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Apr 30, 2013 at 14:37 vote accept Alireza
Apr 30, 2013 at 8:04 answer added Salvatore Siciliano timeline score: 2
Apr 30, 2013 at 2:18 comment added MTS I think this still needs to be clarified. Take $g$ to be a Lie algebra, $I \subseteq g$ an ideal. Are you asking whether $U(I)$ (the universal enveloping algebra of $I$) is an ideal in the universal enveloping algebra $U(g)$?
Apr 30, 2013 at 1:59 comment added Alireza @Sam Gunningham: Exactly!
Apr 30, 2013 at 1:55 comment added Sam Gunningham Lie ring = universal enveloping algebra?
Apr 30, 2013 at 1:52 comment added Alireza Take a Lie ring and build a quotient out of it! Here I mean if $g$ is a Lie algebra and $L$ its Lie ring (an associative ring equipped with a bracket operator $[,]$ can be made into a Lie ring), then $L/I$ defines a quotient ring where I is a two-sided ideal in $L$.
Apr 30, 2013 at 1:51 comment added MTS I don't understand your question. What quotient ring do you mean? What does a Lie group have to do with anything?
Apr 30, 2013 at 1:03 comment added Fernando Muro You need not prove that the quotient of a Lie algebra by an ideal is a Lie algebra, that's done in books. BTW, what do you mean by 'quotient ring'?
Apr 29, 2013 at 23:55 history asked Alireza CC BY-SA 3.0