Let $X$ be a projective variety and let $L$ be a line bundle on $X$. Suppose for all locally free sheaves $M$ on $X$, $ H^i(X,{L^*}^{\otimes r} \otimes M)=0 $ for $i<\dim X$ and $r$ sufficiently big.

Does it follow that $L$ is an ample line bundle? Here $L^*$ denotes the dual of $L$.

This is of course clear if $X$ is smooth using Serre duality, but how is it in general?

After reading Laurent Moret-Bailly and Karl Schwede's comments, below I changed the condition '$M$ coherent' to '$M$ locally free'.

  • 2
    $\begingroup$ It follows that $X$ is finite: if $x$ is a closed point of $X$ and $M$ is the skyscraper sheaf $\mathcal{O}_{\{x\}}$, then $H^0(X,L^{\otimes r}\otimes M)\neq0$ for all $r$, hence your assumption implies that $\dim X=0$. $\endgroup$ – Laurent Moret-Bailly Apr 18 '11 at 15:52
  • $\begingroup$ Perhaps you wanted $M$ to be a line bundle or maybe a vector bundle? $\endgroup$ – Karl Schwede Apr 18 '11 at 15:55

As was pointed out in the comment above by Laurent Moret-Bailly, you can't choose any $M$. Let me try to answer a slightly different question which I hope is close to what you intended.

Edit: This should also answer the revised question, in the negative.

The answer is no, ample line bundles do not satisfy the condition you want, even for $M = \mathcal{O}_X$ (or more generally for $M$ a vector bundle). In other words, suppose what you wanted was true, then if $L$ satisfied the condition it would be ample (this is what you wanted), but as I'll show below, this will also imply that $L$ cannot be ample.

If $X$ is Cohen-Macaulay, then what you want holds (see the version of Serre-duality in Hartshorne).

Suppose now that $X$ is not Cohen-Macaulay (CM) but its non-CM-locus is isolated at a point $z \in X$ (for example, if that point looks locally analytically like the cone over an Abelian variety). Take $M = \mathcal{O}_X$ and suppose that

By assumption, $H^i(X, L^{-r} ) = 0$ for all $i < d = \dim X$ and all $r \gg 0$. However, $$H^i(X, L^{-r}) = \mathbb{H}^{d - i}(X, L^r \otimes \omega_X^{\bullet}[-d])^{\vee}.$$ Here $\omega_X^{\bullet}[-d]$ is the dualizing complex of $X$ shifted over by $d = \dim X$. Being CM is equivalent to $\omega_X^{\bullet}$ just being a sheaf (and not a complex).

Now we wish to compute $\mathbb{H}^{d - i}(X, L^r \otimes \omega_X^{\bullet}[-d])^{\vee}$. This is done using a Grothendieck-spectral sequence (see for example Weibel's homological algebra). Since $r \gg 0$ and $L$ is ample, this implies that the spectral sequence degenerates at the $E^2$ stage (in other words, degenerates immediately). The upshot of this is that $$ \mathbb{H}^{d - i}(X, L^r \otimes \omega_X^{\bullet}[-d]) = H^0(X, L^r \otimes h^{d-i}(\omega_X^{\bullet}[-d]) ). $$ Where the $h^{d-i}(\omega_X^{\bullet}[-d])$ just means that $(d-i)$th cohomology of the shifted dualizing complex. Since $X$ was not Cohen-Macaulay with isolated non-CM-locus, some these are supported at the non-CM point $z \in X$. In other words, some of the $L^r \otimes h^{d-i}(\omega_X^{\bullet}[-d])$ for $i < d$, are skyscraper sheaves (finite dimensional vector spaces even), so some of the $\mathbb{H}^{d - i}(X, L^r \otimes \omega_X^{\bullet}[-d]) \neq 0$.


Your condition for a single locally free $M$ is equivalent to the same condition for all coherent $M$ with $\mathrm{Supp}M=X$ (so in particular coherent or locally free does not make a difference) is equivalent to $X$ being Cohen-Macaulay by Serre's vanishing. See the appropriate statement on page 182 (Cor 5.72) of KollárMori98.


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.