The Power of Compact Manifolds

We now turn our attention to apply topology to other higher-level fields in mathematics. A first discussion would be on manifolds, which introduces us to the field of study called differentiable manifolds, and even more generally, differentiable geometry.

We live in $3$ -dimensional space, commonly modelled as $\mathbb R^3$ . For finite brains, $\mathbb R^n$ still makes some mathematical sense, in that intuitions therein mostly align with intuitions in $\mathbb R^3$ .

In particular, if topological spaces resemble $\mathbb R^n$ , we can apply meaningful intuitions on abstract spaces by porting properties of $\mathbb R^n$ over to them.

Definition 1. Fix $m \in \mathbb N$ and topological space $K$ . We say that $K$ is an $m$ –manifold if it is Hausdorff, second countable, and locally $\mathbb R^m$ in the following sense: for any $x \in K$ , there exists a neighbourhood $V$ of $x$ that is homeomorphic to an open subset of $\mathbb R^m$ .

Example 1. Consider the continuous map $\phi : [-\pi, 3\pi] \to \mathbb R^2$ defined by $\phi(t) = (\cos(t), \sin(t))$ , so that $S^1 = \phi([-\pi, 3\pi])$ . Since $\phi|_{[0, 2\pi]}$ and $\phi|_{[-\pi,\pi]}$ are homeomorphisms onto its image $S^1$ , $S^1$ is a $1$ -manifold.

Notice that $S^1$ is not just a manifold, but a compact manifold, since it is closed and bounded in $\mathbb R^2$ by the Heine-Borel theorem. Compact manifolds are of special interest to us due to their effectively finite nature, which we will formalise later on.

Definition 2. For any map $\phi : K \to \mathbb R$ , the support of $\phi$ , denoted $\mathrm{supp}(\phi)$ , is the closure of the set $\phi^{-1}(\mathbb R \backslash \{0\})$ . Here, we do not require $\phi$ to be continuous.

Definition 3. Let $\{U_1,\dots,U_n\}$ be a finite-indexed open covering of $K$ . A finite-indexed collection $\{\phi_1,\dots,\phi_n\}$ of continuous maps $\phi_i : K \to [0, 1]$ is a partition of unity dominated by $\{U_i\}$ if the following two conditions are met:

$\mathrm{supp}(\phi_i) \subseteq U_i$ for any $i$ ,
$\sum_{i=1}^n \phi_i = 1$ .

Lemma 1. If $K$ is normal and Hausdorff, then for any finite-indexed open cover $\{U_1,\dots,U_n\}$ of $K$ , there exists a partition of unity dominated by $\{U_i\}$ .

Proof. Denote $\mathcal C_0 := \{U_1,\dots,U_n\}$ . For any finite-indexed open cover $\mathcal C = \{L_1,\dots,L_n\}$ of $K$ , define the closed set $M_i(\mathcal C) := K \backslash \bigcup_{j \neq i} L_j$ . For instance, $M_1(\mathcal C_0) = K \backslash \bigcup_{j=2}^n U_j$ . Since $M_1(\mathcal C_0)$ is a closed subset of $U_1$ , use the normal Hausdorffness of $K$ to obtain an open set $V_1$ such that $M_1(\mathcal C_0) \subseteq V_1 \subseteq \bar V_1 \subset U_1$ . Define $\mathcal C_1 := \{V_1,U_2,\dots,U_n\}$ .

Inductively define $\mathcal C_k := \{V_1,\dots,V_k,U_{k+1},\dots,U_n\}$ as follows: given $\mathcal C_{k-1}$ , define the closed set $M_k(\mathcal C_{k-1}) \subseteq U_k$ . Use the normal Hausforffness of $K$ to obtain an open set $V_k$ such that $M_k(\mathcal C_{k-1}) \subseteq V_k \subseteq \bar V_k \subset U_k$ . Define $\mathcal C_k := \{V_1,V_2,\dots,V_k,U_{k+1},\dots,U_n\}$ . In particular, $\mathcal C_n = \{V_1,\dots,V_n\}$ .

Thus, given the finite-indexed open cover $\{U_1,\dots,U_n\}$ of $K$ , we have obtained another finite-indexed open cover $\{V_1,\dots,V_n\}$ of $K$ such that $\bar V_i \subseteq U_i$ . Repeat the process to obtain a finite-indexed open cover $\{W_1,\dots,W_n\}$ of $K$ such that $\bar W_i \subseteq V_i$ .

We now take advantage of Urysohn’s lemma to construct our partition of unity. For each $i$ , since $\bar W_i$ and $K\backslash V_i$ are closed and disjoint, there exists a continuous function $\phi_i : K \to [0, 1]$ such that $\phi_i(W_i) = \{1\}$ and $\phi_i(K \backslash V_i) = \{0\}$ . Then $\phi_i^{-1}(\mathbb R \backslash \{0\}) \subseteq V_i$ , which means $\mathrm{supp}(\phi_i) \subseteq \bar V_i \subseteq U_i$ .

Now define the continuous map $\Phi := \sum_{i=1}^n \phi_i : K \to [0, 1]$ . Since $\Phi(x) \geq 1 > 0$ for any $x \in K$ , $\Phi > 0$ so that the maps $\psi_i := \phi_i/\Phi$ are well-defined. Then $\mathrm{supp}(\psi_i) = \mathrm{supp}(\phi_i) \subseteq U_i$ and

$\displaystyle \sum_{i=1}^n \psi_i = \sum_{i=1}^n \frac{\phi_i}{\Phi} = \frac{\sum_{i=1}^n \phi_i}{\Phi} = \frac{\Phi}{\Phi} = 1.$

By Definition 3, $\{\psi_i\}$ is the desired partition of unity dominated by $\{U_i\}$ .

Theorem 1. If $K$ is a compact $m$ -manifold, then $K$ can be imbedded in $\mathbb R^N$ for some positive integer $N$ . That is, local $\mathbb R^m$ -ness implies global $\mathbb R^N$ -ness.

Proof. Since $K$ is an $m$ -manifold, for any $x \in K$ , let $U_x$ be a neighbourhood of $x$ that is homeomorphic to $\mathbb R^m$ . Since $\{U_x : x \in K\}$ forms an open cover for the compact space $K$ , extract a finite sub-cover $\{U_1,\dots,U_n\}$ . For each $i$ , let $g_i : U_i \to \mathbb R^m$ be a given imbedding.

Since $K$ is compact and Hausdorff, it is normal, so that by Lemma 1 it has a partition of unity $\{\phi_i\}$ dominated by $\{U_i\}$ . Abbreviate $S_i := \mathrm{supp}(\phi_i) \subseteq U_i$ . For each $i$ , define the map $h_i : K \to \mathbb R^m$ by

$h_i(x) := \begin{cases} \phi_i(x) \cdot g_i(x) & \text{for}\, x \in U_i, \\ \mathbf 0 & \text{for}\, x \in K \backslash S_i. \end{cases}$

The map is well-defined since for any $x \in U_i \cap K \backslash S_i$ ,

$\phi_i(x) \cdot g_i(x) = 0 \cdot g_i(x) = \mathbf 0.$

Furthermore, it is continuous since it is defined piece-wise by continuous functions.

We claim that the map

$F : K \to \underbrace{ \mathbb R \times \cdots \times \mathbb R }_n \times \underbrace{ \mathbb R^m \times \cdots \mathbb R^m }_n \cong \mathbb R^{(m+1)n}$

defined by

$F(x) = (\phi_1(x),\dots,\phi_n(x), h_1(x),\dots,h_n(x))$

is the desired imbedding, so the conclusion of the theorem holds with $N = (m+1)n$ . To justify this claim, it is clear that $F$ is continuous, and since $F$ is compact, if we can prove that $F$ is injective, then $F : K \to F(K)$ is a continuous bijection, and $F^{-1} : F(K) \to K$ is continuous since for any closed set $C \subseteq K$ ,

$(F^{-1})^{-1}(C) = F(C)$

is compact and thus closed, since $C$ is compact.

Now we prove that $F$ is injective. Suppose $F(x) = F(y)$ . By definition, for any $i$ , $\phi_i(x) = \phi_i(y)$ and $h_i(x) = h_i(y)$ . Now, $x \in U_i$ for some $i$ so that $\phi_i(y) = \phi_i(x) > 0$ and thus $y \in U_i$ . Hence,

$\displaystyle g_i(x) = \frac{ h_i(x) }{ \phi_i(x) } = \frac{ h_i(y) }{ \phi_i(y) } = g_i(y).$

Since $g_i$ is an imbedding, we have $x = y$ , as required.

For this reason and many others, compact spaces are lovely, and our preceding discussions on compactness establish nontrivially vast situations in which we obtain compactness.

—Joel Kindiak, 8 Jun 25, 1854H

KindiakMath

The Power of Compact Manifolds

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The Power of Compact Manifolds

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