De Rham-Weil theorem

Question

I am having trouble understanding a couple of points with regard to the De Rham-Weil theorem and was hoping that someone might be able to shed some light.

Let $X$ be a smooth manifold and $\mathcal{F}$ the sheaf of abelian groups on $X$. The theorem states that $H^q(X,\mathcal{F}) \cong H^q((A^{\bullet}(X),d_x))$, where $(A^{\bullet}(X),d_x)$ is the complex of global sections for an acyclic resolution $0 \rightarrow \mathcal{F} \rightarrow A^{\bullet}$. But, if I understand the definition correctly (I suppose I don't), an acyclic resolution has trivial higher cohomology groups. Then the theorem would imply that for for any sheaf the higher cohomology groups are trivial?

Further, I have a proposition in my notes saying that any soft resolution of $\mathcal{F}$ is acyclic. But, for example, the Canonical resolution is soft, but higher cohomology groups are zero only if $\mathcal{F}$ is soft itself, correct?

At the risk of sounding dumb, what is wrong with this picture? Thanks.

Answer 1

I think the misunderstanding comes from the different types of cohomology involved, the cohomology arising from taking an acyclic resolution, the cohomology arising from taking the canonical resolution, and possibly others. We'll write $H^k(\mathcal{A}^{\bullet}(X),\mathcal{F})$ for the cohomology obtained from an acyclic resolution, $H^k(X,\mathcal{F})$ for the cohomology obtained from the canonical resolution. It is possible you are using different definitions, involving Cech cohomology or derived functors, in which case let me know and I'll edit accordingly.

What you are remembering is that $\mathcal{F}$ is $\textit{defined}$ to be acyclic if $H^k(X,\mathcal{F}) \cong 0$ for all $k \geq 1$. A resolution $0 \rightarrow \mathcal{F} \rightarrow \mathcal{A}^\bullet$ is $\textit{defined}$ to be acyclic if $H^k(X,\mathcal{A^j}) \cong 0$ for all $j$ and for all $k\geq 1$. In other words, if all of the $\mathcal{A}^j$ are acyclic. In this case, the sequence formed by taking global sections of an acyclic resolution need not be acyclic (which is, I imagine, where the confusion arises). This certainly is confusing because in algebra we say a chain complex is acyclic if it's cohomology is $0$.

It's also true that any soft sheaf (on a paracompact, Hausdorff space) is acyclic, if $\mathcal{F}$ is soft then $H^k(X,\mathcal{F}) \cong 0 $ for all $k \geq 1$. It is true more generally that any flabby (or perhaps flasque, depending on how French you're feeling) sheaf is soft and acyclic (I believe they are acyclic on any topological space). It is not true in general that an acyclic sheaf is soft, i.e. vanishing higher cohomology doesn't imply that $\mathcal{F}$ is soft.

The De Rham-Weil theorem states that if $0 \rightarrow \mathcal{F} \rightarrow \mathcal{A}^\bullet$ is an acyclic resolution of $\mathcal{F}$, then $H^k(X,\mathcal{F}) \cong H^k(\mathcal{A}^{\bullet}(X),\mathcal{F})$. (I assume this is the version you are referring to).

In fact, one approach to defining sheaf cohomology is to show that $H^k(\mathcal{A}^{\bullet}(X),\mathcal{F})$ is well-defined, independent of choice of acyclic resolution (this theorem may also be referred to as the De Rham-Weil theorem, since proving that two acyclic resolutions have the same cohomology is equivalent to showing that any acyclic resolution has the same cohomology as the canonical resolution) and then uses this to define $H^k(X,\mathcal{F})$. You can also define the cohomology to be the derived functor associated to "taking global sections", and then show that this may be computed with acyclic rather than injective resolutions. In the context of sheaves, this may also be known as the De Rham-Weil theorem.

If you're thinking of the same canonical resolution I am (I hope so otherwise it wouldn't be particularly canonical) then it is soft, and in fact is flabby. This is what means we can use it to define the cohomology of $\mathcal{F}$. The De Rham-Weil theorem says we can use any acyclic resolution, but the canonical one is canonical, so it is a sensible one to choose as a definition, so that the cohomology is a group, rather than an isomorphism class of groups.

Answer 2

1) Even if the complex of sheaves $(A^{\bullet}(X),d^\bullet)$ is exact in some degree, this does not imply that the corresponding complex of global sections is exact in the same degree.
For example on the punctured plane $\mathbb C^*$ the complex of sheaves $\mathcal O \stackrel {d}{\to} \mathcal O\to 0$ (where $df=f'$) is exact but the corresponding complex of global sections $\mathcal O (\mathbb C^*)\stackrel {d}{\to} \mathcal O(\mathbb C^*)\to 0$ is not exact since no holomorphic function on $\mathbb C^*$ has derivative $\frac 1z$: the function $\log$ cannot be defined globally on $\mathbb C^*$.

2) This explains that a sheaf that has a soft resolution can very well have non-vanishing cohomology in any degree.
However your assertion that a sheaf can have vanishing cohomology only if it is soft is wrong.
For example the sheaf $\mathcal O$ of holomorphic functions on $\mathbb C$ is definitely not soft: you cannot extend the function $\frac {1}{z-2}$ defined on the closed unit disc $\mid z \mid \leq 1$ to $\mathbb C$.
But nevertheless $H^i(\mathbb C,\mathcal O)=0$ for all $i\geq 1$ .