## When fine just ain’t enough February 2, 2010

Posted by David Speyer in Algebraic Geometry, complex analysis, homological algebra, things I don't understand.

If you use sheaves to study differential geometry, one of the basic lemmas you’ll want is the following: Let $X$ be a smooth manifold and let $\mathcal{E}$ be a sheaf of modules over $C^{\infty}(X)$. (For example, $\mathcal{E}$ might be the sheaf of sections of a vector bundle.) Then all higher sheaf cohomology of $\mathcal{E}$ vanishes.

The proof of this theorem is basically homological algebra plus the existence of partitions of unity. This gives rise to a slogan “when you have partitions of unity, sheaf cohomology vanishes.” One way to make this definition precise is through the technology of fine sheaves.

As Wikipedia says today, “[f]ine sheaves are usually only used over paracompact Hausdorff spaces”. That means they are not used when working with the Zariski topology on schemes, for example. When I started digging into this, I realized there were good reasons: The technology of fine sheaves (and the closely related technology of soft sheaves) does not include the scheme theory cases which we would want it to.

However, there are theorems of the form “when you have partitions of unity, sheaf cohomology vanishes” on schemes and on complex manifolds. I put up a question at MathOverflow asking whether there were better formulations that included these examples, but I probably didn’t formulate it well. I think spelling out all my issues would be too discursive for MathOverflow, so I’m bringing it over here.

## What does it mean to vanish?

Let me start with a technical point that caused me a great deal of confusion. Let $X$ be a topological space, $U$ an open subset of $X$, and $x$ be a point in $U$. Let $\mathcal{E}$ be a sheaf of abelian groups on $X$ and $f$ a section in $\mathcal{E}(U)$. In what sense could we say that $f$ vanishes at $x$?

In this generality, there is only one reasonable definition: That the image of $f$ in the stalk $\mathcal{E}_x$ is zero. Unpacking the definition of the stalk, this means that there is an open set $V$, with $x \in V \subset U$, such that $f|_V=0$.

Now, think about the case where $\mathcal{E}$ is a sheaf of functions on $X$, with restriction meaning honest-to-God-restriction of functions. The above definition is not what we mean when we say $f$ vanishes at $x$! Rather, it is the concept we would express as “$f$ vanishes on a neighborhood of $x$.”

In order to get a concept which generalizes the ordinary meaning of vanishing at a point, we need to restrict to the case where $X$ is a locally ringed space, and $\mathcal{E}$ a sheaf of $\mathcal{O}_X$-modules. In that case, $\mathcal{E}_x$ is a module over the local ring $\mathcal{O}_x$. And the image of $f$ in $\mathcal{E}_x \otimes_{\mathcal{O}_x} k(x)$ is the best analogue to “the value of $f$ at $x$“, where $k(x)$ is the residue field of $\mathcal{O}_x$.

Therefore, in this blog post, I make the following definitions:

With the above notation, I say that $f$ vanishes on a neighborhood of $x$ if the image of $f$ in $\mathcal{E}_x$ is zero or, equivalently, if there is an open set $V \ni x$ such that $f|_V=0$.

I say that $f$ vanishes at $x$ if the image of $f$ in $\mathcal{E}_x \otimes_{\mathcal{O}_x} k(x)$ is zero.

Let $K$ be a closed subset of $U$. We have the following, analogous definitions:

The function $f$ vanishes on a neighborhood of $K$ if either of the equivalent definitions holds: (1) There is an open set $V$ containing $K$, such that $f|_U$ is zero or (2) For every $x \in K$, the function $f$ vanishes in a neighborhood of $x$.

The function $f$ vanishes on $K$ if, for every $x \in K$, the function $f$ vanishes in at $x$.

If more books had adopted this terminology, I would have spent far less time confused about exactly what they meant when they claimed some space had partitions of unity.

## Partitions of unity implies vanishing sheaf cohomology, the standard version

With these definitions out of the way, we can show that the existence of partitions of unity implies vanishing sheaf cohomology.

Theorem 1: Let $(X, \mathcal{O})$ be a locally ringed space, and assume that $X$ is paracompact. (Every cover has a locally finite subcover refinement.) Suppose that, for every open cover $U_i$ of $X$, there are global functions $f_i$ so that $f_i$ vanishes in a neighborhood of $X \setminus U_i$ and $\sum f_i=1$. Let $\mathcal{E}$ be any sheaf of $\mathcal{O}_X$ modules. Then $H^i(X, \mathcal{E})=0$ for all $i >0$.

Proof Sketch: Our proof is by induction on $i$. Let $\mathcal{I}$ be an injective sheaf with an injection $\mathcal{E} \to \mathcal{I}$; and let $\mathcal{K}$ be the cokernel of $\mathcal{E} \to \mathcal{I}$. For $i \geq 2$, the long exact sequence gives $H^{i}(\mathcal{E}) = H^{i-1}(\mathcal{K})$, the right hand side of which is zero by induction. So we simply must establise the base case, that $H^1(\mathcal{E})=0$.

We know that $H^0(\mathcal{I}) \to H^0(\mathcal{K}) \to H^1(\mathcal{E}) \to 0$ is exact, so it is enough to show $H^0(\mathcal{I}) \to H^0(\mathcal{K})$ is surjective. Let $k$ be a global section of $\mathcal{K}$. Since $\mathcal{I} \to \mathcal{K}$ is surjective, there is an open cover $U_i$ of $X$, and functions $h_i \in \mathcal{I}(U_i)$ such that $h_i \mapsto k|_{U_i}$.

Take $f_i$ as in the hypothesis. For each index $i$, let $V_i$ be an open set containing $X \setminus U_i$ such that $f_i|_{V_i}$ vanishes. Let $m_i$ be the section of $\mathcal{I}$ which is $f_i h_i$ on $U_i$ and is $0$ on $V_i$. Such a section exists by the gluing axiom, applied to the open cover $\{ U_i, V_i \}$. Let $m = \sum m_i$. (By paracompactness, we may assume that the cover $\{ U_i \}$ is locally finite, so the sum makes sense.) We claim that $m \mapsto k$.

It is enough to check this claim on stalks. Near any point $u$, we have $m = \sum m_i = \sum_{U_i \ni u} m_i|_{U_i} = \sum_{U_i \ni u} f_i h_i$. By construction, this maps to $\sum_{U_i \ni u} f_i k = \left( \sum f_i \right) k = k$. QED

I would feel guilty if I never defined a fine sheaf in this post. The idea of fine sheaves is that, rather than starting with the sheaf of rings $\mathcal{O}$, we can start with the sheaf $\mathcal{E}$ and define $\mathcal{O}:= \mathcal{H}om(\mathcal{E}, \mathcal{E})$. The sheaf $\mathcal{E}$ is called fine if $\mathcal{O}$, defined in this manner, has partitions of unity in the above sense. Of course, $\mathcal{O}$ may not be commutative and the stalks of $\mathcal{O}$ may not be local, but it turns out that we can still prove Theorem 1 in this setting: A fine sheaf on a paracompact space has no cohomology. Unfortunately, in the examples I discuss below, the extra elements of $\mathcal{H}om(\mathcal{E}, \mathcal{E})$ still don’t create partitions of unity.

## The Regularity Trick

The above proof asks for $f_i$ to vanish on a neighborhood of $X \setminus U_i$. When $X$ is nice enough, we can ask that $f_i$ just vanish on $X \setminus U_i$.

Theorem 2 Let $X$ be a paracompact regular topological space and $(X, \mathcal{O})$ be a locally ringed space. Suppose that, for any open cover $V_i$, there are global functions $f_i$ such that $f_i$ vanishes on $X \setminus V_i$ and $\sum f_i=1$. Then, for any open cover $U_i$, there are global functions $f_i$ so that $f_i$ vanishes in a neighborhood of $X \setminus U_i$ and $\sum f_i=1$.

Proof sketch: Take your open cover $U_i$. For every point $u$ in $X$, let $U_i$ contain $u$. Choose disjoint open sets $V_i$ and $W_i$ such that $u \in V_i$ and $X \setminus U_i \subset W_i$. Find functions $f_i$ such that $\sum f_i=1$ and $f_i$ vanishes on $X \setminus V_i$. Then $X$ will vanish on $W_i$ and hence on a neighborhood of $X \setminus U_i$. QED

Because of the above argument, mathematicians who work on metrizable spaces don’t worry very much about the distinction between vanishing on a closed set and in the neighborhood of a closed set. But the Zariski topology is not metrizable…

## The Zariski world: Cause for caution!

Let us begin, right away, by pointing out that there are affine schemes and sheaves of modules on them which have nontrivial cohomology.

Let $X$ be the scheme $\mathrm{Spec} k[x]$. Let $\mathcal{E}$ be the following sheaf: If $0 \in U$ then $\mathcal{E}(U)$ is the local ring $\mathcal{O}_0$, otherwise $\mathcal{E}(U)=0$. The obvious map $\mathcal{O}_X \to \mathcal{E}$ is a surjection of sheaves (exercise!), yet the map on global sections is not surjective. So, if $\mathcal{F}$ is the kernel of $\mathcal{O}_X \to \mathcal{E}$, then $H^1(X, \mathcal{F}) \neq 0$.

So any theorem about sheaf cohomology vanishing on affine spaces needs to be phrased carefully.

## The Zariski world: Cause for hope!

Affine schemes, with the Zariski topology, do not have partitions of unity in the sense of Theorem 1. Indeed, there do not exist two polynomials on the affine line adding to $1$, the first of which is zero on a neighborhood of $0$ and the other on a neighborhood of $1$. (Since any polynomial which is zero in a neighborhood of a point must be identically zero.)

Nonetheless, we have the following theorem, originally due to Serre:

Theorem 3: (Hartshorne III.3.5, EGA III.1.3.1) Coherent sheaves on an affine scheme have no cohomology.*

Proof Sketch: As before, we reduce to the case of showing that, if $\mathcal{I} \to \mathcal{F}$ is a surjective map of coherent sheaves, then $H^0(\mathcal{I}) \to H^0(\mathcal{K})$ is surjective.

Let $k$ be a global section of $\mathcal{K}$. Let $U_i$ be a basic open cover of $X$. The adjective basic means that $U_i = \{ x : f_i(x) \neq 0 \}$ for some global function $f_i$. Let $h_i \in \mathcal{I}(U_i)$ be a preimage of $k$. There is some $n_i$ such that $f_i^{n_i} h_i$ extends to a section of $\mathcal{I}$. (Exercise! This is the point that the coherence hypothesis is used.)

Since the $U_i$ are a cover; the functions $f_i$ have no common zero, and the functions $f_i^{n_i}$ also have none. So, by the Nullstellansatz, there are global functions $g_i$ such that $\sum f_i^{n_i} g_i =1$. So we can find
global sections $m_i$ extending $f_i^{n_i} g_i h_i$ and, as in the proof of Theorem 1, $\sum m_i$ is a preimage of $k$. QED

This proof used two important facts. In order to avoid the language of basic opens, I’ll phrase them in terms of ideal sheaves; the reader might enjoy rephrasing the above proof in this language.

Fact 1: Let $U$ be an open set in $X$, let $\mathcal{E}$ be a coherent sheaf and $h$ a section in $\mathcal{E}(U)$. Let $I \subset \mathcal{O}(X)$ be the ideal of $f$ such that $fh$ extends to $X$. Then $Z(I) \subset X \setminus U$.

Fact 2: If $\mathcal{I}_1$, $\mathcal{I}_2$, …, $\mathcal{I}_r$ is a collection of coherent ideal sheaves on $X$ such that $\bigcap Z(\mathcal{I}_j)=\emptyset$ then $\sum \mathcal{I}_j = \mathcal{O}$.

Fact 2, to my mind, is a good generalization of the existence of partitions of unity. It is weaker than asking for partitions of unity in the sense of vanishing on neighborhoods of closed sets, but stronger than just asking for partitions of unity in the sense of vanishing on closed sets. I had hoped that the correct generalization of “partitions of unity implies sheaf cohomology vanishing” would be “Facts 1 and 2 imply vanishing of cohomology for coherent sheaves”. But, when I started reading about complex manifolds, I realized this was not the way to go.

## The Stein World: Cause for puzzlement!

A Stein space is a closed complex-submanifold of $\mathbb{C}^n$.
They are the analogues of (smooth) affine schemes for complex analysis. We can talk about the sheaf $\mathcal{O}$ of holomorphic functions on any Stein space.

Stein spaces have Fact 2; this is a consequence of Rückert’s Nullstellansatz. I am willing to consider this a good generalization of the existence of partitions of unity. Of course, Stein Spaces don’t have partitions of unity in the sense of Theorem 1 for the same reason polynomials don’t: An analytic function that vanishes on a neighborhood of a point must be identically $0$.

Stein spaces also have Theorem 3. This is called Cartan’s Theorem B.

But Stein spaces don’t have Fact 1! Consider $X = \mathbb{C}$, let $U = \mathbb{C} \setminus \{ 0 \}$. Consider the section $h=e^{1/z}$ of $\mathcal{O}(U)$. There is no holomorphic function $f$ such that $fh$ is holomorphic! So we can’t use Fact 2 to prove Theorem 3.

I assumed that there was some minor trick which was used to get around this. But I just read through the proof of Cartan’s Theorem B in Grauert and Remmert’s Theory of Stein Spaces and it looks nothing like the proofs of Theorems 1 and 3.

This is where I run out of ideas. But I know we have readers who think about sheaves and homological methods on a much deeper level than I do. So, what is the version of “Partitions of unity imply cohomology vanishing” which works for Stein Spaces?

* Two footnotes for experts: Yes, this also holds for quasi-coherent sheaves. I stated the weaker version because I want to make the analogy to Stein spaces, and I’m not sure if the corresponding result is true for quasi-coherent on Stein spaces. Second, I am implicitly assuming noetherianness, in order to make sure my sums are finite. But the theorem is true without this.

1. Steven Sam - February 2, 2010

I’m confused by this:

The function f vanishes on a neighborhood of K if either of the equivalent definitions holds: (1) There is an open set V containing K, such that f|_U is zero or (2) For every x \in K, the function f vanishes in a neighborhood of x.

The function f vanishes on K if, for every x \in K, the function f vanishes in a neighborhood of x.

The second thing is just repeating part of the first definition?

Nitpick: in your parenthetical remark about paracompact spaces, we need to say that all open covers have locally finite open refinements, not locally finite subcovers (according to wikipedia http://en.wikipedia.org/wiki/Paracompact , requiring locally finite subcovers is equivalent to compactness, but here’s a counterexample to your definition, take the space to be the real line, and take the open cover to be as follows: for each point x, take U_x to be the interval of length 1 around x union (-1,1). No locally finite subcover!

Another correction (while I’m here):

proof of Theorem 2: Then X will vanish on W_i and hence on a neighborhood of U_i

shouldn’t it be

Then f_i will vanish on W_i and hence on a neighborhood of X \setminus U_i?

2. Emmanuel Kowalski - February 5, 2010

Theorem 3 is a result of Serre, that in fact characterizes affine schemes (i.e., the converse is true). I think proper attribution should be given…
Also, isn’t it the case that Theorems “A and B” for analytic geomtery are due to Cartan and Serre?

3. David Speyer - February 6, 2010

Thanks for all the corrections. Regarding the attribution of Theorems A and B: Wikipedia thinks that the analytic case is solely due to Cartan, and that Serre’s contribution was the algebraic case. I must admit that I did not check this with a more reliable source.

4. Emmanuel Kowalski - February 6, 2010

You’re right, theorems A and B are indeed due to Cartan (I looked in books of Gunning-Rossi and Demailly); Hartshorne’s book suggests that Serre proved the converse (vanishing of higher cohomology implies Stein), but doesn’t give a reference.
Amusingly (?), one of the links in the Wikipedia page for Theorems A and B to “Cartan’s Theorem” brings up a page about a theorem of Cartan père… (I’ve seen other pretty bad mistakes in attributions in Wikipedia when it comes to fairly recent mathematics).