Timeline for What is the universal enveloping algebra?
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 18, 2010 at 15:51 | comment | added | Theo Johnson-Freyd | @Bugs: Torsten's and my discussions I think prove that if you have access to the symmetric algebra, then you have the universal enveloping algebra, in characteristic 0. (In char=p, all hell breaks loose, but your question was in char=0.) So however you want to get your symmetric algebra, it gives you a UAE too. | |
| May 18, 2010 at 15:37 | comment | added | Theo Johnson-Freyd | @Torsten Ekedahl: That's a good point. In characteristic 0, you win if idempotents have (co)kernels. And, no, the star product above does not use a basis; I was speaking imprecisely, because it's easiest to understand what the construction is doing if you had honest "monomials". | |
| May 18, 2010 at 9:13 | comment | added | Bugs Bunny | Theo, my friend, I have no cokernels. How much easier would the life be if I had cokernels:-))! I doubt one can complete the category by adding cokernels either. Tell me if I am wrong here. Thanks for Deligne-Morgan reference. I will check it and report back... | |
| May 17, 2010 at 20:27 | comment | added | Torsten Ekedahl | I don't understand your comment about existence of the symmetric algebra. As far as I can see the sum over all n of the symmetric tensors in $V^{\otimes n}$ has a product given by the product in the tensor algebra followed by symmetrisation. This only requires that idempotents have kernels. On the other hand a formula for the star product can not use bases as it is supposed to make sense in an arbitrary symmetric monoidal category. | |
| May 17, 2010 at 18:00 | history | answered | Theo Johnson-Freyd | CC BY-SA 2.5 |