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I have a two-part question:

(1) First and foremost: I have been going through the paper by Dijkgraaf and Witten "Group Cohomology and Topological Field Theories." (See Here they give a general definition for the Chern-Simons action for a general $3$-manifold $M$. My question is if anyone knows of any follow-up to this, or notes about their paper?

(2) To those who know the paper: They say that they have no problem defining the action modulo $1/n$ (for a bundle of order $n$) as $n\cdot S = \int_B Tr(F\wedge F)$ $(mod 1)$, but that this has an $n$-fold ambiguity consisting of the ability to add a multiple of $1/n$ to the action - What do they mean here? Also, later on they re-define the action as $S = 1/n\left(\int_B Tr(F\wedge F) - \langle \gamma^\ast(\omega),B\rangle\right)$ $(mod 1)$ - How does this get rid of the so-called ambiguity?

Basically my question is if anyone can further explain the info between equations 3.4 and 3.5 in their paper. Thanks.

Update: I'm fine with re-defining the action as $S = 1/n\left(\int_B Tr(F\wedge F) - \langle \gamma^\ast(\omega),B\rangle\right)$ $(mod 1)$. But, does anyone know how they came to discover that this is what to add to the action to remove the ambiguity in the previous definition the action? I mean, if you only know $S$ modulo $1/n$, and if you think it's $S_0 = \int_B Tr(F\wedge F)$ it means that the real action is $S = S_0 + a_{M,B}\big/n$, where $a_{M,B}$ is some integer that possibly depends on $M$, $B$ and the extension of $E$ over $B$. So to remove the ambiguity you have to find this integer for the specific data, but how do they FIND the expression for this integer; that is, how do they calculate the $a_{M,B}$ to be $\langle \gamma^\ast(\omega),B\rangle$. where $\omega \in H^4(BG,\mathbb{Z})$?

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They are trying to define the Chern-Simons action over a manifold $M$ by writing it as the integral of $\int F \wedge F$ over a bounding manifold $B$. When the bundle is nontrivial, they consider a more general cochain and show that there exists a $B$ over which the bundle extends such that $\partial B$ = $n$ copies of $M$. So, you can define

$$ n S = \int_B F \wedge F $$

But, because actions enter into imaginary exponentials in the path integral, this is really only defined mod 1 (once you reenter all the coefficients that I omitted). So, the action $S$ is only defined mod $1/n$.

They show how the second formula resolves the ambiguity in the text that follows, but it's probably best to think of it as a differential character or in terms of differential cohomology. A more rigorous presentation might be .


Let me try another explanation. We know from the above that $$ n S_{CS} = \int_B F \wedge F \quad \mbox{mod 1} $$

Thus, $$ S_{CS} = \frac{1}{n}\int_B F \wedge F + \frac{q(B,E)}{n} $$ with $q(B,E) \in \mathbb{Z}$. The simplest guess is that $q = 0$, but it's easy to see that the resulting action is not independent of the choice of $B$. In particular, we would want, for closed $B$, that $\frac{1}{n} \int_B F \wedge F \in \mathbb{Z}$, but it's only in $\frac{1}{n}\mathbb{Z}$.

So, the goal is to choose a $q(B,E)$ such that the action makes sense. Since you want something that is an integer when applied to a closed $B$, it's not too hard to guess something like DV's action.

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Thanks for the explanation! I think I'm fine with everything except for how they arrive at the correction they use to remove the ambiguity (see the update at the top). Yes, I've also been looking at this paper by Freed and Quinn, but they assume finite $G$ which implies the general action reduces to $\langle \gamma_M^\ast(\omega),[M]\rangle$ and they just state this to be the action without explaining why (unless I'm missing something). – Kevin Wray Apr 27 '11 at 14:13
As I recall, it's justified post facto by the fact that it works. I wouldn't be surprised if they knew about the differential character and worked backwards, but it's also possible they just worked it out themselves. – Aaron Bergman Apr 27 '11 at 16:47
So, you think it was more of a guess and check, rather than starting at point A and arrive at point B? – Kevin Wray Apr 27 '11 at 17:31
If it would make you feel better, you can think of this as a definition of the Chern-Simons action. It's sort of the only thing you can really write down that makes sense. – Aaron Bergman Apr 27 '11 at 19:22
Do you mean that adding $\langle \gamma^\ast(\omega), B\rangle$ will give you an action that doesn't depend on bounding $B$, homotopy, etc.? If so, then how does this immediately get rid of the ambiguity? Thanks. – Kevin Wray Apr 27 '11 at 19:43

Takefumi Nosaka has written a paper with Eri Hatakenaka that describes connections between DW and quandle cocycle invariants.

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I don't think they actually find the constant you refer to in the update. As they remark, the choice of $\omega$ which maps to $\frac{k}{8\pi^2}Tr(F\wedge F)$ under $H^4(BG,\mathbb Z)\rightarrow H^4(BG, \mathbb R)$ is only defined up to torsion. The Chern-Simons invariant depends on the choice of $\omega$. At least that's what Dan Freed seems to indicate at the end of the appendix in this paper . He also has a follow up paper which discusses more the Chern-Simons theory where $G$ isn't necessarily simply connected.

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Yes, perhaps constant is not the most accurate word. Do you know why the correct term to add is $\langle \gamma^\ast(\omega),B\rangle$? – Kevin Wray Apr 27 '11 at 15:00
I rephrased the Update part to be a bit more precise. – Kevin Wray Apr 27 '11 at 17:18
I can't say I understand why that's necessarily the only sort of term that can be added, but it seems the most natural. To make everything work out you need something in terms of characteristic classes there. And that choice makes $S$ independent of the choice of bounding manifold. – charris Apr 27 '11 at 18:19
There's a typo in your update: $\omega \in H^4(BG,\mathbb Z)$. And I don't think things should be put that way. I wouldn't say you're looking for something ($a_{\omega}$) depending on $\omega$ from the outset. You're looking for something that depends on $M$ and $B$ and how $E$ is extended over $B$.. What I was getting at in my answer above is that once they choose to add the term of the form $\langle \gamma^* \omega, B \rangle$ this sets $\omega$ as an extra parameter. – charris Apr 27 '11 at 18:34
Thanks, I fixed it. – Kevin Wray Apr 27 '11 at 19:52

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