I am trying to make sense of the following proof showing that $\mathbb{P}\simeq \text{O}(n+1)/(\text{O}(1)\times \text{O}(n)),$ where $\mathbb{P}:=\mathbb{P}^n(\mathbb{R})$ is the $n$-dimensional real projective space.
Let $n\ge 0.$ Consider $\mathbb{P}:=\mathbb{P}^n(\mathbb{R})$ is the $n$-dimensional real projective space, which is defined to be the space of $1$-dimensional linear subspaces of $\mathbb{R}^{n+1}.$ It has a structure of smooth manifold characterized by the requirement that the natural map $\pi:\mathbb{R}^{n+1}\setminus \{0\}\to \mathbb{P}, v\mapsto \mathbb{R}v$ is a smooth immersion.
Consider the natural smooth action of $G:=GL(n+1,\mathbb{R})$ on $\mathbb{R}^{n+1}\setminus \{0\}$ given by $(g,x)\mapsto gx.$ Then $G$ maps fibers of $\pi$ onto fibers, hence the action induces an action $G\times \mathbb{P}\to \mathbb{P}.$ Since $\pi$ is a submersion, it follows that the action of $G$ on $\mathbb{P}$ is smooth. Let $m\in \mathbb{P}$ be the line spanned by the first standard basis vector $e_1$ of $\mathbb{R}^{n+1}.$ Then the stabilizer of $m$ in $G$, denoted by $G_m$ equals the group of invertible $(n+1)\times (n+1)$ matrices with first column a multiple of $e_1.$ This action can be easily checked to be transitive. Therefore, the induced map $G/G_m\to \mathbb{P}$ is a diffeomorphism of manifolds.
Now consider the subgroup $K:=\text{O}(n+1)$ of $G.$ One readily sees that $K$ already acts transitively on $\mathbb{P}.$ Hence the action induces a diffeomorphism from $K/K_m$ onto $\mathbb{P}.$ Note that $K_m=K\cap G_m$ consists of matrices $$\begin{bmatrix} a & 0 \\ 0 & B \end{bmatrix}$$ with $a=\pm1$ and $B\in \text{O}(n).$ Thus $K_m \simeq \text{O}(1)\times \text{O}(n)$ and it follows that $$\mathbb{P}\simeq \text{O}(n+1)/(\text{O}(1)\times \text{O}(n)).$$
My questions: I was able to comprehend most of the argument, except at the following places.
- How does $G$ map fibers of $\pi$ onto fibers and what is the induced action $G\times \mathbb{P}\to \mathbb{P}$?
- This might be straightforward, but how is the action of $K$ already on $\mathbb{P}$ transitive?
I wasn't able to come up with a satisfactory explanation to both these questions, so any help will be useful. Thanks in advance.
The action $G \times \mathbb{P} \to \mathbb{P}$ is what we might expect: $(g, \mathbb{R}v) \mapsto \mathbb{R}gv$, but we have to verify that it's well-defined. For this consider any nonzero multiple $w = tv \in \mathbb{R}^{n+1} \setminus \{0\}$, i.e. $t \neq 0$. By definition of spanning, $\mathbb{R}w = \mathbb{R}v \in \mathbb{P}$, so we need to see that $G$ acts the same on either one. Since the group $G$ acts linearly on $\mathbb{R}^{n+1}$, $$ gw = g(tv) = t(gv), $$ so $\mathbb{R}gw = \mathbb{R}gv \in \mathbb{P}$, as desired.
This is all neatly summarized for each $g \in G$ by the commutative diagram $\require{AMScd}$ \begin{CD} \mathbb{R}^{n+1} @>{g\cdot{}}>> \mathbb{R}^{n+1} \\ @V{\pi}VV @VV{\pi}V \\ \mathbb{P} @>{g\cdot{}}>> \mathbb{P} \end{CD}
As far as retracting to the orthogonal group $K$ (which is compact, showing easily that the projective space is compact as well, as the continuous image), we need to see that the orbit of a point in $\mathbb{P}$ consists of all of $\mathbb{P}$. Since $K$ preserves the lengths of vectors, it will act by shuffling around points on the unit sphere $\mathbb{S}^n \subset \mathbb{R}^{n+1}$, but that's good enough since each line through the origin pierces the unit sphere (twice actually).
Explicitly, given $\mathbb{R}v, \mathbb{R}w \in \mathbb{P}$, we need to find some $g \in K$ such that $g(\mathbb{R}v) = \mathbb{R}w$. Without loss of generality, we can assume that $\lvert v \rvert = \lvert w \rvert = 1$. We just need to write down a rotation matrix $A_v$ mapping $e_1 \mapsto v$ (first column given by $v$). If you like, extend to a basis $\{v, b_1, \dots, b_n\}$, and apply the Gram–Schmidt process. We similarly find an orthogonal matrix $A_w$ mapping $e_1 \mapsto w$. Now, since $K$ is a group, $$ g = A_w (A_v)^{-1} \in K $$ as well, and it sends $v \mapsto w$.