Isotropy group of $SO(n)$

Question

I have found in some books that the isotropy group of the action of $SO(n)$ over a manifold is $SO(n-1)$. In what book can I find the proof of this fact?

Many thanks!

Answer 1

I'll assume that you are talking about the action of $\operatorname{SO}(n)$ on the unit sphere $S^{n-1} \subseteq \mathbb{R}^n$. Let $q = (1,0,\dots,0) \in S^{n-1}$ be the north pole. If $A \in \operatorname{SO}(n)$ satisfies $Aq = q$, this means that $q \in \mathbb{R}^n$ is an eigenvector of $A$ associated to the eigenvalue $1$. Since $A$ is orthogonal, the subspace $\operatorname{span} \{ q \}^{\perp}$ is $A$ invariant so $A$ has the block-diagonal form

$$ A = \begin{pmatrix} 1 & 0 \\ 0 & B \end{pmatrix} $$

with $B \in M_{n-1}(\mathbb{R})$. Since $A|_{\operatorname{span} \{ q \}^{\perp}}$ must preserve the inner product, we see that $B \in \operatorname{O}(n-1)$. Finally, since $\det(A) = \det(B) = 1$, we see that $B \in \operatorname{SO}(n-1)$. Going the other direction, it is readily seen that any matrix of the form above with $B \in \operatorname{SO}(n-1)$ belongs to the istoropy group of $q$. Since the action is transitive, this means that all the isotropy groups are isomorphic to $\operatorname{SO}(n-1)$.