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The way I understand things, roughly speaking, the importance of smooth manifolds is that they form the category of topological spaces on which we can do calculus. The definition of smooth manifolds requires that they be paracompact. I've looked all over, but I haven't found a clean statement for how paracompactness is a necessary condition to do calculus.

I understand that, by a theorem of Stone, every metric space is paracompact, but I'm not sure why we need global metrizability either.

Question: In what sense is paracompactness exactly the right condition to impose on a topological manifold to allow us to do calculus on it? Is there some theorem of the form "X has [some structure we strictly need in calculus] if and only if it is paracompact"?
Daniel Moskovich
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  • What is the definition of a smooth manifold that you're working with? I've learned that smooth manifolds are topological manifolds with a smooth structure by a certain construction. All topological manifolds are metrizable, and as you said thus paracompact. Paracompactness as it self never was implicitly in the definition of a smooth manifold in my textbooks though. – T. Eskin Jan 11 '12 at 09:41
  • Indeed: A locally Euclidean Hausdorff (LEH) space is metrisable iff it is paracompact, and if it's second countable then it's paracompact. Hence there are at least 3 different definitions of topological manifold commonly in use: LEH, paracompact LEH and second countable LEH. (Wikipedia). – kahen Jan 11 '12 at 10:01
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    Dear @kahen: formally the distinction between paracompact LEH and second countable LEH is correct but it is also completely artificial: paracompact LEH is equivalent to second countable LEH unless the manifold has more than countably many connected components, in which case it obviously cannot be second countable but might be paracompact. – Georges Elencwajg Jan 11 '12 at 14:43
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    You do not need paracompactness to have well-defined derivatives, notably Serre develops the theory of analytic manifolds and Lie groups without this requirement. – Aleksei Averchenko May 13 '12 at 15:32

3 Answers3

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A) For a differential manifold $X$ the following are equivalent:

a) X is paracompact
b) X has differentiable partitions of unity
c) X is metrizable
d) Each connected component of X is second countable
e) Each connected component of X is $\sigma$-compact

Partitions of unity are a fundamental tool in all of differential geometry (cf. kahen's answer) and would suffice to justify these conditions but the other equivalent properties can also be quite useful .

B) However occasionally non paracompact manifolds have been studied too. For example:

1) In dimension $1$ you have the long line obtained roughly by taking the first uncountable ordinal set and adding open segments $(0,1)$ between its successive points.
2) In dimension $2$ there exist non paracompact differentiable surface ( Prüfer and Radò). However every Riemann surface, that is a holomorphic manifold of complex dimension $1$ and thus real dimension $2$, is automatically paracompact.
3) Calabi and Rosenlicht have introduced a complex manifold of complex dimension $2$ which is not paracompact .

Edit As an answer to Daniel's question in the comments below, here are a few random examples of consequences of the existence of partitions of unity on a differential manifold $M$ of dimension $n$.

$\bullet$ If $M$ is orientable it has an everywhere non-vanishing differential form $\omega\in \Omega^n(M)$ of degree $n$.
$\bullet$ If $M$ is oriented you can define the integral $\int_M\eta$ of any compactly supported differential form $\eta\in \Omega^n_c(M)$ of degree $n$.
$\bullet$ The manifold $M$ can be endowed with a Riemannian metric.
$\bullet$ Every vector bundle on $M$ is isomorphic to its dual bundle.
$\bullet$ Every subbundle of a vector bundle on $M$ is a direct summand.

A sophisticated point of view (very optional !)
All sheaves of $C^\infty_M$-modules (for example locally free ones, which correspond to vector bundles) are acyclic in the presence of partitions of unity.
This has as a consequence that paracompact manifolds behave like affine algebraic varieties or Stein manifolds in that you can apply to them the analogue of Cartan-Serre's theorems A and B.
This is, in my opinion, the deep reason for the usefulness of partitions of unity on a manifold. (The last bullet for example was directly inspired from its analogue on affine varieties or Stein manifolds)

Georges Elencwajg
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    Just to be clear: In your definition of a differential manifold $X$, you are requiring that $X$ be Hausdorff and locally euclidean, but not necessarily second-countable, is that right? – Jesse Madnick Jan 11 '12 at 13:27
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    Dear @Jesse: that's exactly right. – Georges Elencwajg Jan 11 '12 at 14:30
  • This is what I was looking for- thanks! Is there an easily-stated reason that partitions of unity are an essential tool for doing calculus on manifolds? (and couldn't be traded for something weaker?) – Daniel Moskovich Jan 11 '12 at 23:45
  • Dear @Daniel, I have added an edit giving examples of applications of partitions of unity. – Georges Elencwajg Jan 12 '12 at 00:29
  • Thanks for this! I'm thinking about this, and I'm not fully understanding yet. Why is a volume form essential? Aren't there other ways we might construct a measure with respect to which we could integrate? And couldn't the integral of aa compactly supported function be defined as the integral over its support? My thought is along the lines of the Tietze Extension Theorem... somehow extending functions over the whole space is essential if we ever want to do anything global. Does this depend on partitions of unity in an essential way? Further thinking about the "sophisticated point of view"... – Daniel Moskovich Jan 12 '12 at 01:11
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    Dear @Daniel, the pleasantly compact booklet Calculus on Manifolds by Michael Spivak explains why you have to use differential forms if you want to arrive at Stokes' theorem. You will see there in detail how partitions of unity are used to that end. The crucial point is that differential forms have built into them the change of variable formula for integrals on open subsets of $\mathbb R^n$. – Georges Elencwajg Jan 12 '12 at 01:33
  • Dear @GeorgesElencwajg, what are some easy consequences of the vanishing of higher sheaf cohomology of $C^\infty_M$-modules? – Arrow Dec 18 '16 at 00:32
  • @Arrow: for example that smooth complex line bundles are classified by $H^2(M,\mathbb Z)$ [Long exact sequence in cohomology associated to the exponential sequence $0\to \mathbb Z \to C^\infty_M \stackrel {exp 2i\pi(\cdot)}{\to} (C^\infty _M) ^*\to 0)$] – Georges Elencwajg Dec 18 '16 at 08:17
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Because paracompactness is needed to prove the existence of smooth partitions of unity subordinate to any open covering.

For example as stated in Theorem 2.25 (Existence of Partitions of Unity) in Lee's "Introduction to Smooth Manifolds":

If $M$ is a smooth manifold and $\mathcal X = \{X_\alpha\}_{\alpha \in A}$ is any open cover of $X$, there exists a smooth partition of unity subordinate to $\mathcal X$.

EDIT: It occurs to me that I should probably also state the definition of partition of unity (from earlier on the same page):

Now let $M$ be a topological space, and let $\mathcal X = \{X_\alpha\}_{\alpha \in A}$ be an arbitrary open cover of $M$. A partition of unity subordinate to $\mathcal X$ is a collection of continuous functions $\{\psi_\alpha: M \to \mathbb R\}_{\alpha \in A}$ with the following properties:
(i) $0 \leq \psi_\alpha(x) \leq 1$ for all $\alpha \in A$ and all $x\in M$.
(ii) $\operatorname{supp} \psi_\alpha \subset X_\alpha$.
(iii) The set of supports $\{\operatorname{supp} \psi_\alpha\}_{\alpha\in A}$ is locally finite.
(iv)$\sum_{\alpha\in A}\psi_\alpha(x) = 1$ for all $x \in M$.

He then goes on to prove the extension lemma and the existence of bump functions as well as exhaustion functions.

kahen
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  • Thanks- this must be it. It turns out that some authors define paracompact to be "admits a partition of unity"- e.g. http://www.math.utk.edu/~dydak/myWebPage/myinfo/preprints/CalcOfPU.pdf I'm still not entirely clear on why partitions of unity are essential though- see my comment to Georges Elencwajg's answer. – Daniel Moskovich Jan 11 '12 at 23:47
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I've thought about this a bit myself, and I think that, calculus-wise, a key point might be that a locally Euclidean space has to be paracompact in order to be completely metrizable. I think that a complete metric is indispensible to any attempt to do calculus, because without it, the analogues of the IVT (which is a direct consequence of completeness of the reals) and MVT, and Stokes Theorem (if you could even formulate it!), would fail.

Daniel Moskovich
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