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I keep reading that the Reshetikhin-Turaev construction actually yields a 3-2-1 tqft. I know the construction that associates to a suitably decorated surface a vector space built up from a hom-space in a modular tensor category and to a decorated 3-manifold a linear map between these vector spaces as described in the book by Bakalov and Kirillov or in Turaev's blue book.

But to obtain a 3-2-1 theory, I would need a 2-category first. I suppose in this case this would be the category of all small linear categories. What I don't see is, how the Reshetikhin-Turaev construction associates a functor to surfaces and a natural transformation to 3-cobordisms. Or is this the wrong way to think about this?

I have the feeling that this was probably already asked and answered here, but I could not find it. If so, sorry.

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Maybe 4-3-2-1. The Crane-Yetter invariant of four manifolds can be extended down to recapture the Witten Reshetikhin Turaev invariant. Kevin Walker has some incomplete notes where he shows how to do it. –  Charlie Frohman Jan 19 '12 at 18:11
    
By the way, I think a great person to e-mail about this is Benjamin Balsam, who's a graduate student at Stony Brook. His three pre-prints are probably the kind of thing you're looking for. –  Hiro Lee Tanaka Jan 20 '12 at 7:45
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3 Answers 3

up vote 13 down vote accepted

Hi Ulrich,

I think that's a good way to think about it. There's also a good reason why we have the associations

1-manifold <--> Linear Category

2-manifolds F <--> Functors <--> Vector Spaces (when F is closed)

3-manifolds X <--> Natural Transformations <--> Numbers (when X is closed),

which I can try to answer now. In what follows, I'll let $Z$ be the topological field theory. It's a functor

$Z: Cob_1^3 \to LinearCat$

between 2-categories, in your language. What the Reshetikhin-Turaev TFT assigns to the circle should be the linear category of fixed-level, positive energy representations of a loop group, or the linear category of (certain) representations of a quantum group. But those details won't matter for these general comments:

(Surfaces) First you should pin down what you mean by a morphism in the category of linear categories. (i.e., What kinds of functors you want to allow.) If you impose, for instance, the condition that all morphisms must preserve finite colimits, then you see that a functor

$Z(F): Vect --> Vect$

is determined completely by what $Z(F)$ does to the one-dimensional vector space $k$. So $Z(F)(k) \cong V$ for some vector space $V$, and one can simply think of $Z(F)$ as $V$ itself. Vect is the unit in linear categories, so a closed surface (a cobordism from the empty manifold to the empty manifold) is hence assigned a vector space.

(3-manifolds) Now a natural transformation from $Z(F)$ to $Z(F')$ is given by a map of vector spaces. When the three manifold is closed, its boundaries F and F' are both empty manifolds. These are assigned the 1-dimensional vector space $k \cong Z(\emptyset)$, which is the unit in the category of functors from Vect to Vect. (Tensor product of vector spaces gives rise to a monoidal structure on the category of functors from Vect to itself.) Hence a natural transformation corresponds to a a linear transformation

$Z(X): k \to k$,

and this is just an element of $k$, since it's determined by what it does on the unit $1 \in k$. (This is the analogue of the statement I made above about how a functor $Vect \to Vect$ is given by what it does to the unit $k \in Vect$.)

(On 3-2-1-0 and 4-3-2-1 and 4-3-2-1-0) Chern-Simons is interesting because it might extend up and it might extend down. First, to extend to zero-manifolds, there are some notes by Dan Freed from a lecture he gave at UPenn Strings this year. He says that, since Chern-Simons has an anomaly, one can think of Chern-Simons as a 3-2-1-0 TFT with a twist, and he gives a concrete mathematical definition of this notion.

Another fun direction is seeing Chern-Simons as coming from a field theory in higher dimensions. The tip of this iceberg is visible through Khovanov homology--in Khovanov, we recover a link invariant whose Euler characteristic recovers the Jones poynomial. Why is this a suggestion of extending "up?" On the one hand, the Jones polynomial comes out of $SU(2)$ Chern-Simons theory, and on the other hand, Khovanov homology assigns morphisms to cobordisms between links, where a 2-manifold inside a 4-manifold is a cobordism between embedded links. I think Witten has written about how this comes from a 5- or 6-dimensional field theory in this pre-print, and also in a lecture at the Simons center: Video, Slides. Charlie above commented about the Crane-Yetter invariant, but I don't know enough to talk about connections to that.

PS I am using Chern-Simons and Resthetikhin-Turaev interchangeably because in my mind they are philosophically the same, but some schools may complain about this.

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That's a great answer that untied some knots in my brain. Thank you, Hiro. –  Ulrich Pennig Jan 19 '12 at 19:17
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Your brain must be four dimensional. –  Spice the Bird Jan 22 '12 at 4:28
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The traditional way to define Reshetikhin–Turaev's version of Chern–Simons TQFT as a 321 theory assigns to the circle the category of representations of the quantum group at fixed root of unity (depending on the level), which is equivalent to the category of positive-energy representations of the loop group (at the fixed level). Let me call this category $C$. From either picture, $C$ is a braided-monoidal category. (From the loop-group side, I believe the monoidal structure is what is called "Connes' fusion".) This monoidal structure $\otimes : C \boxtimes C \to C$ is what the functor that the TQFT assigns to a pair of pants. If I am not mistaken, the braiding corresponds to the bordism which is the mapping cylinder for the map that crosses the pant legs.

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What's the best reference for the equivalence “to the category of positive-energy representations of the loop group (at the fixed level)” –  Dmitri Pavlov Jan 20 '12 at 11:11
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I don't think there is anything self-contained. Finkelberg's "An equivalence of fusion categories" proves it, but uses results from a series of papers by Kazhdan-Lusztig. –  Pavel Safronov Jan 20 '12 at 15:05
    
@Dmitri: What Pavel said. I've certainly never seen the equivalence written down fully. As evidence of the believability, what you can see directly is that as categories they are the same --- you can pretty explicitly classify the irreducible positive-energy representations and the irreducible quantum group representations (for the semisimplified quantum group at root of unity). Moreover, you can I think check directly that the Grothendieck rings of both categories are the same. But of course this is far from good enough: the braiding is important data. And I don't know how to check that. –  Theo Johnson-Freyd Jan 20 '12 at 17:36
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I really like Hiro Lee Tanaka's answer. This is just a long comment.

The Witten-Reshitkhin-Turaev TQFT needs more data than a three-manifold, or a surface. You can decorate the three-manifolds with a framing, or a $p_1$ structure, and then decorate surfaces with a Lagrangian subspace of their first homology and an integer, and put colors on one manifolds, and the whole thing works... in theory. Actual computations are a bear.

The most successful modes of computation descend from coloring surgery diagrams to define morphisms between reduced skein modules as can be found in the seminal paper of Habegger, Masbaum and Vogel on TQFT derived from the Kauffman bracket.

Blanchet, C.; Habegger, N.; Masbaum, G.; Vogel, P. Topological quantum field theories derived from the Kauffman bracket. Topology 34 (1995), no. 4, 883–927.

In practice all the extra data can be accounted for quite elegantly by accepting that the TQFT underlying WRT is four dimensional. I think this idea is due to Kevin Walker.

The four dimensional theory is so trivial, that you can account for it with homological data, and hence decategorify the theory to get a less elegant $3$-dimensional TQFT. The ambiguities you are eliminating are in fact just scalar multiplication by a power of the central charge, which is a root of unity.

Here is what I think the extension looks like from top down.

Choose a level $r$

The Crane-Yetter invariant is assigned to four manifolds.
The reduced Kauffman bracket skein module at level $r$ is assigned to three-manifolds. The surface $F$ is assigned the category of left modules over the reduced Kauffman bracket skein algebra of the surface. The morphisms in the category correspond to four dimensional cobordisms. A one manifold is assigned a $2$-category. The objects are relative reduced skein modules of three manifolds, so that there is a disk in the boundary of the three-manifold for each circle in the one manifold, inside of which is an oriented arc that is the boundary of relative skeins. The morphisms come from gluing three manifolds along the complement of the disk, and are induced by including relative skein modules. The two morphisms are given by morphisms between the reduced skein modules induced by surgery diagrams corresponding to the four manifolds.

Pretty sketchy.

Practically speaking, current interest in quantum invariants is focused on their asymptotic behavior as the level goes to infinity. Generally people are focused on the magnitude of the invariants not the phase, so all the fancy book keeping to keep track of phase to get a three-dimensional theory is wasted effort.

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Re the 4d theory: I would say that a 2-manifold $F$ is assigned a category whose objects are collections of labeled ribbon endpoints in $F$ and whose morphisms are the skein module of $F\times I$. A circle $S$ is assigned a 2-category whose 2-morphisms are the skein module of $S\times D^2$, whose 1-morphisms are labeled ribbon endpoints in $S\times I$, and whose 0-morphisms are trivial. A point is assigned the Temperley-Lieb/Kauffman 3-category (braided category) we started with. Strictly speaking, we should dualize all of the above, e.g. $F$ is assigned ... –  Kevin Walker Jan 21 '12 at 20:28
    
... the category of representations of the skein category above, and similarly for 3- 1- and 0-manifolds. –  Kevin Walker Jan 21 '12 at 20:29
    
My thinking is the reduced skein module only depends on the boundary, so as long as $\partial M=\partial F\times I$ it fits the bill, so the morphisms in the category assigned to a one manifold can be made larger with no loss. –  Charlie Frohman Jan 22 '12 at 16:05
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