Vtyjkyuk

Let $S^n subset mathbbR^n+1$ 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 $(0,0,0,vecv),~ vecv in S^n-3$.

So one must compute $dim SO(n-2) - dim SO(n-2)_vecx$, where $SO(n-2)$ denotes the isotropy group at $vecx$. I know that the isotropy group has dimension at least $1$, since given any vector $vecv in S^n-3$, we can take a rotation whose rotation axis is $vecv$, 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 $S^2$.

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 $(0,0,0,vecv)$. Indeed, for any $(a,b,c,vecv)in S^n$ with $vecvneq 0$, you will again get an $(n-3)$-sphere as the orbit for the same reason, just a scaled down $(n-3)$-sphere.)