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?

  • 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

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.

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