THE MULTILINEAR KAKEYA INEQUALITY

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1 THE MULTILINEAR KAKEYA INEQUALITY 1. The dscusso from last tme, heurstcs ad memores Suppose that {T} s a Kakeya set of tubes R. Each tube has radus 1 ad legth N, ad there are N 1 tubes. Suppose that T N γ. The umber of ut cubes from the ut cube lattce that tersect T s N γ. We use the polyomal ham sadwch theorem to choose a polyomal P so that Z(P) bsects each of these ut cubes. The degree of Z(P) s N 1 γ/. What does such a polyomal tell us? Let Q(K) be ths set of cubes. Cosder oe of the tubes, T. Let l be a le parallel to the axs of T, a radomly chose parallel le the tube T. For almost every choce of l, we have l Z(P) deg(p) N 1 γ/. O the other had, the tube T cotas N cubes of Q(K), ad Z(P) bsects each of them. Let Q(T ) be ths lst of cubes. They are dsjot, so we get Averageq Q(T ),l l Z(P) q N deg(p) N. For a typcal cube q, we kow that Z(P) bsects q, ad yet Average l Z(P) q N γ/ s much smaller tha 1. Ths s oly possble f the surface Z(P) q s approxmately parallel to the tube T. We ca make a revsed pcture of the surface Z(P) the tube T. So the geometry of the surface Z(P) s coected wth the geometry of the tubes T. If we try to mtate Dvr s proof of the fte feld Nkodym or Kakeya cojectures, we are led to the followg questo. Exted each tube T a further legth N, ad let q be a ut cube the exteso. Is t true that Z(P) approxmately bsects q? Ths type of questo looks dffcult, ad t may be ulkely. The surface Z(P) s approxmately taget to the tube T sde of T, but t s hard to kow whether Z(P) wll bed sharply as soo as t leaves T ad come owhere ear to q. I spet a whle tryg to force Z(P) to ht q, ad t was pretty frustratg. I would charge dow oe of the tubes T, tryg to p Z(P) ad carry t dow to q, ad Z(P) would stay wth me for a whle ad the swg out of the way, whle I wet chargg harmlessly by... However, the structure that we observed above does say somethg terestg about Kakeya sets. We otced that for a typcal cube q T, the surface Z(P) s approxmately taget to T. But there are may dfferet tubes T j cotag q. Wth the method above, we ca argue that T j s approxmately taget to Z(P) for 1 1 γ/

2 2 THE MULTILINEAR KAKEYA INEQUALITY most of the tubes. I fact, there must be a hyperplae π(q), ad the tubes T j must usually be almost taget to π(q). Ths s a somewhat surprsg structure, called plaess. Plaess was frst dscovered by Katz, Laba, ad Tao, the paper A mproved boud o the Mkowsk dmeso of Bescovtch sets R 3. (A. of Math. (2) 152 (2000), o. 2, ) Plaess was oe of the observatos/tools that allowed 3 them to prove that a Kakeya set of tubes R (wth mld addtoal hypotheses) has volume at least N 2.5+ǫ. Later, Beett, Carbery, ad Tao proved stroger ad more geeral plaess estmates the paper O the multlear restrcto ad Kakeya cojectures Acta Math. 196 (2006), o. 2, We wll come to ther work below. If we had a hypothetcal Kakeya set of tubes, a typcal cube would le may tubes T j. Wthout ay experece, we mght guess that the dfferet tubes T j q would pot a bush of drectos that was pretty dese o the ut sphere. Suprsgly, they eed to cocetrate ear to a plae. Aother way to say ths s that they do t form a whole lot of jots. Durg the course, we met may theorems about the cdece patters of les space. Each of these questos ca be adapted to a questo about log th tubes stead of les. Usually the adapted questo s wde ope. But for the jots problem, the adapted questo has a ce aswer based o the deas we have just bee dscussg. 2. The geeralzed Looms-Whtey equalty We prove here a aalogue of the jots theorem wth log th tubes stead of perfect les. Theorem 2.1. (Beett-Carbery-Tao, Guth) Suppose that T j,a are cylders R for 1 j ad 1 a A. Each cylder has radus 1 ad fte legth. The axs of a cylder T j,a makes a agle of < (100) 1 wth the x j -axs. Let I be the pots whch le oe cylder for each value of j = 1... I equatos I := j=1( A a=1t j,a ). The the volume of I s A 1. Remarks. If the tubes T j,a are parallel to the x j -axs, the ths estmate follows from the Looms-Whtey equalty. We see that the projecto of I to ay coord ate hyperplae les A ut balls, ad the Looms-Whtey gves I A 1. The case of axs-parallel cylders s bascally equvalet to the Looms-Whtey equalty. The problem here s to see that the equalty remas true f we are allowed to tlt the tubes a few degrees. Hstory. BCT proved a ty bt weaker estmate usg mootocty formulas for the heat equato. G proved ths estmate usg the polyomal method. Ths

3 THE MULTILINEAR KAKEYA INEQUALITY 3 theorem ca be thought of as a verso of jots for early-orthogoal tubes. It mples, partcular, the jots theorem for early orthogoal les. The proof volves the dea of the drected volume of a surface. Suppose S s a smooth hypersurface R wth ormal vector N. If v s a ut vector, we defe the drected volume of S perpedcular to V by the formula V S (v) := N v dvol S. S Notce that f the taget plae of S s perpedcular to v, we have N v = 1, ad f the taget plae cotas v, we have N v = 0. For example, we cosder the drected volume of the ut crcle the drecto v = (0, 1). The drected volume of a arc of the upper sem-crcle drecto v s exactly the chage the x-coordate over the arc. Therefore, the drected volume of the whole upper sem crcle s 2, ad the drected volume of the whole crcle s 4. The computato for the crcle geeralzes as follows. Let π be the orthogoal projecto from R to v R. Lemma 2.2. V S (v) = S π 1 (y) dvol(y). v As a corollary, we ca mmedately estmate the drected volume of a degree d varety a cylder T. Lemma 2.3. (Cylder estmate) Let T be a fte cylder R of radus r. Let v be a ut vector parallel to the axs of T. Let Z(P) be the vashg set of a polyomal P. The V (v) r 1 deg(p). Z(P) T Proof. Let π be the projecto from T to the cross-secto v T. Ths crosssecto s just a (-1)-dmesoal ball of radus r. For almost every y ths ball, π 1 (y) Z(P) deg(p). By the last lemma, V Z(P) T (v) s bouded by deg(p) tmes the volume of the cross-secto, whch s r 1. Lemma 2.4. If S s a hypersurface R, ad v 1,..., v are ut vectors ad the agle from v j to the x j -axs s (100) 1, the V ol 1 S 2 j V S(v j ). Proof. At a gve pot of S wth ormal vector N, we have to prove that j N v j 1/2. If e j are the coordate vectors, the t s straghtforward to check that j N v j 1 for ay ut vector N. The vectors v j are very close to e j, ad so the error has sze j e j v j (1/100). Now we ca do the proof of the theorem. Proof. Cosder the ut cubcal lattce. Let Q 1,..., Q V be all the ut cubes the lattce whch tersect the set I. We wll prove V A 1.

4 4 THE MULTILINEAR KAKEYA INEQUALITY Let P be a o-zero polyomal so that Z(P) bsects each cube Q 1,..., Q V ad degp V 1/. Ths bsecto requres a certa amout of area, therefore: V ol 1 Z(P) Q 1. Let T j (Q ) be a tube from our lst, drecto j, whch tersects Q. Let v j, be the drecto of ths tube. The drectos v 1,,..., v, are early orthoormal, ad so VZ(P) Q (v j,) V ol 1Z(P) Q 1. j=1 For each cube, choose oe drecto j so that V Z(P) Q v j, 1, ad assg the cube Q to the tube T j (Q ). We have V cubes ad A tubes, so oe of the tubes has V/A cubes assged to t. Let T be ths tube, ad let v be ts drecto. We have V/A cubes Q obeyg the followg codtos: The cube Q tersects T. V Z(P) Q (v) 1. Let T be a wder cylder wth radus 2 ad wth the same cetral axs as T. All of the cubes Q le T. Therefore, we have V/A V Z(P) T The last equalty s by the cylder estmate. Rearragg we get V A 1. 1/ (v) V. 3. Multlear Kakeya The strogest verso of the Kakeya cojecture s the L p verso. If T are a Kakeya set of tubes of radus 1 ad legth N, the L p Kakeya cojecture says that for each ǫ > 0, χ T 1 C ǫ N ǫ N. (1) R Remarks: If we arrage the tubes a dsjot way, the left had sde s N. If we arrage them all cetered at the org, the the left had sde s (log N)N. If true, ths cojecture gves essetally sharp bouds for χ T p for every p. It mples that the uo of tubes has volume at least cǫn ǫ for ay ǫ > 0. Ths cojecture s stll wde ope. The multlear Kakeya cojecture allows us to cotrol a postve fracto of all the terms - a certa sese. Frst we rewrte the left had sde of (1).

5 THE MULTILINEAR KAKEYA INEQUALITY 5 χ T 1 = 1 χ T χ T 1 1 O the rght had sde we have a product of detcal copes of χ T 1. Now we edt the formula, keepg oly a costat fracto of the terms each copy of 1 χ T 1. Let I(j) be the subset of tubes T where the agle betwee v(t ) ad the x j axs s (100) 1. For each j, the umber of such tubes s N 1 - they form a postve fracto of all of the tubes. Theorem 3.1. (Beett-Carbery-Tao) For ay ǫ > 0, there exsts a costat C ǫ so that for ay Kakeya set of tubes, 1 χ 1 C ǫ N ǫ T N. j=1 I(j) (I ths equalty, the N ǫ factor ca actually be removed, see my paper O the edpot case of the Beett-Carbery-Tao multlear Kakeya equalty. But ths takes a lot of extra work.) Ths equalty s a geeralzato of the last theorem. We expla how they are related ad we sketch the extra steps eeded to prove the theorem. For ay tegers µ 1,..., µ 0, cosder the set of pots: The left had sde s I(µ) := {x R 2 µ j χ < 2 µ j+1 T for all j.} I(j) I(µ) 2 µ j/( 1). µ j Therefore, the theorem follows from the followg lemma: Lemma 3.2. For each µ as above, I(µ) N j 2 µ j/( 1). The lemma shows that each term the sum above has sze N, ad the umber of terms s (log N), ad so we get a boud for the total of N (log N), whch proves the theorem. If µ = 0, we have I(0) cotaed the -fold tersecto set I defed above, ad the equalty follows from the Theorem the last secto. The other values of µ are farly smlar. Let us radomly choose I (j) I(j), cludg each tube wth probablty 2 µ j. Let I be the pots lyg oe tube T, I (j) for each j. A pot of I(µ) les

6 6 THE MULTILINEAR KAKEYA INEQUALITY I wth probablty 1. Wth hgh probablty, the sze of I (j) s N 1 2 µ j. Therefore, our boud for I(µ) follows from the followg lemma. Lemma 3.3. Let T j,a a = 1...A j be cylders of radus 1 early parallel to the x j axs. The the volume of the set of pots lyg at least oe tube of each drecto 1 s j=1 A 1 j. If all the A j happe to be equal, ths lemma s exactly the theorem from the last secto. The case of uequal A j requres a extra refemet the proof. We cut each cube Q to may smaller peces, ad we choose P to bsect each smaller pece. The smaller peces are arraged to a grd, cut more fely the drectos j where A j s small ad more coarsely the drectos where A j s large. Detals the exercses... (More detals. Take a cube Q. Pck tubes T j (Q ). Chage coordates so that the vectors v(t j (Q )) become exactly orthogoal. I these coordates, Q s ot qute a cube, but cotas a slghtly smaller cube Q. Chop Q to a grd, where the j th drecto s cut subdvded to j =j A j. Choose Z(P) to bsect each of these peces....) 4. Sharp turs of algebrac varetes? So far, the polyomal method has ot led to ay progress o the Kakeya problem. There are major dffcultes applyg the methods we have see to log th tubes stead of perfect les. I the proof of fte feld Kakeya or Nkodym, we use parameter coutg to fd a polyomal that vashes some places, ad the we argue that the polyomal also must vash somewhere else. Ths step plays a key role most of the proofs we have see ths course. It s hard to see whether somethg lke ths ca work the settg of tubes. Suppose as the frst secto that K s the uo of a Kakeya set of 1 N tubes wth surprsgly small volume, ad that P s a polyomal so that Z(P) bsects each cube of the ut lattce that tersects K. Pck a tube T from the Kakeya set, ad mage extedg t to twce ts legth, ad let q be a ut cube ths exteso. Is there ay hope that Z(P) also roughly bsects q? We kow that Z(P) bsects all the cubes T, ad we ve also see that most of these cubes Z(P) s roughly parallel to T. If Z(P) keeps gog the drecto of ts taget plae, t wll come reasoably close to q (although t s stll ot clear t wll really ht q). But t s ot at all clear whether Z(P) wll cotue the drecto of ts taget plae. Perhaps Z(P) wll curve dramatcally ad go owhere ear q. It mght be helpful to uderstad better how may sharp beds there ca be a degree d algebrac surface. Here s a toy problem that gets at some of these ssues.

7 THE MULTILINEAR KAKEYA INEQUALITY 7 Let P be a polyomal two varables. Let Pos(P) := {x R 2 P(x) > 0}. For a gve degree d, how closely ca Pos(P) look lke the square [ 1, 1] 2? Recall that the Hausdorff dstace from Pos(P) to [ 1, 1] 2 s < ǫ f [ 1, 1] 2 les the ǫ- eghborhood of Pos(P) ad Pos(P) les the ǫ-eghborhood of the square. Let ǫ(d) be the fmum over all degree d polyomals P of dst Haus (Pos(P), [ 1, 1] 2 ). Ca we descrbe the order of magtude of ǫ(d)? Very lttle s kow about ths. We kow that ǫ(d) > 0 for each d. The reaso s that dst Haus (Pos(P), [ 1, 1] 2 ) vares lower sem-cotuously as P moves V (d) \ {0}. Multplyg P by a postve costat does ot chage Pos(P), ad so we ca restrct atteto to polyomals the ut sphere of V (d). By compactess the fmum s attaed. But f dst Haus (Pos(P), [ 1, 1] 2 ) were zero, we would have P = 0 o the boudary of the square. The P would vash o the le x = 1, ad (x 1) would factor out of P. Wrte P as (1 x) a P 1 (x, y), where (1 x) does ot dvde P 1. The polyomal P 1 vashes at oly ftely may pots of the le x = 1. If a s eve, the we see that P 1 eeds to vash o the sde of the square where x = 1, ad the 1 x dvdes P 1, ad we get a cotradcto. If a s odd, the we see that P 1 eeds to vash o the part of the le x = 1 where y > 1. Ths stll mples that 1 x dvdes P 1, ad we get a cotradcto. If d s eve, a ce example s the polyomal P d = 1 x d y d. The set Pos(P d ) s the ut ball the L d orm. As d, t approaches the square, whch s the ut ball the L orm. For every eve d, Pos(P d ) [ 1, 1] 2, ad P d > 0 o the square [ (1/2) 1/d, (1/2) 1/d ]. Now 1 (1/2) 1/d 2 1/d, ad so dst Haus (Pos(Pd), [ 1, 1] ) 1/d. Hece ǫ(d) 1/d. It seems plausble that P d s ear-optmal ad that ǫ(d) 1/d. The hard problem s to gve quattatve lower bouds o ǫ(d). I do t kow of ay explct lower boud the lterature. I worked o the problem, ad I had a pla for a lower boud of the form e ed... I thk the moral ssue s to gve quattatve bouds o how sharply a degree d curve ca make a certa type of tur. It s mportat to keep md the followg example. The zero set of the hyperbola xy = ǫ makes a very sharp tur ear the org, whch looks somethg lke the corer of a square. But the hyperbola has two braches, ad so stead of beg postve o approxmately oe quartat, t s postve o two opposte quartats, ad ts postve set does ot really look lke the eghborhood of a corer of a square. A algebrac curve ca make a arbtrarly sharp tur f t looks locally lke a hyperbola wth two braches, but t s harder for t to make a sharp tur wth oly oe brach. I mght have goe o too log about ths toy problem. A soluto to ths problem would ot drectly lead to ay bouds o Kakeya. Tryg to go further wth the polyomal method ad tubes, ths type of estmate seems to come up. I geeral, t

8 8 THE MULTILINEAR KAKEYA INEQUALITY mght be helpful to have more quattatve estmates about the geometry of degree d algebrac surfaces.

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