2
$\begingroup$

Serre's vanishing theorem (SV) states that, on a projective variety $X$ with a choice of ample line bundle $\mathcal{O}_X(1)$, for any coherent sheaf $F$, we have $$H^i(X,F(m))=0,\quad m>>0$$ for every $i>0$ (i.e. there's an $m$ big enough such that the vanishing occurs for any bigger multiple $m'>m$ of the polarization and for every index $i$), where $F(m)$ denotes the twist $F\otimes\mathcal{O}_X(m)$.

If $D_*:=\{ D_m\}_{m\in\mathbb{N}}$ is a sequence of divisors on the projective variety $X$, we say that $D_*$ satisfies SV on $X$ if, for every coherent sheaf $F$ on $X$, $$H^i(X,F(D_m))=0,\quad m>>0$$ for every $i>0$, where as usual $F(D_m)=F\otimes\mathcal{O}_X(D_m)$.

I'm mostly interested in the case in which:

($\star$) $X$ is a normal surface over $\mathbb{C}$.

Q1. Is there some "general" and "nontrivial" criterion for a sequence $D_*$ to satisfy SV? E.g. $D_*:=$ a subsequence of $\{m\cdot D_0\}$ for a fixed ample $D_0$ clearly satisfies it, but it's a trivial consequence of the usual SV.

Also, is there some such property that is valid for every $X$? (For example, could it be something like: $D_*$ eventually forms a subsequence of a sequence of ample divisors which is "increasingly positive" in some specified sense?)

Narrowing down the question (and assuming ($\star$), if it is of any use):

Q2. Assume that $D_*$ is a sequence of (not necessarily effective) ample divisors such that $D_{m+1}-D_m$ is effective. Does $D_*$ satisfy SV?

$\endgroup$
3
  • 1
    $\begingroup$ I think Q2 is easily seen to be false, eg. $X$ the blow up of $P^2$ at a point $p$ with exc. div $E$ and $D_m=(m+1)H-E$ where $H$ is the pullback of a general hyperplane. Then we have a short exact sequence $0\to O_X(D_m+2E)\to O_X(D_m+3E)\to O_{P^1}(-2)\to 0$ and since $H^1(O(-2))\ne 0$, then either $H^1(O_X(D_m+3E))\ne 0$ or $H^2(O_X(D_m+2E))\ne 0$. I guess that one could hope that if $D_m$ are ample and $\lim D_m^{\dim Z}\cdot Z\to \infty$ for any subvariety $Z$, then $D_*$ satisfies SV (?). $\endgroup$
    – Hacon
    Jul 7 '16 at 23:32
  • $\begingroup$ @Hacon: thank you for the counterexample. I suspected that Q2 might not be true. What if we require that $D_{m+1}-D_m$ has strictly positive self intersection? It still does not imply (say by Nakai-Moishezon) that it is ample right? $\endgroup$
    – Qfwfq
    Jul 7 '16 at 23:39
  • 1
    $\begingroup$ Not sure I follow. In my example $(D_{m+1}-D_m)^2=H^2>0$. If you require $(D_m^{\dim Z}\cdot Z / H^m\cdot Z)>\epsilon m$ for any subvariety, then using Anhern-Siu type arguments you should be able to show that $D_m=mH+P$ where $H$ is an ample Q-divisor and P is nef and then deduce that $D_*$ is SV by a combination of SV and Fujita vanishing....not sure if this is useful (?). $\endgroup$
    – Hacon
    Jul 7 '16 at 23:56
6
$\begingroup$

Claim: $D_*$ is SV iff for any ample divisor $H$ there exists $m(H)$ such that if $m\geq m(H)$, then $D_m-H$ is nef.

Suppose $D_*$ is SV. Fix $A$ very ample. Since $D_*$ is SV, then $H^i(D_m-H-jA)=0$ for all $i>0$ and $0\leq j\leq \dim X +1$. By Castelnuovo-Mumford regularity $D_m-H$ is generated by global sections and in particular nef.

Suppose that for any ample divisor $H$ there exists $m(H)$ such that if $m\geq m(H)$, then $D_m-H$ is nef. Let $F$ be a coherent sheaf on $X$ and $A$ an ample line bundle on $X$, then by Fujita vanishing there exists an integer $t_0$ such that $H^i(F(tA+P))=0$ for any $t\geq t_0$ and $P$ a nef line bundle. But then, for any $m\geq m(t_0A)$, we have $D_m-t_0A$ is nef and hence $H^i(F(D_m))=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.