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.

  • $\begingroup$ I'm not quite sure I see the problem; according to EGA IV, 1.8.4. (or see e.g. wikipedia), Chevalley's theorem is true for any morphism of schemes of finite presentation. There is no mention of fields. So the same proof should work over any field to give the result you want. $\endgroup$ Sep 18, 2014 at 8:19
  • 1
    $\begingroup$ @Daniel: I think that the "$G$-orbits" here are the orbits of $G(\mathbb{R})$ acting on $X(\mathbb{R})$. $\endgroup$ Sep 18, 2014 at 9:52
  • $\begingroup$ @Laurent: I see. As the OP talked about the Zariski topology, I assumed he was interested in $X$ as a variety, rather than its real points. It would be useful if he could clarify. $\endgroup$ Sep 18, 2014 at 9:59
  • $\begingroup$ Yes, if one thinks of $G$ and $X$ as schemes over $\mathbb{R}$, then my $G$-orbit is indeed an orbit of $G(\mathbb{R})$ on $X(\mathbb{R})$. I had been thinking purely in terms of classical algebraic geometry. $\endgroup$ Sep 18, 2014 at 14:08

1 Answer 1


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 $Y$.

  • $\begingroup$ This is a fantastic answer! Thank you! Do you happen to know of any assumptions on $G$ and $X$ which guarantee that $Y(\mathbb{R})$ and $\Omega$ necessarily coincide? For instance, will they coincide when $G$ is semisimple? $\endgroup$ Sep 20, 2014 at 16:31
  • $\begingroup$ Take $G=SL_2$ and $Y=SL_2/(\pm1)=PGL_2$, with $x_0=1$. Then $\Omega=PSL_2(\mathbb{R})=SL_2(\mathbb{R})/(\pm1)$, while $Y(\mathbb{R})$ has another component, containing the class of $\begin{pmatrix}0&i\\ i&0\end{pmatrix}$ (the matrix is not real but its image in $Y$ is). I doubt very much that you can find interesting conditions on $G$ alone (well, "$G$ unipotent" would do, I guess). By the above discussion, the key point is the structure of stabilizers. $\endgroup$ Sep 20, 2014 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.