It is well know that the differential calculus is developed with respect open sets but a manifold not necessarily is an open set and actually many times it not. So in Differential Geometry it is usual give the following definition of $C^r$-function.


A function $f:S\rightarrow\Bbb R^n$ where $S\subseteq\Bbb R^k$ is of class $C^r$ if it can be extended to a $C^r$-function defined in an open set $U$ containing $S$.

Naturally if $\tilde {f_1}, \tilde{f_2}$ are two different extension of $f$ it is possible that even their derivatives are different in $S$ but fortunately this do not happens if $S$ is a manifold as Victor Gullemin and Peter Heine show in the text Differential forms part of which I summarise to follow but if you like you can read it here.

So let be $f:X\rightarrow Y$ a $C^r$-mapping from a k-manifold $X$ of $\Bbb R^n$ and an $l$-manifld $Y$ of $\Bbb R^m$ and thus let be $\tilde f$ any $C^r$-extension. So we define the derivative of $f$ in $X$ to be the restriction of the derivative of $\tilde f$ in $X$, that is $$ Df(x):=D\tilde f(x) $$ for any $x\in X$. So let's prove that this definition is consistent, that is if does not depend from the choice of a particular extension. Therefore let be $\phi:U\rightarrow V$ a local patch of $X$ and thus let be $h:=\tilde f\circ\phi$. Now it is possible to prove that $$ D\tilde f(\phi(x))=Dh(x) $$ for any $x\in U$ and since $h=\tilde f\circ\phi=f\circ\phi$ we conclude that in $X$ the derivatives of two different extension of $f$ are equal.

So I observe that if $\phi:U_1\rightarrow V_1$ and $\phi_2:U_2\rightarrow V_2$ are two different charts then what above showed says that $$ D(f\circ\phi_2)(x_1)=D(f\circ\phi_2)(x_2) $$ where $x_2=\phi_2^{-1}(\phi(x_1))$ but unfortunately to me it seems false this. Anyway since this was not clear to me I refer to the text Differential Topology by Victor W. Gullemin and Alan Pollack where I found this definition of derivative of a mapping between two manifold that I completely understand. However I do not understand if what showed in the link I posted implies that two different extesions of $f$ have the same derivative at $X$. So since well understand the last approach I primarily ask if with respect this it is true that two different extesions of $f$ have the same derivative at $X$ and then (only if you like) I ask to prove the equivalence of this approach with respect the first. So could anyone help me, please?

  • $\begingroup$ The fact that you want to use in questions like this is the following: Given a submanifold $S^k$ in a manifold $M^n$ and $p \in S$, there exists local coordinates $(x^1, \dots, x^n)$ on $M$ such that $S$ is given by $x^{k+1} = \cdots = x^n = 0$. In that case, given a function $f$ on $M$, the differential of $f$ restricted to $S$ clearly depends only on $f$ restricted to $S$ and not on how it's extended away from $S$. $\endgroup$ – Deane Feb 20 at 21:47
  • $\begingroup$ Sorry, but I do not understand what you mean: unfortunately what you claim it is not clear and obvious to me. Excuse my ignorance. $\endgroup$ – Antonio Maria Di Mauro Feb 20 at 21:55
  • $\begingroup$ I'd be happy to elaborate. There are two parts to my comment. The first is about $S$ itself, and the second is about $f$. Would you like me to provide more details on both? $\endgroup$ – Deane Feb 20 at 22:13
  • $\begingroup$ So, first of all I try to explain the formalism I know. If $M$ is a $k$-manifold in $\Bbb R^n$ to me a coordinate patch is a function $\phi$ from an open set of $\Bbb R^k$ or $H^k$ (the upper-half space) to an open set $V$ of $M$. So are your $x^1,...,x^n$ the coordinate functions individuate by the local chart as I above described? Then I do not know the existence of a coordinate patch such that $[\phi(x)](i)=0$ for $i=1,...,(n-k)$ and thus I can not accept this explanation if you do not prove first the result. Moreover why the existence of this local chart proves what I ask? $\endgroup$ – Antonio Maria Di Mauro Feb 20 at 22:20
  • $\begingroup$ If you like to know it I say to you that I am studying by the text Analysis on Manifold by James Munkres and I refer to the texts that Munkres put in the bibliography that use substantially the same formalism. So I am sure you realise that I can accept only some particular explanation: precisely those are consistent with respect Munkres or affine formalims. $\endgroup$ – Antonio Maria Di Mauro Feb 20 at 22:29

Hints (pretty much the same as Deane's):

Step 1. First prove the claim in the case when $S\subset {\mathbb R}^k$ is actually an open subset of ${\mathbb R}^m\subset {\mathbb R}^k$.

Step 2. Prove that if $S$ is an $m$-dimensional submanifold in ${\mathbb R}^k$ then for each $x\in S$ there exists a neighborhood $U$ of $x$ in ${\mathbb R}^k$ and a diffeomorphism $h: U\to {\mathbb R}^k$ such that $h(U\cap S)$ is an open subset of ${\mathbb R}^m\subset {\mathbb R}^k$.

Step 3. Use the chain rule plus Step 1 in order to get the desired conclusion.

  • $\begingroup$ Well, I tried to attempt the following arguments. So let us to distinguish the case where $S$ is a $m$-manifold of $\Bbb R^k$ with empty or not empty boundary. So in the first case as you substantially stated $S$ is diffeomorphic to an open subset of $\Bbb R^m$ and thus for any $y\in S$ there exist a local patch $\alpha:U\rightarrow V$ defined in an open set $U$ of $\Bbb R^m$ such that $y=\alpha(x)$ and thus $$f(y)=\big(f\circ\alpha\big)(x)$$ for any $y\in V$ where $x:=\alpha^{-1}(y)$. $\endgroup$ – Antonio Maria Di Mauro Feb 22 at 9:05
  • $\begingroup$ So if $\tilde f_1$ and $\tilde f_2$ are two different $C^r$ extension of $f$ then $$D\tilde f_1(y)=D\big(f\circ\alpha\big)(x)=D\tilde f_2(y)$$ for any $y\in V$ and thus it seems that we could claim that $D\tilde f_1(y)=D\tilde f_2(y)$ for any $y\in V$ and so for any $y\in S$. $\endgroup$ – Antonio Maria Di Mauro Feb 22 at 9:05
  • $\begingroup$ However if $\alpha':U'\rightarrow V'$ is another local patch about some $y\in V$ then it would be $$ D\big(f\circ\alpha'\big)(x')=D\tilde f_1(y)=D\big(f\circ\alpha\big)(x)$$ but generally $$ D\big(f\circ\alpha')(x')=D\big(\tilde f\circ\alpha'\big)(x')=D\big(\tilde f\circ\alpha\big)(x)\cdot D\big(\alpha^{-1}\circ\alpha'\big)(x')\neq D\big(\tilde f\circ\alpha\big)(x)$$ and thus this (it seems) invalid my argument that is the same of Gullemin and Heine. $\endgroup$ – Antonio Maria Di Mauro Feb 22 at 9:06
  • $\begingroup$ Then if the boundary of $S$ is not empty I did not able to implement your (second) hint and I did not able to prove prove the third point knowing the first and second. So could you help me, please? $\endgroup$ – Antonio Maria Di Mauro Feb 22 at 9:06

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