THE POLYNOMIAL METHOD LECTURE 2 SZEMERÉDI-TROTTER THEOREM AND POLYNOMIAL PARTITIONING

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1 THE POLYNOMIAL METHOD LECTURE 2 SZEMERÉDI-TROTTER THEOREM AND POLYNOMIAL PARTITIONING MARCO VITTURI Abstract. This series of notes is intended to provide an introduction to the polynomial method through examples of its successful application - mainly in Harmonic Analysis, but other fields will also be considered (e.g. combinatorics, transcendence theory, etc). In this second set of notes we prove the Szemerédi- Trotter theorem by means of the polynomial method. First we discuss very briefly some incidence geometry results; then the polynomial tools needed for the proof are introduced. These are the concepts of smooth and singular points, and even more importantly the polynomial partitioning method. Finally the proof of the theorem combining these with a weaker estimate is given. A different proof - unrelated to polynomials - is given for comparison in the appendix for the curious reader. Contents 1. Incidence geometry 1 2. Smoothness 5 3. Polynomial partitioning 7 4. Proof of the Szemerédi-Trotter theorem using polynomial partitioning 9 Appendix A. Alternative proof by the Crossing Number inequality 11 References 13 Notation. Throughout these notes, E will denote the cardinality of the set E when this is discrete and finite, and its Lebesgue measure otherwise. Context will usually suffice to determine which is the case. The expression X À Y will mean that there is a constant C ą 0 s.t. X ď CY. We might specify the dependence of this constant on additional parameters e.g. α by writing X À α Y when this is the case. By X Y we will mean X À Y and Y À X. If A, B are sets, by A \ B we will mean the same as A Y B but with the implicit understanding that the sets A, B are disjoint. If P is a polynomial with coefficients in F 1, we denote by Z F2 pp q its zero set in the field F 2 ě F 1 (i.e. an algebraic extension of F 1). If ψ is a logical statement then 1pψq 1 if ψ is true and 0 otherwise. 1. Incidence geometry Incidence geometry is the branch of combinatorics concerned with counting the number of pairs pp, lq P A ˆ B with p l, where is a specified (geometric) incidence relation and A, B are collections of objects. For example, we might want to count the number of tangencies in a collection of circles. In the case that will interest us here p will be a geometric point and l a geometric line. 1

2 2 MARCO VITTURI Possibly the best known result of incidence geometry is the celebrated Szemerédi- Trotter theorem Theorem 1 (Szemerédi-Trotter, [ST83]). Let P be a finite collection of points in the plane R 2, L be a finite collection of lines in the plane, and IpP, L q their set of incidences, i.e. IpP, L q : tpp, lq P P ˆ L s.t. p P lu. Then IpP, L q À P 2{3 L 2{3 ` P ` L. The exponents given in the statement of the theorem are sharp, as can be seen by the following extremal examples. Example 1. Let N ą 0 be a large integer and let our collections be P tpa, bq P Z 2 s.t. a P r1, Ns, b P r1, 2N 2 su, L tx ÞÑ px, mx ` cq s.t. m, c P Z, m P r1, Ns, c P r1, N 2 su. Then there are P 2N 3 points and L N 3 lines; for every x P r1, Ns the expression mx ` c gives a different integer in the range r1, 2N 2 s, and so every line contains N points in P and therefore there are N 4 incidences all together. Thus the two sides of the Szemerédi-Trotter inequality are comparable in this case: IpP, L q N 4 p2n 3 q 2{3 pn 3 q 2{3 P 2{3 L 2{3. Figure 1. The critical configuration in example 1 when N 3 (with the x, y axes reversed). Example 2. Let N " 1 be a large even integer and let 1 ă R! N be another integer, and let our collections be P tpa, bq P Z 2 s.t. pa, bq P r N{2, N{2s 2 u, L tlines that contain between R and 2R points in Pu. Notice P is a regular grid. We estimate how many lines pass through a given point of this grid. If l is a line in L and p P P is a point that belongs to l, then the closest p 1 P P s.t. p 1 p and p 1 P l must be in a square centered in p and of sidelength OpN{Rq. This is because there are R points of P in l, and thus the projections of these points onto the x or y axes can be separated by at most OpN{Rq units. Thus we conclude that there can be at most OpN 2 {R 2 q lines in L through a given point p. On the other hand, we claim there are at least N 2 {R 2 distinct such lines. Indeed, it suffices by symmetry to consider points in the upper-right quadrant of the grid P; we also restrict to those lines that form an angle θ with the x-axis such that 1{2 ď tan θ ď 2. A moment s reflection reveals that in order for such a line to contain R points of P it must be tan θ r q P Q, with r, q coprimes and r, q P r N 2R, N R s. There are at least N 2 {R 2 distinct such pairs, since the fraction of pairs that share a factor of 2 is 1{2 2 and the fraction of pairs that shares a factor of 3 is 1{3 2, and so on, and ř kě2 1 k ă ă 1. Thus there are at least N 2 {R 2

3 LECTURE 2 3 distinct lines in L through each point. We then have P N 2 L P N 2 1 R 2 R N 4 R 3 IpP, L q P N 2 R 2 N 4 R 2, and thus the two sides of the Szemerédi-Trotter inequality are comparable: IpP, L q N 4 R 2 pn 2 q 2{3 2{3 ˆN4 R 3 P 2{3 L 2{3. The original proof in [ST83] is somewhat complicated, but we can describe very approximately a strategy that is close in spirit to the original one and was used in later proofs that simplified the original argument. The idea is to use a divide-etimpera approach in terms of a cell decomposition. Indeed, one can imagine drawing M lines in general positions (these lines are not in L in general); these subdivide the plane in M 2 connected components - the cells - and any line in L can intersect at most M ` 1 of these cells (each cell has one of the M lines as one of its sides). Now, if we can choose the M lines in such a way that some of the relevant statistics of the points and lines are roughly uniform across different cells, then we could use trivial estimates for incidences inside every cell and sum them up using the fact that there are few cells along every line. Here we use statistics in the sense that for example most points in a cell should be made to have a comparable number of lines in L that pass through them, or that the number of lines of L that enters a given cell should be roughly constant. If one chooses the correct statistics to preserve, this can roughly be achieved (with additional effort needed to remove some logarithmic losses). In these notes we will prove the Szemerédi-Trotter theorem by a different cell decomposition, obtained through polynomial partitioning. This cell decomposition is better behaved, and thus it s easier to reach the desired estimates; overall though, the approach is similar to the divide-et-impera one outlined above. We would like to mention that incidence estimates have a number of applications, as many problems can be reformulated at least partially in terms of incidences. We give one simple example to support this point, namely an estimate of Elekes in sum-product theory. Theorem 2 (Elekes, [Ele97]). Let A be a finite subset of Z. Then 1 maxt A ` A, A A u Á A 5{4. Proof. Let P pa`aqˆpa Aq and let L be the set of lines of equation y apx bq for a, b P A. Thus every such line is incident to all the points pc ` b, acq P P, and there are A of them. Thus by Szemerédi-Trotter theorem A A 2 ď IpP, L q À p A ` A A A q 2{3 p A 2 q 2{3 ; rearranging one obtains A ` A A A Á A 5{2 and therefore one of the factors on the left is larger than a A 5{2 A 5{4. 1 Here we are using the Minkowski sum and the Minkowski product: A`B : ta`b s.t. a, b P Au and A B : ta b s.t. a, b P Au.

4 4 MARCO VITTURI Erdős and Szemerédi conjectured that the inequality should be valid for any exponent smaller than 2, i.e. for all ε ą 0 maxt A ` A, A A u? Á ε A 2 ε. Stronger inequalities than the Elekes one have been proven, but the conjecture on Z is still open. Heuristically, one can interpret the conjecture as saying that the multiplicative structure is not really compatible with the additive one, in the sense that finite subsets of Z are always far from being even approximately a (sub)ring. Remark 1. Let us mention that estimates in the same spirit as the one above have found applications in the Kakeya problems. Before Dvir s solution of the full Kakeya conjecture in the finite fields, progress had been made by using a sumproduct estimate like the above in such fields (see [BKT04]). The use of sumproduct estimates in the study of Kakeya problems was a development of the earlier idea of using sumset estimates, which originated in Bourgain [Bou99]. The above remark leads us to consider what happens in terms of incidences in the case of the finite fields. As mentioned in the previous notes, the fundamental difference between the finite fields and R is the trivial topology of the former. In particular, there are fewer topological obstacles (none) and as such we should expect the finite fields case to be worse behaved than the real one, in the sense that the expected number of incidences should be greater. We ll show this is the case. Let F F q be a finite field with q p k elements, p prime, and let P be a collection of points in F 2, and L a collection of lines in F 2, i.e. sets of the form l tp 0 ` tv s.t. t P Fu, where p 0 P F 2 and 0 v P F 2 {Fˆ is a direction. We can rewrite IpP, Lq as IpP, Lq ÿ ÿ 1pp P lq; ppp lpl by Cauchy-Schwarz (in l) then IpP, Lq 2 ď L ÿ ` ÿ ÿ 1pp P lq 2 L lpl ppp lpl ÿ p,p 1 PP The last sum is exactly the cardinality of the set of triplets tpp, p 1, lq P P ˆ P ˆ L s.t. p P l, p 1 P lu; 1pp P lq1pp 1 P lq. when p p 1 there are exactly IpP, Lq such triplets, and when p p 1 there can be at most 1 line passing through both p and p 1, and therefore there are at most P 2 such triplets. All together we ve proven IpP, Lq 2 ď IpP, Lq L ` P 2 L ď 2 maxt IpP, Lq L, P 2 L u; if the maximum is given by the first term then IpP, Lq À L, otherwise IpP, Lq À P L 1{2, and therefore in general we have proven Theorem 3 (Szemerédi-Trotter in the finite fields). Let P, L be as above. Then IpP, Lq À P L 1{2 ` L. (1) Observe that we could run the same argument above exchanging the roles of P and L, and we d get IpP, Lq À P 1{2 L ` P. This is due to the general fact that there is a duality between points and lines in the plane 2. Indeed, the statement that two (distinct) points uniquely determine a line is the dual statement to the one that says two (distinct, non parallel) lines intersect at a unique point, and every 2 More precisely, in the projective plane.

5 LECTURE 2 5 incidence statement about points and lines can be rewritten accordingly exchanging their roles. One might ask at this point, in view of the comment that preceded the last proof, if the estimate above is sharp; it s easy to see it is. See the following example originally due to Erdős. Example 3. Take P F 2 (the entire plane) and L to be the set of all lines in F 2. Thus P F 2, and L F 2 (you need two parameters to specify a line), and every line contains exactly F points in F 2 P, thus giving F 3 incidences. This proves that the inequality is sharp in the exponents: IpP, Lq F 3 p F 2 qp F 2 q 1{2 ` F 2. Compare this example with the real plane example 2 for N F. In there, a general line can only contain far fewer than F points, and more precisely F {R. This is not the case for the finite fields, since any line will contain exactly F points of the grid F 2. The difference, heuristically speaking, is that the lines of F 2 wrap around the grid F 2, while in R 2 they escape to infinity. Remark 2. Observe that the proof of (1) works as well if points and lines are taken in the real plane R 2, thus providing a first estimate on the number of incidences in that case. This estimate is quantitatively worse than the one offered by the Szemerédi-Trotter theorem though: in (1) the sum of the exponents of the main term is 1 ` 1{2 3{2, while in the Szemerédi-Trotter one it is 2{3 ` 2{3 4{3, which is smaller. Thus there can only be much fewer incidences in the real case, since there are topological obstructions - the sort of obstructions that make the Kakeya problem much more complicated in R. Estimate (1) will be used in the proof of the Szemerédi-Trotter theorem given in section 4. We ll refer to it as the trivial estimate. 2. Smoothness In this section we present a definition regarding the geometric behaviour of the zero sets of polynomials. We state things in terms of a general field F, and we denote by F its algebraic closure. Remember the Hasse derivatives are defined on monomials by # j1 D i1,...,in X 1 0 if j k ă i k for some k, Xjn n `j1 i1 `jn 1 i1 in Xj Xn jn in otherwise, and then extended to arbitrary polynomials by linearity. We write bold i for a multi-index pi 1,..., i n q P N n for shortness. We list without proof some properties of the Hasse derivatives for completeness: i) D i D k `i 1`k 1 i 1 `in`k n i Di 1`k 1,...,i n`k n n [composition]; ii) D i pp Qq ř j,k s.t. D j P D k Q [product rule]; j`k i iii) P pxq ř ipn n s.t. i ďdeg P Di P p0qx i1 1 Xin n [Taylor s formula]. Remark 3. By looking at the definition of D i1,...,in we see that it coincides up to a factor of pi 1! i n!q with the standard derivative B i1 X 1 B in X n when F R. The reason for this slightly different definition is that if the field F has finite characteristic p then there are polynomials with formal standard derivative 0 that are kx not identically constant: namely, P pxq : X p d is not constant but p dx

6 6 MARCO VITTURI ppp 1q pp k ` 1qX p k 0 P F for all k 1,..., p. With the Hasse derivative though, we have D p P p0q `p p Xp p 1, so that by Taylor s formula X p pÿ D k P p0qx k 0 `... ` 0 ` k 0 We can form the Hasse gradient as P : pd x1 P,..., D xn P q, ˆp X p X p. p where D x k is the Hasse derivative with index whose components are all 0 except the k-th one, which is 1. Let P be a polynomial. For a point p P Z F pp q it can either be P ppq 0 or not; it turns out that geometrically the two cases behave rather differently. Indeed, when the field is R we have the implicit function theorem that tells us that near a smooth point the hypersurface is diffeomorphic to a hyperplane (the tangent space). We introduce Definition 4. Let P P FrX 1,..., X n s be a polynomial and let p P ZpP q. Then p is said to be smooth if P ppq 0, and it is said to be singular 3 otherwise. We can show for general fields the familiar result that if p is a smooth point then necessarily P ppq is orthogonal to any lines contained in the hypersurface. Notice where we need to refer to F rather than F (but check remark 4 for as to why this is somewhat arbitrary). Proposition 5. Let P P FrX 1,..., X n s and p P F n be a smooth point of Z F pp q. Let l tp ` tv s.t. t P Fu be a line that is contained in Z F pp q (notice that p P l). Then v P ppq 0. Proof. Restricting the polynomial P to l gives polynomial P l ptq P pp ` tvq; it is P l ptq 0 for all t P F by assumption, and since F is infinite this implies that the polynomial vanishes identically, and thus DP l p0q vanishes too. But DP l ptq P pp ` tvq v, and the claim follows. Remark 4. The only property we needed of F was that its cardinality is infinite. Thus any field extension of F F q of infinite cardinality would do as well. This fact looks rather innocent but it can actually be used to give a solution to the conjecture of Sharir on joints, which we now state. Theorem 6 (joints conjecture). Let L be a finite set of lines in F n, where F is an arbitrary field. A joint is a point p that belongs to at least n lines in L that are non-coplanar 4. Then if we denote by J the set of joints, J À n L n{pn 1q. The proof was independently given in [KSS10] and [Qui10], and in [GK10] for n 3. Although simple we don t include it in here. We also mention that a multilinear version of this problem was solved in [Ili15]. 3 Some authors use critical instead. 4 i.e. their directions span R n as a vector space.

7 LECTURE Polynomial partitioning In the following we will only work with polynomials with real coefficients. Remember from the previous lecture that the proof of Dvir s theorem relied crucially on the following lemma Lemma. If E Ă F n has cardinality E ă `d`n n then there exists a (non-zero) polynomial P P FrX 1,..., X n s of degree degpp q ď d such that E Ă Z F pp q. Behind this lemma was the trivial linear algebra fact that if T : F n Ñ F m is a linear map with n ą m, then there exists a non-trivial zero, i.e. x 0 with T x 0. This is the trivial version of a deeper and famous result, the Borsuk-Ulam theorem, which we now state. We say that a function f is an antipodal map if it is continuous and odd, in the sense that fp xq fpxq for all x in the domain. Theorem 7 (Borsuk-Ulam). Let f : S n Ñ R n be an antipodal map. Then there exists a point y P S n such that fpyq 0 P R n. In the following we will use an equivalent restatement of this theorem where f is replaced by an antipodal map T : R n zt0u Ñ R m, with n ą m; the Borsuk-Ulam theorem then says that this map has a zero. We won t provide a proof of the Borsuk-Ulam theorem here, but for a proof that is palatable to analysts see [Cara]. Intuitively we expect that the Borsuk-Ulam theorem should give us stronger tools than the ones like the lemma above. This is indeed the case, as we ll see shortly. A first application of the Borsuk-Ulam theorem is the Theorem 8 (ham sandwich theorem, [ST42]). Let A 1,..., A M be M bounded open subsets of R n, with M ď n. Then there exists a hyperplane π such that π bisects A i for all i 1,..., M. It s easy to convince oneself that M ď n is necessary in this case. Proof. A hyperplane is specified up to a multiplicative constant by a choice of coefficients a 0 P R, a P R n (the hyperplane is given by those points x P R n s.t. a 0 ` a x 0). Define the function ż M T pa 0, aq : sgnpa 0 ` a xq dx ; A i i 1 thus the i-th component of T is the difference between the volume of the upper half A i Xtx s.t. a 0`a x ą 0u and that of the lower half A i Xtx s.t. a 0`a x ă 0u. Then T is continuous 5 from R n`1 zt0u to R M, and M ă n ` 1; moreover T is clearly an odd map, and thus it s antipodal and there is a hyperplane such that T pa 0, aq 0. This hyperplane clearly bisects all sets at once. This theorem is little more than a mathematical curiosity, but if we substitute hyperplanes with polynomial hypersurfaces of higher degree, things start to become interesting. Theorem 9 (the polynomial ham sandwich theorem). Let A 1,..., A M be M bounded open sets of R n. Then there exists a non-trivial polynomial P of degree at most OpnM 1{n q such that the hypersurface given by the zero set ZpP q bisects A j for all j 1,..., M. Notice that now we are not restricted to n sets, but we can have arbitrarily many in principle - at the price of making the degree of P eventually very large. 5 By the dominated convergence theorem, since sgnpa0 ` a xq ď χ Ai pxq on A i.

8 8 MARCO VITTURI Proof. A polynomial is specified by its coefficients, so in particular a polynomial of degree d can be specified by a point in R pd`n n q. Thus we can define similarly as before ż M T pp q : sgnpp pxqq dx, A i i 1 and we can reach the same conclusion provided that `d`n n ą M. Since `d`n n ě pd{nq n, we can achieve this by choosing d to be OpnM 1{n q. Remark 5. As mentioned in the previous notes, considerations involving the polynomial ham sandwich theorem have been of uttermost importance in proving the endpoint case of the Multilinear Kakeya conjecture in [Gut10], along with other algebraic-topological notions; but actually the polynomial ham sandwich theorem turns out to be the only algebraic topology needed to prove it (see [CV13]). There is not enough room in here to add details, but the interested reader may find a nice exposition in [Carb]. In these notes the polynomial ham sandwich theorem will be important to us only through the following consequence - the polynomial partitioning technique. Theorem 10 (polynomial partitioning). Let P be a collection of N points in R n and M 2 K ě 1 a given integer. Then there exists a polynomial P of degree at most Opn 2 M 1{n q such that we can partition R n ZpP q \ Ω 1 \ \ Ω M, where the elements of the union are all disjoint from each other, Ω j is open, the boundary of Ω j is contained in ZpP q for all j, and moreover for all j 1,..., M. P X Ω j ď P M The proof relies on the following discretized version of the polynomial ham sandwich theorem - which we state as a lemma and prove first. Lemma 11. Let E 1,..., E M be M ě 1 finite sets of points in R n. Then there exists a non-trivial polynomial P of degree at most OpnM 1{n q such that P bisects the sets E i, i.e. E i X tx P R n s.t. P pxq ą 0u ď E i 2, and analogously for the intersection with tx s.t. P pxq ă 0u. Proof. Let ɛ ą 0 and let Ei ɛ be the ɛ-neighbourhoods of E i; then by the polynomial ham sandwich theorem for every ɛ ą 0 there exists a polynomial P ɛ of degree at most OpnM 1{n q such that ZpP ɛ q bisects all the sets Ei ɛ. Since the degree is bounded, all the norms on these polynomials are equivalent, so choose any and assume that P ɛ is normalized in this norm (since normalizing leaves the zero set invariant). Then the P ɛ belong to the sphere, which is compact, and therefore we can extract a subsequence P ɛk that converges to P as k Ñ 8. The zero set ZpP q then bisects the sets E i. Proof of Theorem 10. The proof is by an iterative procedure. We start with one collection only, namely P : P 0. Then we can find a polynomial P 1 of degree 6 to us. 6 The minimal degree with which we can achieve this is 1 obviously, but this will be irrelevant

9 LECTURE 2 9 Opnq that bisects P in the sense described above. Denote by P 1 ε for ε P t 1, `1u the sets P 1` : P 0 X tx P R n s.t. P 1 pxq ą 0u, P 1 : P 0 X tx P R n s.t. P 1 pxq ă 0u. Now suppose that the procedure has been carried over for k steps; we describe the pk`1q-th step. Apply the polynomial ham sandwich theorem to the collections P k ε, where ε P t 1, `1u k, thus obtaining a polynomial P k`1 of degree at most Opn2 k{n q such that ZpP k`1 q bisects all the collections P k ε. Then for ε 1 pε 1,..., ε k`1 q pε, ε k`1 q P t 1, `1u k`1 define P k`1 ε 1 : Pk ε X tx P R n s.t. sgnpp pxqq ε k`1 u; iterate the procedure until k K. Thus the desired polynomial is clearly seen to be the product and its degree is Opn P : P 1... P K, Kÿ 2 k{n q Opn 2 M 1{n q. k 0 We will refer to the regions Ω j as the cells of the polynomial partition, and we will sometime refer to such a partition as the cellular decomposition. Notice the cells are not necessarily simply connected, and moreover some of the points might lie on ZpP q in general. Remark 6. We conclude this section by mentioning that the polynomial ham sandwich theorem has another useful variant which will be used in the notes at some point (when discussing Guth s proof of a restriction estimate by polynomial partitioning). Namely, we don t have to restrict ourselves to sets, but we can also bisect absolutely continuous measures by a polynomial of controlled degree. Indeed, let f 1,..., f M be M positive integrable functions, then by considering the mapping ż M T pp q : sgnpp pxqqf i pxq dx R n i 1 we see as above that this map has a zero for some P of degree OpnM 1{n q. Then we can start with a positive integrable function f and proceed as before by applying the above iteratively to obtain a polynomial partitioning of the mass of f, in the sense that we have for some polynomial P of degree OpnM 1{n q the cellular decomposition where for all i 1,..., M. R n ZpP q \ Ω 1 \... \ Ω M, ż Ω i fpxq dx ď ş f dx 4. Proof of the Szemerédi-Trotter theorem using polynomial partitioning We are finally ready to prove the Szemerédi-Trotter theorem. M

10 10 MARCO VITTURI Proof. Let M 2 K be a fixed parameter to be chosen later in the proof. We apply polynomial partitioning to the collection P with parameter M, thus obtaining a polynomial P of degree d : degpp q at most OpM 1{2 q, M cells Ω j such that P X Ω j ď P M for all j 1,..., M, and the cell decomposition R n ZpP q \ Ω 1 \... \ Ω M. We will also assume that P is taken of minimal degree, and therefore it is squarefree. We may split the collection of points into those contained in the cells and those contained in the zero set of P ; thus we can write Mÿ IpP, L q IpP X ZpP q, L q ` IpP X Ω j, L q. We estimate the second term in the right hand side first. Fix a cell Ω j and denote by L j the lines that pass through that cell, i.e. j 1 L j : tl P L s.t. l X Ω j Hu, so that IpP X Ω j, L q IpP X Ω j, L j q. We can apply the trivial estimate (1) for point-line incidences to this term, obtaining IpP X Ω j, L j q À P X Ω j L j 1{2 ` L j À P M L j 1{2 ` L j. Now, observe that a line can intersect at most d distinct cells, and therefore by double counting Mÿ L j ÿ Mÿ 1pl X Ω j Hq ď d L ; (2) j 1 lpl j 1 on the other hand, by this and Cauchy-Schwarz ÿ M L j 1{2 2 Mÿ ď M L j ď Md L. (3) j 1 Therefore, by (2) and (3) we can estimate Mÿ IpP X Ω j, L j q À P M M 1{2 d 1{2 L 1{2 ` d L À P M L 1{4 1{2 ` M 1{2 L. (4) j 1 This is superficially similar to the trivial incidence estimate in (1) but we have gained one degree of freedom - namely, we can choose M. Doing so will give us the estimate we want, but we have to deal with the incidences coming from the zero set ZpP q first, which we do next. It will turn out that these give a minor contribution. First of all, notice that if l P L is not contained in ZpP q then it can intersect with ZpP q in at most d points. Therefore, if we define the set of lines contained in ZpP q, we have j 1 L Z : tl P L s.t. l Ă ZpP qu, IpP X ZpP q, L zl Z q ď d L. This term is already contained in (4), so by worsening the constant there we need not worry about it. Thus it suffices to estimate the quantity IpP X ZpP q, L Z q. Now, the points in P X ZpP q are either smooth or singular. In the former case, suppose p smooth is incident to l, l 1 distinct lines in L Z ; then P ppq 0 is orthogonal to both l and

11 LECTURE 2 11 l 1 by Proposition 5, but in the plane this is impossible. Thus every smooth point is incident to at most one line in L Z, and therefore the contribution of the smooth points to IpP X ZpP q, L Z q is at most P. Finally, suppose that p is a singular point, so P ppq 0 and P ppq p0, 0q. Notice that P is square-free, and therefore it doesn t have factors in common with D i P for any i 1, 2. If l Ă ZpP q then l cannot be contained in Zp P q, since otherwise P and D i P (i 1, 2) would share a common factor (the equation of the line); therefore l P L Z may intersect Zp P q in at most d 1 points, and the incidences between singular points and lines in L Z are therefore bounded by d L too, which is harmless. Summing up, we have proved that IpP, L q À P M M 1{2 d 1{2 L 1{2 ` d L ` P ď P M 1{4 L 1{2 ` M 1{2 L ` P, and we can optimize this in M. This is achieved by taking M 3{4 P L 1{2, which gives IpP, L q À P 2{3 L 2{3 ` P, and if P L 1{2 ă 1 then we set M 1 and the trivial estimate gives instead IpP, L q À L, thus proving the theorem. Remark 7. Observe how we have used the rigidity provided by the partitioning hypersurface to amplify the trivial estimate (1) into a much more powerful one, where there is a parameter that can be optimized. Appendix A. Alternative proof by the Crossing Number inequality In this appendix we present a second proof (by Székely, [Szé97]) of the Szemerédi- Trotter theorem that relies on the Crossing Number inequality from graph theory. The proof has nothing to do with the polynomial method, but we decided to present it here to show another amplification trick like the one used in section 4. Let G pv, Eq be a graph, with V its set of vertices and E its set of edges. We define the crossing number CrpGq of the graph G to be the minimum number of edge crossings over all possible proper 7 drawings of G. Theorem 12 (Crossing Number inequality). If E ą 4 V then CrpGq Á E 3 V 2. Once one has this fact, the Szemerédi-Trotter theorem is easily proven. Indeed, given P and L as before, we can split L into L 1 and L 2, where L 1 are the lines in L that contain at most one point in P, and L 2 is the complement. Then IpP, L q IpP, L 1 q ` IpP, L 2 q and IpP, L 1 q ď L, so it suffices to estimate the second term, where every line is incident to at least two points in P. The collections P, L 2 generate a graph 7 A drawing is proper if any point that is not a vertex belongs to at most 2 edges, i.e. every crossing is between only two edges at once.

12 12 MARCO VITTURI G pv, Eq in the plane (with some edges connected to 8). There are exactly P ` 1 vertices in V, since V P Y t8u. If a line contains k points then it gives rise to k ` 1 edges, and this is the only way edges arise; let L k be the number of lines in L 2 that contain exactly k points in P, then IpP, L 2 q ÿ ką1 kl k, and the number of edges is therefore ÿ pk ` 1qL k IpP, L 2 q ` L 2. ką1 Finally, the number of crossings is at most L 2 2, trivially. Therefore, by Crossing Number inequality we have Rearranging this we have L 2 ě CrpGq Á p IpP, L 2q ` L 2 q 3 p P ` 1q 2. IpP, L 2 q À P 2{3 L 2{3 ` L, and summing up with IpP, L 1 q we obtain the desired inequality. Proof of the Crossing Number inequality. The proof relies on the Euler formula for planar graphs. The idea is that if a graph violates Euler s formula, then it can t be planar, and then we can use this information as a wedge to make our way to the inequality. First we can assume that the graph G is connected, since we can work on single connected components. If the graph is planar, we can identify the set F of its faces (including the face surrounding 8); then Euler s formula states that V E ` F 2. Since every face has at least 3 sides (edges) and every edge is adjacent to exactly 2 faces, we have by double counting that 3 F ď 2 E ; therefore in a planar graph F 2 3 E 2 ` 1 E V ď 0, 3 or equivalently 3 V ě E ` 6. (5) It follows that if E ą 4 V then the graph must be non-planar and thus have at least one crossing: E ą 4 V ñ CrpGq ą 0. Now we amplify this inequality by exploiting the freedom to delete edges. To turn G into a planar graph it suffices to delete CrpGq edges, thus obtaining a graph for which (5) does hold. This graph has E CrpGq edges, so we have E CrpGq ă 6 ` E CrpGq ď 3 V, which we rewrite as CrpGq ą E 3 V. (6) Now we amplify this again by exploiting the freedom to delete vertices. When you delete a vertex you must also delete all edges connected to that vertex; since it s hard to figure out what happens to the crossings when you delete edges, we will give an estimate using the probabilistic method. Indeed, take a binomial distribution on the vertices, removing a vertex v P V and the associated edges with probability 1 p; call G 1 pv 1, E 1 q the (random) graph thus obtained. For the random graph G 1 we will then have CrpG 1 q ą E 1 3 V 1,

13 LECTURE 2 13 and if we average over all the random graphs obtained in this way, we have ErCrpG 1 qs ą Er E 1 s 3Er V 1 s. The expectation for the number of points is clearly Er V 1 s p V ; an edges survives if and only if both its vertices survive, and therefore Er E 1 s p 2 E. Finally, suppose that G is drawn with exactly CrpGq crossings; a crossing consists of two edges, and to survive the two edges must survive. Since these are independent events, this happens with probability p 4 for any particular crossing; on the other hand, the resulting drawing for G 1 might not be the optimal one (i.e. it might have unnecessary crossings), so the best we can say is that ErCrpG 1 qs ď p 4 CrpGq. Thus we have shown that p 4 CrpGq ą p 2 E 3p V, having gained a degree of freedom over the estimate (6). Now we can optimize for p P r0, 1s, and it turns out that we need to take p 1 E { V, so that the inequality becomes CrpGq Á E 2 V 2 E, as desired. Notice how we have used the almost trivial inequality (6) and amplified it into the same inequality but with a parameter of freedom, in which we have optimized. This is analogous to what has been done in the proof of the Szemerédi-Trotter theorem by polynomial partitioning, in which we amplified the trivial estimate (1). References [BKT04] J. Bourgain, N. Katz, and T. Tao. A sum-product estimate in finite fields, and applications. Geom. Funct. Anal., 14(1):27 57, [Bou99] J. Bourgain. On the dimension of Kakeya sets and related maximal inequalities. Geom. Funct. Anal., 9(2): , [Cara] Anthony Carbery. The Brouwer Fixed Point theorem and the Borsuk-Ulam theorem. Accessed: [Carb] Anthony Carbery. Multilinear Kakeya, factorization and algebraic topology. Accessed: [CV13] Anthony Carbery and Stefán Ingi Valdimarsson. The endpoint multilinear Kakeya theorem via the Borsuk-Ulam theorem. J. Funct. Anal., 264(7): , [Ele97] György Elekes. On the number of sums and products. Acta Arith., 81(4): , [GK10] Larry Guth and Nets Hawk Katz. Algebraic methods in discrete analogs of the Kakeya problem. Adv. Math., 225(5): , [Gut10] Larry Guth. The endpoint case of the Bennett-Carbery-Tao multilinear Kakeya conjecture. Acta Math., 205(2): , [Ili15] Marina Iliopoulou. Incidence bounds on multijoints and generic joints. Discrete Comput. Geom., 54(2): , [KSS10] Haim Kaplan, Micha Sharir, and Eugenii Shustin. On lines and joints. Discrete Comput. Geom., 44(4): , [Qui10] René Quilodrán. The joints problem in R n. SIAM J. Discrete Math., 23(4): , 2009/10. [ST42] A. H. Stone and J. W. Tukey. Generalized sandwich theorems. Duke Math. J., 9: , [ST83] Endre Szemerédi and William T. Trotter, Jr. Extremal problems in discrete geometry. Combinatorica, 3(3-4): , [Szé97] László A. Székely. Crossing numbers and hard Erdős problems in discrete geometry. Combin. Probab. Comput., 6(3): , Marco Vitturi, Room 4606, James Clerk Maxwell Building, University of Edinburgh, Peter Guthrie Tait Road, Edinburgh, EH9 3FD. address: m.vitturi@sms.ed.ac.uk