Every Manifold is Paracompact

Introduction

In their paper Betancourt et al. (2014), the authors give a corollary which starts with the phrase “Because the manifold is paracompact”. It wasn’t immediately clear why the manifold was paracompact or indeed what paracompactness meant although it was clearly something like compactness which means that every cover has a finite sub-cover.

It turns out that every manifold is paracompact and that this is intimately related to partitions of unity.

Most of what I have written below is taken from some hand-written anonymous lecture notes I found by chance in the DPMMS library in Cambridge University. To whomever wrote them: thank you very much.

Limbering Up

Let \{U_i : i \in {\mathcal{I}}\} be an open cover of a smooth manifold M. A partition of unity on M, subordinate to the cover \{U_i : i \in {\mathcal{I}}\} is a finite collection of smooth functions

\displaystyle   X_j : M^n \longrightarrow \mathbb{R}_+

where j = 1, 2, \ldots N for some N such that

\displaystyle   \sum_{j = 0}^N X_j(x) = 1 \,\mathrm{for all}\, x \in M

and for each j there exists i(j) \in {\mathcal{I}} such that

\displaystyle   {\mathrm{supp}}{X_j} \subset U_{i(j)}

We don’t yet know partitions of unity exist.

First define

\displaystyle   f(t) \triangleq  \begin{cases}  0            & \text{if } t \leq 0 \\  \exp{(-1/t)} & \text{if } t > 0 \\  \end{cases}

Techniques of classical analysis easily show that f is smooth (t=0 is the only point that might be in doubt and it can be checked from first principles that f^{(n)}(0) = 0 for all n).

Next define

\displaystyle   \begin{aligned}  g(t) &\triangleq \frac{f(t)}{f(t) + f(1 - t)} \\  h(t) &\triangleq g(t + 2)g(2 - t)  \end{aligned}

Finally we can define F: \mathbb{R}^n \rightarrow \mathbb{R} by F(x) = h(\|x\|). This has the properties

  • F(x) = 1 if \|x\| \leq 1
  • 0 \leq F(x) \leq 1
  • F(x) = 0 if \|x\| > 2

Now take a point p \in M centred in a chart (U_p, \phi_p) so that, without loss of generality, B(0,3) \subseteq \phi_p(U_p) (we can always choose r_p so that the open ball B(0,3r_p) \subseteq \phi'_p(U_p) and then define another chart (U_p, \phi_p) with \phi_p(x) = \phi'_p(x)/\|x\|).

Define the images of the open and closed balls of radius 1 and 2 respectively

\displaystyle   \begin{aligned}  V_p &= \phi_p^{-1}(B(0,1)) \\  W_p &= \phi_p^{-1}\big(\overline{B(0,2)}\big) \\  \end{aligned}

and further define bump functions

\displaystyle   \psi_p(y) \triangleq  \begin{cases}  F(\phi_p(y)) & \text{if } y \in U_p\\  0            & \text{otherwise} \\  \end{cases}

Then \psi_p is smooth and its support lies in W_p \subset U_p.

By compactness, the open cover \{V_p : p \in M\} has a finite subcover \{V_{p_1},\ldots,V_{p_K}\}. Now define

\displaystyle   X_j : M^n \longrightarrow \mathbb{R}_+

by

\displaystyle   X_j(y) = \frac{\psi_{p_j}(y)}{\sum_{i=1}^K \psi_{p_i}(y)}

Then X_j is smooth, {\mathrm{supp}}{X_j} = {\mathrm{supp}}{\psi_{p_j}} \subset U_{p_j} and \sum_{j=1}^K X_j(y) = 1. Thus \{X_j\} is the required partition of unity.

Paracompactness

Because M is a manifold, it has a countable basis \{A_i\}_{i \in \mathbb{N}} and for any point p, there must exist A_i \subset V_p with p \in A_i. Choose one of these and call it V_{p_i}. This gives a countable cover of M by such sets.

Now define

\displaystyle   L_1 = W_{p_1} \subset V_{p_1} \cup V_{p_2} \cup \ldots \cup V_{p_{i(2)}}

where, since L_1 is compact, V_{p_2}, \ldots, V_{p_{i(2)}} is a finite subcover.

And further define

\displaystyle   L_n = W_{p_1} \cup W_{p_2} \cup \ldots \cup W_{p_{i(n)}}        \subset        V_{p_1} \cup V_{p_2} \cup \ldots \cup V_{p_{i(n+1)}}

where again, since L_n is compact, V_{p_{i(n)+1}}, \ldots, V_{p_{i(n+1)}} is a finite subcover.

Now define

\displaystyle   \begin{aligned}  K_n &= L_n \setminus {\mathrm{int}}{L_{n-1}} \\  U_n &= {\mathrm{int}}(L_{n+1}) \setminus L_n  \end{aligned}

Then K_n is compact, U_n is open and K_n \subset U_n. Furthermore, \bigcup_{n \in \mathbb{N}} K_n = M and U_n only intersects with U_{n-1} and U_{n+1}.

Given any open cover {\mathcal{O}} of M, each K_n can be covered by a finite number of open sets in U_n contained in some member of {\mathcal{O}}. Thus every point in K_n can be covered by at most a finite number of sets from U_{n-1}, U_n and U_{n+1} and which are contained in some member of {\mathcal{O}}. This is a locally finite refinement of {\mathcal{O}} and which is precisely the definition of paracompactness.

To produce a partition of unity we define bump functions \psi_j as above on this locally finite cover and note that locally finite implies that \sum_j \psi_j is well defined. Again, as above, define

\displaystyle   X_j(y) = \frac{\psi_{j}(y)}{\sum_{i=1} \psi_{i}(y)}

to get the required result.

Bibliography

Betancourt, M. J., Simon Byrne, Samuel Livingstone, and Mark Girolami. 2014. “The Geometric Foundations of Hamiltonian Monte Carlo,” October, 45. http://arxiv.org/abs/1410.5110.

Advertisements

One thought on “Every Manifold is Paracompact

  1. Interesting. Every time I saw the definition of manifold at undergrad it was always paracompact by definition, but paracompact and second-countable were the kind of terms that weren’t examined too closely and lecturers often stated it wasn’t quite a precise definition. Admittedly I never took the actual “manifolds” course. Thanks for the detail!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s