# Notes on Lattice Topological Field Theory in Three Dimensions, Part I

I have recently been interested in constructing topological field theories for loop quantum gravity, with a specific eye towards the topspin construction. With this in mind I was lead to the classic paper by Dijkgraaf and Witten, “Topological Gauge Theories and Group Cohomology” (1990). I am working my way through it to try and understand – which usually requires taking notes on the paper and working on things I don’t understand. I thought maybe I would post my notes here.

Introduction

For an oriented, smooth 3-manifold $M$ we can specify a principal bundle $E=(M,\pi,G)$ with a gauge group $G$. An action on this bundle must be constructed from some kind of invariant, to make sure it is somehow independent of the choice of connection $A$. If $E$ is trivial, the connection is a Lie-algebra valued one-form $A\in \Omega^1(M)\otimes \mathfrak{g}$ and the action is the Chern-Simons functional

$S(A)=\frac{k}{8pi ^2}\int_M Tr(A\wedge dA + \frac{2}{3}A\wedge A\wedge A).$

The path integral is $Z(M)=\int\mathcal{D}A\exp (2\pi i S(A)$. The integer $k$ is called the level of the theory, and must be an integer. This is because when we take $G$ to be non-abelian, under a gauge transformation the action goes to $S(A^g)=S(A)-2\pi k N$ for some integer $N$ (http://arxiv.org/abs/hep-th/9902115). Thus in order for the amplitude $\exp(S(A))$ to remain a gauge-invariant, $k$ must at least be an integer.

If $E$ is not a trivial bundle, the action is not a global gauge-invariant. However, since we know curvatures are good gauge invarients (in 4 dimensions), we can use them to define a global Chern-Simons action. So, assume we can extend the bundle $E$ over a smooth four-manifold $B$. Then we could write

$S(A)=\frac{k}{8\pi^2}\int_B Tr(F\wedge F)$,

where $F$ is the usual curvature of the connection $A$. If we have a trivial extension of the connection $A(x)=\partial_t A dt+A(\vec{x})$ then the curvature is only a function of the coordinates $\vec{x}$ on $M$:

$F(x)=dA+A\wedge A=dA(\vec{x})+A(\vec{x})\wedge A(\vec{x}).$ The question remains how to do this in general, since one cannot always find such an extension of the connection to the boundary (I basically chose $B=\mathbb{R}\times M$, but we cannot guarantee that $E$ can be extended in the way that I extended $A$).