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 $i>0$.
 A: 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.
A: 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 $H^{*}(U,\mathcal{S}^{\vee})$.
