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Apr 13, 2017 at 12:58 history edited CommunityBot
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Apr 24, 2014 at 4:56 comment added André Henriques I was a bit too quick when I wrote "everything" above. Let me scale down my claim to: the statement about the center can be formulated using the purely algebraic notion of center. The conjectural 3-category in which the ⊗-category of representations of $\widetilde{\Omega G}$ lives is certainly not a purely algebraic 3-category.
Apr 24, 2014 at 4:13 comment added André Henriques Hi Noah... there was some progress since my talk. First of all, I realized that I didn't really know what "continuity" means in that context (or rather, that the notion of continuity that I had in mind was fatally flawed). Secondly, I figured out that everything can be formulated purely algebraically, so I stand by what I wrote in the above comment: no need to modify the notion of center.
Apr 22, 2014 at 18:06 comment added Noah Snyder @AndréHenriques: I'm confused, don't you have to modify the notion of center? In your talk you make the point that you need to restrict to continuous half-braidings. Or did I misunderstand?
Apr 16, 2014 at 16:59 comment added David Ben-Zvi @AndréHenriques: I'm not claiming one has to modify the notion of center, only where that center lives -- it's an object of the invertible 2-category attached to the circle by the 4d anomaly theory, which is given as $E_2$ (or "chiral") module categories over the MTC --- one of which is your result, namely the MTC itself.
Apr 16, 2014 at 4:04 comment added André Henriques @David Ben-Zvi. I'll have to think about your reformulation... The tensor category that I build doesn't require to modify the notion of center: its center (just as a tensor category) is the MTC.
Apr 15, 2014 at 16:23 comment added David Ben-Zvi So the claim is roughly there's an invertible 3-category (modules over modules over the modular tensor category MTC) and a 3-dualizable object in there (in the sense of cobordism hypothesis with tangles), whose center (as a braided module category over the MTC) is the MTC itself. Sounds like Andre has found a candidate for this!
Apr 15, 2014 at 15:41 comment added David Ben-Zvi I wouldn't say the existence of the anomaly means CS doesn't fit into the cobordism hypothesis framework or that there's no object to assign to a point - just that we have to look for a fully dualizable object relative to a simple (invertible) 4-dimensional TFT. (In particular the anomaly can often be trivialized on a fixed low-dim manifold I believe, though not functorially - eg to a 3-manifold it assigns a line, and a line on its own can be trivialized!)
Apr 14, 2014 at 18:24 comment added Adrien But modular categories are believed to be the fully dualizable objects in a certain 4-cat, hence leads to a extended 4d TFT. So the claim, which I think has been folklore for a long time, is that RT is secretly really an extended 4d TFT with boundaries. The anomaly indeed means roughly that the invariant RT attach to a 3-manifold $X$ depends on the choice of a bordism class of 4-manifolds whose boundary is $X$.
Apr 14, 2014 at 18:18 comment added Adrien Qiaochu> Well, my point was precisely that RT (hence Chern-Simons) doesn't quite fit into the framework of the cobordism hypothesis, so there isn't even a target 3-category in this picture. You might ask if there's some 3-cat, in which modular categories would be fully dualizable and such that the corresponding TFT coming from the cobordism hypothesis coincide with the RT construction. I think that's precisely the sort of things that the anomaly rules out.
Apr 14, 2014 at 17:46 comment added Qiaochu Yuan @Adrien: I'm willing to believe you can't go down to the point with the target $3$-category I described above, but that doesn't preclude the possibility of switching to a more sophisticated target $3$-category, right? I also admit that I don't understand this anomaly issue at all; going to have to read more about that.
Apr 14, 2014 at 14:01 comment added Theo Johnson-Freyd A too-coarse but perhaps useful analogy is with $\sqrt{2}$. The rational number $2$ does not have a rational square root (a fact for which one Pythagorean famously committed suicide by drowning, since this discovery was so heretical). But you can find an analytic context, namely $\mathbb R$, where it does have a square root. The reason this analogy is not too far off is that $Z(C)$ is something like "$C^2$", and in particular $\dim Z(C) = (\dim C)^2$, and $\dim (SL(2)_1) = 2$ (loops-$sl(2)$-reps at level $1$).
Apr 14, 2014 at 13:57 comment added Theo Johnson-Freyd This is correct, although with the following caveat. The statement "The category of $U_q$-modules you mention is certainly not the center of any other category." depends on what "any other category" may range over. The statement is true if you work with "just categories". Andre's proposal in a different answer involves switching to a more analytical setting, where no longer are hom-sets "just sets" (or "just abelian groups"), but rather topological vector spaces, and where the statement may fail.
Apr 14, 2014 at 11:45 history answered Adrien CC BY-SA 3.0