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$$\Omega_{ij}(k)^{\pm} = \frac{v^2cos(\theta)\left(b\pm\frac{vk_z^2}{\sqrt{k_y^2\sin^2(\theta) +k_z^2}}\right)}{4\left(v^2k_x^2+v^2k_y^2cos^2(\theta)+\left( b\pm v\sqrt{k_y^2\sin^2(\theta) +k_z^2}\right)^2)\right)^{\frac{3}{2}}}$$ $$\sigma_{ij}=\frac{e^2}{h2\pi^3}\int d^3k \Omega_{ij}(k)$$ where $\sigma_{ij}$ is the anomalous Hall conductivity. The integration is over the Brillouin zone. The anomalous Hall conductivity turns out to be $$\sigma_{ij} = \frac{e^2b}{v\pi}sgn(cos(\theta))$$ How to solve the above integration? This is the link to the paper in which the above calculations are done. I have not been able to solve this particular integral.

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I have not tried to do the integral in this case. But usually these topological things boil down to writing the hamiltonian in the form $d_x\sigma_x+d_y\sigma_y+d_z\sigma_z$ and then computing (by simply looking at the formula) the winding number of the map $\hat {\bf d}:{\rm Brillouin{\phantom i}zone} \to S^2$ where $\hat {\bf d}$ is the unit vector in the direction ${\bf d}$. There should be no need to actually do the integration.

In you case you have both $\sigma$'s and $\tau$'s, and a three-d integral so the connection between Chern number and winding number is a bit more complicated, but it should still exist.

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