action of $O(n,\mathbb{R})$ on $S^n$

Question

I need to know what is the action of $O(n,\mathbb{R})$ on $S^n$, and $O(n,\mathbb{R})/O(n-1,\mathbb{R})\cong S^{n-1}$, how does $O(n-1,\mathbb{R})$ sit inside $O(n,\mathbb{R})$?

The obvious action may be $\phi:O(n,\mathbb{R})\times S^n\rightarrow S^n:(A,x)\mapsto Ax$?

Answer 1

Yes, the action is $(A,x) \mapsto Ax$.

Now fix $x_0 \in S^{n-1}$. Verify that the closed subgroup $G = \{A \in O(n) \mid A x_0 = x_0\}$ fixes the hyperplane $(x_0)^{\perp}$ and that it can be identified with $O(n-1)$. For example, if you take $x_0 = (0,\dots,0,1)$ then $G$ consists of the matrices of the form $$ \begin{bmatrix} B & 0 \\ 0 & 1 \end{bmatrix} \in O(n), \quad B \in O(n-1). $$ The to see that $f \colon O(n)/G \to S^{n-1}$ is a homeomorphism, observe that it is a continuous bijection since $O(n)$ acts transitively, then use that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.