1
$\begingroup$

Given a smooth map $f:M\to N$ between smooth manifolds how do you show that the differential map $df:TM\to TN$ is smooth?

$\endgroup$
  • 4
    $\begingroup$ Express it locally, as usual. $\endgroup$ – Mariano Suárez-Álvarez Mar 26 '12 at 16:58
  • $\begingroup$ I'm confused as to how. I have local charts for $TM$ and $TN$ but I don't know what to do with them. $\endgroup$ – 09867 Mar 26 '12 at 17:16
  • $\begingroup$ if $x$ is a chart for $M$, $y$ one for $N$, then you have $Tx$ a chart for $TM$, $Ty$ a chart for $TN$. Untangle the definitions and write down $Ty \circ df \circ (Tx)^{-1}$ $\endgroup$ – Blah Mar 26 '12 at 20:07
  • $\begingroup$ All this chart business is confusing me. Could anyone just write a specific proof so that I can see exactly what goes on? $\endgroup$ – 09867 Mar 27 '12 at 15:15
1
$\begingroup$

This is proved in Proposition $3.21$ of Lee's Introduction to Smooth Manifolds (second edition) for example.

If $v \in TM$, then choosing a coordinate chart $(U, \varphi)$ containing $p = \pi(v)$ where $\pi : TM \to M$ is the projection, one obtains a coordinate neighbourhood $(\pi^{-1}(U), \widetilde{\varphi})$ containing $v$. More precisely, in local coordinates,

$$v = \left.v^1\frac{\partial}{\partial x^1}\right|_p + \dots + \left.v^m\frac{\partial}{\partial x^m}\right|_p$$

and $\widetilde{\varphi}(v) = (x^1(p), \dots, x^m(p), v^1, \dots, v^m)$. Once you have such charts, the local expression for $df$ becomes

$$df(x^1, \dots, x^m, v^1, \dots, v^m) = \left(f^1(x), \dots, f^n(x), \frac{\partial f^1}{\partial x^i}(x)v^i, \dots, \frac{\partial f^n}{\partial x^i}(x)v^i\right).$$

As $f$ is smooth, all the components of $df$ (in these local coordinates) are smooth, and therefore $df : TM \to TN$ is smooth.

| cite | improve this answer | |
$\endgroup$

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.