The spin resolved Chern number is the integer prefactor C of $$\sigma_{\text{Hall}}$$ of the Haldane model with 4 orbitals and fluxes. Kubo gives C=2, that is even parity. It corresponds to the 2 edge states per spin that propagate in the **same **direction. 2 edge states per spin that propagate in the** opposite** direction would become gapful and thus a trivial insulator. Does the argument -even parity of spin-resolved Chern number = trivial insulator, odd parity = non-trivial- hold only for the latter situation?

Kramers degenerate eigenstates, defined via $$[H(\textbf{k}),T]=0$$ do exist at the 4 TRIM $$\textbf{k}=\Gamma_{i}$$.

The parity operator does not commute anymore, $$[H(\textbf{k}),P]\neq0$$ for $$\textbf{k}=\Gamma_{i}$$ (nor for any other k).

Is it obvious from the unit cell that inversion symmetry is violated with fluxes?

So the Fu-Kane formula for the invariant can not be used!

I computed the invariant via the Pfaffians at $$\textbf{k}=\Gamma_{i}$$. I checked it using a 3rd neighbour hopping that induces a trivial insulator in the plain KM model. This gives $$\nu=1$$ (non-trivial) for the flux model. (But there is an inconsistency in my calculation, so I do not trust it fully at this moment)