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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 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.

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  • $\begingroup$ What is an almost geometric quotient? $\endgroup$
    – Friedrich Knop
    Commented Aug 10, 2016 at 23:53
  • $\begingroup$ @FriedrichKnop See my edit. $\endgroup$
    – Avi Steiner
    Commented Aug 10, 2016 at 23:58
  • $\begingroup$ So a quotient is almost geometric if the generic orbit is closed? Affine varieties satsfying this conditions are usually called "stable". $\endgroup$
    – Friedrich Knop
    Commented Aug 11, 2016 at 0:20
  • $\begingroup$ @FriedrichKnop I'm using terminology from Cox, Little, and Schenck's "Toric Varieties". $\endgroup$
    – Avi Steiner
    Commented Aug 11, 2016 at 0:26

1 Answer 1

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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.

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  • $\begingroup$ Do you know if there's anything known about the less restrictive condition that the inclusion of rings $R^G\to R$ is pure injective? $\endgroup$
    – Avi Steiner
    Commented Aug 11, 2016 at 0:49
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    $\begingroup$ The injection $R^G\hookrightarrow R$ is always pure since $R^G$ is a direct summand of $R$ as an $R^G$-module. So "pure injective" is no restriction at all, at least in characteristic zero. $\endgroup$
    – Friedrich Knop
    Commented Aug 11, 2016 at 0:53
  • $\begingroup$ Do you have a reference for this fact about direct summands? $\endgroup$
    – Avi Steiner
    Commented Aug 11, 2016 at 1:01
  • $\begingroup$ That $R^G$ is a direct summand of $R$ follows from complete reducibility. The projection $R\to R^G$ is sometimes called the Reynolds operator. The purity of the inclusion of a direct summand is considered "obvious" by Hochster-Roberts in section 6 of their famous paper "Rings of invariants [$\ldots$] are Cohen-Macaulay". $\endgroup$
    – Friedrich Knop
    Commented Aug 11, 2016 at 11:39

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