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From Do Carmo:

Let $S_1$, $S_2$ be regular surfaces. Suppose $S_1\subset V\subset \mathbb{R}^3$ and $\varphi:V\rightarrow \mathbb{R}^3$ is a differentiable map such that $\varphi(S_1)\subset S_2$. Then the restriction \begin{align*} \varphi\big|_{S_1}:S_1\rightarrow S_2 \end{align*} is differentiable.

The statement seems obvious but I am having trouble proving it. Perhaps I am thinking about it too hard. I am comfortable with teh statement: 'The restriction of a smooth map is also smooth' - but this is in terms of smooth maps from Euclidean space to Euclidean space.

In order to show $\hat{\varphi}:=\varphi|_{S_1}$ is differentiable at some point $p\in S_1$ (when considered as a map between two regular surfaces), we must show that given parameterisations \begin{align*} \textbf{x}_1:U_1\subset\mathbb{R}^2&\rightarrow S_1\\ \textbf{x}_2:U_2\subset\mathbb{R}^2&\rightarrow S_2 \end{align*} such that $p\in\textbf{x}_1(U_1)$ and $\varphi(p)\in\textbf{x}_2(U_2)$, that the map \begin{align*} \textbf{x}_2^{-1}\circ {\varphi}\circ \textbf{x}_1:U_1\rightarrow\mathbb{R}^2 \end{align*} is a differentiable map between Euclidean space at $\textbf{x}_1^{-1}(p)$.

So how would one show this? The immediate 'argument' I think of is chain rule - 'the composition of differentiable maps is differentiable' -- We have that $\textbf{x}_1$ is a differentiable map between Euclidean space by definition, as is $\varphi$. However, the map $\textbf{x}_2^{-1}$ is differentiable in the sense of a map between two regular surfaces ($S_2$ and the $xy$-plane). So can we really employ the chain rule here given that $\textbf{x}_2^{-1}$ is defined as only a map from a regular surface to $\mathbb{R}^2$?

What am I missing?


Edit: From the advice below, I have tried to adapt the argument on Pg 70 of Do Carmo: is this a suitable justification that \begin{align*} \textbf{x}_2^{-1}\circ\varphi\circ\textbf{x}_1 \end{align*} is differentiable at $\textbf{x}_1^{-1}(p )$?

Given that $\varphi(\textbf{x}_1(U_1))\subset\textbf{x}_2(U_2)$, it follows there exists $q\in U_2$ such that $\varphi(p )=\textbf{x}_2(q)$.

Extend the map $\textbf{x}_2=(x_2(u,v),y_2(u,v),z_2(u,v))$ to a map $\tilde{\textbf{x}}_2:U_2\times \mathbb{R}\rightarrow \mathbb{R}^3$ defined by \begin{align*} \tilde{\textbf{x}}_2:(u,v,t)\rightarrow (x_2,y_2,z_2+t). \end{align*} Then $\tilde{\textbf{x}}_2$ is differentiable and $\tilde{\textbf{x}}_2|_{U_2\times\{0\}}=\textbf{x}_2$. The determinant of $d\tilde{\textbf{x}}_2(q)\neq 0$. Thus, by the inverse function theorem, there exists a neighbourhood $M$ ($\subset\mathbb{R}^3$) of $\textbf{x}_2(q)=\varphi(p )$ such that $\tilde{\textbf{x}}_2^{-1}$ exists AND is differentiable on $M$ (as a map between Euclidean spaces). Define $N:=\textbf{x}_2(U_2)\cap M$. Then $\tilde{\textbf{x}}_2^{-1}|_N$ is smooth (as a map between a subset of $\mathbb{R}^3$ and $U_2\times \mathbb{R}$ (restriction of smooth map is smooth)). If $\pi$ is the projection on to the first two factors, then $\pi$ is smooth and thus so is $\pi\circ \tilde{\textbf{x}}_2^{-1}|_N=\textbf{x}_2^{-1}$. Thus we have \begin{align*} \textbf{x}_2^{-1}\circ\varphi\circ\textbf{x}_1=\pi\circ\tilde{\textbf{x}}_2^{-1}\circ\varphi\circ\textbf{x}_1 \end{align*} This map on the right hand side is a composition of smooth maps between Euclidean spaces. Thus, by the chain rule, it is smooth.

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I'm assuming you are working with the definitions given in Do Carmo's "Curves and Surfaces" book and that $V \subseteq \mathbb{R}^3$ is an open set. First, let us write precisely the domains and ranges of the maps involved:

  1. The map $\mathbf{x}_1 \colon U_1 \rightarrow \mathbb{R}^3$ is classically differentiable. Here, $U_1 \subseteq \mathbb{R}^2$ is an open set and $\mathbf{x}_1(U_1) \subseteq S_1$.
  2. The map $\varphi \colon U \rightarrow \mathbb{R}^3$ is classically differentiable. Here, $U \subseteq \mathbb{R}^3$ is an open set and $\varphi(S_1) \subseteq S_2$.
  3. The map $\mathbf{x}_2 \colon U_2 \rightarrow \mathbb{R}^3$ is classically differentiable. Here, $U_2 \subseteq \mathbb{R}^2$ is an open set and $\mathbf{x}_2(U_2) \subseteq S_2$.

As you have noted, $\mathbf{x}_2^{-1} \colon \mathbf{x}_2(U_2) \rightarrow U_2$ is not classically differentiable but is differentiable in the sense of a map between a regular surface $\mathbf{x}_2(U_2)$ and an open set in $\mathbb{R}^2$ (if you want, you can embed $\mathbb{R}^2$ in $\mathbb{R}^3$ and consider $U_2$ as a regular surface in the sense of Do Carmo and then $\mathbf{x}_2^{-1}$ will be differentiable in the sense of a map between two regular surfaces but you won't gain much from it).

Thus, you have two options:

  1. Prove that $\varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))} \colon U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2))) \rightarrow \mathbf{x}_2(U_2)$ is a differentiable map between an open subset in $\mathbb{R^2}$ and a regular surface. Then, if you already proved this, apply the chain rule for maps between open subsets and regular surfaces to deduce that $\mathbf{x}_2^{-1} \circ \varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))}$ is smooth as a map between open subsets (or as a map between regular surfaces - the two notions concide).
  2. Show that one can extend the map $\mathbf{x}_2^{-1}$ to a map $\hat{\mathbf{x}}_2^{-1} \colon V \rightarrow \mathbb{R}^2$ where $\mathbf{x}_2(U_2) \subseteq V$ and $V$ is an open subset of $\mathbb{R}^3$ such that $\hat{\mathbf{x}}_2^{-1}$ is differentiable. Then, use the $\mathbb{R}^n$ version of the chain rule to deduce that the composition of $\varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))} \colon U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2))) \rightarrow V$ and $\hat{\mathbf{x}}_2^{-1} \colon V \rightarrow \mathbb{R}^2$ is smooth, thus showing that $\mathbf{x}_2^{-1} \circ \varphi \circ \mathbf{x}_1|_{U_1 \cap \mathbf{x}_1^{-1}(\varphi^{-1}(\mathbf{x}_2(U_2)))}$ is also smooth. This is a standard argument done using the inverse function theorem and you can see an example of it in page 70 of Do Carmo's book.
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  • $\begingroup$ Thanks for your response. 'The chain rule for maps between open subsets and regular surfaces' - where am I able to find this? $\endgroup$ – beedge89 Aug 24 '15 at 9:42
  • $\begingroup$ I have added an edit above and have tried to answer using option 2. Would this suffice? $\endgroup$ – beedge89 Aug 24 '15 at 10:42
  • $\begingroup$ Looks good. Note that you don't want to restrict $\tilde{\mathbf{x}}_2^{-1}$ to $N$ because $N$ is not an open subset of $\mathbb{R}^3$ and then you can't use the regular chain rule. Also, you need to make the open subset on which $\mathbf{x}_1$ is defined possibly smaller so that the image of $\varphi \circ \mathbf{x}_1$ will land in $M$. $\endgroup$ – levap Aug 25 '15 at 9:11
  • $\begingroup$ But N is the intersection of two open sets and is therefore open? $\endgroup$ – beedge89 Aug 26 '15 at 1:34
  • $\begingroup$ $\mathbf{x}_2(U_2)$ is an open subset of the regular surface $S_2$, while $M$ is an open subset of $\mathbb{R}^3$. The intersection is an open subset of $S_2$ but not an open subset of $\mathbb{R}^3$. $\endgroup$ – levap Aug 26 '15 at 11:48

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