Structure group of tangent bundle of Riemannian manifold

Question

In the article "Spin structures on manifolds" by J Milnor, the author begins as follows -

Let $M$ be an oriented, Riemannian manifold. Then the tangent bundle of $M$ has the rotation group $SO(n)$ as structural group.

I was wondering if anyone could tell me where I could find a proof of this fact (or give me hints to prove it myself). Google didn't help much.

EDIT : Definition of Oriented manifold being used -

A smooth manifold $M$ is oriented if it admits an orientable atlas. That is an atlas whose all transition functions have positive Jacobian determinant.

Thank you.

Answer 1

If $E\to M$ is a rank-$k$ vector bundle, a reduction of the structure group of $\boldsymbol E$ to a subgroup $G\subset GL(k,\mathbb R)$ is a covering of $M$ by local trivializations of $E$ such that the transition functions are of the form $(x,v) \mapsto (x,\tau(x)v)$, where $\tau$ takes its values in $G$.

Thus a reduction of the structure group of $TM$ to $SO(n)$ is a covering by local frames (NOT necessarily frames arising from coordinate charts) such that the change-of-frame matrix takes its values in $SO(n)$.

If $M$ is an oriented Riemannian manifold, you just take a covering by oriented orthonormal frames, and Bob's your uncle.