# cohomology and $j_!$

I have a projective variety $X$ and an open immersion $j : U \to X$.

Say I have a sheaf, locally free in my case of interest, $\mathcal{S}$ on $U$. Is there any reasonable relationship between $H^i(X,j_! \mathcal{S})$ and $H^i(U,\mathcal{S})$? What if I add that I know that $H^i(U,\mathcal{S}) = 0$ for $i>0$. I'm hopeful the latter can imply that $H^i(X,j_! \mathcal{S}) = 0$ for $i>0$.

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Consider $X = \mathbb{P}^{1}$, $U =\mathbb{A}^{1}$ and j the inclusion of a affine chart in the projective line. The complement of the affine chart is a point P, let i be the inclusion of this point in the projective line. Consider the sheaf $\mathcal{F} = \mathcal{O}(-2)$ on the projective line. One has $dim H^{1}( \mathbb{P}^{1}, \mathcal{F} ) =1$. One has an exact sequence :

$0 \to j_{!} ( \mathcal{F}_{U} ) \to \mathcal{F} \to i_{*} ( \mathcal{F}_{P} ) \to 0$

$i_{*} ( \mathcal{F}|_{P} )$ is a skyscraper sheaf over P so its $H^{1}$ is 0.

$H^{1}$ of $\mathcal{F}$ is not 0. By long exact sequence in cohomology, we have $H^{1}$ of $j_{!}( \mathcal{F}|_{U} )$ is not 0.

But on the other hand $\mathcal{F}|_{U}$ is a trivial sheaf over a affine scheme so its $H^{1}$ is 0.

This example seems to show that the expected relation is not true.

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unknown (google) -- I've edited your answer so that the formulae are shown correctly. In general, to avoid problems with the asterisks one could use back ticks (); apostrophes (') do not seem to help. – algori Aug 10 '12 at 19:14
Thanks for the example. This is the sort of set-up I've been working with. I guess some of the particulars of my sheaves and spaces will need to come into play. – Andy B Aug 10 '12 at 21:24
algori--thanks for your edit. I confused  and '. – user25309 Aug 11 '12 at 10:41

If $X$ is smooth and we consider sheaves of $k$-modules, $k$ a commutative ring, then $H^*(X, j_!\mathcal{S})\cong H^*_c(U,\mathcal{S})$; the latter is equipped with a non-degenerate pairing $$H^*_c(U,\mathcal{S})\otimes H^{2d-*}(U,\mathcal{S}^{\vee})\to k,$$

where $\mathcal{S}^\vee$ is the Verdier dual local system (i.e., the local system that is constructed from the representation of $\pi_1(U)$ dual to the one that gives $\mathcal{S}$) and $d=\dim X$. So if $k$ is a field, we do get a statement relating the vanishing of $H^*(X, j_!\mathcal{S})$ and $H^{*}(U,\mathcal{S}^{\vee})$.

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This is great, thanks! For the case I'm interested in, $X$ is not smooth. However, it is a (normal) toric variety. Does this buy me anything? – Andy B Aug 10 '12 at 18:56
You seem to consider local system whereas in the question, I think that locally free means locally free as sheaf of O_X-modules which is a different kind of sheaf. – user25309 Aug 10 '12 at 19:15
Andy -- I'm not sure: for a non-smooth $X$ both the dualizing sheaf of $X$ and the dual of $\mathcal{S}$ are in general not so easy to describe: e.g. for all I know, there is nothing to prevent them from being concentrated in several degrees etc. Maybe it's easier to look directly at $H^*_c(U,\mathcal{S})$. – algori Aug 10 '12 at 19:20
unknown (google) -- yes, I was considering sheaves of $k$-modules, not $\mathcal{O}_X$-modules. Andy -- if it is the coherent case you're interested in then please let me know. – algori Aug 10 '12 at 19:29
I did mean, although I did not say, locally free sheaves of $\mathcal{O}_X$-modules. – Andy B Aug 10 '12 at 20:12