# 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 $E$.

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