# Smooth in codimension-k and the weight filtration

Let $X$ be an algebraic variety. Then $H_{et}^k(X)$ has a filtration whose associated graded pieces are labeled by "weights", certain integers between $0$ and $2k$. If $X$ is smooth, then the weights are between $k$ and $2k$.

If we know that $X$ is smooth in codimension $c$, do we get a nontrivial bound on the weights of $H^k_{et}(X)$?

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Maybe you already tried this: the long exact sequence induced by decomposing $X$ into the smooth part and the rest. This seems not of too much help in the general situation, or at least one needs, I guess, to do some dévissage, as the sequence involves $H^k_Z(X)$ and purity doesn't apply directly. But if you have some concrete situation in mind, this might be of a little help. –  shenghao May 10 '11 at 23:20
Unfortunately, I want to use such a fact inside a proof by contradiction, so the concrete situation is one that I am trying to show is impossible and thus can't look at examples of. I can tell you that I am looking at projective varieties (but that should affect the other end of the weight filtration) and that they are normal (so automatically smooth in codimension 1) if that helps. –  David Speyer May 10 '11 at 23:32
Sorry, but I'm afraid I didn't see what you are hoping for, e.g. what is the "fact" you mentioned? Do you have a conjectural nontrivial bound for the weight? BTW, if char. $k=0$ one can reduce to $\mathbb C$ and use Hodge theory to get the weight filtration, but if char. $k=p,$ I don't really see how to reduce to $k=\overline{F}_p$ to get the wt. filtr. (unless you start with finite fields): somewhere when doing spreading out, we cannot compare the cohom. of the special fiber with that of the generic fiber, as $X$ is not smooth. Or maybe I'm wrong...? –  shenghao May 10 '11 at 23:46
uh... maybe one uses de Jong's alteration to reduce to the proper smooth case, and for that we may apply the proper smooth base change to reduce to $k=\overline{\mathbb F}_p.$ Not sure if it's too difficult to check the abutment filtr. is indep. of the choice of the alteration though... –  shenghao May 11 '11 at 0:11

I'll keep this short. If I understood your question correctly then you might want to use something like Gabber purity for intersection cohomology. In certain ranges this gives you information for ordinary cohomology. E.g. if $X$ is proper with isolated singularities, then $H^i(X)=IH^i(X)$ is pure of weight $i$ for $i> \dim X$. The reference for this is BBD. But perhaps http://www.math.purdue.edu/~dvb/preprints/delgab.pdf would be helpful as well.

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Thanks for the link! I was unaware of the history you discuss in the introduction. –  Mike Skirvin May 11 '11 at 14:01

Assume for simplicity that X is projective. Then you have a Gysin sequence $H^i_c(X_{smooth})\to H^i_c(X)\to H^i_c(X_{sing})\to\dots$

If $X_{sing}$ has small dimension then for large $i$ you find isomorphisms $H^{2n-i}(X_{smooth})^*(-n)\cong H^i_c(X)=H^i(X)$. This limits the weights.

For non-proper X you probably need alterations in order to compare $H^i_c(X)$ and $H^i(X)$. (In char 0 you could use resolution of singularities to compare $H^i_c(X)$ and $H^i(X)$.)

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Can you explain to me how $H^i(X)=H^i_c(X_{sm})$ will limit weights? Thanks. –  shenghao May 12 '11 at 0:11
The weights for $H^{2n−i}(X_{smooth})$ lie in $[2n−i,4n−2i]$, therefore the weights for $H^{2n−i}(X_{smooth})^∗(n)$ lie in $[2i−2n,i]$. If $i>2 dim X_{sing}+1$ and $i>n$ this yields a non-trivial bound on the weights for $H^i_c(X)$ and $H^i(X)$. –  Remke Kloosterman May 12 '11 at 5:22

The geometry certainly puts restrictions on the weights. For proper varieties, the allowed weights on $H^m(X)$ are between $0$ and $m$. So on $H^1$, the weights are $0$ and $1$; but if $X$ is normal, $H^1$ is pure of weight one (reflected in the fact that the Picard variety is an abelian variety). This is an example of the type of restrictions you seek, if I am not mistaken.

A general result may be in the literature (this is very probable). The suggestion of Shenghao of using devissage and using induction on the dimension of $X$ seems right on target. The weight filtration in characteristic zero and over a finite field are related and even compatible; the argument is by spreading out to Spec $Z[1/N]$ for some positive integer $N$ and this works even for singular varieties. See Deligne's ICM 1974 address (the examples in section 3, and sections 7, 13 and 14 are the relevant ones).

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I think the following is a common point to the previous answers. It is a very general and important step in many "devissages" of mixed sheaves (perverse sheaves, D-modules, mixed Hodge modules etc...) and illustrates why resolution of singularities is so important. This is really standard but it was never clearly explained to me and I don't know a good reference for it so I thought it was worth a post. If anyone has a good reference for it I'd be glad to read it.

Let's consider a proper morphism $\pi: Y\to X$ that is a isomorphism outside $i:Z\hookrightarrow X$. Basically $\pi$ is a resolution of the singularities. Then we have a distinguished triangle $$F \to \pi_{*} \pi^{*} F \to i_* Q \to +1$$ where $Q = Cone(i^* F \to i^* \pi_* \pi^* F)$ (using proper base change).

Taking cohomology (by that I mean $H^i(a_*-)$ with $a:X\to pt$) we get $$H^i(X,F) \to H^{i}(Y,F) \to H^i(Z,Q) \to H^{i+1}(X,F)$$

If $\pi$ is a blowup with smooth center $Z$, we have $Q = \bigoplus_{q=1}^{c-1} i^*F(-q)[-2q]$ so this is really easy. By iterating, we get results in the case we have a nice resolution of singularities (sequence of blow-ups with smooth centers giving a normal crossing divisor). So we reduced the problem to computing cohomology on a normal crossing divisor. And by induction on the number of irreducible components this reduces to the case of a a smooth variety.

In your case what you want is to control the weights of $H^i(Z,Q)$ so the suggestion is to track the weight filtration in the previous reasoning which (in theory) isn't so hard since it is strict.

I don't know enough about alterations to comment about positive caracteristic but I think the principles are the same.

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