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Contents lists available atScienceDirect

Journal of Pure and Applied Algebra

www.elsevier.com/locate/jpaa

Rank 2 Ulrich bundles on general double plane covers

Ronnie Sebastiana,∗, Amit Tripathib

a DepartmentofMathematics,IndianInstituteofTechnologyBombay,Powai,Mumbai400076, Maharashtra,India

bDepartmentofMathematics,IndianInstituteof TechnologyHyderabad,Kandi,Sangareddy,502285, Telangana,India

a r t i cl e i n f o a b s t r a c t

Articlehistory:

Received4December2020

Receivedinrevisedform2May2021 Availableonline19June2021 CommunicatedbyV.Suresh

MSC:

14E20;14J60;14H50

Keywords:

Ulrichbundles Doubleplanes Cayley-Bacharach

Weprove thata doublecoverof P2 ramified alongageneralsmoothcurveB of degree2s,fors3,supportsarank2 specialUlrichbundle.

©2021ElsevierB.V.Allrightsreserved.

1. Introduction

ThroughoutthispaperweworkoverthefieldofcomplexnumbersC.LetX be ad-dimensionalsmooth projectivevariety.Unlessmentionedotherwise,OX(1) willalwaysdenoteanample andgloballygenerated linebundle onX.

Definition 1.1.A locallyfree sheaf(vectorbundle) E onX is saidto be Ulrichwith respect toOX(1) (or simplyUlrichwhenthebundleOX(1) isunderstood)ifthefollowingtwoconditionsaresatisfied

(1) Hi(X,E(−i))= 0 foralli>0, (2) Hj(X,E(−j−1))= 0 forallj < n.

We referthe reader to [3, §2] forbasic definitions. Inthe literature authors usually define Ulrich with respecttoaveryamplelinebundle,kindlyseethenextsectionforsomeremarksrelatedtothis.Aconjecture

* Correspondingauthor.

E-mailaddresses:[email protected](R. Sebastian),[email protected](A. Tripathi).

https://doi.org/10.1016/j.jpaa.2021.106823 0022-4049/©2021ElsevierB.V.Allrightsreserved.

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ofEisenbudandSchreyer[11] statesthateverysmoothprojectivevarietysupportsanUlrichbundle.Several people have constructed Ulrich bundles on particular varieties and we list a few. Theyhave been shown to exist on completeintersections by [14], oncurvesand del Pezzo surfaces by[11], onK3 surfaces by[2]

and [12],onabeliansurfaces by[4], onruledsurfaces by[1], onregularsurfaceby[6],[8],onsurfaces with pg = 0 and q = 1 by[7], onsurfaces of maximal Albanese dimensionand someirregular surfaces by[15], [16].Theabovelistisfarfrombeing completeandwereferthereadertotheabovepapers,forexample[7], andthereferencesthereinformoreresults.In[10,Theorem4.3] theauthorsshow,overthefieldofcomplex numbers,theexistenceofUlrichbundlesofranktwoinasufficientlyample embedding.

RecentlyNarayananandParameswaran[17] studiedtheexistenceofUlrichlinebundlesonadoubleplane π:X P2 branchedalongasmoothcurveB P2ofdegree2s.In[17,Theorem1.5],theyprovethatfor eachs≥3,therearespecialclassesofdoubleplaneswhichadmitUlrichline bundles.In[17,Theorem 1.4]

theyshow thatadouble planebranchedalongagenericsmoothcurveof degree2s,where s≥3,doesnot supportanUlrichlinebundle.

LetX beasurfaceandletKX bethecanonicallinebundle.AnUlrichbundleEofrank2,withrespectto OX(1),iscalledspecialUlrichifitalsosatisfiesdet(E)=KX⊗ OX(3).Letπ:X P2beadegree2cover which isbranchedalongasmoothcurveB P2 ofdegree2s.Forsuchamap denoteOX(1):=πOP2(1) andbyanUlrich(respectively,specialUlrich)bundleonX wewillalwaysmeanUlrich(respectively,special Ulrich)with respecttoOX(1).Inthisnoteweshow thefollowing.

Theorem 1.2. Letπ:X P2 be adouble cover branched along a genericsmooth curve B⊂P2 of degree 2s,wheres≥3.ThenX admits aspecialrank2Ulrich bundle.

To provethe aboveresult we usetwo inputs. The firstis the well knowncorrespondence betweenzero dimensional subschemes satisfying the Cayley-Bacharach property and global sections of a rank 2vector bundle, see [19, §5].Let F be thedegree2shomogeneouspolynomialwhich definesB.Using [19] wefirst prove

Theorem 1.3 (Theorem 2.7).Letπ:X P2 be adegree 2 cover whichis branched along a smoothcurve B P2 of degree 2s,where s≥3.LetF denotethepolynomial ofdegree 2swhichdefines B.Assume that there are twopolynomials F1 andF2 of degree s suchthat F (F1,F2).Then X supportsa specialUlrich bundleof rank2.

Thesecond inputisthefirstpointin[9,Theorem 5.1] whichenablesusto concludethatforthegeneral degree2shypersurfaceF wecanfinddegreeshypersurfacesF1and F2 suchthatF (F1,F2).Wedonot know ifthis holdsforallsmoothdegree2shypersurfaces.

It hasbeen brought to our attentionthat Mohan Kumar, P. Narayanan and A.J. Parameswaran have proved,using adifferentmethod,thateverydouble planecoversupportsarank2Ulrichbundle.

Acknowledgements. We thank Enrico Carlini and Luca Chiantini for several helpful discussions related to their article [9]. We thank Gianfranco Casnati for several useful comments. We thank the referee for an extremely carefulreading of this article andfor numerous usefulsuggestions whichhaveimproved the exposition.

2. ExistenceofUlrichbundles

To showthatabundleE onX isUlrichwewill usethefollowing criterion.

Lemma 2.1. LetX be a d-dimensionalsmooth projective variety and let π : X Pd be a surjective and finite mapofdegreee.AbundleE onX isUlrichwithrespecttoπOPd(1)ifandonlyifπE∼=OPrank(E)d .

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Proof. LetusfirstassumethatπE∼=OePrank(E)d .Sincethemapisfinite,andusingprojectionformula,we haveHi(X,E(k))=Hi(PdE(k)) foralli,k∈Z.Since πE =OePrank(E)d itisclear thattheconditions inDefinition1.1are satisfied.

Conversely,assumethatEsatisfiestheconditionsinDefinition1.1.ThenitfollowsthatπEand(πE) are0-regularofrankrank(E).Thus,bothofthemarem-regularforallm≥0.Fromthisiteasilyfollows that Hi(PdE(k)) = 0 for allk Z and for all 1 ≤i ≤d−1, thatis, πE is an ACM bundle. Now applying Horrockscriterion, see [13] or [18, Theorem 2.3.1] where itis stated moreprecisely, we get that πE isadirectsumoflinebundles.IfOPd(a) isasummandofπEthenwegetthatHi(Pd,OPd(a−i))= 0 foralli>0 andHj(Pd,OPd(a−j−1))= 0 forallj < d.Inparticular,bytakingi=dandj= 0 itfollows thata= 0.ThisshowsthatπE∼=OPe·drank(E).

As mentioned before, authors usually define Ulrich bundles with respect to very ample line bundles.

However, the existence of an Ulrich bundle with respect to an ample and globally generated line bundle L ensuresthat there are Ulrich bundleswith respect to Ln for alln >0,see [3, Proposition3] and the remarksfollowing it.

2.2. Double covers of P2. We briefly recall the main properties of double covers which we will use. A generalreferenceforthis is[5,§17].LetB P2be asmoothcurveofdegree2sdefinedbyahomogeneous polynomialF. Letπ:X P2 bethedouble coverofP2 branchedalongB, theconstructionofwhichwe briefly explain forthe benefitof the reader.Let A denote thetotal space ofthe line bundle A=OP2(s), π:AP2theprojectionand

T ∈H0(A, πA) (2.3)

be the tautological section. Define X to be the subvariety of A defined by the section T2 −πF H0(AA2) = H0(AOP2(2s)). We will abuse notation and denote the composite X A P2 alsobyπ.Thenπisafinite mapofdegree2betweenprojectivevarieties.Welisttheimportantproperties ofdoublecoversthatwewilluse,see[5,Lemma17.1,Lemma17.2].

(1) IfB issmooththenX issmooth,thisisexplainedinthesentencepreceding[5,Lemma17.1].

(2) Let R⊂X denotethereduceddivisorπ1(B).ThenπOP2(s)=OX(R).

(3) Thecanonicalbundle KX =OX(s−3).

(4) πOX =OP2⊕ OP2(−s).

2.4. Existence ofUlrichbundles.Inthis sectionweshall provethatageneraldouble planecoversupports aspecialUlrichbundle ofrank2.Wewill usetheresultin[19, Theorem10] toconstruct arank2bundle onX.Forthebenefitofthereaderwestatethemainresultfrom[19] thatweneed,butfirstwerecallsome definitionsfrom[19,page2].GivenasubschemeZ2⊂Z1,the“complement”Z ofZ2 inZ1isthecanonical closedsubschemeZ ⊂Z1with sheafofidealsIZ = [IZ1 :IZ2], thatis,forany opensetU ⊂X,wedefine

IZ(U) :={g∈ OX(U)|gIZ2(U)⊂IZ1(U)}.

Wecall Z theresidualsubscheme ofZ2 inZ1 anddenote itbyZ =Z1−Z2 inthestatementof thenext theorem,whichis[19,Theorem10]. Therearethreeequivalences, butwestateonlytwoofthese.

Theorem 2.5 (Theorem 10, [19]). Let X be a complex smooth projective variety of dimension n 2. Let Z ⊂X be asubschemeof purecodimension2.Thenthefollowingareequivalent:

(1) Z isthezero subschemeof asectionof arank2vectorbundleE.

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(2) There arehypersurfacesD1,D2,D3 suchthat D1andD2havenocommoncomponents,Z =D1∩D2 D1∩D2∩D3 andsuchthat D1∩D2∩D3 isof purecodimension2 andisCohen-Macaulay.

Further,if (1)and(2) holdthendet(E)≡ OX(D1+D2−D3).

WithnotationasaboveletE bethebundlewhichsitsinthefollowingshortexactsyzygysequence,this is explainedjustbefore[19,Theorem10].

0→ E(−D1−D2)

i

OX(−Di)→ID1D2D3 0. (2.6)

Theorem2.7. Letπ:X P2 beadegree 2coverwhichisbranchedalongasmoothcurve B⊂P2of degree 2s, where s 3. Let F denote the polynomial of degree 2s which defines B. Assume that there are two polynomials F1andF2 ofdegree ssuchthat F∈(F1,F2).ThenX supportsaspecial Ulrichbundleofrank 2.

Proof. Firstletusnote thatthere isnonon-constantpolynomialH whichdividesboth F1 andF2,orelse H willdivideF,contradictingthesmoothnessofB.Thus,thesubschemeofP2definedbytheideal(F1,F2) is 0-dimensionaland iscontainedinB.Wedenote this byZ thesubscheme definedby theideal(F1,F2).

ConsidertheschemetheoreticinverseimageZ1:=π1(Z).

Fori= 1,2 takeHi=πFi∈H0(X,OX(s)).IfDi denotesthedivisordefinedbyHi thenZ1=D1∩D2 in the notation of Theorem 2.5. Take H3 =T ∈H0(X,OX(s)) (see equation (2.3), byabuse of notation we denote the restriction of T to X also by T) and Z2 to be the subscheme of Z1 defined by H3, thus, Z2=D1∩D2∩D3,where D3 isthedivisordefinedbyH3. Letus computetheidealIZ := [IZ1:IZ2].

If x,y,z are homogeneous coordinates on P2 then let C[x,y] (we abuse notation hereand denote the affine coordinates also by xand y) denote the coordinatering of the open set {z = 0}. Denote byf the equation F in C[x,y], similarly, for the other polynomials.The inverse image of this open set inX has coordinatering

C[x, y, t]/(t2−f).

TheidealIZ1 = (f1,f2) andtheidealIZ2 = (f1,f2,t).IfI⊂J aretwoidealsinaringthenitisclearthat I⊂[I:J].Thus,IZ1 ⊂IZ.Weclaimthatt∈[IZ1:IZ2].AnelementofIZ2 lookslikeλf1+μf2+θt.Since t2 =f (f1,f2) itfollows that t(λf1+μf2+θt)∈IZ1 thatis, t [IZ1 : IZ2]. Thus, IZ = (f1,f2,t). In particular, Z=Z2=D1∩D2∩D3.Thus,inthenotationfrom[19,§1,page2],wemaywrite

Z =D1∩D2−D1∩D2∩D3.

Moreover,D1∩D2∩D3 isofpurecodimension2andisCohen-Macaulay(sincebothdepthanddimension are 0).Thus,there isabundleE whichsitsintheshortexactsequence(2.6) andhasaglobalsection

OX → E (2.8)

whose vanishing gives Z. The line bundles OX(Di) are allisomorphic to OX(s). Thus, det(E) = OX(s).

Twisting theshortexactsequence(2.6) byOX(2s) wegetashortexactsequence

0→ E → OX(s)3→IZ(2s)0. (2.9) Considerthecommutativediagram

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0 IZ

i

OP2 π#

OZ 0

0 πIZ πOX πOZ 0.

Inthenotationweused above,thecoordinateringofZ is C[x,y]/(f1,f2) and thecoordinateringofZ is C[x,y,t]/(f1,f2,t)=C[x,y]/(f1,f2).Thus, theright vertical arrow isanisomorphism, whichproves that thecokernelofπ# andiareisomorphic.

Let σ denote the involution of X interchanging the two sheets of the double cover. One hasthe trace map Tr :πOX → OP2 definedas follows. LetU P2 be openand letf ∈ OX(π1(U)).Define Tr(f):=

f+f◦σ∈ OP2(U).Now iff ∈ OX(π1(U)) andf vanishesat apoint p∈Z,thensinceZ ⊂B andσis theidentityonB,itfollows thatf◦σalsovanishesat p.From thisitisclearthatTr mapsπIZ toIZ.

ThemapTr givesasplittingofthemapπ#.SinceTr mapsπIZ toIZ itfollowsthatTr givesasplitting of themap i.Thus, πIZ isisomorphic tothe directsumofIZ andthe cokernelof i. Wesawabovethat thecokernelofπ#andthecokernelofiareisomorphic.SinceπOX =OP2⊕ OP2(−s),seeproperty(4)in subsection2.2,itfollowsthatthecokernelofπ# isisomorphictoOP2(−s).Thus,itfollowsthat

πIZ =IZ⊕ OP2(−s). (2.10)

Thus,wehave

h0(X, IZ(2s)) =h0(P2, IZ(2s)) +h0(P2,OP2(s)). (2.11) Usingtheshort exactsequence

0→ OP2(2s)→ OP2(−s)2→IZ 0, (2.12) wegetthath0(P2,IZ(2s))= 2h0(P2,OP2(s))1.Usingequation(2.11) weget

h0(X, IZ(2s)) = 3h0(P2,OP2(s))1. NextwewillcomputeH0(X,E).Takingdualof(2.8) wegetanexactsequence

0det(E)→ E→IZ0.

Sincedet(E)=OX(s) andE isof rank2,wegetE=E ⊗det(E),whichgives 0→ OX → E →IZ(s)0.

Applyingπto thisweget

0→πOX→πE →πIZ(s)0. (2.13) Using(2.12) wegeth0(P2,IZ(s))= 2 andusing(2.10) wegetthat

h0(P2, πIZ(s)) = 3.

Fromthis itfollowsthath0(P2E)= 4.Applyingπ to(2.9) andtakingcohomology weget

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h1(P2, πE) =h0(P2, πIZ(2s)) +h0(P2, πE)33h0(P2,OP2(s))

= 3h0(P2,OP2(s))1 + 433h0(P2,OP2(s))

= 0. Considerthecommutativediagram

H0(P2, πE) H0(P2, πIZ(s))

H0(P2,OP2)

The verticalarrowis theprojectioninequation(2.10) and issurjective.The horizontalarrow issurjective as is easily seen by taking cohomology of the sequence (2.13). Thus, we have a map πE → OP2 which induces asurjectiononglobalsections. Thisshowsthatthis mapissplit.Thus,

πE =G ⊕ OP2, where G isalocally freesheafandsitsinashortexactsequence

0→πOX → G →IZ(s)0.

Wewill nowshowthatGistrivial.Considerthefollowingpullbackdiagram.

0 πOX F

a

OP22 b

0

0 πOX G IZ(s) 0

From this itfollows thatF =OP23⊕ OP2(−s) sinceExt1(OP2OX)= 0.Wemaysplitthe toprowand compose thesplittingwithato getadiagram

0 OP2(−s)

d

OP22 c

IZ(s) 0

0 πOX G IZ(s) 0

SupposeKerc= 0,thentheimageofcisasheafofrank1,whichsurjectsontoIZ(s).Thisforcesthatthe image isisomorphictoIZ(s),whichdefinesasplittingofthebottomrow.However,sinceGislocally free, this isnotpossible.Thus,Kerc= 0.

Now letus consider theleft vertical arrow d:OP2(−s)→ OP2⊕ OP2(−s). Ifthe cokernelis OP2 then we getthatG is thetrivialbundle. Theonlyother possibility forthecokernel isOC⊕ OP2(−s),where C is ahypersurfaceofdegreesinP2.Inthiscase,G sitsinasequence

0→ OP22 → G → OC⊕ OP2(−s)0.

Since Gis asummandof πE andH1(P2E)= 0,itfollows thatH1(P2,G)= 0.This forcesthat H1(P2,OC) = 0.

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But now using 0 → OP2(−s) → OP2 → OC 0 we get 0 = H1(P2,OC) = H2(P2,OP2(−s)) = H0(P2,OP2(s−3)), which is not possible if s 3. Thus, the cokernel of d is OP2 and so G and πE aretrivial.Thisproves thatE isanUlrichbundleonX.

Recallfrom (3)thatthecanonicalline bundleofX isOX(s−3).Sincedet(E)=OX(s)=OX(s−3) OX(3), itfollowsthatE isaspecialUlrichbundle.

Proof of Theorem1.2. Fortheconvenienceofthereaderwerecallthestatementfrom[9] thatweareusing.

Their main result givesa descriptionof all the possible complete intersections of codimension r thatcan be found ona general hypersurfaceof degree din Pn when 2r ≤n+ 2. In our situationn =r = 2 and d= 2s.Inthissituationthefirstpointin[9,Theorem5.1] enablesustoconcludethatforthegeneraldegree 2s hypersurface F we can find degree s hypersurfaces F1 and F2 such that F (F1,F2). Now we apply Theorem2.7 andgetthatX supportsaspecialUlrichbundleofrank2.This provesTheorem1.2.

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