Timeline for Is it possible to reconstruct a finitely generated group from its category of representations?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 12, 2021 at 6:22 | comment | added | PULITA ANDREA | @Johannes Thanks for your help ! | |
| May 10, 2021 at 12:32 | comment | added | Johannes Hahn | @PULITAANDREA I've expanded the proof. | |
| May 10, 2021 at 12:31 | history | edited | Johannes Hahn | CC BY-SA 4.0 |
added 770 characters in body
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| May 10, 2021 at 11:28 | comment | added | PULITA ANDREA | @Johannes Hahn can you expand the last part of the proof please ? Why is $u$ a group element ? | |
| Mar 3, 2019 at 15:55 | comment | added | Johannes Hahn | I've edited my post to correct my previous argument as you suggested. | |
| Mar 3, 2019 at 15:14 | history | edited | Johannes Hahn | CC BY-SA 4.0 |
Incorporated the monodial structure because of Simon Henry's comment
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| Mar 3, 2019 at 14:47 | comment | added | Johannes Hahn | Oooh. So that was what I missing! $\tau^{k[G]}(1)$ can be any unit of $k[G]$, not just a group element. I should have seen that. | |
| Mar 3, 2019 at 12:56 | comment | added | Simon Henry | The answer as stated seems false to me: The fiber functor functor is representable (as a $k$-linear functor) by the regular representation. So the $k$-enriched natural transformation are exactly the automorphisms of the regular representations. And there are already a lot more of these than justs $G$. A more reasonable statement would be that $G$ identifies with the enriched & monoidal natural transformation, meaning that you also needs the monoidal structure. | |
| Mar 3, 2019 at 1:04 | vote | accept | Grisha Papayanov | ||
| Mar 2, 2019 at 23:31 | history | answered | Johannes Hahn | CC BY-SA 4.0 |