Does delta-function always enter the real part of conductivity? A well-known general form of electrical conductivity for a free system is $$\sigma_{\mu\nu}(\omega)=\frac{i\hbar}{V}\sum_{mn} \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n} \frac{\langle n \lvert j_{\mu}\rvert m\rangle  \langle m \lvert j_{\nu}\rvert n\rangle}{\hbar(\omega +i\eta)+\varepsilon_n-\varepsilon_m},$$
in which the infinitesimally small positive $\eta$ will lead to $\delta(\omega+\varepsilon_n-\varepsilon_m)$ in the real part $\mathrm{Re}\,\sigma$ (note the $\mathrm{i\hbar}/V$ factor).
In this paper, Eqs. (7,8) are just the $\mathrm{Re}\,\sigma_H$ and $\mathrm{Im}\,\sigma_H$ of a BdG system calculated from equivalent Kubo formula of the Hall conductivity $\sigma_H=(\sigma_{xy}-\sigma_{yx})/2$. But the $\eta$-induced $\delta$-functions obviously only enter the imaginary part $\mathrm{Im}\,\sigma_H$ instead. In the more detailed version of that paper, Eq. (10) of $\sigma_H(\omega)$ shows that it follows from terms like (other factors are real) $$\frac{1-n_F(\varepsilon_1)-n_F(\varepsilon_2)}{(\varepsilon_1+\varepsilon_2)[(\varepsilon_1+\varepsilon_2)^2-(\omega+i\eta)^2]}.$$
But it is nothing but the Kubo formula of conductivity applied to a specific case. It looks hard to imagine the subtraction $\sigma_{xy}-\sigma_{yx}$ can make such a huge difference. How to understand this apparent discrepancy?
 A: 
A well-known general form of electrical conductivity for a free system is $$\sigma_{\mu\nu}(\omega)=\frac{i\hbar}{V}\sum_{mn} \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n} \frac{\langle n \lvert j_{\mu}\rvert m\rangle  \langle m \lvert j_{\nu}\rvert n\rangle}{\hbar(\omega +i\eta)+\varepsilon_n-\varepsilon_m},$$

I will use some abbreviations and conventions to re-write this expression as:
$$
\sigma_{\mu\nu}(\omega)=
\frac{i}{V}
\sum_{mn} \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n} \frac{ j_{\mu}^{nm}j_{\nu}^{mn}}{\omega +\varepsilon_n-\varepsilon_m+i\eta}\;,
$$
where I use superscript $nm$ to avoid writing out bras and kets, and I set $\hbar=1$.
To make the notation even easier I will also define:
$$
A_{nm} = \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n}\;,
$$
which is symmetric and which I assume to be real (and I'll drop the indices).
And define:
$$
\alpha_{nm} = \omega + \varepsilon_n - \varepsilon_m
$$
(and I'll drop the indices).
So, with this further abbreviation:
$$
\sigma_{\mu\nu}(\omega)=
\frac{i}{V}
\sum_{mn} A \frac{ j_{\mu}^{nm}j_{\nu}^{mn}}{\alpha+i\eta}\;.
$$
--
Or, re-writing a little more:
$$
\sigma_{\mu\nu}=
\frac{i}{V}
\sum_{mn} A \frac{ j_{\mu}^{nm}j_{\nu}^{mn}(\alpha - i\eta)}{\alpha^2+\eta^2}
$$
We also have:
$$
\sigma_{\nu\mu}=
\frac{i}{V}
\sum_{mn} A \frac{ j_{\nu}^{nm}j_{\mu}^{mn}(\alpha - i\eta)}{\alpha^2+\eta^2}
$$
and
$$
\sigma_{\mu\nu}^*=
\frac{-i}{V}
\sum_{mn} A \frac{ j_{\mu}^{mn}j_{\nu}^{nm}(\alpha + i\eta)}{\alpha^2+\eta^2}
$$
and
$$
\sigma_{\nu\mu}^*=
\frac{-i}{V}
\sum_{mn} A \frac{ j_{\nu}^{mn}j_{\mu}^{nm}(\alpha + i\eta)}{\alpha^2+\eta^2}
$$
and
$$
\sigma_{\mu\nu} - \sigma_{\nu\mu}
=
\frac{i}{V}
\sum_{mn} A \frac{(\alpha - i\eta)}{\alpha^2+\eta^2} 
\left(j_{\mu}^{nm}j_{\nu}^{mn}
-
j_{\nu}^{nm}j_{\mu}^{mn}\right)
$$
and
$$
(\sigma_{\mu\nu} - \sigma_{\nu\mu})^*
=
\frac{-i}{V}
\sum_{mn} A \frac{(\alpha + i\eta)}{\alpha^2+\eta^2} 
\left(j_{\mu}^{mn}j_{\nu}^{nm}
-
j_{\nu}^{mn}j_{\mu}^{nm}\right)
$$
and finally:
$$
iIm(\sigma_{\mu\nu} - \sigma_{\nu\mu})
=
\frac{1}{V}
\sum_{mn}  \frac{A\eta}{\alpha^2+\eta^2} 
\left(j_{\mu}^{nm}j_{\nu}^{mn}
-
j_{\nu}^{nm}j_{\mu}^{mn}\right)\;.
$$
Restoring all the abbreviations and letting $\eta \to 0$ gives:
$$
Im(\sigma_{\mu\nu} - \sigma_{\nu\mu})
=
\frac{\hbar}{iV}
\sum_{mn}  \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n}\delta(\hbar\omega + \varepsilon_n - \varepsilon_m)
\left(\langle n \lvert j_{\mu}\rvert m\rangle \langle m \lvert j_{\nu}\rvert n\rangle
-
\langle m \lvert j_{\mu}\rvert n\rangle
\langle n \lvert j_{\nu}\rvert m\rangle
\right)\;.
$$
$$
=
\frac{2\hbar}{V}
\sum_{mn}  \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n}\delta(\hbar\omega + \varepsilon_n - \varepsilon_m)
{Im}\left(\langle n \lvert j_{\mu}\rvert m\rangle \langle m \lvert j_{\nu}\rvert n\rangle
\right)\;.
$$
