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In differential geometry, how can I prove that the differentiability of a function is independent of the choice of the admissible atlas?

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  • $\begingroup$ Hint: look at the conditions that have to be satisfied for an atlas. Now see if you can get your result from it $\endgroup$ – jdoicj Jun 3 '19 at 13:58
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Let $M$ and $N$ two manifolds and a function $f: M \rightarrow N$

By definition of differentiability between manifolds,

A function $f$ is differentiable according to the associated atlas $A_{M}$ and $A_{N}$ if only if for all pair of charts $(U_{\alpha}, > \phi_{\alpha}) \in A_{M}$ and $(V_{\phi}, \omega_{\phi}) \in A_{N}$ the function $\omega_{\phi}\circ f \circ (\theta)^{-1} $ is differentiable.

Now, to prove that this definition is independent of the chose atlas we must consider two atlas $A^{'}_{M}$ and $A^{'}_{N}$ different but compatibles with the previous atlas.

We want to show that for any given two charts ($U^{'}_{\beta} $, $\theta^{'}_{\beta}) \in A^{'}_{M}$ and ($V^{'}_{\rho}, \omega^{'}_{\rho} ) \in A^{'}_{M}$ the function: $(\theta^{'}_{\beta})\circ f \circ (\omega^{'}_{\rho})^{-1}$

$(\theta^{'}_{\beta})\circ f \circ (\omega^{'}_{\rho})^{-1}$ is differenciable.

Since $A_{M}$ and $A_{N}$ are atlas, we have:

  • $U^{'}_{\beta}= \bigcup_{\alpha\in I}(U_{\alpha}\cap U^{'}_{\beta}))$ where $I=\alpha U_{\alpha}\in A_{M}$ and ($U_{\alpha}\cap U^{'}_{\beta})\neq \emptyset $

  • $U^{'}_{\beta}= \bigcup_{\phi\in K}(U_{\phi}\cap U^{'}_{\rho}))$ where $K=\phi V_{\phi}\in A_{M}$ and ($V_{\phi}\cap V^{'}_{\rho})\neq \emptyset $

Moreover, since the atlas is compatible and differentiable with each other we have that:

  • $\theta_{\alpha}\circ (\theta^{'}_{\beta})^{-1}: (\theta^{'}_{\beta} (U_{\alpha} \circ (U^{'}_{\beta})) \rightarrow \theta_{\alpha}(U_{\alpha} \circ (\theta^{'}_{\beta}))$
  • $\omega^{'}_{\rho}\circ (\omega_{\phi})^{-1}: (\omega_{\phi}( V_{\phi} \circ (U^{'}_{\rho})) \rightarrow \omega_{\rho}(V_{\phi} \circ (V^{'}_{\rho}))$
  • $\omega_{\phi} \circ f \circ (\phi_{\alpha})^{-1}$

Use the above 3 conditions (function composition) we have:

$\omega^{'}_{\rho} \circ f \circ \theta^{'}_{\rho}$

As we using differential function composition, we get a differentiable function:

$\theta^{'}_{\rho}(U_{\alpha}\cap U^{'}_{\beta}) \rightarrow f (\omega^{'}_{\rho} (V_{\theta}\cap V^{'}_{\rho} )) $

Since this is for every $\alpha \in I $ and every $\phi \in K$, we have that $\omega^{'}_{\rho} \circ f \circ \theta^{'}_{\rho}$ is infinite differentiable on all $\theta^{'}_{\beta}$ to all $f(\omega^{'}_{\rho}(V^{'}_{\rho}))$

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  • $\begingroup$ On the definition I wanted to write ...all pair of charts $(U_{\alpha}, \phi_{\alpha}) \in A_{M}$ and $(V_{\phi}, \omega_{\phi}) \in A_{N}$... $\endgroup$ – Lorenzo Castagno Jun 2 '19 at 17:18

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