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:

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


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

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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – SM Kravec
    Jun 28 '14 at 19:26
  • $\begingroup$ 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? $\endgroup$
    – Gary B
    Jun 28 '14 at 20:43
  • 3
    $\begingroup$ 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. $\endgroup$
    – Gary B
    Jun 28 '14 at 22:02
  • 1
    $\begingroup$ Related mathoverflow post: mathoverflow.net/q/31905/13917 $\endgroup$
    – Qmechanic
    Jun 30 '14 at 13:49

Consider the Chern-Simons Lagrangian density $$ \mathcal{L}_{CS} = \kappa \varepsilon^{\mu\nu\rho} \text{Tr}\left(A_\mu \partial_\nu A_\rho + \frac{2}{3} A_\mu A_\nu A_\rho \right) $$ under the gauge transformation $$ A_\mu \to g^{-1} A_\mu g + g^{-1} \partial_\mu g \,. $$ To derive the gauge transformation let us consider both terms separatly. You need to use the cyclic property of the trace, the anti-symmetry of the Levi-Civita tensor and the product rule $$ \partial_\mu g^{-1}g = \partial_{\mu}(g^{-1}g) - g^{-1}\partial_\mu g = -g^{-1}\partial_\mu g $$ to get the result. I am not going to give the detailed path of the calculation but I can give some steps in between s.t. you can check your calculations for mistakes. For the first term you should obtain $$ \varepsilon^{\mu\nu\rho}\text{Tr}\left(\frac{2}{3}A_\mu A_\nu A_\rho\right) \to \varepsilon^{\mu\nu\rho}\left[\frac{2}{3}\text{Tr}(A_\mu A_\nu A_\rho) + \frac{2}{3}\text{Tr}(g^{-1}\partial_\mu g g^{-1} \partial_\nu g g^{-1} \partial_\rho g) \\ + 2\text{Tr}(A_\mu A_\nu \partial_\rho g g^{-1}) - 2\text{Tr}(A_\mu \partial_\nu g \partial_\rho g^{-1}) \right] $$ For the second one you get $$ \varepsilon^{\mu\nu\rho}\text{Tr}(A_\mu \partial_\nu A_\rho) \to \varepsilon^{\mu\nu\rho} \left[\text{Tr}(A_\mu \partial_\nu A_\rho) - 2\text{Tr}(A_\mu A_\nu \partial_\rho g g^{-1}) + 3\text{Tr}(A_\mu \partial_\nu g \partial_\rho g^{-1}) \\ + \text{Tr}(\partial_\mu A_\nu \partial_\rho g g^{-1}) + \text{Tr}(g^{-1}\partial_\mu g \partial_\nu g \partial_\rho g^{-1}) \right] $$ Combining both results returns (some terms cancel out directly)

\begin{align} \mathcal{L}_{CS} & \to \mathcal{L}_{CS} + \kappa \varepsilon^{\mu\nu\rho} \left[ \text{Tr}(A_\mu \partial_\nu g \partial_\rho g^{-1}) + \text{Tr}(\partial_\mu A_\nu \partial_\rho g g^{-1}) + \text{Tr}(g^{-1} \partial_\mu g \partial_\nu g^{-1} \partial_\rho g) \\ + \frac{2}{3}\text{Tr}(g^{-1} \partial_\mu g g^{-1} \partial_\nu g g^{-1} \partial_\rho g) \right] \\ & = \mathcal{L}_{CS} + \kappa \varepsilon^{\mu\nu\rho} \left[ \text{Tr}((\partial_\mu g^{-1} A_\nu + g^{-1} \partial_\mu A_\nu) \partial_\rho g) + \text{Tr}(g^{-1} \partial_\mu g \partial_\nu g^{-1} \partial_\rho g) \\ - \frac{2}{3} \text{Tr}(g^{-1} \partial_\mu g \partial_\nu g^{-1} \partial_\rho g) \right] \\ & = \mathcal{L}_{CS} + \kappa \varepsilon^{\mu\nu\rho} \left[ \text{Tr}(\partial_\mu( g^{-1} A_\nu) \partial_\rho g) + \frac{1}{3} \text{Tr}(g^{-1} \partial_\mu g \partial_\nu g^{-1} \partial_\rho g) \right] \\ & = \mathcal{L}_{CS} + \kappa \varepsilon^{\mu\nu\rho} \left[ \partial_\mu\text{Tr}(g^{-1} A_\nu \partial_\rho g) - \frac{1}{3} \text{Tr}(g^{-1} \partial_\mu g g^{-1} \partial_\nu g g^{-1} \partial_\rho g) \right] \\ & = \mathcal{L}_{CS} - \kappa \varepsilon^{\mu\nu\rho} \partial_\mu\text{Tr}(\partial_\nu g g^{-1} A_\rho) - \varepsilon^{\mu\nu\rho} \frac{\kappa}{3} \text{Tr}(g^{-1} \partial_\mu g g^{-1} \partial_\nu g g^{-1} \partial_\rho g) \end{align}

So we get the same result as in "Aspects of Chern-Simons Theory", Gerald V. Dunne.


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.