# 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.

• What do you already know? For example, do you already know that a vector bundle is orientable iff the structure group reduces from $Gl(n,\mathbb{R})$ to $Gl^+(n,\mathbb{R})$? – Jason DeVito Nov 15 '16 at 14:52
• The Jacobian of the coordinate transformation is an element of the structure group of the tangent bundle, so saying "transition functions have positive Jacobian determinant" is precisely the same as saying "structure group reduces to $Gl^+(n,\mathbb{R})$. In other words, one equivalent version as to what Milnor is claiming is that on an oriented Riemannian manifold, there is an open cover for which all transition functions have Jacobians in $SO(n)$. – Jason DeVito Nov 15 '16 at 15:15
• I just realized that what I said is the same as saying that the structure group of the tangent bundle of $M$ reduces to $GL^+(n,\mathbb R)$. How do I go from there to $SO(n)$? – R_D Nov 15 '16 at 15:18
• This is turning out to be harder than I was suspecting. In particular, when I wrote "one equivalent version..." above - I don't know how to prove its equivalent. – Jason DeVito Nov 15 '16 at 18:03
• @JasonDeVito: If I understood you correctly, you claimed that on an oriented Riemannian manifold, there should be a covering by coordinate charts whose transition functions have Jacobians in $SO(n)$. This is certainly not true, unless the metric is flat. – Jack Lee Nov 15 '16 at 19:42

## 1 Answer

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.

• I'm sorry, but as I understand the change of frame matrix takes values in $GL^+(n,\mathbb R)$. Why is it $SO(n)$? I understand why the determinant of this matrix should be positive (follows from the definition) but not why it should be $+1$ – R_D Nov 16 '16 at 4:29
• Since all of the local frames are orthonormal, the change-of-frame matrix always takes values in $O(n)$. Since they are all positively oriented, the change-of-frame matrix takes values in $GL^+(n,\mathbb R)$. Now use the fact that $GL^+(n,\mathbb R) \cap O(n) = SO(n)$. – Jack Lee Nov 17 '16 at 0:00
• Actually, I don't understand how we get an orthonormal frame when the manifold is oriented. In general the tangent bundle of a manifold has structure group $GL(n,\mathbb R)$ right? If it oriented then I understand that the structure group reduces to $GL^+(n,\mathbb R)$. I don't understand where $O(n)$ comes from. – R_D Nov 17 '16 at 6:16