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This is a very basic beginners question about Chern-Simons theory. The configurations that we sum over to get the partition function are given by a Lie-algebra valued 1-form $A$ on a topological 3-manifold, called the connection. More precisely, $A$ is a connection of a principal Lie-group bundle over the manifold.

What is the role of this principal bundle?

I didn't find this spelled out explicitly anywhere, but I assume one does not sum over different choices of principle bundles when calculating the partition function? Or are different choices of principle bundle already encoded in different choices of $A$?

So is Chern-Simons theory not only a topological field theory that one can put on any topological manifold, but one that we can additionally put on any topological manifold with a Lie-group principle bundle? Can one restrict to trivial principle bundles without loosing any of the physical interpretation? If one talks about the "3-manifold invariants" of the Chern-Simons theory, does that refer to the partition function on the trivial principle bundle over those manifolds? Or is the partition function independent of the choice of bundle?

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  • $\begingroup$ Dan Freed: arxiv.org/abs/hep-th/9206021 $\endgroup$
    – AHusain
    Jun 21 '19 at 16:10
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    $\begingroup$ One thing to remember is that a principal $G$-bundle on a 3-manifold is trivializable when $G$ is connected and simply-connected. So, the partition function where you restrict to the trivial $G$-bundle is simply the partition function of Chern--Simons for the universal cover of $G$. $\endgroup$ Jun 21 '19 at 16:44
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In quantum Chern-Simons theory with gauge group $G$ (compact Lie), a field on a 3-manifold $M$ is a principal $G$-bundle with a connection $A$. The partition function/path integral associated to $M$ is supposed to be the integral over the (generally infinite-dimensional) space ("stack") $\mathcal{F}$ of all principal $G$-bundles $P$ over $M$ equipped with a choice of connection $A$ (modulo gauge equivalence) of $\exp(iS(A))$, where $S$ is the classical Chern-Simons action.

Perhaps it is easier for pedagogical purposes to explore the simpler case of Chern-Simons theory when $G$ is finite; this is Dijkgraaf-Witten theory. In this case, one can make the notion of an integral over $\mathcal{F}$ perfectly rigorous. If $G$ is finite, then each bundle has a unique connection, so $\mathcal{F}$ is precisely $[M, BG]$, where $BG$ is the classifying space of $G$. The Lie algebra of $G$ is also trivial, so the action vanishes, and we're left with integrating (summing) $1$ over the space $[M,BG]$ with respect to some measure; the weight of a principal $G$-bundle $P$ in this case is $1/|\mathrm{Aut}(P)|$. In other words, the finite-group version of the Chern-Simons partition function is $$Z(M) = \sum_{P\in [M,BG]} \frac{1}{|\mathrm{Aut}(P)|}.$$

This is the literal analogue of Chern-Simons theory for a finite gauge group; however, a more interesting analogue is twisted Dijkgraaf-Witten theory, which might be what you were trying to get at. Recall that the action associated to the field $(P,A)$ over $M$ is $S(A) = \int_M q(A)$; what matters is that the integrand $q(A)$ is a $3$-form. Since a Chern-Simons field on $M$ is a principal $G$-bundle with a choice of connection $A$, one might attempt to "canonically" associate to each principal $G$-bundle a $3$-form on $M$ in the finite group case; this $3$-form would be the replacement of $A$. Note that in this story, $A$ plays a somewhat different role.

Think of the $3$-form as a singular cochain on $M$, so integration is pairing the cochain with the fundamental class of the manifold $M$ (assume it's closed and oriented). Since a field on our manifold is still a principal $G$-bundle, determined by a map $f_P:M\to BG$, we can associate to each bundle a $3$-dimensional cohomology class if we fix a choice of $\alpha\in \mathrm{H}^3(BG;\mathbf{C}^\times)$; then, the $3$-form "$q(A)$" associated to $P$ is $f_P^\ast(\alpha)\in \mathrm{H}^3(M;\mathbf{C}^\times)$. (Really, one should fix a $3$-cocycle in $Z^3(BG;\mathbf{C}^\times)$.) The action associated to $P$ is then $\langle [M], f_P^\ast(\alpha)\rangle$, and to obtain the quantum theory, one can now integrate over the space of all $G$-bundles (with the same measure as the untwisted case). Note that $A$ by itself doesn't appear in this story; only the analogue of the associated $3$-form does.

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    $\begingroup$ Wait, what's the gauge group and the level now in the case of twisted Dijkgraaf-Witten theory? Is the former still the finite group $G$? (Thought it is something like $U(1)\times U(1)$ and $k$ being some matrix, at least for $G=Z_2$.) $\endgroup$
    – Andi Bauer
    Jun 22 '19 at 8:21
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    $\begingroup$ @AndiBauer the gauge group is G in (twisted) Dijkgraaf-Witten theory. $\endgroup$
    – skd
    Jun 22 '19 at 14:39
  • $\begingroup$ @sdk Ah ok thanks! But then I thought the only inputs for Chern-simons theory are the gauge group and the level (which is usually an integer $k$ but then also I have seen it to be a matrix). So does that mean the different twisted Dijkgraaf-Witten theories for the same group $G$ only differ by their level? So then, what are the different levels for the different group $3$-cocycles? $\endgroup$
    – Andi Bauer
    Jun 22 '19 at 18:05
  • $\begingroup$ Yes, because to each gauge group G and choice of $\alpha\in \mathrm{H}^3(BG;\mathbf{C}^\times)$ is associated a twisted Dijkgraaf-Witten theory $Z_\alpha$. There might be examples of manifolds for which $Z_\alpha$ agrees with $Z_{\alpha'}$ for $\alpha\neq \alpha$, but the TQFTs are distinct. $\endgroup$
    – skd
    Jun 23 '19 at 2:12
  • $\begingroup$ Hmm, let me reformulate: Dijkgraaf-Witten takes as input a finite group $G$ and a $3$-cocycle $\alpha$. Chern-Simons takes as input a gauge Lie group $H$ and an integer level $k$. If I got you right, you're saying $H=G$, even in the twisted case. So how do I obtain $k$ from $G$ and $\alpha$? $\endgroup$
    – Andi Bauer
    Jun 23 '19 at 10:12

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