# A simple conjecture on the Chern number of a 2-level Hamiltonian $H(\mathbf{k})$?

For example, let's consider a quadratic fermionic Hamiltonian on a 2D lattice with translation symmetry, and assume that the Fourier transformed Hamiltonian is described by a $2\times2$ Hermitian matrix $H(\mathbf{k})=a(\mathbf{k})\sigma_x+b(\mathbf{k})\sigma_y+c(\mathbf{k})\sigma_z$ and has a finite energy gap, then the Chern number $N$ can be determined.

If $H(-\mathbf{k})=H(\mathbf{k})$ holds for all $\mathbf{k}\in BZ$, then the Chern number $N$ is always an even number, am I right? This seems to be true from the geometrical interpretation of Chern number as a winding number covering a unit sphere, but I have not yet found a rigorous mathematical proof.

Remark: The necessary condition finite energy gap ($\Leftrightarrow$ The map $(a(\mathbf{k}),b(\mathbf{k}),c(\mathbf{k}))/\sqrt{a(\mathbf{k})^2+b(\mathbf{k})^2+c(\mathbf{k})^2}$ from BZ(2D torus) to the unit sphere is well defined) is to ensure that the Chern number/winding number is well defined.

## 2 Answers

I just found a relative rigorous argument supporting my conjecture:

The Chern number $N=\frac{1}{2\pi}\int _{BZ}b(\mathbf{k})$, where $b(\mathbf{k})$ is the Berry curvature. Since $H(-\mathbf{k})=H(\mathbf{k})$, it's easy to show that $b(-\mathbf{k})=b(\mathbf{k})$, accordingly, we can divide the $BZ$ into two halfs called $BZ_1$ and $BZ_2$, therefore, $N=\frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})+\frac{1}{2\pi}\int _{BZ_2}b(\mathbf{k})=2\times \frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})$. Now there are two ways to show that $\frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})$ is an integer,

(1)Due to the relation $b(-\mathbf{k})=b(\mathbf{k})$ and periodic structure of the $BZ$, the half Brillouin zone $BZ_1$ is topologically equivalent to a sphere, and the 'flux' through a sphere(closed surface) $\frac{1}{2\pi}\int _{BZ_1}b(\mathbf{k})$ must be quantized as an integer;

(2)Since $H(-\mathbf{k})=H(\mathbf{k})$, the eigenfunction $\psi(\mathbf{k})$ is also even(i.e. $\psi(-\mathbf{k})=\psi(\mathbf{k})$), then the Berry connection $\mathbf{a(k)} \propto \left \langle \psi(\mathbf{k})\mid \bigtriangledown_{\mathbf{k}} \psi(\mathbf{k})\right \rangle$ is odd, thus, it's easy to show that $\int _{BZ_1}b(\mathbf{k})=\oint _ {\partial BZ_1}\mathbf{a(k)}\cdot d\mathbf{k}=0$(where $\partial BZ_1$ is the boundary of $BZ_1$), however, to be consistent, the number '0'(Berry phase) here should be understood as $2\pi\times integer$.

Remark: The key point in our argument (1) is that the points $\mathbf{k}$ and $-\mathbf{k}$ are equivalent, and hence the half $BZ$ is topologically equivalent to a sphere which is a closed surface.

• See the Supplemental Material in arXiv:1604.04781 – Kai Li Apr 25 '16 at 7:36

I agree with your argument, but I thought I would just rephrase the same logic in a slightly different way, similar to how one would prove it in an algebraic topology course. (I would have done this as a comment, but it's a bit too big for that.)

Basically the Chern number measures the topology of the map $\newcommand{\ra}{\kern-1.5ex\xrightarrow{ \ \ #1\ \ }\phantom{}\kern-1.5ex}T^2 \;\ra{n} \; \;S^2$, where $n(\mathbf k) = (a(\mathbf k),b(\mathbf k),c(\mathbf k))$. More exactly the map $n$ induces a map of the homology groups $\newcommand{\ra}{\kern-1.5ex\xrightarrow{ \ \ #1\ \ }\phantom{}\kern-1.5ex}H_2(T^2) \; \;\ra{n} \; \;H_2(S^2)$ and the Chern number is given by $n ([T^2]) = C_1 *[S^2]$, where $[X]$ is the generator of the group $H_2(X)$.

Now because of the property of $n$, we can write down the following commutative diagram:

$\newcommand{\ra}{\kern-1.5ex\xrightarrow{ \ \ #1\ \ }\phantom{}\kern-1.5ex} \begin{array}{ccccc} T^2 && \ra{\qquad\quad n \qquad\quad} & &S^2 \\ &\searrow_g&&\nearrow_h& \\ && (T^2/\sim) \;\simeq S^2& \end{array}$

where $g$ is the map that identifies the points $\mathbf k$ and $-\mathbf k$ on the torus. As you point out, this quotient space is homotopically equivalent to $S^2$. The above also implies a commutative diagram for the homology groups, such that $n([T^2]) = h \circ g ([T^2])$, but it is clear that $g ([T^2]) = 2 [S^2]$. (One can justify that rigorously using the fact that every point in $S^2$ has two pre-images in $T^2$.) Hence we have proven that $C_1 [S^2] = 2 \; h([S^2])$, i.e. $C_1$ is even.