vote up 16 vote down
star
5

There are several theorems I know of the form "Let $X$ be a locally ringed space obeying some condition like existence of partitions of unity. Let $E$ be a sheaf of $\mathcal{O}_X$ modules obeying some nice condition. Then $H^i(X, E)=0$ for $i>0$."

What is the best way to formulate this result? I ask because I'm sure I'll wind up teaching this material one day, and I'd like to get this right.

I asked a similar question over at nLab. Anyone who really understands this material might want to write something over there. If I come to be such a person, I'll do the writing!


Two versions I know:

(1) Suppose that, for any open cover $U_i$ of $X$, there are functions $f_i$ and open sets $V_i$ such that $\sum f_i=1$ and $\mathrm{Supp}(f_i) \subseteq U_i$. Then, for $E$ any sheaf of $\mathcal{O}_X$ modules, $H^i(X,E)=0$. Unravelling the definition of support, $\mathrm{Supp}(f_i) \subseteq U_i$ means that there exist open sets $V_i$ such that $X = U_i \cup V_i$ and $f_i|_{V_i}=0$.

Notice that the existence of partitions of unity is sometimes stated as the weaker condition that $f_i$ is zero on the closed set $X \setminus U_i$. If $X$ is regular, I believe the existence of partitions of unity in one sense implies the other. However, I care about algebraic geometry, and affine schemes have partitions of unity in the weak sense but not the strong.

(2) Any quasi-coherent sheaf on an affine scheme has no higher sheaf cohomology. (Hartshorne III.3.5 in the noetherian case; he cites EGA III.1.3.1 for the general case.) There is a similar result for the sheaf of analytic functions: see Cartan's Theorems.

I have some ideas about how this might generalize to locally ringed spaces other than schemes, but I am holding off because someone probably knows a better answer.


It looks like the answer I'm getting is "no one knows a criterion better than fine/soft sheaves." Thanks for all the help. I've written a blog post explaining why I think that fine sheaves aren't such a great answer on non-Hausdorff spaces like schemes.

flag
Dear David, I don't quite understand the statement of (1). There seem to no conditions stated that involve the $V_i$; so how do conclude that $f_{| V_i}$ vanishes? Maybe $X = U_i \cup V_i$ is a condition, not a conclusion? – Emerton Jan 12 at 18:32
@Emerton The statement "$\mathrm{Supp}(f) \subset U$" is equivalent to "there exists an open set $V$ such that $X = U \cap V$ and $f|_V=0$." I'll see if I can make this clearer. – David Speyer Jan 12 at 19:01
@Anweshi: that's certainly relevant. In particular, there is a nice idea there of using looking at the sheaf $End(E)$ instead of some ambient sheaf of commutative rings. But I worry about two things. (1) Fine sheaves are "usually only used over paracompact Hausdorff spaces": so not affine schemes. – David Speyer Jan 12 at 19:05
I'll unpack my second worry as an edit up above; it's overflowing the character limit. – David Speyer Jan 12 at 19:08

4 Answers

vote up 4 vote down
check

Although we clearly all have more or less the same answers, here is how I like to organize things.

I) Let $\mathcal F$ be a sheaf of abelian groups on the topological space $X$. It is said to be soft if every section $s \in \Gamma (A,\mathcal F)$ over a closed subset $A\subset X$ can be extended to $X$. Notice carefully that the definition of $s$ is NOT that it is the restriction to $A$ of some section of $\mathcal F$ on an open neighbourhood of $A$ [but that it is an element $ s\in \prod \limits_{x\in X} \mathcal F_x$ satisfying some more or less obvious conditions]

II) Consider the following condition on the [not necessarily locally] ringed space $(X, \mathcal O)$ :

The space $X$ is metrizable and given an inclusion $A\subset U \subset X$ with $A$ closed and $U$ open there exists a global section $s\in \Gamma (X,\mathcal O)$ such that $s|A=1$ and $ s|X \setminus U=0 \quad \quad (SOFT)$.

We then have the

$\textbf{Theorem }$ : If the ringed space satisfies (SOFT), then every sheaf of $\mathcal O_ X -Modules$ is soft.

III) A metrizable space endowed with its sheaf of continuous functions satisfies $(SOFT)$. A metrizable differential manifold endowed with its sheaf of smooth functions satisfies $(SOFT)$.

IV) On a metrizable space every soft sheaf is acyclic

Put together these results yield all standard acyclicity results on functions,vector bundles, distributions,etc.

It is interesting to notice that you use partitions of unity only once: in the proof of III). But never more afterwards; you just check that your sheaves are $\mathcal O -Modules$. I like this approach (which I learned from Grauert-Remmert) more than the usual one, where a proof of acyclicity is given for the sheaf of smooth functions, followed by the ( correct!) assertion that you have to repeat it with minor changes for, say, vector bundles. Moreover fine sheaves needn't even be mentioned if you follow this route.

link|flag
vote up 9 vote down

This answer takes a different point of view to that expressed in the body of your question, but is relevant to the question in the title.

A sheaf is called soft if any section over a closed subset of $X$ extends to a section over $X$. As the linked wikipedia article states, on a paracompact Hausdorff space, soft implies acyclic. The typical partitions of unity arguments in differential topology can be interpreted as using showing that sheafs of smooth functions, smooth sections of bundles, and so on, are soft. (One can always extend smooth functions locally from a closed set to a neighbourhood, and the partitions of unity allow one to patch these extensions.) Another way to phrase this is that fine implies soft .

link|flag
vote up 4 vote down

Regarding point (2) in the question, I believe that Grothendieck's proof involves choosing a $f_i$ and $g_i$ in $A$ such that $\sum_i f_i g_i = 1$ (in other words, a cover of Spec $A$ by open sets of the form Spec $A_{f_i}$), computing the cohomology Cech-wise (which amounts to consider a Koszul complex), and then taking the limit over all such covers and using the Cech-to-sheaf-cohomology spectral sequence to conclude. So this is certainly an argument in the spirit of partitions of unity arguments.

link|flag
vote up 3 vote down

On a paracompact Hausdorff space with a structure sheaf of functions $f:U\to k$ which satisfies partition of unity as in your version (1), the structure sheaf $O_X$ is fine (i.e. the sheaf $\operatorname{Hom}(O_X,O_X)$ is soft). A basic fact, easy to prove, is that if $O_X$ is fine and $E$ is a sheaf of $O_X$-modules then $E$ is fine.

(See for example Godement's book on sheaves, which is the basic source for these definitions and proofs for paracompact spaces, which for him include "Hausdorff").

Now fine implies soft, and soft on a paracompact Hausdorff space is acyclic (again the proof is in Godement, and it is exactly the same as the standard proof for flasque sheaves).

The upshot: on a paracompact Hausdorff space with a partition of unity, any sheaf of $O_X$-modules is acyclic.

The prime example of this is the sheaves $\Omega^i$ of differential forms on a real manifold. Since $O_X$ is fine, each $\Omega^i$ is fine, hence acyclic. Thus, the Poincare lemma gives an acyclic resolution $$ 0\to \mathbb R\to \Omega^1 \to \Omega^2\to \dots $$ of the constant sheaf $\mathbb R$. In particular, the cohomologies $H^i(X,\mathbb R)$ can be computed as the cohomologies of the complex $$ 0\to \Gamma(\Omega^1) \to \Gamma(\Omega^2)\to \dots $$ where $\Gamma$ means global sections; i.e. by using global differential forms (a well-known fact).

Version (2) should not be called "partition of unity" in my opinion. (Maybe a "weak partition of unity"?) It sure does look similar to a partition of unity, but it is not true that $Supp(f_i)\subset U_i$. More importantly, analyzing the proof of Serre's vanishing theorem for acyclicity of quasicoherent sheaves on affine schemes/varieties, this proof does not look similar (to me) to the proof in the case (1), so the analogy seems tenuous. To me, it looks like a different circle of ideas.

link|flag
1 
You write: "On a paracompact space with a partition of unity, the structure sheaf O_X is fine" 1) All paracompact spaces have partitions of unity. 2) What does "the structure sheaf" mean for a given topological space? In any case it is certainly not true that for a sheaf of rings on X, O_X is always acyclic: think of a complex manifold and its sheaf of holomorphic functions, which is not acyclic in general, although the underlying topological space is paracompact. 3) On the next line you write "if O_X is fine...". That is indeed what you should ASSUME: a space + a fine sheaf of rings. – Georges Elencwajg Jan 12 at 23:30
Nice post by the way: I was allowed too few characters in the preceding comment to tell you ! – Georges Elencwajg Jan 13 at 0:00
Georges: thank you for your comment. Hopefully, I cleared up the confusion in this edit. – VA Jan 13 at 0:34

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.