How to understand Chern-Simons action

Question

Hi all. The question I have should be a rather simple one, but I just can't think it through.

So the Chern-Simons action reads \begin{equation} S = \int_M {\rm tr} (A\wedge dA + \frac{2}{3} A\wedge A \wedge A) \end{equation} where $M$ is 3-fold, and similarly for higher dimensional manifold.

Now, my question is:

*since $A$, the connection 1-form is only defined patch by patch, what do we really mean by doing the integration? *

It would be understandable if I write \begin{equation} S = \int_M {\rm tr} \left[(A-A_0)\wedge d(A-A_0) + \frac{2}{3} (A-A_0)\wedge (A-A_0) \wedge (A-A_0)) \right ] \end{equation} where $A_0$ is some reference connection, since $A-A_0$ is globally defined 1-form valued in ${\rm Lie}G$.

I see that under gauge transformation (or different chart), \begin{equation} CS(A^g) - CS(A) = d\alpha(A,g) + Q(g) \end{equation} where $Q(g)$ is closed. But I don't know how I can infer the validity of doing the integration from this gauge transformation.

Thank you!

Answer 1

Often in the literature by "Chern-Simons theory" is meant by default $G$-Chern-Simons theory whose gauge group is a connected and simply connected semisimple compact group $G$, such as $G = SU$. In this case it so happens that all $G$-principal bundles on a 3-manifold $\Sigma_3$ are trivializable, and hence one can identify the space of G-principal connections on $\Sigma_3$ just with that of $\mathfrak{g}$-valued differential forms. So one gets away with the naive formula that you recall above.

In stark contrast to this is what may seem to be a simpler example, namely $U(1)$-Chern-Simons theory. Since $U(1)$ is not simply connected, clearly, there are of course non-trivial $U(1)$-principal bundles on $\Sigma_3$, in general, and hence the above naive approach fails, as you notice.

In this case the correct Chern-Simons action is instead obtained this way: given a field configuration $\nabla$ which is a circle-principal connection, we can form its differential cup-product square in ordinary differential cohomology. This yields a $\mathbf{B}^2 U(1)$-principal 3-connection $\nabla \cup \nabla$, often known as a bundle 2-gerbe with connection or else as a degree-4 cocycle in Deligne cohomology. This now has a connection 3-form and hence has a volume holonomy over $\Sigma_3$. And this now is the correct action functional for Chern-Simons theory. For more on this see at nLab:higher dimensional Chern-Simons theory.

Secretly this higher principal connection structure also governs the first, seemingly simpler case. The action functional of Chern-Simons theory is always the volume holonomy of a 3-connection, the Chern-Simons circle 3-connection.

This is in fact the general abstract characterization of Chern-Simons theories and all its higher (and lower) dimensional variants. A Chern-Simons-type action functional is always the volume holonomy of a refinement of a universal characteristic class to ordinary differential cohomology. Further remarks along these lines are for instance in

Domenico Fiorenza, Hisham Sati, Urs Schreiber, A higher stacky perspective on Chern-Simons theory.

Answer 2

Here is a mid 1970s point of view, courtesy of Atiyah-Patodi-Singer.

Suppose you have a complex vector bundle $E$ of rank $r$ over a smooth manifold $M$. A polynomial function $P$ on the space of $r\times r$ matrices is called invariant if $P(T AT^{-1})=P(A)$ for any $r\times r$ complex matrix $A$ and any invertible $r\times r$ matrix $T$. If you look at

$$ \Delta(x)=\det(1+ xA)=\sum_{k=0}^r c_k(A) x^k, $$

then the coefficient $c_k(A)$ is a homogeneous invariant polynomial function of degree $k$. For example

$$c_1(A)={\rm tr}\; A,\;\;c_r(A)=\det A. $$

To a connection $\nabla$ on $E$ with curvature $F(\nabla)$, we can associate the degree $2k$ form on $E$

$$ c_k(\nabla) = c_k\bigl(\; F(\nabla)\;\bigr), $$

where in the above equality one thinks of $F(\nabla)$ as an $r\times r$-matrix whose entries are $2$-forms. For example

$$ c_1(\nabla)= {\rm tr}\; F(\nabla)= F_{11}(\nabla)+\cdots +F_{rr}(\nabla). $$

Chern-Weil theory proves two things:

• The form $c_k(\nabla)$ is closed.

• If $\nabla^1$, $\nabla^0$ are two connections on $E$, then there exists a canonical form of degree $(2k-1)$, called the transgression form and denoted by $Tc_k(\nabla^1,\nabla^0)$, which satisfies

$$ d Tc_k(\nabla^1, \nabla^0)= c_k(\nabla^1)-c_k(\nabla^0). $$

In other words, the cohomology class determined by $c_k(\nabla)$ is independent of $\nabla$. This cohomology class is the $k$-th Chern class of $E$.

Suppose now that $\dim M= 2k-1$. Then, on account of dimension, $c_k(\nabla)=0$, yet $Tc_k(\nabla^1,\nabla^0)$ is a top degree form well defined for any choices of $\nabla^0,\nabla^1$.

Suppose additionally that $E$ is trivial and we have fixed a trivialization. Then we can choose $\nabla^0$ to be the trivial connection on $E$ and then we set

$$ CS_k(\nabla):= Tc_k(\nabla,\nabla^0). $$

The usual Chern-Simmons theory is a special case of this construction when $k=2$, i.e., $E$ is a trivial complex vector bundle of rank $r\geq 2$ over a $3$-manifold.