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

Let $X$ be a finite-dimensional, locally compact topological space, and consider the dualizing complex $K_X \in \mathbf{D}^b(X,k)$ (bounded derived category of $k$-sheaves, where $k$ is a noetherian ring). We can define the dualizing functor $$C \mapsto D(C) = \mathbf{R}\mathcal{H}om(C, K_X),$$ (derived internal hom), which leads to a biduality map $C \to D^2(C)$. In SGA 4.5 "Th. de finitude," Deligne shows that, when $k = \mathbb{Z}/n$, the analogous biduality morphism is an isomorphism on the constructible bounded derived category (so, an anti-involution of said category) when one is working with a scheme of finite type over a field or DVR (with $n$ prime to the characteristic). I have heard that the same is true for topological spaces under certain conditions, although I'm not sure what the statement (or proof) should be: first, presumably we are going to want with a nice (Gorenstein?) ring like $\mathbb{Z}/n$, and second, probably there needs to be some analog of the constructible derived category. What is this statement?

I had a look at Kashiwara-Schapira's "Sheaves on Manifolds," but I can't parse the biduality statement given in chapter 3. It's not clear to me how to adapt Deligne's argument to the present case, anyway.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

For an analytic space, you can find this on page 118 of Verdier's article "Classe d'homologie d'un cycle" in Asterisque 36-37. And yes this is on an appropriate constructible derived category with $\mathbb{Z}$-coefficients. I seem to recall that Borel, in his book on intersection cohomology, also discusses this for pseudomanifolds, in case you need something more general.

share|improve this answer
    
Thanks! . –  Akhil Mathew Jul 14 '11 at 13:53
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.