Is a vector bundle orientable if and only if its dual bundle is orientable?

Question

I was reading up on my dual spaces today and I made the following hypothesis:

A vector bundle $\xi$ is orientable if and only if $\xi^*$ is orientable.

This seems rather intuitive, and although it doesn't seem too hard to prove, I'm not sure how to formally prove it. Any help?

EDIT: Also, I guess another assertion that would fall along these lines would be whether the following is true:

If the ordered bases $v_1,...,v_n$ and $v'_1,...,v'_n$ for $V$ are equally oriented, then the same is true of the bases $v^*_1,...,v^*_n$ and $v^{'*}_1,...,v^{'*}_n$ for $V^*$.

Can anyone help me on the proofs (if they're true)?

Answer 1

Alternatively, we can make two observations:

• A bundle $E$ of dimension $n$ is orientable iff $\Lambda^n E$ is trivial.

• If we denote $F^*$ the bundle dual to a vctor bundle $F$, then there is a isomorphism $\Lambda^n(E^*)\to(\Lambda^nE)^*$.

Answer 2

If $\xi$ is a real vector bundle then $\xi$ and $\xi^*$ will actually always be isomorphic; you can always put a metric on a bundle, which gives an isomorphism between $\xi$ and $\xi^*$. The complex case follows from the fact that all complex vector bundles are orientable.

EDIT: to put a metric on $\xi$ requires the base space to be paracompact (e.g. a manifold).

EDIT 2: Another way to see this is via transition functions. $\xi$ is orientable means you can reduce the structure group to $GL_n^+$, the group of matrices with positive determinant. Since the transition functions for $\xi^*$ will be given by the conjugate inverse and since $A \in GL_n^+$ if and only if $A^{-t} \in GL_n^+$, it follows that $\xi$ is orientable if and only if $\xi^*$ is.