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Nov 2, 2023 at 3:48 history edited GH from MO CC BY-SA 4.0
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May 9, 2012 at 9:07 comment added GH from MO Dear David Roberts: Thank you, that was very kind of you!
May 9, 2012 at 5:46 comment added David Roberts And welcome to the 10k club, GH!
Jan 15, 2012 at 15:03 answer added Alain Valette timeline score: 1
Jan 14, 2012 at 1:21 answer added GH from MO timeline score: 3
Jan 12, 2012 at 6:54 comment added Alexander Chervov Let us take R=Identity. Do we agree that trace( pi(f)) is eigendistribution ? It seems it is character. Center of U(g) should act on it by scalars.
Jan 12, 2012 at 6:53 answer added Alain Valette timeline score: 3
Jan 12, 2012 at 3:29 comment added GH from MO Thanks, Mariano, that would make sense. I hope someone can clarify this with proof.
Jan 12, 2012 at 2:59 comment added Mariano Suárez-Álvarez I have not looked at the paper, but I guess an eigendistribution is justa distribution which is an eigenvalue. That is, when you act on it by a central element in the enveloping algebra, it gets multiplied by a scalar.
Jan 12, 2012 at 2:44 history asked GH from MO CC BY-SA 3.0