Timeline for Drinfeld center of a braided category
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
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Jan 6, 2020 at 15:11 | comment | added | Calvin McPhail-Snyder | That sounds right: it was a topics course and we were discussing Lie bialgebras and Poisson geometry. Thanks for your helpful comments. | |
Jan 5, 2020 at 23:55 | comment | added | Adrien | If you remember something about direct sums then Reshetikhin was probably talking about Lie bialgebras. | |
Jan 5, 2020 at 23:43 | comment | added | Adrien | Well in that example the functor $C\rightarrow Z(C)$ comes from an Hopf algebra morphism $D(H) \rightarrow H$ so this would be an equivalence iff this map was an isomorphism. What is true is that the double of a factorizable f.d. Hopf algebra $H$ is isomorphic as an algebra to $H\otimes H$. Categorically it implies $Z(H-mod)\simeq H-mod \boxtimes H-mod$ where $\boxtimes$ is an appropriate tensor product of categories. In particular, the double of an arbitrary f.d. Hopf algebra is factorizable, hence the double of a double is a tensor square of the original double. | |
Jan 5, 2020 at 20:19 | comment | added | Calvin McPhail-Snyder | The equivalence definitely won't be isomorphism, but instead some kind of Morita (?) equivalence. But it could be completely wrong as well: my only basis for this is Reshektihin once saying that he proved something along the lines of "the double of a double is a direct sum of the original double." | |
Jan 4, 2020 at 10:04 | comment | added | Adrien | Note that by definition, as a vector space $D(H)$ is $H \otimes H^*$ so $H$ and $D(H)$ are not even the same size, so they're basically never isomorphic (except when $H$ is trivial). | |
Jan 3, 2020 at 17:35 | vote | accept | Calvin McPhail-Snyder | ||
Jan 3, 2020 at 0:08 | history | became hot network question | |||
Jan 2, 2020 at 16:58 | answer | added | Sam Gunningham | timeline score: 11 | |
Jan 2, 2020 at 16:00 | history | asked | Calvin McPhail-Snyder | CC BY-SA 4.0 |