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Jan 7, 2017 at 15:23 comment added Andreas Cap Thanks for the plan to acknoledge the answer. I am fine with this and don't have preferences.
Jan 7, 2017 at 12:40 comment added Asaf Shachar I am going to publish a paper which uses vector valued-forms quite heavily. For completeness, I am reproducing your argument in the appendix (I need this commutation property as a small lemma inside a much vaster argument...). Of course, I plan to give you credit for this (at least in the acknowledgment, but I am also thinking to actually cite your answer). If you have any objections or preferences, please let me know. Thanks again for your help.
Oct 24, 2016 at 12:11 vote accept Asaf Shachar
Oct 24, 2016 at 7:56 comment added Andreas Cap Thanks for pointing out the typo, I have edited the answer. The alternative to using the global formula is similar to the answer of @Sebastian : Write forms as $\alpha\otimes s$ for $\alpha\in\Omega^k(M)$ and $s\in\Gamma(V)$ and use that $d^{\nabla^V}(\alpha\otimes s)=d\alpha\otimes s+\alpha\wedge\nabla^V s$. But in both versions, it is not really necessary to involve an induction by the degree of forms.
Oct 24, 2016 at 7:53 history edited Andreas Cap CC BY-SA 3.0
added 151 characters in body
Oct 24, 2016 at 7:41 comment added Asaf Shachar Thanks! nice answer. I guess you have a minor typo in the statement of the compatibility condition; It is supposed to be $ \nabla_{\xi}^W (\Phi(s))=\Phi(\nabla_{\xi}^V s)$. The use of the global formula is of course very reasonable. I am still wondering however if this property can be deduced directly from the Leibnitz rule via induction (that is, from the defining property of the exterior derivative, without using the explicit formula).
Oct 24, 2016 at 7:18 history answered Andreas Cap CC BY-SA 3.0