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, 99-160). 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 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 1 + 80Ld 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.
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 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 1 + 80Ld 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.
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 2. 4. 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 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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