Timeline for How weird can Modular Tensor Categories be over non-algebraically closed fields?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 17, 2014 at 7:09 | vote | accept | Chris Schommer-Pries | ||
| Jul 16, 2014 at 20:46 | answer | added | Victor Ostrik | timeline score: 9 | |
| Jul 16, 2014 at 20:42 | comment | added | Noah Snyder | There's a bunch of great stuff related to Andre's comment in this short paper of Greg's. | |
| Jul 16, 2014 at 20:34 | comment | added | Noah Snyder | Modularity should be thought of as being as far from symmetry as possible. In the symmetric case the S-matrix has rank 1, and in the modular case it has full rank. On the other hand, Rep(G) is contained in its center which is modular, so @DavidSpeyer's example still works. | |
| Jul 16, 2014 at 20:17 | comment | added | David E Speyer | @JamieVicary Yes, it would be. Is that a problem? I looked through the list of properties at ncatlab.org/nlab/show/modular+tensor+category , all of them seem to apply to representations of a finite group. | |
| Jul 16, 2014 at 19:24 | comment | added | Jamie Vicary | Wouldn't that example be symmetric monoidal? | |
| Jul 16, 2014 at 19:10 | comment | added | David E Speyer | I'm not very familiar with the language here, but if $G$ is a finite group and $k$ is a field with characteristic not dividing $|G|$, are the representations of $G$ over $k$ a modular tensor category? Because you can certainly get other division algebras that way: Take $k=\mathbb{R}$ and $G$ the quaternion $8$-group; the simple representations are four $1$-dimensional reps and a $4$-dimensional rep, and the $4$-dimensional rep has endomorphism algebra the quaternion algebra. | |
| Jul 16, 2014 at 17:30 | answer | added | Jamie Vicary | timeline score: 0 | |
| Jul 16, 2014 at 14:39 | comment | added | Jamie Vicary | Building on André's comment: I guess that a monoidal Ab-category is automatically $k$-linear for $k:=\text{End}(1)$. | |
| Jul 16, 2014 at 14:28 | comment | added | André Henriques | Concerning your: "What if we drop the requirement $End(1) = k$?" I have the feeling that one should define $k$ to be $End(1)$. So, by definition, you then always have $End(1) = k$, and the question is what happens when you take $k$ to be a ring that looks further and further from an algebraically closed field. | |
| Jul 16, 2014 at 14:18 | comment | added | Jamie Vicary | An important point here is that the definition of modularity becomes a bit more complicated. The vector space associated to the torus is no longer generated in general by a basis given by the isomorphism classes of simple objects. So the S-matrix acts on a possibly larger space: the product of the centres of endomorphism algebras of the simple objects. | |
| Jul 16, 2014 at 13:59 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
corrected spelling
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| Jul 16, 2014 at 13:35 | history | asked | Chris Schommer-Pries | CC BY-SA 3.0 |