4
$\begingroup$

Suppose $\{\phi_\alpha\}_{\alpha\in\mathcal{A}}$ and $\{\phi_\beta\}_{\beta\in\mathcal{B}}$ are two smooth atlases on a topological manifold $M$. My definition of two such atlases being equivalent is that their union $\{\phi_\alpha\}_{\alpha\in\mathcal{A}\cup\mathcal{B}}$ is also a smooth atlas, that is, the transition maps between the charts of different atlases are smooth.

I have shown that if two atlases are equivalent, then they determine the same set of smooth functions $f:M\rightarrow\mathbb{R}$ - i.e. $f$ is smooth with respect to one chart if and only if it is smooth with respect to the other. But I do not know how to prove the converse statement. I would like to do something along the lines of $(f\circ\phi_\beta^{-1})^{-1}\circ(f\circ\phi_\alpha^{-1})=\phi_\beta\circ\phi_\alpha^{-1}$ and conclude from that that the RHS is smooth since both bracketed functions on the LHS are, but I know I can't take the inverse on the LHS like this,since there is no guarantee $f$ is invertible. Any help would be appreciated.

$\endgroup$
2
$\begingroup$

Suppose $\varphi_\beta\circ\varphi_\alpha^{-1}$ isn't smooth. Then one of $x_i\circ\varphi_\beta$ isn't smooth in $(\varphi_\alpha)_{\alpha\in A}$, while it's clearly smooth in $(\varphi_\beta)_{\beta\in B}$. Choose a sufficiently small compact neighborhood $K$ of a point in which it isn't smooth. Then $x_i\circ\varphi_\beta|_K$ can be smoothly extended to the whole $M$ and is the function you want.

$\endgroup$
  • $\begingroup$ I guess you mean "Choose a sufficiently small compact neighborhood $K$ of a point in which it wasn't smooth ... " $\endgroup$ – KonKan Sep 25 '16 at 1:54
  • $\begingroup$ @KonKan Thanks, corrected. $\endgroup$ – user326572 Sep 26 '16 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.