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Let $K$ be an algebraically closed field of characteristic zero, and $X$ be an affine $K$-variety (identify $X$ with its set of $K$-points). Let $G$ be group acting "abstractly" on $X$, by which I mean there is simply a group homomorphism $\rho:G \to \operatorname{Aut}(X)$. Then $G$ also acts on the coordinate ring $K[X]$ by $K$-algebra automorphisms (there is an associated group homomorphism $\rho^*:G \to \operatorname{Aut}(K[X])$.

It is a classical fact that if $G$ is an algebraic $K$-group (identify $G$ with the group of points $G(K$)) acting algebraically on $X$ (i.e. the map $G \times X \to X$ is a morphism of varieties), then the orbits in $K[X]$ span finite-dimensional $K$-subspaces of $K[X]$.

Are there examples in which a group $G$ acts on an affine variety $X$ where this finite-dimensionality property for orbits in $K[X]$ holds for some other reason than the action being algebraic? In particular, I am interested in the case when $G$ is the elementary subgroup of a Chevalley group with coefficients in a ring of $S$-integers, something like $G = \operatorname{SL}_3(\mathbb{Z})$ or $G = \operatorname{SL}_3(\mathbb{Z}[\sqrt{d}])$.

So far, the only example I have is incredibly trivial, namely $X = \mathbb{A}^1_K$ is the affine line with coordinate ring $K[x]$. Since automorphisms of $K(x)$ are möbius transformations, $K$-algebra automorphisms of $K[x]$ are affine transformations. In particular, I believe this forces orbits to be finite-dimensional for any group acting on $K[x]$.

I tried extending this example to $X = \mathbb{A}^2_K$, using the result that automorphisms of $K[x,y]$ are tame. However, tameness is still not particularly close to being affine, even for automorphisms of finite order. I think this question is related.

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A locally finite action on an affine variety is basically algebraic. More precisely, it factors through an algebraic group action. Proof: By assumption there is a $G$-stable finite dimensional subspace $V\subseteq K[X]$ containing a set of generators of $K[X]$ (just take $V=\sum_i\langle Gf_i\rangle_K$ with generators $f_i$). This corresponds to a closed embedding $X\hookrightarrow V^*$. Let $\overline G:=\{g\in GL(V^*)\mid g(X)=X\}$. This is a Zariski closed subgroup of $GL(V^*)$ acting algebraically on $X$. Since $G$ acts on $V$ it also acts on $V^*$ leaving $X$ invariant. This yield a homomorphism $G\to\overline G$ such that the $G$-action is induced by the $\overline G$-action.

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  • $\begingroup$ This argument is exactly my motivation for the question. My goal is to find an example where this implication tells you something meaningful about an action which you did not already know was algebraic. $\endgroup$ Commented Apr 7, 2022 at 14:51
  • $\begingroup$ Your question was "Are there examples ... where this finite-dimensionality property ... holds for some other reason than the action being algebraic?" and the answer in "no" in the sense of my answer. Could it be that you meant to ask "Given a subgroup $G$ of an algebraic group $\overline G$ and $\overline G$ acts algebraically on $X$ does the induced $G$-action on $X$ have special properties?"?. $\endgroup$ Commented Apr 8, 2022 at 9:27
  • $\begingroup$ No, this is not what I meant to ask. I am studying the situation in which $G$ is an elementary subgroup of a Chevalley group, e.g. $\operatorname{SL}_n(\mathbb{Z})$. I am interested in finding an example of such $G$ acting on an affine variety in a locally finite way, but that we have deduced local-finiteness in some other way than already knowing it is just the restriction of an algebraic action of $\operatorname{SL}_n(k)$. $\endgroup$ Commented Apr 9, 2022 at 21:43
  • $\begingroup$ I wanted to know if there were situations where this could be deduced perhaps just working with $K[X]$ or the fact that $\operatorname{SL}_3(\mathbb{Z})$ is generated by elements of finite order. $\endgroup$ Commented Apr 9, 2022 at 21:52

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