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