$z$ axis. Since $|J_z\rangle \rightarrow e^{i \pi J_z/2} |J_z\rangle$ under such a rotation, the operators $\tau^\pm$ carry a net angular momentum of $\pm 2$, and therefore must couple to the “d-wave” combinations of $p_x$ and $p_y$, which are precisely given by $d_3(p)$ and $d_4(p)$.

Though the quadratic dispersion normal to the field is due to symmetry, the touching itself has a topological character. To see it, it is convenient to define the reduced Hamiltonian in the form $$\label{eq:4}
  \mathcal{H}_{2}^\pm = \vec{b}_\pm(\vec{p}) \cdot \vec\tau,$$ with $$\label{eq:5}
  \vec{b}_\pm(\vec{p}) = \left( \begin{array}{c} -\frac{\sqrt{3}}{2}(p_x^2-p_y^2) \\
    \sqrt{3} p_x p_y \\ \mp v p_z \end{array}\right) .$$ From this, one can define the $U(1)$ Berry flux, which is analogous to a magnetic field in momentum space, $$\label{eq:6}
  \mathcal{B}_\pm^\mu = \frac{1}{8\pi}\epsilon^{\mu\nu\lambda} \hat{b}_\pm
  \cdot \partial_\nu \hat{b}_\pm \times \partial_\nu \hat{b}_\pm,$$ where $\hat{b}_\pm = \vec{b}_\pm/|\vec{b}_\pm|$, and the derivatives are with respect to $\vec{p}$. By construction, the magnetic field is divergenceless, $\partial_\mu \mathcal{B}_\pm^\mu=0$, away from points of singularity where $\vec{b}_\pm$ vanishes. However, the band touching point is such a singularity, and it is in fact a source of Berry flux. To see this, one may compute the integral of the flux of $\mathcal{B}^\mu$ through a sphere around one of the touching points. 