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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!

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1 Answer 1

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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.

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

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