TITLE: Transitive embedding of the projective plane $\Bbb R P^2$ into the $4$-sphere
QUESTION [7 upvotes]: Is there an embedding (i.e. injective continuous map) 
$$\phi:\Bbb R \Bbb P^2\hookrightarrow S^4\subseteq\Bbb R^5$$ 
of the projective plane  $\Bbb R\Bbb  P^2$ into the $4$-sphere, that is transitive, i.e. for any two $x,y\in \Bbb R\Bbb  P^2$ there is an orthogonal transformation $T\in\mathrm{O}(\Bbb R^5)$ that fixed the image $\mathrm{im}(\phi)$ set-wise, and has $Tx=y$?
Is $\Bbb R^5$ the lowest dimensional space in which such an embedding is possible, or do we need even more dimensions?
I may ask the same question for $\Bbb R\Bbb  P^n$: what is the lowest dimensional Euclidean space needed for such an embedding.

REPLY [8 votes]: Since you may ask about $\mathbb{RP}^n$, I may point out that you can embed $\mathbb{RP}^n$ into $\mathbb{R}^{(n+2)(n+1)/2-1}$ with a transitive action of $O(n+1)$.
From e.g. Exercise 5-C of Milnor-Stasheff, $\mathbb{RP}^n$ is the space of symmetric idempotent $(n+1)\times (n+1)$ matrices with trace $1$. This lives inside $\mathbb{R}^{(n+2)(n+1)/2-1}$ as the space of symmetric $(n+1)\times (n+1)$ matrices with trace $1$. $O(n+1)$ acts by conjugation on this space as isometries, and preserving the subspace of idempotent matrices. 
An  exceptional case is $\mathbb{RP}^3 \cong SO(3)$. The group $SO(3) \subset \mathbb{R}^9$ as $3\times 3$ matrices. But we can do a bit better: we may think of $SO(3)$ as pairs of orthogonal unit vectors $(v_1,v_2)\in (\mathbb{R}^3)^2$. This gives an embedding of $SO(3)\subset S^5\subset \mathbb{R}^6$ with a transitive group action. I believe that this special embedding exists since $so(4)=so(3)\oplus so(3)$. 
One might be able to construct similar smaller dimensional embeddings using fibrations $S^1\to \mathbb{RP}^{2n+1}\to \mathbb{CP}^n$ and $\mathbb{RP}^3\to \mathbb{RP}^{4n+3}\to \mathbb{HP}^n$, but I haven't checked if they give smaller embeddings with isometric actions. 
However, $\mathbb{RP}^{2n}$ is not a fibration (for the same reason that $S^{2n}$ is not a fibration). Hence, if we have an embedding $\mathbb{RP}^{2n}\subset \mathbb{R}^k$ and a transitive action by isometries $G\leq O(k)$, then the representation of the compact group $G$ must be irreducible. If not, then there is a splitting $\mathbb{R}^k=\mathbb{R}^{k_1}\times \mathbb{R}^{k_2}$ which is invariant under $G$. In this case, we get $v=(v_1,v_2)\in \mathbb{RP}^n\subset \mathbb{R}^k$, $v_i\in \mathbb{R}^{k_i}$, and $\mathbb{RP}^n = G\cdot v \to G\cdot v_1$. Hence we have a fibration $\mathbb{RP}^{2n} \to G\cdot v_1$, a contradiction unless $G\cdot v_1=\mathbb{RP}^{2n}$, in which case $k$ was not minimal.