Locally Closed Orbits in Real Algebraic Geometry Let $G$ be a real algebraic group, and let $X$ be a real affine $G$-variety. I am looking for conditions on $G$ and $X$ for which the $G$-orbits are known to be locally closed in the Zariski topology on $X$. Please feel free to offer any conditions you desire. References would also be much appreciated. Of particular interest to me is the case in which $G$ is semisimple and $X$ is a finite-dimensional $G$-representation.
Over $\mathbb{C}$, one usually establishes that orbits are locally closed by appealing to Chevalley's Theorem. It tells us that the an orbit contains a non-empty open subset of its closure. If we act on this open set by elements of $G$, then we realize the orbit as being open in its closure. Unfortunately, I believe Chevalley's Theorem fails over non-algebraically closed fields.  
ADDED: I would be perfectly happy with conditions under which a $G$-orbit is the unique orbit of maximal dimension in its Zariski closure. This is weaker than being locally closed. 
 A: Let $x_0\in X(\mathbb{R})$, and consider the orbit map $f:g\mapsto g.x_0$, as a morphism of $\mathbb{R}$-schemes. Denote by $S\subset G$ the stabilizer of $x_0$. Chevalley's therorem asserts that $f$ factors as
$$G\longmapsto G/S \xrightarrow{\sim} Y \hookrightarrow X$$
where the first map is the (faithfully flat) canonical projection, the second is an isomorphism, and the third is a (locally closed) immersion. The $G(\mathbb{R})$-orbit of $x_0$ is $\Omega:=f(G(\mathbb{R}))\subset Y(\mathbb{R})\subset X(\mathbb{R})$. Of course, $Y$ is the "algebraic" orbit, and $Y(\mathbb{C})=G(\mathbb{C}).x_0$. 

Proposition. If $Y$ is connected (in particular if $G$ is), then $\Omega$ is Zariski-locally closed in $X(\mathbb{R})$ if and only if it is equal to $Y(\mathbb{R})$.  

The "if" part is trivial. Conversely, if $\Omega$ is Zariski-locally closed in $X(\mathbb{R})$, the same holds in $Y(\mathbb{R})$, so we may forget about $X$ from now on.
The key point is that (without any locally closed assumption) $\Omega$ is always open and closed in $Y(\mathbb{R})$, for the real topology. More generally:

Lemma. Let $S$ be a real algebraic group, and let $f:U\to V$ be an $S$-torsor (= principal homogeneous $S$-bundle), where $U$ and $V$ are $\mathbb{R}$-varieties. Then $f(U(\mathbb{R}))$ is open and closed in $V(\mathbb{R})$, for the real topology.  

(This is certainly well known, but I don't have a reference; in fact, it also works over $p$-adic fields).  
Proof of Lemma: since $f$ is automatically a smooth morphism, the induced map on real points is open (implicit function theorem), so $f(U(\mathbb{R}))$ is open. To see that it is closed, consider the characteristic map $c:V(\mathbb{R})\to H^1(\mathbb{R},S)$ sending $v\in V(\mathbb{R})$ to the class of $f^{-1}(v)$ as an $S$-torsor. Now $f(U(\mathbb{R}))=c^{-1}(e)$ where $e$ is the trivial class, so it suffices to see that $c$ is locally constant. For $\xi\in H^1(\mathbb{R},S)$, consider the twist $f^\xi:U^\xi\to V$ of $f$ by $\xi$: this is a torsor under the inner twist $S^\xi$ of $S$, and it has the property that $c^{-1}(\xi)=f^\xi(U^\xi(\mathbb{R}))$, which is open as we have seen. QED  
Now let us finish the proof of the proposition. Note that $Y$ is a smooth connected (hence irreducible) variety. Since $\Omega$ is open for the real topology it is Zariski-dense in $Y(\mathbb{R})$, so if it is Zariski-locally closed it must be Zariski-open. But then its complement in $Y(\mathbb{R})$ is Zariski-closed, hence has empty interior for the real topology (again because $Y$ is smooth) so it must be empty since it is (strongly) open. QED  
Remark. If $Y$ is not connected and $\Omega$ is locally closed, the conclusion is just that $\Omega=Y_1(\mathbb{R})$ where $Y_1$ is a union of components of $Y$.
