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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?

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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.)

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  • $\begingroup$ 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? $\endgroup$
    – R. Rankin
    Apr 13, 2021 at 23:59
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    $\begingroup$ 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. $\endgroup$
    – Deane
    Apr 14, 2021 at 0:01
  • $\begingroup$ @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. $\endgroup$
    – R. Rankin
    Jul 30, 2021 at 22:24
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    $\begingroup$ @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 $SO(4)\simeq (SU(2)\times SU(2))/\mathbb{Z}_2$. $\endgroup$
    – Deane
    Jul 31, 2021 at 23:27

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