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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)?

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The comment in your edit gets to the heart of a fairly elementary proof. Saying $v_i$ and $v_i'$ are equivalently oriented says there's a map $f : V \to V$ with $f(v_i) = v_i'$ and $Det(f) > 0$. The adjoint to $f$ is the map $f^* : V^* \to V^*$ which sends one dual basis to another, and $Det(f^*) = Def(f)$, because the determinant is invariant under taking transposes. – Ryan Budney Oct 17 '11 at 2:53
up vote 3 down vote accepted

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 $\Lambda^n(E^*)\to(\Lambda^nE)^*$.

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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.

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