0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Why do you say that the CS-form is non-topological? It only makes use of connection forms and exterior algebra. $\endgroup$ – NDewolf Oct 28 at 12:55
  • $\begingroup$ The original paper by Chern and Simons states that it is a conformal invariant. Indeed, it shows that even deformations to the choice of metric changes the invariant. $\endgroup$ – Alonso Perez Lona Oct 28 at 13:01
  • $\begingroup$ @NDewolf, how come when we choose G=GLn as in the paper we only have conformal invariance, but we have topological invariance if we phrase the theory in terms of any principal bundle? $\endgroup$ – Alonso Perez Lona Oct 28 at 17:49
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thank you for your answer. I am still confused as to how topological invariance agrees with the answers given in Examples 1 and 2 in section 6 of the Chern and Simons paper. $\endgroup$ – Alonso Perez Lona Oct 29 at 10:12
  • 2
    $\begingroup$ The 2+1 gravitational Chern-Simons action does depend on the metric. Its functional derivative wrt to $g_{\mu\nu}$ is the Cotton tensor which is not zero. In the context of the 1+1 diemsnional energy-momentum anomaly it is the Cotton tensor that provides the energy-momentum inflow to the boundary. The statement that the gravitational CS theory does not depend on the metric is (modulo the framing anomaly) because we have integrated over the metric. $\endgroup$ – mike stone Oct 29 at 13:52
  • $\begingroup$ I was suspecting that this was the case (there aren't a lot of other possibilities). Are there other cases in which CS-theory (or any other related constructions with characteristic classes) is not topologically invariant? $\endgroup$ – NDewolf Oct 29 at 14:12
  • 1
    $\begingroup$ I think they always mean the gauge theory case. Where the action is obviously metric independent. $\endgroup$ – mike stone Oct 29 at 15:36
  • 1
    $\begingroup$ The correct answers in that problem address a different issue. The gravitation CS action contains $\omega$ or $\Gamma$'s that are the Levi-Civita connection induced from from the metric. The action density is therefore metric dependendent. In the gravity theory we then integrate over the metric. The result is metric independent just as $\int_a^b f(x) dx$ does not depend on $x$. The first answer, although approved by a number of readers, seems incorrect to me.. $\endgroup$ – mike stone Oct 29 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.