Question

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 $\mathbb{Z}$.

Answer 1

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$}.

Answer 2

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.