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Syzygies of adjoint linear series on projective varieties

Purnaprajna Bangere  and  Justin Lacini Department of Mathematics, University of Kansas, 1450 Jayhawk Blvd. , Lawrence, KS 66045, USA [email protected] Department of Mathematics, University of Kansas, 1450 Jayhawk Blvd. , Lawrence, KS 66045, USA [email protected]
Abstract.

Let XX be a smooth complex projective variety of dimension nn and let AA be an ample and basepoint free divisor. We prove KX+mAK_{X}+mA satisfies property NpN_{p} for mn+1+pm\geqslant n+1+p. We also show the graded ring of sections R(X,KX+mA)R(X,K_{X}+mA) is Koszul for mn+2m\geqslant n+2.

1. Introduction

Equations defining algebraic varieties have been a topic of interest to geometers for a long time. In the early eighties Mark Green [Gre84a, Gre84b] brought a new perspective to the subject by viewing classical results on projective normality and normal presentation as particular cases of a more general phenomenon involving minimal free resolutions of the homogeneous coordinate ring. Green and Lazarsfeld proved beautiful results for the case of algebraic curves connecting the geometry of the embedding with the structure of the minimal resolution [Gre84a, GL88, GL86, GL87]. More recently, much progress has been made in this direction (see for example [Voi02, Voi05]).

A result of M. Green has attracted particular interest as it provides a path for generalizing the syzygy results on curves to higher dimensions. Let LL be a line bundle on a curve CC of genus gg with deg(L)2g+1+p\operatorname{deg}(L)\geqslant 2g+1+p. Then LL satisfies property NpN_{p}, which is defined as follows. Let S=SymH0(X,L)S=\operatorname{Sym}^{\bullet}H^{0}(X,L) and consider the graded ring of sections R=R(X,L)=kH0(X,Lk)R=R(X,L)=\oplus_{k}H^{0}(X,L^{\otimes k}) with the natural SS-module structure. Let EE_{\bullet} be a minimal graded free resolution of RR.

Definition 1.1.

The line bundle LL satisfies property NpN_{p} if

  1. (1)

    E0=SE_{0}=S if p0p\geqslant 0.

  2. (2)

    Ei=S(i1)biE_{i}=S(-i-1)^{\oplus b_{i}} for 1ip1\leqslant i\leqslant p.

Green’s result therefore shows that a divisor that is as positive as KC+mAK_{C}+mA satisfies NpN_{p} if AA is ample and mp+3m\geqslant p+3. A few years later Reider [Rei88] proved that KS+mAK_{S}+mA is very ample for any m4m\geqslant 4 if AA is an ample divisor on a smooth algebraic surface SS. Following Reider’s work, Mukai conjectured that KS+mAK_{S}+mA satisfies NpN_{p} for any mp+4m\geqslant p+4 if AA is ample. Mukai’s conjecture is in general open even for p=0p=0. Some work has been done in this direction. A stronger version of the conjecture has been proved for anti-canonical rational surfaces in [GP01], a generic version has been proved for surfaces of general type when AA is ample and basepoint free in [GP99, Pur05] and weaker bounds have been obtained for surfaces with Kodaira dimension zero in [GP99]. For ruled varieties see [But94, Par06, AS91, GP96] among others.

Fujita famously conjectured that KX+mAK_{X}+mA is very ample for any mn+2m\geqslant n+2 if AA is an ample divisor on a projective variety XX of dimension nn. Motivated by these circle of ideas and conjectures, Ein and Lazarsfeld [EL93] proved the following elegant result: if AA is a very ample line bundle on XX then KX+mAK_{X}+mA satisfies NpN_{p} for any mn+1+pm\geqslant n+1+p. It has been an open question ever since (see [EL93, Section 4]) whether the analogous result holds if AA is just ample and basepoint free. The purpose of this article is to give a positive answer to this question:

Theorem 1.2.

Let XX be a smooth complex projective variety of dimension nn and let AA be an ample and basepoint free divisor. Then the line bundle Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) satisfies property NpN_{p} for any mn+1+pm\geqslant n+1+p.

Relaxing the positivity assumption on AA from very ample to ample and basepoint free poses significant challenges and our methods are very different from those of [EL93]. Results concerning N0N_{0} and N1N_{1} were obtained in [MR19] for varieties with KXK_{X} nef with methods similar to [Pur05], and optimal bounds for property NpN_{p} have been proved for abelian varieties [Par00] and Calabi-Yau varieties [GP98, Niu19].

Whenever property N1N_{1} holds, it is a natural question to ask if the homogeneous ring of sections is Koszul. This topic has long been of interest to algebraists and geometers alike. Let RR be a graded \mathbb{C}-algebra, and let EE_{\bullet} be the minimal resolution of \mathbb{C} as an RR-module.

Definition 1.3.

The ring RR is called Koszul if Ei=R(i)biE_{i}=R(-i)^{\oplus b_{i}} for any i1i\geqslant 1.

Pareschi [Par93] adapted the methods of [EL93] to prove that if AA is very ample, then R(X,KX+mA)R(X,K_{X}+mA) is Koszul for any mn+2m\geqslant n+2. It has been an open question if the same holds when AA is just ample and basepoint free. As a corollary of the methods used to prove the main theorem, we obtain the following:

Theorem 1.4.

Let XX be a smooth complex projective variety of dimension nn and let AA be an ample and basepoint free divisor. Then the graded ring of sections R(X,KX+mA)R(X,K_{X}+mA) is Koszul for any mn+2m\geqslant n+2.

Based on Fujita’s conjecture, [EL93] and the present results, we conclude by asking the following natural question.

Question 1.5.

Let XX be a smooth complex projective variety of dimension nn and let AA be an ample divisor. Does the line bundle Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) satisfy property NpN_{p} for all mn+2+pm\geqslant n+2+p?

We point out that at the moment this seems to be completely out of reach, and it is not even known in the special case when n=2n=2 and p=0p=0.

Acknowledgements. We thank Lawrence Ein for his interest, helpful discussions and encouragement.

2. Notation and conventions

We work over the field of complex numbers \mathbb{C}. Of course, everything holds verbatim over any algebraically closed field of characteristic zero. We use Definition 1.1 to define the property NpN_{p} even when LL is just globally generated. A pair (X,Δ)(X,\Delta) is the datum of a normal variety XX and a \mathbb{Q}-Weil divisor Δ\Delta such that KX+ΔK_{X}+\Delta is \mathbb{Q}-Cartier. If Δ0\Delta\geqslant 0, we say that (X,Δ)(X,\Delta) is a log pair. A log resolution of (X,Δ)(X,\Delta) is a projective morphism f:YXf:Y\rightarrow X such that YY is smooth and Ex(f)Supp(f1(Δ))\operatorname{Ex}(f)\cup\operatorname{Supp}(f^{-1}(\Delta)) is a simple normal crossings divisor. A \mathbb{Q}-Cartier divisor DD is nef if DC0D\cdot C\geqslant 0 for any curve CXC\subseteq X. If L1L_{1} and L2L_{2} are two sheaves on XX, we denote by L1L2L_{1}\boxtimes L_{2} the sheaf pr1L1pr2L2\operatorname{pr}_{1}^{*}L_{1}\otimes\operatorname{pr}_{2}^{*}L_{2} on X×XX\times X. We use analogous notation in the case of multiple products and in the case of divisors.

3. Preliminaries

In this section we collect a couple of basic results that we use in the proof of Theorem 1.2 and Theorem 1.4. We start with some elementary commutative algebra.

3.1. Products and diagonals.

Let AA be a local Noetherian \mathbb{C}-algebra of dimension nn. Let A(r)=AAA(r)=A\otimes_{\mathbb{C}}\cdots\otimes_{\mathbb{C}}A be the tensor product of rr copies of AA. For 1jr1\leqslant j\leqslant r, let ij:AA(r)i_{j}:A\rightarrow A(r) be the natural inclusion and let mj=ij(m)A(r)m_{j}=i_{j}(m)\cdot A(r).

Lemma 3.1.

Fix 1jkr1\leqslant j\neq k\leqslant r. In the above notation, we have:

  1. (1)

    ToriA(r)(A(r)/mj,mk)=0\operatorname{Tor}_{i}^{A(r)}(A(r)/m_{j},m_{k})=0 for any i>0i>0.

  2. (2)

    Tor0A(r)(A(r)/mj,mk)=mkA(r)A(r)/mjmkA(r)/mj\operatorname{Tor}_{0}^{A(r)}(A(r)/m_{j},m_{k})=m_{k}\otimes_{A(r)}A(r)/m_{j}\cong m_{k}\cdot A(r)/m_{j}.

  3. (3)

    ToriA(r)(A(r)/mj,A(r)/mk)=0\operatorname{Tor}_{i}^{A(r)}(A(r)/m_{j},A(r)/m_{k})=0 for any i>0i>0.

  4. (4)

    ToriA(r)(mj,mk)=0\operatorname{Tor}_{i}^{A(r)}(m_{j},m_{k})=0 for any i>0i>0.

Proof.

First, notice that

A(r)/mj(A/m)A(r1)A(r1)A(r)/m_{j}\cong(A/m)\otimes_{\mathbb{C}}A(r-1)\cong A(r-1)

Now let FF_{\bullet} be a free resolution of mm as an AA-module. Since ik:AA(r)i_{k}:A\rightarrow A(r) is flat, we have that FAA(r)F_{\bullet}\otimes_{A}A(r) is a free resolution of mAA(r)mkm\otimes_{A}A(r)\cong m_{k} as an A(r)A(r)-module. Therefore, we may compute ToriA(r)(A(r)/mj,mk)\operatorname{Tor}_{i}^{A(r)}(A(r)/m_{j},m_{k}) by taking the tensor product with A(r)/mjA(r)/m_{j}. In light of the isomorphism A(r1)A(r)/mjA(r-1)\cong A(r)/m_{j}, we get

(FAA(r))A(r)A(r)/mjFAA(r1)\left(F_{\bullet}\otimes_{A}A(r)\right)\otimes_{A(r)}A(r)/m_{j}\cong F_{\bullet}\otimes_{A}A(r-1)

Since this is exact, we get (1)(1) and (2)(2). Consider now the short exact sequence

0mjA(r)A(r)/mj00\rightarrow m_{j}\rightarrow A(r)\rightarrow A(r)/m_{j}\rightarrow 0

Taking tensor products with A(r)/mkA(r)/m_{k} and with mkm_{k} gives (3)(3) and (4)(4) respectively. ∎

Corollary 3.2.

In the above notation, we have

mjA(r)mkmjmk=mjmkm_{j}\otimes_{A(r)}m_{k}\cong m_{j}\cdot m_{k}=m_{j}\cap m_{k}
Proof.

Consider once again the short exact sequence

0mjA(r)A(r)/mj00\rightarrow m_{j}\rightarrow A(r)\rightarrow A(r)/m_{j}\rightarrow 0

Then mjmk=mjmkm_{j}\cdot m_{k}=m_{j}\cap m_{k} follows by taking the tensor product with A/mkA/m_{k} and using Lemma 3.1 (3), whereas mjA(r)mkmjmkm_{j}\otimes_{A(r)}m_{k}\cong m_{j}\cdot m_{k} follows by taking the tensor product with mkm_{k} and using Lemma 3.1 (1). ∎

Lemma 3.3.

Fix 1sr1\leqslant s\leqslant r and set mr(s)=m1A(r)A(r)msm_{r}(s)=m_{1}\otimes_{A(r)}\cdots\otimes_{A(r)}m_{s}. Then

  1. (1)

    ToriA(r)(A(r)/mr(s),mr)=0\operatorname{Tor}_{i}^{A(r)}(A(r)/m_{r}(s),m_{r})=0 for all sr1s\leqslant r-1 and i>0i>0.

  2. (2)

    ToriA(r)(mr(s),A(r)/mr)=0\operatorname{Tor}_{i}^{A(r)}(m_{r}(s),A(r)/m_{r})=0 for all sr1s\leqslant r-1 and i>0i>0.

  3. (3)

    ToriA(r)(A(r)/mr(s),A(r)/mr)=0\operatorname{Tor}_{i}^{A(r)}(A(r)/m_{r}(s),A(r)/m_{r})=0 for all sr1s\leqslant r-1 and i>0i>0.

  4. (4)

    ToriA(r)(mr(s),mr)=0\operatorname{Tor}_{i}^{A(r)}(m_{r}(s),m_{r})=0 for all sr1s\leqslant r-1 and i>0i>0.

  5. (5)

    mr(s)m1ms=m1msm_{r}(s)\cong m_{1}\cdots m_{s}=m_{1}\cap\cdots\cap m_{s} for all srs\leqslant r.

Proof.

We prove (1)(5)(1)-(5) by induction on rr. The case r=2r=2 is settled in Lemma 3.1 and Corollary 3.2. Suppose therefore that the statement holds for rr, and let us show it holds for r+1r+1. If sr1s\leqslant r-1, then (1)(4)(1)-(4) follow immediately from the fact that A(r+1)A(r+1) is flat over A(r)A(r). Similarly, (5)(5) follows if srs\leqslant r. Assume now that s=rs=r and consider the following short exact sequence.

0A(r+1)mr+1(r1)mrA(r+1)mr+1(r1)A(r+1)mrA(r+1)(mr+1(r1),mr)00\rightarrow\frac{A(r+1)}{m_{r+1}(r-1)\cap m_{r}}\rightarrow\frac{A(r+1)}{m_{r+1}(r-1)}\oplus\frac{A(r+1)}{m_{r}}\rightarrow\frac{A(r+1)}{(m_{r+1}(r-1),m_{r})}\rightarrow 0

First notice that we have a natural isomorphism:

A(r+1)(mr+1(r1),mr)A(r)mr(r1)\frac{A(r+1)}{(m_{r+1}(r-1),m_{r})}\cong\frac{A(r)}{m_{r}(r-1)}

Furthermore, by (5)(5) we have that

A(r+1)mr+1(r1)mrA(r+1)mr+1(r)\frac{A(r+1)}{m_{r+1}(r-1)\cap m_{r}}\cong\frac{A(r+1)}{m_{r+1}(r)}

Therefore, (1)(1) and (3)(3) follow by taking tensor products in the above short exact sequence with mr+1m_{r+1} and A(r+1)/mr+1A(r+1)/m_{r+1} respectively, and by using the inductive hypothesis.

The rest follow as in the proofs of Lemma 3.1 and Corollary 3.2. ∎

Corollary 3.4.

Let CC_{\bullet} be a free resolution of m=C0m=C_{0} as an AA-module. For any 1jr1\leqslant j\leqslant r, let CjC_{\bullet}^{j} be CAA(r)C_{\bullet}\otimes_{A}A(r) where the module structure is given by the map ij:AA(r)i_{j}:A\rightarrow A(r). Let C(s)=C1A(r)A(r)CsC(s)=C^{1}\otimes_{A(r)}\cdots\otimes_{A(r)}C^{s}, and let TC(s)TC(s) be the associated total complex. Then C(s)C(s) and TC(s)TC(s) are exact.

Proof.

Immediate from Lemma 3.3. ∎

Our next goal is to introduce some notation and globalize Lemma 3.3. Let 2sr2\leqslant s\leqslant r and 1ijr1\leqslant i\neq j\leqslant r be positive integers. Let XX be a projective variety of dimension nn. We denote by

X(r)=X××XX(r)=X\times\cdots\times X

the product of rr copies of XX. We denote by

pri:X(r)X\operatorname{pr}_{i}:X(r)\rightarrow X

and by

pri,j=pri×prj:X(r)X×X\operatorname{pr}_{i,j}=\operatorname{pr}_{i}\times\operatorname{pr}_{j}:X(r)\rightarrow X\times X

the corresponding projections. We denote by

ΔX(r)i,j=pri,jΔX\Delta_{X(r)}^{i,j}=\operatorname{pr}_{i,j}^{*}\Delta_{X}

the diagonal relative to the entries ii and jj. We set

ΔX(r)1(s)=2isΔX(r)1,i\Delta_{X(r)}^{1}(s)=\bigcup_{2\leqslant i\leqslant s}\Delta_{X(r)}^{1,i}

and

ΔX(r)2(s)=2isΔX(r)i1,i\Delta_{X(r)}^{2}(s)=\bigcup_{2\leqslant i\leqslant s}\Delta_{X(r)}^{i-1,i}

Similarly, we set

ΔX(r)1(s)=ΔX(r)1,2ΔX(r)1,3ΔX(r)1,s\mathcal{I}_{\Delta_{X(r)}}^{1}(s)=\mathcal{I}_{\Delta_{X(r)}^{1,2}}\otimes\mathcal{I}_{\Delta_{X(r)}^{1,3}}\otimes\cdots\otimes\mathcal{I}_{\Delta_{X(r)}^{1,s}}

and

ΔX(r)2(s)=ΔX(r)1,2ΔX(r)2,3ΔX(r)s1,s\mathcal{I}_{\Delta_{X(r)}}^{2}(s)=\mathcal{I}_{\Delta_{X(r)}^{1,2}}\otimes\mathcal{I}_{\Delta_{X(r)}^{2,3}}\otimes\cdots\otimes\mathcal{I}_{\Delta_{X(r)}^{s-1,s}}

Of course,

ΔX(r)1(2)=ΔX(r)2(2)=ΔX(r)1,2\mathcal{I}_{\Delta_{X(r)}}^{1}(2)=\mathcal{I}_{\Delta_{X(r)}}^{2}(2)=\mathcal{I}_{\Delta_{X(r)}^{1,2}}
Corollary 3.5.

Assume that r3r\geqslant 3, s<trs<t\leqslant r and i>0i>0. Then:

  1. (1)

    𝒯ori𝒪X(r)(𝒪X(r)/ΔX(r)1(s),ΔX(r)1,t)=0\mathscr{T}\kern-0.5ptor_{i}^{\mathcal{O}_{X(r)}}(\mathcal{O}_{X(r)}/\mathcal{I}_{\Delta_{X(r)}}^{1}(s),\mathcal{I}_{\Delta_{X(r)}^{1,t}})=0.

  2. (2)

    𝒯ori𝒪X(r)(ΔX(r)1(s),𝒪X(r)/ΔX(r)1,t)=0\mathscr{T}\kern-0.5ptor_{i}^{\mathcal{O}_{X(r)}}(\mathcal{I}_{\Delta_{X(r)}}^{1}(s),\mathcal{O}_{X(r)}/\mathcal{I}_{\Delta_{X(r)}^{1,t}})=0.

  3. (3)

    𝒯ori𝒪X(r)(𝒪X(r)/ΔX(r)1(s),𝒪X(r)/ΔX(r)1,t)=0\mathscr{T}\kern-0.5ptor_{i}^{\mathcal{O}_{X(r)}}(\mathcal{O}_{X(r)}/\mathcal{I}_{\Delta_{X(r)}}^{1}(s),\mathcal{O}_{X(r)}/\mathcal{I}_{\Delta_{X(r)}^{1,t}})=0.

  4. (4)

    𝒯ori𝒪X(r)(ΔX(r)1(s),ΔX(r)1,t)=0\mathscr{T}\kern-0.5ptor_{i}^{\mathcal{O}_{X(r)}}(\mathcal{I}_{\Delta_{X(r)}}^{1}(s),\mathcal{I}_{\Delta_{X(r)}^{1,t}})=0.

  5. (5)

    𝒯ori𝒪X(r)(𝒪X(r)/ΔX(r)2(s),ΔX(r)s,s+1)=0\mathscr{T}\kern-0.5ptor_{i}^{\mathcal{O}_{X(r)}}(\mathcal{O}_{X(r)}/\mathcal{I}_{\Delta_{X(r)}}^{2}(s),\mathcal{I}_{\Delta_{X(r)}^{s,s+1}})=0.

  6. (6)

    𝒯ori𝒪X(r)(ΔX(r)2(s),𝒪X(r)/ΔX(r)s,s+1)=0\mathscr{T}\kern-0.5ptor_{i}^{\mathcal{O}_{X(r)}}(\mathcal{I}_{\Delta_{X(r)}}^{2}(s),\mathcal{O}_{X(r)}/\mathcal{I}_{\Delta_{X(r)}^{s,s+1}})=0.

  7. (7)

    𝒯ori𝒪X(r)(𝒪X(r)/ΔX(r)2(s),𝒪X(r)/ΔX(r)s,s+1)=0\mathscr{T}\kern-0.5ptor_{i}^{\mathcal{O}_{X(r)}}(\mathcal{O}_{X(r)}/\mathcal{I}_{\Delta_{X(r)}}^{2}(s),\mathcal{O}_{X(r)}/\mathcal{I}_{\Delta_{X(r)}^{s,s+1}})=0.

  8. (8)

    𝒯ori𝒪X(r)(ΔX(r)2(s),ΔX(r)s,s+1)=0\mathscr{T}\kern-0.5ptor_{i}^{\mathcal{O}_{X(r)}}(\mathcal{I}_{\Delta_{X(r)}}^{2}(s),\mathcal{I}_{\Delta_{X(r)}^{s,s+1}})=0.

Furthermore, we have:

ΔX(r)1(s)ΔX(r)1,2ΔX(r)1,s=ΔX(r)1,sΔX(r)1,s=ΔX(r)1(s)\mathcal{I}_{\Delta_{X(r)}}^{1}(s)\cong\mathcal{I}_{\Delta_{X(r)}^{1,2}}\cdots\mathcal{I}_{\Delta_{X(r)}^{1,s}}=\mathcal{I}_{\Delta_{X(r)}^{1,s}}\cap\cdots\cap\mathcal{I}_{\Delta_{X(r)}^{1,s}}=\mathcal{I}_{\Delta_{X(r)}^{1}(s)}

and

ΔX(r)2(s)ΔX(r)1,2ΔX(r)s1,s=ΔX(r)1,2ΔX(r)s1,s=ΔX(r)2(s)\mathcal{I}_{\Delta_{X(r)}}^{2}(s)\cong\mathcal{I}_{\Delta_{X(r)}^{1,2}}\cdots\mathcal{I}_{\Delta_{X(r)}^{s-1,s}}=\mathcal{I}_{\Delta_{X(r)}^{1,2}}\cap\cdots\cap\mathcal{I}_{\Delta_{X(r)}^{s-1,s}}=\mathcal{I}_{\Delta_{X(r)}^{2}(s)}
Proof.

Immediate from Lemma 3.3 after a change of coordinates. ∎

3.2. Koszul cohomology and property NpN_{p}.

We briefly recall here some of the basic definitions and properties of Koszul cohomology groups. We refer to Green’s original papers [Gre84a] and [Gre84b] and to Lazarsfeld’s expository notes [Laz89] for a more detailed treatement. Let VV be a finite dimensional vector space, let S(V)S(V) be the symmetric algebra on VV and let B=qBqB=\oplus_{q\in\mathbb{Z}}B_{q} be a graded S(V)S(V)-module. Then there is a natural Koszul complex

p+1VBq1dp+1,q1pVBqdp,qp1VBq+1\cdots\rightarrow\wedge^{p+1}V\otimes B_{q-1}\xrightarrow{d_{p+1,q-1}}\wedge^{p}V\otimes B_{q}\xrightarrow{d_{p,q}}\wedge^{p-1}V\otimes B_{q+1}\rightarrow\cdots

We define the Koszul cohomology groups of BB as

𝒦p,q(B,V)=Ker(dp,q)Im(dp+1,q1)\mathcal{K}_{p,q}(B,V)=\frac{\operatorname{Ker}(d_{p,q})}{\operatorname{Im}(d_{p+1,q-1})}

Now let XX be a smooth projective variety, LL a line bundle, VH0(X,L)V\subseteq H^{0}(X,L) a vector space and \mathcal{F} a sheaf. Then we set

𝒦p,q(X,,L,V)=𝒦p,q(B,V)\mathcal{K}_{p,q}(X,\mathcal{F},L,V)=\mathcal{K}_{p,q}(B,V)

where B=qH0(X,Lq)B=\oplus_{q\in\mathbb{Z}}H^{0}(X,\mathcal{F}\otimes L^{\otimes q}). It is common to simply write 𝒦p,q(X,,L)\mathcal{K}_{p,q}(X,\mathcal{F},L) if V=H0(X,L)V=H^{0}(X,L), and to write 𝒦p,q(X,L)\mathcal{K}_{p,q}(X,L) if furthermore \mathcal{F} is the structure sheaf. Notice that Definition 1.1 may be rephrased as

ToriS(V)(R,)d=0\operatorname{Tor}_{i}^{S(V)}(R,\mathbb{C})_{d}=0

for any 0ip0\leqslant i\leqslant p and di+2d\geqslant i+2. By using the standard Koszul resolution of \mathbb{C} as an S(V)S(V)-module and the fact that ToriS(V)\operatorname{Tor}_{i}^{S(V)} is a balanced functor, we see that a globally generated line bundle LL satisfies property NpN_{p} if and only if

𝒦r,q(X,L)=0\mathcal{K}_{r,q}(X,L)=0

for all 0rp0\leqslant r\leqslant p and q2q\geqslant 2. With this in mind, in the sequel we will show property NpN_{p} via the following theorem.

Theorem 3.6.

Let XX be a smooth projective variety, let LL be a line bundle and let EE be a vector bundle. Then

𝒦p,q(X,E,L)=0\mathcal{K}_{p,q}(X,E,L)=0

for all q2q\geqslant 2 provided that

H1(X(r),ΔX(r)1(r)(ELk)LL)=0H^{1}(X(r),\mathcal{I}_{\Delta_{X(r)}}^{1}(r)\otimes(E\otimes L^{\otimes k})\boxtimes L\boxtimes\cdots\boxtimes L)=0

for all 2rp+22\leqslant r\leqslant p+2 and k1k\geqslant 1.

Proof.

Follow the proof of [Gre84b, Theorem 3.2] and use Corollary 3.5 when needed. See also [Nor93]. ∎

Remark 3.7.

The proof given by Green in [Gre84b] contains a subtle mistake, in that the diagonal sheaves ΔX2(r)\mathcal{I}_{\Delta_{X}}^{2}(r) are used, rather than the correct ΔX1(r)\mathcal{I}_{\Delta_{X}}^{1}(r). Unfortunately, this mistake seems to have gone unnoticed in several papers; for example it appears in [Ina97] as well.

3.3. Koszul rings.

Let RR be a positively graded \mathbb{C}-algebra. Let

Pn+1ϕn+1PnϕnP1ϕ1P00\cdots\rightarrow P_{n+1}\xrightarrow{\phi_{n+1}}P_{n}\xrightarrow{\phi_{n}}\cdots\rightarrow P_{1}\xrightarrow{\phi_{1}}P_{0}\rightarrow\mathbb{C}\rightarrow 0

be a minimal graded free resolution of \mathbb{C}. Notice that this resolution is finite if and only if RR is a polynomial ring. The ring RR is called Koszul if all the entries in the matrix representing ϕn\phi_{n} have degree one. This is equivalent to requiring that ToriR(,)d=0\operatorname{Tor}_{i}^{R}(\mathbb{C},\mathbb{C})_{d}=0 for all i>0i>0 and did\neq i. We will use the following criterion.

Theorem 3.8.

Let XX be a projective variety and let LL be a line bundle. The graded ring of sections R(X,L)R(X,L) is Koszul if

H1(X(r),ΔX(r)2(r)LLLk)=0H^{1}(X(r),\mathcal{I}_{\Delta_{X(r)}}^{2}(r)\otimes L\boxtimes L\boxtimes\cdots\boxtimes L^{\otimes k})=0

and

H1(X(r),ΔX(r)r1,rLLLk)=0H^{1}(X(r),\mathcal{I}_{\Delta_{X(r)}^{r-1,r}}\otimes L\boxtimes L\boxtimes\cdots\boxtimes L^{\otimes k})=0

for all r2r\geqslant 2 and k1k\geqslant 1.

Proof.

This is Proposition 1.9 in [IM94]. ∎

3.4. Resolutions of the diagonal.

Let M=Ωn(1)M=\Omega_{\mathbb{P}^{n}}(1). Beilinson’s resolution of the diagonal Δn\Delta_{\mathbb{P}^{n}} of n×n\mathbb{P}^{n}\times\mathbb{P}^{n} is the exact sequence:

(1) 0n(𝒪n(1)M)𝒪n(1)MΔn00\rightarrow\wedge^{n}(\mathcal{O}_{\mathbb{P}^{n}}(-1)\boxtimes M)\rightarrow\cdots\rightarrow\mathcal{O}_{\mathbb{P}^{n}}(-1)\boxtimes M\rightarrow\mathcal{I}_{\Delta_{\mathbb{P}^{n}}}\rightarrow 0

This resolution will play a crucial role in the proof of Theorem 1.2.

3.5. Duality for finite morphisms.

We recall here a well-known duality statement. Let f:YXf:Y\rightarrow X be a finite surjective morphism of normal projective schemes. Let ωX\omega_{X} be a dualizing sheaf for XX and assume that ωX\omega_{X} is invertible. By [Har77, Chapter III, Exercise 7.2 (a)], we have that ωY=f!ωX=omX(f𝒪Y,ωX)\omega_{Y}=f^{!}\omega_{X}=\mathscr{H}\kern-0.5ptom_{X}(f_{*}\mathcal{O}_{Y},\omega_{X}) is a dualizing sheaf for YY (see also [Har77, Chapter III, Exercise 6.10]). Therefore we get

fωYomX(f𝒪Y,ωX)f_{*}\omega_{Y}\cong\mathscr{H}\kern-0.5ptom_{X}(f_{*}\mathcal{O}_{Y},\omega_{X})

so that

fωY/X(f𝒪Y)f_{*}\omega_{Y/X}\cong(f_{*}\mathcal{O}_{Y})^{*}

Since XX and YY are normal, we have that f𝒪Yf_{*}\mathcal{O}_{Y} is split by the trace morphism. In particular, the structure sheaf 𝒪X\mathcal{O}_{X} is canonically a direct summand of fωY/Xf_{*}\omega_{Y/X}.

4. The very ample case

In this section we present a first simple approach to the problem in the special case where AA is very ample. Our hope is that this preliminary case may serve as a motivating example, and highlight some of the main difficulties of the problem, but we do not aim here for optimal bounds. Since the rest of the discussion does not rely on the methods of this section, the reader who wishes to do so may directly skip to Section 5.

Recall that our aim is to show property NpN_{p} and property Koszul via Theorem 3.6 and Theorem 3.8 respectively. A first natural approach is then to try using well known generalizations of Kodaira’s vanishing theorem (i.e. Kawamata-Viehweg and Nadel vanishing theorems). Let us briefly recall the definition of multiplier ideals.

Definition 4.1.

Let (X,Δ)(X,\Delta) be a log pair with XX smooth, and let μ:YX\mu:Y\rightarrow X be a log resolution. We define the multiplier ideal sheaf of Δ\Delta to be

(X,Δ)=μ𝒪Y(KY/XμΔ)𝒪X\mathcal{I}(X,\Delta)=\mu_{*}\mathcal{O}_{Y}(K_{Y/X}-\lfloor\mu^{*}\Delta\rfloor)\subseteq\mathcal{O}_{X}
Theorem 4.2 (Nadel’s vanishing theorem).

Let XX be a smooth complex projective variety and Δ0\Delta\geqslant 0 a \mathbb{Q}-divisor on XX. Let LL be any integral divisor such that LΔL-\Delta is big and nef. Then

Hi(X,𝒪X(KX+L)(X,Δ))=0H^{i}(X,\mathcal{O}_{X}(K_{X}+L)\otimes\mathcal{I}(X,\Delta))=0

for i>0i>0.

The above vanishing closely resembles the one needed in Theorem 3.6 and 3.8. In fact, one only has to arrange for the given diagonal ideals to be multiplier ideals of appropriate divisors. To this end, let M1,,Mn+1M_{1},\cdots,M_{n+1} be general divisors in

2H0(X,A)H0(X,A)H0(X,A)H0(X×X,AA)\wedge^{2}H^{0}(X,A)\subseteq H^{0}(X,A)\otimes H^{0}(X,A)\cong H^{0}(X\times X,A\boxtimes A)

and let M=(M1++Mn+1)/(n+1)M=(M_{1}+\cdots+M_{n+1})/(n+1). An easy check shows that

(X×X,nM)=ΔX\mathcal{I}(X\times X,n\cdot M)=\mathcal{I}_{\Delta_{X}}

Furthermore, if we take

M(r)=j=2rpr1,jMM(r)=\sum_{j=2}^{r}\operatorname{pr}_{1,j}^{*}M

we see that

(X(r),nM(r))=ΔX(r)1(r)\mathcal{I}(X(r),n\cdot M(r))=\mathcal{I}_{\Delta_{X(r)}}^{1}(r)

By Nadel’s vanishing theorem and by Theorem 3.6, we get:

Theorem 4.3.

Let XX be a smooth projective variety of dimension nn, let AA be a very ample divisor and let BB be a nef divisor. Set Lm=𝒪X(KX+mA+B)L_{m}=\mathcal{O}_{X}(K_{X}+mA+B) for some m0m\geqslant 0. Then LmL_{m} satisfies property NpN_{p} if mnp+n+1m\geqslant np+n+1.

By taking

M(r)=j=2rprj1,jMM(r)=\sum_{j=2}^{r}\operatorname{pr}_{j-1,j}^{*}M

a similar discussion shows:

Theorem 4.4.

Let XX be a smooth projective variety of dimension nn, let AA be a very ample divisor and let BB be a nef divisor. Set Lm=𝒪X(KX+mA+B)L_{m}=\mathcal{O}_{X}(K_{X}+mA+B) for some m0m\geqslant 0. Then R(X,Lm)R(X,L_{m}) is Koszul if m2n+1m\geqslant 2n+1.

Theorem 4.3 and Theorem 4.4 recover the main results of [EL93] and [Par93] respectively, with weaker bounds. We will show below how to extend these results to the case in which AA is only ample and basepoint free and how to strengthen them to optimal bounds. For the moment, let us just remark that the above method no longer works. A first problem is that we can no longer hope to cut down to the diagonal ΔX\Delta_{X} by using sections in 2H0(X,A)\wedge^{2}H^{0}(X,A). If we try to do this, we instead get an \sayenlarged diagonal, say ΓX\Gamma_{X}. A more serious problem is that the associated multiplier ideals give rise to non-reduced schemes. In fact, by Skoda’s theorem [Laz04, Theorem 9.6.21], we have that

(X×X,nM)=ΓX(X×X,(n1)M)\mathcal{I}(X\times X,n\cdot M)=\mathcal{I}_{\Gamma_{X}}\cdot\mathcal{I}(X\times X,(n-1)\cdot M)

In general, however,

(X×X,(n1)M)𝒪X×X\mathcal{I}(X\times X,(n-1)\cdot M)\neq\mathcal{O}_{X\times X}

and therefore it is not clear how to relate (X×X,nM)\mathcal{I}(X\times X,n\cdot M) with ΔX\mathcal{I}_{\Delta_{X}}. We show below how to circumvent both problems via certain resolutions of the diagonal and duality theory.

5. Proof

We are now ready to start the proof of Theorem 1.2. For the reader’s convenience, we divide the proof in several smaller steps.

5.1. The setup.

Let XX be a smooth projective variety of dimension nn and let AA be an ample and basepoint free divisor. Let VH0(X,A)V\subseteq H^{0}(X,A) be a basepoint free subspace of dimension n+1n+1. Let α:Xn\alpha:X\rightarrow\mathbb{P}^{n} be the corresponding finite flat morphism. Let β:YX\beta:Y\rightarrow X be the Galois closure of α\alpha and let δ=αβ\delta=\alpha\circ\beta. Let G=Gal(Y/n)G=\operatorname{Gal}(Y/\mathbb{P}^{n}) and H=Gal(Y/X)H=\operatorname{Gal}(Y/X). As usual, we denote by ΔXX×X\Delta_{X}\subseteq X\times X and ΔYY×Y\Delta_{Y}\subseteq Y\times Y the respective diagonals. Let ΓY\Gamma_{Y} and ΓX\Gamma_{X} be the schemes defined by Δn𝒪Y×Y\mathcal{I}_{\Delta_{\mathbb{P}^{n}}}\cdot\mathcal{O}_{Y\times Y} and Δn𝒪X×X\mathcal{I}_{\Delta_{\mathbb{P}^{n}}}\cdot\mathcal{O}_{X\times X} respectively. We have ΓY=Y×nY\Gamma_{Y}=Y\times_{\mathbb{P}^{n}}Y and ΓX=X×nX\Gamma_{X}=X\times_{\mathbb{P}^{n}}X. We set ΓY(r)j(s)=Δ(r)j(s)𝒪Y(r)\mathcal{I}_{\Gamma_{Y(r)}}^{j}(s)=\mathcal{I}_{\Delta_{\mathbb{P}(r)}}^{j}(s)\cdot\mathcal{O}_{Y(r)} and ΓX(r)j(s)=Δ(r)j(s)𝒪X(r)\mathcal{I}_{\Gamma_{X(r)}}^{j}(s)=\mathcal{I}_{\Delta_{\mathbb{P}(r)}}^{j}(s)\cdot\mathcal{O}_{X(r)} for j=1,2j=1,2, and we denote by ΓY(r)j(s)\Gamma_{Y(r)}^{j}(s) the corresponding schemes. Finally, we denote by ΔY(r)j(s)\Delta_{Y(r)}^{j}(s) and ΔX(r)j(s)\Delta_{X(r)}^{j}(s) the schemes defined by ΔY(r)j(s)\mathcal{I}_{\Delta_{Y(r)}}^{j}(s) and ΔX(r)j(s)\mathcal{I}_{\Delta_{X(r)}}^{j}(s) for j=0,1j=0,1.

5.2. Enlarged diagonals.

Here we use Beilinson’s resolution (1)(1) to get resolutions of the sheaves Δn(r)j(s)\mathcal{I}_{\Delta_{\mathbb{P}^{n}(r)}}^{j}(s). Although these \sayresolutions do not consist of locally free sheaves, they will prove to be equally useful for our purposes. We start our study by pulling back Beilinson’s resolution (1)(1) to n(r)\mathbb{P}^{n}(r) via prj1,j2\operatorname{pr}_{j_{1},j_{2}} for 1j1<j2r1\leqslant j_{1}<j_{2}\leqslant r to get the following exact sequences

0prj1,j2n(𝒪n(1)M)\displaystyle 0\rightarrow\operatorname{pr}_{j_{1},j_{2}}^{*}\wedge^{n}(\mathcal{O}_{\mathbb{P}^{n}}(-1)\boxtimes M)\rightarrow\cdots prj1,j2𝒪n(1)M\displaystyle\rightarrow\operatorname{pr}_{j_{1},j_{2}}^{*}\mathcal{O}_{\mathbb{P}^{n}}(-1)\boxtimes M\rightarrow
(j1.j2j_{1}.j_{2}) Δnj1,j20\displaystyle\rightarrow\mathcal{I}_{\Delta_{\mathbb{P}^{n}}^{j_{1},j_{2}}}\rightarrow 0

We consider each sequence (j1.j2)(j_{1}.j_{2}) as an exact complex positively graded and with the ideal sheaf in degree zero, which we call C(j1,j2,n,r)C(j_{1},j_{2},n,r). Let

C1(s,n,r)=C(1,2,n,r)C(1,3,n,r)C(1,s,n,r)C^{1}(s,n,r)=C(1,2,n,r)\otimes C(1,3,n,r)\otimes\cdots\otimes C(1,s,n,r)

and

C2(s,n,r)=C(1,2,n,r)C(2,3,n,r)C(s1,s,n,r)C^{2}(s,n,r)=C(1,2,n,r)\otimes C(2,3,n,r)\otimes\cdots\otimes C(s-1,s,n,r)

Let TC1(s,n,r)TC^{1}(s,n,r) and TC2(s,n,r)TC^{2}(s,n,r) be the corresponding total complexes. An application of Corollary 3.4 gives:

Lemma 5.1.

Let 0<sr0<s\leqslant r and nn be positive integers. Then C1(s,n,r)C^{1}(s,n,r), C2(s,n,r)C^{2}(s,n,r), TC1(s,n,r)TC^{1}(s,n,r) and TC2(s,n,r)TC^{2}(s,n,r) are exact.

Write MV=αMM_{V}=\alpha^{*}M. Pulling back Beilinson’s resolution (1)(1) to XX via α\alpha gives a resolution of ΓX\Gamma_{X}:

0n(𝒪X(A)MV)\displaystyle 0\rightarrow\wedge^{n}(\mathcal{O}_{X}(-A)\boxtimes M_{V})\rightarrow\cdots 𝒪X(A)MV\displaystyle\rightarrow\mathcal{O}_{X}(-A)\boxtimes M_{V}\rightarrow
(2) ΓX0\displaystyle\rightarrow\mathcal{I}_{\Gamma_{X}}\rightarrow 0

Similarly, pulling back the complexes C(j1,j2,n,r)C(j_{1},j_{2},n,r), C1(s,n,r)C^{1}(s,n,r) and C2(s,n,r)C^{2}(s,n,r) to X(r)X(r) gives exact complexes, which we denote by C(X,j1,j2,r)C(X,j_{1},j_{2},r), C1(X,s,r)C^{1}(X,s,r) and C2(X,s,r)C^{2}(X,s,r) respectively.

5.3. Vanishing.

Here we prove the crucial vanishing results in view of Theorem 3.6.

Lemma 5.2.

Let Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) for some mn+1m\geqslant n+1. Then

Hi(X,Lmk𝒪X(lA))=0H^{i}(X,L_{m}^{\otimes k}\otimes\mathcal{O}_{X}(lA))=0

for any k1k\geqslant 1, m+l1m+l\geqslant 1, and 1in1\leqslant i\leqslant n.

Proof.

Immediate application of Kodaira’s vanishing theorem. ∎

Lemma 5.3.

Let Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) for some mn+1m\geqslant n+1. Then

Hi(X,jMVLmk𝒪X(lA))=0H^{i}(X,\wedge^{j}M_{V}\otimes L_{m}^{\otimes k}\otimes\mathcal{O}_{X}(lA))=0

if k1k\geqslant 1 and either

  1. (1)

    i>ji>j and m+l1m+l\geqslant 1, or

  2. (2)

    i1i\geqslant 1 and m+ln+2jm+l\geqslant n+2-j.

Proof.

For (1)(1), we proceed by induction on jj, starting with the case j=1j=1. Consider the short exact sequence

0MVV𝒪X𝒪X(A)00\rightarrow M_{V}\rightarrow V\otimes_{\mathbb{C}}\mathcal{O}_{X}\rightarrow\mathcal{O}_{X}(A)\rightarrow 0

Then the result follows immediately from Lemma 5.2. Suppose therefore that the result holds for j1j-1 and let us prove it for jj. Consider the short exact sequence

(*) 0jMVjV𝒪Xj1MV𝒪X(A)00\rightarrow\wedge^{j}M_{V}\rightarrow\wedge^{j}V\otimes_{\mathbb{C}}\mathcal{O}_{X}\rightarrow\wedge^{j-1}M_{V}\otimes\mathcal{O}_{X}(A)\rightarrow 0

The statement follows then immediately by the inductive hypothesis. Now we prove (2)(2) by descending induction on jj, starting with the case j=nj=n. In this case nMV𝒪X(A)\wedge^{n}M_{V}\cong\mathcal{O}_{X}(-A) and the statement follows from Kodaira’s vanishing theorem. Suppose then that the (2)(2) holds for jj, and let us prove it for j1j-1. Consider again the short exact sequence ()(*). Then statement follows then by Lemma 5.2 and the inductive hypothesis. ∎

Lemma 5.4.

Fix r2r\geqslant 2. Let Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) for some mn+r1m\geqslant n+r-1. Let it0i_{t}\geqslant 0 for 1tr1\leqslant t\leqslant r and at1a_{t}\geqslant 1 for 1tr11\leqslant t\leqslant r-1 be two sequences of positive integers. Let t=1rit=i\sum_{t=1}^{r}i_{t}=i and t=1r1at=d\sum_{t=1}^{r-1}a_{t}=d. Assume that idr+2i\geqslant d-r+2. Then

Hi1(X,𝒪X(dA)Lmk)t=2rHit(X,at1MVLm)=0H^{i_{1}}(X,\mathcal{O}_{X}(-dA)\otimes L_{m}^{\otimes k})\otimes\bigotimes_{t=2}^{r}H^{i_{t}}(X,\wedge^{a_{t-1}}M_{V}\otimes L_{m})=0

for any k1k\geqslant 1.

Proof.

If it>0i_{t}>0 for any t2t\geqslant 2, then we are done by Lemma 5.3 (2)(2). Therefore we may assume that i1=ii_{1}=i. Since dr1d\geqslant r-1, we have that i11i_{1}\geqslant 1. If i1>ni_{1}>n, then there is nothing to prove, so we may assume that i1ni_{1}\leqslant n. Therefore,

dn+r2d\leqslant n+r-2

and we may conclude by Lemma 5.2. ∎

Lemma 5.5.

Fix r1r\geqslant 1. Let Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) for some mn+2m\geqslant n+2. Let it1i_{t}\geqslant 1 for 1tr1\leqslant t\leqslant r and 0atn0\leqslant a_{t}\leqslant n for 0tr0\leqslant t\leqslant r be two sequences of positive integers. Let t=1rit=i\sum_{t=1}^{r}i_{t}=i and t=0rat=d\sum_{t=0}^{r}a_{t}=d. Assume that idri\geqslant d-r. Then

t=1rHit(X,at1MVLmkt𝒪X(atA))=0\bigotimes_{t=1}^{r}H^{i_{t}}(X,\wedge^{a_{t-1}}M_{V}\otimes L_{m}^{\otimes k_{t}}\otimes\mathcal{O}_{X}(-a_{t}A))=0

for any kt1k_{t}\geqslant 1.

Proof.

If it>at1i_{t}>a_{t-1} for any 1tr1\leqslant t\leqslant r, then we are done by Lemma 5.3 (1)(1). Therefore, we may assume that itat1i_{t}\leqslant a_{t-1} for all 1tr1\leqslant t\leqslant r. Similarly, if atat1a_{t}\leqslant a_{t-1} for any 1tr1\leqslant t\leqslant r, then we are done by Lemma 5.3 (2)(2). Therefore, we may assume that at>at1a_{t}>a_{t-1} for all 1tr1\leqslant t\leqslant r. In particular, arr+1a_{r}\geqslant r+1. Putting everything together, we have

t=1rat1t=1ritt=0ratr\sum_{t=1}^{r}a_{t-1}\geqslant\sum_{t=1}^{r}i_{t}\geqslant\sum_{t=0}^{r}a_{t}-r

Therefore, we get arra_{r}\leqslant r, contradiction. ∎

Lemma 5.6.

Fix r2r\geqslant 2. Let Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) for some mn+2m\geqslant n+2. Let it0i_{t}\geqslant 0 for 1tr1\leqslant t\leqslant r and at1a_{t}\geqslant 1 for 1tr11\leqslant t\leqslant r-1 be two sequences of positive integers. Choose two integers 0a0n0\leqslant a_{0}\leqslant n and 0arn0\leqslant a_{r}\leqslant n. Let t=1rit=i\sum_{t=1}^{r}i_{t}=i and t=0rat=d\sum_{t=0}^{r}a_{t}=d. Let zz be the cardinality of the set {t{0,r}|at>0}\{t\in\{0,r\}|a_{t}>0\}. Assume that idrz+2i\geqslant d-r-z+2. Then

t=1rHit(X,at1MVLmkt𝒪X(atA))=0\bigotimes_{t=1}^{r}H^{i_{t}}(X,\wedge^{a_{t-1}}M_{V}\otimes L_{m}^{\otimes k_{t}}\otimes\mathcal{O}_{X}(-a_{t}A))=0

for any kt1k_{t}\geqslant 1.

Proof.

We may assume that atna_{t}\leqslant n for all 0tn0\leqslant t\leqslant n. Let ss be the cardinality of the set {1tr|it=0}\{1\leqslant t\leqslant r|i_{t}=0\}. We proceed by induction on ss. The case s=0s=0 is settled in Lemma 5.5. Assume therefore that the statement holds for ss and let us show it for s+1s+1. Let t0=min{t|it=0}t_{0}=\operatorname{min}\{t|i_{t}=0\}. If t0=1t_{0}=1 or t0=rt_{0}=r, then the statement follows by the inductive hypothesis after discarding the index t0t_{0}. Assume therefore that 1<t0<r1<t_{0}<r. Let j=t=1t01itj=\sum_{t=1}^{t_{0}-1}i_{t} and e=t=0t01ate=\sum_{t=0}^{t_{0}-1}a_{t}. If jet0+1j\geqslant e-t_{0}+1, then we are done by Lemma 5.5. Therefore we may assume that ij(de)(rt0)z+2i-j\geqslant(d-e)-(r-t_{0})-z+2. Since zz can only increase after cuts, we are done by inductive hypothesis. ∎

Lemma 5.7.

Fix r2r\geqslant 2. Let Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) for some m0m\geqslant 0. Let at0a_{t}\geqslant 0 for 1tr11\leqslant t\leqslant r-1 be a sequence of integers and set t=1r1at=d\sum_{t=1}^{r-1}a_{t}=d. Let ss be the cardinality of the set {t|at=0}\{t|a_{t}=0\}. Assume that id+sr+2i\geqslant d+s-r+2. If mn+2m\geqslant n+2 and k1k\geqslant 1, then

Hi(X(r),t=1r1C(X,t,t+1,r)atLmLmLmk)=0H^{i}\left(X(r),\bigotimes_{t=1}^{r-1}C(X,t,t+1,r)_{a_{t}}\otimes L_{m}\boxtimes L_{m}\cdots\boxtimes L_{m}^{\otimes k}\right)=0

If mn+r1m\geqslant n+r-1 and k1k\geqslant 1 instead, then

Hi(X(r),t=1r1C(X,1,t+1,r)atLmkLmLm)=0H^{i}\left(X(r),\bigotimes_{t=1}^{r-1}C(X,1,t+1,r)_{a_{t}}\otimes L_{m}^{\otimes k}\boxtimes L_{m}\cdots\boxtimes L_{m}\right)=0
Proof.

We proceed by induction on ss. The case s=0s=0 follows immediately from Lemma 5.4, Lemma 5.6 and Künneth’s formula. Suppose therefore that the statement holds for ss, and let us show it for s+1s+1. Let t0t_{0} be an index such that at0=0a_{t_{0}}=0. By Lemma 5.1, it is then enough to show that

Hi+j(X(r),C(X,t0,t0+1,r)jtt0C(X,t,t+1,r)atLmLmk)=0H^{i+j^{\prime}}\left(X(r),C(X,t_{0},t_{0}+1,r)_{j}\otimes\bigotimes_{t\neq t_{0}}C(X,t,t+1,r)_{a_{t}}\otimes L_{m}\boxtimes\cdots\boxtimes L_{m}^{\otimes k}\right)=0

and

Hi+j(X(r),C(X,1,t0+1,r)jtt0C(X,1,t+1,r)atLmkLm)=0H^{i+j^{\prime}}\left(X(r),C(X,1,t_{0}+1,r)_{j}\otimes\bigotimes_{t\neq t_{0}}C(X,1,t+1,r)_{a_{t}}\otimes L_{m}^{\otimes k}\boxtimes\cdots\boxtimes L_{m}\right)=0

for any j1j\geqslant 1 and jj1j^{\prime}\geqslant j-1. This holds by inductive hypothesis, and we are done. ∎

Theorem 5.8.

Fix r2r\geqslant 2. Let Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) for some m0m\geqslant 0. If mn+2m\geqslant n+2, then

H1(X(r),ΓX(r)2(r)LmLmLmk)=0H^{1}(X(r),\mathcal{I}_{\Gamma_{X(r)}}^{2}(r)\otimes L_{m}\boxtimes L_{m}\boxtimes\cdots\boxtimes L_{m}^{\otimes k})=0

for any k1k\geqslant 1. If mn+r1m\geqslant n+r-1 instead, then

H1(X(r),ΓX(r)1(r)LmkLmLm)=0H^{1}(X(r),\mathcal{I}_{\Gamma_{X(r)}}^{1}(r)\otimes L_{m}^{\otimes k}\boxtimes L_{m}\boxtimes\cdots\boxtimes L_{m})=0

for any k1k\geqslant 1.

Proof.

Immediate consequence of Lemma 5.7 applied to the complexes C1(X,r,r)C^{1}(X,r,r) and C2(X,r,r)C^{2}(X,r,r). ∎

5.4. Projective normality.

Here we prove:

Theorem 5.9.

Let XX be a smooth complex projective variety of dimension nn and let AA be an ample and basepoint free divisor. Then the line bundle Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) satisfies property N0N_{0} for any mn+1m\geqslant n+1.

We fix notation at in Subsection 5.1. First, notice that ΓX\Gamma_{X} is reduced since α\alpha is flat, n\mathbb{P}^{n} is reduced and ΓX\Gamma_{X} is generically reduced. Furthermore, ΓX\Gamma_{X} is Cohen-Macaulay since α\alpha is flat and n\mathbb{P}^{n} is smooth. Let ωY\omega_{Y} be the dualizing sheaf of YY. Let Y0Y_{0} be the smooth locus of YY. Notice that YY0Y\setminus Y_{0} has codimension at least two since YY is normal; in particular ωYiωY0\omega_{Y}\cong i_{*}\omega_{Y_{0}}. Up to shrinking Y0Y_{0} while keeping the codimension at least two, we may assume that X0=β(Y0)X_{0}=\beta(Y_{0}) is open and β:Y0X0\beta:Y_{0}\rightarrow X_{0} is finite and flat. Let ΓY0=Y0×nY0\Gamma_{Y_{0}}=Y_{0}\times_{\mathbb{P}_{n}}Y_{0}. Consider

F=(ωYβ𝒪X(mA))(ωYβ𝒪X(mA))F=(\omega_{Y}\otimes\beta^{*}\mathcal{O}_{X}(mA))\boxtimes(\omega_{Y}\otimes\beta^{*}\mathcal{O}_{X}(mA))

There is an injective map

F|ΓY0gGF|gΔY0F|_{\Gamma_{Y_{0}}}\rightarrow\bigoplus_{g\in G}F|_{g^{*}\Delta_{Y_{0}}}

where we let GG act on the second entry. By taking invariants we get an isomorphism

(F|ΓY0)G(gGF|gΔY0)G\left(F|_{\Gamma_{Y_{0}}}\right)^{G}\cong\left(\bigoplus_{g\in G}F|_{g^{*}\Delta_{Y_{0}}}\right)^{G}

Therefore we have a surjective map

H0(ΓY0,F|ΓY0)H0(ΔY0,F|ΔY0)H^{0}(\Gamma_{Y_{0}},F|_{\Gamma_{Y_{0}}})\rightarrow H^{0}(\Delta_{Y_{0}},F|_{\Delta_{Y_{0}}})

Equivalently, this also follows by noticing that the ramification formula applied to the morphism Y0δ(Y0)Y_{0}\rightarrow\delta(Y_{0}) gives rise to a canonical isomorphism

pr1(ωYβ𝒪X(mA)|ΓY0pr2(ωYβ𝒪X(mA)|ΓY0\operatorname{pr}_{1}^{*}(\omega_{Y}\otimes\beta^{*}\mathcal{O}_{X}(mA)|_{\Gamma_{Y_{0}}}\cong\operatorname{pr}_{2}^{*}(\omega_{Y}\otimes\beta^{*}\mathcal{O}_{X}(mA)|_{\Gamma_{Y_{0}}}

The Galois case. If XX is Galois over n\mathbb{P}^{n}, then Y=XY=X. Therefore Lemma 5.10 below follows immediately.

For the general case, consider the trace map

Tr:(β×β)FLmLm\operatorname{Tr}:(\beta\times\beta)_{*}F\rightarrow L_{m}\boxtimes L_{m}

We have the following diagram, which commutes up to multiplication by |H||H| on the sections pulled back from H0(ΔY0,F|ΔY0)H^{0}(\Delta_{Y_{0}},F|_{\Delta_{Y_{0}}}) to H0(ΓY0,F|ΓY0)H^{0}(\Gamma_{Y_{0}},F|_{\Gamma_{Y_{0}}}) via the Galois action.

H0(ΓX0,(β×β)F|ΓX0){H^{0}(\Gamma_{X_{0}},(\beta\times\beta)_{*}F|_{\Gamma_{X_{0}}})}H0(ΓX0,LmLm){H^{0}(\Gamma_{X_{0}},L_{m}\boxtimes L_{m})}H0(ΔX0,(β×β)F|ΔX0){H^{0}(\Delta_{X_{0}},(\beta\times\beta)_{*}F|_{\Delta_{X_{0}}})}H0(ΔX0,LmLm){H^{0}(\Delta_{X_{0}},L_{m}\boxtimes L_{m})}Tr|ΓX0\scriptstyle{\operatorname{Tr}|_{\Gamma_{X_{0}}}}r\scriptstyle{r}r\scriptstyle{r}Tr|ΔX0\scriptstyle{\operatorname{Tr}|_{\Delta_{X_{0}}}}

The left vertical map and the trace maps are both surjective. Therefore the right vertical map also is surjective. Since ΔX\Delta_{X} and ΓX\Gamma_{X} are both reduced and Cohen-Macaulay, we have proved:

Lemma 5.10.

The restriction map

H0(ΓX,LmkLm)H0(ΔX,LmkLm)H^{0}(\Gamma_{X},L_{m}^{\otimes k}\boxtimes L_{m})\rightarrow H^{0}(\Delta_{X},L_{m}^{\otimes k}\boxtimes L_{m})

is surjective.

Theorem 5.9 then follows by combining Lemma 5.10 and Theorem 5.8.

5.5. Property NpN_{p} and Koszul.

Here we prove:

Theorem 5.11.

Let XX be a smooth complex projective variety of dimension nn and let AA be an ample and basepoint free divisor. Then the line bundle Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) satisfies property NpN_{p} for any mn+1+pm\geqslant n+1+p.

We keep notation as in the previous subsection, and we furthermore define by analogy ΓX0(r)j(s)\Gamma_{X_{0}(r)}^{j}(s) and ΔX0(r)j(s)\Delta_{X_{0}(r)}^{j}(s). Consider

F=LmLmLmF=L_{m}\boxtimes L_{m}\boxtimes\cdots\boxtimes L_{m}

We show by double induction on 1sr1\leqslant s\leqslant r that

H0(ΓX(r)1(r),F)H0(ΓX(r)1(s)ΔX(r)1,s+1ΔX(r)1,r,F)H^{0}(\Gamma_{X(r)}^{1}(r),F)\rightarrow H^{0}(\Gamma^{1}_{X(r)}(s)\cup\Delta_{X(r)}^{1,s+1}\cup\cdots\cup\Delta_{X(r)}^{1,r},F)

is surjective. The case r=2r=2 and s=1s=1 was done above, whereas the case r=2r=2 and s=2s=2 is trivial. Assume therefore that the result holds for any 1sr11\leqslant s\leqslant r-1, and let us prove it for rr. We use descending induction on ss. The case s=rs=r is obvious. Suppose therefore that 1s<r1\leqslant s<r and pick a section

uH0(ΓX(r)1(s)ΔX(r)1,s+1ΔX(r)1,r,F)u\in H^{0}(\Gamma^{1}_{X(r)}(s)\cup\Delta_{X(r)}^{1,s+1}\cup\cdots\cup\Delta_{X(r)}^{1,r},F)

By inductive hypothesis, there is a surjection

H0(ΓX(r)1(s)ΓX(r)1,s+2ΓX(r)1,r,F)H0(ΓX(r)1(s)ΔX(r)1,s+2ΔX(r)1,r,F)H^{0}(\Gamma_{X(r)}^{1}(s)\cup\Gamma_{X(r)}^{1,s+2}\cup\cdots\cup\Gamma_{X(r)}^{1,r},F)\rightarrow H^{0}(\Gamma_{X(r)}^{1}(s)\cup\Delta_{X(r)}^{1,s+2}\cup\cdots\cup\Delta_{X(r)}^{1,r},F)

By Theorem 5.8, there is a surjection

H0(ΓX(r)1(r),F)H0(ΓX(r)1(s)ΓX(r)1,s+2ΓX(r)1,r,F)H^{0}(\Gamma_{X(r)}^{1}(r),F)\rightarrow H^{0}(\Gamma_{X(r)}^{1}(s)\cup\Gamma_{X(r)}^{1,s+2}\cup\cdots\cup\Gamma_{X(r)}^{1,r},F)

Therefore, we may assume that uu maps to zero in

H0(ΓX(r)1(s)ΔX(r)1,s+2ΔX(r)1,r,F)H^{0}(\Gamma_{X(r)}^{1}(s)\cup\Delta_{X(r)}^{1,s+2}\cup\cdots\cup\Delta_{X(r)}^{1,r},F)

Consider the short exact sequence

0\displaystyle 0 H0(ΓX(r)1(s+1)ΔX(r)1,s+2ΔX(r)1,r,F)\displaystyle\rightarrow H^{0}(\Gamma_{X(r)}^{1}(s+1)\cup\Delta_{X(r)}^{1,s+2}\cup\cdots\cup\Delta_{X(r)}^{1,r},F)\rightarrow
H0(ΓX(r)1(s)ΔX(r)1,s+2ΔX(r)1,r,F)H0(ΓX(r)1,s+1,F)\displaystyle\rightarrow H^{0}(\Gamma_{X(r)}^{1}(s)\cup\Delta_{X(r)}^{1,s+2}\cup\cdots\cup\Delta_{X(r)}^{1,r},F)\oplus H^{0}(\Gamma_{X(r)}^{1,s+1},F)\rightarrow
H0((ΓX(r)1(s)ΔX(r)1,s+2ΔX(r)1,r)ΓX(r)1,s+1,F)\displaystyle\rightarrow H^{0}((\Gamma_{X(r)}^{1}(s)\cup\Delta_{X(r)}^{1,s+2}\cup\cdots\cup\Delta_{X(r)}^{1,r})\cap\Gamma_{X(r)}^{1,s+1},F)

We may extend u|ΔX(r)1,s+1u|_{\Delta_{X(r)}^{1,s+1}} to ΓX(r)1,s+1\Gamma_{X(r)}^{1,s+1} via the Galois action as in Subsection 5.4. By construction, this patches with

u|ΓX(r)1(s)ΔX(r)1,s+2ΔX(r)1,r=0u|_{\Gamma_{X(r)}^{1}(s)\cup\Delta_{X(r)}^{1,s+2}\cup\cdots\cup\Delta_{X(r)}^{1,r}}=0

Therefore, it yields a section

u~H0(ΓX(r)1(s+1)ΔX(r)1,s+2ΔX(r)1,r,F)\tilde{u}\in H^{0}(\Gamma_{X(r)}^{1}(s+1)\cup\Delta_{X(r)}^{1,s+2}\cup\cdots\cup\Delta_{X(r)}^{1,r},F)

Finally, we may lift u~\tilde{u} to ΓX(r)1(r)\Gamma_{X(r)}^{1}(r) by inductive hypothesis. Of course, an entirely analogous proof holds for ΓX(r)2(r)\Gamma_{X(r)}^{2}(r). By the particular case s=1s=1, we have a surjective map

H0(ΓX(r)j(r),F|ΓX(r)j(r))H0(ΔX(r)j(r),F|ΔX(r)j(r))H^{0}(\Gamma_{X(r)}^{j}(r),F|_{\Gamma_{X(r)}^{j}(r)})\rightarrow H^{0}(\Delta_{X(r)}^{j}(r),F|_{\Delta_{X(r)}^{j}(r)})

Theorem 5.11 follows by combining the above surjection in the case j=1j=1 with Theorem 5.8. By using the case j=2j=2 instead, we get the following.

Theorem 5.12.

Let XX be a smooth complex projective variety of dimension nn and let AA be an ample and basepoint free divisor. Then the graded ring of sections of the line bundle Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA) is Koszul for any mn+2m\geqslant n+2.

Ein and Lazarsfeld showed that if AA is very ample then one may get a slightly stronger bound provided that (X,A)(n,H)(X,A)\neq(\mathbb{P}^{n},H). More precisely, they showed that KX+mAK_{X}+mA satisfies property NpN_{p} for all mn+pm\geqslant n+p if p>0p>0. It is then natural to ask:

Question 5.13.

Let XX be a smooth complex projective variety of dimension nn and let AA be an ample and basepoint free divisor. Suppose that (X,A)(n,H)(X,A)\neq(\mathbb{P}^{n},H), where HH is a hyperplane section. Does 𝒪X(KX+mA)\mathcal{O}_{X}(K_{X}+mA) satisfy property NpN_{p} for mn+pm\geqslant n+p if p>0p>0?

5.6. Examples.

We conclude by discussing some examples which show that the hypothesis of Theorem 5.11 may not be strengthened beyond Question 5.13.

Fix integers nn, dd and rr. Let α:Xn\alpha:X\rightarrow\mathbb{P}^{n} be the degree dd simple cyclic cover ramified along a general hypersurface of degree drdr. Let HH be a hyperplane section and let A=αHA=\alpha^{*}H. We have

KX=((n+1)+(d1)r)AK_{X}=(-(n+1)+(d-1)r)A

and

α𝒪X=𝒪n𝒪n(r)𝒪n((d1)r)\alpha_{*}\mathcal{O}_{X}=\mathcal{O}_{\mathbb{P}^{n}}\oplus\mathcal{O}_{\mathbb{P}^{n}}(-r)\oplus\cdots\oplus\mathcal{O}_{\mathbb{P}^{n}}(-(d-1)r)

As usual, we denote Lm=𝒪X(KX+mA)L_{m}=\mathcal{O}_{X}(K_{X}+mA). By the above, we have

Lm=𝒪X(KX+mA)=𝒪X((mn1+(d1)r)A)L_{m}=\mathcal{O}_{X}(K_{X}+mA)=\mathcal{O}_{X}((m-n-1+(d-1)r)A)

Now choose m=nm=n, d=2d=2 and r=2r=2. Then we get Ln=𝒪X(A)L_{n}=\mathcal{O}_{X}(A),

H0(X,Ln)=H0(n,𝒪n(1))H^{0}(X,L_{n})=H^{0}(\mathbb{P}^{n},\mathcal{O}_{\mathbb{P}^{n}}(1))

and

H0(X,Ln2)=H0(n,𝒪n(2))H0(n,𝒪n)H^{0}(X,L_{n}^{\otimes 2})=H^{0}(\mathbb{P}^{n},\mathcal{O}_{\mathbb{P}^{n}}(2))\oplus H^{0}(\mathbb{P}^{n},\mathcal{O}_{\mathbb{P}^{n}})

Therefore LnL_{n} is not projectively normal even when XnX\neq\mathbb{P}^{n}.

Suppose now that |A||A| realizes XX as a double cover of YY with YnY\neq\mathbb{P}^{n}. It is an open question whether 𝒪X(KX+nA)\mathcal{O}_{X}(K_{X}+nA) is projectively normal or not. It was shown in [BCG] however that for Horikawa varieties |KX||K_{X}| realizes XX as a double cover of a variety of minimal degree YY and nKXnK_{X} is not projectively normal. Therefore the bound cannot be lowered to n1n-1 even when YnY\neq\mathbb{P}^{n}.

On the other hand, the bounds of Question 1.5 are optimal if one drops the assumption that AA is basepoint free, as already seen in the case of curves [GL88]. Going up in dimension, one may construct examples for which the bound on projective normality and N1N_{1} is optimal in the case of elliptic ruled surfaces (see [GP99]).

Finally, going back to the construction for cyclic covers, choose n=2n=2, m=2m=2, d=3d=3 and r=3r=3. It was shown in [Pur05, Example 5.2] that L2=𝒪X(KX+2A)L_{2}=\mathcal{O}_{X}(K_{X}+2A) satisfies property N0N_{0} but not N1N_{1}.

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