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Let $\pi:A\longrightarrow M$ be a vector bundle and $E\subseteq A$ a vector sub-bundle. Recall, a $k$-form on $A$ is a section of $\Lambda^k A^*$. Let us write $\Omega^k(A):=\Gamma(\Lambda^k A^*)$.

The inclusion $\jmath: E\longrightarrow A$ is a vector bundle map and therefore induces $\jmath^*:\Omega^k(A)\longrightarrow \Omega^k(E)$ so that we can ''restrict" forms on $A$ to forms on $E$.

Conversely, is it possible to extend forms on $E$ to forms on $A$?

Thanks.

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    $\begingroup$ With a metric on $A$ you can create the sub bundle that is the orthogonal complement to $E$. Then extend by zero. $\endgroup$ – Matthew Leingang Dec 14 '16 at 12:56
  • $\begingroup$ Cool, thanks =) $\endgroup$ – PtF Dec 14 '16 at 13:09
  • $\begingroup$ @MatthewLeingang The inclusion map $\jmath$ is injective, so $\jmath^*$ is surjective, isn't it? This would do because I don't need to find the explicit extension.. $\endgroup$ – PtF Dec 14 '16 at 13:25
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To answer your original question: If $A$ has a metric on it, you can let $E^\perp$ be the subbundle of $A$ orthogonal to $E$. Then $A = E \oplus E^\perp$. The map $\pi \colon A \to E$ which is the identity on $E$ and zero on $E^\perp$ induces $\pi^* \colon \Omega^k(E) \to \Omega^k(A)$ (for any $k$). Since $\pi \circ \jmath = \operatorname{id}_E$, $\jmath^*\circ\pi^* = \operatorname{id}_{\Omega^k(E)}$, so $\pi^*$ extends forms on $E$ to forms on $A$.

To answer your follow up question: I think you're right that if $\jmath$ is an injection than $\jmath^*$ is a surjection. The short exact sequence $0 \to E \to A \to A/E \to 0$ gives the contravariant short exact sequence in forms.

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