Submanifold

Submanifold of ${\mathbb{R}}^n$

A subset $S\subseteq {\mathbb{R}}^n$ is a $k$-dimensional submanifold of ${\mathbb{R}}^n$, if for every $x\in S$ there exists an open set $U\subseteq {\mathbb{R}}^n$ containing $x$ and a diffeomorphism $\phi :U\to {\mathbb{R}}^n$ such that $\phi (U\cap S)=\phi (U)\cap ({\mathbb{R}}^k\times \{0\})$, with $0\in {\mathbb{R}}^{n-k}$. We call such a map $\phi$ a submanifold chart for $S$ at $x$.

e.g.

Proposition

Any $k$-dimensional submanifold $S$ of ${\mathbb{R}}^n$ is a smooth manifold, if equipped with the subspace topology inherited from ${\mathbb{R}}^n$ and the atlas of charts given as follows: If ${\Psi }_{\alpha }:U_{\alpha }\to V_{\alpha }$ are a collection of submanifold charts for $\Sigma$ with ${\bigcup }_{\alpha }{U}_{\alpha }\supset \Sigma$, then we take $\mathcal{A}=\{{\phi }_{\alpha }:=\pi \circ {\Psi }_{\alpha }{|}_{\Sigma \cap U_{\alpha }}\}$, where $\pi$ is the projection from ${\mathbb{R}}^k\times \{0\}$ to ${\mathbb{R}}^k$.

Proof The given maps are homeomorphisms to their images: ${\Psi }_{\alpha }$ is a diffeomorphism, so the restriction to $\Sigma \cap U_{\alpha }$ is a homeomorphism to its image which is a subset of ${\mathbb{R}}^k\times \{0\}$. $\pi$ is a homeomorphism from ${\mathbb{R}}^k\times \{0\}$ to ${\mathbb{R}}^k$ since it is smooth and has a smooth right-inverse ${\pi }^{-1}(x)=(x,0)$. Therefore ${\phi }_{\alpha }$ is a homeomorphism to its image.

Next we check smoothness: Note that ${\phi }_{\alpha }^{-1}={\Psi }_{\alpha }^{-1}{|}_{({\mathbb{R}}^k\times \{0\})\cap V_{\alpha }}\circ {\pi }^{-1}$, which is smooth as a map from ${\mathbb{R}}^k$ into ${\mathbb{R}}^n$. Therefore the transition map ${\phi }_{\beta }\circ {\phi }_{\alpha }^{-1}=\pi \circ {\Psi }_{\beta }\circ {\Psi }_{\alpha }^{-1}\circ {\pi }^{-1}$ is a smooth map into ${\mathbb{R}}^k$. Since this is also true with $\alpha$ and $\beta$ exchanged, the inverse map is also smooth and the transition map is therefore a diffeomorphism.

There are several equivalent ways to define submanifolds:

Proposition

Let $\Sigma$ be a subset of ${\mathbb{R}}^n$. The following are equivalent:

1. $\Sigma$ is a $k$-dimensional submanifold of ${\mathbb{R}}^n$;
2. $\Sigma$ is locally the graph of a smooth function. That is, for every $x\in \Sigma$, there is a neighbourhood $U$ of $x$ in ${\mathbb{R}}^n$, a linear injection $\iota :{\mathbb{R}}^k\to {\mathbb{R}}^n$ and a complementary linear injection ${\iota }^{\perp }:{\mathbb{R}}^{n-k}\to {\mathbb{R}}^n$,
   an open subset $A\subset {\mathbb{R}}^k$, and a smooth map $u:A\to {\mathbb{R}}^{n-k}$ such that
   $\Sigma \cap U=\{\iota (z)+{\iota }^{\perp }(u(z)):z\in A\}.$
3. $\Sigma$ is locally the level set of a submersion. That is, for every $x\in \Sigma$, there is a neighbourhood $U$ of $x$ in ${\mathbb{R}}^n$ and a submersion $G:U\to {\mathbb{R}}^{n-k}$ such that
   $\Sigma \cap U=G^{-1}(0)=\{y\in U:G(y)=0\}.$
4. $\Sigma$ is locally the image of an embedding. That is, for every $x\in \Sigma$, there exists a neighbourhood $U$ of $x$ in ${\mathbb{R}}^n$, an open set $A\subset {\mathbb{R}}^k$, and a smooth embedding $F:A\to U$ such that $\Sigma \cap U=F(A)$.

Embedded Submanifold

Suppose $M$ is a smooth manifold with or without boundary. An embedded (or regular) submanifold of $M$ is a subset $S\subset M$ that is a manifold (without boundary) in the subspace topology, endowed with a smooth structure with respect to which the inclusion map $S\hookrightarrow M$ is a smooth embedding.

Smooth Chart Lemma

Let $M$ be a set, and $\mathcal{A}=\{{\phi }_{\alpha }:U_{\alpha }\to V_{\alpha }{\}}_{\alpha \in A}$ a collection of maps, where each ${\phi }_{\alpha }$ is a map from a subset $U_{\alpha }$ to an open subset $V_{\alpha }$ of ${\mathbb{R}}^n$. Suppose that the following conditions hold: (i). Each ${\phi }_{\alpha }$ is a bijection from $U_{\alpha }$ to $V_{\alpha }$; (ii). For each $\alpha ,\beta \in A$ the set ${\phi }_{\alpha }(U_{\alpha }\cap U_{\beta })$ is open in ${\mathbb{R}}^n$; (iii). For each $\alpha ,\beta \in A$, the transition map ${\phi }_{\beta }\circ {\phi }_{\alpha }^{-1}:{\phi }_{\alpha }(U_{\alpha }\cap U_{\beta })\to {\phi }_{\beta }(U_{\alpha }\cap U_{\beta })$ is smooth; (iv). There is a countable subset $A_0\subset A$ such that ${\bigcup }_{\alpha \in A_0}{U}_{\alpha }=M$; (v). For any distinct points $p$ and $q$ in $M$, either there exists $\alpha \in A$ with $p,q\in U_{\alpha }$, or there exist $\alpha ,\beta \in A$ with $U_{\alpha }\cap U_{\beta }=\varnothing$ and $p\in U_{\alpha }$, $q\in U_{\beta }$. Then there is a unique differentiable structure on $M$ in which $\mathcal{A}$ is a smooth atlas.