Return to Answer

added 155 characters in body

Source Link

edited Apr 21, 2016 at 8:14

Johannes Huisman

1.4k
9
11

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: for any finite disjoint unionLet $U$ be a finite union of topological surfaces, each of which is either nonorientable or orientable and satisfying one of genus $\leq1$,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.

.

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: for any finite disjoint union $U$ of topological surfaces, each of which is either nonorientable or orientable and of genus $\leq1$, 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.

.

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.

.

Source Link

created Apr 21, 2016 at 7:56

Johannes Huisman

1.4k
9
11

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: for any finite disjoint union $U$ of topological surfaces, each of which is either nonorientable or orientable and of genus $\leq1$, 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.

.