 1αγ2/31-\alpha\leq\gamma\leq 2/3 such that \mathcal{L} contains Ω(n1γ/logn)\Omega(n^{1-\gamma}/\log n) disjoint families of Θ(nγ)\Theta(n^{\gamma}) concurrent lines.

(b) Assume that we are in the case of Ω(n1β/logn)\Omega(n^{1-\beta}/\log n) families of Θ(nβ)\Theta(n^{\beta}) parallel lines. There exists n2βtn3βn^{2\beta}\leq t\leq n^{3\beta} such that, for Ω(n1β/log2n)\Omega(n^{1-\beta}/\log^{2}n) of these families, the additive energy of the yy-intercepts is Θ(t)\Theta(t). Let SS be the set of slopes of these families. Then

E×(S)t=Ω(n3α/log12n).E^{\times}(S)\cdot t=\Omega(n^{3-\alpha}/\log^{12}n).
Proof.

We remove from \mathcal{L} every line that is not incident to Θ(n1/3)\Theta(n^{1/3}) points of A×BA\times B. By Lemma 5.1, this does not change the asymptotic size of \mathcal{L} and of I(A×B,)I(A\times B,\mathcal{L}). By definition, \mathcal{L} contains at most nα+n1αn^{\alpha}+n^{1-\alpha} axis-parallel lines. These axis-parallel lines form at most 2n2n incidences with the points of A×BA\times B. We may thus discard these lines from \mathcal{L}, without changing the asymptotic size of |||\mathcal{L}| and I(A×B,)I(A\times B,\mathcal{L}). We write I(A×B,)=cn4/3+o(n4/3)I(A\times B,\mathcal{L})=cn^{4/3}+o(n^{4/3}) for some positive cc\in\mathbb{R}.

An iterative process. We repeat the following process until I(A×B,)<cn4/3/2I(A\times B,\mathcal{L})<cn^{4/3}/2. Combining Theorem 1.5 with the assumption I(A×B,)=Θ(n4/3)I(A\times B,\mathcal{L})=\Theta(n^{4/3}) leads to

n4/3=O(|B|1/2|A|2/3E()1/6||1/3+|B|1/2||).n^{4/3}=O(|B|^{1/2}|A|^{2/3}E(\mathcal{L})^{1/6}|\mathcal{L}|^{1/3}+|B|^{1/2}|\mathcal{L}|). (16)

Since |B|=n1α=o(n2/3)|B|=n^{1-\alpha}=o(n^{2/3}), we get that |B|1/2||=o(n4/3)|B|^{1/2}|\mathcal{L}|=o(n^{4/3}). In other words, the right-hand side of (16) is dominated by the first term. We rewrite (16) as

n4/3=O(n(1α)/2n2α/3E()1/6n1/3)=O(n(5+α)/6E()1/6).n^{4/3}=O(n^{(1-\alpha)/2}n^{2\alpha/3}E(\mathcal{L})^{1/6}n^{1/3})=O(n^{(5+\alpha)/6}E(\mathcal{L})^{1/6}).

Rearranging gives

E()1/6=Ω(n1/2α/6),E(\mathcal{L})^{1/6}=\Omega(n^{1/2-\alpha/6}),

which implies that

E()=Ω(n3α).E(\mathcal{L})=\Omega(n^{3-\alpha}). (17)

We imitate the notation of Theorem 1.6: Let MM be the maximum number of lines of \mathcal{L} that are incident to the same point. Let mm be the maximum number of lines of \mathcal{L} that have the same slope. Theorem 1.6 implies that

E()=O(m1/2n5/2+Mn2).E(\mathcal{L})=O\left(m^{1/2}n^{5/2}+Mn^{2}\right). (18)

Combining (17) and (18) leads to

n3α=O(m1/2n5/2+Mn2).n^{3-\alpha}=O\left(m^{1/2}n^{5/2}+Mn^{2}\right).

Thus, we have that m=Ω(n12α)m=\Omega(n^{1-2\alpha}), or that M=Ω(n1α)M=\Omega(n^{1-\alpha}), or both. When m=Ω(n12α)m=\Omega(n^{1-2\alpha}), there exists a set of Ω(n12α)\Omega(n^{1-2\alpha}) parallel lines of \mathcal{L}. When M=Ω(n1α)M=\Omega(n^{1-\alpha}), there exists a set of Ω(n1α)\Omega(n^{1-\alpha}) concurrent lines of \mathcal{L}. In either case, we remove those lines from \mathcal{L}. If we still have that I(A×B,)cn4/3/2I(A\times B,\mathcal{L})\geq cn^{4/3}/2, then we begin another iteration of the above process.

Studying the removed lines. Let \mathcal{L}^{\prime} be the set of lines of \mathcal{L} that were removed in the iterative process above. By definition, we have that I(A×B,)>cn4/3/2I(A\times B,\mathcal{L}^{\prime})>cn^{4/3}/2. Every line of \mathcal{L}^{\prime} was removed as part of a family of Ω(n12α)\Omega(n^{1-2\alpha}) parallel lines or as part of a family of Ω(n1α)\Omega(n^{1-\alpha}) concurrent lines. Let par\mathcal{L}^{\prime}_{\text{par}} be the set of the former lines and let con\mathcal{L}^{\prime}_{\text{con}} be the set of the latter lines. We note that I(A×B,par)>cn4/3/4I(A\times B,\mathcal{L}^{\prime}_{\text{par}})>cn^{4/3}/4 or that I(A×B,con)>cn4/3/4I(A\times B,\mathcal{L}^{\prime}_{\text{con}})>cn^{4/3}/4 (or both).

We first assume that I(A×B,par)>cn4/3/4I(A\times B,\mathcal{L}^{\prime}_{\text{par}})>cn^{4/3}/4. For log2n12βjlog2n\log_{2}n^{1-2\beta}\leq j\leq\log_{2}n, let par,j\mathcal{L}^{\prime}_{\text{par},j} be the set of lines that joined \mathcal{L}^{\prime} as part of a family of at least 2j2^{j} parallel lines, and of fewer than 2j+12^{j+1} such lines. By the pigeonhole principle, there exists a jj for which I(A×B,par,j)=Ω(n4/3/logn)I(A\times B,\mathcal{L}^{\prime}_{\text{par},j})=\Omega(n^{4/3}/\log n). We denote one such par,j\mathcal{L}^{\prime}_{\text{par},j} as ′′\mathcal{L}^{\prime\prime}. Since every line is incident to Θ(n1/3)\Theta(n^{1/3}) points of A×BA\times B, we have that |′′|=Ω(n/logn)|\mathcal{L}^{\prime\prime}|=\Omega(n/\log n). By writing β=logn2j\beta=\log_{n}2^{j}, we get that there exist Ω(n1β/logn)\Omega(n^{1-\beta}/\log n) families of Θ(nβ)\Theta(n^{\beta}) parallel lines in \mathcal{L}.

Consider a family of Θ(nβ)\Theta(n^{\beta}) parallel lines from ′′\mathcal{L}^{\prime\prime}. Since these lines are parallel, every point of A×BA\times B is incident to at most one line. Thus, the family forms at most nn incidences with A×BA\times B. Since ′′\mathcal{L}^{\prime\prime} consists of O(n1β)O(n^{1-\beta}) such families, we get that I(A×B,′′)=O(n2β)I(A\times B,\mathcal{L}^{\prime\prime})=O(n^{2-\beta}). Since I(A×B,′′)=Ω(n4/3/logn)I(A\times B,\mathcal{L}^{\prime\prime})=\Omega(n^{4/3}/\log n), we conclude that β2/3\beta\leq 2/3.

Next, we assume that I(A×B,par)cn4/3/4I(A\times B,\mathcal{L}^{\prime}_{\text{par}})\leq cn^{4/3}/4. In this case, I(A×B,con)>cn4/3/4I(A\times B,\mathcal{L}^{\prime}_{\text{con}})>cn^{4/3}/4. By repeating the argument from the parallel case, we obtain 1αγ11-\alpha\leq\gamma\leq 1 such that lines from families of Θ(nγ)\Theta(n^{\gamma}) concurrent lines form Ω(n4/3/logn)\Omega(n^{4/3}/\log n) incidences. Let ′′\mathcal{L}^{\prime\prime} be the set of lines of con\mathcal{L}^{\prime}_{\text{con}} that belong to such a family. Since every line of ′′\mathcal{L}^{\prime\prime} is incident to Θ(n1/3)\Theta(n^{1/3}) points of A×BA\times B, we have that |′′|=Ω(n/logn)|\mathcal{L}^{\prime\prime}|=\Omega(n/\log n).

Consider a family of Θ(nγ)\Theta(n^{\gamma}) concurrent lines from ′′\mathcal{L}^{\prime\prime}. Excluding the point of intersection, every point of A×BA\times B is incident to at most one such line. Thus, the family forms at most n+Θ(nγ)n+\Theta(n^{\gamma}) incidences. Since ′′\mathcal{L}^{\prime\prime} consists of O(n1γ)O(n^{1-\gamma}) such families, we get that I(A×B,′′)=O(n2γ)I(A\times B,\mathcal{L}^{\prime\prime})=O(n^{2-\gamma}). Since I(A×B,′′)=Ω(n4/3/logn)I(A\times B,\mathcal{L}^{\prime\prime})=\Omega(n^{4/3}/\log n), we conclude that γ2/3\gamma\leq 2/3.

Proof of part (b). In this part, we assume that ′′\mathcal{L}^{\prime\prime} consists of Ω(n1β/logn)\Omega(n^{1-\beta}/\log n) families of Θ(nβ)\Theta(n^{\beta}) parallel lines. Let S′′S^{\prime\prime} be the set of slopes of the lines in ′′\mathcal{L}^{\prime\prime}. Then |S′′|=Ω(n1β/logn)|S^{\prime\prime}|=\Omega(n^{1-\beta}/\log n) and |S′′|=O(n1β)|S^{\prime\prime}|=O(n^{1-\beta}). For sS′′s\in S^{\prime\prime}, we set Y′′sY_{\mathcal{L}^{\prime\prime}}^{s} to be the set of yy-intercepts of the lines with slope ss in ′′\mathcal{L}^{\prime\prime}.

We recall that I(A×B,′′)=Ω(n4/3/logn)I(A\times B,\mathcal{L}^{\prime\prime})=\Omega(n^{4/3}/\log n). By definition, for every sS′′s\in S^{\prime\prime}, we have that |Y′′s|=Θ(nβ)|Y_{\mathcal{L}^{\prime\prime}}^{s}|=\Theta(n^{\beta}). This implies that E+(Y′′s)=Ω(n2β)E^{+}(Y_{\mathcal{L}^{\prime\prime}}^{s})=\Omega(n^{2\beta}) and E+(Y′′s)=O(n3β)E^{+}(Y_{\mathcal{L}^{\prime\prime}}^{s})=O(n^{3\beta}). For n2βtn3βn^{2\beta}\leq t\leq n^{3\beta}, we define

t={′′:E+(Y′′s)=Θ(t)}.\mathcal{L}_{t}=\{\ell\in\mathcal{L}^{\prime\prime}\ :\ E^{+}(Y_{\mathcal{L}^{\prime\prime}}^{s})=\Theta(t)\}.

By another dyadic pigeonhole argument, there exists a tt that satisfies I(A×B,t)=Ω(n4/3/log2n)I(A\times B,\mathcal{L}_{t})=\Omega(n^{4/3}/\log^{2}n). We fix such a value of tt. Since every line is incident to Θ(n1/3)\Theta(n^{1/3}) points of A×BA\times B, we get that |t|=Ω(n/log2n)|\mathcal{L}_{t}|=\Omega(n/\log^{2}n).

Let SS be the set of slopes of lines of t\mathcal{L}_{t}. We have that

|S||S′′|=O(n1β) and that |S|=Ω(n1β/log2n).|S|\leq|S^{\prime\prime}|=O(n^{1-\beta})\quad\text{ and that }\quad|S|=\Omega(n^{1-\beta}/\log^{2}n).

To obtain an upper bound for E(t)E(\mathcal{L}_{t}), we imitate the proof of Theorem 3.1. While t\mathcal{L}_{t} may not be a Cartesian product of lines, we can instead rely on the property E+(Y′′s)=Θ(t)E^{+}(Y_{\mathcal{L}^{\prime\prime}}^{s})=\Theta(t). We think of E(t)E(\mathcal{L}_{t}) as the number of solutions to (4) and (5).

Equation (4) depends only on the elements of SS. The number of distinct solutions to (4) from SS is Θ(E×(S))\Theta(E^{\times}(S)). This does not take into account the number of lines that have the same slope. We fix a solution to (4) and consider the number of corresponding solutions to (5). In other words, we fix the slopes c1,c2,c3,c4Sc_{1},c_{2},c_{3},c_{4}\in S and consider valid values for the yy-intercepts d1,d2,d3,d4d_{1},d_{2},d_{3},d_{4}. We set s=c1/c3s=c_{1}/c_{3} and rephrase (5) as

d2d1=s(d4d3).d_{2}-d_{1}=s(d_{4}-d_{3}). (19)

Similarly to rA(x)r_{A}^{-}(x), for sets AA and BB we define

rA,B(x)\displaystyle r_{A,B}^{-}(x) =|{(a,b)A×B:ab=x}|.\displaystyle=|\{(a,b)\in A\times B\ :\ a-b=x\}|.

By the Cauchy-Schwarz inequality, the number of solutions to (19) is

krYtc1,Ytc2(k)rYtc3,Ytc4(k/s)\displaystyle\sum_{k\in\mathbb{R}}r_{Y_{\mathcal{L}_{t}}^{c_{1}},Y_{\mathcal{L}_{t}}^{c_{2}}}^{-}(k)r_{Y_{\mathcal{L}_{t}}^{c_{3}},Y_{\mathcal{L}_{t}}^{c_{4}}}^{-}(k/s) (krYtc1,Ytc2(k)2)1/2(krYtc3,Ytc4(k))1/2\displaystyle\leq\left(\sum_{k\in\mathbb{R}}r_{Y_{\mathcal{L}_{t}}^{c_{1}},Y_{\mathcal{L}_{t}}^{c_{2}}}^{-}(k)^{2}\right)^{1/2}\left(\sum_{k\in\mathbb{R}}r_{Y_{\mathcal{L}_{t}}^{c_{3}},Y_{\mathcal{L}_{t}}^{c_{4}}}^{-}(k)\right)^{1/2}
=E+(Ytc1,Ytc2)1/2E+(Ytc3,Ytc4)1/2.\displaystyle\hskip 113.81102pt=E^{+}(Y_{\mathcal{L}_{t}}^{c_{1}},Y_{\mathcal{L}_{t}}^{c_{2}})^{1/2}\cdot E^{+}(Y_{\mathcal{L}_{t}}^{c_{3}},Y_{\mathcal{L}_{t}}^{c_{4}})^{1/2}.

Combining this with (1) implies that the number of solutions to (19) is at most

E+(Ytc1)1/4E+(Ytc2)1/4E+(Ytc3)1/4E+(Ytc4)1/4.E^{+}(Y_{\mathcal{L}_{t}}^{c_{1}})^{1/4}\cdot E^{+}(Y_{\mathcal{L}_{t}}^{c_{2}})^{1/4}\cdot E^{+}(Y_{\mathcal{L}_{t}}^{c_{3}})^{1/4}\cdot E^{+}(Y_{\mathcal{L}_{t}}^{c_{4}})^{1/4}.

By definition, each of these four energies is Θ(t)\Theta(t). We conclude that the number of solutions to (19) is O(t)O(t).

To recap, there are Θ(E×(S))\Theta(E^{\times}(S)) solutions to (4), each with O(t)O(t) corresponding solutions to (19). We thus have that

E(t)=O(E×(S)t).E(\mathcal{L}_{t})=O(E^{\times}(S)\cdot t). (20)

Combining Theorem 1.5 with I(𝒫,t)=Ω(n4/3/log2n)I(\mathcal{P},\mathcal{L}_{t})=\Omega(n^{4/3}/\log^{2}n) leads to

n4/3/log2(n)=O(|B|1/2|A|2/3E(t)1/6|t|1/3+|B|1/2|t|).n^{4/3}/\log^{2}(n)=O(|B|^{1/2}|A|^{2/3}E(\mathcal{L}_{t})^{1/6}|\mathcal{L}_{t}|^{1/3}+|B|^{1/2}|\mathcal{L}_{t}|). (21)

Since |B|=n1α|B|=n^{1-\alpha} and α>1/3\alpha>1/3, we have that |B|1/2|t|=o(n4/3/log2n)|B|^{1/2}|\mathcal{L}_{t}|=o(n^{4/3}/\log^{2}n). This implies that the right-hand side of (21) is dominated by the first term. We rewrite (21) as

n4/3/log2n=O(n(1α)/2n2α/3E(t)1/6n1/3)=O(n(5+α)/6E(t)1/6).n^{4/3}/\log^{2}n=O(n^{(1-\alpha)/2}n^{2\alpha/3}E(\mathcal{L}_{t})^{1/6}n^{1/3})=O(n^{(5+\alpha)/6}E(\mathcal{L}_{t})^{1/6}).

Rearranging gives

E(t)1/6=Ω(n1/2α/6/log2n),E(\mathcal{L}_{t})^{1/6}=\Omega(n^{1/2-\alpha/6}/\log^{2}n),

which implies that

E(t)=Ω(n3α/log12n).E(\mathcal{L}_{t})=\Omega(n^{3-\alpha}/\log^{12}n).

Combining the bound above with (20) leads to

E×(S)t=Ω(n3α/log12n).E^{\times}(S)\cdot t=\Omega(n^{3-\alpha}/\log^{12}n).

6 Elekes’s structural problem

This concluding section contains the proof of Theorem 1.10. We first recall the statement of this result.

Theorem 1.10. Let AA be a set of nn reals and let α<1\alpha<1 satisfy α=Ω(n1/2log2k)\alpha=\Omega(n^{-1/2}\log^{2}k). Let \mathcal{L} be a set of kk non-axis-parallel lines, each incident to at least αn\alpha n points of A×AA\times A. Consider 0β10\leq\beta\leq 1 such that \mathcal{L} contains Ω(k1β/logk)\Omega(k^{1-\beta}/\log k) families of Θ(kβ)\Theta(k^{\beta}) parallel lines. There exists k2βtk3βk^{2\beta}\leq t\leq k^{3\beta} such that, for Ω(k1β/log2k)\Omega(k^{1-\beta}/\log^{2}k) of these families, the additive energy of the yy-intercepts is Θ(t)\Theta(t). Let SS be the set of slopes of these families. Then

k=O(n1/4t1/4E×(S)1/4log3nα3/2).k=O\left(\frac{n^{1/4}t^{1/4}E^{\times}(S)^{1/4}\log^{3}n}{\alpha^{3/2}}\right).
Proof.

We imitate the proof of Theorem 1.4(b). Let \mathcal{L}^{\prime} be the set of lines that belong to the Ω(k1β/logn)\Omega(k^{1-\beta}/\log n) families of Θ(kβ)\Theta(k^{\beta}) parallel lines. Let SS^{\prime} be the set of slopes of the lines of \mathcal{L}^{\prime}. Then |S|=Ω(k1β/logn)|S^{\prime}|=\Omega(k^{1-\beta}/\log n) and |S|=O(k1β)|S^{\prime}|=O(k^{1-\beta}). For sSs\in S^{\prime}, we denote as YsY_{\mathcal{L}^{\prime}}^{s} the set of yy-intercepts of the lines of \mathcal{L}^{\prime} with slope ss.

By definition, for every sSs\in S^{\prime}, we have that |Ys|=Θ(kβ)|Y_{\mathcal{L}^{\prime}}^{s}|=\Theta(k^{\beta}). This implies that E+(Ys)=Ω(k2β)E^{+}(Y_{\mathcal{L}^{\prime}}^{s})=\Omega(k^{2\beta}) and that E+(Ys)=O(k3β)E^{+}(Y_{\mathcal{L}^{\prime}}^{s})=O(k^{3\beta}). For n2βtn3βn^{2\beta}\leq t\leq n^{3\beta}, we define

t={:E+(Ys)=Θ(t)}.\mathcal{L}_{t}=\{\ell\in\mathcal{L}^{\prime}\ :\ E^{+}(Y_{\mathcal{L}^{\prime}}^{s})=\Theta(t)\}.

By the pigeonhole principle, there exists tt that satisfies |t|=Ω(k/log2k)|\mathcal{L}_{t}|=\Omega(k/\log^{2}k). We fix such a value of tt.

Since every line of \mathcal{L} is incident to at least αn\alpha n points of A×AA\times A, we get that

I(A×A,t)=Ω(αnk/log2k).I(A\times A,\mathcal{L}_{t})=\Omega(\alpha nk/\log^{2}k).

Theorem 1.5 implies that

I(A×A,t)=O(n7/6E(t)1/6|t|1/3+|t|n1/2)=O(n7/6E(t)1/6k1/3+kn1/2).I(A\times A,\mathcal{L}_{t})=O(n^{7/6}E(\mathcal{L}_{t})^{1/6}|\mathcal{L}_{t}|^{1/3}+|\mathcal{L}_{t}|n^{1/2})=O(n^{7/6}E(\mathcal{L}_{t})^{1/6}k^{1/3}+kn^{1/2}).

Combining the two above bounds for I(A×A,t)I(A\times A,\mathcal{L}_{t}) leads to

αnk/log2k=O(n7/6E(t)1/6k1/3+kn1/2).\alpha nk/\log^{2}k=O(n^{7/6}E(\mathcal{L}_{t})^{1/6}k^{1/3}+kn^{1/2}).

Since α=Ω(n1/2log2k)\alpha=\Omega(n^{-1/2}\log^{2}k), we have that

αnk/log2k=O(n7/6E(t)1/6k1/3).\alpha nk/\log^{2}k=O(n^{7/6}E(\mathcal{L}_{t})^{1/6}k^{1/3}). (22)

By repeating the argument from the proof of Theorem 1.4(b), we obtain that

E(t)=O(tE×(S)).E(\mathcal{L}_{t})=O(t\cdot E^{\times}(S)).

Combining this with (22) and noting that logk=O(logn)\log k=O(\log n) leads to

αnk=O(n7/6E(t)1/6k1/3log2k)\displaystyle\alpha nk=O(n^{7/6}E(\mathcal{L}_{t})^{1/6}k^{1/3}\log^{2}k) =O(n7/6k1/3t1/6E×(S)1/6log2n).\displaystyle=O(n^{7/6}k^{1/3}t^{1/6}E^{\times}(S)^{1/6}\log^{2}n).

Rearranging leads to

k=O(n1/4t1/4E×(S)1/4log3nα3/2).k=O\left(\frac{n^{1/4}t^{1/4}E^{\times}(S)^{1/4}\log^{3}n}{\alpha^{3/2}}\right).

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