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?
[Math] Sheaves with no cohomology
ag.algebraic-geometrysheaf-cohomology
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About Jon Woolf's answer, it seems to me that the condition that "$x$ is a closed point" was implicitly used: the extension by zero $Z_A$ is only defined for a locally closed subset $A$ (see e.g. Tennison "Sheaf theory," 3.8.6). So $X-x$ must be locally closed. How about the following trivial modification: instead of $Z_X$, consider the sheaf $i_\ast Z$, where $i$ is the inclusion of a point $x$ into $X$.
Suppose that $x$ does not have the smallest open neighborhood and $x$ has a basis of connected neighborhoods. Then $i_\ast Z$ is not a quotient of a projective sheaf $P$. Suppose otherwise. Then for any connected open neighborhood $U$ of $x$, the homomorphism $P(U) \to i_\ast Z(U)$ is zero. This implies that the homomorphism $P \to i_\ast Z$ is zero since it is equivalent to a homomorphism from the stalk $P_x$ to $Z$. Indeed, pick a neighborhood $V$ which is smaller than $U$. We have a surjection $Z_V \to i_\ast Z$. The homomorphism $P \to i_\ast Z$ must factor through $Z_V$, so $P(U) \to i_\ast Z(U)$ must factor through $Z_V(U)$. But $Z_V(U)=0$. This equality may fail if $U$ is not connected however.
So to summarize Jon Woolf and David Treumann, the category of sheaves of abelian groups on a locally connected topological space $X$ has enough projectives iff $X$ is an Alexandrov space.
Surely this must appear in some standard text. Anybody knows a reference? And what about non-locally connected spaces?
For ringed spaces $(X,\mathcal{O}_X)$ one direction is still clear: $X$ being an Alexandrov space implies you'll have enough projectives. But on reflection the other direction, $X$ being a locally connected space without minimal open neighborhoods implies you don't have enough projectives, appears to be rather tricky. One can think of some weird structure sheaves for which the above argument does not go through, in particular $\mathcal{O}_V(U)\ne 0$. So I still wonder what the answer is for ringed spaces.
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 $O_X$.
Best Answer
The example of Alex above is a special case of the Borel-Weil-Bott 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:
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.