Given a smooth proper real algebraic variety can you find a smooth proper real algebraic variety defined over $\mathbb{Q}$ that is diffeomorphic to it?
$\begingroup$
$\endgroup$
20
-
7$\begingroup$ Probably the fake projective planes, which have no deformations and are not even defined over $\mathbb R$, provide counterexamples. $\endgroup$– Will SawinCommented Jun 9, 2020 at 18:27
-
1$\begingroup$ Do you mean defined over $\bar{{\mathbb Q}}$ rather than ${\mathbb Q}$? $\endgroup$– Moishe KohanCommented Jun 9, 2020 at 18:34
-
1$\begingroup$ I believe that Will's comment settles your question. As Moishe implicitly suggests, your question has a positive answer if you replace $\mathbb{Q}$ by $\bar{\mathbb{Q}}$. $\endgroup$– ChrisCommented Jun 9, 2020 at 19:20
-
8$\begingroup$ Wait, if it's not defined over $\bf R$ how can it be a "real algebraic variety"? $\endgroup$– Noam D. ElkiesCommented Jun 9, 2020 at 19:27
-
2$\begingroup$ @vrz The issue is that perturbing the equations a small amount may lead to a much smaller-dimensional variety if the original equations have some redundancy, but that equations with redundancy are necessary (not everything is a complete intersection). There is no way the proof can be made to work. $\endgroup$– Will SawinCommented Jun 9, 2020 at 21:30
|
Show 15 more comments
1 Answer
$\begingroup$
$\endgroup$
8
Edoardo Ballico and Alberto Tognoli proved in their paper "Algebraic models defined over $\mathbb{Q}$ of differential manifolds" (Geom. dedicata 42, 155-161, 1992) that every compact differential manifold is diffeomorphic to the real points of a regular affine variety defined over $\mathbb{Q}$.
For non-smooth algebraic varieties there are obstructions to descend from $\mathbb{R}$ to $\mathbb{Q}$, there is a recent paper by Adam Parusinski and Guillaume Rond "Algebraic varieties are homeomorphic to varieties defined over number fields" arXiv 1810.00808 on this subject.
Edit : let me recall some basic facts about real algebraic sets (I refer to "Real algebraic geometry" Bochnak, Coste and Roy).
A complete nonsingular affine real algebraic variety is projective (see BCR 3.4 p.74-75)
An algebraic subset of $\mathbb{R}^n$ is complete if and only if it is closed and bounded (3.4.9 and 3.4.10).
-
$\begingroup$ but we want a proper variety not affine one $\endgroup$– user145520Commented Jun 10, 2020 at 5:57
-
$\begingroup$ As in the comment I just wrote to the OP, here by "real algebraic variety" is it meant something for which $1/(1+x^2)$ is a globally defined regular function on the real line or not? $\endgroup$– QfwfqCommented Jun 10, 2020 at 12:02
-
$\begingroup$ @Qfwfq I expect not (or it would be a quite exotic notion of "regular function") $\endgroup$– YCorCommented Jun 10, 2020 at 17:29
-
$\begingroup$ @YCor: I agree it'd be exotic, but see Example1.5 here [ web.math.unifi.it/gruppi/algebraic-geometry/RealAG_basic.pdf ] - if that injection of real points (which I think is regarded there as a "map of real varieties") came from a map of schemes, it would imply that $x/(1+x^2)$ was a regular function on the real line (consider the case $n=1$, take a component of that map and read it in the affine chart $x_1\neq 0$ of $\mathbb{P}^1$, and let $x=x_0/x_1$ be the homogeneous coordinate for that chart). Or am I misunderstanding something? $\endgroup$– QfwfqCommented Jun 10, 2020 at 19:04
-
$\begingroup$ @Qfwfq I don't think the authors uses the word "regular". I don't know if there's a standard name for a rational map that is regular on the set of real points. $\endgroup$– YCorCommented Jun 10, 2020 at 19:15