Is it possible to realize $ \mathbb{RP}^2 $ as a linear group orbit?

Question

Does there exists a Lie group $ G $ a representation $ \pi: G \to Aut(V) $ and a vector $ v \in V $ such that the orbit $$ \mathcal{O}_v=\{ \pi(g)v: g\in G \} $$ is diffeomorphic to the real projective plane $ \mathbb{RP}^2 $?

Answer 1

We can embed $\mathbb{RP}^2$ into the space $M_3$ of $3\times 3$ matrices via the map which takes a line $[v]$ to the orthogonal projection onto $[v]$: $$ [v]\mapsto\frac{vv^T}{\|v\|^2} $$ This gives a diffeomorphism from $\mathbb{RP}^2$ to the space of rank $1$ orthogonal projection operators: $$ \mathbb{RP}^2\cong\{A\in M_3:\operatorname{rank}(A)=1,\ A^2=A=A^T\} $$ One can show that this submanifold is an orbit of the linear $O(3)$ action on $M_3$ defined by $O\cdot A=OAO^T$.