Timeline for How fine an invariant of a representation is its quotient singularity?

6 events

when toggle format	what		by	license	comment
Apr 13, 2017 at 12:19	history	edited	CommunityBot		replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Mar 28, 2016 at 18:54	vote	accept	benblumsmith
Mar 26, 2016 at 15:20	answer	added	Ben Webster♦		timeline score: 6
Mar 26, 2016 at 14:41	comment	added	Jason Starr		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.
Mar 26, 2016 at 14:37	comment	added	Jason Starr		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 $A$.
Mar 26, 2016 at 14:09	history	asked	benblumsmith	CC BY-SA 3.0