Timeline for Is it possible to reconstruct a finitely generated group from its category of representations?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 3, 2019 at 1:04 | vote | accept | Grisha Papayanov | ||
| Mar 2, 2019 at 23:31 | answer | added | Johannes Hahn | timeline score: 14 | |
| Mar 2, 2019 at 5:30 | comment | added | Grisha Papayanov | @JohannesHahn That answers my question, thank you very much! Could you please write an answer so I could mark the question as closed? | |
| Mar 2, 2019 at 2:47 | comment | added | Robert Furber | @JohannesHahn Actually, scratch what I just said. I think I misinterpreted your last sentence. If you mean that (for $G$ a discrete group, $k$ a field) the (non-monoidal) endmorphisms of $F : \mathbf{Rep}(G)_k \rightarrow \mathbf{Vect}_k$ correspond to elements of $kG$ by the enriched Yoneda lemma (twice), and then the grouplike elements of $kG$ are exactly those that map to monoidal natural transformations, and therefore recover $G$, then I agree that this answers the question, and if you write it as an answer I will upvote it. | |
| Mar 2, 2019 at 1:55 | comment | added | Robert Furber | @JohannesHahn As stated in the question, there are infinite finitely-presented discrete groups that have no finite-dimensional representations on vector spaces except the trivial rep (and the zero rep). Any finitely-presented infinite simple group, such as Thompson's groups $T$ and $V$, is like this, because any representation is faithful, and any group with a faithful representation on a finite-dimensional vector space is residually finite (Mal'cev's theorem) which contradicts the group being simple. | |
| Mar 1, 2019 at 17:56 | comment | added | Johannes Hahn | What am I missing here? Isn't $t: G\to Aut(F), g\mapsto (t_V^g)_{V\in Rep_k(G)}$ an isomorphism, where $F: Rep_k(G)\to k-mod$ is the fibre functor and $t_V^g$ is $F(V)\to F(V), v\mapsto gv$ ? $t$ is easily seen to be bijective by evaluation on the regular representation. Isn't that the whole point of the Tannakian theorem that one does not need the regular representation if one has all the finite-dimensional ones? | |
| Mar 1, 2019 at 17:47 | comment | added | Ben Wieland | You cannot recover the group from the group ring as a ring, but you can recover it from the group ring as a Hopf algebra. A Hopf algebra contains a subset of grouplike elements: $\{x\;|\;\Delta x=x\otimes x\}$. Over a field, the nonzero grouplike elements recover the group. Invertible grouplike elts for general nonzero ring? Alternately, Tannaka's reconstruction theorem identifies the group with the group of natural automorphisms of the fiber functor. Krein's hypotheses about dualizability are only necessary to identify a category of representations, but you assume that you have one. | |
| Mar 1, 2019 at 11:12 | comment | added | Robert Furber | You can then tensor the group rings by any ring $R$ you like to get an isomorphism of the corresponding $R$-algebras. However, finite groups are compact groups, so we can use the Tannakian theory for compact groups (equivalently affine proalgebraic groups over $\mathbb{R}$). The key point is that we need the symmetric monoidal structure to recover the group multiplication. This doesn't affect the question as asked, because the monoidal structure was explicitly included. | |
| Mar 1, 2019 at 11:07 | comment | added | Robert Furber | There are groups $G_1,G_2$ such that $\mathbb{Z}G_1 \cong \mathbb{Z}G_2$, as described here: jstor.org/stable/3062112 | |
| Mar 1, 2019 at 8:58 | comment | added | YCor | I read "over any ordinary field"; I guess it means any field in which the order of the groups is invertible. | |
| Mar 1, 2019 at 8:53 | comment | added | Grisha Papayanov | @YCor do you mean isomorphic over any field? Over algebraically close field of char 0, group algebra of a finite group is just direct sum of matrix algebras, so the group algebra alone indeed doesn't give much information. I have to remember, at the very least, the usual Tannakian stuff (tensor product and fiber functor). | |
| Mar 1, 2019 at 8:31 | comment | added | YCor | There are non-isomorphic (finite) groups with isomorphic group algebras over some field (Berman). So you have to cheat at some point to keep track of the group. | |
| Mar 1, 2019 at 6:50 | history | asked | Grisha Papayanov | CC BY-SA 4.0 |