Problems in Discrete Geometry and Extremal Combinatorics

Zilin Jiang Department of Mathematical Sciences Carnegie Mellon University

A thesis submitted in partial fulﬁllment for the degree of Doctor of Philosophy

April 2016 .Ÿ⌘Ñ6Õ Acknowledgements

IamespeciallygratefultomyadvisorandcoauthorBorisBukh,withoutwhom IwouldnothavebeenabletofinishmyPhD.IwouldliketothankPo-Shen Loh, who first introduced extremal combinatorics to me. I would also like to thank other faculty members of math, physics and computer science departments, Tom Bohman, James Cummings, Alan Frieze, Robert Griths, Rami Grossberg, Bill Hrusa, Gautam Iyer, Giovanni Leoni, Gary Miller, Ryan O’donnell, Ernest Schimmerling and Dejan Slepˇcev, from whom I have learnt a lot. It is a pleasure to acknowledge the positive influence to various aspects of my life from my fellow graduate students, Devi Borg, Joe Briggs, Sam Cohn, Chris Cox, Nicolás Garc´ıa, Nate Ince, Slav Kirov, Jing Liu, Pan Liu, Clive Newstead, Yue Pu, Matteo Rinaldi, Daniel Rodr´ıguez, Thames SaeSue, Xiaofei Shi, Sebastien Vasey, Yan Xu, Jing Zhang, Linan Zhang and Andy Zucker. In addition, I would like to thank my friends Ti↵any Bao, Tianyuan Deng, Duo Ding, Wenying Gan, Staren Gong, Henry Hu, Yi Hu, Tao Huang, Yangyang Huang, Zhijun Huang, Ruizhang Jin, Aoxi Li, Jason Li, Yun Ling, Bin Liu, Lexi Liu, Mingqian Liu, Victoria Lv, Xuele Ma, Yutian Nie, Neal Pei, Sharon Romero, Lily Shen, Roger Shen, Rujun Song, Vincent Sun, Xuening Sun, Maigo Wang, Hong Wang, Zhiyu Wang, Vena Wei, Susan Weng, Liang Wu, Ben Yang, Emma Yang, Yaoqing Yang, Jenkin Xia, Tianying Xia, Zhukai Xu, Gaochi Zhang, Ruixiang Zhang, Ruixun Zhang, Yu Zhao, Yufei Zhao, Hai Zhou, Ye Zhou, Catherine Zhu, Charlene Zhu, Ted Zhu and Ye Zhu, for their advices and encouragements through my ups and downs. Finally, I would like to thank my parents for their unconditional love. Chapter 2 is a version of the paper A slight improvement to the colored Bárány’s theorem, which has been published in Electron. J. Combin.. Chapter 3 is a version of the paper A bound on the number of edges in graphs without an even cycle co-authored with Boris Bukh, which has been accepted to Comb. Probab. Comp.. Chapter 4 is a version of the paper Bipartite algebraic graphs without quadrilaterals co-authored with Boris Bukh, which has been submitted. All these works were carried out under the supervision of Boris Bukh.

4 Abstract

We study several problems in discrete geometry and extremal combinatorics. Discrete geometry studies the combinatorial properties of finite sets of simple geometric objects. One theme of the field is geometric Ramsey theory. Given m geometric objects, we want to select a not too small subset forming a configuration that is “regular” in some sense. The first problem that we study is a colored variant of the point selection problem d in discrete geometry. Given d+1 disjoint n-point sets, P0,...,Pd,inR ,acolorful simplex is the convex hull of d +1 pointseachofwhichcomesfroma distinct Pi. We establish a “positive-fraction theorem” that asserts the existence of a point 2d d+1 common to at least (d+1)(d+1)! n colorful simplices. Extremal combinatorics studies the maximum or minimum size of discrete structures under given constraints. For a graph H,definetheTuránnumberex(n, H) as the maximum number of edges that an H-free graph on n vertices can have. When H is bipartite, the problem of pinning down the order of magnitude of ex(n, H)remainsingeneralasoneofthecentralopenproblemsincombina- torics. The second and the third problems respectively consider the Tuánnumber for two classes of graphs: cycles of even length and complete bipartite graphs. For cycles of even length C ,weshowthatforex(n, C ) 80pk log k n1+1/k +O(n). 2k 2k  · For complete bipartite graphs Ks,t,motivatedbythealgebraicconstructionsof the extremal Ks,t-free graphs, we restrict our attention algebraically constructed graphs. We conjecture that every algebraic hypersurface that gives rise to a Ks,t- free graph is equivalent, in a suitable sense, to a hypersurface of low degree. We establish a version of this conjecture for K2,2-free graphs. Contents

1 Introduction 1

2 A slight improvement to the colored B´ar´any’s theorem 3 2.1 Introduction ...... 3 2.2 Proof of the Theorem 2.4 ...... 5

3 A bound on the number of edges in graphs without an even cycle 11 3.1 Introduction ...... 11 3.2 Graph exploration ...... 13 3.2.1 Trilayered graphs ...... 13 3.3 ⇥-graphs ...... 14 3.3.1 Well-placed ⇥-graphs ...... 15 3.3.2 Finding a trilayered subgraph of large minimum degree ...... 15 3.3.3 Locating well-placed ⇥-graphs in trilayered graphs ...... 19 3.4 Proof of Theorem 3.2 ...... 22

4 Bipartite algebraic graphs without quadrilaterals 25 4.1 Introduction ...... 25 4.2 Conjectures on the (s, t)-grid-free case ...... 28 4.3 Results on the (1,t)-grid-free case ...... 30 4.4 Results on the (2, 2)-grid-free case ...... 31 4.5 Fieldsofﬁnitecharacteristic ...... 34

Bibliography 36

i List of Figures

2.1 3 red points, 3 green points and 3 blue points are placed in the plane. The point marked by a square is contained in 6 (= 2 33) colorful triangles. . . . 5 9 · 2.2 The bird’s-eye view of a triangulation of S2 with a 2-simplex containing 1 and the cone over part of the triangulation...... 6 2.3 An illustration of an 1-simplex , @, cone(@) and a homeomorphism from tocone(@)...... 7 2.4 The illustration shows a cone over part of @cone()withapexc and a point v on the boundary, and how a point w on the line segment [v, c)aremapped under h...... 8 2.5 An illustration of the partition, the result of ﬁlling in against c,andf(cone())mod2. 9

3.1 Aschematicdrawingofatyilayeredgraph...... 14 3.2 An arbitrary long path P ...... 19

ii Chapter 1

Introduction

Discrete geometry studies the combinatorial properties of finite sets of points, lines, circles, planes, or other simple geometric objects. For example, one can ask, “what is the maximum number of incidences between m points and n lines?”, or “what is the minimum possible number of distinct distances occurring among n points in the plane?”. We refer the interested readers to the book of Matouˇsek [Mat02]. One theme of the field is geometric Ramsey theory. Given m geometric objects, we want to select a not too small subset forming a configuration that is “regular” in some sense. In many cases, we obtain “positive-fraction theorems”: the regular configuration has d n size at least ⌦(m). Consider n points in R in general position, and draw all the m = d+1 d simplices with vertices at the given points. The point selection problem asks for a point of R common to as many simplices as possible. In the early eighties, Bárány [Bár82] established a“positive-fractiontheorem”thatassertstheexistenceofapointcommontoatleastc m d · of these simplices, where cd > 0isaconstantdependingonlyond. Gromov [Gro10] made a major breakthrough in improving the lower bound of cd by introducing a topological proof method. Later Krasev [Kar12] found a very short and elegant proof of Gromov’s bound. The problem that we study in Chapter 2 is a colored variant of the point selection d problem. Given d +1 disjoint n-point sets, P0,...,Pd,inR ,acolorfulsimplexistheconvex hull of d+1 points each of which comes from a distinct Pi.Thecoloredpointselectionproblem d d+1 asks for a point of R covered by at least c0 n of the colorful simplices. As it generalizes d · the original point selection problem, one can show that cd cd0 .Karasev[Kar12]indeed 1  proved that cd0 (d+1)! .Basedonhismethodthatcombinesprobabilisticandtopological 2d arguments, we improve his result to c0 matching Gromov’s bound for c . d (d+1)(d+1)! d Extremal combinatorics studies the maximum or minimum size of discrete structures under given constraints. The basic statement of extremal combinatorics is Mentel’s theorem [Man07], proved in 1907, which states that any graph on n vertices with no triangle

1 contains at most n2/4 edges. The natural generalization of this theorem to clique is proved by Turán [Tur41] in 1941. Turán’s theorem states that, on a given vertex set, the graph with the most edges with no clique of size t is the complete and balanced (t 1)-partite graph, in that the part sizes are as equal as possible. For a graph F ,definetheTuránnumberex(n, F )asthemaximumnumberofedgesthat agraphonn vertices can have without containing a copy of F .ForgeneralgraphsF ,we still do not known how to compute the Turánnumber exactly, but if we are satisfied with an approximate answer, the theory becomes quite simple: Erd˝osand Stone [ES46] showed in 1 n 2 1946 that if the chromatic number (H)=t,thenex(n, Kr) 1 t 1 2 + o(n ). When  F is not bipartite, this gives asymptotic result for the Turán number. When F is bipartite, the problem of pinning down the order of magnitude of ex(n, F )remainsingeneralasone of the central open problems in combinatorics. Most of the study of ex(n, F )forbipartiteF has been concentrated on trees, complete bipartite graphs, and cycles of even length. We address the cycles of even length C in Chapter 3. A general bound of ex(n, C ) 2k 2k  n1+1/k + O(n), for some unspecified constant ,wasassertedbyErd˝os[Erd64,p.33]. k · k The first proof was by Bondy and Simonovits [BS74, Lemma 2], who showed the bound for =20k.ThiswasimprovedbyVerstraëte[Ver00]to =8(k 1) and by Pikhurko [Pik12] k k to k = k 1. Pikhurko asked whether limk k/k can be 0. Inspired by the proof of !1 Pikhurko, we answer this question in the positive by improving k to 80pk log k. In the final chapter, Chapter 4, we further study the Turánnumber for complete bipartite graphs Ks,t.SowhichgraphsareKs,t-free with a maximum number of edges? The question was considered by Füredi in his unpublished manuscript [Für88] asserting that every K2,2- free graph with q vertices (for q q )and 1 q(q +1)2 edges is obtained from a projective 0 2 plane via a polarity with q + 1 absolute elements. Motivated by the algebraic constructions that match the upper bounds in the cases that ex(n, Ks,t) has been solved asymptotically, we restrict our attention to algebraic bipartite graphs defined over algebraically closed fields. We conjecture that every algebraic hypersurface that gives rise to a Ks,t-free graph is equivalent, in a suitable sense, to a hypersurface of low degree. We establish a version of this conjecture for K2,2-free graphs. In the following chapters, backgrounds and motivations of the problems will be discussed in more details in their own introductions.

2 Chapter 2

A slight improvement to the colored B´ar´any’s theorem

2.1 Introduction

n Let P Rd be a set of n points. Every d +1ofthemspanasimplex,foratotalof ⇢ d+1 simplices. The point selection problem asks for a point contained in as many simplices as possible. Boros and Füredi [BF84] showed for d = 2 that there always exists a point in R2 contained in at least 2 n O(n2)simplices.Ashortandcleverproofofthisresultwas 9 3 given by Bukh [Buk06]. Bárány [Bár82] generalized this result to higher dimensions:

Theorem 2.1 (Bárány [Bár82]). There exists a point in Rd that is contained in at least c n O(nd) simplices, where c > 0 is a constant depending only on the dimension d. d d+1 d This general result, the Bárány’s theorem, is also known as the first selection lemma. We will henceforth denote by cd the largest possible constant for which the Bárány’s theorem holds true. Bukh, Matouˇsek and Nivasch [BMN10] used a specific construction called the 2 stretched grid to prove that the constant c2 = 9 in the planar case found by Boros and Füredi [BF84] is the best possible. In fact, they proved that c d! .Ontheotherhand, d  (d+1)d d Bárány’s proof in [Bár82]implies that cd (d +1) ,andWagner[Wag03]improveditto 2 c d +1 . d (d+1)d+1 Gromov [Gro10] further improved the lower bound on cd by topological means. His method gives c 2d . Matouˇsek and Wagner [MW14] provided an exposition of the d (d+1)(d+1)! combinatorial component of Gromov’s approach in a combinatorial language, while Karasev [Kar12] found a very elegant proof of Gromov’s bound, which he described as a “decoded and refined” version of Gromov’s proof.

3 The exact value of cd has been the subject of ongoing research and is unknown, except for the planar case. Basit, Mustafa, Ray and Raza [BMRR10] and successively Matouˇsek and Wagner [MW14] improved the Bárány’s theorem in R3.Král’,MachandSereni[KMS12] used flag algebras from extremal combinatorics and managed to further improve the lower bound on c3 to more than 0.07480, whereas the best upper bound known is 0.09375. However, in this chapter, we are concerned with a colored variant of the point selection d problem. Let P0,...,Pd be d +1disjointfinitesetsinR .Acolorful simplex is the convex hull of d +1pointseachofwhichcomesfromadistinctPi. For the colored point selection problem, we are concerned with the point(s) contained in many colorful simplices. Karasev proved:

Theorem 2.2 (Karasev [Kar12]). Given a family of d +1absolutely continuous probability d 1 measures m =(m0,...,md) on R , an m-simplex is the convex hull of d+1 points v0,...,vd with each point vi sampled independently according to probability measure mi. There exists d 1 a point of R that is contained in an m-simplex with probability pd . In addition, if (d+1)! two probability measures coincide, then the probability can be improved to p 2d . d (d+1)(d+1)! By a standard argument which we will provide immediately, a result on the colored point selection problem follows:

Corollary 2.3. If P0,...,Pd each contains n points, then there exists a point that is contained in at least 1 nd+1 colorful simplices. (d+1)! · Our result drops the additional assumption in theorem 2.2, hence improves corollary 2.3:

d Theorem 2.4. There is a point in R that belongs to an m-simplex with probability pd 2d (d+1)(d+1)! .

Corollary 2.5. There exists a point that is contained in at least 2d nd+1 colorful (d+1)(d+1)! · simplices.

d Proof of corollary 2.5 from the Theorem 2.4. Given d+1 sets P0,...,Pd in R each of which d contains n points. Let : R R be the bump function deﬁned by (x1,...,xd)= ! d 1/(1 x2) d i=1 (xi), where (x)=e 1 x <1,andset n(x1,...,xd)=n (nx1,...,nxd)for | | n N.Itisastandardfactthat and n are absolutely continuous probability measures Q2 supported on [ 1, 1]d and [ 1/n, 1/n]d respectively. For each n and 0 k d,deﬁnem(n)(x):= 1 (x p)forx d. Note that N k n p Pk n R (n) 2   2 2 mk is an absolutely continuous probability measure supportedP on the Minkowski sum of Pk 1An m-simplex is actually a simplex-valued random variable.

4 Figure 2.1: 3 red points, 3 green points and 3 blue points are placed in the plane. The point marked by a square is contained in 6 (= 2 33) colorful triangles. 9 · and [ 1/n, 1/n]d.Letm(n) be the family of d +1probabilitymeasuresm(n),...,m(n).By 0 d the Theorem 2.4, there is a point p(n) of Rd that belongs to an m(n)-simplex with probability 2d at least (d+1)(d+1)! . Because no point in a certain neighborhood of inﬁnity is contained in any m(n)-simplex, the set p(n) : n N is bounded, and consequently the set has a limit point p.Suppose { 2 } p is contained in N colorful simplices. Let ✏>0bethedistancefromp to all the colorful simplices that do not contain p.Choosen large enough such that 1/n ✏ and p(n) p ✏. ⌧ ⌧ By the choice of n,ifp is not contained in a colorful simplex spanned by v0,...,vd,then (n) d p is not contained the convex hull of v0 ,...,v0 for all v0 v +[ 1/n, 1/n] .Thisimplies 0 d i 2 i (n) (n) N that the probability that p is contained in an m -simplex is at most nd+1 . Hence p is the desired point contained in N 2d nd+1 colorful simplices. (d+1)(d+1)! · Readers who are familiar with Karasev’s work [Kar12] would notice that our proof of the Theorem 2.4 heavily relies on his arguments. The author is deeply in debt to him.

2.2 Proof of the Theorem 2.4

In this section, we provide the proof of the Theorem 2.4. The topological terms in the proof are standard, and can be found in [Mat03]. In addition to the notion of an m-simplex, in the proof, we will often refer to an (m ,...,m )-face which means the convex hull of d k +1 k d points vk,...,vd with each point vi sampled independently according to probability measure mi. An m-simplex and an (mk,...,md)-face are both set-valued random variables.

Proof of the Theorem 2.4. To obtain a contradiction, we suppose that for any point v in d 2d R ,theprobabilitythatv belongs to an m-simplex is less than pd := (d+1)(d+1)! . Since this

5 Figure 2.2: The bird’s-eye view of a triangulation of S2 with a 2-simplex containing and 1 the cone over part of the triangulation. probability, as a function of point v,iscontinuousanduniformlytendsto0asv goes to infinity, there is an ✏>0suchthatv is contained in an m-simples with probability at most d pd ✏ for all v in R . Let Sd := Rd be the one-point compactification of the Euclidean space Rd.Take [{1} = ✏/d.Chooseafinitetriangulation2 of Sd with one of the d-simplices containing T 1 such that for 0 simplicial complex K, homeomorphic to X, T together with a homeomorphism h: K X. Since the finite triangulation of interest is an k k! extension of the triangulation of a d-simplex X in Rd and h is an identity map, we will freely use topological notions such as “a k-face (as a subset of Sd)” instead of “the image of a k-face in K under h”. With such abuse of language, we can avoid going back and forth between the simplicial complex and the topological space. 3The cone over a space X is the quotient space cone(X):=(X [0, 1]) / (X 1 ). The apex ⇥ ⇥{ } is the equivalence class (x, 1) : x X . { 2 } 4The k-skeleton of a simplicial complex consists of all simplices of of dimension at most k.

6 the k-faces of inductively on dimension k of while we maintain the property that the T image of the boundary of cone()underf,thatisf(@cone()), intersects an (mk,...,md)- face with probability at most (k +1)!(p ✏ + k). We say f is economical over a k-face d d 1 of  if f and satisfy the above property. Unlike Karasev [Kar12], our inductive T construction of f follows the same pattern until k = d 2insteadofd 1. The main innovation of this proof is a di↵erent construction for k = d 1, which enables us to remove the additional assumption in theorem 2.2. Note that for any 0-face in , f(@cone()) = f( ,O )= , . According to the T { } { 1} assumption at the beginning of the proof, f(@cone()) intersects an (m0,...,md)-face, that is, an m-simplex, with probability at most p ✏.Thereforef is economical over 0-faces of d .Thisfinishesthefirststep. T k Suppose f is already defined on cone( ) and it is economical over k-faces of .We T T k+1 are going to extend the domain of f to cone( ) .Indeed,weonlyneedtodefinef on T cone()foreveryk-face of . T Take any k-face of .Supposeconvexhullofv ,...,v ,denotedbyconv(v ,...,v ), T k d k d is an (mk,...,md)-face. Notice that the following statements are equivalent:

1. f(@cone()) intersects conv(vk,...,vd); 2. for some v f(@cone()), the ray with initial point v in the direction vv v intersects 2 k conv(vk+1,...,vd).

We call the union of such rays the shadow of f(@cone()) centered at vk.Sincef is economical over ,theprobabilityforan(m ,...,m )-face to meet f(@cone()) is at most (k +1)!(p k d d ✏ + k), and so there exists v Rd such that the shadow of f(@cone()) centered at v k 2 k intersects conv(v ,...,v ) with probability at most (k +1)!(p ✏ + k). k+1 d d Now, we define f on cone(). First, let g be the homeomorphism from cone()onto the cone over @cone()withapexc such that g is an identity on @cone(). This can be done because cone()ishomeomorphictoa(k +1)-simplexanditiseasytofinda homeomorphism from to cone(@) that keeps @fixed.

e1 e1 e1 e1

e0 e0 e0 e0 c c

Figure 2.3: An illustration of an 1-simplex , @, cone(@) and a homeomorphism from to cone(@).

Next, note that every point w in cone()exceptc is on a line segment [v, c)foraunique point v on @cone(). If t = vw/wc [0, ), then put h(w)=vf(v)+t vvf(v). In addition, 2 1 · k 7 set h(c)= .Thefunctionh maps [v, c)onto[f(v),v)linearlyandthentakestheinversion 1 k centered at vk with radius vk f(v)sothat[f(v),vk )getsmappedontotheraywiththeinitial point f(v)inthedirectionvvk f(v). Evidently, h is a continuous map from cone(@cone()) onto the shadow of f(@cone) centered at vk that coincides with f on @cone().

c v v1 w w0 f(v) h(w)

@cone() f(@cone())

Figure 2.4: The illustration shows a cone over part of @cone()withapexc and a point v on the boundary, and how a point w on the line segment [v, c)aremappedunderh.

Deﬁne f on cone() to be the composition of g and h:

= f @cone() / @cone() / f(@cone()) _ _ _

✏ ✏ ✏ g h cone() / cone (@cone()) / the shadow of f(@cone()) centered at vk .

According to the commutative diagram above, f is well-defined on cone() in the sense that k it is compatible with its definition on cone( ) .Weusethephrase“fillintheboundaryof T cone()againstthecentervk ” to represent the above process that extends the domain of f from @cone()tocone(). To complete the inductive step, we must demonstrate that f is economical over (k +1)- faces of .Pickany(k +1)-face ⌧ of .Let ,..., be the k-faces of ⌧.Observingthat T T 0 k+1 f(@cone(⌧)) = f(⌧ cone(@⌧)) = ⌧ f(cone( )) ... f(cone( )) and that f(cone( )) [ [ 0 [ [ k+1 i i is the shadow of f(@cone(i)) centered at vk which intersects an (mk+1,...,md)-face with probability at most (k+1)!(p ✏+k), we obtain that the probability for an (m ,...,m )- d k+1 d face to intersect f(@cone(⌧)) is dominated by +(k +2)(k +1)!(p ✏ + k) (k +2)!(p d  d ✏ +(k +1)). d 1 We have so far defined a continuous map f on cone( ) such that for any (d 1)- T face of the probability for an (md 1md)-face to intersect D := f(@cone()) is at most T 1 d!(p ✏ +(d 1)). We write f(X)mod2 := y f(X): f (y) X =1(mod2) d { 2 | \ | } for the set of points in f(X)whosefibersinX have an odd number of points. Setm ¯ :=

(md 1 + md)/2. We are going to deﬁne f on cone()suchthat¯m (f(cone())mod2) is less 1 than d+1 .

8 A 3

B B 2 2

A 1

c c

Figure 2.5: An illustration of the partition, the result of ﬁlling in against c,and f(cone())mod2.

Fix a point s in Rd D.Foranypointt in Rd D, if a generic piecewise linear path from \ \ s to t intersects with D an odd number of times, then put t in B,otherwiseputitinA. Here the number of intersections of a piecewise linear path L and D might not be the cardinality 1 of L D. Instead, the number of intersections is precisely x L D f (x) @cone() ,that \ 2 \ | \ | is, it takes the multiplicity into account. Thus we have partitioned Rd D into A and B P \ such that any generic piecewise linear path from a point in A to a point in B meets D an odd number of times. Suppose a := md 1(A), b := md(A)andx :=m ¯ (A)=(a + b)/2. The probability that an (md 1md)-face intersects with D is at least a(1 b)+(1 a)b. Hence 1 1 a(1 b)+(1 a)b

f(@cone(⌧))mod2 = [⌧ f(cone( )) ... f(cone( ))] mod2 [ 0 [ [ d ⌧ f(cone( ))mod2 ... f(cone( ))mod2. ⇢ [ 0 [ [ d

9 1 Thereforem ¯ (f(@cone(⌧))mod2) is less than +(d +1)d+1 =1,andsothedegreeoff on @cone(⌧), denoted by deg (f,@cone(⌧)), is even. Because

deg (f,@cone(⌧)) = 2 deg (f,cone()) + deg (f, ) = deg (f, )(mod2), T T ⌧ X X where the ﬁrst sum and the second sum are over all d-faces and all (d 1)-faces of T respectively, we know that deg (f, )iseven,whichcontradictswiththefactthatf is T identity on . T

10 Chapter 3

A bound on the number of edges in graphs without an even cycle

3.1 Introduction

Let ex(n, F )bethelargestnumberofedgesinann-vertex graph that contains no copy of afixedgraphF . The systematic study of ex(n, F )wasstartedbyTurán[Tur41]over70 years ago, and it has developed into a central problem in extremal graph theory (see surveys [FS13, Kee11, Sid95]). The function ex(n, F )exhibitsadichotomy:ifF is not bipartite, then ex(n, F )grows quadratically in n,andisfairlywellunderstood;ifF is bipartite, ex(n, F )issubquadratic, and for very few F the order of magnitude is known. The simplest classes of bipartite graphs are trees, complete bipartite graphs, and cycles of even length. Most of the study of ex(n, F ) for bipartite F has been concentrated on these classes. In this chapter, we address the even cycles. For an overview of the status of ex(n, F )forcompletebipartitegraphssee[BBK13]. For a thorough survey on bipartite Turán problems see [FS13].

The first bound on the problem is due to Erd˝os[Erd38] who showed that ex(n, C4)= O(n3/2). Thanks to the works of Erd˝os and Rényi [ER62], Brown [Bro66, Section 3], and Kövari, Sós and Turán [KST54] it is now known that

3/2 ex(n, C4)=(1/2+o(1))n .

The current best bounds for ex(n, C6)forlargevaluesofn are

0.5338n4/3 < ex(n, C ) 0.6272n4/3 6  due to F¨uredi, Naor and Verstra¨ete [FNV06].

11 Ageneralboundofex(n, C ) n1+1/k,forsomeunspecifiedconstant ,wasasserted 2k  k k by Erd˝os [Erd64, p. 33]. The first proof was by Bondy and Simonovits [BS74, Lemma 2], who showed that ex(n, C ) 20kn1+1/k for all suciently large n.Thiswasimprovedby 2k  Verstraëte [Ver00] to 8(k 1)n1+1/k and by Pikhurko [Pik12] to (k 1)n1+1/k + O(n). The principal result of the chapter is an improvement of these bounds:

Theorem 3.1. If G is an n-vertex graph that contains no C and n (2k)8k2 , then 2k ex(n, C ) 80 k log k n1+1/k +10k2n. 2k  · p It is our duty to point out that the improvement o↵ered by the Theorem 3.1 is of uncertain 1+1/k value because we still do not know if n is the correct order of magnitude for ex(n, C2k). 1+1/k Only for k =2, 3, 5areconstructionsofC2k-free graphs with ⌦(n )edgesknown[Ben66, Wen91, LU95, MM05]. We stress again that the situation is completely di↵erent for odd cycles, where the value of ex(n, C2k+1)isknownexactlyforalllargen [Sim68]. Our proof is inspired by that of Pikhurko [Pik12]. Apart from a couple of lemmas that we quote from [Pik12], the present chapter is self-contained. However, we advise the readers to at least skim [Pik12] to see the main idea in a simpler setting. Pikhurko’s proof builds a breadth-first search tree, and then argues that a pair of adjacent levels of the tree cannot contain a ⇥-graph1.Itisthendeducedthateachlevelmustbeat least /(k 1) times larger than the previous, where is the (minimum) degree. The bound on ex(n, C )thenfollows.Theestimateof/(k 1) is sharp when one restricts one’s 2k attention to a pair of levels. In our proof, we use three adjacent levels. We find a ⇥-graph satisfying an extra technical condition that permits an extension of Pikhurko’s argument. Annoyingly, this extension requires a bound on the maximum degree.Toachievesuchaboundweuseamodification of breadth-first search that avoids the high-degree vertices. What we really prove in this chapter is the following:

Theorem 3.2. Suppose k 4.IfG is a biparite n-vertex graph of minimum degree at least 2d +5k2, where d max(20 k log k n1/k, (2k)8k), (3.1) · then G contains C2k. p

Theorem 3.1 follows from Theorem 3.2 and two well-known facts: every graph contains abipartitesubgraphwithhalfoftheedges,andeverygraphofaveragedegreedavg contains 1 asubgraphofminimumdegreeatleast 2 davg. 1We recall the deﬁnition of a ⇥-graph in Section 3.3

12 The rest of the chapter is organized as follows. We present our modification of breadth- first search in Section 3.2. In Section 3.3, which is the heart of the chapter, we explain how to find ⇥-graphs in triples of consecutive levels. Finally, in Section 3.4 we assemble the pieces of the proof.

3.2 Graph exploration

Our aim is to have vertices of degree at most d for some k d1/k.Theparticular ⌧ ⌧ choice is fairly ﬂexible; we choose to use := k3. Let G be a graph, and let x be any vertex of G.Westartourexplorationwiththeset

V0 = x , and mark the vertex x as explored. Suppose V0,V1,...,Vi 1 are the sets explored { } in the ﬁrst i steps respectively. We then deﬁne Vi as follows:

1. Let Vi0 consist of those neighbors of Vi 1 that have not yet been explored. Let Bgi be the set of those vertices in Vi0 that have more than d unexplored neighbors, and let

Sm = V 0 Bg . i i \ i 2. Deﬁne 1 Vi0 if Bgi > 2k Vi0 , Vi = | | | | 1 8Sm if Bg V 0 . < i | i| 2k | i | The vertices of Vi are then marked: as explored. 1 We call sets V ,V ,... levels of G.AlevelV is big if Bg > V 0 ,andisnormal otherwise. 0 1 i | i| 2k | i | Lemma 3.3. If d, and G is a bipartite graph of minimum degree at least , then each  v V has at least neighbors in V V 0 . 2 i+1 i [ i+2 Proof. Fix a vertex v V (G). We will show, by induction on i,thatifv V V ,then 2 62 1 [···[ i v has at least neighbors in V (G) (V1 Vi 1). The base case i =1isclear.Suppose \ [···[ i>1. If v Bg ,thenv has d neighbors in the required set. Otherwise, v is not in V 0 2 i i and hence has no neighbors in Vi 1. Hence, v has as many neighbors in V (G) (V1 Vi 1) \ [···[ as in V (G) (V1 Vi 2), and our claim follows from the induction hypothesis. \ [···[ If v V ,thentheneighborsofv are a subset of V V V 0 . Hence, at least 2 i+1 1 [···[ i [ i+2 of these neighbors lie in V V 0 . i [ i+2

3.2.1 Trilayered graphs

We abstract out the properties of a triple of consecutive levels into the following deﬁnition. A trilayered graph with layers V ,V ,V is a graph G on a vertex set V V V such 1 2 3 1 [ 2 [ 3 that the only edges in G are between V and V ,andbetweenV and V .IfV 0 V , 1 2 2 3 1 ⇢ 1

13 V 0 V and V 0 V ,thenwedenotebyG[V 0,V0,V0]thetrilayeredsubgraphinducedby 2 ⇢ 2 3 ⇢ 3 1 2 3 three sets V10,V20,V30.BecausethegraphG that has been explored is bipartite, there are no edges inside each level. Therefore any three sets Vi 1,Vi,Vi0+1 from the exploration process naturally form a trilayered graph; these graphs and their subgraphs are the only trilayered graphs that appear in this chapter. We say that a trilayered graph has minimum degree at least [A : B,C : D]ifeachvertex in V1 has at least A neighbors in V2,eachvertexinV2 has at least B neighbors in V1,each vertex in V2 has at least C neighbors in V3,andeachvertexinV3 has at least D neighbors in V2.

A C ! !

V1 V2 V3

B D

Figure 3.1: A schematic drawing of a tyilayered graph.

3.3 ⇥-graphs

A⇥-graph is a cycle of length at least 2k with a chord. We shall use several lemmas from the previous works.

Lemma 3.4 (Lemma 2.1 in [Pik12], also Lemma 2 in [Ver00]). Let F be a ⇥-graph and 1 l V (F ) 1. Let V (F )=W Z be an arbitrary partition of its vertex set into  | | [ two non-empty parts such that every path in F of length l that begins in W necessarily ends in W . Then F is bipartite with parts W and Z.

Lemma 3.5 (Lemma 2.2 in [Pik12]). Let k 3. Any bipartite graph H of minimum degree at least k contains a ⇥-graph.

Corollary 3.6. Let k 3. Any bipartite graph H of average degree at least 2k contains a ⇥-graph.

For a graph G and a set Y V (G), let G[Y ] denote the graph induced on Y .For ⇢ disjoint Y,Z V (G), let G[Y,Z]denotethebipartitesubgraphofG that is induced by the ⇢ bipartition Y Z. [

14 3.3.1 Well-placed ⇥-graphs

Suppose G is a trilayered graph with layers V ,V ,V .Wesaythata⇥-graphF G is well- 1 2 3 ⇢ placed if each vertex of V (F ) V is adjacent to some vertex in V V (F ). The condition \ 2 1 \ ensures that, for each vertex v of F in V2 there exists a path from the root x to v that avoids F .

Lemma 3.7. Suppose G is a trilayered graph with layers V1, V2, V3 such that the degree of 2 every vertex in V2 is at least 2d +5k , and no vertex in V2 has more than d neighbors in V . Suppose t is a nonnegative integer, and let F = d e(V ,V )/8k V . Assume that 3 · 1 2 | 3| F 2, (3.2a) e(V ,V ) 2kF V , (3.2b) 1 2 | 1| 2 2k 1 e(V ,V ) 8k(t +1) (2k) V , (3.2c) 1 2 | 1| e(V ,V ) 8(et/F )tk V , (3.2d) 1 2 | 2| e(V ,V ) 20(t +1)2 V . (3.2e) 1 2 | 2| Then at least one of the following holds:

1. There is a ⇥-graph in G[V1,V2].

2. There is a well-placed ⇥-graph in G[V1,V2,V3].

The proof of Lemma 3.7 is in two parts: ﬁnding trilayered subgraph of large minimum degree (Lemmas 3.8 and 3.9), and ﬁnding a well-placed ⇥-graph inside that trilayered graph (Lemma 3.10).

3.3.2 Finding a trilayered subgraph of large minimum degree

The disjoint union of two bipartite graphs shows that a trilayered graph with many edges need not contain a trilayered subgraph of large minimum degree. We show that, in contrast, if a trilayered graph contains no ⇥-graph between two of its levels, then it must contain a subgraph of large minimum degree. The next lemma demonstrates a weaker version of this claim: it leaves open a possibility that the graph contains a denser trilayered subgraph. In that case, we can iterate inside that subgraph, which is done in Lemma 3.9.

Lemma 3.8. Let a, A, B, C, D be positive real numbers. Suppose G is a trilayered graph with 2 layers V1, V2, V3 and the degree of every vertex in V2 is at least d +4k + C. Assume also that a e(V ,V ) (A + k +1) V + B V . (3.3) · 1 2 | 1| | 2| Then one of the following holds:

15 1. There is a ⇥-graph in G[V1,V2].

2. There exist non-empty subsets V 0 V , V 0 V , V 0 V such that the induced 1 ⇢ 1 2 ⇢ 2 3 ⇢ 3 trilayered subgraph G[V10,V20,V30] has minimum degree at least [A : B,C : D]. 3. There is a subset V V such that e(V , V ) (1 a)e(V ,V ), and V D V /d. 2 ⇢ 2 1 2 1 2 2  | 3| Proof. We suppose that alternative 1 does not hold. Then, by Corollary 3.6, the average e e e degree of every subgraph of G[V1,V2]isatmost2k. Consider the process that aims to construct a subgraph satisfying 2. The process starts with V10 = V1, V20 = V2 and V30 = V3, and at each step removes one of the vertices that violate the minimum degree condition on G[V10,V20,V30]. The process stops when either no vertices are left, or the minimum degree of G[V10,V20,V30] is at least [A : B,C : D]. Since in the latter case we are done, we assume that this process eventually removes every vertex of G.

Let R be the vertices of V2 that were removed because at the time of removal they had fewer than C neighbors in V30.Put

E0 := uv E(G):u V ,v V , and v was removed before u , { 2 2 2 2 3 } S := v V : v has at least 4k2 neighbors in V . { 2 2 1}

Note that E0 D V .WecannothaveS V /k,forotherwisetheaveragedegreeof | | | 3| | || 1| the bipartite graph G[V ,S]wouldbeatleast 4k 2k.So S V /k. 1 1+1/k | || 1| The average degree condition on G[V1,S]impliesthat

e(V ,S) k( V + S ) (k +1) V . 1  | 1| | |  | 1| Let u be any vertex in R S.Sinceitisconnectedtoatleast(d +4k2 + C) 4k2 = d + C \ vertices of V3, it must be adjacent to at least d edges of E0.Thus,

R S E0 /d D V /d. | \ || |  | 3| Assume that the conclusion 3 does not hold with V = R S.Thene(V ,R S) < 2 \ 1 \ (1 a)e(V1,V2). Since the total number of edges between V1 and V2 that were removed due e to the minimal degree conditions on V and V is at most A V and B V respectively, we 1 2 | 1| | 2| conclude that

e(V ,V ) e(V ,S)+e(V ,R S)+A V + B V 1 2  1 1 \ | 1| | 2| < (k +1) V +(1 a)e(V ,V )+A V + B V , | 1| 1 2 | 1| | 2| implying that a e(V ,V ) < (A + k +1) V + B V . · 1 2 | 1| | 2| The contradiction with (3.3) completes the proof.

16 Remark 3.1. The next lemma can be eliminated at the cost of obtaining the bound ex(n, C2k)= O(k2/3n1+1/k)inplaceofex(n, C )=O(pk log k n1+1/k). To do that, we can set B k2/3, 2k · ⇡ D k1/3 and a =1/2. One can then show that when applied to trilayered graphs arising ⇡ from the exploration process the alternative 3 leads to a subgraph of average degree 2k. The two remaining alternatives are dealt by Corollary 3.6 and Lemma 3.10. However, it is possible to obtain a better bound by iterating the preceding lemma.

Lemma 3.9. Let C be a positive real number. Suppose G is a trilayered graph with layers V1, V , V , and the degree of every vertex in V is at least d+4k2+C. Let F = d e(V ,V )/8k V , 2 3 2 · 1 2 | 3| and assume that F and e(V ,V ) satisfy (3.2) for some integer t 1. Then one of the 1 2 following holds:

1. There is a ⇥-graph in G[V1,V2].

2. There exist numbers A, B, D and non-empty subsets V 0 V , V 0 V , V 0 V 1 ⇢ 1 2 ⇢ 2 3 ⇢ 3 such that the induced trilayered subgraph G[V10,V20,V30] has minimum degree at least [A : B,C : D], with the following inequalities that bind A, B, and D:

D 1 B 5, (B 4)D 2k, A 2k(D) . (3.4) Proof. Assume, for the sake of contradiction, that neither 1 nor 2 hold. With hindsight, set 1 (0) (1) aj = t j+1 for j =0,...,t 1. We shall deﬁne a sequence of sets V2 = V2 V2 (t) ◆ ◆···◆ V2 inductively. We denote by

(i) (i) di := e(V1,V2 )/ V2 (i) (0) (1) (t) the average degree from V2 into V1. The sequence V2 ,V2 ,...,V2 will be constructed so as to satisfy

e(V ,V(i+1)) (1 a )e(V ,V(i)), (3.5) 1 2 i 1 2 i d d Fa (1 a ). (3.6) i+1 i · i j j=0 Y

Note that (3.5) and the choice of a0,...,ai imply that e(V ,V ) e(V ,V(i)) 1 2 . (3.7) 1 2 t +1

(0) (i) The sequence starts with V2 = V2. Assume V2 has been deﬁned. We proceed to deﬁne (i+1) V2 .Put

a e(V ,V(i)) 8k A = i 1 2 k 1,B= 1 a d +5,D=min 2k, . 2 V 4 i i a d | 1| ✓ i i ◆ 17 With help of (3.7) and (3.2c) it is easy to check that the inequalities (3.4) hold for this choice of constants. In addition,

(A + k +1) V + B V (i) = 3 a e(V ,V(i))+5 V (i) | 1| 2 4 i 1 2 2 (3.2e) 3 a e(V ,V(i))+ 1 e(V ,V )  4 i 1 2 4(t+1) 2 1 2 (3.7) a e(V ,V(i)).  i 1 2

(i) So, the condition (3.3) of Lemma 3.8 is satisﬁed for the graph G V1,V2 ,V3 .ByLemma3.8 there is a subset V (i+1) V (i) satisfying (3.5) and V (i+1) DhV /d. i 2 ⇢ 2 2  | 3| (i+1) Next we show that the set V2 satisﬁes inequality (3.6). Indeed, we have (i+1) (i) e(V1,V2 ) (1 ai)e(V1,V2 ) d (i) di+1 = (1 ai)aidi e(V1,V2 ) (i+1) D V3 /d 8k V3 V2 | | | | i 1 i (3.5) d e(V ,V ) (1 a )a d · 1 2 (1 a )=d Fa (1 a ). i i i 8k V j i · i j 3 j=0 j=0 | | Y Y

Iterative application of (3.6) implies

t 1 t 1 1 t t t j t e (F/e) d d F a (1 a ) d F = d . (3.8) t 0 j j 0 t j +1 0 (t +1)! j=0 j=0 Y Y If we have V (t) < V ,thentheaveragedegreeofinducedsubgraphG V ,V(t) is 2 | 1| 1 2 greater than h i (3.7) (3.2c) (t) e(V ,V )/ V e(V ,V )/(t +1) V 2k, 1 2 | 1| 1 2 | 1| which by Corollary 3.6 leads to outcome 1. If V (t) V and d 4k,thentheaveragedegreeofG V ,V(t) is at least d /2 2k 2 | 1| t 1 2 t (t) h i because dtis the average degree of V2 into V1,againleadingtotheoutcome1.So,wemay assume that d < 4k.Since(t +1)! 2tt we deduce from (3.8) that t  d < 4k(t +1)!(e/F )t 8k(et/F )t. 0  This contradicts (3.2d), and so the proof is complete.

18 3.3.3 Locating well-placed ⇥-graphs in trilayered graphs

We come to the central argument of the chapter. It shows how to embed well-placed ⇥- graphs into trilayered graphs of large minimum degree. Or rather, it shows how to embed well-placed ⇥-graphs into regular trilayered graphs; the contortions of the previous two lemmas, and the factor of plog k in the ﬁnal bound, come from authors’ inability to deal with irregular graphs.

Lemma 3.10. Let A, B, D be positive real numbers. Let G be a trilayered graph with layers

V1, V2, V3 of minimum degree at least [A : B,d + k : D]. Suppose that no vertex in V2 has more than d neighbors in V3. Assume also that

B 5(3.9) (B 4)D 2k 2(3.10) D 1 A 2k(D) . (3.11) Then G contains a well-placed ⇥-graph.

Proof. Assume, for the sake of contradiction, that G contains no well-placed ⇥-graphs. Leaning on this assumption, we shall build an arbitrary long path P of the form

V • • • • • • 3

V • • • • • • • • 2

v0 v1 v2 V • • • 1

Figure 3.2: An arbitrary long path P .

where, for each i,verticesvi and vi+1 are joined by a path of length 2D that alternates between V2 and V3.Sincethegraphisﬁnite,thiswouldbeacontradiction. While building the path, we maintain the following property:

Every v P V has at least one neighbor in V P. (3.12) 2 \ 2 1 \ We call a path satisfying (3.12) good.

We construct the path inductively. We begin by picking v0 arbitrarily from V1.Suppose agoodpathP = v0 ! v1 ! ! vl 1 has been constructed, and we wish to ﬁnd a path ··· extension v0 ! v1 ! ! vl 1 ! vl. ···

19 There are at least about A ways to extend the path by a single vertex. The idea of the following argument shows that many of these extensions can be extended to another vertex, and then another, and so on. For each i =1, 2,...,2D 1weshalldeﬁneafamily of good paths that satisfy Qi 1. Each path in i is of the form v0 ! v1 ! ! vl 1 ! u,wherevl 1 ! u is a Q ··· path of length i that alternates between V2 and V3.Thevertexu is called a terminal of the path. The set of terminals of the paths in is denoted by T ( ). Note that Qi Qi T ( ) V for odd i and T ( ) V for even i. Qi ⇢ 2 Qi ⇢ 3 2. For each i,thepathsin have distinct terminals. Qi 3. For odd-numbered indices, we have the inequality

1 i j 3k + A 1 . (3.13) |Q2i+1| D ✓ ◆ j i ✓ ◆ Y 4. For even-numbered indices, we have the inequality

e(T ( 2i),V2) d 2i 1 . (3.14) Q |Q | Let t := B/2 .Wewillrepeatedlyusethefollowingstraightforwardfact,whichwecall d e the small-degree argument:wheneverQ is a good path and u V Q is adjacent to the 2 2 \ terminal of Q,thenu is adjacent to fewer than t vertices in V Q.Indeed,ifvertexu were 1 \ adjacent to v ,v ,...,v V Q with j

(3.10) 2D(t 2) + 2 2D(B/2 2) + 2 2k.

As uvj3 is a chord of the cycle, and u is adjacent to vj1 that is not on the cycle, that would contradict the assumption that G contains no well-placed ⇥-graph. The set consists of all paths of the form Pu for u V P .Letuscheckthatthe Q1 2 2 \ preceding conditions hold for 1.Vertexvl 1 cannot be adjacent to k or more vertices in Q P V2,forotherwiseG would contain a well-placed ⇥-graph with a chord through vl 1.So, \ 1 A k. Next, consider any u V2 P that is a neighbor of vl 1. By the small-degree |Q | 2 \ argument vertex u cannot be adjacent to t or more vertices of P V ,andPu is good. \ 1

Suppose 2i 1 has been deﬁned, and we wish to deﬁne 2i.Consideranarbitrarypath Q Q Q = v0 ! v1 ! ! vl 1 ! u 2i 1.Vertexu cannot have k or more neighbors ··· 2Q in Q V ,forotherwiseG would contain a well-placed ⇥-graph with a chord through u. \ 3 Hence, there are at least d edges of the form uw,wherew V Q. As we vary u we obtain 2 3 \ afamilyofatleastd 2i 1 paths. We let 2i consist of any maximal subfamily of such |Q | Q

20 paths with distinct terminals. The condition (3.14) follows automatically as each vertex of

T ( 2i 1)hasatleastd neighbors in T ( 2i). Q Q Suppose has been deﬁned, and we wish to deﬁne .Consideranarbitrarypath Q2i Q2i+1 Q = v0 ! v1 ! ! vl 1 ! u 2i. An edge uw is called long if w P ,andw ··· 2Q 2 is at a distance exceeding 2k from u along path Q.Ifuw is a long edge, then from u to Q there is only one edge, namely the edge to the predecessor of u on Q,forotherwisethere is a well-placed ⇥-graph. Also, at most i neighbors of u lie on the path vl 1 u. Since ! deg u D,itfollowsthatthereareatleast(1 i/D)degu short edges from u that miss vl 1 ! u.Thusthereisaset of at least (1 i/D)e(T ( 2i),V2)walks(notnecessarily W Q paths!) of the form v0 ! v1 ! ! vl 1 ! uw such that vl 1 ! uw is a path and w ··· occurs only among the last 2k vertices of the walk. From the maximum degree condition on V it follows that walks in have at least 2 W (1 i/D)e(T ( ),V )/d distinct terminals. A walk fails to be a path only if the terminal Q2i 2 vertex lies on P . However, since the edge uw is short, this can happen for at most 2k possible terminals. Hence, there is a of size Q2i+1 ⇢W (1 i/D)e(T ( ),V )/d 2k (3.15) |Q2i+1| Q2i 2 that consists of paths with distinct terminals. It remains to check that every path in Q2i+1 is good. The only way that Q = v0 ! ! vl 1 ! uw 2i+1 may fail to be good is if ··· 2Q w has no neighbors in V Q.Bythesmall-degreeargumentw has fewer than t neighbors 1 \ in V .Sincew has at least B neighbors in V and B t +2,weconcludethatw has at 1 1 least two neighbors in V1 outside the path. Of course, the same is true for every terminal of apathin .Thecondition(3.13)for follows from (3.15), (3.14) and from validity Q2i+1 Q2i+1 of (3.13) for 2i 1. Q Note that 2D 1 is non-empty. Let Q = v0 ! ! vl 1 ! u 2D 1 be an arbitrary Q ··· 2Q path. Note that since 2D 1isodd,u V .BythepropertyofterminalsofV (odd i) 2 2 i that we noted in the previous paragraph, there are two vertices in V Q that are neighbors 1 \ of u.Letvl be any of them, and let the new path be Qvl = v0 ! ! vl 1 ! uvl.This ··· path can fail to be good if there is a vertex w on the path Q that is good in Q,butisbad in Qv .Bythesmall-degreeargument,w is adjacent to fewer than t vertices in Q V that l \ 1 precede w in Q.ThesameargumentappliedtothereversalofthepathQvl shows that w is adjacent to fewer than t vertices in Q V that succeeds w in Q.Since2t 2

21 Lemma 3.7 follows from Lemmas 3.9 and 3.10 by setting C = d + k,inviewofinequality 4k2 + k 5k2.Welosek2 k here for cosmetic reason: 5k2 is tidier than 4k2 + k. 

3.4 Proof of Theorem 3.2

Suppose that G is a bipartite graph of minimum degree at least 2d +5k2 and contains no

C2k.Pickarootvertexx arbitrarily, and let V0,V1,...,Vk 1 be the levels obtained from the exploration process in Section 3.2.

Lemma 3.11. For 1 i k 1, the graph G[Vi 1,Vi,Vi+1] contains no well-placed ⇥-graph.   Proof. The following proof is almost an exact repetition of the proof of Claim 3.1 from [Pik12] (which is also reproduced as Lemma 3.12 below).

Suppose, for the sake of contradiction, that a well-placed ⇥-graph F G[Vi 1,Vi,Vi+1] ⇢ exists. Let Y = V V (F ). Since F is well-placed, for every vertex of Y there is a path i \ avoiding V (F )oflengthi to the vertex x. The union of these paths forms a tree T with x as a root. Let y be the vertex farthest from x such that every vertex of Y is a T -descendant of y.Pathsthatconnectx to Y branch at y.Pickonesuchbranch,andletW Y be the ⇢ set of all the T -descendants of that branch. Let Z = V (F ) W .FromW = V V (F )it \ 6 i \ follows that Z is not an independent set of F ,andsoW Z is not a bipartition of F . [ Let ` be the distance between x and y.Wehave`

Lemma 3.12 (Claim 3.1 in [Pik12]). For 1 i k 1, neither of G[V ] and G[V ,V ]   i i i+1 contains a bipartite ⇥-graph.

The next step is to show that the levels V0,V1,V2,... increase in size. We shall show by induction on i that

22 To prove Theorem 3.2, we only need (3.20); the remaining inequalities play auxiliary roles in derivation of (3.20). Clearly, these inequalities hold for i =0sinceeachvertexofV1 sends only one edge to V0.

Proof of (3.16). By Lemma 3.3 the degree of every vertex in Vi is at least 2d +4k,andso

induc. e(Vi,Vi0+1) (2d +4k) Vi e(Vi 1,Vi) (2d +2k) Vi . | | | |

We next distinguish two cases depending on whether Vi+1 is big (in the sense of the deﬁnition from Section 3.2). If Vi+1 is big, then e(Vi,Vi+1)=e(Vi,Vi0+1), and (3.16) follows. If Vi+1 is normal, then Corollary 3.6 and Lemma 3.12 imply that

1 1 e(V , Bg ) k( V + Bg ) k V + V 0 k V + e(V ,V0 ) i i+1  | i| i+1  | i| 2k i+1  | i| 2 i i+1 and so

1 e(V ,V )=e(V ,V0 ) e(V , Bg ) e(V ,V0 ) k V d V i i+1 i i+1 i i+1 2 i i+1 | i| | i| implying (3.16).

Proof of (3.17). Consider the graph G[Vi,Vi+1]. Inequality (3.16) asserts that the average degree of V is at least d 2k. If (3.17) does not hold, then the average degree of V is at i i+1 least 2k as well, contradicting Corollary 3.6 and Lemma 3.12.

Proof of (3.18). The argument is the same as for (3.17) with G[Vi,Vi0+1]inplaceofG[Vi,Vi+1].

Proof of (3.19). This follows from (3.17) and (3.16).

Proof of (3.20) in the case Vi is a normal level. We assume that (3.20) does not hold and will derive a contradiction. Consider the trilayered graph G[Vi 1,Vi,Vi0+1]. Let t =2logk. Suppose momentarily that the inequalities (3.2) in Lemma 3.7 hold. Then since Vi is normal, each vertex in Vi has at most d neighbors in Vi0+1, and so Lemma 3.7 applies. However, the lemma’s conclusion contradicts Lemmas 3.11 and 3.12. Hence, to prove (3.20) it suces to verify inequalities (3.2a–3.2d) with F = d e(Vi 1,Vi)/8k Vi0+1 . · We may assume that F 2e2 log k, (3.21) and in particular that (3.2a) holds. Indeed, if (3.21) were not true, then inequality (3.16) 2 2 would imply Vi0+1 (d /16e k log k) Vi 1 ,andthus | | 1 2 2 Vi+1 (1 ) Vi0+1 (d /32e k log k) Vi 1 , | | k | | 23 and so (3.20) would follow in view of 32e2 400.  Inequality (3.2b) is implied by (3.19). Indeed,

(3.19) (3.19) 1 e(Vi 1,Vi)=8k Vi0+1 F/d 8k Vi+1 F/d 4F Vi 2k dF Vi 1 , | | | | | | and d k2 by the deﬁnition of d from (3.1). Inequality (3.2c) is implied by (3.1) and (3.16). Next, suppose (3.2d) were not true. Since F/t e2 by (3.21), we would then conclude (3.19) 1 2 1 t Vi+1 (2k) d Vi d(16k ) (F/et) e(Vi 1,Vi) | | | | (3.16) 2 1 2logk 1 2 d(16k ) e e(Vi 1,Vi) d Vi 1 , 16 | | and so (3.20) would follow. Finally, (3.2e) is a consequence of (3.16).

Proof of (3.20) in the case Vi is a big level. We have

(3.18) 1 1 1 1 V V 0 (4k) e(V ,V0 ) (4k) e(Bg ,V0 ) (4k) d Bg | i+1| 2 i+1 i i+1 i i+1 | i| (3.19) 2 1 3 1 2 1 2 (8k )d Vi (16k ) d Vi 1 = d Vi 1 , | | | | 16 | | and so (3.20) holds.

Proof of Theorem 3.2. If k is even, then k/2applicationsof(3.20)yield

dk V . | k|(400k log k)k/2

If k is odd, then (k 1)/2applicationsof(3.20)yield k 1 k d d Vk (k 1)/2 V1 (k 1)/2 . | |(400k log k) | |(400k log k)

Either way, since V

24 Chapter 4

Bipartite algebraic graphs without quadrilaterals

4.1 Introduction

The Turánnumber ex(n, F )isthemaximumnumberofedgesinanF -free graph1 on n vertices. The first systematic study of ex(n, F ) was initiated by Turán [Tur41], who solved the case when F = Kt is a complete graph on t vertices. Turán’s theorem states that, on agivenvertexset,theKt-free graph with the most edges is the complete and balanced (t 1)-partite graph, in that the part sizes are as equal as possible. For general graphs F , we still do not know how to compute the Turán number exactly, but if we are satisfied with an approximate answer, the theory becomes quite simple: Erd˝osand

Stone [ES46] showed that if the chromatic number (F )=t,thenex(n, F )=ex(n, Kt)+ 2 1 n 2 o(n )= 1 t 1 2 + o(n ). When F is not bipartite, this gives an asymptotic result for the Turánnumber. On the other hand, for all but few bipartite graphs F ,theorder of ex(n, F )isnotknown.Mostoftheresearchonthisproblemfocusedontwoclassesof graphs: complete bipartite graphs and cycles of even length. A comprehensive survey is by Füredi and Simonovits [FS13]. Suppose G is a K -free graph with s t.TheKövari–Sós–Turántheorem[KST54] s,t  1 s 2 1/s 2 1/s implies an upper bound ex(n, K ) pt 1 n + o(n ), which was improved by s,t  2 · Füredi [Für96b] to

1 s 2 1/s 2 1/s ex(n, K ) pt s +1 n + o(n ). s,t  2 · Despite the lack of progress on the Tur´anproblem for complete bipartite graphs, there are certain complete bipartite graphs for which the problem has been solved asymptotically, or 1We say a graph is F -free if it does not have a subgraph isomorphic to F .

25 even exactly. The constructions that match the upper bounds in these cases are all similar to one another. Each of the constructions is a bipartite graph G based on an algebraic 2 s hypersurface H.BothpartitesetsofG are Fp and the edge set is defined by: u v if and ⇠ s s only if (u, v) H.Inshort,G = Fp, Fp,H(Fp) ,whereH(Fp)denotestheFp-points of H. 2 s Note that G has n := 2p vertices. In the previous works of Erd˝os,Rényi and Sós[ERS66], Brown [Bro66], Füredi [Für96a], Kollár, Rónyai and Szabó[KRS96] and Alon, Rónyai and Szabó[ARS99], various hypersurfaces were used to define Ks,t-free graphs. Their equations were

x1y1 + x2y2 =1, for K2,2;(4.1a) (x y )2 +(x y )2 +(x y )2 =1, for K ;(4.1b) 1 1 2 2 3 3 3,3 (N ⇡ )(x + y ,x + y ,...,x + y )=1, for K with t s!+1; (4.1c) s s 1 1 2 2 s s s,t (Ns 1 ⇡s 1)(x2 + y2,x3 + y3,...,xs + ys)=x1y1, for Ks,t with t (s 1)! + 1, (4.1d) s where ⇡s : Fp Fps is an Fp-linear isomorphism and Ns(↵) is the ﬁeld norm, Ns(↵):= ! (ps 1)/(p 1) ↵ . Clearly, the coecients in (4.1a) and (4.1b) are integers and even independent of p. With some work, one can show that both (4.1c) and (4.1d) are polynomial equations of degree s with coecients in Fp.Thereforeeachequationin(4.1)canbewrittenas  F (x, y):=F (x1,...,xs,y1,...,ys)=0forsomeF (x, y) Fp[x, y]ofboundeddegree.The 2 previous works directly count the number of Fp solutions to F (x, y)=0andyield H(Fp) = | | 2s 1 2 1/s 3 ⇥(p )=⇥(n ), for each prime p.

Deﬁnition 1. Given two sets P and P , a set V P P is said to contain an (s, t)-grid 1 2 ⇢ 1 ⇥ 2 if there exist S P ,T P such that s = S , t = T and S T V . Otherwise, we say ⇢ 1 ⇢ 2 | | | | ⇥ ⇢ that V is (s, t)-grid-free.

Observe that every F (x, y)derivedfrom(4.1)issymmetricinxi and yi for all i.We s know that (u, v) H if and only if (v, u) H for all u, v Fp.Theresultingbipartitegraph 2 2 2 s s G = Fp, Fp,H(Fp) would be an extremal Ks,t-free graph if H(Fp)hadbeen(s, t)-grid-free. So which graphs are Ks,t-free with a maximum number of edges? The question was considered by ZoltánFüredi in his unpublished manuscript [Für88] asserting that every K -free graph with q vertices (for q q )and 1 q(q +1)2 edges is obtained from a projective 2,2 0 2 2An algebraic hypersurface in a space of dimension n is an algebraic subvariety of dimension n 1. The terminology from algebraic geometry used throughout the article is standard, and can be found in [Sha13]. 3 We need p 3 (mod 4) for (4.1b) to get the correct number of Fp points on H.Ifp 1 ⌘ ⌘ (mod 4), then the right hand side of (4.1b) should be replaced by a quadratic non-residue in Fp.

26 plane via a polarity with q +1absoluteelements.Thislooselyamountstosayingthatall extremal K2,2-free graphs are deﬁned by generalization of (4.1a).

However, classification of all extremal Ks,t-free graphs seems out of reach. We restrict our attention to algebraically constructed graphs. Given a field F and a hypersurface H defined over F,itisnaturaltoaskwhenH(F)is(s, t)-grid-free. Because the general case is dicult, we work with algebraically closed fields K in this chapter. Denote by Ps(K)the s-dimensional projective space over K.WeareinterestedinhypersurfaceH in Ps(K) Ps(K). ⇥ Since standard machinery from model theory, to be discussed in Section 4.5, allows us to transfer certain results over C (the field of complex numbers) to algebraically closed fields of large characteristic, our focus will be on the K = C case. We use Ps for the s-dimensional s s complex projective space and A := P x0 =0 for the s-dimensional complex ane space. \{ } Note that even if H contains (s, t)-grids, one may remove a few points from the projective space to destroy all (s, t)-grids in H.Forexample,thehomogenizationof(4.1b)is

(x y x y )2 +(x y x y )2 +(x y x y )2 = x2y2. 1 0 0 1 2 0 0 2 3 0 0 3 0 0

3 3 2 2 2 The equation deﬁnes hypersurface H in P P .LetV := x0 = x1 + x2 + x3 =0 be a ⇥ { } variety in P3.SinceV P3 H, H contains a lot of (3, 3)-grids. However, H (A2 A2) ⇥ ⇢ \ ⇥ is (3, 3)-grid-free.

Definition 2. A set V Ps Ps is almost-(s, t)-grid-free if there are two nonempty Zariski- ⇢ ⇥ open sets X, Y Ps such that V (X Y ) is (s, t)-grid-free. ⇢ \ ⇥ Suppose the defining equation of H,sayF (x, y), is of low degree in y. Heuristically, for s generic distinct u1,...,us P ,byBézout’stheorem,onewouldexpect F (u1, y)= = 2 { ··· F (u , y)=0 to have few points. So we conjecture the following. s } Informal conjecture. Every almost-(s, t)-grid-free hypersurface is equivalent, in a suitable sense, to a hypersurface whose degree in y is bounded by some constant d := d(s, t).

The right equivalence notion depends on X and Y in Definition 2. We shall discuss possible notions of equivalence in Section 4.2, and make three specific conjectures. Results in support of these conjectures can be found in Section 4.3 and Section 4.4. Before we make our conjectures precise, we note that an analogous situation occurs 1+1/t for C2t-free graphs. The upper bound ex(n, C2t)=O(n ) first established by Bondy– Simonovits [BS74] has been matched only for t =2, 3, 5. The t =2casewasalready mentioned above because C4 = K2,2.Theconstructionsfort =3, 5arealsoalgebraic(see [Ben66, FNV06] for t =3and[Ben66,Wen91]fort = 5). Also, a conjecture in a similar

27 spirit about algebraic graphs of girth eight was made by Dmytrenko, Lazebnik and Williford [DLW07]. It was recently resolved by Hou, Lappano and Lazebnik [HLL15]. The chapter is organized as follows. In Section 4.2 we flesh out the informal conjecture above, in Section 4.3 we briefly discuss the s =1case,inSection4.4wepartiallyresolve the s = t =2case,andfinallyinSection4.5,weconsideralgebraicallyclosedfieldsoflarge characteristic.

4.2 Conjectures on the (s, t)-grid-free case

Given a ﬁeld F,wedenotebyF[x]thesetofhomogeneouspolynomialsinF[x]andby Fhom[x, y] the set of polynomials in F[x, y]thatareseparatelyhomogeneousinx and y. We might be tempted to guess the following instance of the informal conjecture.

False conjecture A. If H is almost-(s, t)-grid-free, then there exists F (x, y) Chom[x, y] 2 of degree d in y for some d = d(s, t) such that H = F =0 .  { } Unfortunately, Conjecture A is false because of the following example. Example 4.1. Consider H := x y + x y + x y =0 and H deﬁned by 0 { 0 0 1 1 2 2 } 1 d d 1 d 1 d x0y0 + x1y0 y1 + x2 y0 y2 + y0 f(y1/y0) =0, (4.2) where f is a polynomial of degree d.Onecancheckthatboth H and H y =0 are 0 1 \{ 0 } (2, 2)-grid-free, whereas equation (4.2) can be of arbitrary large degree in y.

Behind Example 4.1 is the birational automorphism : P2 99K P2 deﬁned by

d d 1 d 1 d (y0 : y1 : y2):= y0 : y0 y1 : y0 y2 + y0 f(y1/y0) .

Note that id is a biregular map4 from H y =0 to H y =0 .Compositionwiththe ⇥ 1\{ 0 } 0\{ 0 } automorphism increased the degree of H0 in y while preserving almost-(2, 2)-grid-freeness. Here is another example illustrating the relationship between birational automorphisms and (s, t)-grid-free hypersurfaces. Example 4.2. Deﬁne H := x y y + x y y + x y y =0 .OnecanalsocheckthatH 2 { 0 1 2 1 0 2 2 0 1 } 2 \ y y y =0 is (2, 2)-grid-free. Behind this example is the standard quadratic transformation { 0 1 2 } 2 from P to itself given by (y0 : y1 : y2)=(y1y2 : y0y2 : y0y1). Note that id is a biregular ⇥ map from H y y y =0 to H y y y =0 . 2 \{ 0 1 2 } 0 \{ 0 1 2 } Let Cr (Ps)bethegroupofbirationalautomorphismsonPs,alsoknownasthe Cremona group.Evidently,thealmost-(s, t)-grid-freeness is invariant under Cr (Ps) Cr (Ps). ⇥ 4A biregular map is a regular map whose inverse is also regular.

28 s s Proposition 4.1. If V1 P P is an almost-(s, t)-grid-free set, then so is V2 := (X ⇢ ⇥ ⇥ s Y )V1 for all X ,Y Cr (P ). 2 Remark 4.1. Though little is known about the structure of the Cremona group in 3 dimensions and higher, the classical Noether–Castelnuovo theorem says that the Cremona group Cr (P2) is generated by the group of projective linear transformations and the standard quadratic transformation. The proof of this theorem, which is very delicate, can be found in [AC02, Chapter 8].

s s We say that sets V1,V2 P P are almost equal if there exist nonempty Zariski-open ⇢ ⇥ s sets X, Y P such that V1 (X Y )=V2 (X Y ). We believe that the only obstruction ⇢ \ ⇥ \ ⇥ to Conjecture A is the Cremona group.

Conjecture B. Suppose H is an irreducible hypersurface in Ps Ps.IfH is almost-(s, t)- ⇥ s grid-free, then there exist Cr (P ) and F (x, y) Chom[x, y] of degree d in y for some 2 2  d = d(s, t) such that H is almost equal to F (id )=0 . { ⇥ }

Remark 4.2. The conjecture is false if the irreducibility of H is dropped. Take H0 and H1 from Example 4.1 and set f(y)=yd in (4.2), where d can be arbitrarily large. Because both H and H are almost-(2, 2)-grid-free, we know that H H is almost-(2, 3)-grid-free. 0 1 0 [ 1 5 s However, one can show that for any Cr (P ), the degree of (id )(H0 H1)iny is d. 2 ⇥ [ In fact, we believe in an even stronger conjecture.

Conjecture C. Suppose H is an irreducible hypersurface in Ps Ps. Let X, Y be nonempty ⇥ s s Zariski-open subsets of P .IfH (X Y ) is (s, t)-grid-free, then there exist Y 0 P ,a \ ⇥ ⇢ biregular map : Y Y 0 and F (x, y) Chom[x, y] of degree d in y for some d = d(s, t) ! 2  such that H (X Y )= F (id )=0 (X Y ). \ ⇥ { ⇥ }\ ⇥ We prove Conjecture C if s =1andifs = t =2,Y = P2 (see Section 4.3 and 4.4 respectively). One special case is when H (As As)is(s, t)-grid-free. In this case, H can be seen as \ ⇥ an ane algebraic hypersurface in 2s-dimensional ane space. The group of automorphisms of As, denoted by Aut (As), is a subgroup of the Cremona group. In this special case, we make a stronger conjecture.

5 1 Let R1 = P1/Q1,R2 = P2/Q2 be rational functions such that =(1:R1 : R2) and set d =degP =degQ ,d =degP =degQ . On the one hand, H := (id )H is deﬁned by 1 1 1 2 2 2 00 ⇥ 0 x +x R +x R = 0, and so deg H d . On the other hand, H is deﬁned by x +x R +x (R + 0 1 1 2 2 y 00 2 10 0 1 1 2 2 d d d 1 d d R ) = 0 and so deg H = dd + d deg G,whereG = gcd Q Q ,P Q Q ,Q P + P Q . 1 y 10 1 2 1 2 1 1 2 1 2 1 2 d 1 d d ⇣ 2 ⌘ Since G = gcd Q1 Q2,Q1P2 + P1 Q2 , it follows that G divides Q2. Hence we estimate that deg G 2d2 and⇣ deg H0 dd1 d2 d⌘ d2. So, deg (H0 H0 ) d.  y 1 y 0 [ 1 29 Conjecture D. Suppose H is an irreducible ane hypersurface in As As.IfH is (s, t)- ⇥ grid-free, then there exist Aut (As) and F (x, y) C[x, y] of degree d in y for some 2 2  d = d(s, t) such that H = F (id )=0 . { ⇥ } Remark 4.3. An automorphism Aut (As)iselementary if it has a form 2

:(x1,...,xi 1,xi,xi+1,...,xs) (x1,...,xi 1,cxi + f,xi+1,...,xs), 7! where 0 = c C,f C[x1,...,xi 1,xi+1,...,xs]. The tame subgroup is the subgroup 6 2 2 of Aut (As)generatedbyalltheelementaryautomorphisms,andtheelementsfromthis subgroup are called tame automorphisms,whilenon-tameautomorphismsarecalledwild.In Example 4.1, we used a tame automorphism to make a counterexample to Conjecture A. It is known [Jun42, vdK53] that all the elements of Aut(A2) are tame. However, in the case of 3 dimensions, the following automorphism constructed by Nagata (see [Nag72]):

(x, y, z)= x +(x2 yz)z,y +2(x2 yz)x +(x2 yz)z,z was shown [SU03, SU04] to be wild. See also [Kur10]. We note that the question on the existence of wild automorphisms remains open for higher dimensions.

4.3 Results on the (1,t)-grid-free case

As for the s =1case,oneisabletofullycharacterize(1,t)-grid-free hypersurfaces. We always assume that H is a hypersurface in P1 P1 and X, Y are nonempty Zariski-open ⇥ subsets of P1 throughout this section.

Theorem 4.2. Suppose H = F =0 , where F (x, y) Chom[x, y]. Let { } 2 r1 r2 rn F (x, y)=f(x)g(y)h1(x, y) h2(x, y) ...hn(x, y) (4.3) be the factorization of F such that h1,h2,...,hn are distinct irreducible polynomials depending on both x and y. Let d be the degree of h in y. Then H (X Y ) is (1,t)-grid-free if i i \ ⇥ and only if f =0 X = and g =0 Y + d + d + + d

deg hi(u, y)

hi(u, y)=g(y)=0,i=1, 2,...,n; h (u, y)=h (u, y)=0,i= j. i j 6

30 Since hi’s are irreducible and distinct, B´ezout’s theorem tells us that each of these systems has no solution in P1 for a generic u. So for a generic u P1, F (u, y)=0hasexactly 2 M := g =0 Y + d + d + + d distinct solutions in Y .Theconclusionfollowsas |{ }\ | 1 2 ··· n M is the maximal number of distinct solutions.

The informal conjecture thus holds when s =1asTheorem4.2implies:

Corollary 4.3. If H (X Y ) is (1,t)-grid-free, then there exists F (x, y) Chom[x, y] of \ ⇥ 2 degree

F˜(x, y):=g1(y)g2(y) ...gm(y)h1(x, y)h2(x, y) ...hn(x, y) is of degree m + d + d + ...d

4.4 Results on the (2, 2)-grid-free case

Throughout the section we assume that H is a hypersurface in P2 P2 and X is a nonempty ⇥ Zariski-open subset of P2.

2 Theorem 4.4. If H (X P ) is (2, 2)-grid-free, then there exists F (x, y) Chom[x, y] of \ ⇥ 2 degree 2 in y such that H (X P2)= F =0 (X P2).  \ ⇥ { }\ ⇥ The theorem resolves Conjecture C for s = t =2,Y = P2. Note that the birational map 2 : Y Y 0 in the conjecture becomes trivial since biregular automorphisms of P are linear. ! Our argument uses a reduction to an intersection problem of plane algebraic curves. The key ingredient is a theorem by Moura [Mou04] on the intersection multiplicity of plane algebraic curves.

Theorem 4.5 (Moura [Mou04]). Denote by Iv(C1,C2) the intersection multiplicity of algebraic curves C1 and C2 at v. For a generic point v on an irreducible algebraic curve C1 of 2 degree d1 in P ,

1 2 2 (d2 +3d2) if d1 >d2; max Iv(C1,C2):C1 C2, deg C2 d2 = C2 { 6⇢  } 8d d 1 (d2 3d +2) if d d . < 1 2 2 1 1 1  2 31 : 2 Corollary 4.6. For a generic point v on an algebraic curve C in P , any algebraic curve C0 with v C0 intersects with C at another point unless C is irreducible of degree 2. 2 

Proof. Suppose C has more than one irreducible components. Let C1 and C2 be any two of them. Since C C is ﬁnite, we can pick a generic point v on C C . Now any algebraic 1 \ 2 1 \ 2 curve C0 containing v intersects C at another point on C2.So,C is irreducible.

Let d and d0 be the degrees of C and C0 respectively. By Theorem 4.5, one can check that I (C, C0) 2. From B´ezout’s theorem, we v 2 deduce that C intersects C0 at another point unless d 2.  In our proof of Theorem 4.4, we think of H as a family of algebraic curves in P2,each of which is indexed by u X and is deﬁned by C(u):= v P2 :(u, v) H .Wecall 2 { 2 2 } algebraic curve C(u)thesection of H at u.AhypersurfaceH is (2, 2)-grid-free if and only if C(u)andC(u0)intersectat 1pointforalldistinctu, u0 X.Thelastpiecethatwe  2 need for our proof is a technical lemma on generic sections of irreducible hypersurfaces.

2 2 Lemma 4.7. Suppose H1 and H2 are two di↵erent irreducible hypersurfaces in P P ⇥ deﬁned by h1(x, y),h2(x, y) Chom[x, y] (Chom[x] Chom[y]) respectively. Denote the section 2 \ [ 2 of Hi at u by Ci(u) for i =1, 2. For generic u P , C1(u) and C2(u) share no common 2 6 irreducible components, and moreover, each Ci(u) is a reduced algebraic curve.

Proof of Theorem 4.4 assuming Lemma 4.7. Suppose H (X P2)is(2, 2)-grid-free. Take \ ⇥ an arbitrary u X and consider algebraic curve C(u)inP2.Weclaimthateveryv C(u)is 2 2 an intersection of C(u)andC(u0)forsomeu0 X u .DefineD(v):= F (x, v)=0 X. 2 \{ } { }\ Since P2 X is Zariski-closed, the set D(v) is either empty or infinite. However, u D(v) \ 2 and the claim is equivalent to D(v) 2. | | Now pick a generic v C(u). We know that point v is an intersection of C(u)andC(u0) 2 2 for some u0 X u and it is the only intersection because H (X P )is(2, 2)-grid-free. 2 \{ } \ ⇥ We apply Corollary 4.6 to C(u)andC(u0)andgetthatC(u)isirreducibleofdegree 2.  Suppose H is defined by F (x, y) Chom[x, y]and 2

r1 r2 rn F (x, y)=f(x)g(y)h1(x, y) h2(x, y) ...hn(x, y) is the factorization of F such that h1,h2,...,hn are distinct irreducible polynomials in

Chom[x, y] (Chom[x] Chom[y]). The set f =0 X is either empty or inﬁnite. So, \ [ { }\ for H (X P2)tobe(2, 2)-grid-free we must have f =0 X = . Similarly, we know \ ⇥ { }\ ; that g =0 = ,thatis,g(y)isanonzeroconstant. { } ; 6 The algebraic curve Ci(u) is reduced in the sense that its deﬁning equation hi(u, y) is square- free.

32 Without loss of generality, we may assume that f(x)=g(y)=1andthatF (x, y)is square-free, that is, r = r = = r =1.LetC (u)bethesectionofH := h =0 1 2 ··· n i i { i } at u for i =1, 2,...,n.FromLemma4.7,weknowthat,foragenericu X, C (u)and 2 i C (u)havenocommonirreduciblecomponentsforalli = j.ThereforeC(u)= n C (u) j 6 [i=1 i has at least n irreducible components, and so n = 1. Now C(u)=C (u)= h (u, y)=0 1 { 1 } for all u X.ByLemma4.7,h (u, y)issquare-freeforgenericu.Thisandthefactthat 2 1 C(u)isirreducibleofdegree 2implythatdegh (u, y) 2 for a generic u X,andso  1  2 deg h (x, y) 2. y 1 

Proof of Lemma 4.7. Let d1,d2 be the degrees of h1,h2 in y respectively. Suppose on the 2 contrary that C1(u)andC2(u) share common irreducible components for a generic u P . 2 So, h1(u, y)andh2(u, y)haveacommondivisorinC[y]. Therefore there exist two nonzero u u polynomials g1 (y) Chom[y]ofdegree

u u By treating the coecients of g1 (y)andg2 (y)asvariables,wecanviewequation(4.4) d1+d2+1 d1+1 d2+1 as a homogeneous system of M := 2 linear equations involving N := 2 + 2 variables. Note that the coecient in the ith equation of the jth variable, say cij,isa polynomial of u,thatis,cij = cij(u)forsomecij(x) C[x]thatdependsonh1,h2 only. 2 Because the system of linear equations has a nontrivial solution and clearly M>N,the rank of its coecient matrix (cij(u)) is

Pk(cij(x)) = 0 in C[x], for all k [L], (4.6) 2 Reversing the argument above, we can deduce that the rank of matrix (cij(x)), over the x quotient ﬁeld C(x), is

x x Multiplying (4.7) by the common denominator of g1 (y)andg2 (y), we get two nonzero polynomials g1(x, y) C[x, y]ofdegree

h1(x, y)g1(x, y)+h2(x, y)g2(x, y)=0, (4.8)

33 which is impossible as gcd (h1,h2)=1anddegy h1(x, y)=d1 > degy g2(x, y). It remains to prove that C1(u) is reduced for generic u.Becauseh1(x, y) / Chom[x], the 2 polynomial h10 (x, y):=@h1(x, y)/@y0 might be assumed to be nonzero. Again, we assume, on 2 the contrary, that h1(u, y)isnotsquare-freeforagenericu P .Thisimpliesthath1(u, y) 2 and h10 (u, y) have a common divisor. The same linear-algebraic argument, applied to h1 and h10 instead of h1 and h2,thenyieldsacontradiction.

We can adapt the proof of Theorem 4.4 to the case when P2 Y is ﬁnite. In this case, \ we obtain a weaker result though.

2 Proposition 4.8. Suppose P Y = v1, v2,...,vn .IfH (X Y ) is (2, 2)-grid-free, then \ { } \ ⇥ either

1. there exists F (x, y) Chom[x, y] of degree 2 in y such that H (X Y )= F = 2  \ ⇥ { 0 (X Y ), }\ ⇥ 2 2. or there exists i [n] such that P vi H. 2 ⇥{ }⇢ Sketch of a proof. We follow the proof of Theorem 4.4 up to the point where we apply 2 Corollary 4.6. Note that Xi := u P : vi C(u) is Zariski-closed for all i [n]. If none { 2 2 } 2 2 2 of those Xi’s equals P ,thenforagenericu P , C(u) does not pass through any of the 2 points in P2 Y .TherestoftheproofofTheorem4.4stillholdsandweendintheﬁrst \ 2 case. Otherwise Xi = P for some i [n], which corresponds to the second case. 2

4.5 Fields of ﬁnite characteristic

Astandardmodel-theoreticargumentallowsustotransferstatementsoverﬁeldsofcharac- teristic 0 to the ﬁelds of large characteristic.

Theorem 4.9. Let be a sentence in the language of rings. The following are equivalent. 1. is true in complex numbers. 2. is true in every algebraically closed ﬁeld of characteristic zero. 3. is true in all algebraically closed ﬁelds of characteristic p for all suciently large prime p.

The theorem is an application of the compactness theorem and the completeness of the theory of algebraically closed field of fixed characteristic. We refer the readers to [Mar02, Section 2.1] for further details of the theorem and related notions. As quantifiers over all polynomials are not part of the language of rings, one has to limit the degree of hypersurface H and the complexity of the open set X in Theorem 4.4. We now formulate the analog over the fields of large characteristic.

34 Theorem 4.10. Let K be an algebraically closed ﬁeld of large characteristic, let H be a hypersurface in P2(K) P2(K) of bounded degree, and let X be a Zariski-open subset of ⇥ P2(K) of bounded complexity (i.e. X is a Zariski-open subset of P2(K) that can be described by some ﬁrst order predicate in the language of rings of bounded length). If H is (2, 2)- 2 grid-free in X P (K), then there exists F (x, y) Khom[x, y] of degree 2 in y such that ⇥ 2  H (X P2)= F =0 (X P2). \ ⇥ { }\ ⇥ The proof essentially rewrites Theorem 4.4 as a sentence in the language of rings to which Theorem 4.9 is applicable. We skip the tedious but routine proof.

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