So, we've been asked to show, given a real vector bundle equipped with a metric, that there is a canonical isomorphism from the vector bundle and its dual.

Now, there's a theorem that says two vector bundles are isomorphic iff their transition functions satisfy $\mu_i g_{ij} = f_{ij} \mu_j$, where $\mu_i$ is a map $U_i \rightarrow GL(r, \mathbb{R})$ and $g_{ij}$ and $f_{ij}$ are the two bundles transition functions.

I was going to say the following: Given a metric and Gram-Schmid one can always arrange for orthonormal frames, and hence orthogonal transition functions. Therefore (since the dual bundle transition functions equal the inverse transpose of the bundles transition functions), the above is satisfied trivially.

My question is: Is this still 'canonical'? I mean, the fact you can do this is a 'universal property' of vector bundles equipped with metrics, so it should be a canonical iso (according to wikipedia). But, I'm confused by the meaning of 'canonical isomorphism' as meaning "independent of a basis". In the above, I'm specifying a basis.


Yes, if you have a vector bundle $E \to M$ with a fiber metric $h$, then there is a natural isomorphism induced by the usual identification of a vector space with its dual in the presence of an inner product (not necessarily of definite signature): The map $\Phi: E \mapsto E^*$ is given fiberwise by $$\Phi_p: v \mapsto h_p(v, \,\cdot\,),$$ which is manifestly independent of basis, and as a section of $\text{Hom}(E, E^*)$ it obviously is as smooth as $h$ is, because it is $h$, viewed dually.

(Of course, all of this requires a metric or an appropriate substitute, in general there is no natural isomorphism between a vector space and its dual, or between a vector bundle and its dual.)

  • $\begingroup$ this is a nice argument, but is there a more precise way of stating that $\Phi$ is smooth? I understand that since $h$ is a metric on $E$, it can be seen as a smooth section of $E^*\otimes_{\mathbb{R}}E^*$ (which we identify with $Hom(E,E^*)$), but how does that imply the smoothness of $\Phi$- we haven't talked too much about the actual structure of vector bundles as manifolds, so this doesn't seem as obvious as I'd like it too $\endgroup$ – CWsl2 Oct 21 '14 at 2:21
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    $\begingroup$ Here's a very concrete way of seeing it: In any (smooth) local frame on $E$, for example, the components of the (smooth) fiber metric $h$ are smooth, but by these are precisely the components of $\Phi$, so $\Phi$ is itself smooth. $\endgroup$ – Travis Willse Oct 21 '14 at 3:18
  • $\begingroup$ Thank you this helps a lot $\endgroup$ – CWsl2 Oct 21 '14 at 3:31
  • $\begingroup$ You're welcome, I'm glad you found it useful. $\endgroup$ – Travis Willse Oct 21 '14 at 7:29

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