Coset space and transitiviy

I have a question regarding coset space or homogeneous space $SO(n+1)/SO(n)$ which is simply $S^n$. I need some intuition regarding this result.

As everyone knows that for a simple case of $SO(3)/SO(2)$, one can have $SO(3)$ as a group acting on $\mathbb{R}^3$ and $SO(2)$ as an isotropy group of $x\in\mathbb{R}^3$, then the group $SO(3)$ acts transitively on $S^2$ and we get $S^2$ as the coset.

Since the result is just 2-sphere or $n$-sphere, is there an intuitive way of seeing it?

I am not sure if I am just restating your question so correct me if I am.

My answer is basically that any rotation in $SO(n+1)$ can be written as a rotation that moves the north pole of $S^n$ to some new point on $S^n$, and then a rotation about this new point. The rotations about the new point just form $SO(n)$, so $S^n$ is what you get when you take $SO(n+1)$, and you say you only care about the final position of the north pole and not the rotation about that final position.

Look at the action of $SO(n+1)$ on the sphere $S^n$ embedded in $\mathbb{R}^{n+1}$. Let's take the north pole $p \in S^n$ and look at the orbit of $p$ under the action of $SO(n+1)$. Since the action is transitive, we know that the orbit is the whole $S^n$. So in your example of a two-sphere, this statement becomes that the north pole $p$ can be taken to any point on the two-sphere by a rotation.

Now consider the stabilizer of $p$. The stabilizer is simply rotations that keep $p$ fixed. These are rotations in hyperplane orthogonal to $p$. In other words, the stabilizer of $p$ is $SO(n)$.

Now consider a coset $rSO(n)$, were $r \in SO(n+1)$. Consider the action of the elements of this set on the north pole. Since the north pole is invariant under $SO(n)$, the set $rSO(n)p$ is just the set $\{rp\}$. Since the action is transitive, we know that the cosets map onto $S^n$. On the other hand if we have two different cosets, $r_aSO(n)$ and $r_bSO(n)$, that map to the same point on $S^n$, then $r_ap = r_bp$ so that $r_a^{-1}r_bp=p$ and so $r_a^{-1}r_b \in SO(n)$, but then $r_aSO(n) = r_ar_a^{-1}r_bSO(n) = er_bSO(n) = r_bSO(n)$, so the two cosets are not different after all. Thus there is a bijection between cosets $SO(n+1) / SO(n)$ and the sphere $S^n$.

• I actually figured out a simple explanation for SO(3)/SO(2) using equivalence classes, but thanks for the explanation. Commented Apr 26, 2014 at 19:52
• For everyone wondering what he is talking about, have a look at page 590 of Tony Zees' Einstein Gravity in a Nutshell: books.google.de/…
– jak
Commented Jan 6, 2017 at 16:51

You can think of an element of $SO(n+1)$ a collection of $(n+1)$ ordered orthonormal vectors $(v_1, \ldots, v_{n+1})$. In such a case, we have that $SO(n)$ fits inside $SO(n+1)$ by choosing a fixed vector $v \in S^n$, and choosing $n$ orthonormal vectors in the orthogonal complement of the line spanned by $v$.

It is not too hard to imagine then that the cosets are in bijection with the initial choice of $v$, i.e. the coset space is exactly $S^n$.