Timeline for Drinfeld center of a Deligne tensor product
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26, 2021 at 5:56 | vote | accept | JeCl | ||
| Aug 25, 2021 at 16:42 | answer | added | Chris Schommer-Pries | timeline score: 5 | |
| Aug 25, 2021 at 15:54 | answer | added | Dave Penneys | timeline score: 7 | |
| Aug 25, 2021 at 5:01 | comment | added | JeCl | I still think there ought to be a more elementary proof though?!? | |
| Aug 25, 2021 at 5:01 | comment | added | JeCl | I wondered whether there was a proof using the cobordism hypothesis. Thanks for explaining it! | |
| Aug 25, 2021 at 0:14 | comment | added | Theo Johnson-Freyd | ... Then $Z(\mathcal{C}) \cong \mathcal{C} \boxtimes_{C^e} \mathcal{C}^\vee$. This gives formulas for both sides of your equation entirely in terms of (balanced) tensor products. | |
| Aug 25, 2021 at 0:12 | comment | added | Theo Johnson-Freyd | There is a different way to say this which doesn't use the word "cobordism hypothesis". Working in the bicategory $\mathrm{Mod}(\mathcal{C}^e)$ of $\mathcal{C}^e$-bimodules, where $\mathcal{C}^e := \mathcal{C} \boxtimes \mathcal{C}^{\mathrm{mop}}$ and "$\mathrm{mop}$" means the monoidal opposite, we have a canonical isomorphism $Z(\mathcal{C}) \cong \mathrm{End}_{C^e}(\mathcal{C})$. But, again quoting arxiv.org/abs/1312.7188, $\mathcal{C}$ is dualizable as a $\mathcal{C}^e$-module. Call its dual "$\mathcal{C}^\vee := \mathrm{hom}_{C^e}(\mathcal{C}, \mathcal{C}^e)$".... | |
| Aug 25, 2021 at 0:08 | comment | added | Theo Johnson-Freyd | According to arxiv.org/abs/1312.7188, fusion categories are at least 2-dualizable. Thus for any fusion category $\mathcal{C}$, the cobordism hypothesis provides a TQFT which assigns $\mathcal{C}$ to a point. This TQFT assigns $Z(\mathcal{C})$ to a circle with outward boundary framing (i.e. to $\partial D^2$). The cobordism hypothesis is monoidal in the sense that the TQFTs depend monoidally on the value of a point. So, yes, in the fusion case this is an equivalence. | |
| Aug 24, 2021 at 18:59 | comment | added | მამუკა ჯიბლაძე | You might as well add that injective = fully faithful and surjective = every simple object of the codomain is a subquotient of an object in the image of the functor | |
| Aug 24, 2021 at 18:57 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 4.0 |
link added (Author's final version made available with permission of the publisher, American Mathematical Society. License or copyright restrictions may apply to redistribution; see http://www.ams.org/publications/ebooks/terms)
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| Aug 24, 2021 at 18:06 | history | asked | JeCl | CC BY-SA 4.0 |