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

Myshkin
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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.

Martin
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  • could you please explain how the transitive action of $O(n)$ says that is a continous bijection?please – Myshkin Jan 26 '13 at 07:13
  • If $G$ acts on a set $S$ and $H$ is the stabilizer of $s \in S$ then $G/H$ is in bijection with the orbit $Gs$ (this is just algebra). In particular, if $G$ acts transitively on $S$ then $G/H$ is in bijection with $Gs = S$. That the map is continuous follows from the definition of the quotient topology. – Martin Jan 26 '13 at 07:15