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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 $\mathbb{Z}$.

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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$}.

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  • $\begingroup$ I did not ask whether it is singular or not. I asked whether the closure of orbit is a sub scheme or not. $\endgroup$ – Mohammad Farajzadeh-Tehrani Sep 6 '11 at 17:07
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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.

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

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