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

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.

• 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. – SM Kravec Jun 28 '14 at 19:26
• 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? – Gary B Jun 28 '14 at 20:43
• 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 $(g g^{-1}) - \int (\partial_{\mu} g) g^{-1} d^{3}x$. Then $g g^{-1} =0$ since g is unitary. This allows added freedom in moving the unitary matrices, $g$ and $g^{-1}$, around. – Gary B Jun 28 '14 at 22:02
• Related mathoverflow post: mathoverflow.net/q/31905/13917 – Qmechanic Jun 30 '14 at 13:49