1 Outline. 2 Kakeya in Analysis. Kakeya Sets: The Paper: The Talk. I aim to cover roughly the following things:

Transcription

1 Kakeya Sets: The Paper: The Talk 1 Outlie I aim to cover roughly the followig thigs: 1 Brief history of the Kakeya problem i aalysis 2 The ite-eld Kakeya problem >> Dvir's solutio 3 Kakeya over o-archimedea local rigs >> my ad Marci's work 2 Kakeya i Aalysis A Kakeya (eedle) set is a subset of the plae R 2 iside which it is possible to rotate a eedle of legth 1 completely aroud Obvious example: A circle of diameter 1, area π 4 Less obvious example: A deltoid, area π 8 Questio (Kakeya eedle problem): How small (i measure) ca a Kakeya eedle set be? Origially asked by Soichi Kakeya i 1917, for a covex set (That problem was also solved, although I do't kow what the aswer is) Aswer (Besicovitch, 1919): Arbitrarily small measure Costructio is fairly geometric: geeral idea is to divide up the regio ito smaller pieces, the traslate the pieces so they overlap a lot Ed result is a sort of spiky-triagle shape It is also possible to adjust the costructio so as to keep the Kakeya set simply-coected ad still have arbitrarily small measure If we reformulate the problem slightly, to say a Kakeya set is oe cotaiig a uit lie segmet i every possible directio, the oe ca costruct a Kakeya set of measure zero Followup questio 1: Are there Kakeya sets of measure zero i R? (yes: just take the product of a 2- dimesioal set with a uit segmet) Followup questio 2 (Kakeya cojectures): How small of a measure zero set ca a Kakeya set be? I other words, must a Kakeya set i R actually be -dimesioal (Hausdor or Mikowski), or could its dimesio be less tha? Aswer: I the plae, a Kakeya set must be 2-dimesioal (This is a ot easy theorem) For higher dimesios, oly lower bouds are kow Example: Terry Tao ad (Nets) Katz showed that a Kakeya set i R must have dimesio at least (2 2)( 4) + 3 Some other remarks: Kakeya sets have some uses i harmoic aalysis, ad are of iterest to aalysts There is also a related Kakeya maximal fuctio, which I do't kow aythig about, that is also iterestig Example: A theorem of Feerma uses Kakeya sets to show that certai trucated Fourier itegrals eed ot coverge i L p orm for p 2 Barry Mazur told me that Feerma's result partly explais why L 2 is the best L p space

2 3 Fiite-Field Kakeya I 1999, Thomas Wol posed a reformulatio of the Kakeya problem i a ite eld settig Wol's ite-eld Kakeya problem asks: For F q a eld, we dee a Kakeya set i (F q ) to be a (ite) set which cotais a lie i every possible directio Here, a lie through x i the directio of y is the set of poits x + t y as t rus through the elemets of F q So i this cotext, a Kakeya set is a set K such that, for every y F q there exists x F q such that x + ty K for all t F q Wol cojectured that a Kakeya set i F q must cotai at least c F q poits, for some costat c depedig oly o (ad ot o q) Several authors proved lower bouds o the order of q 4/7 usig fairly dicult methods i additive umber theory, before Zeev Dvir proved the full cojecture i roughly oe page i 2008 usig very elemetary algebraic geometry Theorem (Dvir): If K is a Kakeya set i F q, the K 1! q Proof: Suppose otherwise, ad take K to be a Kakeya set of size less tha The the collectio of polyomials i F q [x 1,, x ] of degree at most q 1 is a vector space of dimesio > K, so there exists a ozero polyomial P F q [x 1,, x ] of degree at most q 1 such that P (x) = 0 for all x K q 1 Write P = P i, where P i is homogeeous of degree i, ad x y F q i=0 Sice K is a Kakeya set, for ay y F q there exists b F q for which P (b + ty) = 0 for all t F For xed b ad y, this is a polyomial of degree q 1 i the variable t which vaishes for q dieret values of t Hece this polyomial i t is idetically zero, ad so i particular its coeciet of t q 1 is zero Expadig shows this is just P q 1 (y) Therefore, P q 1 (y) = 0 for every y F q However, this ca oly happe if P q 1 is the zero polyomial to see this, just use the divisio algorithm with respect to x 1, the x 2, up through x, ad the fact that P q 1 has degree less tha q Now repeat the above argumet for each of P q 2,, P 1 to see that P must be a costat, hece the zero polyomial Cotradictio (q + 2)(q + 3) (q 1) Hece K = 1!! q for 4 4 No-Archimedea Kakeya, iterlude I 2010, Jorda alog with Richard Oberli ad Terry Tao wrote a paper reviewig the Kakeya problem over ite ( elds ) Others had show that the costat i Dvir's estimate is ot sharp it ca be improved to o(1), which EOT thought likely to be optimal, perhaps up to removig the o(1) If we take the probability measure o F q (so that the whole space has measure 1) the Dvir's result says that the measure of a Kakeya set i F q is at least 1!, which is positive (This is i oppositio to the case of Kakeya sets over the reals, which ca have zero measure)

3 I sectio 420 of their paper, EOT discuss the aalogy betwee the ite-eld Kakeya problem ad the real oe Part of the reaso that the ite-eld results might ot capture useful iformatio about the problem over R is the lack of multiple scales: over F q there is o useful idea of a distace betwee poits ad lies either two poits are the the same, or they are ot while over R there is a very strog otio of distace EOT suggested cosiderig Kakeya sets over other rigs which do have some otio of distace, such as the ite rigs Z/p Z ad F q [t]/t Over these rigs we ca pose the Kakeya cojecture sice we have a otio of dimesio Eve better would be to cosider the iite rigs Z p ad F q [[t]], which are complete ad thus seem much closer to R EOT also observed that although the measure of a Kakeya set i F q is always positive, the best kow boud evertheless goes to zero as goes to They therefore asked whether, perhaps, there might be a Kakeya set of measure zero lurkig i the completio F q [[t]] 5 My actual stu Let R be a iite rig admittig a Haar measure µ such that µ(r) is ite A lie with directio vector v R through the poit x R cosists of the elemets i R of the form x + tv as t rus through the elemets of R A Kakeya set i R is a subset of R which cotais (all the poits o) at least oe lie with each possible directio vector For a ite rig R, the Mikowski dimesio of a set E R log E is deed as The atural aalogue log R of the Mikowski dimesio of a compact subset E F q [[t]] log E k is lim k log F q, where k E k is the image of the projectio of E oto F q [[t]]/t k (Similarly, for E iside Z p ) Theorem 1 (, Habliscek): For all > 1, there exists a Kakeya set E F q [[t]] of measure 0 Theorem 2 (, Habliscek): The Mikowski dimesio of ay Kakeya set i F q [[t]] 2 or Z 2 p is 2 51 Sketch of Theorem 1 (Measure Zero) I costruct a Kakeya set K of measure zero i F q [[t]] 2 ; to get oe i a higher dimesio just take K F q [[t]] 2 I refer to a ozero directio vector v = (a, b) i F q [[t]] 2 as oreduced if t divides both a ad b, ad as reduced otherwise It is obvious that ay lie with oreduced directio vector v passig through (x, y) is cotaied i the lie with directio vector v/t through (x, y); thus, we eed oly cosider reduced directio vectors Every ozero reduced directio vector is of the form (1, b) or (b, 1) So we eed oly d a set E of measure zero cotaiig a lie with directio vector (1, b) for each b F q [[t]]; the K = {(x, y) : (x, y) or (y, x) E} still has measure zero, ad is Kakeya Notatio: For ay a F q [[t]], let a i deote the coeciet of t i For ay a F q [[t]], dee the elemet a F q [[t]] by { a 0 if i = 2 k 2 for some atural umber k, i = a i+1 otherwise

4 Here is the costructio: dee E = {(x, y) F q [[t]] 2 : ax + y = a for some a F q [[t]]} To see that this cotais the required lies, observe that, for ay b F q [[t]], the poits (x, y) = (0, b )+ s(1, b) are cotaied i E, as s rages over F q [[t]], sice ( b)s + ( b + bs) = ( b) It oly remais to prove that E has measure zero (which is, obviously, the hard part) We ca see that (x, y) E if ad oly if there exist a i F q for all i 0 such that the coeciets of x ad y satisfy the followig iite system: a 0 x 0 + y 0 = 0, [0] a 1 x 0 + a 0 x 1 + y 1 = a 2, [1] a 2 x 0 + a 1 x 1 + a 0 x 2 + y 2 = 0, [2] a x 0 + a 1 x a 0 x + y = a, [] For a arbitrary elemet (x, y) F q [[t]] 2, dee s (x, y) to be the umber of tuples (a 0,, a ) F +1 q satisfyig the rst equatios; observe that s (x, y) oly depeds o {x 0, y 0,, x, y } Clearly if s (x, y) = 0 for ay iteger, the (x, y) E, ad s k (x, y) = 0 for all k >, so µ ({(x, y) s (x, y) = 0}) is o-decreasig as We show this measure teds to 1 as Observe that the equatios at ay stage are liear i the a i Moreover, for i ot of the form 2 k 2 for some iteger k, Equatio [i] states a i+1 = a i x 0 + a i 1 x a 0 x i + y i, ad so we may reduce our system of equatios by elimiatig a i+1 Basic liear algebra the implies that s 2 k 2(x, y) is either zero or q l for some iteger l k Now we do a somewhat ucoscioable amout of polyomial arithmetic to show the followig: Lemma: If for a give {x 0, x 1,, x 2 2} ad {y 0, y 1,, y 2 2} we have s 2 2(x, y) = q l, ad {x 2 1, y 2 1,, x , y 2 are radomly ad uiformly chose from F q, the 0 with probability q 1, q l+2 s (x, y) = q l with probability 1 1 q l+1 with probability 1, q l+1 q l+2 What this says is that we have a Markov chai [DRAW PICTURE!] o the poits 0, 1, q, q 2, such that a positive proportio of measure, idepedet of time, at each ozero poit is set to 0 Oe ca also easily check that the expected value does ot chage over time Sice the expected value at time zero is 1, ad evetually (by a trivial iductio) the measure at q l poits goes to zero, we see that the measure cocetrated at 0 must go to 1 as Remark: Usig a more itricate aalysis (which is surprisigly dicult despite the fact that everythig is totally explicit!), oe ca prove that at time t, the measure away from 0 is l(t), for a explicitly computable t costat that depeds oly o q 52 Sketch of Theorem 2 (Mikowski Dimesio) Theorem 2 (, Habliscek): The Mikowski dimesio of ay Kakeya set i F q [[t]] 2 or Z 2 p is 2 Propositio: Let E be a Kakeya set i R 2 where R = F q [t]/t k or Z/p k Z The E R 2 2k

5 Eumerate the lies by their coeciets take L i to be ay lie with equatio α i x + y = b i where α i i mod p k (Do the same thig for R = F q [t]/t k by readig polyomials i base q) We prove the propositio essetially by a coutig argumet Give two lies i R 2 with equatios L i : α i x+y = b i ad L j : α j x+y = b j with i j, we see that if (x 0, y 0 ) L i L j the (α i α j )x 0 = b i b j Hece if v(α i α j ) = l, the the umber of possible values for x 0 caot exceed m l, where v is the m- adic valuatio I particular, sice the value of x 0 determies the value of y 0, the size of the itersectio L i L j is at most m l = m v(αi αj) Lemma: For the fuctio f(u) := u i=1 mv(αi), we have f(u) u log m (u) I particular, for u m k /k = R /k, we have f(u) R Proof: Just cout the umber of terms for which v(α i ) = w Proof of Prop: Apply iclusio-exclusio by writig our Kakeya set E as the uio of a buch of lies, ad subtractig away their possible itersectios For l = R k we have ( ) l+1 E l+1 j=1 L j 1 j L j L i L j j=1 Now look at the terms i the sum We have L j j 1 i=1 L i L j = R j 1 i=1 mv(αj αi) = R f(j 1) Now just sum ad use the upper boud o f from the lemma The result follows Proof of Theorem: Suppose E is a Kakeya set i F q [[t]] or Z p The its projectio E k to R = F q [t]/t k or Z/p k Z is also Kakeya By the propositio, we have E k R 2k The the Mikowski dimesio of E k is at least 2 log(2k) k log( m ) So for xed p or q, as k this boud goes to 2 Hece result i=1