Partitions of Unity

In this section, we develop bump functions and partitions of unity, which are tools for patching together local smooth objects into global ones.

Partition of Unity

Partition of Unity

Let $M$ be a manifold, and $\mathcal{A}=\{U_{\alpha }{\}}_{\alpha \in A}$ an open cover of $M$. A (smooth) partition of unity subordinate to the cover $\mathcal{A}$ is a collection of functions $\{{\rho }_{\alpha }{\}}_{\alpha \in A}\subset C^{\infty }(M)$ satisfying:

1. For every $x\in M$ and every $\alpha \in A$, $0\le {\rho }_{\alpha }(x)\le 1$.
2. The support $\mathrm{supp}({\rho }_{\alpha })=\{x\in M:{\rho }_{\alpha }(x)\ne 0\}$ is contained in $U_{\alpha }$ for each $\alpha \in A$.
3. The set of supports $\{\mathrm{supp}({\rho }_{\alpha }):\alpha \in A\}$ is locally finite. That is, for every point $x\in M$, there exists an open neighborhood $U$ of $x$ in $M$ such that $\{\alpha :\mathrm{supp}({\rho }_{\alpha })\cap U\ne \varnothing \}$ is finite.
4. For each $x\in M$, ${\sum }_{\alpha \in A}{\rho }_{\alpha }(x)=1$.

Partition of Unity Theorem

Let $M$ be a manifold and $\mathcal{A}$ an open cover. Then there exists a smooth partition of unity subordinate to $\mathcal{A}$.

First, we will develop the basic building blocks for the construction, which are smooth compactly supported functions on Euclidean space:

Lemma

If $0<r_1<r_2$, then there exists $H_{r_1,r_2}\in C^{\infty }({\mathbb{R}}^n)$ such that

• $H(x)=1$ for $x\in \overline{B_{r_1}(0)}$,
• $0<H(x)<1$ for $x\in B_{r_2}(0)\setminus B_{r_1}(0)$, and
• $H(x)=0$ for $x\notin B_{r_2}(0)$.

Proof Let $f:\mathbb{R}\to \mathbb{R}$ satisfy $f(t)=\{\begin{array}{ll}{e}^{-1/t},& t>0,\\ 0,& t\le 0.\end{array}$Then the function $H:{\mathbb{R}}^n\to \mathbb{R}$ such that $H(x):=\frac{f(r_2-|x|)}{f(r_2-|x|)+f(|x|-r_1)}$will do the job. $\square$

Proof of theorem Step 1: Find a countable, locally finite refinement Find a countable, locally finite refinement of $\mathcal{A}$ consisting of coordinate balls of radius 2:

${\mathcal{B}}_2={\left\{{\phi }_i:U_i\to {\mathbb{R}}^n\right\}}_{i=1}^{\infty }$

where each $({\phi }_i,U_i)$ is a chart on $M$ such that ${\phi }_i(U_i)={\mathring{B}}_2(0)\subseteq {\mathbb{R}}^n$.

Let $B_i={\phi }_i^{-1}({\mathring{B}}_1(0))$. The collection of these inner balls, ${\mathcal{B}}_1=\{B_i{\}}_{i=1}^{\infty }$, is also an atlas (i.e., it covers $M$).

Because ${\mathcal{B}}_2$ is locally finite, for any $x\in M$, the number of coordinate balls containing $x$ is finite:

$\#\{i\mid x\in U_i\}<\infty$

Step 2: Verify the existence of such a refinement (using Compact Exhaustion)

• (i) If $M$ is compact: We can choose ${\mathcal{B}}_2$ such that it consists of finitely many charts whose corresponding inner balls $B_i$ cover $M$.

• (ii) If $M$ is not compact: We need to construct a compact exhaustion of $M$, which is a sequence of compact sets $K_1,K_2,\dots$ such that $M={\bigcup }_{j=1}^{\infty }{K}_j$ and $K_j\subseteq {\mathring{K}}_{j+1}$.

We already have a countable open cover of $M$ by precompact coordinate balls $\{B_i{\}}_{i=1}^{\infty }$. Define the first compact set as the closure of the first ball:

$K_1={\overline{B}}_1$ Because $\{B_i{\}}_{i=1}^{\infty }$ covers the entire manifold $M$, it naturally covers $K_1$. Since $K_1$ is compact, it admits a finite subcover. Let $m_1$ be the smallest integer such that this inclusion holds: $K_1\subseteq {\bigcup }_{i=1}^{m_1}{B}_i$ Now, we define the next compact set $K_2$ by $K_2={\bigcup }_{i=1}^{m_1}{\overline{B}}_i$ then $K_1\subseteq {\mathring{K}}_2$ We repeat this process inductively. Assume $K_j$ has been defined and is compact. Let $m_j$ be the smallest integer such that $K_j\subseteq {\bigcup }_{i=1}^{m_j}{B}_i$. We define the next layer as:

$K_{j+1}={\bigcup }_{i=1}^{m_j}{\overline{B}}_i,K_j\subseteq {\mathring{K}}_{j+1}$ Furthermore, since $m_j\ge j$, we are guaranteed that $B_j\subseteq K_{j+1}$, ensuring that the union ${\bigcup }_{j=1}^{\infty }{K}_j$ exhausts all balls, and therefore covers the entire manifold $M$. Step 3: Construct local bump functions For each chart $({\phi }_i,U_i)$, we find a standard smooth bump function $\rho \in C^{\infty }({\mathbb{R}}^n,\mathbb{R})$ such that:

$\rho (x)=\{\begin{array}{ll}1,& \Vert x\Vert \le 1\\ 0,& \Vert x\Vert \ge 2\end{array}$

and $\rho (x)\in [0,1]$ everywhere else. The actual function can be taken from previous lemma.

Step 4: Pull back the bump functions to the manifold Compose $\rho$ with the coordinate maps ${\phi }_i$ to get smooth functions on $M$, extended by 0 outside $U_i$

${\tilde{\rho }}_i=\rho \circ {\phi }_i\in C^{\infty }(M)$ These functions satisfy:

• ${\tilde{\rho }}_i(x)\in [0,1]\forall x\in M$

• ${\tilde{\rho }}_i{|}_{B_i}\equiv 1$

• $\text{supp}({\tilde{\rho }}_i)\subset U_i=2B_i$

Step 5: Normalize to obtain a partition of unity for the refinement

Define the normalized functions:

${\rho }_i(x)=\frac{{\tilde{\rho }}_i(x)}{{\sum }_{j=1}^{\infty }{\tilde{\rho }}_j(x)}$

This ${\rho }_i(x)$ is a partition of unity subordinated to ${\mathcal{B}}_2$.

(Note: The sum in the denominator is well-defined and smooth because the cover is locally finite, meaning only finitely many terms are non-zero near any point $x$, which is ensured by the construction of compact exhaustion for non-compact manifold. It is also strictly positive because the inner balls $B_i$ cover $M$, ensuring at least one ${\tilde{\rho }}_k(x)=1$.)

Step 6: Regroup for the original open cover

For the original open cover $\mathcal{A}=\{U_{\alpha }{\}}_{\alpha \in A}$, since ${\mathcal{B}}_2$ is a refinement, each $U_i$ is contained in some $U_{\alpha }$. We assign each index $i$ to exactly one such $\alpha$. Let $I_{\alpha }$ be the set of indices $i$ assigned to $U_{\alpha }$:

$U_{\alpha }={\bigcup }_{i\in I_{\alpha }}{U}_i$

Define the regrouped functions:

${\tilde{\rho }}_{\alpha }(x)={\sum }_{i\in I_{\alpha }}{\rho }_i(x)\in C^{\infty }(M)$

This collection $\{{\tilde{\rho }}_{\alpha }{\}}_{\alpha \in A}$ is the desired smooth partition of unity subordinated to the original cover $\mathcal{A}$.

Applications of Partitions of Unity

Existence of Bump Functions

Let $M$ be a smooth manifold. For any closed subset $A\subseteq M$ and any open subset $U$ containing $A$, there exists a smooth bump function for $A$ supported in $U$. That is, a smooth function $\rho :M\to \mathbb{R}$ such that $\rho (x)=1$ for $x\in A$ and $\text{supp}\rho \subset U$.

Proof Let $U_0=U$, $U_1=M\setminus A$, and $\mathcal{A}=\{U_0,U_1\}$. Then we can find a partition of unity $\{{\rho }_0,{\rho }_1\}$ subordinate to $\mathcal{A}$. Since ${\rho }_1(x)=0$ for $x\in A$, we have that ${\rho }_0(x)=1$ for $x\in A$. Moreover, the support of ${\rho }_0$ is contained in $U_0=U$. $\square$

Proposition

Smooth functions on a closed submanifold are restrictions. That is, if $\Sigma$ is a closed $k$-dimensional submanifold of $M$, then for any smooth function $f\in C^{\infty }(\Sigma )$, there exists $g\in C^{\infty }(M)$ such that $g{|}_{\Sigma }=f$.

Proof Step 1: Choose Slice Charts (Submanifold Charts) for $\Sigma$ Because $\Sigma$ is a regular submanifold, for every point $x\in \Sigma$, there exists a slice chart $(U_x,{\phi }_x)$ on $M$ such that:

${\phi }_x(U_x\cap \Sigma )={\phi }_x(U_x)\cap ({\mathbb{R}}^k\times \{0\})$ We can define a natural chart for the submanifold itself, denoted as ${\psi }_x:U_x\cap \Sigma \to {\tilde{A}}_x\subseteq {\mathbb{R}}^k$. This ${\psi }_x$ is essentially the restriction of ${\phi }_x$ to the submanifold, followed by a standard projection down to the first $k$ coordinates.

Step 2: Construct an Open Cover of $M$ and a Partition of Unity We need to cover the entire manifold $M$. To cover the submanifold $\Sigma$, we use the collection of local chart domains $\{U_x{\}}_{x\in \Sigma }$. To cover the rest of $M$, we define $U_0=M\setminus \Sigma$. Since $\Sigma$ is assumed to be a closed submanifold, $U_0$ is an open set in $M$. So, our complete open cover for $M$ is: $\mathcal{U}=\{U_x\mid x\in \Sigma \}\cup \{U_0\}$ By the Partition of Unity theorem, there exists a smooth PoU subordinated to this cover: $\{{\rho }_x{\}}_{x\in \Sigma }\cup \{{\rho }_0\}$ This guarantees that $\text{supp}({\rho }_x)\subseteq U_x$, $\text{supp}({\rho }_0)\subseteq U_0$, and for all $y\in M$: ${\sum }_{x\in \Sigma }{\rho }_x(y)+{\rho }_0(y)=1$ Step 3: Define Local Extensions of $f$ Given our function $f\in C^{\infty }(\Sigma )$, its restriction $f{|}_{U_x\cap \Sigma }:U_x\cap \Sigma \to \mathbb{R}$ is smooth. We want to extend this locally to the whole open set $U_x$. We define a local extension $f_x\in C^{\infty }(U_x)$ using the sequence of maps drawn in the first image: $U_x\stackrel{{\phi }_x}{\to }{\phi }_x(U_x)\subseteq {\mathbb{R}}^n\stackrel{\pi }{\to }{\tilde{A}}_x\subseteq {\mathbb{R}}^k\stackrel{{\psi }_x^{-1}}{\to }\Sigma \cap U_x\stackrel{f{|}_{\Sigma \cap U_x}}{\to }\mathbb{R}$ Here, $\pi$ is the projection that drops the last $n-k$ coordinates. So, $f_x(y)=f({\psi }_x^{-1}(\pi ({\phi }_x(y))))$. Crucial property: If $y$ is already on the submanifold (i.e., $y\in \Sigma \cap U_x$), its last $n-k$ coordinates in ${\mathbb{R}}^n$ are already zero. The projection $\pi$ does nothing to its position, meaning ${\psi }_x^{-1}(\pi ({\phi }_x(y)))=y$. Therefore, for any $y\in \Sigma \cap U_x$: $f_x(y)=f(y)$ Step 4: Glue the Local Extensions Together Now we define our global function $g:M\to \mathbb{R}$ by gluing the local extensions together using the bump functions:

$g(y)={\sum }_{x\in \Sigma }{\rho }_x(y)\cdot f_x(y)$ Smoothness check: For each term, ${\rho }_x(y)f_x(y)$ is smoothly defined on $U_x$. Because $\text{supp}({\rho }_x)$ is closed inside $U_x$, we can smoothly extend this term to the rest of $M$ by defining it to be $0$ outside $U_x$. Since the PoU is locally finite, this sum is locally finite, meaning $g\in C^{\infty }(M)$. (Note: We do not include ${\rho }_0$ in this sum because we want the extension to naturally drop to 0 outside the influence of $\Sigma$). Step 5: Verify that $g{|}_{\Sigma }=f$ Take any point $y\in \Sigma$. Because $y\in \Sigma$ and $\text{supp}({\rho }_0)\subseteq M\setminus \Sigma$, we must have ${\rho }_0(y)=0$. From the PoU property ($\sum {\rho }_x(y)+{\rho }_0(y)=1$), it follows that for $y\in \Sigma$, ${\sum }_{x\in \Sigma }{\rho }_x(y)=1$. If ${\rho }_x(y)\ne 0$, then $y\in \text{supp}({\rho }_x)\subseteq U_x$. Since $y$ is also in $\Sigma$, we have $y\in U_x\cap \Sigma$. As shown in Step 3, this means $f_x(y)=f(y)$. Now, substitute this into the definition of $g(y)$: $g(y)={\sum }_{x\in \Sigma }{\rho }_x(y)f_x(y)$ Since $f_x(y)=f(y)$ for all non-zero terms in the sum at point $y$: $g(y)={\sum }_{x\in \Sigma }{\rho }_x(y)f(y)=\left({\sum }_{x\in \Sigma }{\rho }_x(y)\right)f(y)$ Because the sum of the bump functions equals 1 at $y$: $g(y)=1\cdot f(y)=f(y)$ This confirms that $g{|}_{\Sigma }=f$.

Embedding Theorem

It is an important question whether an abstract manifold can always be realized as a submanifold of some Euclidean space. A simple case is when the manifold is compact. In this case, we can use the following theorem:

Whitney Embedding Theorem

Let $M$ be a compact smooth manifold of dimension m. Then for sufficiently large $N$ there exists a smooth embedding $F:M\to {\mathbb{R}}^N$.

Proof Step 1: Construct a finite atlas and a subordinate PoU Because $M$ is compact, we can cover it with a finite number of charts $\mathcal{A}=\{({\phi }_i,U_i){\}}_{i=1}^k$ such that each chart maps to a ball of radius 2: ${\phi }_i(U_i)={\mathring{B}}_2(0)\subseteq {\mathbb{R}}^m$. We choose this finite atlas such that the smaller pre-images $\{{\phi }_i^{-1}({\mathring{B}}_1(0)){\}}_{i=1}^k$ still form an open cover of $M$. Take a smooth Partition of Unity $\{{\rho }_i{\}}_{i=1}^k$ subordinate to this cover. Crucially, ${\rho }_i$ is strictly positive (nowhere vanishing) on the inner ball ${\phi }_i^{-1}({\mathring{B}}_1(0))$, and its support is contained within $U_i$. Step 2: Define the global map $F$ Define $F:M\to ({\mathbb{R}}^m{)}^k\times {\mathbb{R}}^k\cong {\mathbb{R}}^{k(m+1)}$ by packaging the charts and bump functions together: F(x) = \big( \rho_1(x)\varphi_1(x), \dots, \rho_k(x)\varphi_k(x), \rho_1(x), \dots, \rho_k(x) \big)$$$F$ is clearly $C^\infty$. Because $M$ is compact, [[Differential Calculus#^bc039a|any injective immersion is an embedding]]. Therefore, we only need to prove $F$ is injective and that its differential $DF|_x$ is injective. **Step 3: Prove $F$ is injective** Suppose $F(x) = F(y)$ for some $x, y \in M$. Matching the last $k$ components of $F$, we get $\rho_i(x) = \rho_i(y)$ for all $i$. Since the inner balls cover $M$, there is some index $i$ where $x \in \varphi_i^{-1}(\mathring{B}_1(0))$, meaning $\rho_i(x) \neq 0$. Thus, $\rho_i(x) = \rho_i(y) \neq 0$, which ensures both $x$ and $y$ are in $\text{supp}(\rho_i) \subset U_i$. Matching the corresponding first-half components gives $\rho_i(x)\varphi_i(x) = \rho_i(y)\varphi_i(y)$. Since $\rho_i(x) \neq 0$, we divide it out to get $\varphi_i(x) = \varphi_i(y)$. Because $\varphi_i$ is a coordinate chart (a bijection on $U_i$), it follows that $x = y$. **Step 4: Prove $F$ is an immersion (Injective Differential)** Suppose $DF|_x(v) = 0$ for some tangent vector $v \in T_xM$. We need to show $v = 0$. Since the differential applied to $F$ yields zero in all components, looking at the last $k$ components gives $D\rho_i|_x(v) = 0$ for all $i$. Again, choose an index $i$ such that $\rho_i(x) \neq 0$. Looking at the corresponding chart component, we have $D(\rho_i \cdot \varphi_i)|_x(v) = 0$. Applying the product rule: D\rho_i|_x(v) \cdot \varphi_i(x) + \rho_i(x) D\varphi_i|_x(v) = 0Substitute $D\rho_i|_x(v) = 0$ into the equation, and the first term vanishes. We are left with: \rho_i(x) D\varphi_i|_x(v) = 0$$Since ${\rho }_i(x)\ne 0$, it must be that $D{\phi }_i{|}_x(v)=0$. Because ${\phi }_i$ is a valid local coordinate chart, it is a local diffeomorphism, meaning its differential $D{\phi }_i{|}_x$ is strictly injective. This forces $v=0$. Therefore, $DF{|}_x$ is injective, making $F$ an immersion.