According to Manifolds and differential Geometry (J. M. Lee, 2009), a Riemannian metric on a vector bundle $\pi:E\rightarrow M$ is a smooth assignment to each point $p\in M$ a scalar product $g_p$ on the fibre $\pi^{-1}(p)$, so this is the standard definition.
However, he wants to show that a choice of Riemannian metric for $E$ is a reduction of structure group from $GL(V)$ to $(O(V)$. And he will do it using charts. Let's go.
First, he says:
If the vector bundle has such a structure, it is convenient to assume that a fixed inner product has is chosen on the typical fibre and that a distinguished orthonormal basis $(e_1,\dots,e_k)$ has been chosen.
(Here $k$ is the rank). Now he continues:
If $(U,\Psi)$ is a local trivialization, then define a local frame field $(X_1,\dots,X_k)$ simply as $$ X_a(p) = \Psi^{-1}(p,e_i) \qquad \forall p\in U, $$ for all $a=1,\dots,k$.
This is also good because he has already proven the relation between local frame fields and vector bundle charts, so it is okay. But now, he says:
One can perform a Gram-Schmidt proccess in the basis $\{X_1(p),\dots,X_k(p)\}$ simultaneously for all $p\in U$ so that we have a new orthonormal basis $\{E_1(p),\dots,E_k(p)\}$ (...). Thus, $\{E_1(p),\dots,E_k(p)\}$ is a local frame called orthonormal frame.
HERE the point is. Some comments:
First, yo haven't a metric on $\pi^{-1}(p)$, but in $V$, so you must to import it. I mean, Gram-Schmidt process does not make sense at this moment.
Second, since you really haven't such a metric, I think Gram-Schimdt process would look like
$$ E_a(P) = \Psi^{-1}(p,W_a), $$
where $W_a$ is defined as
$$ W_p = \frac{\psi\left(X_a(p)\right)-\displaystyle \sum_{b=1}^{a-1}g\bigg(\psi\left(X_b(p)\right),\psi\left(X_a(p)\right) \bigg)\psi\left(X_a(p)\right)}{\psi\left(X_a(p)\right)-\displaystyle \sum_{b=1}^{a-1}g\bigg(\psi\left(X_b(p)\right),\psi\left(X_a(p)\right) \bigg)\psi\left(X_a(p)\right)} $$ (here $\Psi$ is supposed to be $\Psi=(\pi,\psi)$.
- If the above process is right, Lee's words don't not make sense, since $\Psi(X_a(p))=e_a$ exactly, and then it is already orthonormal.
So, what is happening here? Lee's book is wrong? Is all that a mistake? Or what is he actually thinking in?
Thanks
EDIT
Reading Riemannian Geometry and Geometric Analysis (Jürgen Jost, 2011) I realize that I am wrong:
Each (real) rank $k$ vector bundle $\pi:E\rightarrow M$ with a bundle metric $g$ has structure group $(On)$. In particular, there exists bundle charts $(U,F)$, $F:\pi^{-1}\rightarrow U\times\mathbb{R}^k$, for which for each point $p\in U$, $F^{-1}(p,(e_1,\dots,e_k))$ is an orthonormal basis of $\pi^{-1}(p)$.
Since my bundle charts was arbitrary, $\{X_1,(p),\dots,X_k(p)$ cannot be an orthonormal frame, but I don't know why. I'm missing something.