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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.

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    $\begingroup$ In the last sentence, I think you mean a map from a manifold to some exterior power of the cotangent bundle. $\endgroup$ Commented Dec 6, 2017 at 16:36

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If you want to view both sides consistently, use

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

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  • $\begingroup$ 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? $\endgroup$
    – user405156
    Commented Dec 7, 2017 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. $\endgroup$ Commented Dec 7, 2017 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 $\endgroup$
    – user405156
    Commented Dec 8, 2017 at 9:27
  • $\begingroup$ I don't think so, and precisely for the reason you mentioned: you can't pushforward vector fields. $\endgroup$ Commented Dec 8, 2017 at 13:36

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