Let $X$ be a smooth projective variety of dimension $d$ over a field $k$. Suppose $\mathcal F$ is a coherent sheaf on $X$ such that $H^i(X,\mathcal F) = 0$, for all $i$. What can one say about $\mathcal F$? Does it necessarily mean that $\mathcal F = 0$? If not, can such sheaves be classified? What if $\mathcal F$ is assumed to be locally free? What if we take for $X$ the projective space or a product of projective spaces?

$\begingroup$ $\mathcal{O}(1)$ on $\mathbb{P}^1$. Plenty of similar examples. $\endgroup$– Alex DegtyarevCommented Feb 17, 2014 at 14:13

4$\begingroup$ You may want to look here mathoverflow.net/questions/43586/… $\endgroup$– SashaCommented Feb 17, 2014 at 14:20

4$\begingroup$ There are plenty of such sheaves  for example on $P^1\times P^1$ you can take the kernel of the evaluation morphism $H^0(O(a,b))\otimes O \to O(a,b)$. It will be a nontrivial vector bundle with no cohomology. $\endgroup$– SashaCommented Feb 17, 2014 at 14:22

1$\begingroup$ Such complexes of sheaves exist on any smooth Fano variety over $\mathbb{C}.$ $\endgroup$– user36931Commented Feb 17, 2014 at 16:36

1$\begingroup$ A non trivial torsion line bundle on an abelian variety has no cohomology. $\endgroup$– Damian RösslerCommented Feb 17, 2014 at 21:24
4 Answers
The example of Alex above is a special case of the BorelWeilBott Theorem applied to $\operatorname{SL}_2/B = \Bbb{P}^1$ with $B$ the standard Borel subgroup in $\operatorname{SL}_2$. The general case is this:
Let $G$ be a semisimple complex Lie group with Weyl Group $W$. If for no $w \in W$ we have $w\ast \lambda$ dominant then $H^i(G/B, L_\lambda) = 0 $ for all $i$.
The $L_\lambda$ are all line bundles over $G/B$ which are not necessarily zero, giving a negative answer to one of your questions above.

3$\begingroup$ This example also carries over to algebraic groups (in any characteristic). A typical example of such a $\lambda$ is when there is a simple root $\alpha$ with $\langle\lambda,\alpha^{\vee}\rangle = 1$ (this is one of the reasons the "dot" action shows up so often, as that translates to having $\rho$ as the new "$0$", and $\rho$ behaves like a $0$ should when we look at the cohomology). $\endgroup$ Commented Feb 18, 2014 at 9:28

2$\begingroup$ For the transition to algebraic geometry (in characteristic 0), see Demazure: gdz.sub.unigoettingen.de/dms/load/img/?PPN=GDZPPN002092379 $\endgroup$ Commented Feb 18, 2014 at 12:30

1$\begingroup$ @JimHumphreys: Thanks a lot for the nice references. $\endgroup$– CarlCommented Feb 18, 2014 at 15:41
If $X$ is Fano, the characteristic is zero, and $\mathscr L$ is an ample invertible sheaf contained in, but not equal to $\omega_X^{1}$, then $H^i(X,\mathscr L^{1})=0$ for all $i$. This is Kodaira vanishing for $i\neq \dim X$ and for $n=\dim X$, $H^n(X,\mathscr L^{1})$ is dual to $H^0(X, \mathscr L\otimes \omega_X)$ which is $0$ by the choice of $\mathscr L$.
The examples $\mathscr O_{\mathbb P^n}(d)$ with $1\leq d\leq n$ are covered by this.
In general, if $\mathscr L$ is contained in $\omega_X$, then $H^n(X,\mathscr L)\neq 0$, so if $K_X=0$, then a negative ample invertible sheaves will not work. However, it is relatively easy to find invertible sheaves with this property on K3's. 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 $\mathscr L=\mathscr O_X(L_1L_2)$. Then $\mathscr L$ has no cohomology for any $i$.
This can be seen the following way: if $H^0(X,\mathscr L)\neq 0$, then adding $L_2$ would produce an effective divisor linearly equivalent to $L_1$, but not equal to it (which then would contradict $L_2^2=2$). Similarly, $h^2(X,\mathscr L^{1})=h^0(X, \mathscr O_X(L_2L_1))=0$. Finally, then RiemannRoch 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.
Yet another possibility in characteristic zero is an ample invertible sheaf on a variety with $K_X=0$ that does not have a section. The higher cohomology is zero by Kodaira vanishing.
There is one obvious requirement for a sheaf to have no cohomology: its Euler characteristic must be zero. This imposes a numerical condition on the Chern character of such a sheaf. It is sometimes the case that this is the only obstruction to cohomology vanishing for a "general" sheaf.
For an explicit example, suppose $\xi$ is the Chern character of a semistable sheaf on $\mathbb{P}^2$. There is an irreducible moduli space $M(\xi)$ parameterizing $S$equivalence classes of sheaves with character $\xi$. If $\xi$ has Euler characteristic $0$ then by a theorem of Hirschowitz and Laszlo the locus of sheaves with some nonzero cohomology forms a (usually nonempty) "theta" divisor. Thus in this case the general sheaf in the moduli space has no cohomology, but some interesting sheaves do have cohomology. Describing the sheaves with no cohomology then essentially amounts to describing a dense open subset of the moduli space of sheaves.
In general, the following two questions are interesting, say for a surface $X$ (or generalize as you wish):
(1) Given a Chern character $\xi$ of Euler characteristic $0$, is there a reasonable moduli space $M(\xi)$?
(2) Does the general sheaf $F\in M(\xi)$ have no cohomology? Or is the locus of sheaves with nonzero cohomology the whole space?

$\begingroup$ Do you have a reference for the theorem of Hirshowitz and Laszio? I tried some googling and failed. $\endgroup$– DrewCommented Aug 5, 2015 at 20:12
Two more remarks:
 If a coherent sheaf $\mathcal{F}$ on $X$ has vanishing cohomology, so will be $\mathcal{F}\boxtimes \mathcal{G}$ on $X\times Y$, for any variety $Y$ and coherent sheaf $\mathcal{G}$ on $Y$, by the Künneth formula.
 On a curve of genus $g$, a general line bundle of degree $g1$ has vanishing cohomology.
With these you can construct a huge family of examples...