# Prove that the differentiability of a function is independent of the choice of the admissible atlas

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

• Hint: look at the conditions that have to be satisfied for an atlas. Now see if you can get your result from it – jdoicj Jun 3 '19 at 13:58

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 $$f(\omega^{'}_{\rho}(V^{'}_{\rho}))$$

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