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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)

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https://arxiv.org/abs/1205.5792 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.

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  • $\begingroup$ When I say total number of orbit, I mean the sum of number of orbits of all sites in a unit cell. Then the 3 layered Haldane model has the appearence of 6 orbits per unit cell. However, indeed if they're able to restore the translation invariance in the stacked direction, then indeed the unit cell is reduced and again possess 2 orbits again. However, the our problem becomes 3D, and there are 3 available momentums in Brillouin Zone in the stacked direction. $\endgroup$
    – An Zhou
    Commented May 20, 2015 at 16:56
  • $\begingroup$ I still think this is kind of artificial. Suppose we have a finite system in the 2D plane. So then our B.Z actually also has 3 layer. It would be strange to say which is the total Chern-number, rather, it only makes sense to say what is the Chern-number for each layer of them. And I guess each one of them does not has C>=3. $\endgroup$
    – An Zhou
    Commented May 20, 2015 at 16:59
  • $\begingroup$ Sorry the 'finite' in the last reply should be infinite. In such stacking trick, stacking layers behave as a internal degree of freedom essentially is adding orbitals.Keeping unit cell having only 2 cites is more or less an meaningless appearance for Chern-number purpose since it is not defined for 3D material. We're forced to view this stacked system as a three band tight-binding and each of them I guess could still have Chern-number 2 at most. $\endgroup$
    – An Zhou
    Commented May 20, 2015 at 17:03
  • $\begingroup$ To make a cleaner discussion, let's add a further restriction: a 2D free fermion model. Are there examples that violate the statement then? $\endgroup$
    – An Zhou
    Commented May 20, 2015 at 17:05
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    $\begingroup$ Forget about how they get their model (through 3-layers, whatever), there is a clear, well-defined question: how many orbitals per unit cell? And the answer is 2. The model is purely 2D. I don't think there is anything ambiguous about these statements. So this is a counterexample to your conjecture. $\endgroup$
    – Meng Cheng
    Commented May 20, 2015 at 17:17

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