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

This post continues the series on TQFT – the first post can be found here. In this edition I will illustrate why we might be interested in the TQFT of a lattice. In short, manifolds generally admit triangulations, and triangulations are generally lattices.

Triangulations

If you don’t know what a triangulation is, whatever you are thinking it probably is, is probably correct. You basically draw a bunch of triangles (of appropriate dimension) in your manifold – and the more you draw the better you understand the characteristics of the manifold in question. More precisely, it is a homeomorphism $\mathcal{K}\to M$ of a simplicial complex $\mathcal{K}$ into a manifold $M$. Since we are physicists, we will always take our manifolds smooth and orientated. The existence of triangulations is an open question in mathematics, particularly in higher dimensions, but luckily for us all 3 dimensional manifolds can be triangulated. Further, given a triangulation, each cell is homeomorphic to a sphere and a triangulation can be upgraded to a PL-structure, which can further be upgraded to a smooth structure (this process works in dimensions less then 5, if I remember correctly). So studying triangulations in dimension 3 is akin to studying 3-manifolds.

A triangulation is a lattice in 3-dimensions exactly as you expect. There are vertices $v_i$ with links $l_{ij}$ between them. Around these links are additionally faces $F_{ijk}$ (which will be important later, but not so much for the lattice theory). Lattice gauge theory assigns a group element $g_{ij}\in G$ to each link, with a gauge-transformation $g_{ij}\to h_i\cdot g_{ij}\cdot h^{-1}_j$ for $h_i,h_j\in G$. We will discuss how to find the partition function for a TQFT on such a lattice, thinking of it as a triangulation.

Lattice Gauge Theory with Finite Gauge Group

First we need to motivate why our finite gauge theory in $M$ is naturally a lattice gauge theory. Consider the classifying map $\gamma: M\to BG$, and pick a point $p\in BG$. Now triangulate $M$ – every vertex $v_i$ is mapped to a point $p_i\in BG$ under $\gamma$. Now, since $BG$ is connected, we can deform $\gamma$ to  move each point $p_i$ along a path to $p$. Then each link in the triangulation is a loop in the classifying space based at $p$. Each of these loops is an element of the fundamental group, which is isomorphic to the gauge group $G$ itself. Thus the classifying map assigns to each link $l_{ij}$ a group element $g_{ij}\in G$, exactly as one expects for a lattice gauge theory. In addition, this theory admits only flat connections. The curvature of a lattice gauge theory is the holonomy along loops, or $f_{ijk}=g_{ij}\cdot g_{jk}\cdot g_{ki}$. However, since $g_{ij}$ is an element of the fundamental group that bounds a simplex, it is contractible and $f_{ijk}=1$.

Recall that we used differential characters to define the action of a TQFT. These guys lived in $H^3(BG,U(1))$, so for a 3-simplex $T$ we want an action which assigns $W(T)\in U(1)$, and the product of all of these will be the action on $M$. We need to deal a little with orientations – the orientation of a 2-simplex is positive in the counterclockwise direction when looking into the 3 simplex, which must by the same for all faces of the 3-simplex. In addition, we need an ordering of the independent edges to define the action (shortly…). To each $T$ assign an ordering  of the vertices $(v^{(0)}_i,v^{(1)}_j,v^{(2)}_k,v^{(3)}_l)$. The action can then be defined as

$W=\Pi_i W(T_i)^{\epsilon_i}$,

where $\epsilon_i=\pm 1$ encodes the orientation. I have not discussed it, but this action is independent of the choice of classifying map $\gamma$, and also works for manifolds with boundary.

Of course, we expect this action to be a group cocycle, following our previous discussions. In addition, it should only depend on the gauge fields on the boundary of the simplex, since the differential characters are evaluated on boundaries. So by way of “best guess”, let’s assume that on a single simplex, the weight for the partition function is given by

$W(T)=\alpha(g,h,k)$,

where $\alpha$ is a generic function (which we hope is a group cocycle), and $g,h,k\in G$ are the gauge fields on the edges of the simplex $T$. The specific order $g,h,k$ is given by the increasing numbering of the vertices $v^{(i)}_j$.

We will show this is a cocycle by acting on it with the coboundary operator (we have used such techniques before):

$\delta\alpha(g,h,k,l)=\alpha(g,h,k)\alpha(h,k,l)\alpha(gh,k,l)\alpha(g,hk,l)\alpha(g,h,kl)$

Of course, if this is the action then it must vanish on the boundary, so $\delta \alpha=1$ and we have our cocycle condition. In addition, since $W$ is supposed to be invariant under a gauge transformation $g_{ij}\to h_i g_{ij}h_j^{-1}$, we can have $\alpha\to \alpha \delta\beta$, with $\beta$ an 2-cocycle with $\delta\beta=1$. For each triangulation, we will need to form a product of these cocycles with the correct orientations a la the discussion above.

Example Calculations

Finally we can work with some examples! I will first do several simple cases which are not covered in Dijkgraaf and Witten. First, $\mathbb{S}^3$. We need to find a triangulation of the 3-sphere, so first notice that the 3-sphere can be decomposed into two 3-balls with their boundaries identified (in fact, this generally works for higher spheres too). A 3-ball is triangulated by a single 3-simplex, so we have the following triangulation:

If those crazy arrows are confusing to you ignore them; they are just trying to emphasize how the identification is being done if you are not familiar. Assigning a positive orientation (outward normal) to the left simplex, to have a consistent orientation for the entire sphere we need to make the right simplex negative. Therefore

$W(\mathbb{S}^3)=\alpha(g,h,k)\alpha(g,h,k)^{-1}=1,$

Not very interesting, but certainty illustrative.

I will pause very briefly here to mention that triangulating 3-sphere is apparently something of a cottage industry. There are many (MANY!) ways to go about doing it, and it’s not even clear how many there are. There is a nice little blog post mentioning just one of the interesting features at the Low Dimension Topology Blog: The 3-sphere has interesting triangulations. For us, this doesn’t matter much since the action is supposed to be invariant under triangulations – no matter how interesting they get, we will always get $W(\mathbb{S}^3)=1$.

Next we will take a much more complicated example, $D^2\times \mathbb{S}^1$. $D^2$ is the 2-simplex, so crossing that with the 3-sphere looks like a solid triangular prism with the ends identified. Easy, but finding a triangulation is a little subtle. The first thing you draw is a prism with 3 simplices, but when you work out the orientations, you see that the top and bottom both have positive orientation – that is, you cannot identify them (see the left side of the image below for how this works – I also asked question about this on MathStacks here, which has gotten very little attention). A solution (although maybe not the best one) is to combine a prism with positive orientation and a prism with negative orientation, and triangulate $D^2\times \mathbb{S}^1$ with six simplices (right figure):

As you can see, this gets the right orientation of the top and bottom after adding some more vertices and edge generators. At the end, we get a complicated expression but one with some nice symmetry:

$W(D^2\times \mathbb{S}^1)=\alpha(k,h,\bar{l}_1)\alpha(k,h,l_1)^{-1}\alpha(m_2\bar{l}_3,m_1,\bar{m}_2\bar{m}_1)\alpha(m_2l_3,m_1,\bar{m}_2\bar{m}_1)^{-1}\alpha(k,l_2,m_1)\alpha(k,\bar{l}_2,m_1)^{-1}$

The difference in orientation is being encoded by the change in direction of the azimuthal generators $l_i$. Using this, we can determine another example, $\mathbb{S}^2\times \mathbb{S}^1$. By decomposing the 2-sphere into 2-balls glued at their boundary, we get two copies of the above terrible picture with the boundaries identified. However, just like in the case of the 3-sphere, the orientations all cancel pairwise, so we find

$W(\mathbb{S}^2\times \mathbb{S}^1)=1$.

I will stop with these simple examples, and present the examples in the Dijkgraaf and Witten paper next time – in what will hopefully the last post in this series!