Let $X$ be an affine variety over the field $\mathbb{R}$ of real numbers. And suposse the group $GL(n,\mathbb{R})$ acts smoothly on $X$. By the Closed Orbit Lemma, the orbit $Gx$ of a point $x \in X$ is locally closed in $X$. Is the Zariski closure of $Gx$ equals to its closure in the Euclidean topology? In the case of algebraic closed fields the answer is yes, but I guess the answer is negative in the real case.
2
-
$\begingroup$ @DietrichBurde Think not... Just have a look at the lecture notes on Linear Algebraic Groups by Brian Conrad $\endgroup$– gal16Oct 13, 2022 at 11:07
-
$\begingroup$ @DietrichBurde the particular context is the one of the closure of $GL(n,\mathbb{R})$ orbits on the affine variety of Lie algebra laws. I know how to prove that the two closures are equal in the case of complex Lie algebras, but I don't know how to proceed in the real case. Do you know any references? Thanks in advance. $\endgroup$– gal16Oct 13, 2022 at 11:16
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
As to your last sentence, no, the claim is also correct for real numbers.
For a reference, see for example the paper by Maryna Nesterenko and Roman Popovych here, on page $8$.
The only claim in the closed orbit lemma requiring an algebraically closed field is that then the orbits of minimal dimension are closed. Otherwise the proof for complex numbers might be still working, or at least one can assume first that the field is $\Bbb C$, and then conclude further.
References for the complex case:
If an orbit $G\cdot x$ is closed in the standard topology, is it Zariski-closed?
The Zariski closure of a constructible set is the same as the standard closure?
answered Oct 13, 2022 at 11:32