All varieties here are over $\Bbb C$. Let $G$ be a reductive algebraic group acting algebraically on affine $n$-space $\Bbb A^n$. Let $R$ be the coordinate ring of $\Bbb A^n$. Assume that the natural morphism $\pi\colon \Bbb A^n \to X=\operatorname{Spec}(R^G)$ is an almost geometric quotient. (**EDIT:** See [this question](https://math.stackexchange.com/q/1614319) for the definition of almost geometric quotient. )


**Question:** Are there "useful" (interpret as you will) conditions on $G$ which imply that $\pi$ is flat? The cases I care about are when $G$ is a closed subgroup of $(\Bbb G\_m)^n$ acting via the induced action. 



When is an almost geometric quotient flat?
Let $G$ be a reductive group acting linearly on affine space $\mathbb A^n$ and let $\pi:\mathbb A^n\to X$ be the categorical quotient. Then the following conditions are equivalent:


1. $\pi$ is flat.
2. $\mathcal O(\mathbb A^n)$ is a free $\mathcal O(X)$-module.
3. The morphism $\pi$ is equidimensional and $X$ is smooth.


These are observations by G. Schwarz (Lifting smooth homotopies). Because of 2, actions of this type are called "cofree". These are very rare. In fact, many classes of cofree representations have been classified (e.g. $G$ torus, $G$ simple, or $\mathbb A^n$ irreducible). In general, there are two extreme cases to consider:


If $G$ is finite then $\pi$ is automatically equidimensional. Thus $\mathbb A^n$ is cofree iff $X$ is smooth iff $G$ is generated by reflections (Shepherd-Todd).


If $G$ is connected then Popov conjectured that the condition on $X$ to be smooth is superfluous. So cofree and equidimensional should be equivalent. This conjecture has been verified in all cases where a classification is available. In particular, the case the PO is most interested in, namely that $G$ is a torus, has been settled by Wehlau (A proof of the Popov conjecture for tori). In fact he classifies all cofree actions of tori.


PS1: The condition that the action is linear is not essential. All one needs is that the $G$-variety is smooth and affine. One can reduce that case to the linear one using Luna's slice theorem.


PS2: The assumption that the action is stable does not simplify the problem.


