# 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.

## 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!