3d TQFTs constructed from modular tensor categories don't in general give an honest 3d TQFT, instead they have an "anomaly." My vague understanding from Kevin Walker's talks and from skimming Freed-Hopkins-Lurie-Teleman is that what's really going on is that its a 4d TQFT that's almost boring on the 4d part and that's what the "anomaly" means. Does anyone know how to make this more precise?



What's the right way to think about "anomalies" in 3d TQFTs?
You can find some of the details in a draft version of my TQFT notes on my web page. Here's a summary.


Given an n-category with strong duality (by which I mean, more or less, pivotal if $n=2$ and a higher dimensional version of pivotal for $n > 2$; this is stronger than what Lurie means by an n-category "with duals"), there's a standard procedure to construct an n+1-dimensional TQFT. This procedure works for free for the 0-through n-dimensional parts of the (extended) n+1-dimensional TQFT. In these dimensions, we need no additional assumptions on the n-cat, and there is no need to choose a decomposition of the manifolds, so there are no combinatorial topology moves to check. The construction is manifestly invariant.


To construct the top, n+1-dimensional part of the TQFT, we need to make some finiteness assumptions on the n-category. (This corresponds to the top-dimensional part of what Lurie et al mean by "fully dualizable".) If the n-category satisfies these assumption, then we get, for each handle decomposition of the n+1-manifold, a state sum type expression for the path integral of the n+1-manifold. It is not hard to show that this state sum is invariant under handle slides and handle cancelation, so we get a well-defined invariant of n+1-manifolds that interacts with the rest of the TQFT (via gluing formulas) in the correct way.


(Small technicality: the path integral construction depends on a choice of element in the Hilbert space of the n-sphere, corresponding to the path integral of the n+1-ball. Multiplying this choice by $\lambda$ changes the path integral by $\lambda^\chi$, where $\chi$ is the Euler characteristic of the n+1-manifold.)


A modular tensor category is a 3-category with strong duality and the right sort of finiteness properties, so we can apply the above construction to get a 3+1-dimensional TQFT. 


In dimension 3 the vector space we construct is an old-fashioned skein module: finite linear combinations of ribbon graphs in M^3 modulo local relations. (Actually, the dual of this vector space.) If M is closed this in 1-dimensional. More generally, if M has boundary then it has the same dimension as the Witten-Reshetikhin-Turaev vector space associated to the boundary of M.


In dimension 4, the type of state sum we get depends on the type of handle decomposition. For a general handle decomposition we get the Crane-Yetter state sum. For 2-handles attached to a 4-ball along a framed link L we get the Reshetikhin-Turaev surgery formula for L. For a 4-dimensional neighborhood of a 2-complex we get the Turaev "shadow" state sum. For a closed 4-manifold we find that the path integral is equal $a^\chi b^\sigma$, where $\chi$ is the Euler characteristic and $\sigma$ is the signature of the 4-manifold. By choosing $\lambda$ above appropriately we can make a=1. b is related to the central charge of the MTC (or to the value of the RT surgery formula on framing +-1 unknot). For a 4-manifold with boundary we find that the state sum computes the Witten-Reshetikhin-Turaev invariant of the boundary of the 4-manifold.


In dimension 2, the 1-category associated to a closed surface is a full matrix category; i.e. it is Morita trivial. For a surface with boundary k circles the category is Morita equivalent to k copies of the MTC thought of as a 1-category.


In all of the above cases, we find that the TQFT invariant of $Z(X)$, where dim(X) = 2 or 3 or 4, depends strongly on the boundary of $X$ but only weakly (i.e. only up to bordism) in the interior of $X$. So we can define a new 2+1 dimensional TQFT $Z'$ via the formula


$$Z'(Y) := Z(\mathrm{boundary}^{-1}(Y))$$


This TQFT has an anomaly, since we need to enhance Y with enough extra structure to pick out an inverse-boundary, up to bordism.


