# Chern-Simons term in Coulomb or radiation gauge

In some of the literature (for example, below Eq. (A3) of this paper), the following is claimed to be the Chern-Simons term in the Coulomb gauge: $$$$2a_0(\partial_1a_2-\partial_2a_1)$$$$ and it cited this paper, which below its Eq. (5) states that the above term is the Chern-Simons term in radiation gauge.

My questions are:

1. Is the radiation gauge the same as the Coulomb gauge, where $$\partial_1a_1+\partial_2a_2=0$$.

2. Why is the above term the Chern-Simons in the Coulomb gauge? The standard Chern-Simons term is $$\epsilon_{\mu\nu\lambda}a_\mu\partial_\nu a_\lambda=a_0(\partial_1a_2-\partial_2a_1)+a_1(\partial_2a_0-\partial_0a_2)+a_2(\partial_0a_1-\partial_1a_0)$$ After integrating by parts it becomes $$\epsilon_{\mu\nu\lambda}a_\mu\partial_\nu a_\lambda\rightarrow 2a_0(\partial_1a_2-\partial_2a_1)+2a_2\partial_0a_1$$ which still differs from $$2a_0(\partial_1a_2-\partial_2a_1)$$ by the last term that does not seem to vanish in the Coulomb gauge.

2. The question now is to show $$\int d^2x\epsilon^{\mu\nu\sigma}a_{\mu} \partial_{\mu}a_{\sigma}=\int d^2x [2 a_0(\partial_1 a_2-\partial_2 a_1)+(a_2\partial_0 a_1-a_1\partial_0 a_2)] \\=\int d^2x 2 a_0(\partial_1 a_2-\partial_2 a_1) \tag{1}.$$ That is, $$\int d^2x (a_2\partial_0 a_1-a_1\partial_0 a_2)=\int d^2x \vec{a}\times \partial_0 \vec{a}=0 \tag{2}.$$ The Coulomb gauge condition ($$\nabla \cdot \vec{a}=0$$) in momentum space is $$\vec{k}\cdot\vec{a}(\vec{k})=0$$. Differentiating with respect to time and $$k\rightarrow -k$$ gives $$\vec{k}\cdot \partial_0 \vec{a}(-\vec{k})=0.$$. These two equations imply that $$\vec{a}(\vec{k})$$ and $$\partial_0 \vec{a}(-\vec{k})$$ are both perpendicular to $$\vec{k}$$, which in turn imply that they are parallel to each other: $$\vec{a}(\vec{k}) \times \partial_0 \vec{a}(\vec{-k})=0 \tag{3}.$$ Now, representing (2) in momentum space gives $$\int d^2x \vec{a}\times \partial_0 \vec{a}=\int \frac{d^2k}{(2\pi)^2}\vec{a}(\vec{k})\times \partial_0 \vec{a}(-\vec k)=0,\tag{4}$$ and the claim is proved.