In the first section of Freed-Hopkins-Lurie-Teleman they construct a one-dimensional Topological Quantum Field Theory.

  • $F(\circ_+)$ is a vector space and $F(\circ_-)$ is the dual.
  • $F(\circ-\circ)$ is the duality pairing
  • $F(S^1) = \dim F(\circ_+)$ is a number

Then it as the 1D gauge theory has an action determined by an element of the classifying bundle $H^2(BG, \mathbb{Z})$. How is the action an element of the cohomology?

$\mathrm{Hom}(G, \mathbb{T}) \simeq H^2(BG, \mathbb{Z})$ is a canonical isomorphism

While I don't know anything about the K-theory of the universal bundle, I do know about characters of compact groups.

Why is Gauss Law compared to the fixed points of Abelian characters $\lambda: G \to \mathbb{T}$?

Is there a more elementary description of the relationship between 1D tqft and harmonic analysis on groups? What happened to the non-abelian characters?

I don't know category theory. Instead, just know the character theory of finite groups and some about compact Lie groups. 1D TQFT looks like Fourier analysis with some very fancy jargon attached.

If my notation is not conventional that is because it reflects my own understanding of the material.


1D TQFT looks like Fourier analysis with some very fancy jargon attached.

The example in FHLT is just an example. There are many others. In general, an oriented 1D TQFT with values in a symmetric monoidal category is precisely a dualizable object in that category, e.g. a finite-dimensional vector space in the case of $(\text{Vect}, \otimes)$. The particular example in FHLT amounts to a roundabout way of picking either a $0$-dimensional or a $1$-dimensional vector space, but the point is not the TQFT itself but the way in which the TQFT is a simpler version of more complicated TQFTs.

How is the action an element of the cohomology?

The short answer is that you can use this cohomology class to produce a complex number which you can think of as an analogue of the exponentiated action $e^{ \frac{i}{\hbar} S(\varphi)}$ of a classical field; this is the number you use to weight path integrals in quantum field theory, and similarly here it is used to weight a certain "finite path integral."

The long answer:

The $1$-dimensional case is in some sense too degenerate to get a feel for what's going on here, so let me tell you what the $n$-dimensional case looks like. The TQFT under consideration here is in general and for finite groups called Dijkgraaf-Witten theory. The $n$-dimensional theory takes as input a finite group $G$, the "gauge group," and a cohomology class $\alpha \in H^n(BG, \text{U}(1)) \cong H^{n+1}(BG, \mathbb{Z})$, the "action." I'll explain why this deserves the name in a bit. To a closed, oriented $n$-manifold, it assigns the "path integral"

$$Z_G(M) = \sum_{\varphi \in [M, BG]} E(\varphi)$$


$$E(\varphi) = \frac{1}{|\text{Aut}(\varphi)|} \int_M \varphi^{\ast}(\alpha).$$

Here $[M, BG]$ denotes the finite set of homotopy classes of maps $M \to BG$, or equivalently isomorphism classes of $G$-covers of $M$, or equivalently (when $M$ is connected) conjugacy classes of homomorphisms $\pi_1(M) \to G$. This finite set is really a groupoid, namely the groupoid of isomorphism classes of $G$-covers, and $\text{Aut}(\varphi)$ refers to the automorphism group of a given $\varphi : M \to BG$ regarded as a $G$-cover. This groupoid is a toy model of the moduli space of principal $G$-bundles on $M$ for $G$ a Lie group, and the sum above is a toy model for integrating on this moduli space, which is the sort of thing you'd want to do in order to quantize a gauge theory like Chern-Simons.

Okay, so why does this cohomology class deserve the name "action"? The starting point is that we want to think of the above finite sum as a toy model of a path integral. In general, a path integral is an integral over a space of fields $\varphi$, which each field weighted by an exponential $E(\varphi) = e^{ \frac{i}{\hbar} S(\varphi)}$ where $S(\varphi)$ is the classical action of the field $\varphi$.

In Dijkgraaf-Witten theory, a field is a $G$-cover, so the exponentiated classical action, whatever that means, ought to assign to each $G$-cover of a manifold $M$ a unit complex number. One way to get a unit complex number is to get a cohomology class in $H^n(M, \text{U}(1))$ and integrate it over $M$, so we can reduce to the problem of assigning to each $G$-cover a cohomology class in $H^n(M, \text{U}(1))$. Once we require in addition that this assignment is natural in $M$, we are essentially asking for a characteristic class of $G$-covers, and once we take the further step of allowing $M$ to be an arbitrary space then the usual Yoneda lemma argument shows that such characteristic classes are naturally in bijection with homotopy classes of maps $BG \to B^n \text{U}(1)$, or cohomology classes in $H^n(BG, \text{U}(1))$ as desired.

What happened to the non-abelian characters?

The fact that characters showed up was in some sense a low-dimensional accident; in general for higher $n$ we use something more complicated, namely cohomology classes in $H^n(BG, \text{U}(1))$.

Why is Gauss Law compared to the fixed points of Abelian characters $\lambda : G \to \mathbb{T}$?

This I genuinely don't know. I always found this remark completely mysterious.

Edit: I asked Freed what this remark meant. Apparently there is a general principle called the Gauss law in quantum field theory to the effect that in a gauge theory the result of a path integral is gauge invariant (so if the naive thing is not gauge invariant one should identify gauge equivalent results, or something like that). There should be some connection to what is classically called Gauss's law, since electromagnetism is a $\text{U}(1)$ gauge theory, but Freed said he wasn't sure off the top of his head what that connection is.

  • $\begingroup$ By the way, in everything above $\text{U}(1)$ has the discrete topology. $\endgroup$ – Qiaochu Yuan Jul 8 '14 at 7:07
  • $\begingroup$ There is a nice discussion of gauge theory and Gauss's law in eduardo.physics.illinois.edu/phys582/582-chapter9.pdf Basically, $\nabla\cdot \mathbf{E}$ generates infinitesimal gauge transformations $\endgroup$ – Vít Tuček Jul 8 '14 at 7:39

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