TITLE: Zariski closure and Euclidean closure.
QUESTION [0 upvotes]: 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.

REPLY [1 votes]: 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?
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