Polynomial Method Lecture 1 the Kakeya Conjecture in Finite Fields

THE POLYNOMIAL METHOD LECTURE 1 THE KAKEYA CONJECTURE IN FINITE FIELDS

MARCO VITTURI

Abstract. This series of notes is intended to provide an introduction to the polynomial method through examples of its successful application - mainly in Harmonic Analysis, but other elds will also be considered (e.g. combinatorics, transcendence theory, etc). In this rst set of notes we introduce the Kakeya problem and then the analogous Kakeya problem in the nite elds. This introduction will necessarily be brief and won't do justice to this vast and beautiful eld. For a more comprehensive introduction we refer the reader to e.g. [KT02]. We then give the explicit construction of a Besicovitch set. Finally we present Dvir's solution to the Kakeya conjecture in the nite elds by use of the polynomial method.

Contents 1. The Kakeya conjecture 1 2. The construction of Besicovitch 9 3. Dvir's proof 11 Appendix A. Proof of lemma 2 14 Appendix B. Multiplicity 15 References 16

Notation. throughout these notes, |E| will denote the cardinality of the set E when this is discrete and nite, and its Lebesgue measure otherwise. Context will usually suce to determine which is the case. The expression X À Y will mean that there is a constant C ¡ 0 s.t. X ¤ CY . We might specify the dependence of this constant on additional parameters e.g. α by writing X Àα Y when this is the case. By X Y we will mean X À Y and Y À X.

1. The Kakeya conjecture In 1917 Soichi Kakeya asked what was the minimal area of a region in which a unit segment could be turned around continuously by 360¥. This question became known as the Kakeya needle problem, and an answer came from a construction of Besicovitch which is described in section 2 (originally appearing in [Bes19]). Namely, for every ¡ 0 there exists a set that contains a unit segment in every direction but has area smaller than , and therefore there is no minimal area to such a region - a result that was somewhat surprising for the time1. A generalization of the denitions leads to2

1It was believed a deltoid curve was a sharp example. 2We drop the requirement that the unit segment can be turned continuously because it won't be relevant to us. 1 2 MARCO VITTURI

Denition 1. A Kakeya set is a set E Rn that contains a unit segment in every direction.

Thus the construction of Besicovitch shows that for n 2 these sets can have arbitrarily small Lebesgue measure - and indeed there are Kakeya sets with Lebesgue measure exactly zero3. Moreover, by taking a solid of revolution with cross section a Besicovitch set it can be seen that we can build such sets with Lebesgue measure zero in dimensions n ¡ 2 too. A question arises then: what's the dimension of these sets? this question can of course be asked with respect to various notions of dimension - e.g. Hausdor, upper/lower Minkowski, packing, etc. In 1971 Davies proved in [Dav71] that, although the Lebesgue measure can be zero, when n 2 a Kakeya set must always have Hausdor dimension 2. This result led to the conjecture that the (Hausdor or Minkowski) dimension would still be n for any Kakeya set:

Conjecture 1 (The Kakeya Conjecture (for Hausdor dimension)). Let K be a Kakeya set in n. Then p q . R dimHaus K n Conjecture 2 (The Kakeya Conjecture (for Minkowski dimension)). Let K be a Kakeya set in n. Then p q . R dimMink K n There is an analogous conjecture with upper Minkowski dimension, of course. The conjectures are still wide open for n ¡ 2, although a huge number of partial results exist in the literature. We don't attempt to survey them in here. See [KT02] for a summary up to the year 2000 (thus somewhat outdated, as it doesn't contain the result of Dvir presented in these notes). It is still open even in the weaker case4 of upper or lower Minkowski dimension. Recall that the upper Minkowski dimension is dened as

log |Eδ| dimMinkpEq : lim sup n ¡ , δÑ0 log δ where Eδ is the δ-neighbourhood of E, and analogously the lower Minkowski dimension is dened as | | p q ¡ log Eδ dimMink E : lim inf n ; δÑ0 log δ when they coincide, they are commonly referred to as the Minkowski dimension of E. Here we assume that 0 δ ! 1. If K is a Kakeya set in Rn then its δ-neighbourhood contains a δ-tube in every direction, where by δ-tube we mean exactly the δ- neighbourhood of a unit segment. Since the tubes have cross section of radius δ it doesn't make sense to distinguish directions that dier by an angle less than δ (they become indistinguishable); therefore we might restrict ourselves to sets that are unions of N distinct δ-tubes that have δ-separated directions, in the sense that 1 1 ¡ 1 whenever T,T are δ-tubes with distinct directions ω, ω P Sn 1 then =ω, ω Á δ. Notice the δ-separation condition forces N to be Opδ1¡nq, and so we can assume - and we will do so from now on - that the collection is critical in the sense that N δ1¡n. With this setup we have that conjecture 2 is equivalent to

3They can be built by specializing the construction in section 2 to nested collections of paral- lelograms, for example, and then taking intersections. 4Weaker because p q ¤ p q ¤ p q when they are all dened. dimHaus E dimMink E dimMink E THE KAKEYA PROBLEM 3

Conjecture 2 (The Kakeya Conjecture (for Minkowski dimension)). Let t uN Tj j1 be a collection of δ-tubes in Rn with δ-separated directions. Then for every ε ¡ 0 § ¤N § § § ε Tj Áε δ . (1) j1 Remark 1. At the heuristical level, conjecture 1 is saying that the δ-separated δ-tubes are essentially disjoint, up to a logarithmic error. Indeed, if they were exactly disjoint, then we'd have

Now, it's a simple geometric fact that if the angle between Tj and Tk is θ then δ2 |T X T | À , j k |θ| so that

Opδ¡1q Opδ¡1q Opδ¡1q ¸ ¸ ¸ δ2 ¸ 1 |T X T | À N log δ¡1. j k kδ k k1 j,k : k1 k1 |=ωj ,ωk|kδ Thus by Cauchy-Schwarz ¸ ¸ § ¤ § 2 2 2 1 χ ¤ χ χ À log δ¡1§ T §, Tj L1 Tj L2 j Tj L2 j j j j which gives the desired inequality.

There is also a stronger variant of the Kakeya conjecture(s) that is more directly relevant to harmonic analysts, namely the Kakeya maximal function conjecture. This can be stated in the language of δ-tubes too: Conjecture 3 (The Kakeya maximal function conjecture). Let t uN be a col- Tj j1 lection of δ-tubes in Rn with δ-separated directions (as above). Then for every ε ¡ 0 ¸N À ¡ε (2) χTj n{pn¡1qp nq ε δ . L R j1 The conjecture can also be put in the more suggestive form

¸N ¸N À ¡ε | |qpn¡1q{n χTj n{pn¡1qp nq ε δ Tj , L R j1 j1 4 MARCO VITTURI which is another assertion of essential disjointness of the tubes. A few observations are in order. Firstly, for 1 d ¤ n (not necessarily integer) notice that (2) implies the inequality

¸N ¡ n ¡ À 1 d ε (3) χTj d{pd¡1qp nq ε δ . L R j1 Indeed, we have the (sharp) trivial inequality

¸N À 1¡n χTj 8p nq δ , L R j1 thus if P r s is such that d¡1 p ¡ q n¡1 1 , then by logarithmic Hölder δ 0, 1 d 1 θ n θ 8 inequality

¸N ¡ ¡ ¡ ¡ p q ¡ n À p εq1 θp 1 nqθ O ε 1 d χTj d{pd¡1qp nq ε δ δ δ δ . L R j1 Notice that n d for , so n is the endpoint of a family of estimates n¡1 d¡1 d n n¡1 really. Secondly, it is stronger than conjecture 1 because it implies it; notice in particular that the expression on the left in the conjectured inequality is not just measuring the Lebesgue measure of the union but is taking into account the multiplicity of overlapping too, so this is to be expected. The implication goes as follows (im- plicitely already used above): by Hölder

¸N ¸N § ¤N § ¡ε§ §1{n ¤ j À χTj 1p nq χTj n{pn¡1qp nq χ T np nq ε δ Tj ; L R L R j L R j1 j1 j1 but then ¸N ¸N | | χTj 1p nq Tj 1, L R j1 j1 and thus (1) is implied. As it turns out, conjecture 3 also implies the Kakeya conjecture for Hausdor dimension, but we won't comment further on this here. Thirdly, the name of the conjecture refers to an actual maximal function that is hidden in the statement through duality. The Kakeya maximal function is given by » ¦p q 1 | p q| fδ ω : sup | | f y dy, T δ¡tube s.t. T ||ω T T ¡ where ω P Sn 1 and T ||ω means that the tube T has direction parallel to that of ω. We shall show that conjecture 3 is equivalent to the boundedness (up to a logarithm) of this maximal function in an appropriate range; but in order to do so we introduce a small lemma rst. Lemma 2. The estimate 3 is equivalent to

¸N À ¡ε (4) ajχTj d{pd¡1qp nq ε δ , L R j1 ° uniformly in t uN P d{pd¡1q s.t. d{pd¡1q À 1¡n. aj j1 ` j aj δ

Proof. See appendix A for the ñ implication. For the full proof see [Mat15]. THE KAKEYA PROBLEM 5

We see that by duality the estimate (4) is equivalent to » ¸N ¡ε n | | À } } np nq @ P aj g dy ε δ g L R g L ; j1 Tj since this must hold for all collections of δ-separated δ-tubes then it's equivalent to » ¸N ¡ε | | À } } np nq sup aj g dy ε δ g L R , t u Tj j j1 Tj where the supremum is taken over such collections. If Ω is a maximal set of δ- ¡ separated direction in Sn 1 then the last expression on the left hand side is com- parable to ¸ » aω sup |g| dy, ¡tube s.t. || ωPΩ T δ T ω T where we have written a for a , with ω direction of T ; notice that ω j » j | | n¡1 ¦p q sup g dy δ gδ ω , T δ¡tube s.t. T ||ω T so taking the supremum in the sequences aj the above is equivalent to ¸ ¨ ¦ n 1{n ¦ g pωq }g } np n¡1q, δ δ L S ωPΩ and therefore conjecture 3 is equivalent to

Conjecture 3 (the Kakeya maximal function conjecture). For all ε ¡ 0 ¦ ¡ε }f } np n¡1q À δ }f} np nq. δ L S ε L R Remark 2. By testing it against the characteristic function of a Besicovitch set as constructed in section 2 we see that the inequality would certainly be false without the logarithmic factor δ¡ε. Finally, conjecture 3 can be restated equivalently in terms of just characteristic functions of sets that are supported inside a family of tubes. We include it for completeness.

Conjecture 3 (the Kakeya maximal function conjecture). Let t uN be a collec- Tj j1 tion of -tubes in n with -separated directions. Let ; if t uN is a δ R δ 0 λ 1 Ej j1 collection of sets s.t. |Ej| ¡ λ|Tj| for all j, then for all ε ¡ 0 and all 1 ¤ d ¤ n § ¤N § § § d n¡d ε Ej Áε λ δ . j1 The advantage of these formulations involving δ-tubes is that the combinatorial nature of the problem is immediately evident: we want to study the overlapping of tubes in space and prove that under our conditions they don't overlap much in one sense or another.

The Kakeya maximal function conjecture was solved for n 2 by Córdoba in [Cor77] by means of geometric arguments a little more sosticated than those in [Dav71], and in a dierent way by Bourgain in [Bou91]. A fundamental feature of the two dimensional problem is that the endpoint exponent n{pn ¡ 1q is just 2, which provides a nice geometrical interpretation as used above, namely that ¸ ¸ 2 | X | χTj L2 Tj Tk . j j,k 6 MARCO VITTURI

In subsequent years two conceptually simple but clever arguments appeared that provided lowerbounds for the dimension of a Kakeya set in any dimension n. These are § § i) the Bush argument of Bourgain [Bou91], that provided the lowerbound § T § Á § § j j pn¡1q{2; the idea behind the argument is that if § § is small then there δ j Tj must be a point covered by a high number of δ-tubes; but these tubes have δ-separated directions and thus at distance 1{10 from the point they must be disjoint from each other - thus having large total mass; ii) §the Hairbrush§ argument of Wol [Wol95], that improved the lowerbound to § § Á pn¡2q{2; the idea is that instead of looking for just a point of high j Tj δ multiplicity we look for a δ-tube of high multiplicity, in the sense that a large fraction of its points have high multiplicity. Then if the tubes that intersect such a tube are arranged in bushes we might argue somewhat as in the Bush argument (only with way more bushes now); if the tubes in distinct bushes are not essentially disjoint the argument doesn't work as well, but one can rule out this case separately. Notice that by denition of Minkowski dimension the latter result implies that a Kakeya set has (lower) Minkowski dimension at least ¡ n¡2 n 2 . This gives n 2 2 5 for a -dimensional Kakeya set. Improving upon the Hairbrush argument proved 2 3 extremely dicult - e.g. Katz, aba and Tao improved the lowerbound for the dimension to 5 ¡10 in [KT00] with a huge amount of eort (compared to 2 10 the simplicity of the Hairbrush argument). Since it seemed that the techniques available at the time were not able to push this much further, Wol came up with a brilliant idea: to ask the analogous questions in the simpler setting of nite elds. Indeed, it is possible to dene Kakeya sets in n (where is a power of a prime) as Fq q sets that contain an -line in every direction. Lines in n are of the form Fq Fq t s.t. P u `x,v x tv t Fq for P n P n{ ¢ (i.e. is identied with 1 i 1 for some P ¢). x Fq , 0 v Fq Fq v v v tv t Fq Thus the analogous of the Kakeya conjecture in this setting takes the form

Conjecture 4 (the Kakeya conjecture in nite elds). Let K be a Kakeya set in n. Then for every ε ¡ 0 Fq § § § § Á | |n¡ε K ε Fq . Since there are no non-trivial neighbourhoods in n (it has the discrete topology), Fq this is the only possible form of a Kakeya conjecture in the nite elds - there are no tubes-equivalents because there are no tubes. An equivalent way to look at this is to say that in the nite elds there are no scales - as opposed to the Rn case, where there is a continuum of scales. The insight behind this proposal of Wol is that many of the arguments given for the real Kakeya case adapt to the nite elds case nicely and become actually somewhat cleaner, while mantaining the same role for the main ideas used. The hope is then that an improvement or solution in the simpler nite elds case would shed light on the real case and perhaps translate to a corresponding solution too. To support this point we present briey the aforementioned Bush and Hairbrush arguments in n. We write in place of with the understanding that the nite Fq F Fq eld is xed. i) the Bush argument: let µ be a xed multiplicity parameter to be chosen later; then either there is a point p P K such that there are µ lines in K passing through p or for every point in K there are fewer than µ lines passing through it. In the former case, since the lines have distinct directions, they must become THE KAKEYA PROBLEM 7

disjoint when the point p is removed; this implies |K| Á µ|F|. In the latter case, by double counting ¸ ¸ n¡1 |F||F| 1pp P `q ¤ µ|K|, pPK `PK which gives the lower bound | |n |K| ¥ F . µ p ¡ q{ By optimizing in µ we get (for µ |F| n 1 2) pn 1q{2 |K| Á |F| . ii) the Hairbrush argument: rst notice that by repeating almost verbatim5 the argument of Davies given above for the Kakeya conjecture (with Minkowski dimension) in dimension n 2 we get that if K is a Kakeya set in F2 then 2 |K| Á |F| . By being slightly more careful we might see that if is a set of 2 that Km F contains a line in m distinct dimensions then actually | | Á | | (5) Km m F . Let then µ be a xed multiplicity parameter to be chosen later. We say that a line ` has high multiplicity if for at least |F|{2 points p P ` there are at least µ lines (distinct from `) in K passing through p. Then either there exists a line of high multiplicity or it doesn't. In the former case, let `˜ be such a line of high multiplicity, and consider the family Π of 2-dimensional planes passing through `˜. If a line ` intersects with `˜then there is a unique π P Π s.t. `, `˜ π. ˜ Let Lπ denote the set t` π such that ` intersects `u. Then K X π is a set in (a isomorphic copy of) 2 that contains at least | | lines, and thus by the F Lπ 2-dimensional Kakeya estimate (5) above | X | Á | || | K π Lπ F ; therefore ¸ ¸ | | ¥ |p X qz˜| Á | | | | Á | |2 K K π ` F Lπ µ F , πPΠ πPΠ where the last inequality is due to the fact that since `˜ has high multiplicity then it intersects at least µ|F|{2 lines (they must all be distinct). In the case there are no lines of high multiplicity, let K1 : tp P K such that p belongs to at most µ lines in Ku; 1 notice that, by assumption, for any line ` K we have |K X `| ¡ |F|{2. Therefore, by double counting as before ° ° P 1 P 1pp P `q |K| ¥ |K1| ¥ p K ` K µ 1 ¸ 1 |K1 X `| Á | |n¡1| |. µ µ F F `PK Thus we have the estimates

2 n |K| Á µ|F| or |K| Á |F| {µ;

5 The bound on the intersection of tubes becomes simply |` X `1| ¤ 1. 8 MARCO VITTURI

optimizing in µ we see that we get pn 2q{2 |K| Á |F| . Being very promising, the nite elds case was intensively studied, most notably in [MT04], but still without substantial advancements. It then came as a big surprise that Dvir, in 2008, was able to prove the Kakeya conjecture in nite elds by a simple application of the polynomial method - without logarithmic factors, too. In section 3 we will present his beautifully simple proof as an illustration of the power of the polynomial method. Remark 3. We haven't yet mentioned one of the major results of the eld, namely the solution of the Multilinear Kakeya conjecture. We introduce it briey (for now) because it provided a fertile ground for the study of polynomial method techniques and is thus very much relevant to the topic of these notes. We can imagine writing, in the estimate in the Kakeya maximal function conjecture, the left hand side as » ¡ © ¸ ¹n ¸ 1{pn¡1q n{pn¡1q χTj Ln{pn¡1q χTj dx; n j R k1 j the multilinear Kakeya conjecture may be thought of as a somewhat restricted version of the Kakeya maximal function conjecture, where the factors in the product above are allowed to be distinct but each sub-collection of tubes is assumed to be all oriented in roughly the same direction (within Op1q of that of a coordinate axis) and these directions must be distinct for distinct families. More precisely

Conjecture 5 (the Multilinear Kakeya conjecture). Let t uNj for Tj,a a1 j 1, . . . , n be n families of δ-tubes with δ-separated directions, such that the directions of the tubes in t uNj form an angle of at most p q¡1 with the coordinate vector . Tj,a a1 10n ej Then for every ε ¡ 0 » ¡ N © ¹n ¸j 1{pn¡1q ¹n À ¡ε p 1{pn¡1qq χTj,a dx ε δ δNj . n R j1 a1 j1 Notice repetitions within a family are allowed. As before, this estimate can be seen as the endpoint of a family of estimates for a range of exponents; namely for n ¤ ¤ 8 the corresponding estimate would be n¡1 p ¡ N © ¹n ¸j ¹n À ¡ε p n{p q χTj,a ε δ δ Nj . Lp{n j1 a1 j1 This conjecture was solved by Bennett, Carbery and Tao in [BCT06], except for the endpoint. Amongst the techniques employed in the proof are the heat ow method and induction on scales6. The endpoint case was later solved too, by Guth, in [Gut10] - and the solution was directly inspired by Dvir's proof for the (linear) Kakeya conjecture in nite elds! Now, adapting polynomial methods to the multilinear Kakeya case is absolutely not straightforward and Guth ended up using large amounts of advanced algebraic geometry and algebraic topology tools. Some time later though, Carbery and Valdimarsson were able in [CV13] to successfully reprove the endpoint result with a proof that reduced all of the algebraic topology needed to just the Borsuk-Ulam theorem - which is needed to prove the polynomial Ham Sandwich theorem. Informally, the polynomial Ham Sandwich theorem states

6This turns out to be the crucial point, since recently Guth has given a short proof of the Multilinear Kakeya inequality (except for the endpoint) using nothing but an induction on scales and the Loomis-Whitney inequality. See [Gut15] THE KAKEYA PROBLEM 9 that given bodies in n we can nd a polynomial in r s of degree at N R R X1,...,Xn most OpN 1{nq s.t. the hypersurface given by its zero set ZpP q exactly bisects all of the N bodies. The next set of notes will contain a more extensive discussion of the polynomial Ham Sandwich theorem and its applications, as it's one of the gems of the polynomial method. Remark 4. We have not mentioned the relationship between the Kakeya conjecture and one of the other big open problems in Harmonic Analysis - the Restriction conjecture. The two are deeply related in extremely interesting ways, but a discussion of this topic would take us too much o track. Suces to say that in general the Restriction conjecture implies the Kakeya conjecture, but partial implications in the other direction are also possible, as the Kakeya conjecture can provide useful estimates on oscillatory integrals. We direct the interested reader to the surveys in [Tao01] and [Wol99].

2. The construction of Besicovitch We describe in this section a geometric construction that for every ¡ 0 allows one to produce a set E containing a unit segment in every direction (in a xed arc of S1) with |E| in a nite number of operations. The resulting set is called a Besicovitch set, but the construction presented here is due to Perron7. It's a simplication of Besicovitch's original one.

Fix a parameter 1 ¡ λ ¡ 1{2. Take then a triangle ABC of height 1, and let AB be its basis and M the middle point of AB. Consider the two sub-triangles AMC and MBC and slide one onto the other along the direction of the basis until their respective bases partially overlap by a certain amount specied as follows: the ratio of the total length of the overlapping bases to the length AB is equal to λ. Call the resulting gure 1p q. Notice that 1p q still contains (copies of) Tλ ABC Tλ ABC all the directions of unit segments in ABC. See g. 2 The second iteration goes as

C C' C

F G

A M B A M' M B

Figure 1. The rst application, with ABC on the left and 1p q on the right. Tλ ABC follows. Subdivide into and as before and apply 1 separately ABC AMC MBC Tλ to these two sub-triangles; then slide the resulting 1p q and 1p q onto Tλ AMC Tλ MBC each other along the direction of AB in such a way that the total length of the overlapping bases of the gures is λ times the sum of their separate lengths. Call 2p q the resulting gure. Notice the basis has now length 2 times that of Tλ ABC λ AB. For the n-th iteration, do as follows. Subdivide AB into AMC and MBC as

7the gure produced is also referred to as a Perron tree in the literature. 10 MARCO VITTURI before and apply n¡1 separately to these two sub-triangles; then slide the resulting Tλ n¡1p q and n¡1p q onto each other along the direction of in such Tλ AMC Tλ MBC AB a way that the total length of the overlapping bases of the gures is λ times the sum of their separate lengths. Call np q the resulting gure. And so on. Tλ ABC Now, since it's obvious that the set of directions is invariant under n, it remains Tλ

Figure 2. The resulting gure after the second and third iteration ( 2p q and 3p q respectively; unlabeled). Tλ ABC Tλ ABC to prove that the area of np q goes to as goes to innity. With reference Tλ ABC 0 n to gure 2, consider 1p q. Some simple considerations of euclidean geometry Tλ ABC show that ABE is similar to the original triangle ABC and the sidelength ratio is λ; moreover, the triangles C1EF and CGE are each union of a triangle similar to AMC and one similar to MBC, with ratio p1 ¡ λq (draw a line through E parallel to AB). Therefore we have for the area | 1p q| p 2 p ¡ q2q| | Tλ ABC λ 2 1 λ ABC . Analogous simple geometrical considerations allow one to conclude for the general case that | np q| ¤ p 2n p ¡ 2q p ¡ 2q 2 p ¡ 2q 2n¡2q| | Tλ ABC λ 2 1 λ 2 1 λ λ ... 2 1 λ λ ABC ¤ pλn 2p1 ¡ λqq|ABC|, which can be made arbitrarily small by choosing λ suciently close to 1 and n suciently large. See [Ste93], ch. X, for additional detail. Remark 5. The phenomenal construction of Besicovitch has found many applications in Harmonic Analysis beyond the Kakeya needle problem. Indeed, the construction was originally devised by Besicovitch (in 1917) in order to answer a question pertaining Riemann-integrability: namely, is it true that if f is Riemann- integrable in the plane then we can nd two orthogonal³ axes with respect to which x ÞÑ fpx, yq is Riemann-integrable and y ÞÑ fpx, yq dx is Riemann-integrable? The answer is negative, since one can use the Besicovitch set to build a counterex- ample (see e.g. [Fal86] for details). Another extraordinary application of Besicovitch sets was found by C. Feerman [Fef71], who used one to disprove the ball-multiplier conjecture, proving that the x 8 2 multiplier dened by T f χBp0,1qf is bounded only on L . This in particular implies (though it's not entirely straightforward it does) the following negative result for spherical summation of double (and higher) Fourier series: if f P Lpp 2q and ¸ T f fˆpm, nqe2πipmx nyq P m,n N

8It was believed that would be bounded on p for all 4 . T L 3 p 4 THE KAKEYA PROBLEM 11 is its Fourier series, then ¸ lim fˆpm, nqe2πipmx nyq RÑ8 P m,n N m2 n2 R need not converge to f in Lp-norm when p 2. This is in sharp contrast with the one dimensional case, in which one has Lp convergence for all 1 p 8. Even worse perhaps, by a similar argument, it was shown in [MN73] that for any p 2 one can build a function f P LppT2q s.t. § ¸ § lim sup § fˆpm, nqe2πipmx nyq§ 8 a.e., RÑ8 P m,n N m2 n2 R a result again in sharp contrast with the theory for the one dimensional case, where by Carleson-Hunt theorem we have a.e. convergence for all 1 p 8. Another (much simpler) application of the Besicovitch set is in proving that the maximal function » p q 1 | p q| Mf x : sup | | f y dy, RQx R R where the supremum ranges over the collection R of all rectangles of arbitrary ori- entation and arbitrary sidelength, is unbounded on any Lp except p 8 (see again [Ste93], ch. X for details). This in particular implies the failure of dierentiability of Lp functions along the collection R, i.e. » 1 p ¡ q lim | | f x y dy diampRqÑ0, R R RPR need not tend to fpxq for a.e. x.

3. Dvir's proof Now we nally come to the anticipated result of Dvir. In order to make the presentation clearer, we introduce some lemmas beforehand.

Let be a eld. For a polynomial in r s we denote by p q the F P F X1,...,Xn ZF P zero set of P , i.e. p q t P n s.t. p q u ZF P x F P x 0 . For a general eld F we have ¨ Lemma 3 (Fundamental fact). If n has cardinality | | d n then there E F E n exists a (non-zero) polynomial P r s of degree p q ¤ such that P F X1,...,Xn deg P d p q. E ZF P ¨ Notice that d n is the dimension of the vector space of polynomials in r s n F X1,...,Xn that have degree at most , denoted r s . This simple fact can be seen d F X1,...,Xn ¤d j as follows: the number of monomials 1 ¤ ¤ ¤ jn with degree is the number of X 1 ¨Xn k solutions to , which is n k¡1 . Summing in ¤ one obtains the j1 ... jn k n k d claim by known properties of binomial coecients.

Proof. This is just the consequence of a simple linear algebra fact. Consider the¨ space of polynomials in r s of degree at most ; this has dimension d n . F X1,...,Xn d n Therefore the conditions p q for all p q P dene a system P x1,¨ . . . , xn 0 x1, . . . , xn E of | | linear equations in d n variables (the coecients of the polynomial), and E ¨ n since | | d n the system has a non-trivial solution. E n 12 MARCO VITTURI

Remark 6. Observe the trivial but useful estimates ¢ dn d n pd nqn ¤ ¤ . n! n n! Another fact needed in the proof is the following estimate on the number of zeroes of P r s, where is now a nite eld. It generalizes the one P F X1,...,Xn F dimensional estimate | p q| ¤ p q and agrees with the intuition that p q ZF P deg P Z P should behave like a hypersurface. Lemma 4. Let be a nite eld and P r s be a polynomial of F Fq P F X1,...,Xn degree d that doesn't vanish identically. Then | p q| ¤ | |n¡1 ZF P d F . Proof. The proof is by induction, the case n 1 being well known. Thus, suppose the lemma is true for dimension ¡ . For xed P , if p q n 1 t F Pt X1,...,Xn¡1 : P pX1,...,Xn¡1, tq vanishes then P pX1,...,Xnq pXn ¡ tqQpX1,...,Xnq, where 1 degpQq ¤ d ¡ 1. If t t moreover Pt1 vanishes if and only if Qt1 does. So we can obtain a factorization9 ¹ P pX1,...,Xnq RpX1,...,Xnq pXn ¡ tq, tPE where |E| ¤ d and Rt does not vanish identically for all t P E (notice degpRq ¤ d ¡ |E|). Therefore by inductive hypothesis ¤ ¤ | p q| ¤ | n¡1 ¢ t u| | p q ¢ t u| ZF P F t ZF Rt t P R t E ¸ t E n¡1 n¡2 n¡1 ¤ |F| |E| pd ¡ |E|q|F| ¤ d|F| . tRE With these lemmas we can then prove the Kakeya conjecture in nite elds. Theorem 5 (Dvir, [Dvi09]). Let be a Kakeya set in n. Then K Fq | | Á | |n K n Fq . The constant arising from Dvir's argument is 1{n!. It is worth noticing that this constant can be improved to the much larger 2¡n by repeating the proof below while postulating higher multiplicities for the points in the Kakeya set (i.e. not just contained in ZpP q but with order much greater than 1). This was done by Dvir, Kopparty, Saraf and Sudan in [DKSS13]; a proof is included in appendix B. Proof. We write for for ease of notation. Suppose by contradiction that ¨ F Fq | |¡ | | F 2 n . Therefore by the fundamental Lemma 3 there exists a polynomial K n P r s of degree p q ¤ | | ¡ such that p q. Let P F X1,...,Xn d : deg P F 2 K ZF P j be the coecient of the monomial 1 ¤ ¤ ¤ jn in . cj1,...,jn X1 Xn P Now we add a ctious coordinate to our space and consider the homogeneized polynomial (of homogeneous degree d) ¸ ¡ ¡ ¡ p q d j1 ... jn j1 ¤ ¤ ¤ jn Ph X0,X1,...,Xn : cj1,...,jn X0 X1 Xn . j1,...,jn The homogeneity implies that for P ¢ it is p q d p q. Since λ F Ph λx0, λx λ Ph x0, x p q p q, if P p q it follows that p q P p q for all P ¢. If Ph 1, x P x x ZF P λ, λx ZF Ph λ F ` ta tv s.t. t P Fu is a line contained in the Kakeya set K then the points p q for P P ¢ are all contained in p q. Geometrically this is a λ, λa λtv t F, λ F ZF Ph 9 Note some t's may be repeated. THE KAKEYA PROBLEM 13 plane with a line removed, but we claim the line is contained in the zero set of Ph too. Indeed, restrict the polynomial Ph to this plane, thus obtaining a polynomial Qpλ, tq of degree at most d s.t. the zero set has cardinality at least |F|p|F| ¡ 1q. But |F|p|F| ¡ 1q ¡ d|F| since d ¤ |F| ¡ 2, and thus by the contrapositive of Lemma 4 the polynomial must be the zero polynomial, i.e. p q contains the entire plane Q ZF Ph pλ, λa tvq for t, λ P F. We have thus proven that for all directions the zero set p q contains p q v ZF Ph 0, tv for all P , i.e. the hyperplane t u¢ n is all contained in p q. In other words, t F 0 F ZF Ph p q for all P n; but p q is the sum of the monomials of Ph 0, x 0 x F Ph 0,X1,...,Xn top degree in P . This is a (non-trivial) polynomial of degree d |F| that vanishes everywhere, and this is impossible by Lemma 4. It then follows that ¢ | | ¡ 2 n p| | ¡ 2qn |K| ¥ F ¥ F Á | |n, n n! n F as claimed. Let's briey review the strategy of the proof above, as it contains the essence of the polynomial method: we assumed the cardinality of K was small, and therefore we could nd a polynomial P of small degree that vanished on K; then, using the structure of K, we proved that the polynomial must vanish somewhere else too; nally, we argued that the polynomial can't vanish on such a large set if it has small degree.

The above proof is remarkably simple, ultimately relying on extremely simple facts of linear algebra such as expressed in Lemma 3 and Lemma 4. Thus the Kakeya conjecture in the nite elds turns out to be remarkably simpler than the corresponding question in the real case F R. The dierence here is in the fact that the space n has a trivial topology (the discrete one), as mentioned before, Fq and in particular it lacks scales completely. Also mentioned before was the fact that scales are not an obstacle but can in fact be put to good use, as has been done for Multilinear Kakeya ([BCT06],[Gut15]). Thus we are seeing that the nite elds case and the real case diverge when it comes to their topology - which therefore is suspected to play a big role in understanding the latter. In the next series of notes we will illustrate this point further by considering the fundamental result of incidence theory known as Szemerédi-Trotter theorem10 for both the nite elds and the real plane. In the former case the best estimate possible is far weaker than that one can get in R2, and the reason again is topological - there can be a bigger number of incidences in F2 because there are fewer (none actually) topological obstructions. In particular, we will give a proof of the Szemerédi-Trotter theorem in the real plane by using polynomial partitioning, which is a consequence of the polynomial Ham Sandwich theorem. Remark 7. It is worth mentioning that the analogue of the Kakeya maximal function conjecture in nite elds has been solved too, by Ellenberg, Oberlin and Tao (see [EOT10]). As is to be expected, the proof makes heavy use of the polynomial method. We state the conjecture briey. The Kakeya maximal function in a nite eld is given by F Fq ¸ f ¦pvq : sup |fpxq|, || ` v xP`

10For a given set of points and a set of lines, the theorem bounds the maximum number of point-line incidences - pp, `q s.t. p P ` - in terms of the number of points and the number of lines. 14 MARCO VITTURI

¢ ¡ where v ranges in the directions of Fn, i.e. Fn{F PFn 1 (the projective space). Then the Kakeya maximal function conjecture (or more appropriately theorem) says that

¦ pn¡1q{n } } np nq À | | } } np nq f ` PF n F f ` F .

Appendix A. Proof of lemma 2 First of all observe that there is a slightly more precise version of the Kakeya maximal function conjecture which we have left unstated. If N is the number of δ-separated δ-tubes, then the claim is that for 1 ¤ d ¤ n

¸N À ¡εp n¡1qpd¡1q{d χTj Ld{pd¡1q ε δ Nδ j1

(notice that Nδn¡1 would be the volume of the tubes if they were disjoint). The two are equivalent, although this is not immediate. See [Mat15] for details. Assume the inequality above then. On the one hand

¸N n¡1 À n¡n{d¡ε À δ χTj d{pd¡1qp nq ε δ 1, L R j1

pd¡1q{d ¡pd¡1qpn¡1q{d and on the other hand aj ¤ N δ . Therefore, it suces to n¡1 ¡pn¡1qpd¡1q{d consider those j's such that δ ¤ aj ¤ δ . Thus, dene Apkq : k tj s.t. aj 2 u, so that

¡1 ¸ cn,d ¸log δ ¸ k ajχTj 2 χTj ;

j kcn log δ jPApkq therefore

¡1 ¸N cn,d ¸log δ ¸ À k ajχTj d{pd¡1qp nq 2 χTj d{pd¡1qp nq L R L R j1 kcn log δ jPApkq

¡1 cn,d ¸log δ k ¡ε n¡1 pd¡1q{d Àε 2 δ p|Apkq|δ q .

kcn log δ ° Notice that | p q| kd{pd¡1q À d{pd¡1q ¤ 1¡n, and therefore | p q|pd¡1q{d À A k 2 j aj δ A k δp1¡nqpd¡1q{d2¡k. Therefore the right hand side in the last inequality is bounded by

¡1 ¡1 cn,d ¸log δ cn,d ¸log δ k ¡ε p1¡nqpd¡1q{d ¡k pn¡1qpd¡1q{d ¡ε ¡Opεq 2 δ δ 2 δ δ n,d,ε δ .

kcn log δ kcn log δ THE KAKEYA PROBLEM 15

Appendix B. Multiplicity

In one variable the multiplicity of a zero z of P pXq P FrXs is dened as the m maximal m s.t. pX ¡zq divides P pXq. We denote this fact by ordzpP q m. This is equivalent to say that the Hasse derivatives11 P D0P,DP,D2P,...,DP m¡1 all vanish in z, and this denition has the advantage of being well-posed even when there are multiple variables; therefore we dene the multiplicity of P in p, denoted i1,...,in ordppP q, as the largest m s.t. D P ppq 0 for all i1 ... in m. Notice that since by Taylor expansion ¸ i1,...,in i1 in P pX1,...,Xnq D P ppqpX1 ¡ p1q ¤ ¤ ¤ pXn ¡ pnq ,

i1,...,in we can deduce that the multiplicity has the multiplicative property

ordppPQq ordppP q ordppQq. We can modify the proofs of the Lemmas 3 and 4 very slightly and obtain the following versions with multiplicity: Lemma 3 (with multiplicity). If n is a set with associated t u s.t. ¨ ¨ E F cp pPE ° ¡ cp 1 n ¤ d n , then there exists a (non-zero) polynomial P r s pPE n n P F X1,...,Xn of degree degpP q ¤ d s.t. for all p P E it is ordppP q ¥ cp. Lemma 4 (with multiplicity). Let be a nite eld and P r s F Fq P F X1,...,Xn be a polynomial of degree d that doesn't vanish identically. Then ¸ p q ¤ | |n¡1 ordp P d F . P n p F Proof. Left to the reader. Finally, with these two tools, we can improve the constant in the estimate for the cardinality of a Kakeya set as stated above. Indeed, let K be a Kakeya set in n and suppose that | | | |n{ n; we x ¤ ¤ to be chosen later, and Fq K F 2 1 l m d assume that pm ¡ 1 nqn dn |K| , n! n! so that in particular it holds that ¢ ¢ m ¡ 1 n d n |K| ; n n so by Lemma 3 with multiplicity we can nd a polynomial P r s of P F X1,...,Xn degree at most d s.t. ordppP q ¥ m for all p P K. For the following, we choose d so that the above inequality is tight, i.e. we choose d pm nq|K|1{n. We denote by a multi-index p q P n and by | | . Now, let i i1, . . . , in N i : i1 ... in be an -line contained in the Kakeya set , then for every P and for every ` Fq K p ` i multi-index i s.t. |i| ¤ l we have that ordppD P q ¥ m ¡ |i|. Thus, by Lemma 3 with multiplicity applied to the restriction of P to `, either ¸ | |p ¡ | |q p i q ¤ ¡ | | (6) F m i ordp D P d i pP` or p i q ` ZF D P . 11 Recall that the Hasse derivatives of P P rX ,...,X s are dened on monomials by # F 1 n ¡ © 0 if jk ik for some k, Di1,...,in Xj1 ¤ ¤ ¤ Xjn ¨ ¨ 1 n j1 jn j ¡i jn¡in ¤ ¤ ¤ X 1 1 ¤ ¤ ¤ Xn otherwise, i1 in and then extended to arbitrary polynomials by linearity; thus they coincide with the ordinary derivatives when the eld is R. 16 MARCO VITTURI

By choosing m and l so that (6) is violated then we conclude that K is contained in ZpDiP q for all i s.t. |i| ¤ l. Therefore we can argue as in the proof of Dvir and i conclude that for P0, the top degree part of P , it holds that D P0 vanishes identicallyX \ (in other words, vanishes identically to order ). Now, choose 11 and P0 l m 10 l choose l s.t. (6) is indeed violated. Then by Lemma 4 with multiplicity applied to , P0 ¸ | |n ¤ p q ¤ | |n¡1 l F ordp P0 d F ; P n p F but we claim this is a contradiction. Indeed, by our choice of d this would imply 11 | | l| |n p l nq F | |n¡1, F 10 2 F which is false far suciently large l. Therefore we have just proven Theorem (Improved Dvir bound). Let be a Kakeya set in n, then K Fq 1 |K| ¥ | |n. 2n F Remark 8. In the proof above any choice of m tαlu with 1 α 2 would work.

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Marco Vitturi, Room 4606, James Clerk Maxwell Building, University of Edin- burgh, Peter Guthrie Tait Road, Edinburgh, EH9 3FD. E-mail address: [email protected]