Timeline for What is the universal enveloping algebra?
Current License: CC BY-SA 2.5
8 events
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May 19, 2010 at 8:29 | comment | added | Torsten Ekedahl | I am by the way not quite convinced that the envelopping algebra can be constructed in the way I suggested for a general operad $O$. What I use is that $U(Lie(V)) = S(Lie(V))$, i.e., that there is a $\Sigma$-module $S$ such that the enveloping algebra of a free algebra is isomorphic to $S$ applied to the free algebra. I don't know if that is true in general. | |
May 19, 2010 at 8:13 | comment | added | Torsten Ekedahl | The category of projective (or flat) modules over a commutative ring with tensor product as monoidal operation. | |
May 18, 2010 at 20:10 | comment | added | Victor Protsak | BB: Sorry, I've misunderstood what you were asking, then. If you can decompose the tensor algebra into $S_n$ isotypic components then everything goes through for a general O, as Thorsten's updated post shows in the case of $Lie$. BTW, what is a good example of C w/o cokernels in general, but with (co)kernels of idempotent projectors? | |
May 18, 2010 at 9:15 | comment | added | Bugs Bunny | That is right, no cokernels and I suspect that they may not be required. | |
May 17, 2010 at 22:04 | comment | added | Torsten Ekedahl | But the posed problem assumed that we wouldn't necessarily have cokernels. I get the impression that that was the whole point of the question and I think I have managed to avoid them in my answer. | |
May 17, 2010 at 21:53 | comment | added | Victor Protsak | Why not? Universal enveloping algebra is a quotient of tensor algebra, and in order to be able to form quotients, cokernels are needed. That seems like the natural setting (and is covered by Theo's condition 2). | |
May 17, 2010 at 20:23 | comment | added | Torsten Ekedahl | That is true but it seems that the general construction of the enveloping algebra requires cokernels and we are not allowed to assume that. | |
May 17, 2010 at 19:42 | history | answered | Victor Protsak | CC BY-SA 2.5 |