Differentiability of a function on a manifold

  • #1
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 [itex]f:M\rightarrow N[/itex] is independent of the coordinate chart that we use". He shows this is true in [itex]M[/itex] and then leaves it as an exercise to show that it is also true in [itex]N[/itex]. 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 [itex]f:M\rightarrow N[/itex] be a map from an [itex]m[/itex]-dimensional manifold to an [itex]n[/itex]-dimensional manifold [itex]N[/itex]. A point [itex]p\in M[/itex] is mapped to a point [itex]f(p)\in N[/itex], namely [itex]f:p\mapsto f(p)[/itex]. Take a chart [itex](U,\phi)[/itex] on [itex]M[/itex] and [itex](V,\psi)[/itex] on [itex]N[/itex], where [itex]p\in U[/itex] and [itex]f(p)\in V[/itex]. Then [itex]f[/itex] has the following coordinate presentation: [tex]\psi\circ f\circ\phi^{-1}:\mathbb{R}^{m}\rightarrow\mathbb{R}^{n}[/tex]
If we write [itex]\phi (p)=\lbrace x^{\mu}\rbrace =x[/itex] and [itex]\psi(f(p))=\lbrace y^{\alpha}\rbrace = y[/itex], then [itex]y=\psi\circ f\circ \phi^{-1} (x)[/itex] is a vector-valued function of [itex]m[/itex] variables.


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

Would this be correct?
 

Answers and Replies

  • #2
WWGD
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Yes, that looks correct, but remember that not all manifolds are ## C^{\infty} ##
 
  • #3
Fredrik
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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
Ah, ok. Thanks for your help Fredrik, there's so much to take in, but I think it's slowly starting to become clearer!
 

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