# Is Chern-number for free fermion system always limited by total band number, i.e. number of orbits with a unit cell?

If so, how to see that? Also I think it has been proven that the total Chern-number for free fermion system is 0? If you know how to prove it, please make some comment or hopefully a sketch of proof. I encounter this in a funny way. In fact, I just want to confirm the maximum Chern-number on Honeycomb lattice is 2 which I conjectured from a very peculiar way.

(I know people have already developed method to design bands with arbitrary Chern number. However that does not violate the statement since what they've achieved is to have multilayered system which essentially have N orbits in one unit cell to have a band with Chern-number N)

http://arxiv.org/pdf/1205.5792.pdf The first example in the paper is $C=3$ on a triangular lattice with two orbitals per site. It is essentially three-layers of Haldane's honeycomb lattice model, but stacked together in a clever way so the translation symmetry is restored.
UPDATE: In fact two-band free fermion Hamiltonian on a square lattice can realize higher Chern number. Write $H=\sum_{\mathbf{k}}c_{\mathbf{k}}^\dagger h(\mathbf{k})c_{\mathbf{k}}$, where $h(\mathbf{k})=n(\mathbf{k})\cdot{\sigma}$. $n(\mathbf{k})/|n(\mathbf{k})|$ is a map from $T^2$ to $S^2$ and the Chern number is just the winding number (or the degree of the map). So obviously there is no bound on the winding number. And it is not hard to write down representatives for each winding number. The only problem is that to realize higher Chern number, you need longer and longer range hopping. So the limitation is really how short-range you want the hopping terms to be, not the number of orbitals.