Noncanonical isomorphism of spaces of differential forms

Question

Let $\pi: V \to M$ be a smooth $n$-dimensional vector bundle over $M$. Are the spaces of differential forms $\Omega^i(V)$, $\Omega^i(V^*)$ noncanonically isomorphic? If so, how do I see this? Is there a place where this is proved or disproved?

Answer 1

This post just consolidates the comments.

If $V$ is a real vector bundle, a fiberwise inner product on $V$ determines an isomorphism $V \to V^*$ by $v \mapsto \langle v, -\rangle$. So anything else you could possibly do to $V$ and $V^*$ are isomorphic: in particular, it determines an isomorphism $\Lambda^i T^*M \otimes V \to \Lambda^i T^*M \otimes V^*$. This, then, defines an isomorphism of the spaces of sections (as $C^\infty(M)$-modules) $\Omega^i(V) \to \Omega^i(V^*)$.

However, if $V$ is a complex vector bundle, this is not usually true. One class of obstructions are the Chern classes. We have $c_i(V^*) = (-1)^i c_i(V)$, and so if $V \cong V^*$, then $2c_{2n+1}(V)=0.$ (There might be 2-torsion in the relevant cohomology groups, so that these don't literally vanish, but don't worry too much about this point.) Now, as vector spaces, $\Omega^i(V) \cong \Omega^i(V^*)$ for uninteresting cardinality reasons, but they will usually not be isomorphic as $C^\infty(M;\Bbb C)$-modules.