Structure group of tangent bundle of Riemannian manifold 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.
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.
This is what I've gathered reading the book "Riemannian Geometry and Geometric Analysis, by Jurgen Jost" edition 6 page 45. 

So we have coordinate charts on an orientable manifold $M$ such that ALL chart transitions have positive Jacobian. Any one of these charts (say $x:M\to \mathbb{R}^d$) induces a frame $\partial/\partial x^a|_{m}$ at a point $m\in M$. Consider another chart $y$ and the frame $\partial/\partial y^a|_{m}$. These frames are related by the Jacobian $\partial/\partial y^a=J^b_a\partial/\partial x^b$ (where $J^b_a:=\partial x^b/\partial y^a|_m$). This is precisely the Jacobian that has positive determinant by assumption of orientability. Then consider using the GRAM-SCHMIDT process on the frame $\partial/\partial x^a|_{m}$ to obtain a frame $e_a$ (where $a=1,...,dim(M)$) such that $e_a$ is orthonormal with respect to the metric -- that is $g(e_a,e_b)|_{m}=\delta_{ab}$. Going from the frame $\partial/\partial x^a|_{m}$ to $e_a$ can be written as $e_a=G^b_a\partial/\partial x^b|_{m}$, where $G\in GL(dim(M))$. Similarly, one can GRAM-SCHMIDT orthonormalize $\partial/\partial_y|_{m}$, i.e $\tilde{e}_a=\tilde{G}^b_a\partial/\partial_y|_{m}$ such that $g(\tilde{e}_a,\tilde{e}_b)=\delta_{ab}$. Then consider the matrix $M$ such that $\tilde{e}_a=M^b_ae_b$. 

So to finally address the question: We want to show that this matrix $M$ lies in $SO(dim(M))$. To show that is is orthogonal we write $\delta_{ab}=g(\tilde{e}_a,\tilde{e}_b)=M^c_ag(e_c,e_d)M^d_b=M^c_a\delta_{cd}M^d_b=M^d_aM^d_b=(M^{T}M)_{ab}=\delta_{ab}$. Since it lies in the orthogonal group, $det M=\pm 1$. To show that it has $det M &gt;0$ and therefore $detM=1$, first observe from all the previous definitions that $MG=\tilde{G}J$. Therefore $detM=det \tilde{G} det J/det{G}$. Then FINALLY -- and here is the crux of the matter -- the GRAM SCHMIDT matrix $G$ has $det G&gt;0$ (see Determinant in Gram-Schmidt process for a proof). Then using the fact that $det J&gt;0$ we immediately conclude that $det M&gt;0$.

So I guess the moral of the story is that given an orientable manifold, one can always construct a tangent bundle with structure group SO(d). There always exists one because we can always use the GRAM-SCHMIDT process to construct one, but I don't think there is a unique tangent bundle with structure group SO(d), because you might have another orientation preserving diagonalization procedure.