Timeline for How fine an invariant of a representation is its quotient singularity?
Current License: CC BY-SA 3.0
5 events
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| Apr 27, 2016 at 20:49 | comment | added | benblumsmith | 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. | |
| Apr 22, 2016 at 0:04 | comment | added | Ben Webster♦ | @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. | |
| Apr 21, 2016 at 20:19 | comment | added | benblumsmith | 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? | |
| Mar 28, 2016 at 18:54 | vote | accept | benblumsmith | ||
| Mar 26, 2016 at 15:20 | history | answered | Ben Webster♦ | CC BY-SA 3.0 |