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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'.

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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 $\dim X=0$. –  Laurent Moret-Bailly Apr 18 '11 at 15:52
    
Perhaps you wanted $M$ to be a line bundle or maybe a vector bundle? –  Karl Schwede Apr 18 '11 at 15:55
    
Yes, thanks, I edited. –  user14199 Apr 18 '11 at 16:27

2 Answers 2

up vote 6 down vote accepted

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 $\mathbb{H}^{d - i}(X, L^r \otimes \omega_X^{\bullet}[-d]) \neq 0$.

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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.

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