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Let $X$ be a smooth manifold and $\pi:E\rightarrow X$ a vector bundle of rank $k$ on X. If one manages to redefine $E$ by using a cocycle $\{g_{\alpha,\beta}\}$ whose values are all contained in a subgroup $G$ of $GL(k)$, then one says that $E$ is a $G$-bundle.

Many textbooks say that $E$ comes with a metric (a smoothly-varying fiber-wise metric), the structure group can be reduced to $O(k)$. I am not really convinced with this fact; starting with a transition function $\{g_{\alpha,\beta}\}$ whose values are just in $GL(k)$, is it obvious that one can make a new transition function whose values are just in $O(k)$?

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up vote 7 down vote accepted

If $\{U_{\gamma}\}$ is an open cover of $X$ which trivialises $E$, then over $U_{\alpha}$, there are sections $s_1, \dots, s_m$ such that $\{s_i(x)\ |\ i = 1, \dots, m\}$ is a basis for $E_{x}$ for every $x \in U_{\alpha}$ ($m$ is the rank of $E$). Suppose now that $x \in U_{\alpha}\cap U_{\beta}$, and let $\sigma_1, \dots, \sigma_m$ be the corresponding sections; in particular $\{\sigma_i(x)\ |\ i = 1, \dots, m\}$ is also a basis for $E_x$. Then there is a change of basis matrix $g_{\alpha\beta}(x)$ which transforms $\{\sigma_i(x)\ |\ i = 1, \dots, m\}$ into $\{s_i(x)\ |\ i = 1, \dots, m\}$; that is $s_i(x) = g_{\alpha\beta}(x)\sigma_i(x)$ for $i = 1, \dots, m$.

A choice of metric on $E$ gives an inner product $\langle \cdot, \cdot\rangle_x$ on $E_x$. This allows us to measure angles; in particular, we can apply the Gram-Schmidt process to obtain an orthonormal basis for $E_x$. Therefore, we can take both $\{\sigma_i(x)\ |\ i = 1, \dots, m\}$ and $\{s_i(x)\ |\ i = 1, \dots, m\}$ to be orthonormal bases for $E_x$. Note that

$$\delta_{ij} = \langle s_i(x), s_j(x)\rangle_x = \langle g_{\alpha\beta}(x)\sigma_i(x), g_{\alpha\beta}(x)\sigma_j(x)\rangle_x,$$


$$\delta_{ij} = \langle\sigma_i(x), \sigma_j(x)\rangle_x.$$

Therefore $\langle\sigma_i(x), \sigma_j(x)\rangle_x = \langle g_{\alpha\beta}(x)\sigma_i(x), g_{\alpha\beta}(x)\sigma_j(x)\rangle_x$, so $g_{\alpha\beta}(x)$ preserves the inner product and is therefore an element of the orthogonal group.

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I see. The point is that on sufficiently small open sets one can take sections which form an orthonormal basis. Thank you very much. – Pooya Sep 29 '12 at 6:24
@Pooya: Yes, but sufficiently small only in the sense that the bundle in trivial over those open sets. – Michael Albanese Sep 29 '12 at 6:35
@MichaelAlbanese : Can you explain this part " A choice of metric on $E$ gives an inner product $⟨⋅,⋅⟩_x$ on $E_x$ in more detail? – Babai Mar 27 '15 at 11:28
Also, Your could have directly started your argument by taking two Orthonormal basis for $E_x$. (You are basically saying there exists sections which is a local frame. But is it really required for the proof?) – Babai Mar 27 '15 at 11:49
@Babai: A Riemannian metric is a smoothly varying choice of inner product on each tangent space. I couldn't have started with an orthonormal basis for $E_x$ because the notion of an orthonormal basis requires you to have an inner product on $E_x$. For example, if $V$ is a vector space and $v_1, v_2 \in V\setminus\{0\}$, can you tell me whether or not they are orthonormal? No, because you don't know what the inner product is. Once I tell you which inner product I have chosen, then you can talk about orthonormality. – Michael Albanese Mar 29 '15 at 17:47

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