I am trying to prove that the Chern-Simons term is invariant under an infinitesimal transformation up to first order. I think I have almost solved it but I am completely and utterly stuck now.
I have that the potential transforms as: $A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\alpha+i[A_{\mu},\alpha] $ and the Lagrangina is
$$\mathcal{L}=k\int d^3x\epsilon^{\mu\nu\lambda}Tr\Big(A_{\mu}\partial_{\nu}A_{\lambda}+\frac{2i}{3}A_{\mu}A_{\nu}A_{\lambda} \Big)= $$ where $\epsilon^{\mu\nu\lambda}$ is the Levi-Civita symbol. Now I can rewrite the Lagrangian before I transform,
$$\mathcal{L}= k\int d^3x\epsilon^{\mu\nu\lambda}Tr\Big(A_{\mu}(\partial_{\nu}+\frac{2i}{3}A_{\nu})A_{\lambda}\Big)$$
Then when I transform I get $$\mathcal{L}=k\int d^3x\epsilon^{\mu\nu\lambda}Tr\Big(\delta(A_{\mu})(\partial_{\nu}+2iA_{\nu})A_{\lambda}+A_{\mu}\partial_{\nu}\delta(A_{\lambda})\Big) $$ where I have permuted the index with the help of Levi-cita and the trace so that I can transform the same factor of the potential and I can add them so I lose the factor of 1/3.
Before the I insert the actual transformation I know that the commutator $[A_{\mu},\epsilon]$ is zero as I can use the cyclic properties of the trace. And I can use partial integration to move some derivatives so that I have two on the same term and they cancel. As it is symmetric and multiplied with the Levi-Cevita symbols which is anti-symmetric is zero.
After I still get $$\mathcal{L}=k\int d^3x\epsilon^{\mu\nu\lambda}Tr\Big(2A_{\mu}\partial_{\nu}A_{\lambda}+2iA_{\mu}A_{\nu}A_{\lambda}+2i\partial_{\mu}\alpha A_{\nu}A_{\lambda}\Big) $$ I don't know how to simplify it further.
EDIT: I figured out my mistake. If I only took the variation terms, i.e $\partial_{\mu}\alpha$ and inserted the indices, then used the cyclic property of the trace. All terms vanished.