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I am currently trying to understand intermediate extensions of perverse sheaves, specifically the proof of Gabber's purity theorem, which states that the intermediate extension of a pure perverse sheaf is pure.

As part of the proof, in §5 of "Faisceuax pervers" by Bernstein, Beilinson & Deligne, I've come across the following claim, which is not elaborated on in the paper and which I am struggling to understand.

Let $k$ be a finite field, and suppose that I have an open immersion $j:U\rightarrow X$ of $k$-varieties, and a 'projection' $f:X\rightarrow \mathbb{A}^1_k$. Fix a perverse $\overline{\mathbb{Q}}_\ell$ sheaf $\mathcal{F}$ on $U$. Then the claim is that for almost all closed points $v\in \mathbb{A}^1_k$, taking intermediate extensions of $\mathcal{F}$ commutes with pulling back to the fibre over $v$. In other words, for almost all $v$, looking at the commutative diagram

$f^{-1}(v)\cap U \overset{i}{\rightarrow} U$

$\begin{matrix} &&\downarrow j && \downarrow j \end{matrix}$

$\begin{matrix}f^{-1}(v)&\overset{i}{\rightarrow} &X\end{matrix}$

then $i^*\mathcal{F}[-1]$ and $i^*(j_{!*}\mathcal{F})[-1]$ are both perverse, and $j_{!*}(i^*\mathcal{F}[-1])=i^*(j_{!*}\mathcal{F})[-1]$.

Why is this true?

I also have another closely related question, which comes up in trying to understand Delinge's proof of Weil II in terms of perverse sheaves. Suppose that $Y$ is smooth and connected, and that I have a lisse $\overline{\mathbb{Q}}_\ell$-sheaf $\mathcal{F}$ on some relative curve $X\rightarrow Y$, which admits a good compactification $j:X\hookrightarrow \overline{X}$ into a smooth and proper curve $\overline{X}$ over $Y$ whose complement is finite étale over $Y$. Then does taking intermediate extensions of $\mathcal{F}$ commute with pulling back to closed points of $y$? In other words, if I have a closed point $y\in Y$ then should I expect to have something like $(j_{!*}\mathcal{F}[-\dim X])_y \cong j_{!+}(\mathcal{F}_y[-\dim X_y])$ as perverse sheaves on $\overline{X}_y$? Does this basically follow from the first question, or at least from its method of proof, by repeatedly cutting $Y$ with divisors?

Any help with either of these two questions would be greatly appreciated!

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"in ... "Faisceuax pervers" by Bernstein, Beilinson & Deligne, I've come across the following claim, which is not elaborated on in the paper and which I am struggling to understand" <-- This has been the starting point of so much beautiful mathematics!!! –  Fernando Muro Mar 6 '13 at 14:03
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1 Answer

up vote 7 down vote accepted

Note : [BBD]=Astérisque 100 by Beilinson-Bernstein-Deligne

The answer to the first question is basically Deligne's "generic base change", by which I mean theorem 1.9 of SGA 4 1/2 [Th. finitude].

I suppose that you are in part (b) of the second proof of corollary 5.3.2 in [BBD]. So you're assuming that X is affine, and a "projection" is just an embedding of X into some $\mathbb{A}^n$ followed by a projection on one of the coordinates. So $f^{-1}(v)$ is just the intersection of $X$ and of a hyperplane varying in some pencil.

Let's start with the fact that $i^*\mathcal{F}[-1]$ is perverse for generic $v$. Note that the inclusion of the complement of $f^{-1}(v)$ (in $U$ or $X$) is an open affine embedding, so $i^*\mathcal{F}$ is concentrated in perverse degrees $0$ and $-1$, and saying that $i^*\mathcal{F}[-1]$ is perverse is the same as saying that ${}^p H^0 i^*\mathcal{F}=0$, and it is also the same as saying that the adjunction $k_!k^*\mathcal{F}\rightarrow\mathcal{F}$ is surjective, where $k$ is the embedding of the complement of $f^{-1}(v)$. This is proved in the proof of lemma 3.3 of Beilinson's "On the derived category of perverse sheaves". Basically, you base change the whole situation over the space of all hyperplanes in $\mathbb{A}^n$, you prove that the obvious analogue of the statement you want is true there (it follows easily from smooth base change), and then you use Deligne's generic base change to go back.

Actually, you can apply this proof to a finite family of perverse sheaves on $U$ or $X$, not just one (you'll get a dense open subset of good $v$ for each sheaf, and you just take their intersection). So, after restricting $v$ a bit more, you get the perversity of $i^*(j_{!*}\mathcal{F})[-1]$. In fact, what this really show is that you can make the functor $i^*[-1]$ exact on any finite diagram of perverse sheaves on $X$ or $U$ by taking $v$ generic enough.

Now for the isomorphism. I won't assume that $j$ is affine. First, by generic base change (again !), if $v$ is generic enough then the base change map $i^*j_*\mathcal{F}\rightarrow j_*i^*\mathcal{F}$ is an isomorphism. Using the remark above, we can also assume that $i^*{}^p H^kj_*\mathcal{F}[-1]$ is perverse for every $k$ (and that $i^*\mathcal{F}[-1]$ is perverse), so the two spectral sequences that give the ${}^p H^k i^*j_*\mathcal{F}={}^p H^k j_*i^*\mathcal{F}$ are concentrated on one line or one column, and we easily get $i^*{}^p H^kj_*\mathcal{F}[-1]={}^p H^k j_* (i^*\mathcal{F}[-1])$. In the same way, using base change and the remark above, we can make sure that $i^*{}^p H^kj_!\mathcal{F}[-1]={}^p H^k j_! (i^*\mathcal{F}[-1])$.

Remember that $j_{!*}\mathcal{F}$ is the image of the natural map ${}^p H^0j_*\mathcal{F}\rightarrow {}^p H^0 j_!\mathcal{F}$. Applying $i^*[-1]$ to that map and using what I wrote above, we get for $v$ generic enough the map ${}^p H^0j_* (i^*\mathcal{F}[-1])\rightarrow {}^p H^0 j_!(i^*\mathcal{F}[-1])$, whose image is $j_{!*}(i^*\mathcal{F}[-1])$. Of course $i^*[-1]$ is not exact in general, so why would it preserve images ? But by the remark above, I can make this functor exact on any finite diagram by restricting $v$ a bit more, so that's it.

Second question : You need to know that $j_*$ commutes to restricting to the fibers of $\overline{X}\rightarrow Y$, and you also need to know that restricting to the fibers will preserve perversity (up to a shift) for all the perverse shaves you're using. The first is true by lemma 2.1.10 in SGA 7 XIII if you assume that $\mathcal{F}$ is tamely ramified along $\overline{X}-X$; it's not true in general. If you assume that $\mathcal{F}$ is tamely ramified, then I think that you're okay. (Not sure why you would need to cut $Y$ with divisors.)

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Salut, et bienvenue ! –  Chandan Singh Dalawat Mar 7 '13 at 3:02
    
Merci ! Zut, je suis déjà repérée. ;-) –  user31960 Mar 7 '13 at 22:49
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