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Is there an embedding (i.e. injective continuous map)

$$\phi:\Bbb R P^2\hookrightarrow S^4\subseteq\Bbb R^5$$

of the 2-dimensional projective space $\Bbb R P^2$ into the $4$-sphere, that is transitive, i.e. for any two $x,y\in P^2\Bbb R$ 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 $\Bbb R P^n$: what is the lowest dimensional Euclidean space needed for such an embedding.

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  • $\begingroup$ The notation $P\mathbb{R}^2$ makes it look like the projectivization of $\mathbb{R}^2$, i.e. $\mathbb{RP}^1$. Perhaps $\mathbb{RP}^2$ or $\mathbb{P}^2(\mathbb{R})$ are better notations. $\endgroup$ – Ben McKay Apr 25 at 8:07
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Yes. You take each vector $v \in \mathbb{R}^3$ to the vector $v \cdot v \in \operatorname{Sym}^2\mathbb{R}^3=\mathbb{R}^6$. This takes each unit vector $v$ to the same place as $-v$. So it descends to $S^2/\pm 1=\mathbb{RP}^2$. If we identify each element of $\operatorname{Sym}^2\mathbb{R}^3$ with a symmetric matrix by identifying $\sum a_{ij} e_i \cdot e_j$ with $A=(a_{ij})$, then the action of $\operatorname{SO}(3)$ is by conjugation on these symmetric matrices. The element $e_1 \cdot e_1 + e_2 \cdot e_2 + e_3 \cdot e_3$ is identified with the identity matrix. The action preserves identity matrix, so preserves a 1-dimensional subspace in $\mathbb{R}^6$. We quotient out that subspace to get an action of $\operatorname{SO}(3)$ on $\mathbb{R}^5$. This is well known in representation theory as the unique 5-dimensional irreducible representation of $\operatorname{SO}(3)$. The metric $\left<A,B\right>=\sum_{ij} A_{ij} B_{ij}$ is clearly preserved on $\mathbb{R}^6$, and $\left<v\cdot v,v\cdot v\right>=1$ in this metric. You take the quotient metric on $\mathbb{R}^5$, i.e. project to the orthogonal complement of that 1-dimensional subspace.

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    $\begingroup$ You mean preserving the trace of the symmetric matrix, not $x^2 + y^2 + z^2$. (What are $x, y, z$ anyway?) The only detail left is for which scalar product $G$ on $\mathrm{Sym}^2 \mathbb{R}^3$ the conjugation by matrices from $\mathrm{O}(3)$ belongs to $\mathrm{O}(G)$. $\endgroup$ – Vít Tuček Apr 25 at 9:29
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    $\begingroup$ Here is a little further discussion of the embedding Ben describes: ldtopology.wordpress.com/2012/07/12/… $\endgroup$ – Ryan Budney Apr 25 at 19:19
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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.

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