Why is this Chern-Simons theory gauge invariant?

Question

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}

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.

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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 $\epsilon^{\mu \nu \rho} \partial_\nu \partial_\rho \lambda=0$.

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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

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