# What is Chern-Simons theory expected to assign to a point?

Let $G$ be a compact, connected, (simply connected?) Lie group and let $k \in H^4(BG, \mathbb{Z})$ be a cohomology class. Witten showed, at a physical level of rigor, that this data determines a $3$-dimensional topological quantum field theory (going down to surfaces), Chern-Simons theory.

Later other authors (Reshetikhin-Turaev, ???) described how to extend this theory down to $1$-manifolds. It's known that such a theory is determined by what it assigns to a circle $S^1$, which must be a modular tensor category; this category can be described either as a certain category of representations of the loop group $LG$ at level $k$ or as a certain category of representations of the quantum group $U_q(\mathfrak{g})$, where $q$ is a suitable function of $k$. The relationship between these two descriptions is unclear to me.

My impression is that it's expected that Chern-Simons theory extends all the way down to $0$-manifolds; that is, that it is a fully extended TQFT. By Lurie's classification, such a theory is completely determined by what it assigns to a point, which is a fully dualizable object in a suitable $3$-category.

What are some conjectural descriptions of this object?

The nLab is somewhat vague on this subject. Here's what I know:

The corresponding object for $3$-dimensional Dijkgraaf-Witten theory is known, although I'm not sure exactly who this is due to. Here $G$ is replaced by a finite group and $k$ is thought of as a class in $H^3(BG, \text{U}(1))$. $k$ is used to twist the associator on the monoidal category of $G$-graded vector spaces, giving a monoidal category (in fact a fusion category) of "twisted $G$-graded vector spaces," to be thought of as a fully dualizable object in the $3$-category of monoidal categories and bimodule bicategories over these (Douglas-Schommer-Pries-Snyder?), and I think this is what fully extended $3$-dimensional Dijkgraaf-Witten assigns to a point. Freed-Hopkins-Lurie-Teleman generalized this construction to the case that $G$ is a torus; here $G$-graded vector spaces are replaced by skyscraper sheaves on $G$. I don't know if this is expected to generalize.

• I feel duty-bound to transmit a reaction at the nForum: nforum.mathforge.org/discussion/5885/somewhat-vague/… – Todd Trimble Apr 22 '14 at 17:59
• @Todd: my apologies if my expectations were unreasonable! Based on some comments on G+ I was under the impression that experts (or at least Urs) knew a fairly detailed conjectural answer to this question but I couldn't find one on the nLab page for Chern-Simons. – Qiaochu Yuan Apr 23 '14 at 3:59

I have a proposal for what Chern-Simons should assign to a point:

The $\otimes$-category of (certain) representations of $\widetilde{\Omega G}$.

Here, $\Omega G$ is the based loop group, and the tilde indicates that one should take the central extension inherited from the level $k$ central extensions of $LG$.

Another way of phrasing the proposal, that also works when $G$ is not connected (Dijkgraaf-Witten theory being the special cases thereof when $G$ is finite) is to say that it's the category of (certain) vector bundles over the moduli space of $G$-bundles over $[0,1]$ trivialized at $\{0,1\}$.

The precise definition is spelled out at minute 50 of the following video: http://youtu.be/2imygWqTET8 (and if you're going to watch it, I recommend watching from the beginning)

Added later: Here is a set of notes written by Qiaochu of a talk that I gave on the subject.

• Dear André, will you excuse me if I ask you what your mother tongue is? The only excuse for this indiscrete request is that I am pathologically interested in linguistics [and your English is too perfect to be native:-)] – Georges Elencwajg Apr 14 '14 at 9:42
• I grew up bilingual French and Greek. My school years were a mess (going back and forth between Greek and French schools) and as a result I can't spell in any of my two native languages. – André Henriques Apr 14 '14 at 20:37
• How interesting and how lucky you are to have two native languages: when I think of how much effort one has to make as an adult to learn a new language... (U moet het wel weten: U schijnt veel talen te kennen) And congratulations for your staggering mastery of the hard mathematics you have regaled us with for several years. – Georges Elencwajg Apr 14 '14 at 21:13
• @QiaochuYuan: I certainly don't mind. On the contrary, I should thank you for giving this more visibility. And, of course, I can tell you more in person if you want. – André Henriques Apr 16 '14 at 3:36
• For anyone else thinking of watching the video, Youtube has 1.5x and 2x speed options, and I promise that the video is still understandable at those speeds! – Qiaochu Yuan Apr 16 '14 at 4:44

I wanted to make a few technical comments related to how your question relates to our paper with Chrises Douglas and Schommer-Pries. It got a little long for a comment, so I'm making it an answer.

In our paper the objects in the target 3-category we consider are "finite tensor categories" (over a perfect field) in the sense of Etingof-Ostrik. That is, they are abelian, finite, and rigid. This is for several technical reasons in the construction of the relative version of Deligne's tensor product and elsewhere. In this specific context, we show that the only 3-dualizable objects are the (seperable multi-)fusion categories. So in the setting of finite tensor categories you can only realize 321 theories where the value of the circle is a center of a fusion category, and hence you can't realize Chern-Simons theory with our target. (For some small examples you can just use dimension arguments as in Andre's talk, and for more general examples you can use arguments about the anomaly.) Thus in our restricted setting it is possible to get a negative answer to your question.

However, as my advisor taught me, a negative result is best thought of as a challenge: how can you change your assumptions so that you can avoid the theorem! In this case, one ought to be able to construct a bigger target 3-category allowing more general monoidal categories in the target. This is somewhat tricky, in particular you may need to leave the world of abelian categories and use the Kelly tensor product in the place of Deligne's tensor product. We have not worked out the details so I don't want to try to make a precise definition let alone a claim. In such a more general setting it's important to note that semisimplicity, rigidity, and finiteness are not preserved under Morita equivalence! So certainly there are fully dualizable monoidal categories which are not fusion (e.g. as Dan Freed pointed out to me A-mod-A for any ordinary non-semisimple algebra A). In fact, if you look at our proof that full dualizability implies semisimplicity it goes through two steps: first show that Z(C) is semisimple and then use that to prove C is semisimple. The latter step breaks down without finiteness and rigidity. In fact, what appears to be going on (this is not a theorem since we can't even make the statement precise) is that what's important is that Z(C) be finite, rigid, and semisimple, and that C itself is not as important. Thus there would still be hope to find a monoidal category which isn't fusion but where Z(C) is fusion. Andre's answer gives a concrete suggestion.

Finally let me leave a warning. Some of the results in our paper morally "should" apply in a more general setting (e.g. 2-dualizability should hold more generally) while others use finiteness or rigidity in an essential way (e.g. being a Radford object should not hold more generally). In our setting it was impossible to make such distinctions in theorem statements, but we tried to make this clear in remarks. But you should be careful to remember that finiteness and rigidity are essential assumptions for some results in our work and merely technical assumptions for other results.

• In reference to the last paragraph, a more general 2-dualizability result appears in my joint work with Brochier and Jordan, see arxiv.org/abs/1804.07538 Section 5.4. – Noah Snyder Apr 20 at 3:12

I don't quite know how this answer fits with André's, and there are certainly a bunch of subtleties I'm unaware of, but:

In the spirit of the cobordism hypothesis, to get a 3d TFT you need to attach a monoidal category $C$ to the point. In order to go up to dimension 3, this category should, as you say, satisfies some dualizability conditions which amongs to say that $C$ is actually fusion. This is basically the result of Douglas-Schommer-Pries-Snyder you mention, and what you get is the Turaev-Viro TFT associated with $C$

Then what is attached to the circle has a canonical structure of a modular category, and is nothing but the Drinfeld center $Z(C)$ of $C$.

Roughly, you can get Reshetikhin--Turaev TFT by starting directly from the circle, replacing $Z(C)$ by any modular category, and then reconstructing the higher dimensions "in the same way". This explain why Turaev-Viro of $C$ and Reshetikhin-Turaev of $Z(C)$ essentially coincides.

Therefore, if your modular category is not the center of some fusion category, I think you can't go down to the point.

The category of $U_q$-modules you mention is certainly not the center of any other category. In fact there is a conjecture that these are the building blocks of all modular categories which are not equivalent to the center of a fusion category. Its relation with the category of $LG$-modules at level $k$ has been made precise by Kazdhan--Lusztig and Finkelberg.

All of this is somehow related to the fact that the RT TFT usually have a so-called anomaly and so is strictly speaking not quite an actual 3d TFT. The "right" way to recast this in the framework of the cobordism hypothesis is to see the RT construction as some sort of boundary condition of an extended, honest 4d TFT. See e.g. What's the right way to think about "anomalies" in 3d TQFTs? or Freed-Teleman on relative quantum field theory.

• 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 $1$). – Theo Johnson-Freyd Apr 14 '14 at 14:01
• 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. – Adrien Apr 14 '14 at 18:18
• 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 $X$. – Adrien Apr 14 '14 at 18:24
• 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!) – David Ben-Zvi Apr 15 '14 at 15:41
• 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! – David Ben-Zvi Apr 15 '14 at 16:23