Minimal embedding of real projective surface into $S^4$ I'm reading a paper, and one result was quoted there:
There is a minimal embedding of $\mathbb {RP}^2$ into $S^4$, and the corresponding volume is 6$\pi$.
Any reference is welcomed. And thanks for anyone who could offer help in advance.
(According to the paper, the embedding is called Veronese surface. I googled it that the answer was not satisfactoring. Sorry that I forget to mention it in the beginning)
 A: Not quite an answer, but too long for a comment. I do know about an embedding, but I learned this in a class and I have no idea where the professor took that from. In any case, here it is: consider $$ V = \{ T\colon \Bbb R^3 \to \Bbb R^3 \mid T^\top = T \mbox{ and }\,{\rm tr}(T) = 0  \}.$$It is easy to see that $V$ is a vector space and $\dim V = 5$. The map $(T,S)\mapsto {\rm tr}(TS)$ defines an inner product on $V$. Now define $\Phi\colon \Bbb R P^2 \to V$ by $$\Phi(\Bbb R p) = \sqrt{\frac{3}{2}}\,{\rm pr}_{\Bbb R p, 0},$$where ${\rm pr}_{\Bbb Rp}\colon \Bbb R^3 \to \Bbb R^3$ denotes the projection onto the line $\Bbb R p$ (here $p \in \Bbb R^3\setminus \{0\}$, of course), and for linear $T \colon \Bbb R^3 \to \Bbb R^3$, $T_0 = T - ({\rm tr}(T)/3){\rm Id}_{\Bbb R^3}$ denotes the traceless part of $T$. It turns out that $\Phi$ is a smooth embedding, and the image $\Phi[\Bbb R P^2]$ is contained in the unit sphere of $V$ (the coefficient $\sqrt{3/2}$ is meant to rescale the radius of the sphere to $1$), which is a $\Bbb S^4$. Maybe this is what you want?
