I know a Riemann metric on a manifold $M$ as a smooth colection of inner products on the tangent bundle $T(M)$. Or, in other words a symmetric, positive-definite (0,2) tensor field
I see many times that the existence of a Riemann metric is equivalent to the reduction of the structure group of the tangent bundle from $GL(n)$ to $O(n)$, but I don't understand why
For the first part, if I understand correctly, we need a covering of the bundle by local trivializations, such that the transition functions are all in $O(n)$. here they use the Riemann metric with the Gram-Schmidt process and they made two sets of sections be orthonormal at one point $x$. Don't we need to aply Gram-Schmidt to every point in $U_a \cap U_b$? And even then, how do we find those local trivializations?
For the other direction, I really don't know
Any help on the topic is very welcome
The first thing to note is that a local vector bundle trivialization is the same thing as a local frame: if you have a trivialization $\phi: TM|_U \to U \times \mathbb R^n$ then the corresponding frame is simply $E_k(p)=\phi^{-1}(p,e_k)$ where $e_k$ is the standard basis of $\mathbb R^n.$
With this in mind, the idea for one direction of the equivalence is very easy: if you covering by trivializations where all the transition functions are in $O(n),$ then you can simply declare that its constituent frames are orthonormal, which uniquely determines a metric. In a single trivialization $\phi,$ this metric has the explicit form $$g(v,w) = \phi(v)\cdot\phi(w).$$ Since the transition functions are orthonormal, any overlapping trivializations $\phi,\psi$ must be related by $\phi = R \psi$ for some $O(n)-$valued function $R$; and thus we have $$\phi(v)\cdot\phi(w) = R\psi(v) \cdot R\psi(w) = \psi(v)\cdot\psi(w),$$ so the metric $g$ is globally well-defined.
On the other hand, if you are given a metric, then you can construct a corresponding covering by local orthonormal frames: