# Smooth, irreducible surface with real part containing two projective planes

Let $X$ be a smooth and irreducible projective variety over $\mathbb{R}$ of dimension two. I am looking for an instance of such a variety where two distinct connected components of $X(\mathbb{R})$ are homeomorphic to the real projective plane $\mathbb{RP}^2$. Does someone know how to construct such a surface?

This is of course trivial if we don't want $X$ to be irreducible, since then we can just take the disjoint union of two projective planes.

[Corrected]

Any double cover of ${\bf RP}^2$ whose branch locus has degree $4n$ and no real component should do. An example is the surface $y^2 = x_0^4 + x_1^4 + x_2^4$ in the weighted projective space whose coordinates $(x_0:x_1:x_2::y)$ have degrees $1,1,1,2$. Thus $(x_0:x_1:x_2::y)$ is equivalent to $(\lambda x_0:\lambda x_1:\lambda x_2::\lambda^2 y)$ for nonzero $\lambda \in \bf R$, so there are well-defined components $y>0$ and $y<0$, and the projection to $(x_0:x_1:x_2)$ maps each component homeomorphically to ${\bf RP}^2$.

[Earlier I rashly omitted the "degree $4n$" condition and proposed the two-sheeted hyperboloid $y^2 = x_0^2 + x_1^2 + x_2^2$ as an example. That's clearly wrong because it's isomorphic (as a surface in ${\bf RP}^3$) with a Euclidean sphere.]

• what you call "the two-sheeted hyperboloid" has as real part a sphere rather than the union of two projective planes. For double covers with branch locus of higher degree I don't see why it should be the union of two projective planes. if it's true, can you give an argument? – Hans Apr 19 '16 at 18:08
• I noticed this error and was correcting it as you wrote. – Noam D. Elkies Apr 19 '16 at 18:13

Start with a fibration in conics $Y$ given by the affine equation $$x^2+y^2=-(t+2)(t+1)(t-1)(t-2).$$ It is clear that the set of real points $Y(\mathbf R)$ is a disjoint union of two spheres. Now, blow up $Y$ at two real points, one on each connected component of $Y(\mathbf R)$, and you have your surface $X$.

Such a surface is a geometrically rational surface, i.e., its complexification is rational. However, $X$ is not rational as a real surface, i.e., it cannot be parametrized by real rational functions. The topology of geometrically rational real surfaces has been completely classified by Comessatti: Let $U$ be a finite union of topological surfaces satisfying one of the following conditions:

• every connected component of $U$ is either nonorientable or orientable and homeomorphic to the sphere $S^2$, or
• $U$ is connected and homeomorphic to the torus $S^1\times S^1$.

Then there is a geometrically rational real algebraic surface $X$ such that $X(\mathbf R)$ is homeomorphic to $U$. Moreover, these are the only ones that can be realized by a geometrically rational surface. A general reference on real algebraic surfaces is Silhol's book: Real algebraic surfaces.

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