Identifying antipodal points of an ellipsoid (with axes of different length) defines a Riemannian metric on the real projective plane $\mathbb RP^2$. Is there an explicit global isometric imbedding of this metric into Euclidean space $\mathbb R^N$? By explicit I mean using special functions, ellipsoidal harmonics, etc. The ambient dimension $N$ need not be 5 but perhaps not too large. Of course, the same question can be asked for any $\mathbb RP^n$ with an ellipsoidal metric.
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4$\begingroup$ It might be worth mentioning that it is known that any (smooth) metric on a compact surface can be smoothly isometrically embedded into $\mathbb{R}^5$. (Anton's method below constructs an explicit ellipsoidal metric embedding of $\mathbb{RP}^2$ into $\mathbb{R}^6$.) Meanwhile, it appears still to be unknown whether there is any metric on $\mathbb{RP}^2$ with positive Gauss curvature that can be smoothly isometrically embedded into $\mathbb{R}^4$. Cf., the discussion in M. Gromov's, Partial Differential Relations, Section 3.2.4. $\endgroup$– Robert BryantCommented Aug 29, 2017 at 16:44
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The Veronese embedding provides an isometric embedding $\mathbb{R}\mathrm{P}^n$ into $\mathbb{S}^N\subset \mathbb{R}^{N+1}$. Taking the cone over this map produce an isometric embedding of the quotient space $\mathbb{R}^{n+1}/\iota$ into $\mathbb{R}^{N+1}$, where $\iota$ is the central symmetry $\iota\colon x\mapsto -x$.
(If $n=2$, then the needed length-preserving map $\mathbb{S}^2\to \mathbb{R}^6$ is defined by $$(x,y,z)\mapsto \alpha\cdot(\beta+x^2,\beta+y^2,\beta+z^2,\sqrt{2}{\cdot}x{\cdot}y,\sqrt{2}{\cdot}y{\cdot}z,\sqrt{2}{\cdot}z{\cdot}x)$$ for approprately chosen $\alpha$ and $\beta$.)
So, "yes", for any centrally symmetric surface $\Sigma$ in $\mathbb{R}^n$, the quotient $\Sigma/\iota$ admits an explicit embedding in $\mathbb{R}^{N+1}$.
answered Aug 28, 2017 at 21:11
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2$\begingroup$ I believe that $\alpha = 1/\sqrt2$ and either $\beta=-1$ or $\beta = \frac13$ will give the desired isometric embedding of $\mathbb{RP}^2$ into $S^5\subset\mathbb{R}^6$. It may be worth mentioning that this map is equivariant with respect to actions of $\mathrm{SO}(3)$ on $\mathbb{R}^3$ and $\mathbb{R}^6$. $\endgroup$ Commented Aug 29, 2017 at 12:37
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$\begingroup$ You are very kind. Mainly, I just wanted to convince myself that the roots were real, which was not obvious to me. I suppose it's also worth mentioning that, for the analogous construction for general $n$, the solutions have $\alpha^2= 1/2$ and $\beta = (-1\pm\sqrt{n{+}2})/(n{+}1)$. $\endgroup$ Commented Aug 29, 2017 at 16:31
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$\begingroup$ @RobertBryant The map with $\alpha=1$ and $\beta=0$ is expanding, so moving the center away and rescaling we get a length preserving map. $\endgroup$ Commented Aug 29, 2017 at 16:37