Definition of manifolds as submanifolds of $\mathbb{R}^m$

definitiondifferential-geometrysmooth-manifolds

I'm having trouble understanding the definition of coordinate charts of manifolds given in the book "Analysis II" by Herbert Amann and Joachim Escher. They define manifolds as submanifolds of $\mathbb{R}^m$ (which is equivalent to the abstract aproach by the Whitney embedding theorem right?).

The definition of a submanifold I learned is:

Let $M$ be an $m$-dimensional manifold. We say that a subset $L \subseteq M$ is a submanifold of $M$ of dimension $n$, if for every point $p \in L$ there exists an adapted chart $\phi: U \rightarrow V' \times V''$ with $U \subseteq L$ open in $L$, $V' \subseteq \mathbb{R}^n$ open in $\mathbb{R}^n$ and $V'' \subseteq \mathbb{R}^{m-n}$ open in $\mathbb{R}^{m-n}$ such that $\phi (U \cap L) = V' \times \{0\}$ with $0 \in \mathbb{R}^{m-n}$ Then $L$ becomes an $n$-dimensional manifold of itself with the smooth atlas induced by the restricted charts $\phi : U \cap L \rightarrow V'$.

The authors of the book define a submanifold of $\mathbb{R}^m$ as follows

A subset $L \subseteq \mathbb{R}^m$ is called an $n$-dimensional submanifold of $\mathbb{R}^m$ if for every point $p \in L$ there exists an open set $U \subseteq \mathbb{R}^m$ containing $p$ and an open subset $V\subseteq \mathbb{R}^m$ together with a diffeomorphism $\phi$ from $U$ to $V$ such that $\phi(M \cap U)=V \cap (\mathbb{R}^n \times \{0\})$ with $0 \in \mathbb{R}^{m-n}$.

,which directly coincides with the book's definition if one views $\mathbb{R}^m$ as an $m$-dimensional manifold with the smooth atlas induced by the universal chart $(\mathbb{R}^m, id)$.
Then the authors define coordinate charts of an $n$-dimensional submanifold $L$ of $\mathbb{R}^m$ around a point $p \in L$ as follows

Let $\phi: U \rightarrow V$ be a homeomorphism from an open set $U \subseteq L$ in the subspace topology containing $p$ to an open subset $V$ of $\mathbb{R}^n$ such that $ i_M \ \circ \ \phi ^{-1}$ is a $C^\infty$-immersion, where $i_M$ is the cannonical injection from $L$ to $\mathbb{R}^m$.

How is this equivalent to the restricted charts given in the first definition? Why does $ i_M \ \circ \ \phi^{-1}$ need to be an immersion? I'm really having trouble connecting the abstract aproach to manifolds to the way of defining manifolds as submanifolds of $\mathbb{R}^m$. What's a good way to think about the approach taken by the book?

Best Answer

How is this equivalent to the restricted charts given in the first definition?

It is clear that restrictions from the previous definition satisfy the second definition of charts. Conversely, suppose that $\phi: U \to V$ is as in the second definition: we need to extend $\phi$ as a diffeomorphism on a neighborhood of $p$ in $\mathbb R^m$. For simplicity, let $\psi:=i_M \circ \phi^{-1}$. By the immersion theorem, up to restricting $U$ and $V$, there are diffeomorphisms $f: \tilde U \to \mathbb R^m$ and $g: V \to \mathbb R^n$ such that $\psi(x)=f^{-1}(g(x),0)$, where $\tilde U$ is an open subset of $\mathbb R^m$ such that $\tilde U \cap L=U$. Now it is enough to extend $\psi$ by $\psi(x,t):=f^{-1}(g(x),t)$, for $t \in \mathbb R^{m-n}$ in a ball of small enough radius. You can check that $\psi$ is a diffeomorphism onto its image, and that $\psi^{-1}$ is the required extension of $\phi$.

Why does $i_M \circ \phi^{-1}$ need to be an immersion?

Answer: because it makes the proof above work. If it is not, here's a counter example. Take $m=2$, $n=1$, and $L=\{ (x,y) : x^2=y^3 \}$. Take $U=L$ and $\phi: L \to \mathbb R$ be defined by $\phi(x,y)=x^{1/3}$. Then $i_M \circ \phi^{-1}(t)=(t^3, t^2)$ is $C^\infty$ but not an immersion at $t=0$. If you look at a picture of $L$, you will notice that there is a cusp at $(x,y)=(0,0)$: it is not a smooth submanifold of $\mathbb R^2$ (although it is a topological submanifold, homeomorphic to $\mathbb R$).