How to understand Chern-Simons action 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!
 A: 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.
A: 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.
