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

If $\mathcal F$ is a constructible sheaf (say of $\mathbb C$-modules) on a (real) manifold concentrated in degree $0$ and $i\colon Z \hookrightarrow X$ is a submanifold, can I say anything about $H^j(i^!\mathcal F)$ for $j > \operatorname{codim}_X Z$? Specifically, if $\mathcal F$ is constructible on $\mathbb R^n$ and $i$ is the embedding $\mathbb R^{n-1} \hookrightarrow \mathbb R^n$, does $H^j(i^! \mathcal F)$ vanish for $j > 1$?

(The analoguous statement for coherent sheaves is: if $\mathcal{F}$ is a coherent sheaf on a variety and $Z$ is an l.c.i. subvarity of codimension $n$, then $H^i_Z(\mathcal F)$ vanishes for $i > n$.)

share|improve this question
    
(This was originally posted on math.SE, but didn't get any answers there: math.stackexchange.com/questions/339695) –  Clemens Koppensteiner Apr 4 '13 at 15:57

1 Answer 1

up vote 6 down vote accepted

I think the vanishing you want holds. For simplicity consider the case where $i$ is the inclusion of $Z = \{ z \}$ a point. (One should be able to reduce to this case by taking a normal slice.)

The question is local so we can replace $X$ by a small neighbourhood $U$ of $z$. Let $j$ denote the inclusion of $U - \{ z \}$. Consider the distinguished triangle $i_!i^!\mathcal{F} \to \mathcal{F} \to j_*j^*\mathcal{F} \to$. Now let $S^{n-1}_\epsilon$ be a sphere of radius $\epsilon$ around $z$. By the constructibility assumption the stalk of $j_*j^*\mathcal{F}$ at $z$ is equal, for small enough $\epsilon$, to the cohomology of $S^{n-1}_\epsilon$ with values in the restriction of $\mathcal{F}$. By standard vanishing theorems this vanishes in degrees $\ge n$. Hence the cohomology of $i_!i^!\mathcal{F}$ vanishes in degrees $> n$ as required.

share|improve this answer

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.