0
$\begingroup$

Does this definition of a pullback of a differential form make sense?

$\phi:M\rightarrow N$ and $\alpha\in\Omega^r(N)$ then define $$(\phi^*\alpha(X_1,\dots,X_r))(p) := \alpha(\phi(p))(\phi_{*_p}X_1(p),\dots,\phi_{*_p}X_r(p))$$ If it does make sense, it is annoying because it mixes both interpretations of a differential form. Ie on one side it is a map from vector fields to smooth functions and on the other, it is a map from a manifold to tangent bundle.

$\endgroup$
  • 1
    $\begingroup$ In the last sentence, I think you mean a map from a manifold to some exterior power of the cotangent bundle. $\endgroup$ – Michael Albanese Dec 6 '17 at 16:36
0
$\begingroup$

If you want to view both sides consistently, use

$$(\phi^*\alpha)(p)(X_1(p), \dots, X_r(p)) = \alpha(\phi(p))(\phi_{*_p}X_1(p), \dots, \phi_{*_p}X_r(p)).$$

$\endgroup$
  • $\begingroup$ for the first line, $\phi_*X$ is not necessarily a vector field on the whole of the manifold. Is this just solved by saying that $\alpha$ is local? $\endgroup$ – tomak Dec 7 '17 at 10:23
  • $\begingroup$ @tomak: You're right, I was being silly. The expression $\phi_*X$ need not be well-defined (for example, when $\phi(p) = \phi(q)$ but $\phi_{*_p}X(p) \neq \phi_{*_q}X(q)$). I will remove it. $\endgroup$ – Michael Albanese Dec 7 '17 at 15:05
  • $\begingroup$ And is there then no way of seing both sides as maps from vector fields to smooth functions? I find that interpretation much easier :p $\endgroup$ – tomak Dec 8 '17 at 9:27
  • $\begingroup$ I don't think so, and precisely for the reason you mentioned: you can't pushforward vector fields. $\endgroup$ – Michael Albanese Dec 8 '17 at 13:36

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.