MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a (smooth) algebraic variety (over $\mathbb{C}$). Let $G \subset \operatorname{Aut}(X)$ be a subgroup of automorphisms of $X$. Is it true that for any $x\in X$ the closure $\overline{O_x}$ of the orbit of $x$ is a (possibly singular) subvariety or subscheme of $X$?

If not, can stronger hypotheses be given to guarantee a subscheme structure? In the case I am interested in, $G$ is isomorphic to $\mathbb{Z}$.

share|cite|improve this question
thanks for edit. – Mohammad F. Tehrani Sep 6 '11 at 18:54

Yes, it can be singular. Your question is a little schizophrenic in that $X$ is algebraic, but your $G$ is not, so it becomes unclear whether you want closure in the analytic or the Zariski topology. The analytic closure is very unlikely to be algebraic, so I'm going to assume Zariski.

Let $X = {\mathbb C}^2$, $x = (1,1)$, $G$ generated by $[{4\atop 0} {0 \atop 8}]$.

Then the Zariski closure of the orbit $G\cdot x$ is {$ (a,b) : a^3 = b^2$}.

share|cite|improve this answer
I did not ask whether it is singular or not. I asked whether the closure of orbit is a sub scheme or not. – Mohammad F. Tehrani Sep 6 '11 at 17:07

One interpretation of your question (different from Allen's) is that you are asking about the topological closure of the orbit in the analytification of $X$. In that case, let $X = \mathbb{G}_m$, and let $G$ be generated by an irrational rotation. Then the closure of any point is a circle, which is not the analytification of any subvariety.

share|cite|improve this answer
you are right. this is a counter example. I think I should put more restrictions on $G$ to get a positive answer. – Mohammad F. Tehrani Sep 6 '11 at 17:09
$\epsilon$ simpler, let $\mathbb Z$ act on $\mathbb C$ by translation, obtaining $\mathbb Z$ as an analytically-closed but Zariski-dense subset. – Allen Knutson Sep 6 '11 at 21:51
I should have said that X is compact. – Mohammad F. Tehrani Sep 7 '11 at 1:13
Compactifying to $\mathbb{P}^1$ doesn't change my example substantially. – S. Carnahan Sep 7 '11 at 4:17
Nor mine (it just adds the one point). – Allen Knutson Sep 7 '11 at 12:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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