What is $SO(n+1)/O(n)$ as a topological space?

Question

Consider $SO(n),O(n)$ as topological groups. Find out $SO(n+1)/O(n)$ as a topological space.

My attempt:

Observed the inclusion : $O(n) \hookrightarrow{} SO(n+1)$ by, $$A \mapsto\begin{bmatrix} det(A) &0 \\ 0 &A\end{bmatrix}$$ i.e. basically, $$A \mapsto\begin{bmatrix} 1 &0 \\ 0 &A\end{bmatrix} \text{ or } A \mapsto\begin{bmatrix} -1 &0 \\ 0 &A\end{bmatrix}$$

Now we know that $SO(n+1)$ acts transitively on $S^n$ and moreover, $$SO(n+1)\text{ / }SO(n) \cong S^n \tag{*}.$$

I was trying to imitate the proof of $(*)$, but could not proceed.

Intuitively, it seems to me that $SO(n+1) \text{ / } O(n) \cong \Bbb RP^n$ , but I really don't have any proof or anything,just intuition.

Thanks in advance for help!

Answer 1

You are correct that the quotient can be identified with $\Bbb R \Bbb P^n$.

Hint The usual action of $SO(n + 1)$ on $\Bbb R^{n + 1}$ is transitive and linear, so it descends to a transitive action on the space $\Bbb R \Bbb P^n$ of lines through the origin in $\Bbb R^{n + 1}$. Thus, if we fix an element $\ell \in \Bbb R \Bbb P^n$, we can identify $SO(n + 1) / H$, where $H < SO(n + 1)$ is the stabilizer in $SO(n + 1)$ of $\ell$.

If we take $\ell$ to be the span $[e_0]$ of the first standard basis element, $g \in SO(n + 1)$ fixes $\ell$ iff it has the form $$\pmatrix{\ast & \ast \\ 0 & \ast},$$ and the condition $g \in SO(n + 1)$ implies that any such $g$ has the form $$\pmatrix{\det A & 0 \\ 0 & A}, \qquad A \in O(n); $$ conversely, any matrix of that form is in $H$.