Hi.
Can any one me say if there is a simple proof of this claim which i can prove it by localization and no easy technique of nuclear spaces…
Let $f:X\rightarrow S$ be an open, surjective morphism of complex reduced spaces with constant fiber dimension $n$ (or universally open morphism with $n$-fibers between locally noetherian excellent without embedded componnet schemes on field of charact.0).
Let $G$ be a coherent sheaf on $X$.
Claim: ${\rm R}^{n}f_{*}G=0$ if and only if the support of $G$ satisfies:
${\rm Supp}(G)\cap X_{s}$ is nowhere dense in the fiber $X_{s}$.
Remark:
1) We can assume $X$ $n$-complete space(i.e there exist a smooth strongly $n$-convex exhaustion function on $X$) and then all cohomology group $H^{k}(X, F)$ vanish for all coherent sheaf $F$ and all $k>n$.
2) For a projection $f:S\times U\rightarrow S$ where $U$ is an open polydisque of $C^{n}$, the result is true for ${\rm R}^{n}f_{!}$. By right exactness of the $n$-higher direct image functor, we can localize and deduce from the projection case the general case…. But, it is no simple proof…
Thank you very much.
Best Answer
Are you sure this is what you mean? Take $S$ to be a point, $X=$ affine $n$-space ($n>0$), $G=\mathcal{O}_X$. The support is $X$ and the cohomology vanishes.