For a complex of sheaves $\cal{F}^{\bullet}$ on a variety $X$, a useful fact is that the stalks of the cohomology sheaves of $\mathcal{F}^{\bullet}$ agree with the cohomology groups of the complex of stalks:

$$\mathcal{H}^{i}(\cal{F}^{\bullet})_{x} \cong \mathcal{H}^{i}(\cal{F}^{\bullet}_{x})$$

Consider now the category $D_{c}^{b}(X)$ of bounded constructible complexes on $X$, and the full abelian subcategory $\operatorname{Perv}(X)$ of perverse sheaves. Let us work with sheaves of $\mathbb{Q}$ or $\mathbb{C}$-modules. The perverse cohomology sheaves ${}^{\mathfrak{p}}\mathcal{H}^{i}(\cdot):D_{c}^{b}(X) \to \operatorname{Perv}(X)$ are themselves perverse sheaves.

I'm wondering, are there any nice results helping to compute stalks of ${}^{\mathfrak{p}}\mathcal{H}^{i}(\mathcal{F}^{\bullet})$? Like something analogous to the basic result I recall at the beginning? These stalks should be complexes of vector spaces.

The basic result above for ordinary sheaves can't work as it is. For example, if $X$ is smooth of dimension $n$, then $\mathbb{C}_{X}[n]$ is perverse. And ${}^{\mathfrak{p}}\mathcal{H}^{i}(\mathbb{C}_{X}[n])$ vanishes unless $i=0$ in which case it is just $\mathbb{C}_{X}[n]$. So at any point $q \in X$, the stalk is the skyscrapper sheaf:

$${}^{\mathfrak{p}}\mathcal{H}^{0}(\mathbb{C}_{X}[n])_{q} = \mathbb{C}_{q}[n]$$

But if we first pass to the stalk, and then take perverse cohomology ${}^{\mathfrak{p}}\mathcal{H}^{i}(\mathbb{C}_{q}[n])$, this vanishes unless $i=-n$, in which case it is just $\mathbb{C}_{q}$.

So are there any useful ways of computing stalks of perverse cohomology sheaves, or do you just have to do it directly?

  • 1
    $\begingroup$ Certainly none as nice as for the usual cohomology sheaves, except at a generic point, or more concretely a point where all the cohomology sheaves are lisse - at these points, the perverse filtration is the usual filtration shifted by n. If your complex of sheaves is one to which the decomposition theorem can be applied, the decomposition theorem is a big help. Other than that I don't know any big general principle. $\endgroup$ – Will Sawin Feb 8 at 2:18
  • $\begingroup$ I have been told by various experts in the area that stalks aren’t the right notion for perverse sheaves. Instead, one should be considering nearby and vanishing cycles with respect to various linear forms. These functors (appropriately shifted) do commute with perverse cohomology, as they’re exact on the category of perverse sheaves. $\endgroup$ – Avi Steiner Feb 23 at 15:48

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