Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The derived category $D^{\flat}_{c}(X,R)$ of constructible sheaves of $R$-modules on $X_{et}$ is defined as the full subcategory of $D^b(X,R)$ whose cohomology sheaves are all constructible.

Clearly, given a sequence of constructible sheaves indexed by $\mathbb{Z}$ can be realized as the cohomology of a complex of sheaves consisting entirely of constructible sheaves (just put the desired cohomology sheaf in each degree and make the differential zero). But one might imagine that there is a complex of sheaves, not all of which are constructible, whose cohomology sheaves are constructible, and which is not quasi-isomorphic to a complex consisting entirely of constructible sheaves.

My question: Can such a complex of sheaves exist? If so, under what conditions might it not exist?

I asked this question in a course of N. Katz, who answered that he did not know the answer but that he knew of some results in the affirmative when asking this question for quasi-coherent or coherent sheaves in place of constructible sheaves.

share|improve this question
    
The motivation is reasonable, but your concern is largely irrelevant in practice. It's often unnatural to try to shoehorn your complex into having such properties, even if it could be done: think about the derived Kunneth isomorphism, which would be a nightmare to express with complexes of constructibles (and may give counterexamples you seek). As you know, the exact triangle machinery reduces problems formulated in adequate generality back into the setting of usual sheaves that satisfy whatever "niceness" properties are imposed on the homologies of the complex. –  user28172 Dec 5 '12 at 3:31
    
@Eric: Why does the question reduce to one about maps in the derived category between (bounded below or above?) complexes having constructible terms, as you claim? –  user28172 Dec 5 '12 at 3:33
1  
To show that no such complex exists, it would suffice to show the following: given a map in the derived category between two complexes of constructible sheaves, it be realized by an actual chain map of quasi-equivalent complexes of constructible sheaves, without having to take resolutions that use nonconstructible sheaves. Conversely, if you could construct a counterexample to this condition you could probably construct such a complex from it. –  Eric Wofsey Dec 5 '12 at 3:37
    
@nosr: If you can always realize maps in the derived category using maps of complexes of constructible sheaves, then any extension of complexes of constructible sheaves is again a complex of constructible sheaves. But any constructible complex is an iterated extension of complexes of constructible sheaves, namely those representing its cohomology. –  Eric Wofsey Dec 5 '12 at 3:45
2  
In the article "Constructible sheaves", M. Nori proves that such a complex does not exist in characteristic zero. I do not remember the details now, but perhaps he has also some remarks there on the general case. –  ulrich Dec 5 '12 at 12:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.