= Chern number of the Haldane model
{c}
{title2=$C=\tfrac12[\operatorname{sgn}m_+-\operatorname{sgn}m_-]$}
For $\mathbf d\ne0$, put $\widehat{\mathbf d}=\mathbf d/|\mathbf d|$. The lower-band <orthogonal projector> is $P_-=(I-\widehat{\mathbf d}\cdot\boldsymbol\sigma)/2$. Applying the <Pauli matrix multiplication law> to $i\operatorname{Tr}(P_-[\partial_{k_x}P_-,\partial_{k_y}P_-])$ gives
$$
\Omega_-=\frac12\widehat{\mathbf d}\cdot(\partial_{k_x}\widehat{\mathbf d}\times\partial_{k_y}\widehat{\mathbf d})
=\frac{\mathbf d\cdot(\partial_{k_x}\mathbf d\times\partial_{k_y}\mathbf d)}{2|\mathbf d|^3}.
$$
Thus the <Chern number> is the signed number of times the <Brillouin zone> covers a <two-sphere> under $\widehat{\mathbf d}$. The lower <energy eigenstate>'s actual <Bloch vector> is $-\widehat{\mathbf d}$. The <Berry curvature> has the displayed positive sign for this lower band and the stated <Berry connection>; switching to the upper band changes its sign.
A useful local calculation exposes what each valley contributes. Write $p=\hbar v|\mathbf q|$. The linear <Dirac Hamiltonian> gives
$$
\Omega_{-,\tau}(\mathbf q)=\frac{\tau m_\tau(\hbar v)^2}{2[(\hbar v|\mathbf q|)^2+m_\tau^2]^{3/2}},
\qquad
\frac1{2\pi}\int_{\mathbb R^2}\Omega_{-,\tau}\,d^2q
=\frac\tau2\int_0^\infty\frac{m_\tau p\,dp}{(p^2+m_\tau^2)^{3/2}}
=\frac\tau2\operatorname{sgn}m_\tau.
$$
Each unbounded continuum cone gives a half-integer in this calculation. It is not a separately realizable half-integer <Chern number> of an isolated lattice <energy band>. The full <honeycomb lattice> supplies both valleys and the large-momentum completion. Starting with a trivial large-$|M|$ state, each single-valley mass crossing changes the lattice integer by the corresponding unit. There are no other band touchings. This establishes the exact lattice result
$$
C=\frac12\left[\operatorname{sgn}(M-\mu)-\operatorname{sgn}(M+\mu)\right]
=\begin{cases}-\operatorname{sgn}(\sin\phi),&|M|<3\sqrt3t_2|\sin\phi|,\\0,&|M|>3\sqrt3t_2|\sin\phi|,\end{cases}
$$
for $t_2>0$, excluding the boundaries. This sign belongs to our explicit bond vectors, orientation and occupied-band convention.
For an original worked example, choose $t_1=1$, $t_2=0.08$, $\phi=\pi/2$ and $M=0$. Then $\mu\simeq0.415692$, $(m_+,m_-)=(-\mu,+\mu)$, both valley gaps are $2\mu\simeq0.831384$, and $C=-1$. Choosing instead $t_2=0$, $M=\mu$ gives exactly the same valley-gap magnitudes but $C=0$. The first state's two valley contributions reinforce; the second state's cancel.
\Image[/haldane-valley-berry-curvature.png]
{title=Equal valley gaps can conceal different Berry-curvature topology}
{description=Original calculation of the lower-band Berry curvature of the two explicitly defined lattice Hamiltonians, using the same colour scale. One reciprocal primitive cell is shown in fractional coordinates, with $mathbf g_icdotmathbf a_j=2\pi\delta_{ij}$, $mathbf a_1=\boldsymbol\delta_3-\boldsymbol\delta_1$, and $mathbf a_2=\boldsymbol\delta_2-\boldsymbol\delta_1$. The bond length is $a=1$. The coloured surfaces are model results, not measured curvature or a reconstruction of experimental data.}
{source=haldane-valley-berry-curvature.py}
An independent eigenvector-overlap calculation on periodic $31\times31$, $61\times61$ and $121\times121$ momentum grids gives $C=-1$ for the first example and $C=0$ for the second. Reversing $\phi$ in the first example gives $C=+1$. Direct integration of the analytic lattice <Berry curvature> agrees. For the overlap calculation one can remove the embedding phase from $f$, replacing it by $1+e^{i\mathbf k\cdot\mathbf a_1}+e^{i\mathbf k\cdot\mathbf a_2}$; this makes the cell-basis <Hamiltonian> periodic and prevents a false boundary-seam contribution. The local <Berry curvature> can depend on this basis convention; the integrated <Chern number> agrees. All mass-zero parameters are excluded before evaluating any integer.
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