# Gauge Invariance of the Non-abelian Chern-Simons Term

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I'm trying to prove that, under the gauge transformation $$A_{\mu} \rightarrow A_{\mu}^{\prime} = g^{-1} A_{\mu} g + g^{-1} \partial_{\mu} g,$$ the non-abelian Chern-Simons Lagrangian density:

$$\mathcal{L}_{CS} = \kappa \epsilon^{\mu \nu \rho} tr \left( A_{\mu} \partial_{\nu}A_{\rho} + \dfrac{2}{3} A_{\mu}A_{\nu}A_{\rho} \right)$$

becomes:

$$\mathcal{L}_{CS} ~\longrightarrow~ \mathcal{L}_{CS} - \kappa \epsilon^{\mu \nu \rho} \partial_{\mu} tr \left( \partial_{\nu} g g^{-1} A_{\rho} \right) - \dfrac{\kappa}{3} \epsilon^{\mu \nu \rho} tr \left( g^{-1} \partial_{\mu} g g^{-1} \partial_{\nu} g g^{-1} \partial_{\rho} g \right)$$

as stated in Gerald V. Dunne's lecture notes 'Aspects of Chern-Simons Theory' pages 15-16.

The second term in the last equation can be disregarded as it's a total derivative and the third term can be shown to be some integer multiple of $2\pi$ provided $\kappa$ is an integer.

Now I understand that gauge invariance of the CS-term can be proven using some clever reasoning (see: Gauge invariant Chern-Simons Lagrangian). However I want to show how we can arrive at the second equation above using the `brute force' method of plugging in the gauge transformed vector field into the Lagrangian. Unfortunately I get stuck with a large number of terms that I'm not sure how to combine or cancel.

Does anyone know of a source that goes through the above calculation in more detail, or does anyone have any tips for how to proceed. I've done a rather extensive search and can't find any sources that show some of the steps. I already tried using the cyclic properties of the trace and the cancelation of any symmetric term with the anti-symmetric $\epsilon^{\mu \nu \rho}$.

Thank you in advance for any suggestions.

This post imported from StackExchange Physics at 2014-06-29 09:37 (UCT), posted by SE-user Gary B
retagged Jun 29, 2014

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I'd switch to a notation which uses differential forms. Work out the abelian case first where $A \rightarrow A + d \phi$ and you can just ignore the $A^3$ term. You're going to need to integrate by parts.This post imported from StackExchange Physics at 2014-06-29 09:37 (UCT), posted by SE-user SM Kravec
answered Jun 28, 2014 by (60 points)
Thanks for the feedback. So would you say it's best to practice in the abelian case first? How would integrating by parts play a role - is it simply the case that some terms would disappear under integration over space?

This post imported from StackExchange Physics at 2014-06-29 09:37 (UCT), posted by SE-user Gary B

Just an update to say that a friend explained to me how to use the integration by parts. For the benefit of others who view this page: When performing the gauge transformation we get terms of the form: $g (\partial_{\mu} g^{-1})$ (for example) which are integrated over spacetime. Using integration by parts: $\int g (\partial_{\mu} g^{-1}) d^{3}x$ can be expressed as $- \int (\partial_{\mu} g) g^{-1} d^{3}x$ since $g g^{-1}$ is constant. This allows added freedom in moving the unitary matrices, $g$ and $g^{-1}$, around.

This post imported from StackExchange Physics at 2014-06-29 09:37 (UCT), posted by SE-user Gary B

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