I am reading the lecture notes in https://arxiv.org/abs/hep-th/9902115 and in it, it says that the Lagrangian

$$\mathcal{L}_{\mathrm{CS}}=\frac{\kappa}{2} \epsilon^{\mu \nu \rho} A_{\mu} \partial_{\nu} A_{\rho}-A_{\mu} J^{\mu}$$

is invariant under a gauge transformation because the transformation $A_\mu \to A_\mu + \partial_\mu\lambda$ gives a surface term

$$\delta \mathcal{L}_{\mathrm{CS}}=\frac{\kappa}{2} \partial_{\mu}\left(\lambda \epsilon^{\mu \nu \rho} \partial_{\nu} A_{\rho}\right).$$

For the life of me I can't seem to arrive at that expression. Could somebody please tell me how to arrive at this?? Here's what I’ve tried

\begin{aligned} \delta L &=\frac{\partial L}{\partial A} \delta A+\frac{\partial L}{\partial(\partial A)} \delta(\partial A) \\ &=\frac{\kappa}{2} \epsilon^{\mu \nu \rho}\left[\partial_{\nu}\left(A_{\rho}+\partial_{\rho} \lambda\right)\delta A_\mu + \left(A_{\mu}+\partial_{\mu} \lambda\right) \delta(\partial_\nu A)-J^{\mu} \delta A_\mu \right] \end{aligned}

  • $\begingroup$ Hint: In your last equation, you need to evaluate $\partial L/\partial A$ etc. for the untransformed fields (i.e. without $\lambda$) $\endgroup$
    – Toffomat
    Jun 15, 2020 at 16:45

1 Answer 1


Variation of $\mathcal{L}_{\text{CS}}$ leads to

$$\delta \mathcal{L}_{\text{CS}}= \frac{\kappa}{2} \epsilon^{\mu \nu \rho} \partial_{\mu}\lambda \partial_{\nu} A_{\rho} - \partial_\mu \lambda J^{\mu}. \tag{1}$$ We can express this as

$$\delta \mathcal{L}_{\text{CS}}=\frac{\kappa}{2} \partial_\mu \left( \lambda \epsilon^{\mu \nu \rho} \partial_\nu A_{\rho}\right) + \lambda \partial_\mu J^{\mu},$$

by integrating by parts in the action and noticing that $\epsilon^{\mu \nu \rho} \partial_\mu \partial_{\nu} A_{\rho}=0$. Using $\partial_\mu J^{\mu}$=0, the result follows.

EDIT: \begin{align} \mathcal{L}'_{\text{CS}} &= \frac{\kappa}{2} \epsilon^{\mu \nu \rho} A_{\mu}' \partial_{\nu} A_{\rho}' - A_{\mu}' J^{\mu}\\ &= \frac{\kappa}{2} \epsilon^{\mu \nu \rho}(A_{\mu} + \partial_{\mu}\lambda) \partial_\nu (A_{\rho}+\partial_{\rho} \lambda) - (A_{\mu} + \partial_{\mu} \lambda) J^{\mu}\\ &=\frac{\kappa}{2}\epsilon^{\mu \nu \rho} A_{\mu} \partial_{\nu} A_{\rho} - A_{\mu} J^{\mu}+\frac{\kappa}{2}\epsilon^{\mu \nu \rho}A_{\mu} \partial_\nu \partial_{\rho} \lambda+\frac{\kappa}{2} \epsilon^{\mu \nu \rho} \partial_{\mu}\lambda \partial_{\nu} A_{\rho} - (\partial_\mu \lambda) J^{\mu}\\ &= \mathcal{L}_{\text{CS}} +\frac{\kappa}{2}\epsilon^{\mu \nu \rho}A_{\mu} \partial_\nu \partial_{\rho} \lambda+\frac{\kappa}{2} \epsilon^{\mu \nu \rho} \partial_{\mu}\lambda \partial_{\nu} A_{\rho} - (\partial_\mu \lambda) J^{\mu}. \end{align} Therefore the variation is $$\delta \mathcal{L}_{\text{CS}}=\frac{\kappa}{2}\epsilon^{\mu \nu \rho}A_{\mu} \partial_\nu \partial_{\rho} \lambda+\frac{\kappa}{2} \epsilon^{\mu \nu \rho} \partial_{\mu}\lambda \partial_{\nu} A_{\rho} - (\partial_\mu \lambda) J^{\mu},$$ where the first term vanishes, since $\epsilon^{\mu \nu \rho} \partial_\nu \partial_\rho \lambda=0$.

Let's try to arrive at the same result using your approach.

$$\delta \mathcal{L}_{\text{CS}}=\frac{\partial \mathcal{L}}{\partial A_{\alpha}} \delta A_{\alpha}+ \frac {\partial \mathcal{L}}{\partial (\partial_{\beta} A_{\alpha})} \delta (\partial_{\beta} A_{\alpha}).$$

Here the two variations are

\begin{align} \delta (\partial_{\beta} A_{\alpha}) =\partial_{\beta} \partial_\alpha \lambda \text{ and } \delta A_{\alpha} = \partial_{\alpha} \lambda \end{align} Using the above expressions, we can compute the individual terms of the variation.

$$\frac{\partial \mathcal{L}}{\partial A_{\alpha}} \delta A_{\alpha}=\frac{\kappa}{2} \epsilon^{\alpha \nu \rho} (\partial_{\alpha} \lambda) \partial_\nu A_{\rho} + J^{\alpha}\partial_{\alpha} \lambda \\\ \frac {\partial \mathcal{L}}{\partial (\partial_{\beta} A_{\alpha})} \delta (\partial_{\beta} A_{\alpha})=0,$$ since, again, $\epsilon^{\mu \nu \rho} \partial_\nu \partial_\rho \lambda=0$. Finally, we arrive at (1) again

$$\delta \mathcal{L}_{\text{CS}}=\frac{\kappa}{2} \epsilon^{\mu \nu \rho} \partial_{\mu}\lambda \partial_{\nu} A_{\rho} - (\partial_\mu \lambda) J^{\mu}.$$

  • $\begingroup$ I’m sorry if this is a stupid question but I do not understand how you arrived at the first equation? $\endgroup$
    – Y2H
    Jun 15, 2020 at 17:23
  • 1
    $\begingroup$ @Y2H Edited answer to answer your comment. I hope this helps. $\endgroup$
    – Stratiev
    Jun 15, 2020 at 19:53
  • $\begingroup$ thank you so much!! $\endgroup$
    – Y2H
    Jun 16, 2020 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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