I was reading the paper (https://journals.aps.org/prb/abstract/10.1103/PhysRevB.31.3372) in which Thouless and Wu show that the hall conductivity is a topological invariant. My question is about the Appendix where the authors state that the Hall conductivity for a non-interacting system can be given by
$$ \sigma = \frac{ie^2}{A \hbar} \oint_{\mathcal{C}} \frac{dz}{2\pi i} Tr\left( g \frac{\partial H}{\partial k_x}g \frac{\partial H}{\partial k_y}g\right) $$
where the contour $\mathcal{C}$ surrounds the filled single-particle energy states. I cannot go from this formula to the more well known one given by the equation (A1). I have tried evaluating the trace and inserting a resolution of identity which gave me
$$ \sigma = \frac{ie^2}{A \hbar} \oint_{\mathcal{C}} \sum_{n,m} \frac{dz}{2\pi i} \langle n\vert g \frac{\partial H}{\partial k_x}g \vert m\rangle \langle m \vert \frac{\partial H}{\partial k_y}g \vert n \rangle $$
Furthermore, assuming $g(z) = (z - \hat{H})^{-1}$, I am left with
$$ \sigma = \frac{ie^2}{A \hbar} \oint_{\mathcal{C}} \sum_{n,m} \frac{dz}{2\pi i} \frac{\langle n\vert \frac{\partial H}{\partial k_x} \vert m\rangle \langle m \vert \frac{\partial H}{\partial k_y}\vert n \rangle }{(z - E_n)^2 (z-E_m)} $$
Since only the $(z - E_n)$ contribute to the integral, I have split the $n$ sum into two parts: $E_n < E_F$ and $E_n > E_F$
$$ \sigma = \frac{ie^2}{A \hbar} \oint_{\mathcal{C}} \sum_{E_n < E_F,m} \frac{dz}{2\pi i} \frac{\langle n\vert \frac{\partial H}{\partial k_x} \vert m\rangle \langle m \vert \frac{\partial H}{\partial k_y}\vert n \rangle }{(z - E_n)^2 (z-E_m)} + \frac{ie^2}{A \hbar} \oint_{\mathcal{C}} \sum_{E_n > E_F,m} \frac{dz}{2\pi i} \frac{\langle n\vert \frac{\partial H}{\partial k_x} \vert m\rangle \langle m \vert \frac{\partial H}{\partial k_y}\vert n \rangle }{(z - E_n)^2 (z-E_m)}$$
For the first integral, the contour integration is easy to do: you just replace z with $E_m$. For the second integral, we can take the contour to be anti-clockwise while gaining an additional minus sign. This trick allows us to take the energy levels that are not filled and therefore we can still replace $z = E_m$. However, now I am stuck as I don't know how to argue that $E_m < E_F, E_n < E_F$ and $(E_m > E_F, E_n > E_F) $ parts of the sum vanish. If I can make them vanish, I believe I arrive at the correct answer. Any help would be greatly appreciated.
