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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 ?

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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.

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    $\begingroup$ More generally one can take any $X$ and a proper and generically finite map $f : X -> X$. Then $\mathbb{H}(f_*\mathbb{Q}_X) = H^*(X)$ and so every summand of $f_*\mathbb{Q}_X$ except for $\mathbb{Q}_X$ (an IC by the decomposition theorem) has vanishing cohomology. $\endgroup$
    – Geordie Williamson
    Commented Nov 19, 2012 at 21:02
  • $\begingroup$ @Ben: When you say "non-trivial monodromy" I guess you mean invariants = coinvariants = 0. Of course local systems with unipotent monodromy on $\mathbb{C}^*$ have cohomology. $\endgroup$
    – Geordie Williamson
    Commented Nov 19, 2012 at 21:04
  • $\begingroup$ yes, I should have said "no trivial composition factors." Since we were just looking for one example, I can also just assume rank 1. $\endgroup$
    – Ben Webster ♦
    Commented Nov 19, 2012 at 21:07

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