# Source Berry Curvature Chern Insulator

Why is there non-zero hall conductance for a Chern insulator?

From section 2.3 of Bernevig's book 'Topological insulators and topological superconductors' I learned one can view degeneracies are sources/sinks of berry curvature, i.e. monopoles in parameter space.

In section 3.6 we learn that the non zero Chern number is due to an obstruction in Stokes theorem, one cannot 'gauge away' a degeneracy. Note: the obstruction is a point in k-space here, it is a part of our integration domain.

If we now consider the Hamiltonian $$H=\sin(k_x)\sigma_x + \sin(k_y)\sigma_y + B(2 + M - \cos(k_x) - \cos(k_y))\sigma_z$$ we find degeneracies at $$(M,k_x,k_y)=(0,0,0)$$, $$(M,k_x,k_y)=(-2,\pi,0)$$, $$(M,k_x,k_y)=(-2,0,\pi)$$ and $$(M,k_x,k_y)=(-4,\pi,\pi)$$. One can show using

$$$$\label{curvature formula} \Omega_{ij}^s(\mathbf{k})=\frac{1}{2}\epsilon_{abc}\hat{d}_a\partial_i \hat{d}_b \partial_j \hat{d}_c,$$$$

that we get no hall conductance for $$M>0$$ and $$M<-4$$, but we do get Hall conductance for the other values.

How do I reconcile this with the idea's of 2.3 and 3.6? The monopoles are only there for $$M=0$$, $$M=-2$$ and $$M=-4$$!

Chern insulator is a 2d system you have to think this as a slice of a 3d system for example consider the following hamiltonian $$H(k_x,k_y,k_z)=\sin(k_x)\sigma_x + \sin(k_y)\sigma_y + B(2 - \cos(k_x) - \cos(k_y)-\cos(k_z)+\cos(k_0))\sigma_z$$
which will have a berry curvature sorce at $$k=(0,0,\pm k_0)$$
$$H(k_x,k_y,k_z=k'_z)$$ where $$k'_z$$ is a fixed number and $$k_x,k_y$$ are variables. thus this hamiltonian will be an effectively 2d hamiltonian with $$M=-\cos(k'_z)+\cos(k_0)$$