q1qr1thentheestimateofCorollary2becomestrivialThesecondconsiderablylargerinvolumepartofthepaperdealswiththeupperboundsWegivehereanumberofdi erentconstructionsSomeofthemcanberegardedasre nedan ID: 201247
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2S.KOPPARTY,V.LEV,S.SARAF,ANDM.SUDANWenoticethatTheorem1extends[DKSS,Theorem11]andindeed,thelatterresultisaparticularcaseoftheformer,obtainedforr=1.TheproofofTheorem1usesthepolynomialmethodinthespiritof[DKSS,SS08],themajornoveltyintroducedbeingthatinourpresentsettings,wehavetoconsiderpolynomialsovertheringofrationalfunctions.Usingtheinequality(1+x)�m1�mx;x0;m1;onereadilyderivesCorollary2.Ifnr1areintegers,qaprimepower,andKFnqaKakeyasetofrankr,thenjKj�1�n(q�1)q�rqn:Weremarkthatintheparticularcaser=n�1abetterestimateisknown:namely,[FLS10,Theorem5.1]showsthatifn3isxedandKFnqisaKakeyasetofrankn�1,thenjKjqn�q2+o(q2)asq!1.Tofacilitatecomparisonbetweenestimates,weintroducethefollowingterminology.GiventwoboundsB1andB2forthesmallestsizeofaKakeyasetinFnq(whichareeitherbothupperboundsorbothlowerbounds),wesaythattheseboundsareessentiallyequivalentinsomerangeofnandqifthereisaconstantCsuchthatforallnandqinthisrangewehaveB1CB2;B2CB1;andalsoqn�B1C(qn�B2);qn�B2C(qn�B1):Wewillalsosaythattheestimates,correspondingtothesebounds,areessentiallyequiv-alent.Withthisconvention,itisnotdiculttoverifythatforeveryxed"�0,theestimatesofTheorem1andCorollary2areessentiallyequivalentwhenevern(1�")qr�1.Ifn1+1 q�1qr�1,thentheestimateofCorollary2becomestrivial.Thesecond,considerablylargerinvolumepartofthepaper,dealswiththeupperbounds.Wegivehereanumberofdi erentconstructions.Someofthemcanberegardedasrenedandadjustedversionsofpreviouslyknownones;other,toourknowledge,didnotappearintheliterature,buthavebeen\intheair"forawhile.WerstpresentaKakeyasetconstructiongearedtowardslargeelds.Itisbasedonthe\quadraticresidueconstruction"duetoMockenhauptandTao[MT04](witharenementbyDvir,see[SS08]),the\liftingtechnique"from[EOT09],andthe\tensor 4S.KOPPARTY,V.LEV,S.SARAF,ANDM.SUDANandthattheinequality(1�x)m1�mx+(mx)2;0x1;m1showsthatifrnrqr�1,thenindeed1�q�q 2qrbn r+1c1� n rq�(r�1);withabsoluteimplicitconstants.Therefore,wehaveCorollary5.Letn�r1beintegersandqaprimepower.ThereexistsaKakeyasetKFnqofrankrsuchthatjKjqn� �qn�(r�1);moreover,ifnrqr�1,theninfactjKjqn� n rqn�(r�1)(withabsoluteimplicitconstants).WeremarkthatCorollaries2and5givenearlymatchingboundsonthesmallestpossiblesizeofaKakeyasetofrankrinFnqinthecasewhererisbounded,qgrows,andthedimensionndoesnotgrow\toofast".Incontrast,forqboundedandngrowing,neither(1)norTheorem4givesatisfactoryupperbound.Specically,theO-termin(1)donotevenallowforconstructingKakeyasetsofsizeo(qn),whereastheestimateofTheorem4isnon-trivial,butrelativelyweak.Addressingthissituationrstinthecaser=1,wedevelopfurthertheideabehindtheproofof[SS08,Theorem8]toshowthattheO-termjustmentionedcanbewellcontrolled,makingtheresultnon-trivialintheregimeunderconsideration.Theorem6.Letn1beanintegerandqaprimepower.Thereexistsarank-1KakeyasetKFnqwithjKj8&#x]TJ/;༶ ;.9;Ւ ;&#xTf 1;.42; 31;&#x.681;&#x Td ;&#x[000;&#x]TJ/;༶ ;.9;Ւ ;&#xTf 1;.42; 31;&#x.681;&#x Td ;&#x[000;&#x]TJ ;� -2;.52;&#x Td ;&#x[000;&#x]TJ ;� -2;.52;&#x Td ;&#x[000;:2�1+1 q�1�q+1 2nifqisodd;3 2�1+1 q�1�2q+1 3nifqisanevenpowerof2;3 22(q+p q+1) 3nifqisanoddpowerof2:Theorem6istobecomparedagainstthecaser=1ofTheorem1showingthatifKFnqisarank-1Kakeyaset,thenjKj�q2=(2q�1)n.ForseveralsmallvaluesofqtheestimateofTheorem6canbeimprovedusingacombinationofthe\missingdigitconstruction"andthe\randomrotationtrick"ofwhichwelearnedfromTerryTaowho,inturn,referstoImreRuzsa(personalcommunicationinbothcases).ForaeldF,byFwedenotethesetofnon-zeroelementsofF. 6S.KOPPARTY,V.LEV,S.SARAF,ANDM.SUDANAaanimmediateconsequencewehaveTheorem10.Letnr1beintegersandqaprimepower.ThereexistsaKakeyasetKFnqofrankrsuchthatjKj�1��1�q�bn=qrcqrqn:Usingtheestimatesbn=qrc�n=qr�1and(1�x)m1�mx(appliedwithx=q�bn=qrcandm=qr),weobtainCorollary11.Letnr1beintegersandqaprimepower.ThereexistsaKakeyasetKFnqofrankrsuchthatjKjqn(1�q�r)+r+1:ItisnotdiculttoverifythatCorollary11supersedesCorollary5forn(r+2)qr,andthatforngrowing,Theorem10supersedesTheorem4ifrissucientlylargeascomparedtoq(roughly,r&#x-278;Cq=logqwithasuitableconstantC).AslightlymorepreciseversionofCorollary11isthatthereexistsaKakeyasetKFnqofrankrwithjKjqn�bn=qrc+r;thisisessentiallyequivalenttoTheorem10providedthatn(r+1)qr.(Ontheotherhand,Theorem10becomestrivialifnqr.)TheremainderofthepaperismostlydevotedtotheproofsofTheorems1,6,7,and8,andLemma9.Fortheconvenienceofthereaderandself-completeness,wealsoprove(aslightlygeneralizedversionof)Lemma3intheAppendix.Section6containsashortsummaryandconcludingremarks.2.ProofofTheorem1.AsapreparationfortheproofofTheorem1,webrie yreviewsomebasicnotionsandresultsrelatedtothepolynomialmethod;thereaderisreferredto[DKSS]foranin-depthtreatmentandproofs.Fortherestofthissectionweusemultidimensionalformalvariables,whicharetobeunderstoodjustasn-tuplesof\regular"formalvariableswithasuitablen.Thus,forinstance,ifnisapositiveintegerandFisaeld,wecanwriteX=(X1;:::;Xn)andP2F[X],meaningthatPisapolynomialinthenvariablesX1;:::;XnoverF.ByN0wedenotethesetofnon-negativeintegers,andforXasaboveandann-tuplei=(i1;:::;in)2Nn0weletkik:=i1++inandXi:=Xi11Xinn.LetFbeaeld,n1aninteger,andX=(X1;:::;Xn)andY=(Y1;:::;Yn)formalvariables.ToeverypolynomialPinnvariablesoverFandeveryn-tuplei2Nn0there 8S.KOPPARTY,V.LEV,S.SARAF,ANDM.SUDANLemma15([DKSS,Lemma8]).Letn1beaninteger,Panon-zeropolynomialinnvariablesoveraeldF,andSFaniteset.ThenXz2Sn(P;z)degPjSjn�1:Corollary16.Letn1beaninteger,Panon-zeropolynomialinnvariablesoveraeldF,andSFaniteset.IfPvanishesateverypointofSnwithmultiplicityatleastm,thendegPmjSj.WearenowreadytoproveTheorem1.ProofofTheorem1.Assumingthatmandkarepositiveintegerswithkqrqm�k q�1(5)(notypo:kentersbothsides!),weshowrstthatm+n�1njKjn+kn;(6)andthenoptimizebymandk.Supposeforacontradictionthat(6)fails;thus,byLemma13,thereexistsanon-zeropolynomialPoverFqofdegreeatmostkinnvariables,vanishingateverypointofKwithmultiplicityatleastm.Writel:=lqm�k q�1mandxi=(i1;:::;in)2Nn0satisfyingw:=kikl.LetQ:=P(i),theithHassederivativeofP.SinceKisaKakeyasetofrankr,foreveryd1;:::;dr2Fnqthereexistsb2Fnqsuchthatb+t1d1++trdr2Kforallt1;:::;tk2Fq;hence,(P;b+t1d1++trdr)m;andtherefore,byLemma12,(Q;b+t1d1++trdr)m�wwhenevert1;:::;tr2Fq.ByLemma14,wehave(Q;b+t1d1++trdr)(Q(b+T1d1++Trdr);(t1;:::;tr));whereQ(b+T1d1++Trdr)isconsideredasapolynomialinthevariablesT1;:::;Tr.Thus,foreveryd1;:::;dr2Fnqthereexistsb2FnqsuchthatQ(b+T1d1++Trdr)vanisheswithmultiplicityatleastm�wateachpoint(t1;:::;tr)2Frq.ComparedwithdegQ(b+T1d1++Trdr)degQk�wq(m�w)(asitfollowsfromwl),inviewofCorollary16thisshowsthatQ(b+T1d1++Trdr)isthezeropolynomial. 10S.KOPPARTY,V.LEV,S.SARAF,ANDM.SUDANProof.LetK:=f(x1;:::;xj;t;0;:::;0):0jn�1;t2F;x1;:::;xj2If(t)g:Sincefisnon-linear,wehavejIf(t)j�1foreacht2F,anditfollowsthatjKj=n�1Xj=0Xt2FjIf(t)jj=Xt2FjIf(t)jn�1 jIf(t)j�1:ToshowthatKisarank-1Kakeyasetweprovethatitcontainsalineineverydirectiond=(d1;:::;dn)2Fnnf0g.Withoutlossofgeneralityweassumethat,forsomej2[1;n�1],wehavedj+1=1anddj+2==dn=0,andweletb:=(f(d1);:::;f(dj);0;:::;0):Foreveryt2Fwehavethenb+td=(f(d1)+td1;:::;f(dj)+tdj;t;0;:::;0)2K;completingtheproof.TheassertionofTheorem6forqoddfollowsimmediatelyfromLemma17uponchoos-ingF:=Fqandf(x):=x2,andobservingthatthenjIf(t)j=(q+1)=2foreacht2Finviewofx2+tx=(x+t=2)2�t2=4:InthecaseofqeventheassertionfollowseasilybycombiningLemma17withthefollowingtwopropositions.Proposition18(cf.[W93]).Supposethatqisanevenpowerof2andletf(x):=x3(x2Fq).Thenforeveryt2FqwehavejIf(t)j(2q+1)=3.Proposition19.Supposethatqisanoddpowerof2andletf(x):=xq�2+x2(x2Fq).Thenforeveryt2FqwehavejIf(t)j2(q+p q+1)=3.TocompletetheproofofTheorem6itremainstoprovePropositions18and19.Indeed,Proposition18followsimmediatelyfromthemainresultof[W93];however,weincludebelowaself-containedproofsinceitalsoservesasasimpliedmodelofthemoreinvolvedproofofProposition19.Weneedthefollowingwell-knownfact.Lemma20.Supposethatqisapowerof2,andletTrdenotethetracefunctionfromtheeldFqtoitstwo-elementsubeld.For ;; 2Fqwith 6=0,thenumberofsolutions 12S.KOPPARTY,V.LEV,S.SARAF,ANDM.SUDANinviewof1�1+12=0.Assimplecomputationshowsthatxywithx;y2Fq;x6=yholdsifandonlyif1=(xy)=x+y;thatis,xy2+x2y+1=0.Forx2Fqxed,thisequationinyhas,byLemma20,two(non-zero)solutionsisTr(1=x3)=0,andnosolutionsifTr(1=x3)=1.Itfollowsthateachx2Fqcontainseitherthree,oronenon-zeroelementinitsequivalenceclass,accordingtowhetherTr(1=x3)=0orTr(1=x3)=1.Byaremarkatthebeginningoftheproof,asxrunsoverallelementsofFq,sodoes1=x3.Hence,thereareexactlyq=2�1thosex2FqwithTr(1=x3)=0,andq=2thosex2FqwithTr(1=x3)=1.Consequently,jIf(0)j,whichisthenumberofequivalenceclasses,isequaltoq=2�1 3+q=2=2q�1 3:Fortherestoftheproofweassumethatt6=0.Theequationx�1+x2+tx=t�1iseasilyseentohavethesolutionsetft;1=p tgwhich,therefore,isanequivalenceclass,consistingoftwoelementsift6=1orjustoneelementift=1.Fixx2Fqnft;1=p tg.Fory2Fq;y6=x,wehavexyifandonlyif1=(xy)=x+y+t;equivalently,xy2+x(x+t)y+1=0.Thisequationhastwosolutions(distinctfromxand0)ifTr(1=x(x+t)2)=0,andnosolutionsifTr(1=x(x+t)2)=1.Intheformercasetheequivalenceclassofxcontainsthreenon-zeroelements,and,consequently,ifweletN:=#x2Fqnft;1=p tg:Tr�1=(x(x+t)2)=0 ;thenjIf(t)j(q�2 3Nift=1;q�2 3N�1ift6=1:(7)ToestimateNwenoticethat1 x(x+t)2=1 t2x+1 t2(x+t)+1 t(x+t)2;andthatTr1 t(x+t)2=Tr1 p t(x+t);implyingTr1 x(x+t)2=Tr1 t2x+1 t2+1 p t1 x+t=Trx=p t+1=t x(x+t):Thus,ift=1,thenTr1 x(x+t)2=Tr1 x; 14S.KOPPARTY,V.LEV,S.SARAF,ANDM.SUDANLemma21.Foreveryprimepowerqandfunctionf:Fq!Fq,thereexistsanelementt2FqwithjIf(t)j�q=2:Proof.Forx;y;t2Fqwewritextyiff(x)+tx=f(y)+ty;equivalently,ifeitherx=y,orx6=yand(f(x)�f(y))=(x�y)=�t.ItfollowsfromtherstformofthisdenitionthattisanequivalencerelationonFqandjIf(t)jisthenumberofequivalenceclasses,andfromthesecondformthatforeverypair(x;y)withx6=ythereexistsauniquet2Fqwithxty.Foreacht2Fq,considerthegraph�tonthevertexsetFq,inwhichtwoverticesx6=yareadjacentifandonlyifxty.Bytheremarkjustmade,everyedgeofthecompletegraphonthevertexsetFqbelongstoexactlyonegraph�t.Consequently,thereexistst2Fqsuchthatthenumberofedgesof�t,whichwedenotebye(�t),doesnotexceedq�1�q2=(q�1)=2.Bytheconstruction,thegraph�tisadisjointunionofcliques;letkdenotethenumber,andm1;:::;mkthesizesofthesecliques.Thus,wehavem1++mk=qandjIf(t)j=k;anditremainstoshowthatk�q=2.Wedistinguishtwocases.Ifqiseventhen,usingconvexity,wegetq 2�1e(�t)=m12++mk2kq=k2=1 2qq k�1;whenceq�1�q2 2k;leadingtothedesiredbound.Ifqisodd,welets:=#fi2[1;k]:mi=1gandl:=#fi2[1;k]:mi2g;sothats+l=kands+2lq:(8)Thenq�1 2e(�t)=Xi2[1;k]:mi2mi2l(q�s)=l2=1 2(q�s)q�s l�1=1 2l(q�s)(q�k):Ifwehadkq=2,thiswouldyieldq 2�q�1 21 2l(q�s)q 2; 16S.KOPPARTY,V.LEV,S.SARAF,ANDM.SUDANIfd="1e1++"nen2D0then,lettingb:=Pi2[1;n]:"i=0ei,foreacht2Fqwehave1(b+td)=0(d)+t�1(d)�2n=q�2(n=q)1=2;whenceb+td2A0.Thus,thesetK0:=B[A0containsalineineverypopulardirection.ToestimatethesizeofK0wenoticethat,lettingN:=2n=q�2(n=q)1=2+1,wehavejA0j=nXj=Nnj(q�2)n�j:Assumption(9)impliesthatthesummandsintheright-handsidedecayasjgrows,whencejA0jnnN(q�2)n�N:Consequently,writingH(x):=xln(1=x)+(1�x)ln(1=(1�x));x2(0;1)andusingawell-knownestimateforthebinomialcoecients,wegetjA0jnexp(nH(N=n)+(n�N)ln(q�2)):Now,inviewof(9)wehave1 qN n2 q1�1 q;andtherefore,sinceH(x)isconcaveandsymmetricaroundthepointx=1=2,using(9)onceagain,fromthemeanvaluetheoremwederiveH(N=n)�H(2=q)=O�(N=n�2=q)H0(1=q)=O�(lnq=(nq))1=2H0(1=q)=O�(lnq)3=2=(nq)1=2:HencenH(N=n)+(n�N)ln(q�2)=nH(2=q)+n(1�2=q)ln(q�2)+O�(n=q)1=2(lnq)3=2=nlnq�2 qln2+O�(n=q)1=2(lnq)3=2;implyingjA0jq 22=qnexp�O�(n=q)1=2(lnq)3=2:Sinceq=22=q�2forq3,weconcludethatjK0jjA0j+jBjq 22=qnexp�O�(n=q)1=2(lnq)3=2=q 22=qn+O�p nlnq=q: 18S.KOPPARTY,V.LEV,S.SARAF,ANDM.SUDANOfpossibleimprovementsandresearchdirections,thefollowingtwoseemofparticularinteresttous.First,itwouldbenicetobeattheuniversalsetconstructionintheregimejustmentioned(dimVgrowsandr2),ortoshowthatitproducesanessentiallybestpossiblebound.Eventhecaseq=r=2seemsnon-trivial:wedonotknowanyconstructionofKakeyasetsofrank2inFn2ofsizesmallerthanO(23n=4),theboundsuppliedby4-universalsets.TheseconddirectionstemsfromthefactthattheproductofKakeyasetsofrankrisaKakeyasetofrankrintheproductspace.Itisnotdiculttoderivethat,with(n)q(r)denotingthesmallestpossiblesizeofaKakeyasetofrankrinFnq,thelimitlimn!11 nln(n)q(r)existsforanyxedqandr.Itwouldbeveryinterestingtondthislimitexplicitly,evenforjustoneparticularpair(q;r)6=(2;1).Arguably,mostintriguingistherstnon-trivialcaseq=3;r=1,duetothefactthatlinesinFr3arethree-termarithmeticprogressions.Appendix:proofoftheliftinglemma.Weproveherethefollowinglemma,whichisaslightextensionofLemma3.Lemma22.Letnrr11beintegersandFaeld.SupposethatK1isaKakeyasetofrankr1inFn�(r�r1),consideredasasubspaceofFn,andletK:=K1[(FnnFn�(r�r1)).ThenKisaKakeyasetofrankrinFn.Proof.SupposethatLFnisasubspacewithdimL=r.FromdimL+dimFn�(r�r1)=dim(L+Fn�(r�r1))+dim(L\Fn�(r�r1))itfollowsthateitherL+Fn�(r�r1)isapropersubspaceofFn,ordim(L\Fn�(r�r1))=r1.Observingthatifv=2L+Fn�(r�r1),thenv+LisdisjointwithFn�(r�r1),weconcludethat,ineithercase,thereisatranslateofL,intersectingFn�(r�r1)byasubsetofar1-dimensionalsubspace.Hence,thereisalsoatranslateofL,theintersectionofwhichwithFn�(r�r1)iscontainedinK1.Bytheconstruction,thistranslateofLiscontainedinK.References[ABS09]N.Alon,B.Bukh,andB.Sudakov,DiscreteKakeya-typeproblemsandsmallbases,IsraelJ.Math.174(2009),285{301.[AS08]N.AlonandJ.H.Spencer,Theprobabilisticmethod,Thirdedition,Wiley-InterscienceSeriesinDiscreteMathematicsandOptimization,JohnWiley&Sons,Inc.,Hoboken,NJ,2008.xviii+352pp.[DKSS]Z.Dvir,S.Kopparty,S.Saraf,andM.Sudan,Extensionstothemethodofmultiplici-ties,withapplicationstoKakeyasetsandmergers,ElectronicColloquiumonComputationalComplexity(ECCC)16(2009).[EOT09]J.Ellenberg,R.Oberlin,andT.Tao,TheKakeyasetandmaximalconjecturesforalgebraicvarietiesoverniteelds,Mathematika56(1)(2009),1{25.