Let $S^n \subset \mathbb{R}^{n+1}$ be the standard sphere. Consider the $SO(n-2)$ action on $S^n$ that fixes the first three coordinates. What is the dimension of a regular orbit, i.e, a orbit with smallest isotropy?
My approach was the following: in order to compute the dimension of a regular orbit is enough to take a regular point. This immediately implies that one must take a point of the form $(0,0,0,\vec{v}),~ \vec{v} \in S^{n-3}$.
So one must compute $\dim SO(n-2) – \dim SO(n-2)_{\vec{x}}$, where $SO(n-2)$ denotes the isotropy group at $\vec{x}$. I know that the isotropy group has dimension at least $1$, since given any vector $\vec{v} \in S^{n-3}$, we can take a rotation whose rotation axis is $\vec{v}$, but I was not able to continue. I appreciate any help.
Another thing I've noticed was the following: the action of $SO(n-2)$ on $S^{n-3}$ is transitive, furthermore, $S^{n-3} \cong SO(n-2)/SO(n-3)$. This seems to imply that a regular orbit is precisely $S^{n-3}$. Is this correct?
For instance, for $n = 5$, one has that a regular orbit is $S^2$.
Best Answer
Your latter approach is correct and is the easiest way to understand this. A regular orbit is indeed just $S^{n-3}$, so it has dimension $n-3$.
(Note though that the regular orbits are not restricted to those of points of the form $(0,0,0,\vec{v})$. Indeed, for any $(a,b,c,\vec{v})\in S^n$ with $\vec{v}\neq 0$, you will again get an $(n-3)$-sphere as the orbit for the same reason, just a scaled down $(n-3)$-sphere.)