# Importance of the principal bundle in Chern-Simons theory

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?

• Dan Freed: arxiv.org/abs/hep-th/9206021 Jun 21 '19 at 16:10
• 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 $G$. Jun 21 '19 at 16:44

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.

• 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 $G=Z_2$.) Jun 22 '19 at 8:21
• @AndiBauer the gauge group is G in (twisted) Dijkgraaf-Witten theory.
– skd
Jun 22 '19 at 14:39
• @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? Jun 22 '19 at 18:05
• 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.
– skd
Jun 23 '19 at 2:12
• 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 $\alpha$? Jun 23 '19 at 10:12