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

In this final post I will explain the two primary examples found in the Dijkgraaf and Witten paper, and give some concluding remarks regarding my original interest in this topic – the application to loop quantum gravity.

Example Calculations, Continued.

The first example from the paper is $Y\times \mathbb{S}^1$ (2-sphere with 3 punctures times 1-sphere), which can be used to generate many complicated examples. First we need a representation of this space as a simplex:

The first picture shows that $Y$ can be represented as a 2-complex with the three vertices identified. Taking the product of this with the circle gives us the final image. This can be decomposed into 3 simplices, as shown:

This is certainly confusing, since not only is the top and bottom identified, but all the vertices are being identified (if you recall in the last post orientation was an issue – that is not a problem here due to how this gluing is happening. Just stare at it long enough and it should seem obvious, even if it’s not). In any case, this is the easiest decomposition into tetrahedrons (3D figures with four triangular faces). If the first simplex has positive orientation, the middle one must have negative orientation since their faces are in contact – counterclockwise on the faces of the first results in clockwise on the middle. The third one is again positive, and using the order of the vertices we find the action to be

$c_h(g_1,g_2)=\alpha(h,g_1,g_2)\alpha(g_1,h,g_2)^{-1}\alpha(g_1,g_2,h)$

It’s a messy but very doable calculation to show that $\delta c_h(g_1,g_2)=1$ – it’s a cocycle. In addition, we should check that under $\alpha \to \alpha\delta\beta$ we get $c_h\to c_h\delta\beta_h$ for some 1-cocycle $\beta_h$.. This will get a bit messy as well, but it has at least one subtly so I’ll do it.

\begin{aligned}c_h(g_1,g_2)&\to c_h(g_1,g_2)\frac{\beta (h,g_1g_2)\beta (g_1,g_2)}{\beta (h,g_1)\beta (hg_1,g_2)}\frac{\beta (g_1,g_2h)\beta (g_2,h)}{\beta (g_1,g_2)\beta (g_1g_2,h)}\frac{\beta (g_1h,g_2)\beta (g_1,h)}{\beta (g_1,hg_2)\beta (h,g_2)}\\ &\to c_h(g_1,g_2)\frac{\beta (h,g_1g_2)\beta (g_2,h) \beta (g_1,h)}{\beta (h,g_1)\beta (g_1g_2,h)\beta (h,g_2)}\end{aligned}

To cancel out a bunch of those terms I have used the fact that the $g$s are in the stabilizer subgroup of the $h$ – so $g_i h=g_i$. This can also be seen from the simplicial complex representation above. So we see that the transformation works if $\beta_h(g)=\beta(g,h)\beta(h,g)^{-1}$ is the 1-cocyle which transforms $c_h$, because $\delta \beta_h$ is a 2-cocycle so that:

$\delta \beta_h(g_1,g_2)=\beta_h(g_1)\beta_h(g_2)\beta_h(g_1g_2)^{-1}=\frac{\beta(g_1,h)}{\beta(h,g_1)}\frac{\beta(g_2,h)}{\beta(h,g_2)}\frac{\beta(g_1g_2,h)}{\beta(h,g_1g_2)}$

The first equality comes from the definition of a coboundary, and the second is using the definition of the 1-form $\beta_h(g)$. This matches the expression above.

The next example we will do is $\mathbb{S}^1\times\mathbb{S}^1\times\mathbb{S}^1$ (the three torus). This can be represented as the cube with opposite sides identified, which splits into 6 simplices, a pair of which is shown below.

Each pair of these simplices have differential orientation and order of edges. The ones I have shown contribute $\alpha(g,h,k)\alpha(g,k,h)^{-1}$, and by permutation it’s pretty easy to see the action is

$W(g,h,k)=\frac{\alpha(g,h,k)\alpha(h,k,g)\alpha(k,g,h)}{\alpha(g,k,h)\alpha(h,g,k)\alpha(k,h,g)}.$

As stated above, we can generate a higher-genus surface with the manifold $Y\times \mathbb{S}^1$; in the case of the 3-torus it is easy to see that it takes two $Y\times \mathbb{S}^1$ to make $\mathbb{S}^1\times\mathbb{S}^1\times\mathbb{S}^1$ (just cut the cube along a diagonal and you get a prism). Therefore we can write the action in terms of the 2-cocycles $W(h,g,k)=c_g(h,k)c_g(k,g)^{-1}$.

TQFT and Loop Quantum Gravity

In loop quantum gravity we don’t have a lattice but rather a graph – which is something like an irregular lattice. The fields live in $SU(2)$, and in a 2009 paper, Bianchi constructed something which (at least superficially) looks very much like a TFT by considering the moduli space of flat $SU(2)$ connections. However, the actual topological character is rather hidden in this approach, and in fact in all of the standard approaches to LQG (BF theory, etc). This is where the motivation for topspin networks comes from – you can read more about that elsewhere on this site. The end result of the incorporation of topspin networks is additional $S_n$ labels on the spin networks – which turns LQG into something resembling a gauge theory with finite gauge group on an irregular lattice. The idea is to construct the partition function for a TQFT associated to a topspin network and use it to measure the topological contribution to the correlation function, which is a topic of hot research right now. In addition, this should make the rather tenuous connection between group field theory and LQG more clear.

So…watch out on the archive!

# 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!

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

I will continue this series (part I, part II, part III, part IV) by moving directly to considering topological field theories which have a finite gauge group. These are groups with a finite number of elements – that is, there is a bijection between the elements of $G$ and a subset of the natural numbers (integers greater than zero). Thus, $\mathbb{Z}$, $U(1)$, or $SU(2)$ are not finite groups, whereas $\mathbb{Z}_2$ and the symmetric groups $S_n$ are. I will be in particular interested in the symmetric groups. Do not confused “finite group” with “finitely generated” – $\mathbb{Z}$ is generated by a single element (“1” under addition), but is not a finite group.

Axioms of Quantum Field Theory

There are several different versions of these axioms – I will present those due to Atiyah, because they are more clearly presented than how Witten has done it. Strictly speaking, I won’t be relying too much on these axioms for these notes, but anyone interested in TQFT should know them. These axioms follow the structure of category theory.

Definition (Atiyah): A topological quantum field theory in dimension $d$ over a ring $\Lambda$ is a functor $\Phi: \mathcal{M}_d\to \mathcal{F}(\Lambda)$ from the set of smooth, oriented, closed $d$-manifolds to the set of finitely generated modules over $\Lambda$. This functor is respected on boundaries – for $\Sigma\in \mathcal{M}_d$, if $\partial \Sigma\neq 0$ then $\Phi(\Sigma)\in \Phi(\partial \Sigma)$. This functor satisifes the following axioms:

• Functorial with respect to (orientation-preserving) diffeomorphisms $f:\Sigma\to \Sigma'$: There exist isomorphisms $f(\Phi):\Phi(\Sigma)\to \Phi(\Sigma')$ and $\Phi(fg)=\Phi(f)\Phi(g)$ for $g:\Sigma'\to \Sigma''$.
• Involutory: $\Phi(\bar{\Sigma})=\Phi(\Sigma)^*$ (a reverse of orientation gives the dual module).
• Multiplicative: For $\Sigma=\Sigma_1\cup \Sigma_2$ we have $\Phi(\Sigma)=\Phi(\Sigma_1)\otimes \Phi(\Sigma_2)$.

These properties extend to the boundaries as well, which follows from the initial requirement that $\Phi(\Sigma)\in \Phi(\partial \Sigma)$.

If you are not familiar with rings and modules, I’ll give you the short story. A ring $R$ is an additive group with multiplication defined on it (so $(\mathbb{Z},+)$ where you can also multiply elements) and a module $X$ over a ring is an additive group $X$ on which you can multiply elements of the ring. Good examples of modules are vector spaces, since $\vec{v}+\vec{u}$ is a vector, and so is $a\vec{v}$ for $a$ in a ring (say $\mathbb{C}$). In physics we generally take the modules in the above definition to be Hilbert spaces – which requires that the ring be a field ($\mathbb{R}$, $\mathbb{Z}$, or $\mathbb{C}$) and there be a complete norm on the elements of the module.

Finite Gauge Groups on Riemann Surfaces

The connection on a Lie group $G$ is an element of the tangent space at 0 – which is the Lie algebra $\mathfrak{g}$. For finite groups, the tangent space at the identity is a point so the connections are $A=1$ and are therefore flat for all $G$. They are differentiated only by topology, in the following way: the classifying map $\gamma:M\to BG$ induces a homomorphism on fundamental groups $\pi_1(M)\to \pi_1(BG)$. The fundamental group of the classifying space is equal to the group itself, so the $G$-bundles are classified by a homomorphism

$\gamma_*:=\lambda:\pi_1(M)\to G$.

The partition function is then the sum over all such bundles (since each bundle is a complete state), weighed by an action $W=\exp(2\pi i S)$:

$Z(M)=\frac{1}{|G|}\sum_{\lambda}W(\gamma)$

Here we take $W(\gamma)=h(\partial B)$ for the extension of $M$ to the chain $B$, as discussed in the previous post. Dijkgraaf and Witten call this “$\langle \gamma^*\alpha,[M]\rangle$” since they are identifying the differential character $h$ with its curvature $\alpha$. Now the dimensions of the Hillbert space $\mathcal{H}$ will be given by the number of terms in the partition function. As a practice example, let’s do $\Sigma=\Sigma_g\times \mathbb{S}^1$ where $\Sigma_g$ is a Riemann surface of genus $g$. By picking $\alpha=0$ we have $h=1$ and the partition function will just sum over the representations $\lambda: \pi_1(\Sigma_g\times \mathbb{S}^1)\to G$. $\pi_1(\Sigma_g)$ is generated by pairs of elements $(g_i,h_i)$ for $1\leq g \leq i$ such that $\Pi_i [g_i,h_i]=1$ (see figure).

There is an additional element $k$ which is due to the $\mathbb{S}^1$ part. In the sum, there will be some representations of the form $k\lambda(\delta)=\lambda(\delta)$ for $\delta \in \pi_1(\Sigma_g)$ and $k\in G$ – elements which satisfy this condition are in the stabilizer subgroup $N_{\lambda}\subset G$.  The order (number of copies in $G$) of this subgroup is $|N_\lambda|$, so the size of the Hilbert space is

$\Phi(\Sigma_g\times \mathbb{S}^1)=\frac{|N_\lambda|}{|G|}$

This is exactly what we expect from  quantization – we expect the degrees of freedom to be the moduli space $\mathcal{M}_g$ of $G$ bundles $\mathcal{M}_g=Hom(\pi_1(\Sigma_g),G)/G$, with size $|\mathcal{M}_g|=| N_\lambda|/|G|$.

What about when $\alpha\neq 1$? Well, writing down the explicit answer is not so easy, but finding out some properties that the action satisfies can be instructive. Since we are working with Riemann surfaces, we have the uniformization theorem: there are only three classes of conformally inequivalent Riemann surfaces, a disk (parabolic), a complex plane (flat), or a sphere (spherical). Further, each of these can be generated by considering 2-spheres with punctures:

• $\mathbb{S}^2$ with zero punctures is the Riemann sphere.
• $\mathbb{S}^2$ with 1 puncture is the complex plane (two punctures is also conformally flat)
• $\mathbb{S}^2$ with 3 punctures (which Dijkgraaf and Witten call $Y$) is hyperbolic (a pair of pants!).

So we have all three categories of Riemann surfaces by considering punctured spheres up to 3, and in fact the thrice-punctured version can be used to generate any Riemann surface of genus $g\geq 2$. For instance, by sewing two copies of $Y$ together you get $g=2$. Thus, we can construct any $\Sigma_g\times \mathbb{S}^1$ by sewing together copies of $Y\times \mathbb{S}^1$. Since the boundary of $Y$ is three spheres, we have

$\partial (\Sigma_g\times \mathbb{S}^1)=(\mathbb{S}^1\times \mathbb{S}^1)\cup(\mathbb{S}^1\times \mathbb{S}^1)\cup(\mathbb{S}^1\times \mathbb{S}^1)=\Sigma_1\cup\Sigma_1\cup\Sigma_1$

and our functors are maps $\Phi(\partial (Y\times\mathbb{S}^1)): \mathcal{H}_{\Sigma_1}\times\mathcal{H}_{\Sigma_1}\times\mathcal{H}_{\Sigma_1}\to \mathbb{C}$. A quick note about notation; Dijkgraaf and Witten call this a “path-integral”, which is natural since it represents a transition between the boundaries of a manifold $Y\times \mathbb{S}^1$.

How can we find this map on $Y\times \mathbb{S}^1$? Well,  we need to sum over the representations $\lambda$, but notice that $\pi_1(Y)$ is generated by $g_1,g_2,g_3$ which satisfy $g_1\cdot g_2\cdot g_3=1$ (you have three loops based at the same point going around each hole, so you can deform “around the back of the sphere” to the trivial loop), and $\pi_1(\mathbb{S}^1)$ is just generated by an element $h$. Since $g_3=g_2^{-1}\cdot g_1^{-1}$ we have that our action can only depend on three variables, $W(\gamma)=c_h(g_1,g_2)$. To understand this $c_h$ a little better, consider constructing the sphere with four holes $\mathbb{S}^2_4$ from $Y$. Dijkgraaf and Witten do this diagrammatically:

I think this is pretty clear – gluing a hole means setting the element of the fundamental group equal, taking into account orientations (and the above unity relation is satisfied at each vertex). By chasing around the relations you can see that $g_4=g_3{^-1}g_2^{-1}g_3^{-1}$. By the axioms of TQFT this gluing corresponds to multiplication in the partition function $c(g_1,g_2)c(g_1,g_2,g_3)$. However, this gluing can be done in another way; this is easiest to see by shrinking the $g_1g_2$ line to a four-vertex, and then pulling it apart vertically:

Although it looks like I have some deeper knowledge here, I am really just enforcing the group multiplication at the vertices when playing with these diagrams. One should also check that this is not just a relabeling of the first one, which is pretty easy to see. Anyway, this means that our action satisfies

$c_h(g_1,g_2)c_h(g_1,g_2,g_3)=c_h(g_1,g_2,g_3)c_h(g_2,g_3)$

But notice what happens if we think of our action as a function $c_h:\pi_1(\mathbb{S}^2_4)\to\mathbb{C}$, and apply the coboundary operator to it:

$\delta c_h(g_1,g_2,g_3)=c_h(g_1,g_2)^{-1}c_h(g_2,g_3)c_h(g_1g_2,g_3)^{-1}c_h(g_1,g_2g_3)=1$

The equality to unity comes from applying the relation above (this is actually being relaxed with notation; we had previously defined $c_h$ as “a function of the independent generators”, so it didn’t have a specific number of slot to put things in. Now, we see this is because it lives in the cohomology complex). Thus, our action $c_h$ is a 2-cocycle in $H^2(\mathbb{S}^2_4,\mathbb{C})$. Under the representation, these elements all live in the gauge group $G$, so we have a group 2-cocycle $c_h$ associated to each stabilizer subgroup $N_h\subset G$.

At some level this is the best that can be done with finite groups without further restrictions. We will see in the next post how also making the spacetime finite – that is, a lattice – will finally give us some concrete examples to play with.

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

Now I continue this series (part I, part II, part III) with a discussion of…

Differential Characters

We must consider what happens if the class $\omega$ in cohomology which we choose to represent our Chern-Simons action

$S=\frac{1}{n}[\langle \Omega(F),[B] \rangle-\langle \gamma^*\omega, [B] \rangle ]$

contains torsion. If this element has torsion of order $p$, then our action will have a $\mathbb{Z}_p$ phase. This is a further ambiguity which we want to remove, by specifying an action which detects the presence of torsion in extension to the chain $B$. Dijkgraaf and Witten choose to do this using differential characters. I note that it is not at all obvious to me that this is a unique approach – it seems to work, but that doesn’t mean there aren’t other possible ways of going about doing this.

Differential characters were original described by Cheeger and Simons (1985) – I found that paper a bit difficult to read, but there is a very nice description from an arXiv paper by Baer and Becker (link – chapter 5 in particular). The generalization proceeds by identifying elements in cohomology that, rather then obeying $\partial ^2 \omega=0$, satisfy $h \circ \partial \in \Omega^k$. Specifically, differential characters are elements of the abelian groups

$\hat{H}^k(M;\mathbb{Z}):=\{h\in Hom(C_{k-1}(M;\mathbb{Z}),U(1))| h \circ \partial \in \Omega^k(M)\}$

$h(\partial c)=\exp\left(2\pi i \int_c \alpha(h) \right)$.

Here $\alpha(h)\in \Omega^k(M)$ is called the curvature of $h$ (and is a closed form). Note there is some consistency to the nomenclature here – “characters are exponentials of classes” – which is true for i.e. Chern characters as well. To every $h$ there is a corresponding form $\beta$ which takes values in $\mathbb{R}$, defined by $h(c)=\exp(2\pi i \beta(c))$. We also have the map

$\mu^{\beta}:C_k(M;\mathbb{Z})\to \mathbb{Z},~\mu^\beta(c)=\int_c \alpha(h) - \beta (\partial c).$

Since the curvature is a closed form, it’s easy to show that $\mu^\beta$ is closed, and thus $[ \mu^{\beta}]\in H^k(M;\mathbb{Z})$.

Now, observe what happens when we evaluate this character on a chain $z\in C_{k-1}(M)$ which represents a torsion class in $H_{k-1}(M)$. By some similar arguments as we used before, we can choose $x\in C_k(M)$ such that $z=\partial x /N$ as real cocycles. Then we evaluate the character on the torsion element:

\begin{aligned} h(z)&=\exp(2\pi i \beta (z))=\exp \left(\frac{2\pi i}{N} \beta (\partial x)\right)\\&=\exp\left(\frac{2\pi i}{N}\delta \beta (x)\right)=\exp\left(\frac{2\pi i}{N}\left(\int_x \alpha(h) - \mu^\beta (x)\right)\right)\end{aligned}

To go from the first to the second line we just used the fact that the coboundary operator is the dual to the boundary operator, and the last step uses the definition of $\mu^\beta$ from above. Thus, rather then define the action $S$, we define the exponential of the topological action as the differential character $h$, which is evaluated over the extensions of $M$ to chains $B$.

This reduces to the appropriate notions – if $B$ is closed, we find the action given by $\langle \alpha,[B]\rangle$, which is the same thing as we get when $[B]$ is not a torsion element. When it is, we get the action $\int_B \alpha -\mu^\beta(B)$. By identifying $\alpha \leftrightarrow \Omega(F)$ (they are both curvatures) and $\mu^\beta\leftrightarrow \gamma^*\omega$ (they are both cocycles in $H^4(B;\mathbb{Z})$), we get back to our original topological action. In fact, a pair $(\Omega(F),\omega)$ uniquely determines the reduction of a form $\delta\beta=\Omega(F)-\omega$ if $H^3$ vanishes, which it does for the classifying space. I skip the proof of this, because it will take me too far afield.

One note – the Dijkgraaf and Witten paper looks somewhat different than the above discussion because they often exchange the roles of $\alpha$ (the differential form) and $h$ (the exponential of the integral of that form). Since they uniquely determine each other, there is no real issue here but it does make things slightly confusing to try and work out. I have tried to keep things as consistent as possible.

The Dijkgraaf and Witten paper now continues to discuss manifolds with boundary, CFT, and spin theories. None of these are relevant to my particular interest at this moment, so the next post will continue by fixing a finite gauge group and looking at what happens on the lattice.

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

Continuing my notes from the Dijkgraaf and Witten paper (part I, part II)…

Topological Actions

This will be an extended discussion of the introduction in part I, with the basic goal to understand how to classify these theories by elements in the 3rd cohomology group. Before starting I should say I began a short discussion on MathStax about this – the focus being on the generalization of such actions into the language of higher stacky cohomology. The answer there is very well-written but deviates farther from Chern-Simons then I am interested in at the moment.

When the bundle $E$ over the 3-manifold $M$ is trivial, we can easily extend it to a 4-manifold $B$ which restricts to $M$ on its boundaries. If $E$ is not trivial (not globally $G\times M$), than this is not possible – since we are interested in a generic Chern-Simons form, this is exactly the case we want. First, rather than consider extensions to manifolds Dijkgraaf and Witten consider extensions to a “smooth, singular 4-chain” which is probably just an element $\sum_i n_i \sigma_i\in C_4(B)$ where $\sigma_i:\Delta^4 \to B$ – the “smooth” modifier is probably meant to suggest the map $\sigma_i$ is smooth. In order to have a bundle over this extension, the classifying map must also have an extension $\tilde{\gamma} :B \to BG$ with boundary $\gamma(M)$. This translates into a condition on homology – namely, $\tilde{\gamma }(B)\in C_4(BG)$ is an extension of $\gamma(M)\in C_3(BG)$ if $\partial \gamma(B)=\gamma (M)$. But then $\partial \gamma(M)=0$ and $[\gamma(M)]=0\in H_3(BG,\mathbb{Z})$. So the existence of an extension comes from the vanishing in the 3rd homology group.

This homology group consists entirely of torsion – this is because $H_{odd}(BG,\mathbb{R})=0$ for a Lie group or finite group (and in fact, all $H_*$ vanishes for finite groups). In the previous post we learned that the kernel of the map $\rho:H_k(M,\mathbb{Z})\to H_k(M,\mathbb{R})$ is the torsion elements, so by the first isomorphism theorem $H_{odd}(BG,\mathbb{Z})/Tor=0$. Torsion elements are cyclic of order $p$, so this implies that $p\cdot [\gamma(M)]=0$ (writing the groups multiplicatively). In some way there are “$p$ copies of $E$ in the extension $\tilde{E}$.” Thus we have a natural ambiguity in how we extend the bundle $E$ over $B$ – namely, we can only do so up to an integer $p$.

At this point, it is becoming obvious to me that a lot of the ambiguity coming from this construction has to do with on one hand needing to use differential forms (with coefficients in $\mathbb{R}$) to define the action but using chains (with coefficients in $\mathbb{Z}$) to determine when the construction makes sense. Perhaps this is a partial answer to my inevitable confusion about why torsion is important in homology – but I digress…

The differential form we care about is $\Omega=\frac{k}{8\pi^2}Tr(F\wedge F)\in H^4(B,\mathbb{R})$ for curvature $F$. This form can be described by a integer-valued cochain $\omega$ such that $\rho([\omega])=\Omega$, but only up to a torsion element. The idea is to use this ambiguity to resolve the ambiguity in the action from above. Thus, define the action

$S=\frac{1}{n}[\langle \Omega(F),[B] \rangle-\langle \gamma^*\omega, [B] \rangle ]$

(here I am using the pairing $\langle \alpha, [M]\rangle \to \int_{M}\alpha$). Notice this does not depend on which cocycle $\omega$ we chose to represent the class $[\omega]$, since if we shift it $\omega \to \omega + \delta\epsilon$ where $\epsilon\in C^3(BG,\mathbb{Z})$ the action changes by

$\Delta S=-\frac{1}{n}\langle \gamma^*\delta\epsilon,[B]\rangle=-\frac{1}{n}\langle \gamma^* \epsilon,\partial B\rangle=-\langle \gamma^*\epsilon,M\rangle \in \mathbb{Z}$

(It’s amazing how much mathematics Stoke’s theorem can get you through…). We have used $n M=\partial B$ (the boundary of $B$ consists of $n$ copies of $M$), and the last equality comes from $\epsilon$ having integer coefficients. Since the action changes by an integer, it remains a gauge invariant.

The action above is the topological action for a bundle of order $n$. Dijkgraaf and Witten perform a few more technical consistency checks on it, but we will move on to more interesting things, since my focus is going to be on finite groups and lattice theories. The next short post will be about differential characters, which allow you to define such actions in slightly more generality.

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

Continuing with my last post….

Group Cohomology and the Classifying Space

Recall singular homology: this is the homology of groups of singular chains $C_k(T)$ in a topological space $T$ (note the difference between singular and simplicial – simplicial homology is directly on the simplicial complex, whereas singular homology is on maps $\Delta^n \to T$ which only have to be continuous). With boundaries $B_k(T)$ and cycles $Z_k(t)$ we have the homology groups $H_k(T,\mathbb{Z})=Z_k(T)/B_k(T)$, where the singular chains have coefficients in $\mathbb{Z}$. The dual notion is the cochains $C^k(T)=Hom(C_k(T),\mathbb{Z})$, which give the cohomology groups $H^k(T,\mathbb{Z})$ with the coboundary operator $\delta$. It will be important to use another definition of the cohomology groups with coefficients in abelain group $F$:

$H^k(T,F)=Hom(H_k(T),F).$

where $\alpha \in H^k(T,F)$ vanishes on boundaries since it lives in the homology.

The homology groups are abelian groups which can be written as $\mathbb{Z}_{p1}\times ...\times \mathbb{Z}_{p2}\times \mathbb{Z}^n$, where $p_i$ is the order of the $i$th torsion element and $n$ is the Betti number. Unlike the “$\mathbb{Z}$” part, the torsion cannot be represented by differential forms in de Rham cohomology. Torsion will be very important later – since the Chern-Simons forms are elements of cohomology and we lose information about them when we pass into that.

Torsion is a bit of an archnemesis of mine – I will never fully understand how important it is for homology. However, we can at least see how this last statement might come about. De Rham cohomology satisfies the de Rham theorem, which is

$H_k (M,\mathbb{Z}) \cong H^k_{dR}(M)$

Since the De Rham cohomology consists of differential forms, think of it as having coefficients in $\mathbb{R}$. The problem is that torsion elements are in the kernel of the map $\rho:H^k(T,\mathbb{Z})\to H^k(M,\mathbb{R})$. To see this, take the simplest example where torsion exists – $\mathbb{R}P^2$. It’s simplicial structure is given by the following (thanks Hatcher!):

Ok so I want to calculate $H_1(\mathbb{R}P^2,\mathbb{Z})=Ker(\partial_1)/Im(\partial_2)$. We have

\begin{aligned}pa+qb+rc\in C_1\rightarrow \partial_1 (pa+qb+rc)&=p(w-v)+q(w-v)+r(v-v)=0\\ &\rightarrow p+q=0, r\in\mathbb{Z}\end{aligned}.

\begin{aligned}pU+qL\in C_2\rightarrow \partial_2(pU+qL)&=p(-c+a-b)+q(-c+b-a)\\ & =(p-q)(a-b+c)-2qc\rightarrow Im(\partial_2)=\langle a-b+c,2c\rangle\end{aligned}

Notice that any element in $Ker(\partial_1)$ can be written like $p'(a-b)+q'c=(p'+q')(a-b+c)-q'c$, so these two elements represent the same thing in homology if $2q=q'$ for some $q,q'\in\mathbb{Z}$. If $q'$ is even, we can always find some $q$ which satisfies this equation, and if $q'$ is odd we cannot. Thus there are two classes in this homology group and we find $H_1(\mathbb{R}P^2,\mathbb{Z})=\mathbb{Z}_2$, i.e. a torsion element of order two.

But consider what would happen if we wanted to use $\mathbb{R}$ instead of $\mathbb{Z}$; we would need to satisfy $2q=q'$ for a real numbers.  Given a $q'$, it is always possible to find a $q$ such that this equation is satisfied. This means all elements are the same in homology and $H_1(\mathbb{R}P^2,\mathbb{R})=0$. Of course, this is exactly what the statement “the torsion is in the kernel of the map $H_k(T,\mathbb{Z})\to H_k(T,\mathbb{R})$” means. I think this example typifies what goes on; torsion elements come from periodic conditions on the chains – we have one condition that tells us when two elements are equivalent, so there are two classes. For real numbers, this equation can always be satisfied so there is only one class.

Classifying Spaces

Classifying spaces are key to understanding many aspects of algebraic topology and homology. The classifying space $BG$ is the base space for a $G$-bundle $BG$ called the universal bundle. Basically what one does is start with a bundle $(E,G,M)$ and ask under what conditions it can be obtained as a pullback from a map $f:M \to BG$. You begin by defining the map $f$ over the points (0-cells) of $M$, and extending it over the 1-cells. My understanding is that this is done by assuming the bundle $EG$ is contractible in the following sense – each trivialization is homotopic to a point:

$U_i\times G \sim \{*\}\times G$.

Then the extension to the 1-cells is easy. By making this assumption for all higher k-cells this universal bundle can be constructed for any $G$. The bundle is universal in the sense that any bundle is such a pullback over the classifying map, and any two classifying maps $\gamma, \gamma'$ which are homotopic produce equivalent bundles ($\gamma^*BG \to \gamma'^*BG$ is a homeomorphism). As a sidenote, the characteristic classes $H^*(BG,\mathbb{Z})$ of the classifying space give rise to the homology classes $H^*(M,\mathbb{Z})$ by the pullback of the classifying map, since they depend only on the topology of $E$.

# 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 $A$).