# frame bundle of an n-sphere is $O(n+1)$?

I'm trying to understand frame bundles (mostly from a physics point of view). It is standard for some n-dimensional Riemannian manifold that the frame bundle has an $$O(n)$$ structure. However; It seems there is some exception for an n-sphere where I come across references to it's oriented-orthonormal frame bundle being diffeomorphic to $$SO(n+1)$$. In such cases they say that the matrix with columns $$\left[\phi,\hat{e}_{i}...\hat{e}_{n}\right]\in SO(n+1)$$. Where $$\phi$$ are the coordinates of the n-sphere as embedded within $$\mathbb{R}^{n+1}$$ and the hat's are the unit frame vectors.

Can someone please elaborate upon this for me, I don't quite get it, does this only apply to a sphere of constant curvature? I was trying to construct an explicit example for the 3-sphere but couldn't quite get it. Does this apply to the general topological n-sphere or just a constant curvature one?

## 1 Answer

There are two issues here:

1. The orthonormal frame bundle of a Riemannian $$n$$-manifold has structure group $$O(n)$$.
2. The total space of the oriented orthonormal frame bundle of a round $$n$$-sphere may be identified with $$SO(n+1)$$: As you say, an element of $$SO(n+1)$$ with first column $$\phi$$ is naturally identified with an orthonormal frame of $$T_{\phi} S^{n}$$, the tangent space at $$\phi$$ to the round unit sphere in Euclidean $$(n+1)$$-space.

For a non-round Riemannian manifold diffeomorphic to a sphere, the orthonormal frame bundle is diffeomorphic to $$SO(n+1)$$, but not identified with $$SO(n+1)$$ in the same literal way.

(Separately, take care when speaking of "topological" spheres. In dimensions starting with $$7$$, there exist exotic spheres, smooth manifolds homeomorphic to the sphere but not diffeomorphic to the sphere.)

• If I'm lifting my structure, or equivalently "reducing the structure group" to the double cover, say $spin(n+1)$ then would the set of coordinates necessarily become a spinorial quantity? Apr 13, 2021 at 23:59
• In other words, each fiber $\mathcal{F}_pM$ of the oriented orthonormal frame bundle $\mathcal{F}M$ of an $n$-dimensional Riemannian manifold $M$ can be identified with $SO(n)$. On the other hand, $SO(n+1)$ is the oriented orthonormal frame bundle $\mathcal{F}S^n$ of $S^n$, where the first column is the element in $S^n$ and the remaining $n$ columns are an oriented orthonormal basis of the tangent space at that point. Apr 14, 2021 at 0:01
• @Deane How can the Oriented orthonormal frame bundle $F(S^3)=SO(4)$ when I know the bundle is also trivial and therefore $F(S^3)=SU(2) \times SO(3)$? I know that locally they're the same (Lie algebra), but we're talking about a total space (a principle bundle) here. Jul 30, 2021 at 22:24
• @R.Rankin, that's a good question. Here's a relevant remark: Let $\mathbb{H} \simeq \mathbb{R}^4$ be the space of quaternions. Let $\mathbb{U}$ be the set of unit quaternions. You can show that $\mathbb{U}$ is isomorphic, as a Lie group, to $SU(2)$, by identifying $\mathbb{H} \simeq \mathbb{C}^2$. There is a group action of $\mathbb{U}\times\mathbb{U}$ on $\mathbb{H}$, where $(q_1,q_2)h = q_1h\bar{q}_2$. You can show that this defines a 2-to-1 homomorphism from $SU(2)\times SU(2) \rightarrow SO(4)$, and therefore $SO(4)\simeq (SU(2)\times SU(2))/\mathbb{Z}_2$. Jul 31, 2021 at 23:27