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Let $g_1$ and $g_2$ be two Riemannian metrics on a manifold $M$. These induce two $O(n)$ bundles on $M$, whose fibers over each point $x\in M$ are the groups of orthogonal transformations of $T_x M$ with respect to $g_1$ or $g_2$, respectively. Are these two $O(n)$ bundles necessarily isomorphic?

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    I'm confused. The group of orthogonal transformations of $T_x M$ is indeed isomorphic to $\mathrm{O}(n)$, but not in any canonical way: indeed, choosing an isomorphism amounts to choosing an orthonormal basis for $T_x M$. So although we can construct a group bundle over $M$ with typical fibre $\mathrm{O}(n)$, I don't see how we can get a canonical $\mathrm{O}(n)$-action on it.2012-12-12
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    I don't think we the group of orthogonal transformations to be isomorphic to $O(n)$ in a canonical way, do we? Let $E$ be the bundle whose fiber over each point $x$ is the group $E_x$ of orthogonal transformations of $T_x M$. In an coordinate chart, we can construct a local trivialization $U\times O(n) \cong E_U$ by choosing an orthonormal basis of vector fields over $U$. Does this not give us an $O(n)$ bundle?2012-12-12
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    What do you mean by $\mathrm{O}(n)$-bundle? I mean a fibre bundle with a chosen fibrewise $\mathrm{O}(n)$-action.2012-12-12
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    I guess I just mean a fiber bundle with fiber $O(n)$. Perhaps this is not a standard sort of object to consider, and my question is vacuous. If so, I apologize.2012-12-12

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