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:
\begin{equation}
2a_0(\partial_1a_2-\partial_2a_1)
\end{equation}
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:

*

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


*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.
 A: *

*Yes.


*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.
