Case A. P k = Ni ( 2L i k 1 ) + (# big cells) 10d 2 P k.

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1 THE CELLULAR METHOD In ths lecture, we ntroduce the cellular method as an approach to ncdence geometry theorems lke the Szemeréd-Trotter theorem. The method was ntroduced n the paper Combnatoral complexty bounds for the arrangements of curves and spheres by Clarskon, Edelsbrunner, Gubas, Sharr, and Welzl (Dscrete Comput. Geom. (1990) 5, ). In the next lectures, we wll combne deas from the cellular method wth polynomal method. Our goal here s to descrbe some of the man deas of the cellular method as context. We wll not gve complete proofs but nstead sketch the proofs and do heurstc calculatons. As a model problem, we consder usng the cellular method to prove the Szemeréd-Trotter theorem. (It has many other applcatons, and we wll menton some of them later.) Suppose that we have a set L of L lnes n R 2 and we fx a number k n the range 2 k L 1/2. We wsh to bound the number of k-rch ponts P k. The cellular method s a dvde-and-conquer strategy. We cut the plane nto cells. We use elementary estmates to control the number of k-rch ponts n each cell, and f the ponts and/or lnes are well-dvded among the cells, then we get a stronger estmate by dvdng nto peces n ths way. 1. Good cell decompostons Suppose we take d lnes n R 2 (not necessarly n L). The complement of the d lnes has d 2 connected components, whch we call cells. If the lnes are n general poston, the number of cells s d 2. Each lne enters d + 1 cells. So each lne of L enters only a small fracton of all the cells (at most 1/d). In each cell, we wll employ a smple countng estmate for k-rch ponts, whch we proved n the frst lecture on ncdence geometry. Lemma 1.1. (Countng bound) If L s a set of L lnes, and f L k 2 /4, then P k 2L/k. Our strategy s to emply the countng bound n each cell and add up the results. If all the lnes go through a sngle cell and all the k-rch ponts le n that cell, then we haven t ganed anythng by our cell dvson. The dvde-and-conquer algorthm works well f each cell s roughly equal. We consder two precse condtons. A Even dstrbuton of ponts. We suppose that all the k-rch ponts are n the open cells, and the number of k-rch ponts n each open cell s 10 P k /d 2. 1

2 2 THE CELLULAR METHOD B Even dstrbuton of lnes. We suppose that all the k-rch ponts are n the open cells, and the number of lnes of L that enter each cell s 10L/d. In each case, f we are able to choose d, we wll be able to prove the Szemeréd- Trotter bound P k L 2 k 3. Let us examne how the argument would go n each case. We let O denote the open cells. We let L denote the number of lnes of L whch ntersect O. We let N be the number of k-rch ponts n O. Case A. Lemma 1.2. If d 160Lk 2, and f condton A. holds, then P k 8Ldk 1. Proof. We have the followng bounds for N. By the countng bound, f L (1/4)k 2, then N 2L k 1. Also, by assumpton, N 10 P k d 2 for all. We call a cell bg 2 f L (1/4)k. We can bound P k = N n terms of the number of bg cells as follows: P k = N ( 2L k 1 ) + (# bg cells) 10d 2 P k. We also know that L L(d + 1) because each lne enters d + 1 open cells. We can plug ths n to the frst term of the rght-hand sde. Also, we see that the number of bg cells s at most L /(k 2 /4) 8dLk 2. Therefore we get the followng nequalty: P k 4Ldk Ld 1 k 2 P k. If the coeffcent 80Ld 1 k 2 s 1, ths nequalty s vacuous. But as long as 80Ld 1 k 2 1/2, we can shft the term 80Ld 1 k 2 P k to the other sde. Let us assume that 80Ld 1 k 2 1/2. Ths s equvalent to d 160Lk 2. Under ths assumpton, we see that P k 8Ldk 1. Now suppose we were able to arrange d lnes n the plane obeyng condton A. for any d. We could choose d = 160Lk 2, and we would see that P k 2000L 2 k 3, the Szemeréd-Trotter bound. Case B. Ths case s smlar and even a lttle easer. Lemma 1.3. If d 40Lk 2, and f condton B. holds, then P k 4dLk 1. Proof. Snce we have assumed condton B, L 10L/d. If 10L/d (1/4)k 2, then we can apply the countng bound to deduce that N 2L /k. Then we see that P k = N 1 1 2k L 2L(d + 1)/k 4dLk.

3 THE CELLULAR METHOD 3 Now suppose we were able to arrange d lnes n the plane obeyng condton B. for any d. We could choose d = 40Lk 2, and we would see that P k 200L 2 k 3, the Szemeréd-Trotter bound. Ths rases the queston, can we actually fnd d lnes n the plane obeyng condton A or B? We explore ths n the next secton. 2. Are there good cell decompostons? Can we fnd d lnes obeyng condton A? Morally the answer s no. Here s a more precse related queston. Queston 2.1. Gven a set of N ponts n the plane, and an nteger d N 1/2, can we fnd d lnes whch cut the plane nto cells so that each open cell contans 1000N/d 2 ponts? The answer to ths queston s defntely no. Let γ be a closed strctly convex curve, such as a crcle. Pck N ponts on ths curve. Pck d N 1/2. Consder d lnes n the plane. Each lne contans 2 of our ponts, so only a small fracton of the ponts are n the lnes. More mportantly, each lne ntersects γ n at most 2 ponts. Therefore, the lnes cut γ nto 2d peces. One of those peces must contan (N 2d)/(2d) N 1/2 ponts of our set. Ths badly volates our goal. We wanted to fnd N 1/2 lnes that cut R 2 nto cells wth 1 pont n each cell but n fact one of the cells must have N 1/2 ponts n t. Next we ask: Can we fnd d lnes obeyng condton B? Morally the answer s yes (although I m not sure f the answer s lterally yes). The man dea s to choose a subset of d random lnes from L. If we do ths, a typcal edge of the cell decomposton wll ntersect L/d lnes. To get a rough dea of what s happenng, consder a lne l L. For smplcty, let s suppose that t ntersects all the other lnes of L at dfferent ponts. The ntersecton ponts are L 1 ponts along l. Now we randomly pck d of the L lnes - so essentally we randomly pck d of the ntersecton ponts. Now the lne l s cut nto the segments betweent the selected ponts. The average number of ponts n each segment s L/d, and the probablty that a gven segment has KL/d ponts falls off exponentally n K. So very few edges ntersect more than 1000L/d lnes of L. Next we consder the cells of our decomposton. If a cell has 1000 edges, and each edge ntersects 1000L/d lnes of L, then the number of lnes of L whch ntersect the cell s 10 6 L/d. The cell decomposton may have a few cells wth > 1000 edges, but these are also pretty rare. Here s another perspectve. Suppose we frst choose d/2 random lnes of L and look at the resultng cells. Suppose that one of the cells ntersects > KL/d lnes of L for a very large K. When we choose d/2 more random lnes of L, we are very lkely to choose one of the lnes ntersectng ths popular cell. The probabllty that we wll not choose any of these lnes s exp( ck). As we keep addng random lnes,

4 4 THE CELLULAR METHOD popular cells are lkely to be cut down to sze. If we make ths analyss quanttatve, I beleve we fnd that the fracton of cells O where L KL/d s exp( ck). So ths random decomposton nearly obeys condton B. If the heurstcs above are correct, we can arrange that every cell obeys L C(log L)L/d, and ths mples the S-T estmate up to logarthmc losses. In order to prove the real Szemeréd-Trotter theorem wth the cellular method, one has to subdvde the popular cells by addng some lne segments. Ths requres some care, and we don t dscuss the detals here. 3. Good cell decompostons n three dmensons Havng warmed up n two dmensons, now we consder a set L of L lnes n R 3. We consder d planes n R 3 whch typcally dvde R 3 nto d 3 cells. Each lne can only enter d + 1 open cells, so each lne enters only a small fracton of the cells. If the lnes and/or ponts are evenly dstrbuted among the cells, then we get a good bound for the number of k-rch ponts. We agan consder two precse condtons. A Even dstrbuton of ponts. We suppose that all the k-rch ponts are n the open cells, and the number of k-rch ponts n each open cell s 10 P k /d 3. B Even dstrbuton of lnes. We suppose that all the k-rch ponts are n the open cells, and the number of lnes of L that enter each cell s 10L/d 2. In ether case, we get good bounds for P k especally f we can choose d. Lemma 3.1. If d 13L 1/2 k 1, and f condton A. holds, then P k 8Ldk 1. Proof. We have the followng bounds for N. By the countng bound, f L (1/4)k 2, then N 2L k 1. Also, by assumpton, N 10 P k d 3 for all. We call a cell bg f L (1/4)k 2. We can bound P k = N n terms of the number of bg cells as follows: P k = N ( 2L k 1 ) + (# bg cells) 10d 3 P k. We also know that L L(d + 1) because each lne enters d + 1 open cells. We can plug ths n to the frst term of the rght-hand sde. Also, we see that the number of bg cells s at most L /(k 2 /4) 8dLk 2. Therefore we get the followng nequalty: P k 4Ldk Ld 2 k 2 P k. If the coeffcent 80Ld 2 k 2 s 1, ths nequalty s vacuous. But as long as 80Ld 2 k 2 1/2, we can shft the term 80Ld 2 k 2 P k to the other sde. Let us assume that 80Ld 2 k 2 1/2. Ths s mpled by d 13L 1/2 k 1. Under ths assumpton, we see that P k 8Ldk 1.

5 THE CELLULAR METHOD 5 Now suppose that for any d 1 we could choose d hyperplanes so that A holds. We would choose d L 1/2 k 1, and then we would get the bound P k L 3/2 k 2. Condton B s smlar. If we could fnd d = 20L 1/2 k 1 planes obeyng condton B, then t would agan follow that P k L 3/2 k 2. Ths looks lke a promsng route towards our target theorem: Theorem 3.2. If L s a set of L lnes n R 3 wth L 1/2 lnes n any plane and 3 k L 1/2, then P L 3/2 k k Are there good cell decompostons n three dmensons? So we are led to the followng queston. Queston 4.1. If L s a set of L lnes n R 3 wth L 1/2 lnes n a plane, and f d L 1/2, can we fnd d planes obeyng condton A or condton B? We haven t used or mentoned the restrcton L 1/2 lnes n any plane so far. We take a moment to see how t may be relevant. Suppose we consder L lnes whch all le n a plane π. If we nclude the plane π among the d planes, then the planes wll nclude all the k-rch ponts whch completely volates condton A or B. Each other plane ntersects π n a lne (or not at all). So we are now effectvely parttonng the pont of P k π wth d lnes. But these d lnes only cut π nto d 2 peces, and so an average pece must have P k d 2 k-rch ponts and must meet Ld 1 lnes, volatng ether condton. Now we return to our set of lnes L wth L 1/2 lnes n any plane. Is t possble that wth ths restrcton, we can fnd a good cell decomposton obeyng one of the condtons. The answer s stll no. We have the same problems as before wth condton A. If P s a set of ponts on a convex surface lke a sphere, then for any cell decomposton wth d planes, one of the cells wll have P d 2 ponts. Also, P could be a set of ponts on a curve γ. There are many curves γ that ntersect every plane n 10 ponts - say a typcal trefol knot. If P s a set of ponts on such a curve, then any cell decomposton wth d planes has a cell wth P d 1 ponts on t. We also have a problem wth condton B. Suppose that P s a set of ponts on a convex curve γ n R 2, and L s P R R 3. Suppose that one plane s transverse to the x 3 -axs. Ths plane ntersects the lnes n P ponts lyng on a convex curve. Each other plane ntersects the frst plane n a lne. All together, the other planes cut the frst plane nto d 2 faces. But they cut the convex curve nto d segments. Therefore, we get a 2-dmensonal face of our decomposton whch transversely ntersects Ld 1 lnes. Any open cell borderng ths face must ntersect Ld 1 lnes of L. We could also try usng planes that are all parallel to the x 3 axs, but there are smlar problems, and one of the cells stll contans Ld 1 lnes.

6 6 THE CELLULAR METHOD The cellular method works well for ncdences of codmenson 1 objects, such as planes or spheres n R 3. In ths case, we can buld an nterestng cell decomposton by takng a random subset of the planes or spheres. For objects of codmenson > 1, such as lnes n R 3, t has been dffcult to apply the cellular method (at least drectly). Returnng to our queston, there are many examples where we cannot cut space nto evenly matched cells. It s not clear f these examples share a useful structure or property that we could take advantage of. In our counterexamples, t seems that the ponts or lnes ft onto a nce 2-dmensonal surface or 1-dmensonal curve. Does that always happen, or s t just wshful thnkng? 5. Polynomal cell decompostons A unon of d planes s a specal case of an algebrac surface of degree d. The man dea n ths chapter s to cut space nto peces wth a degree d algebrac surface. Allowng an arbtrary degree d surface nstead of just d planes greatly ncreases our flexblty. (When we pck d planes, we have 3d parameters to play wth, but when we pck a degree d surface we have (1/6)d 3 parameters to play wth!) Wth all ths extra flexblty, we can do a much better job of decomposng space nto evenly matched cells. On the other hand, f Z s a degree d surface, then a lne ether les n Z or ntersects Z n d ponts. Therefore, each lne ntersects d + 1 components of the complement of Z exactly the same bound as f Z was a unon of d planes. n Theorem 5.1. If X s any fnte subset of R and d s any degree, then there s n a non-zero degree d polynomal P so that each component of R \ Z(P) contans C(n) X d n ponts of X. We wll prove ths theorem next tme. The proof s a cousn of fndng a degree d polynomal that vanshes at d n prescrbed ponts, but t uses topology nstead of lnear algebra. We should also gve a caveat. The theorem does NOT guarantee that the ponts of X le n the complement of Z(P). In fact t s possble that X Z(P). There are two extreme cases. If all the ponts of X le n the complement of Z(P), then we get optmal equdstrbuton, and we have a good tool to do a dvde-and-conquer argument. If all the ponts of X le n Z(P), then we see that deg(x) d, and we get a good degree bound on X. Generally, X wll have some ponts n Z(P) and some ponts n the complement. On one part of X we get a degree bound and on the other part of X we get good equdstrbuton.

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