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Is it true that an open restriction to $U \subset X$ induces a surjection on the set of irreducible perverse subquotients of perverse cohomologies (i.e. cohomologies with respect to the perverse t-structure) of an object of the bounded derived category of constructible sheaves? Is it true that this surjection sends to 0 exactly the subquotients of perverse cohomologies that are supported on $X-U$ and is a bijection from the set of irreducible subquotients of perverse cohomologies whose support intersects $U$ to the set of irreducible subquotients of perverse cohomologies of the open restriction?

Is it true that the Fourier transform (Fourier-Sato/Fourier-Deligne transform) induces a bijection on the set of irreducible perverse subquotients of perverse cohomologies of an object of the bounded derived category of constructible sheaves?

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  • $\begingroup$ You should split these into two questions $\endgroup$ Jul 21, 2019 at 18:38

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The answer to both questions in the first paragraph is yes. Here’s why: Let $F$ be such an object. Restricting to an open subset is exact and therefore commutes with taking (perverse) cohomology. So, we may assume that $F$ is perverse. Choose a filtration $0=F_0\subseteq \cdots \subseteq F_n=F$ of $F$ with irreducible (or zero) factor objects $G_1,\ldots G_n$. Since restricting to $U$ is exact, the restriction of this filtration is a filtration of $F|_U$ with (a priori possibly reducible) factor objects $G_1|_U,\ldots,G_n|_U$. Thus, by Jordan–Hölder, it suffices to show that the restriction of an irreducible perverse sheaf to $U$ is still irreducible. But this follows from the characterization of irreducible perverse sheaves via the intermediate extension functor.

As to your question about Fourier transform, you have to be careful, since (if I recall correctly), the Fourier–whatever transform of a perverse sheaf might not be constructible. I might be wrong, though. However, if we restrict to the full subcategory of monodromic (in the sense of Brylinski) perverse sheaves, then the answer to your question is yes: the Fourier–Sato transform is an (exact) equivalence of categories from the category of mondromic perverse sheaves on a complex vector bundle to those on the dual vector bundle. I’m not familiar with the Fourier–Deligne version, though.

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  • $\begingroup$ Is "monodromic" in this case the same thing as having $\mathbb{R}^+$-conic cohomology sheaves, in the sense of Kashiwara-Schapira? $\endgroup$ Oct 21, 2019 at 15:58
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    $\begingroup$ @BrianHepler no. It’s more restrictive, I think. $\endgroup$ Oct 21, 2019 at 17:47

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