Chern-Simons action as a topological invariant It is stated that the Chern-Simons action is a topological invariant that is proportional to the Chern-Simons form. But the latter is just a conformal invariant. How do we reconcile these views? Both require integration, so I don't see how the action can be metric-independent but not the CS invariant.
 A: I'll put my thoughts on this question as an answer to reduce the length of the comments. However, this answer should not(!!) be interpreted as a formal answer, but merely as a lengthy comment (there are undoubtedly more qualified people on this site to give a full answer).
First of all, in the case of Chern-Simons theory with a gauge fields in any other (Lie) group then the O$(p,q)$ structure group [1] of the (orthonormal) frame bundle, it should be immediately clear that the Chern-Simons invariant (so the form itself may transform nontrivially, only the integral should be invariant) is indeed a topological invariant since the only involved quantities are:


*

*Differential forms (with values in a Lie algebra not related to the metric)

*Exterior algebra

*Integration over manifolds (which is perfectly well-defined without reference to a metric)


Edit (thanks to @mikestone) :
The gravitational Chern-Simons theory is not topological as there exists a metric dependency. For example in 3D the variation (under a change of metric) of the Chern-Simons action gives the Cotton tensor.[2] An invariant quantity is only obtained after integrating over the moduli space of metrics.
The information on the following stackexchange page could also be useful: Is gravitational Chern-Simons action "topological" or not?
[1] I used a general Lorentzian signature just for completeness.
[2]Paper by Jackiw and collaborator: https://arxiv.org/abs/gr-qc/0308071
