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The line bundle $O(-1)$ on a projective space or $O(-\rho)$ on a flag variety has a property that all its cohomology vanish. Is there a story behind such sheaves?

Here are more precise questions. Let $X$ be a smooth complex projective surface (say, a nice one like Del Pezzo or K3). Does there always exist a coherent locally free sheaf $M$ whose derived global sections vanish? Can one describe all such sheaves? Is there a coarse moduli space of such sheaves?

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In characteristic p, you can try the cone on the map O --> Frob_* O. I'm confused about whether this works. –  David Treumann Oct 25 '10 at 22:41
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David, I think that this won't work if for example $X$ is supersingular elliptic curve. In this case, the map $H^1(X,O)\to H^1(X,F_*O)$ is $0$, so you would cohomology on the cone. –  Donu Arapura Oct 25 '10 at 23:56
    
Good lord I've been trying to make precisely that computation for like 90 minutes. For a del Pezzo surface it works because H^i(X,O) vanishes for i > 0. Are there "supersingular K3s" on which Frob:H^2(X,O) --> H^2(X,O) vanishes? –  David Treumann Oct 26 '10 at 0:03
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Yes, I think so. If $X$ is a Kummer surface obtained from a product of supersingular elliptic curves $E_i$, I think that $H^2(X,O_X)\cong H^1(E_1,O_{E_1})\otimes H^1(E_2,O_{E_2})$ compatibly with $F$-action. –  Donu Arapura Oct 26 '10 at 1:36
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David, there is an important class of varieties, called ``Frobenius split varieties''. For these the morphism $O \to Frob_* O$ splits, so $O$ is a direct summand of the Frobenius pushforward. The complement summand (isomorphic to the cone of $O \to Frob_* O$) then has no derived global sections. –  Sasha Oct 26 '10 at 3:15
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5 Answers 5

up vote 28 down vote accepted

The bundles with no derived global sections (more generally the objects $F$ of the derived category $D^b(coh X)$ such that $Ext^\bullet(O_X,F) = 0$) form the left orthogonal complement to the structure sheaf $O_X$. It is denoted $O_X^\perp$. This is quite an interesting subcategory of the derived category.

For example, if $O_X$ itself has no higher cohomology (i.e. it is exceptional) then there is a semiorthogonal decomposition $D^b(coh X) =< O_X^\perp, O_X >$. Then every object can be split into components with respect to this decomposition and so many questions about $D^b(coh X)$ can be reduced to $O_X^\perp$ which is smaller. Further, if you have an object $E$ in $O_X^\perp$ which has no higher self-exts (like $O(-1)$ on $P^2$), you can continue simplifying your category --- considering a semiorthogonal decomposition $O_X^\perp = < E^\perp, E >$. For example if $X = P^2$ and $E = O(-1)$ then $E^\perp$ is generated by $O(-2)$, so there is a semiorthogonal decomposition $D^b(coh P^2) = < O_X(-2), O_X(-1), O_X >$ also known as a full exceptional collection on $P^2$. It allows a reduction of many problems about $D^b(coh P^2)$ to linear algebra.

Another interesting question is when $O_X$ is spherical (i.e. its cohomology algebra is isomorphic to the cohomology of a topological sphere). This holds for example for K3 surfaces. Then there is a so called spherical twist functor for which $O_X^\perp$ is the fixed subcategory.

Thus, as you see, the importance of the category $O_X^\perp$ depends on the properties of the sheaf $O_X$.

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This response doesn't quite answer your question, but it may provide some interesting further examples.

A supernatural sheaf on $\mathbb P^n$ (really some twist of such a sheaf) has this property, though supernaturality is much stronger than you are looking for. A supernatural sheaf $\mathcal F$ is defined by two conditions:

  1. For every $j\in \mathbb Z$, there exists at most one such $i$ such that $$ H^i(\mathbb P^n, \mathcal F(j))\ne 0. $$

  2. The Hilbert polynomial $\chi(\mathbb P^n, \mathcal F)$ has distinct integral roots.

Hence, if $\mathcal F$ is supernatural and $e$ is a root of $\chi(\mathbb P^n,\mathcal F)$, then $\mathcal F(e)$ has the property that $H^i(\mathbb P^n, \mathcal F(e))=0$ for all $i$.

This is really the answer to a different question, but Eisenbud and Schreyer show that these sheaves can be used to generate the cohomology table of any sheaf. In addition, you could probably apply their constructions to build further examples of sheaves with the property that you are looking for. See

http://arxiv.org/abs/0712.1843 or http://arxiv.org/abs/0902.1594.

Hope this helps.

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I haven't really thought about it, so I can't say whether there is good description in general. But here a class of examples, which suggests to me that such the set of such sheaves of fixed rank won't always form a bounded family. Take a smooth rational surface $X$, and choose a smooth rational curve $C\subset X$. The ideal sheaf fits into an exact sequence $$0\to \mathcal{O}_X(-C)\to \mathcal{O}_X\to \mathcal{O}_C\to 0$$ Writing out cohomology, one sees that $H^i(X,\mathcal{O}_X(-C))=0$ for all $i$.

Edit It's interesting to read other people's perspectives on this. But, anyway, for some reason I didn't finish my thought yesterday. To get an unbounded family, take $X$ to be a rational surface with infinitely many exceptional curves $C$ (which exists thanks to Nagata). Then the collection $\{\mathcal{O}_X(-C)\}$ gives an infinite set of such sheaves with distinct Chern classes.

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Assume that ${\rm char}=0$. If $X$ is Fano and $\mathcal L$ is an ample invertible sheaf contained in, but not equal to $\omega_X^{-1}$, then $H^i(X,\mathcal L^{-1})=0$ for all $i$. This is Kodaira vanishing for $i\neq \dim X$ and for $n=\dim X$, $H^n(X,\mathcal L^{-1})$ is dual to $H^0(X, \mathcal L\otimes \omega_X)$ which is $0$ by the choice of $\mathcal L$.

I think both of your examples are covered by this.

In general, if $\mathcal L$ is contained in $\omega_X$, then $H^n(X,\mathcal L)\neq 0$, so for a K3 (or more generally if $K_X=0$) negative ample invertible sheaves will not work. It is also relatively easy to see that if ${\rm Pic}\\, X = \mathbb Z$, then there are no such invertible sheaves.

Edit: On the other hand, it is relatively easy to find invertible sheaves with this property on K3's with higher Picard numbers. For instance, let $X$ be a K3 and $L_1, L_2$ two disjoint $(-2)$-curves. This requires the Picard number to be at least $3$, but then it is quite common. For instance, every Kummer has plenty of disjoint $(-2)$-curves. So, let $\mathcal L=\mathcal O_X(L_1-L_2)$. Then $\mathcal L$ has no cohomology for any $i$.

This can be seen the following way: First notice that the sheaf $\mathcal O_X(L_1)$ has a $1$-dimensional $H^0$ since $L_1$ has negative self-intersection, $H^2=0$ since its dual has no $H^0$, and $H^1=0$ by Riemann-Roch. Then in the short exact sequence $$ 0 \to \mathcal L \to \mathcal O_X(L_1) \to \mathcal O_{L_2} \to 0 $$ the two sheaves on the right have the exact same cohomology groups and clearly the induced map on $H^0$ is an isomorphism, so the sheaf on the left cannot have any non-zero cohomology.

Second edit: I just realized that this can be done much simpler: $H^0(X,\mathcal L)=0$ since if it had a non-zero section, that would produce an effective divisor linearly equivalent to $L_1$, but not equal to it. Similarly $h^2(X,\mathcal L^{-1})=h^0(X, \mathcal O_X(L_2-L_1))=0$. Finally, then Riemann-Roch shows that $h^1=0$ as well. This argument has the advantage that it shows that if $L_1,L_2$ intersect, then $h^1\neq 0$.

Using similar ideas one can make up lots of invertible sheaves on K3 surfaces with this property.

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On a K3-surface if a line bundle $\mathcal L$ has vanishing cohomology, then $\mathcal L^2=-4$ by the Riemann-Roch formula. Conversely, if $\mathcal L^2=-4$ and neither $\mathcal L$ nor its inverse has global sections, then also the first cohomology vanishes.

If $X$ is a projective K3-surface with $\mathrm{Pic}(X)=\mathbb Z$, then $\mathcal L^2\geq0$ for all line bundles and hence no line bundle has all its cohomology vanishing. On the other hand there are (plenty of) K3-surfaces with an ample line bundle $\mathcal M$ and a line bundle $\mathcal L$ with $\mathcal M\cdot\mathcal L=0$ and $\mathcal L^2=-4$. Then neither $\mathcal L$ nor its inverse has global sections (as otherwise we would get a strictly effective divisor with zero intersection with $\mathcal M$). Thus for line bundles at least the problem restricted to K3-surfaces depends very much on the surface.

As for the idea of consider the cokernel $B$ of $\mathcal O_X\rightarrow F_\ast\mathcal O_X$. We have that the cohomology of $B$ vanishes if $X$ is ordinary. The general complete intersection is ordinary (by a theorem of Deligne) so there are plenty of those. Raynaud has also shown that if $X$ is a curve then $B\bigotimes \mathcal L$ has vanishing cohomology for the general line bundle $\mathcal L$ of degree $0$.

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