Generalizations of Chevalley–Shephard–Todd's Theorem?

Question

Major Edit

I will reformulate my question signicantly, given Anton Geraschenko's comment. The old version of the question is bellow.

For simplicity, my base field is $\mathbb{C}$. If $G<\operatorname{GL}_n$ is a finite group of automorphisms of $\mathbb{C}^n$, then Chevalley–Shephard–Todd's Theorem says that $\mathbb{C}^n/G \simeq \mathbb{C}^n$ if and only if $G$ is a complex reflection group. My question was, given $G$ a non necessarily linearizable finite group of automorphisms of $\mathbb{C}^n$, to know what should be necessary or sufficient conditions for $\mathbb{C}^n/G$ to be isomorphic to $\mathbb{C}^n$ in this generality.

Clearly, a necessary condition is that the quotient $\mathbb{C}^n/G$ is smooth. However, if $G$ is a subgroup of $\operatorname{GL}_n$ this conditions is also sufficient. A proof of this fact can be found in Bourbaki's 'Groupes et algèbres de Lie', Chapter V.

Moreover, as Anton Geraschenko pointed out, there are known necessary and sufficient conditions on $G$ for $\mathbb{C}^n/G$ to be smooth:

Theorem: The quotient variety $\mathbb{C}^n/G$ is smooth if and only if the action on $T_x$ of the stabilizer $G_x$ is by complex reflections.

As I learned, this follows from Cartan's theorem on local holomorphic linearization.

Updated Question

The updated question, hence, is:

For arbitrary finite $G$, if $\mathbb{C}^n/G$ is smooth, is the quotient variety necessarily isomorphic to $\mathbb{C}^n$?

My guess is that the answer is No but I would not be surprised by a Yes.

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Original Question

For the sake of simplicity, I will only consider fields $\mathsf{k}$ of characteristic $0$.

By $R_n$ I will mean the polynomial algebra over $\mathsf{k}$ with $n$ indeterminates. By $\operatorname{Aut} R_n$ I will denote the group of (algebra) automorphisms of $R_n$.

We can naturally consider $\operatorname{GL}_n$ as a subgroup of $\operatorname{Aut} R_n$. The Chevalley-Shephard-Todd Theorem is a celebrated result in classical invariant theory. It says that if $G< \operatorname{GL}_n$ is a finite group, then the invariant subalgebra $R_n^G$ is polynomial (in necessarly $n$ indeterminates) if and only if $G$ is a pseudo-reflection group (in its natural representation). When $\mathsf{k}$ is the rational numbers, then pseudo-reflection groups are just the Weyl groups; over the reals, the pseudo-reflection groups are the finite Coxeter groups; over $\mathbb{C}$, these are called unitary reflection groups and were classified by Shephard and Todd.

Now, $\operatorname{GL}_n$ is just a subgroup of the full automorphism group of $R_n$. A subgroup $G$ of $\operatorname{Aut} R_n$ is called linearizable if it is conjugated to a subgroup of $\operatorname{GL}_n$ inside $\operatorname{Aut} R_n$.

The study of which groups of automorphisms can be linearized is a very important topic in affine algebraic geometry, and for a classical (but somewhat outdated) survey of this question we have Kraft's 'Challenging problems on affine $n$-spaces'.

Restricting our attention to finite groups of automorphisms, we have that in $R_1$ every one one them is linearizable; the same can be shown for $R_2$, using the Jung–van der Kulk Theorem on the structure of $\operatorname{Aut} R_2$.

However, for $n > 2$ the situation is different. It was shown by Freudenburg and Moser-Jauslin ['A nonlinearizable $S_3$ action on $\mathbb{C}^4$'] that, for every $n \geq 4$, there is non-linearizable algebraic action of $S_3$ on $\mathbb{C}^n$ (and, as far as I know, the question of non-linearizable finite groups of automorphisms for $n=3$ is still open).

So, since finite non-linearizable subgroups of $\operatorname{Aut} R_n$ exists, I am interested in the following problem:

For arbitrary finite subgroups $G < \operatorname{Aut} R_n$, are there are known necessary or sufficiente conditions for the invariant subalgebra $R_n^G$ to be polynomial?

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A second, more general question, I have, is the following: let $X$ be an irreducible complex affine variety, $G$ a finite group of automorphisms of $X$, and $\mathcal{O}(X)$ the algebra of regular functions on $X$. When can we say that $\mathcal{O}(X)^G \simeq \mathcal{O}(X)$?

Answer 1

I have exchanged e-mail with some specialists on invariant theory and affine algebraic geometry, and here is a summary of what I learned.

For the sake of simplicity (although this is by no means necessary), I will assume the base field to be $\mathbb{C}$. Again, I will denote by $P_n$ the polynomial algebra in $n$-indeterminates.

My original question was, given an arbitrary (in general, non-linearizable) finite group $G$ of automorphisms of $P_n$, to find necessary or sufficient conditions for the subalgebra of invariants $P_n^G$ to be again polynomial.

In this generality, with current knowledge, an answer to this question is hopeless. The main reasons are that, for $n>2$, the structure of $\operatorname{Aut} P_n$ is to a large extent unknown, and that very few families of non-linearizable group actions are known.

However, there is something that can be said in this generality. Instead of focusing the atention on $P_n^G$, we can considerer separating invariants, introduced by Derksen and Kemper, which is a weaker concept than generating invariants.

Let $X$ be an affine variety (not necessarily irreducible), and $G$ be a finite group of automorphisms of it. A subset $S \subset \mathbb{C}[X]^G$ is called separating if, for all $x,y \in X$, the following holds: if there is an invariant $f \in \mathbb{C}[X]^G$ such that $f(x) \neq f(y)$ then there exists $h \in S$ such that $h(x) \neq h(y)$. $\gamma_{sep}$ denotes the smallest interger $m$ for which there exists a separating set of size $m$.

An automorphism $\sigma$ of $X$ is called a reflection if the fixed subspace $X^\sigma$ has codimension at most $1$. Then we have a result by F. Reimers (Theorem 2.4 in the paper Separating invariants of finite groups) that says

Theorem: Assume that the $G$-variety $X$ is connected, Cohen-Macauley and that $G$ is generated by elements with a fixed point. If $\gamma_{sep}=\operatorname{dim}X$, then $G$ is generated by reflections.

Remark: If $X=\mathbb{C}^n$ and $G<\operatorname{GL}_n(\mathbb{C})$, every element has the origin as a fixed point.

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My second question was about affine varieties $X$ and finite groups of automorphism $G$ such that $X/G$ is isomorphic to $X$.

When $X$ is an algebraic torus, there is a nice result in this direction, which is a multiplicative invariant theory analogue of Chevalley-Shephard-Todd Theorem.

Let $G$ be a finite group. A $G$-lattice $L \simeq \mathbb{Z}^n$ is a lattice together with a faithful homomorphism $G \rightarrow \operatorname{GL}_n(\mathbb{Z})$. Then $G$ acts on the group algebra $\mathbb{C}[L]$, which is just the algebra of Laurient polynomials $\mathbb{C}[x_1^\pm, \ldots, x_n^\pm]$, which is of course the algebra of regular functions of $\mathbb{T}^n$.

An element $g \in G$ is called a pseudo-reflection if it acts on $\mathbb{Q} \otimes_\mathbb{Z} L$ be pseudo-reflections, and $G$ is called a pseudo-reflection group if it is generated by pseudo-reflections.

Then we have the following theorem (see Theorem 7.1.1 in M. Lorenz book Mutiplicative invariant theory)

Theorem: Let $L$ be a $G$-lattice, for finite $G$. Then the following are equivalent

a) $\mathbb{C}[L]^G$ is regular.

b) $\mathbb{C}[L]$ is a projective $\mathbb{C}[L]^G$-module.

c) $\mathbb{C}[L]^G \simeq \mathbb{C}[\mathbb{Z}_+^r \oplus \mathbb{Z}^s]$, a mixed Laurient polynomial algebra, with $r+s=\operatorname{rank}(L)$.

d) $G$ is a pseudo-reflection group and $\mathbb{Z}[L]^G$ is a unique factorization domain.