Timeline for A question about the Tannaka-Krein reconstruction of finite groups
Current License: CC BY-SA 4.0
9 events
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| Oct 11, 2019 at 18:05 | vote | accept | Zhaoting Wei | ||
| Oct 11, 2019 at 17:16 | comment | added | Noah Snyder | @ZhaotingWei: There's two separate issues here: Tannakian reconstruction which recovers an algebraic structure from a category with a fiber functor, and the relationship between symmetric tensor categories and fiber functors. Rep(G) might have many monoidal fiber functors, and each of them will give you a Hopf algebra such that Rep(G) = H-mod via reconstruction. But it only has one symmetric fiber functor, essentially because you can see what vector space you have to assign by looking at which wedge power vanishes. | |
| Oct 11, 2019 at 17:11 | comment | added | Noah Snyder | Corrected an error. H = End(F) is all endomorphisms, not just the tensor automorphisms. This is why the tensor automorphisms are the grouplike elements (and not just all invertible elements). | |
| Oct 11, 2019 at 17:10 | history | edited | Noah Snyder | CC BY-SA 4.0 |
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| Oct 11, 2019 at 16:47 | comment | added | Zhaoting Wei | @NoahSnyder Thank you! Do you know then why $\text{rep}_{\mathbb{C}}(G)$ as a monoidal category cannot reconstruct the group $G$? | |
| Oct 11, 2019 at 16:01 | comment | added | Noah Snyder | Yes. I edited accordingly and added an additional clarification. | |
| Oct 11, 2019 at 16:01 | history | edited | Noah Snyder | CC BY-SA 4.0 |
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| Oct 11, 2019 at 15:59 | comment | added | S. Carnahan♦ | Just to clarify the last sentence, are you are saying that if $H$ is not isomorphic to a group ring, then Rep(G) will be different from H-mod? | |
| Oct 11, 2019 at 15:33 | history | answered | Noah Snyder | CC BY-SA 4.0 |