Can someone tell me which of the following are true? Let $X$ be a reasonable space. 


Suppose $F$ is a complex whose cohomology groups are constructible sheaves, at least one of which is nontrivial.



> 
> Can $\mathbb{H}(X, F) = 0$?
> 
> 
> 


If so, can it still happen assuming $F$ is really just 


(1) a constructible sheaf 
(2) a local system 
(3) a perverse sheaf 
(4) an intersection cohomology complex ? 



can an nonzero IC sheaf have zero hypercohomology?
Yes. Consider any local system (**EDIT:** of rank 1) over a characteristic 0 field on $\mathbb{C}^\*$ with non-trivial monodromy. This satisfies all of (1), (2), (3) and (4). There are lots of ways to check that this has trivial cohomology; for example, if the monodromy has finite order, it's a summand of the pushforward from the constant sheaf, which has the same cohomology as the constant sheaf. If you want a projective example, an elliptic curve works.


