Let $X$ be a smooth manifold. Let $F$ be a sheaf of $\mathbb{R}$-vector spaces on $X$. I have three closely related questions.

1) Under what sufficient conditions on $F$ for any compact subset $K\subset X$ one has $$R^i\Gamma_K(X,F)=0\mbox{ for all } i>0,\,\,\,\,(1)$$ where $\Gamma_K$ is the functor of global sections supported on $K$, and $R^i\Gamma_K$ are its right derived functors?

2) Under what sufficient conditions on $F$ one has $$R^i\Gamma_{\{x\}}(X,F)=0\mbox{ for all } i>0,\,\,\,(2)$$ for any point $x\in X$.

3) Let $F$ be the sheaf of generalized functions on $X$. Is it true that either (1) or (2) are satisfied for this particular $F$?


2 Answers 2


To complement Sándor's answer, let me focus on (3) since that seems to be what you are interested in. Let $F$ be the sheaf of generalized functions or distributions. Consider the localization sequence $$ H^0(X,F)\to H^0(X-x,F)\to H_x^1(X, F)\to H^1(X,F)$$ To ensure vanishing of the 3rd term, you would need to know that any distribution on $X-x$ extends to $X$. But surely this isn't true. (E.g. take $1/|x|$ viewed as a distribution on $X-x=\mathbb{R}-\{0\}$ and try to extend.; see Igor's comment, below) On the plus side, since $F$ is soft, the continuation of the same sequence shows that $H^i_x(X, F)=0$ when $i\ge 2$.

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    $\begingroup$ The $1/|x|$ example is not a good one, since it does have distributional extensions to the real line (cf. Riesz distributions). To get something that does not extend, I believe, you need growth faster than any power law at $0$, like $e^{1/x^2}$. This is discussed in Gel'fand and Shilov's Generalized Functions. $\endgroup$ Aug 22, 2015 at 13:24

I don't work with smooth manifolds, but in algebraic geometry this could never happen, at least not for a coherent sheaf (which should be the case if your vector spaces are finite dimensional). For simplicity let's assume that $K=\{x\}$, that is, we're in case (2).
Grothendieck's theorem on local cohomology says that those sheaves/groups vanish if $t<i<d$, but do not vanish when $i=t$ or $i=d$ where $t$ is the depth and $d$ is the dimension of $F_x$ as a module over the local ring $\mathscr O_{X,x}$.
I realize this might not cover your situation but I'd say that it suggests that you should not expect a blanket vanishing theorem like that.

I also offer the following experiment that might show that the same phenomenon is happening in your case:

Suppose $X$ and $F$ are such that $F$ has no higher cohomology on $X$, i.e., $$ H^i(X,F)=0, \quad\text{ for $i>0$}. \tag{$\star$} $$ I suppose the euclidean space with a trivial bundle is such. Now, I would expect that it should be easy to find a compact subset $K\subset X$ such that the same does not hold for $X\setminus K$ and $F\left|_{X\setminus K}\right.$. Again, in algebraic geometry this is easy: Choose $X$ to be an arbitrary affine variety (say a euclidean space) of dimension at least $2$, $F$ a trivial bundle and $K=\{x\}$ where $x\in X$ is a smooth point.

So, if you have that setup, then consider the long exact sequence of local cohomology. By the assumptions, your condition cannot hold. (Because it would imply the same vanishing as in $(\star)$ on $X\setminus K$.)


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