My teacher defined submanifold in $\mathbb{R}^n$ as:
a subset $S \subset \mathbb{R}^n$ is a submanifold of dimension $m$ of $\mathbb{R}^n$ if for every $ x $ there is an open neighborhood $U$ of $x$ and an application $f: U \rightarrow \mathbb{R}^m \oplus \mathbb{R}^{n-m}$ such that
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$f(U)$ is open in $\mathbb{R}^n$,
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$f: U \rightarrow f(U)$ is a diffeomorphism,
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$f(U\cap S) =f(U)\cap (\mathbb{R}^m \times \{0\})$ where $0 = (0,\cdots,0)$ $n-m$ times.
then he defined a coordinate system in the submanifold $S$: a coordinate system in $S$ is a pair $ (U, \phi) $ where
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$U \subset \mathbb{R}^m$ is open,
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$\phi: U \rightarrow \mathbb{R}^n$ is immersion,
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Exists open set $V \subset \mathbb{R}^n$ such that $\phi(U) = V \cap S$ and $\phi: U \rightarrow \phi(U)$ is a homeomorphism.
My question is whether with the definitions above the result below remains true, if it does, how to prove it?
Theorem: Let $S \subset \mathbb{R}^n$ is a submanifold of dimension $m$ of $\mathbb{R}^n$. If $x \in S$ and $(U_1, \phi_1)$, $(U_2, \phi_2)$ are coordinates systems such that $\phi_1(U_1) = V_1\cap S$ and $\phi_1(U_1)=V_2\cap S$ with $x \in V_1 \cap V_2$, then
$$\phi_2^{-1} \circ \phi_1: U_1 \cap \phi_1^{-1}(V_2) \rightarrow \mathbb{R}^m$$
is differentiable.
he made a sketch of proof using the definition of submanifold to conclude that in a small
neighborhood of $x$,
$$ \phi_2^{-1} \circ \phi_1 = (f \circ \phi_2)^{-1} \circ (f \circ \phi_1)$$
and its a composite of diferentiable functions. I agree that $(f \circ \phi_1)$ because $f$ is diffeomorphism by definition of submanifold and $\phi_1$ is differentiable. but I don't see why $(f \circ \phi_2)^{-1}$ be differentiable.
So I was wondering if the result is really true, or if there is a flaw in the demonstration, or even if I understood something incorrectly myself. Thanks for the help!
Remark: a manifold was defined as a set $M \neq \emptyset$ with a maximal atlas,
we did not impose that $ M $ be Hausdorff or second-countable, ideas of topology of the manifold came later. The atlas is defined as a collection of pairs $ (U_{\lambda}, x_{\lambda}) $ such that
- $\bigcup_{\lambda} U_{\lambda} = M,$ with $U_{\lambda} \subset M,$
- $x_{\lambda}: U_{\lambda} \rightarrow x_{\lambda}( U_{\lambda}) \subset \mathbb{R}^m $ are bijections with open sets in $\mathbb{R}^m $ and $x_{\lambda}( U_{\lambda} \cap U_{\beta})$ is open $\forall \lambda, \, \beta.$
- $x_{\beta} \circ x_{\lambda}^{-1}: x_{\lambda}( U_{\lambda} \cap U_{\beta}) \rightarrow x_{\beta}( U_{\lambda} \cap U_{\beta})$ are diffeomorphisms $\forall \lambda, \, \beta.$
Best Answer
In your question you define an atlas on a set $M$ as a collection of bijections $x_\lambda : U_\lambda \to V_\lambda \subset \mathbb R^n$ with open $V_\lambda$ such that suitable conditions are satisfied. This induces a unique topology on $M$ such that all $U_\lambda$ are open in $M$ all all $x_\lambda$ are homeomorphisms. Thus an alternative approach is to start with a topological space $M$ and define an atlas on $M$ as a collection of homeomorphisms $x_\lambda : U_\lambda \to V_\lambda$ between open subsets $U_\lambda$ of $M$ and open subsets $V_\lambda$ of $\mathbb R^n$ such that
Note that $x_{\lambda}( U_{\lambda} \cap U_{\beta})$ is automatically open in $\mathbb R^n$ since $x_\lambda$ is a homeomorphism and $U_{\lambda} \cap U_{\beta}$ is open in $M$. Moreover note that the $x_{\beta} \circ x_{\lambda}^{-1}$ are automatically diffeomorphisms because $(x_{\beta} \circ x_{\lambda}^{-1})^{-1} = x_{\lambda} \circ x_{\beta}^{-1}$ is also differentiable.
For $S$ you introduce the concept of a coordinate system $(U,\phi)$. The collection of all $\phi^{-1} : U' = \phi(U) \to U$ forms an differentiable atlas on the space $S$. Clearly the $\phi^{-1}$ are homeomorphisms between open subsets of $S$ and open subsets of $\mathbb R^m$ (recall that $\phi(U) = V \cap S$ with an open $V \subset \mathbb R^n$). It is easy to see that the $\phi(U)$ cover $S$, but that does not belong to your question. So let us show that the transition functions $\phi_{21} = \phi_2^{-1} \circ ( \phi_1^{-1})^{-1} = \phi_2^{-1} \circ \phi_1 : \phi_1^{-1}(U'_1 \cap U'_2) \to \phi_2^{-1}(U'_1 \cap U'_2)$ are differentiable. Recall that the $\phi_i : \phi_i^{-1}(U'_1 \cap U'_2) \to U'_1 \cap U'_2$ are homeomorphisms.
Given a coordinate system $(U,\phi)$ and $x \in U$, consider $p = \phi(x) \in S$. There exist an open neigborhood $O$ of $p$ in $\mathbb R^n$ and a diffeomorphism $f : O \to f(O) \subset \mathbb R^n$ such that $f(O\cap S) = f(O)\cap (\mathbb{R}^m \times \{0\})$; w.l.o.g. we may assume that $O \cap S \subset U' = \phi(U)$. Let $\pi : \mathbb R^n \to \mathbb R^m$ denote the projection onto the first $m$ coordinates and $j: \mathbb R^m \to \mathbb R^n, j(x) = (x,0)$. Define $$\phi^* : \phi^{-1}(O) \stackrel{\phi}{\to} O \stackrel{f}{\to} \mathbb R^n ,$$ $$\phi^\# = \pi \circ \phi^* : \phi^{-1}(O) \to \mathbb R^m ,$$ $$f^* : O \cap S \stackrel{f}{\to} f(O) \cap ( \mathbb R^m \times \{0\}) \stackrel{\pi}{\to} \pi(f(O) \cap ( \mathbb R^m \times \{0\}) = O^* .$$ Note that $f^*$ is a homeomorphism such that $f^*(\phi(\xi)) = \phi^\#(\xi)$. Moreover we have $j \circ \phi^\# = \phi^*$ because $\phi(\phi^{-1}(O)) \subset S$, thus $\phi^*(\phi^{-1}(O)) \subset \mathbb R^m \times \{0\}$. The map $\phi^\#$ is differentiable. Since $\phi$ is an immersion and $f$ is a diffeomorphism, the derivative $d\phi^*(x)$ of $\phi^*$ at $x$ has rank $m$. This implies that also $d\phi^\#(x)$ has rank $m$; if it were $< m$, then also $d\phi^*(x) = d(j \circ \phi^\#)(x)$ would have rank $< m$. This implies that $\phi^\#$ maps some open neigborhood $W$ of $x$ in $\phi^{-1}(O)$ diffeomorphically onto an open $W' \subset \mathbb R^m$.
Now let us come back to $\phi_{21}$. It suffices to find for each $x_1 \in \phi_1^{-1}(U'_1 \cap U'_2) \subset \mathbb R^m$ an open neigborhood $G$ in $\mathbb R^m$ such that $G \subset \phi_1^{-1}(U'_1 \cap U'_2)$ and $\phi_{21} \mid_G$ is differentiable. Let $p = \phi_1(x_1) \in U'_1 \cap U'_2 \subset S$. There exist an open neigborhood $O$ of $p$ in $\mathbb R^n$ and a diffeomorphism $f : O \to f(O) \subset \mathbb R^n$ such that $f(O\cap S) = f(O)\cap (\mathbb{R}^m \times \{0\})$; w.l.o.g. we may assume that $O \cap S \subset U'_1 \cap U'_2$. Because $f^*$ is a homeomorphism, we have $$\phi_{12} = (f^* \circ \phi_2)^{-1} \circ (f^* \circ \phi_1) .$$ Clearly $f^* \circ \phi_1 = \phi_1^\#$ is differentiable. Moreover $f^* \circ \phi_2 = \phi_2^\#$ maps some neigborhood of $x_2 = \phi_{21}(x_1)$ diffeomorphically onto some neigborhood of $(f^* \circ \phi_2)(x_2) = f^*(p)$, thus $(f^* \circ \phi_2)^{-1}$ is differentiable on some neighborhood of $f^*(p) = (f^* \circ \phi_1)(x_1)$. This gives us the desired $G$.