4
$\begingroup$

Show that a perverse sheaf on $\mathbb{A}^1(\mathbb{C})$ (the complex plane with the analytic topology) is a bounded complex $A$ of sheaves of $\mathbb{Q}$-vector spaces with constructible cohomology sheaves $\mathcal{H}^n(A)$ such that the following three conditions hold true:

  1. $\mathcal{H}^n(A)=0$ unless $n\in \{-1,0\}$,
  2. $\mathcal{H}^{-1}(A)$ has no nonzero global sections with finite support,
  3. $\mathcal{H}^0(A)$ is a finite sum of skyscraper sheaves.

Although it seems to be a well-known fact, I have some trouble proving it.

My attempt so far. By definition of a perverse sheaf, the following two conditions hold true: for all integers $q$, we have:

$(i)$ $\dim \operatorname{supp} \mathcal{H}^{-q}(A)\leq q$,

$(ii)$ $\dim \operatorname{supp} \mathcal{H}^{-q}(\mathcal{D}A)\leq q$,

where $\mathcal{D}A$ is the Verdier dual of $A$ (we agree that the dimension of the empty set is $-\infty$). From $(i)$, $q=0$, we deduce point 3. For $q<0$, we deduce half of point 1, that is $\mathcal{H}^{n}(A)=0$ for $n>0$. Because $\mathbb{A}^1_{\mathbb{C}}$ is smooth of dimension $1$, we have $$\mathcal{D}A=R\mathcal{Hom}(A,\mathbb{Q}_{\mathbb{A}^1})[2]$$ where $\mathbb{Q}_{\mathbb{A}^1}$ is the constant sheaf equal to $\mathbb{Q}$ on $\mathbb{A}^1$ placed in degree $0$. Hence $$\mathcal{H}^{-q}(\mathcal{D}A)=\mathcal{H}^{2-q}(R\mathcal{Hom}(A,\mathbb{Q}_{\mathbb{A}^1})).$$ However, $\mathbb{Q}_{\mathbb{A}^1}$ is not an injective complex and nothing ensures me that $\mathcal{H}^{2-q}$ commutes with $R\mathcal{Hom}(-,\mathbb{Q}_{\mathbb{A}^1})$. How should I pursue my computation?

Many thanks!

$\endgroup$

1 Answer 1

3
$\begingroup$

It sounds like you are stuck computing the stalks of $\mathcal D A$. To do this, you can use that the homology of the stalk of $\mathcal D A$ at $p$ is the dual of $$H^*(X, X-p ; A) = H^*(U,U-p;A)$$ for any neighborhood $U$ of $p$. This follows from $i^*\mathcal D A = \mathcal D i^! A$ and the exact triangle $i_* i^! A \to A \to j_* j^* A,$ where $i$ is the inclusion of $p$ and $j$ is the inclusion of the complement.

$\endgroup$
1
  • $\begingroup$ Thank you! It took me some time to complete the argument, but now it is good :) $\endgroup$
    – Stabilo
    Oct 19, 2021 at 5:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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