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This is a refinement of a question asked on MSE.

Let $G$ be a finite group and let $V$ be a finite-dimensional faithful complex representation of $G$. Consider $V$ as an affine complex variety. In general (though not always), the image of the origin in the quotient variety $V/G$ is singular. By the "singularity type" of the pair $G,V$ I refer to the isomorphism class of the ring $\widehat{\mathcal{O}_{V/G,0}}$, the completion of the local ring at this point.

At broadest (soft-question) level, what I want to know is, "how fine an invariant of the pair $G,V$ is the singularity type?" But let me narrow the scope in order to be able to ask something precise:

Suppose $G$ is nonabelian, $V$ has no one-dimensional subrepresentations, and $V/G$ is singular at the image of the origin. Further suppose $A$ is a finite abelian group and $W$ is a faithful representation of $A$. Is it possible for $G,V$ to have the same singularity type as $A,W$?

Comments: (1) The answer is certainly "yes" without the stipulation that $V/G$ be singular at the origin; both abelian and nonabelian groups can have smooth quotients, by Chevalley-Shephard-Todd since both abelian and nonabelian groups can be reflection groups. (2) The stipulation "$V$ has no one-dimensional subrepresentations" is there to rule out a match resulting from $G,V$'s singularity "actually" coming from $G$'s abelianization. This can happen for example if $V$ decomposes as a sum $W\oplus L$ with $W$ such that $W/G$ is smooth and $L$ is a sum of one-dimensional representations.

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    $\begingroup$ It is possible. For instance, the standard action of the symmetric group $G_0=\mathfrak{S}_n$ on $V=\mathbb{C}^n/\mathbb{C}(1,\dots,1)$ has no one dimensional subrepresentations, yet the quotient is smooth since $\mathfrak{S}_n$ is generated by elements (transpositions) that act by pseudoreflections. The quotient $V/G_0$ is smooth by Chevalley-Shephard-Todd. Now do something silly: let $G$ be the product of $G_0$ with the group $A$ of $d^{\text{th}}$ roots of unity acting by scaling. Then $V/G$ is $W/A$ for $W$ the sum or the characters $\overline{2},\dots,\overline{n}$ of $A$. $\endgroup$
    – Jason Starr
    Commented Mar 26, 2016 at 14:37
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    $\begingroup$ Perhaps you already know this, but it make sense to form the quotient of $G$ by the normal subgroup $G_0$ generated by all elements acting by pseudoreflections. Then the quotient group $H=G/G_0$ acts on the smooth space $U=V/G_0$. (Maybe you need to iterate, but I think not). This induced action has no pseudoreflections, so, if the action of $H$ on $U$ is generically free, you can recover $H$ from $U/H$ by forming the fundamental group of the complement of the singular set. $\endgroup$
    – Jason Starr
    Commented Mar 26, 2016 at 14:41

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To cherry-pick the answer from Jason Starr's comment above: the condition you want is that $G$ acts on $V$ without any element giving a pseudo-reflection. Chevalley-Shephard-Todd tells you that you must lose all information about pseudo-reflections, but you don't lose anything else. When there are no pseudo-reflections, the singular set of $V/G$ is exactly the image of the elements in $V$ with non-trivial stabilizer, so the smooth locus of $V/G$ is a quotient of a simply connected space by $G$, and thus its $\pi_1$ is $G$. This should be reconstructable from the singularity type using the Galois theory of the fraction field of the completion at the origin (looking at the Galois group of the largest Galois extension with no ramification at smooth points, I think).

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  • $\begingroup$ Wait - I was totally convinced by this when I read it last month - but, looking at it again just now - what guarantee do we have that the preimage in $V$ of the smooth locus of $V/G$ is simply connected? Is it just that it's a subvariety, so real codimension at least 2? $\endgroup$
    – benblumsmith
    Commented Apr 21, 2016 at 20:19
  • $\begingroup$ @benblumsmith The preimage of the smooth locus is $V$ minus the fixed subspace for each individual group element. All of these have complex codimension $\geq 2$, since none of them are pseudoreflections, so removing them doesn't change the fundamental group. $\endgroup$
    – Ben Webster ♦
    Commented Apr 22, 2016 at 0:04
  • $\begingroup$ Got it. I guess another way to see complex codimension $\geq 2$ is that $k[V]^G$ is integrally closed (since a ratio of invariant polynomials that is itself a polynomial is practically tautologically an invariant polynomial), thus $V/G$ is nonsingular in codimension one; and the preimage in $V$ of the singular locus of $V/G$ will have the same dimension since the map $V\rightarrow V/G$ is finite. $\endgroup$
    – benblumsmith
    Commented Apr 27, 2016 at 20:49

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