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Jul 2, 2010 at 8:01 answer added Bruce Westbury timeline score: 5
Jun 15, 2010 at 10:44 vote accept Bugs Bunny
May 17, 2010 at 19:42 answer added Victor Protsak timeline score: 2
May 17, 2010 at 19:06 answer added Torsten Ekedahl timeline score: 6
May 17, 2010 at 18:00 answer added Theo Johnson-Freyd timeline score: 11
May 17, 2010 at 16:56 history edited Bugs Bunny CC BY-SA 2.5
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May 17, 2010 at 16:54 comment added Bugs Bunny No, not abelian, Torsten, just symmetric monoidal, with homs as vector spaces. To construct $S({\mathfrak g})$, one needs to complete by direct summands (or idempotents, maybe, Karoubian completion - my terminology is wonky). The symmetric group $S_n$ acts on the tensor power $T^n ({\mathfrak g})$, then its group algebra acts and $S^n ({\mathfrak g})$ is a direct summand of $T^n ({\mathfrak g})$ corresponding to the trivial idempotent... I will correct the question.
May 17, 2010 at 16:14 comment added Torsten Ekedahl Exactly what kind of category are we talking about? If it is abelian you can construct the enveloping algebra as a quotient as in the usual case. If not how do you construct the symmetric algebra?
May 17, 2010 at 15:23 history asked Bugs Bunny CC BY-SA 2.5