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Mar 21, 2018 at 19:14 history undeleted Dan Petersen
Mar 21, 2018 at 19:12 history deleted Dan Petersen via Vote
Feb 6, 2011 at 20:11 comment added Donu Arapura Xuanting Cai, perhaps you should have explained your interest in the question. Dan's and the other answers are certainly relevant, but you might find the book by Lubotzky and Magid "Varieties of representations..." gives more specific answers. For the record the (closed) points of the character variety correspond isomorphism classes of semisimple representations, which are the stable points for the action.
Feb 6, 2011 at 19:32 comment added Dan Petersen Here I am assuming you are taking the "GIT quotient" which is the good quotient of $X^{ss}$ that I mentioned in my answer. It has the property that all the stable orbits correspond to actual points of the quotient (since the quotient of $X^s$ was supposed to be geometric) but some semistable orbits will be identified with each other.
Feb 6, 2011 at 19:28 comment added Dan Petersen I know nothing about character varieties; maybe you would have better luck asking a question specifically about them. Anyway you are right that the set of all orbits in your set-up is in a natural way the set of isomorphism classes of representations of G in $SL(2,\mathbb{C})$. Whether or not the orbits correspond bijectively to points in the quotient is a more delicate issue. You will have to linearize the action and figure out which points are stable and semistable (or rather, someone else already has done this and you will have to read it :p).
Feb 6, 2011 at 18:30 comment added Xuanting Cai what I am really interested is following: I have a group G, finitely presented. Consider R(G)=HOM(G,SL(2,C)), the space of all reps of G into SL(2,C). Then SL(2,C) acts on R(G) naturally, i.e. conjugation. So the usual quotient of this action is the space of orbits, which are all conjugacy classes of reps. Now the algebro-geometric quotient is not this. It is the character variety of G. Thanks.
Feb 6, 2011 at 18:12 comment added jlk @Dan Peterson: I am sure you know this, but it might be helpful to emphasize that the semi-stable $X^{\text{ss}}$ locus and the stable locus $X^{\text{s}}$ depend on the choice of projective embedding (i.e., the linearization $L$).
Feb 6, 2011 at 10:50 history edited Dan Petersen CC BY-SA 2.5
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Feb 6, 2011 at 9:34 history answered Dan Petersen CC BY-SA 2.5