subtracting the divergence, we arrive at $$G^R = 2 \Omega - \begin{pmatrix} \frac 1 2 \left(\omega^2-k^2\right) & 0 \\ 0 & \frac 1 2\omega^2\end{pmatrix}\;,$$ When considering modes of definite frequency, the additional electric fields $E_T$ and $E_L$ are $i \omega a_{T(0)}$ and $i \omega a_{L(0)}$, respectively. Therefore the conductivity is given by $$\sigma = \frac{1}{i\omega}G^R = \frac{2}{i\omega} \Omega + \begin{pmatrix} \frac{i}{2\omega}\left(\omega^2-k^2\right) & 0 \\ 0 & \frac{i \omega}2\end{pmatrix}\;.$$

Let us evaluate the conductivity $\sigma$ numerically for the Chern-Simons coupling $\alpha=0.59$. For this value of $\alpha$, the minimum energy occurs at the momentum $k=-2.37 r_+$. We denote the components of the retarded Green’s function and the conductivity respectively as $$G^R=\begin{pmatrix} G_{TT} & G_{TL}\\ G_{LT} & G_{LL} \end{pmatrix}\;,$$ and $$\sigma=\begin{pmatrix} \sigma_{TT} & \sigma_{TL}\\ \sigma_{LT} & \sigma_{LL} \end{pmatrix}\;.$$

This numerical result has several interesting features. The imaginary part of the retarded Green’s function $G_{TT}$ for a pair of $a_T$ fields behaves like $1/\omega$ near $\omega=0$, as shown in Figure \[fig:G11\]. The pole at $\omega=0$ is directly related to the fact that $a_T$ is a Goldstone mode. This can be seen as follows. As $\omega$ goes to 0, the solution becomes static and we can ignore the terms with time derivatives in . In 