8
$\begingroup$

$\def\CC{\mathbb{C}}$Let $K = \CC(x_1, \ldots, x_n)$ and let $G$ be a countable group of automorphisms of $K$; in the cases I care about, $G \cong \mathbb{Z}$. Then the field of $G$-invariants, $K^G$, is an extension of $\CC$ of some transendence degree $d$. I would like to know under what circumstances I can say that the Zariski closure of the $G$-orbit of a generic $n$-tuple has dimension $n-d$.

I want to be a little sloppy about what I mean by generic, but I don't think there is anything deep in that issue. For "generic" $x \in \CC^n$, all the rational maps $g$ are defined at $x$, so we can consider $\{ g(x) \}_{g \in G} \subset \CC^n$ and take its Zariski closure. Every function in $K^G$ is constant on this Zariski closure, so the Zariski closure has dimension $\leq n-d$.

Question Can I conclude that the dimension is "generically" $n-d$?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
7
$\begingroup$

Yes for $G=\mathbb Z$, see Theorem 4.1 of this paper by Amerik-Campana.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks! Following up on your answer, I have found a paper which does the general case; see my answer. $\endgroup$
    – David E Speyer
    Commented Mar 1, 2019 at 5:24
3
$\begingroup$

Following citation links from the paper Jorge directed me to, the case of general $G$ is done in

Bell, Jason; Ghioca, Dragos; Reichstein, Zinovy, On a dynamical version of a theorem of Rosenlicht, Ann. Sc. Norm. Super. Pisa, Cl. Sci. (5) 17, No. 1, 187-204 (2017). ZBL1401.14071. (arXiv link)

Disclaimer -- I have only read the MR review.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .