Differentiability of a function on a manifold

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1. Jan 16, 2015

"Don't panic!"

I am currently working through Nakahara's book, "Geometry, Topology and Physics", and have reached the stage at looking at calculus on manifolds. In the book he states that "The differentiability of a function $f:M\rightarrow N$ is independent of the coordinate chart that we use". He shows this is true in $M$ and then leaves it as an exercise to show that it is also true in $N$. Here is my attempt, please could someone tell me if what I've done is correct, and if not, what the correct way is?

Let $f:M\rightarrow N$ be a map from an $m$-dimensional manifold to an $n$-dimensional manifold $N$. A point $p\in M$ is mapped to a point $f(p)\in N$, namely $f:p\mapsto f(p)$. Take a chart $(U,\phi)$ on $M$ and $(V,\psi)$ on $N$, where $p\in U$ and $f(p)\in V$. Then $f$ has the following coordinate presentation: $$\psi\circ f\circ\phi^{-1}:\mathbb{R}^{m}\rightarrow\mathbb{R}^{n}$$
If we write $\phi (p)=\lbrace x^{\mu}\rbrace =x$ and $\psi(f(p))=\lbrace y^{\alpha}\rbrace = y$, then $y=\psi\circ f\circ \phi^{-1} (x)$ is a vector-valued function of $m$ variables.

The differentiability of $f$ is independent of the coordinate chart in $N$. Indeed, let $(V_{1},\psi_{1})$ and $(V_{2},\psi_{2})$ be two overlapping charts in $N$. Take a point $q=f(p)\in V_{1}\cap V_{2}$, whose coordinates by $\psi_{1}$ are $\lbrace y_{1}^{\mu}\rbrace$, while those $\psi_{2}$ are $\lbrace y_{2}^{\nu}\rbrace$. When expressed in terms of $\lbrace y_{1}^{\mu}\rbrace$, $f$ takes the form $$y_{1} =\psi_{1}\circ f\circ\phi^{-1}$$ while in $\lbrace y_{2}^{\nu}\rbrace$, it takes the form $$y_{2} =\psi_{2}\circ f\circ\phi^{-1} =\psi_{2}\circ\left(\psi^{-1}_{1}\circ\psi_{1}\right)\circ f\circ\phi^{-1}= \left(\psi_{2}\circ\psi^{-1}_{1}\right)\circ\psi_{1}\circ f\circ\phi^{-1}$$ Now, by definition, $\psi_{2}\circ\psi^{-1}_{1}$ is $C^{\infty}$, and therefore if $\psi_{1}\circ f\circ\phi^{-1}$ is $C^{\infty}$ with respect to $y_{1}^{\mu}$ and $\psi_{2}\circ\psi^{-1}_{1}$ is $C^{\infty}$ with respect to $y_{2}^{\nu}$, then $\psi_{2}\circ f\circ\phi^{-1}=\left(\psi_{2}\circ\psi^{-1}_{1}\right)\circ\psi_{1}\circ f\circ\phi^{-1}$ is also $C^{\infty}$ with respect to $y_{2}^{\nu}$.

Would this be correct?

2. Jan 16, 2015

WWGD

Yes, that looks correct, but remember that not all manifolds are $C^{\infty}$

3. Jan 17, 2015

Fredrik

Staff Emeritus
There are some inaccuracies, but you seem to have the right idea. Your post should be talking about an f that's "differentiable at p", and not necessarily "differentiable" (i.e. differentiable at p for all p).

If x and y are coordinate systems with p in their domains, then $f\circ x^{-1}$ is typically not equal to $f\circ y^{-1}\circ y\circ x^{-1}$, because $y^{-1}\circ y$ is the identity map on the domain of $y$, which typically doesn't include the entire domain of $x$. So $f\circ y^{-1}\circ y\circ x^{-1}$ is a restriction of $f\circ x^{-1}$ to a smaller domain. This is irrelevant when we consider the partial derivatives of these functions, but it's a bit of a pain to prove that. (I don't think someone who's learning differential geometry should spend too much time on these subtleties, because they don't contribute much to your understanding).

4. Jan 17, 2015

"Don't panic!"

Ah, ok. Thanks for your help Fredrik, there's so much to take in, but I think it's slowly starting to become clearer!