Spaces of matrices of constant rank and uniform vector bundles

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Contents lists available atScienceDirect

Linear

Algebra

and

its

Applications

www.elsevier.com/locate/laa

Spaces

of

matrices

of

constant

rank

and

uniform

vector

bundles

Ph. Elliaa,∗_, _{P. Menegatti}b

a _Dipartimento_di_Matematica_e_Informatica,₃₅_via_Machiavelli,₄₄₁₀₀_Ferrara,

Italy

b

LaboratoiredeMathématiquesetApplications,UMRCNRS6086,Université de Poitiers,Téléport2,Bd.PetM.Curie,FuturoscopeChasseneuilF-86962, France

a r t i c l e i n f o a bs t r a c t

Article history:

Received2August2015 Accepted9June2016 Availableonline11June2016 SubmittedbyR.Brualdi MSC: 15A30 14J60 Keywords: Spacesofmatrices Constantrank Uniform Vectorbundles

Weconsidertheproblemofdeterminingl(r,a),themaximal dimension of a subspace of a× a matrices of rank r. We ﬁrst review, in the language of vector bundles, the known results.Thenusingknownfactsonuniformbundlesweprove somenewresultsandmakeaconjecture.Finallywedetermine

l(r;a) foreveryr,1≤ r ≤ a,whena≤ 10,showingthatour conjectureholdstrueinthisrange.

0. Introduction

LetA,B bek-vectorspacesofdimensionsa,b (k algebraicallyclosed,ofcharacteristic zero). A sub-vector space M ⊂ L(A,B) is said to be of (constant) rank r if every

f ∈ M,f = 0, has rank r. The question considered in this paper is to determine * Correspondingauthor.

E-mailaddresses:[email protected](Ph. Ellia),[email protected](P. Menegatti).

http://dx.doi.org/10.1016/j.laa.2016.06.019

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l(r,a,b) := max{dim M | M ⊂ L(A,B) has rank r}. This problem has been studied sometime agobyvarious authors [22,20,4,9]and hasbeen recentlyreconsidered, espe-ciallyinits(skew)symmetricversion[17,18,16,5].

Itisknown,atleastsince[20],thattogiveasubspaceM ofconstantrankr,dimension

n+1,isequivalenttogiveanexactsequence:0→ F → a.O(−1)→ b.O → E → 0,ψ onPn_, whereF,E arevectorbundlesofranks(a− r),(b− r).Ourstartingpointistoobserve thatthebundle E := Im(ψ),ofrankr, is uniform,ofsplitting type(−1c_,₀r−c_),_where c:= c1(E) (Lemma 2).This hadbeen previouslyobserved(butnotreally exploited)in

thecasesof spacesofsymmetric or skew-symmetricmaps [17].This allowsus toapply theknownresults(andconjectures)onuniformbundles.

Thispaperis organizedas follows. Inthefirstsectionwerecallsomebasicfactsand fixthe notations.Then inSection two,we set a= b tofix theideasand wesurvey the known results (at least those we are aware of), giving a quick, uniform (!) treatment in the language of vector bundles. In Section three, using known results on uniform bundles, weobtainanewbound onl(r;a) inthe range(2a+ 2)/3> r > (a+ 2)/2 (as well as some other results, see Theorem 18). By the way we don’t expect this bound tobe sharp.Indeedby“translating”(see Proposition 17)alongstandingconjectureon uniformbundles(Conjecture 1),weconjecturethatl(r;a)= a− r + 1 inthisrange(see

Conjecture 2). Finally, with somead hocarguments, we show inthe last section, that ourconjectureholdstruefora≤ 10 (actuallywedeterminel(r;a) foreveryr,1≤ r ≤ a, whena≤ 10).

1. Generalities

Following [20], to give M ⊂ Hom(A,B), a sub-space of constant rank r, with dim(M )= n+ 1,isequivalenttogiveonPn_, _an_exact_sequence:

0 FM a.O(−1)

ψM

b.O EM 0

EM

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where EM = Im(ψM),FM,EM are vectorbundles ofranksr,a− r,b− r (in thesequel wewilldroptheindex M ifnoconfusion canarise).

Indeed the inclusion i : M → Hom(A,B) is an element of Hom(M,A∨⊗ B) M∨⊗ A∨⊗ B and canbe seenas amorphism ψ : A⊗ O → B ⊗ O(1) on P(M) (here

P(M) isthe projective spaceof lines of M ). Ateverypoint of P(M), ψ has rankr, so theimage,thekernelandthecokernel ofψ arevectorbundles.

Adiﬀerent(butequivalent)descriptiongoesasfollows:wecandeﬁneψ : A⊗O(−1)→ B⊗ O onP(M), byv⊗ λf → λf(v).

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Deﬁnition1.Arankr vectorbundle,E,onPn isuniform ifthereexists(a1,...,ar) such thatEL ri=1OL(ai),foreveryline L⊂ Pn ((a1,...,ar) isthesplitting typeofE,it is independentof L).

Thevectorbundle F ishomogeneous ifg∗(F ) F ,foreveryautomorphismofPn_. Clearly ahomogeneousbundleisuniform(but theconverseisnottrue).

Theﬁrstremark is:

Lemma2.Withnotationsasin(1),c1(EM)≥ 0 andEM isauniformbundleofsplitting type (−1c_,₀b_), _where_c_{= c}

1(EM),b= r− c.

Proof. Since E is globally generated, c1(E) ≥ 0 (look at EL). Let EL = OL(ai). We have ai ≥ −1,because a.OL(−1) EL. Wehave ai ≤ 0, becauseEL → b.OL.So −1 ≤ ai ≤ 0,∀i. Since c1(E) = −c1(E) the splitting type is as asserted and does not

depend ontheline L. 2

The classiﬁcationofrank r≤ n+ 1 uniformbundleson Pn_, _n_{≥ 2,}_is _known_[21,10,

12,1];ifwedenote byT,Ω thetangentand cotangentbundle onPn_,_then_we_have: Theorem3.Arankr≤ n+1 uniformvectorbundleonPn_,_n_{≥ 2,}_is_one_of_the_following: r

O(ai), T (a)⊕ k.O(b), Ω(a)⊕ k.O(b) (0≤ k ≤ 1),S2TP2(a).

Wewill usethefollowing result(see [15,8]):

Theorem 4(Evans–Griﬃth).LetF bea rankr vectorbundleon Pn_,_then _{F is}_a _direct sum of linebundlesif andonlyif H_∗i(F)= 0, for1≤ i≤ r − 1.

TheﬁrstpartofthefollowingPropositioniswellknown,thesecondmaybeless. Proposition 5.Assumen≥ 1.

(1) If a≥ b+ n the genericmorphisma.O_Pn → b.O_Pn(1) issurjective.

(2) If a< b+ n no morphisma.O_Pn → b.O_Pn(1) canbe surjective.

Proof. (1)It isenough totreatthe casea= b+ n and,bysemi-continuity, toproduce oneexampleofsurjectivemorphism.Consider

Ψ = ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ x0 · · · xn 0 · · · 0 0 x0 · · · xn 0 · · · 0 .. . ... 0 · · · 0 x0 · · · xn ⎞ ⎟ ⎟ ⎟ ⎟ ⎠

(eachrowcontainsb− 1 zeroes).Itisclearthatthismatrixhasrankb atanypoint.For amoreconceptual(andcomplicated)proofsee[14],Prop. 1.1.

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(2) If n = 1, the statementis clear. Assumen ≥ 2. If ψ is surjective we have0 →

K → a.O → b.O(1)→ 0 and K is avector bundle of rankr = a− b < n. Clearly we haveHi

∗(K∨)= 0 for 1≤ i≤ r − 1≤ n− 2.ByEvans–Griﬃth’stheorem,K splitsasa directsumofline bundles,hencetheexactsequencesplits(n≥ 2) andthisisabsurd.

ThiscanalsobeprovedbyaChernclasscomputation(see[20]). 2 Fromnowonwewill assumeA= B andwritel(r;a) insteadof l(r;a,a).

2. Knownresults

Webeginwithsomegeneralfacts:

Lemma 6. Assume the bundle E corresponding to M ⊂ End(A) of constant rank r,

dim(A)= a, isadirectsum of linebundles.Then dim(M )≤ a− r + 1.

Proof. Let dim(M ) = n+ 1 and assume E = k.O(−1)⊕ (r − k).O. If k = 0, the surjectiona.O(−1) E r.O, showsthata≥ r + n (see Proposition 5).If k > 0, we have0→ k.O(−1)→ (a− r + k).O → E → 0.Dualizingweget:(a− r + k).O k.O(1), hence(alwaysbyProposition 5)a− r + k ≥ k + n.Soinany casea− r ≥ n. 2

Lemma7.Forevery r,1≤ r ≤ a,wehavel(r;a)≥ a− r + 1.

Proof. Setn= a− r.On Pn we haveasurjectivemorphism a.O(−1)→ r.O (ψ Proposi-tion 5).Composingwiththeinclusionr.O → r.O⊕(a−r).O,wegetψ : a.O(−1)→ a.O,

ofconstantrank r. 2 Finallyweget:

Proposition8.(1) Wehavel(r;a)≤ max{r + 1,a− r + 1}. (2)If a≥ 2r,then l(r;a)= a− r + 1.

Proof. (1)Assumer + 1≥ a− r + 1.Ifdim(M )= l(r,a)= n+ 1 andifr < n,then[12]

E is adirectsum of line bundles andn ≤ a− r.But then r < n≤ a− r,against our assumption.Sor + 1≥ n+ 1= l(r;a).

Now assumea− r ≥ r.If n> a− r,then n> r and this impliesthatE is adirect sumoflinebundles.Hencen≤ a− r.

(2)Wehavemax{r + 1,a− r + 1}= a− r + 1 ifa≥ 2r.Sol(r;a)≤ a− r + 1 by (1).

Weconcludewith Lemma 7. 2

Remark9.Proposition 8wasﬁrstproved(byadiﬀerentmethod) byBeasley[4]. Veryfewindecomposablerankr vectorbundleswithr < n areknownonPn(n> 4).

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Lemma 10.Wehave l(t+ 1;2t+ 1)= t+ 2.

Proof. By Proposition 8 we know thatl(t+ 1;2t+ 1) ≤ t+ 2. Soit is enoughto give an example. Set n = t+ 1 and assume ﬁrst n ≥ 3. If T denotes the Tango bundle, then we have: 0→ T (−2) → (2n− 1).O → T → 0. Dualizing we get 0→ T∨(−1)→ (2n− 1).O(−1) → Ω(1) → 0. Combining with the exact sequence: 0 → Ω(1) → (n+ 1).O⊕(n−2).O → O(1)⊕(n−2).O → 0,wegetamorphism(2n−1).O(−1)→ (2n−1).O, of constantrankn.

Ifn= 2,usingthefactthatT (−2) Ω(1),fromEuler’ssequence,weget3.O(−1)→ 3.O,whoseimageisT (−2). 2

Remark11.Lemma 10wasﬁrstprovedbyBeasley [4], byadiﬀerentmethod. Finallyontheoppositeside,whenr isbigcompared witha,wehave: Proposition 12.(Sylvester [20]) Wehave:

l(a− 1; a) =

2 if a is even 3 if a is odd

The proofisaChern classescomputation.Thenextcasea= r− 2 is moreinvolved and thereareonlypartialresults:

Proposition 13.(Westwick [24])Wehave 3≤ l(a− 2;a)≤ 5.Moreover:

(1) l(a− 2;a)≤ 4 except ifa≡ 2,10 (mod 12) where itcould bel(a− 2;a)= 5. (2) If a≡ 0 (mod 3),thenl(a− 2;a)= 3.

(3) If a ≡ 1 (mod 3), we have l(2;4) = 3 and l(8;10) = 4 (so a doesn’t determine l(a− 2;a)).

(4) If a ≡ 2 (mod 3), then l(a − 2;a) ≥ 4. Moreover if a ≡ 2 (mod 4), then l(a− 2;a)= 4.

Proof. WedenotebyC(G)= 1+c1h+...+cnhntheChernpolynomialofG (computations aremadeinZ[h]/(hn+1)).WerecallthatifG isavectorbundleofrankr,thenci= 0 if i> r.WehaveC(F )= C(E)(1−h)a_._Let_{C(F )}_{= 1}_{+ s}

1h+ s2h2,C(E)= 1+ t1h+ t2h2.

Wegets1= t1− a (coeﬃcientof h);s2= a(a− 1)/2− at1+ t2(coeﬃcientof h2).From

the coeﬃcient of h3 _it _follows _that: _t

2 = (a− 1)[3t1− a+ 2]/6. The coeﬃcient of h4

yields after somecomputations: (a+ 1)(a− 2− 2t1) = 0. It follows thatt1 =

(a− 2) 2 and t2 =

(a− 1)(a − 2)

12 if we are on P

n_,_n _{≥ 4.} _Finally _the _coeﬃcient _of _h5 _gives

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IfweareonP4,fromt1= (a−2)/2 weseethata iseven.Fromt2= (a−1)(a−2)/12,

weget a2+ 2− 3a≡ 0 (mod 12). Thisimpliesa≡ 1,2,5,10 (mod 12).Since a is even wegeta≡ 2,10 (mod 12).

Ifa= 3m andifweareonP3_,_then_6t

2= (3m− 1)(3t1− 3m+ 2)≡ 0 (mod 6),which

isneversatisﬁed.Sol(a− 2,a)≤ 3 inthis case.

Theotherstatementsfollow fromtheconstructionofsuitableexamples,see[24]. 2 Remark14.OnP4_a_rank_two_vector_bundle_with_c

1= 0 hastoverifytheSchwarzenberger

conditionc2(c2+ 1) ≡ 0 (mod 12).If l(a− 2;a)= 5 for somea, then a= 12m+ 2 or

a= 12m+ 10.Intheﬁrstcasetheconditionyieldsm≡ 0,5,8,9 (mod 12),inthesecond case m ≡ 2,3,6,11 (mod 12). So, as already noticedin [24], the lowest possible value ofa is a= 34.This wouldgivean indecomposableranktwo vector bundlewith Chern classes c1 = 0,c2 = 24. Indeed if we have an exact sequence (1), E and F cannot be

bothadirectsumoflinebundles(because,byTheorem 4,E wouldalsobe adirectsum oflinebundles, whichis impossible).

Finallywehave:

Proposition15. (Westwick[23]) Foreverya,r,l(r;a)≤ 2a− 2r + 1.

As noticedin[17] (Theorem 1.4)this follows directlyfrom aresultof Lazarsfeldon amplevectorbundles.Wewillcome backlateronthisbound.

3. Furtherresultsandaconjecture

Thereareexamples,foreveryn≥ 2,ofuniformbutnon-homogeneous vectorbundles onPnofrank2n[6].Howeveritisalongstandingconjecturethateveryuniformvector bundle of rank r < 2n is homogeneous. Homogeneous vector bundles of rank r < 2n

onPn _are_classiﬁed_[2]_,_so_the_conjecture_can_be_formulated_as_follows:

Conjecture1.Everyrankr < 2n uniformvectorbundleonPn _is_a_direct_sum_of_bundles chosenamong:S2_T

P2(a),∧2T_P4(b),T_Pn(c),Ω_Pn(d),O_Pn(e);wherea,b,...,e areintegers.

Theconjectureholdstrueifn≤ 3[10,3].

Beforetogoonwepointoutanobviousbutusefulremark.

Remark16.Clearlyanexactsequence(1)existsifandonlyifthedualsequencetwisted by O(−1) exists. So we may replace E by E∨(−1). If E has splitting type (−1c_,₀b_), E∨₍_{−1) has}_splitting_type₍₀c_,₋₁b_).

Proposition 17. (1) Take r,n such that n≤ r < 2n. Assume a− r < n and that every rank r uniformbundleonPnishomogeneous.Thenl(r;a)≤ n,exceptifr = n,a= 2n−1

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(2) AssumeConjecture 1 istrue. Thenl(r;a)= a− r + 1 for r < (2a+ 2)/3, except if r = (a+ 1)/2, inwhichcasel(r;a)= a− r + 2.

Proof. (1) In order to prove the statement it is enough to show that there exists no subspace M ofconstant rankr anddimension n+ 1 underthe assumption a− r < n, n≤ r < 2n (exceptifr = n,a= 2n−1,inwhichcasel(n;2n−1)= n+ 1 byLemma 10). Suchaspacewouldgiveanexactsequence(1)withE uniformofrankr < 2n onPn.If

E isadirectsumoflinebundles,byLemma 6wegetdim(M )= n+ 1≤ a−r+1< n+ 1. HenceE isnotadirectsumofline bundles. Sincethesplitting typeofE is (−1c_,₀r−c₎ (Lemma 2), weseethat:E Ω(1)⊕ k.O ⊕ (r − k − n).O(−1),E T (−2)⊕ t.O ⊕ (r −

t− n).O(−1), or,ifn= 4,E (∧2Ω)(2).

Let’s ﬁrst get rid of this last case. The assumption a− r < n implies a ≤ 9. It is enoughtoshowthatthereisnoexactsequence(1)onP4_,_with_{E = (∧}2_{Ω)(2) and}_a_{= 9.}

From 0→ E → 9.O → E → 0, we get C(E) =C(E)−1. From the Koszul complexwe have0→ E → ∧2V ⊗ O → Ω(2)→ 0. Itfollows thatC(E)=C(Ω(2)).Since rk(E)= 3 and c4(ΩP4(2))= 1,wegetacontradiction.

SowemayassumeE Ω(1)⊕ k.O ⊕ (r − k − n).O(−1) orE T (−2)⊕ t.O ⊕ (r − t−

n).O(−1). Bydualizingtheexactsequence (1),wemayassumeE Ω(1)⊕ k.O ⊕ (r −

k− n).O(−1).Theexactsequence(1)yields:

0→ Ω(1) ⊕ (r − n − k).O(−1) → (a − k).O → E → 0 (2) SinceHi

∗(Ω)= 0 for2≤ i≤ n− 1,fromtheexactsequence(2)wegetH∗i(E)= 0,for 1≤ i≤ n− 2. Sincerk(E)= a− r < n, it followsfrom Evans–Griﬃth’s theorem that

E O(ai).Wehaveai≥ 0,∀i,becauseE isglobally generated.Moreoveroneai at leastmustbeequalto 1(otherwiseh1(E∨⊗ E)= 0 andthesequence(2)splits,whichis impossible).Soa1= 1,ai≥ 0,i> 1.Itfollowsthath0(E)≥ (n+1)+(a−r−1)= n+a−r. Ontheother handh0_(E)_{= a}_{− k from}₍₂₎_.

If k < r− n,we see thatoneof theai’s, i> 1,must be > 0. This implies h0(E)≥ 2(n+ 1)+ (a− r − 2)= 2n+ a− r.Soa− k = h0(E)≥ 2n+ a− r.Since a≥ a− k,it follows thata≥ 2n+ a− r and sor≥ 2n,againstourassumption.

We conclude that k = r − n and E = O(1)⊕ (a− r − 1).O. In particular E = Ω(1)⊕ (r − n).O ((2)isEuler’s sequenceplussomeisomorphisms).Weturnnowto the other exactsequence:

0→ F → a.O(−1) → Ω(1) ⊕ (r − n).O → 0 (3) WehaveC(F )= (1− h)a_._C(Ω(1))−1_._From_the_Euler_sequence _C(Ω(1))−1_{= 1}_{+ h.}_It follows that: C(F ) = (1 + h). _a i=0 a i (−1)ihi

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Sincerk(F )= a− r < n,cn(F )= 0.Since a≥ r ≥ n, itfollows that a n = a n− 1 . Thisimpliesa= 2n− 1.

Observethatr≥ n (becausek = r− n≥ 0).Ifr≥ n+ 1,thenrk(F )≤ n− 2,hence

c_n−1(F )= 0.Thisimplies: 2n− 1 n− 1 = 2n− 1 n− 2 ,whichisimpossible.

Weconcludethatr = n anda= 2n−1,sowearelookingatl(n;2n−1).ByLemma 10

weknowthatl(n;2n− 1)= n+ 1. Thisproves(1).

(2) Now we apply (1) by setting n := a− r + 1. Clearly n > a− r. The condition

n≤ r < 2n translatesin:(a+ 1)/2≤ r < (2a+ 2)/3.So,underthese assumptions, we getl(r;a)≤ n= a− r + 1,exceptifr = n, a= 2n− 1.Inthis lattercaseweknowthat

l(n;2n− 1)= n+ 1 (Lemma 10).WeconcludewithLemma 7. 2

SinceConjecture 1istrueforr≤ n+ 1 andn= 3,r = 5[3],wemaysummarizeour resultsasfollows:

Theorem18.

(1)If r≤ a/2, thenl(r,a)= a− r + 1.

(2)If a isodd,l(a+1₂ ;a)= a+1₂ + 1 (= a− r + 2).

(3)If (2a+2)₃ > r≥a

2+ 1, thenl(r;a)≤ r − 1.

(4)If a iseven:l(a₂+ 1;a)=a₂ (= a− r + 1).

(5)If r≥ (2a+ 2)/3,thenl(r,a)≤ 2(a− r)+ 1.

(6)Wehave l(5;7)= 3 (= a− r + 1). Proof. (1)This isProposition 8.

(2)ThisisLemma 10.

(3)Setn= r− 1.Uniformvectorbundlesofrankr = n+ 1 onPn _are_homogeneous. Wehaven≤ r < 2n if r≥ 3 anda− r < n ifr≥ (a/2)+ 1.Ifr≤ 2 andr≥ (a/2)+ 1, thena≤ 2.Hencer = a= 2 andl(2;2)= 1.Sotheassumptions ofProposition 17, (1) arefulﬁlled.Weconcludethatl(r,a)≤ r − 1.

(4)Followsfrom (3)andLemma 7. (5)ThisisProposition 15.

(6)Sinceuniform vectorbundlesofrank5onP3 _are_homogeneous,_this_follows_from

Proposition 17(1)andLemma 7. 2

Remark19.Point3ofthetheoremimprovesthepreviousboundofBeasleybutwedon’t expect this bound to be sharp (see Conjecture 2). Points 4 and 6 also are new. The boundof(5)is sofarthebestknownboundinthis range.Itisreachedforsomevalues ofa inthe caser = a− 1 (Proposition 12),butalreadyinthecaser = a− 2 we don’t knowifitissharp.

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Conjecture 2.Let a,r be integers such that (2a+ 2)/3> r > (a/2)+ 1,then l(r;a)=

a− r + 1.

Remark20.ThisconjectureshouldbeeasiertoprovethanConjecture 1,indeedinterms of vector bundles ittranslates as follows: everyrankr < 2n uniformvector bundle, E, ﬁttinginanexactsequence (1)onPn ishomogeneous.

Bythewaytheconditionr < 2n seemsnecessary.Ifn= 2 thiscanbeseenasfollows. Considerthefollowingmatrix(takenfrom[20]):

Ψ = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 −x2 0 −x0 0 x2 0 0 −x1 −x0 0 0 0 −x2 −x1 x0 x1 x2 0 0 0 x0 x1 0 0 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

It iseasytoseethatΨ hasrankfourat anypointofP2_,_hence_we_get:

0→ O(b) → 5.O(−1)→ 5.O → O(c) → 0Ψ

withE = Im(Ψ) arankfouruniformbundle.OnthelineL ofequationx2= 0,Ψ= (f,g),

wheref : 3.OL(−1)→ 2.OLisgivenby

x0 x1 0

0 x0 x1

andwhereg : 2.OL(−1)→ 3.OL

is givenby ⎛ ⎜ ⎝ −x0 0 −x1 −x0 0 −x1 ⎞ ⎟

⎠. Clearlyf issurjective andbycomputing theChernclasses the kernel is OL(−3). Also g is injective of constant rank two, hence the cokernel is OL(2).

It follows that b = −3,c = 2 and the splitting type of E is (−12_,₀2_). _Now _rank

four homogeneous bundles on P2 _are _classiﬁed _{(Prop. 3,} _{p. 18 of} _[7]₎ _and _are _direct

sumofbundles chosenamongO(a),T (b),S2T (c),S3T (d).IfE is homogeneoustheonly possibilityisE(1) T (−1)⊕O(1)⊕O,butinthiscasetheexactsequence0→ O(−2)→ 5.O → E(1)→ 0,wouldsplit,whichisabsurd.WeconcludethatE isnothomogeneous. InfactE isoneofthebundlesfoundbyElencwajg[11].

Remark21.Theresultsof thissectionandthepreviousonedeterminel(r;a) fora≤ 8,

1≤ r ≤ a.Togetacompletelistfora≤ 10,wehavetoshow,accordingtoConjecture 2, thatl(6;9)= l(7;10)= 4.This willbedoneinthenextsection.

4. Somepartialresults

In the following lemma we relax the assumption r < 2n in Proposition 17 when

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Lemma22. Assume wehave an exact sequence(1) onPn,with rk(F )= a− r < n and

c1(E(1))= 1. Thena= 2n− 1 andr = n.

Proof. Ifc1(E(1))= 1,E(1) hassplittingtype(1,0r−1).Itfollowsfrom[13],Prop. IV,2.2,

thatE(1)=O(1)⊕ (r − 1).O orE(1)= T (−1)⊕ (r − n).O.Fromtheexactsequence0→

F (1)→ a.O → E(1)→ 0,wegetc1(F (1))=−1.SinceF (1)→ a.O,itfollowsthatF (1)

isuniformofsplittingtype(−1,0a−r−1).Sincerk(F )< n,F (1)=O(−1)⊕(a−r−1).O. This shows thatnecessarily E(1) = T (−1)⊕ (r − n).O.Now from the exactsequence: 0→ T (−1)⊕ (r − n).O → a.O(1)→ E(1)→ 0, wegetC(E(1))= (C(t(−1))−1(1+ h)a_, i.e.C(E(1))= (1− h)(1+ h)a_. _Since_rk(E)_{< n,} _we_have_c

n(E(1))= 0 and arguingas intheproofofProposition 17,wegeta= 2n− 1,r = n. 2

Remark23.Sinceweknowthatl(n;2n− 1)= n+ 1 (Lemma 10),wemay,fromnowon, assumec1(E(1))≥ 2.

SinceE(1) isglobally generated,taking r− 1 generalsectionsweget:

0→ (r − 1).O → E(1) → IX(b)→ 0 (4) HereX isapure codimensiontwosubscheme, whichis smoothifn≤ 5 and whichis irreducible,reduced,withsingularlocusofcodimension≥ 6,ifn≥ 6.

Lemma 24. Assume n ≥ 3 and rk(F ) < n. If X is arithmetically Cohen–Macaulay (aCM),i.e.if H_∗i(IX)= 0 for 1≤ i≤ n− 2,then F isadirectsum of linebundles. Proof. From (4) we get Hi

∗(E) = 0 for 1≤ i ≤ n− 2. BySerre duality H∗i(E∨) = 0, for 2 ≤ i ≤ n− 1. From the exact sequence 0 → E∨ → a.O(1) → F∨ → 0, we get

Hi

∗(F∨)= 0, for1≤ i≤ n− 2.Since F∨ hasrank< n,byEvans–Griﬃth theorem we concludethatF∨ (hencealsoF )isadirectsumoflinebundles. 2

Proposition 25. Assumethat we have an exact sequence(1) onP4 _with _{rk(F )}_{< 4.}_Let

(−1c_,₀r−c_{) be}_the_splitting_type_of_E._If_{r > 4 and}_if_{F is}_not_a_direct_sum_of_line_bundles, thenc,r− c≥ 4;inparticular rk(E)≥ 8.

Proof. Assumec orb:= r− c< 4.Bydualizingtheexactsequence(1)ifnecessary, we mayassumeb< 4.Wehaveanexactsequence(4):

0→ (r − 1).O → E(1) → IX(b)→ 0 whereX ⊂ P4_is_a_smooth_surface_of_degree_d_{= c}

2(E(1)).Ifb< 3,X iseitheracomplete

intersection(1,d) orliesonahyper-quadric. InanycaseX isa.C.M.ByLemma 24,F

isadirectsumoflinebundles.

Assumeb= 3.FromtheclassiﬁcationofsmoothsurfacesinP4weknowthatifd≤ 3,

(11)

to asmooth surface,S, ofdegree 9− d by suchacomplete intersection.If S isa.C.M. the sameholds forX.From theclassiﬁcationofsmoothsurfaces oflow degreeinP4_, _if

X isnota.C.M.wehavetwopossibilities:

(i)X isaVeronesesurfaceandS isanellipticquinticscroll, (ii)X isanellipticquinticscrolland S isaVeronesesurface.

(i)IfX = V isaVeronesesurfacethenwehaveanexactsequence: 0→ 3.O → Ω(2) → IV(3)→ 0

ItfollowsthatC(E(1))=C(Ω(2))= 1+ 3h+ 4h2_{+ 2h}3_{+ h}4_._So_{C(F (1))}_{= (}_C(E(1))−1 ₌

1− 3h+ 5h2_{− 5h}3_._It_follows_that_{F (and}_hence_{E also)}_has_rank_three._From_C(E(1))₌

(1+ h)a.C(F (1)) andc4(E(1))= 0, weget 0= a(a− 5)(a− 6)(a− 7).So a≤ 7.Since

a= rk(E)+ r,wegetacontradiction.

(ii)IfX = E isanellipticquinticscroll,then wehave: 0→ T (−2) → 5.O → IE(3)→ 0

ItfollowsthatC(E(1))=C(T (−2))−1 andC(F (1))=C(T (−2))= 1−3h+4h2_−2h3_+h4_,

incontradictionwithrk(F )< 4. 2

Lemma 26.Assumewehave anexact sequence(1)onP4 _with_a_{− r < 4.} _If_{r > 4 and}_if

F is adirectsum ofline bundles,then rk(E)≥ 8.

Proof. If r = 5 we concludewith Theorem 18, (3), (6). If r = 6,then a ≤ 9 andit is enough to show that l(6;9) ≤ 4 i.e. that there is no exact sequence (1) on P4. In the sameway,ifr = 7 itisenoughtoshowthatl(7;10)≤ 4.

Ifr = 6,wemayassumethatthesplittingtypeofE is(−11_,₀5_),₍₋₁2_,₀4_),₍₋₁3_,₀3_).

BydualizingandbyLemma 22wemaydisregardtheﬁrstcase.Itfollowsthatc1(F )=−7

or−6.Ifr = 7,inasimilarway,wemayassumethatthesplittingtypeofE is(−12,05) or (−13_,₀4_)._So_c

1(F )=−7 or−8.

LetC(F (1))= (1− f1h)(1− f2h)(1− f3h).WehaveC(E(1))= (1+ h)aC(F (1)).From

c4(E(1))= 0 weget: ψ(a) := a3− a2(4s + 6) + a(12d + 12s + 11)− 12d − 8s − 24t − 6 = 0 where s = −c1(F (1))= fi, d = c2(F (1))= i<jfifj, t = −c3(F (1))= fi. We havefi ≥ 0,∀i and3≤ s≤ 5.

We havetocheck thatthis equalitycan’tbe satisﬁedfor a= 9,10.Wehaveψ(9) = 8(42− 28s+ 12d− 37).Ifψ(9)= 0 weget3| s.Itfollowsthats= 3.Sotheconditionis: 4d− t= 14.Ifoneofthefi’siszero,thent= 0 andwegetacontradiction.Sofi> 0,∀i and theonlypossibility is(fi)= (1,1,1),butthen4d− t= 11= 14.

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Fora= 10,wegetψ(10)= 504−288s+ 108d−24t.Iff1= f2= 0,thend= t= 0 and

wegets= 504/288 whichisnotaninteger.Iff1= 0,thent= 0,d= f2f3,s= f2+ f3.

Ifs≥ 4, 504+ 108d= 288s≥ 1152.It followsthatd≥ 6.If d= 6 we havenecessarily

s= 5 andψ(10)= 0.Sos= 3 andd= 2,butalso inthis caseψ(10)= 0.Weconclude thatfi > 0,∀i.So we areleft with (fi) = (1,1,1),(1,1,2),(1,1,3),(1,2,2).In any of thesecasesoneeasily checksthatψ(10)= 0. 2

Corollary 27. We have l(6;9) = l(7;10) = 4.In particular l(r,a) is known for a≤ 10 and1≤ r ≤ a andConjecture 2holds truefora≤ 10.

Proof. We have seenthatl(6;9),l(7;10) ≤ 4, byLemma 7 we have equality. Then all theothervaluesofl(r;a) aregivenbyTheorem 18,Proposition 12and Proposition 13, ifa≤ 10. 2

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