# How does the lack of partitions of unity affect the structure of analytic/holomorphic manifolds?

The standard way to define integration on a smooth manifold is to use partitions of unity, to extend to the case where the form you're integrating isn't supported on just one coordinate patch. Of course, in the analytic/holomorphic case, we don't have partitions of unity. So how do we do integration?

Furthermore, how does this affect the space $\mathcal{T}(M)$ of analytic vector fields (analytic global sections of the tangent bundle)? The usual extension lemma for smooth sections of a vector bundle depends on partitions of unity, so there doesn't seem to be any reason you should always be able to find an nonzero analytic section. Does it ever happen that $\mathcal{T}(M) = 0$?

Also in that vein - I remember needing to use partitions of unity to prove in an exercise that the space of 1-forms $\mathcal{T}^*(M)$ is actually the dual $C^\infty(M)$ module $\, \text{Hom}(\mathcal{T}(M),C^\infty(M))$ - because given a map $\mathcal{T}(M) \to C^\infty(M)$, it seems like you need some kind of extension lemma to construct a 1-form that induces it. Does this fail in the analytic case?

I imagine the answer to the previous two questions will involve sheaves, but I don't quite know enough about sheaves to frame them in the appropriate sheaf-theoretic way. Maybe per this question, I should ask if this means the sheaves are not soft and the sheaf cohomology doesn't vanish?

And in general, any good references (or just explanations you care to give) on the major differences between the smooth and analytic cases? It seems like most differential geometry books pay almost no attention to the analytic case.

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You don't necessarily need partitions of unity to define integrals on smooth manifolds. An alternative approach is outlined here: mathoverflow.net/questions/38439/…. It also works in the analytic case. – Dmitri Pavlov Oct 24 '12 at 14:44
Dmitri, could you explain how this works out in the case M = R, or [0,1]? It seems like de Rham theory at least relies on the fundamental theorem of calculus to do work; I'm a little skeptical that you could build the entire thing without any prior notion of integration, and then, in the C^0 case, say, integrate something awful like Cantor's function. Any enlightenment? – Kevin Casto Oct 25 '12 at 6:39
@Kevin: No, you do not need “the fundamental theorem of calculus” to develop the de Rham theory. See for example Lemma 6.5.3 in Schapiro's notes people.math.jussieu.fr/~schapira/lectnotes/AlTo.pdf. Further details can be found in comments here: mathoverflow.net/questions/43681/motivating-the-de-rham-theorem/…. – Dmitri Pavlov Oct 25 '12 at 14:34

## 1 Answer

(1) A holomorphic manifold is also (or "can also be viewed as") a smooth manifold, and that lets you define integration. To put it another way, you do have partitions of unity, just not holomorphic ones.

(2) Even before you get to tangent bundles, there are well-known cases where local things can't be patched globally. For example, on the Riemann sphere, there are non-trivial holomorphic functions in neighborhoods of every point, but the only global holomorphic functions are the constants. In the case of the tangent bundle, it seems to me (experts please edit if I mess this up) that there are no non-zero global tangent vector fields on Riemann surfaces of genus 2 or more, though of course there are such fields locally everywhere.

(3) You're right that this phenomenon (and related ones) are the beginning of sheaf cohomology.

(4) I won't try to answer your question about 1-forms being the dual of tangent vectors, since it seems to mix pointwise things (the space of 1-forms $T^*(X)$) with global things in a way that I don't understand.

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1) Haha, duh. I guess I assumed that we needed to integrate "analytically", but I guess that isn't really meaningful. 2) Ah okay, interesting. Any quick explanation for why it's g >= 2? And is this true when you view them as analytic 2-manifolds, or just Riemann surfaces? (when does this distinction make a difference?) 3) Does this make things clear? I can sketch a proof if you want. I imagine you think I'm confusing the local duality-as-vector-spaces with global duality-as-C^inf-modules – Kevin Casto Oct 25 '12 at 5:45