Now that we understand what is special about the strong invariant, let’s deal with the weak invariants.

From their definition, we know that the weak invariants don’t change the parity of the number of Dirac cones on any surface. Furthermore, there is a very good reason why the weak invariants are called ‘weak’. Imagine we keep the Hamiltonian the same, but instead we double the unit cell in each direction. This folds the Brillouin zone onto itself such that \(k=\pi\) is mapped to \(k=0\). This doesn’t impact the strong invariant, but all the weak invariants become 0.

As a final illustration of the relation between weak and strong invariants, let’s see how the invariants change as a function of \(M\) as we vary \(M\) on a scale comparable with the band width.

We determine the topological invariant in the same way as for QSHE: we see if the phase of the reflection matrix connects the Pfaffians of \(r(k_y=0)\) and \(r(k_y=\pi)\).

We see the values of the invariants change several times:

• Initially, when \(M>0\), the system is trivial.
• Then, as \(M\) is lowered, the topological invariants become \(\mathcal{Q}(k_x=0) = 1\) and \(\mathcal{Q}(k_x=\pi) = 0\), and there’s a Dirac cone at \(k=0\).
• When \(M\) is lowered further, two new Dirac cones appear at \(k = (0,\pi)\) and \(k = (\pi, 0)\). This changes the invariants to \(\mathcal{Q}(k_x=0) = 0\) and \(\mathcal{Q}(k_x=\pi) = 1\).
• Finally one more Dirac cone appears at \(k = (\pi, \pi)\), accompanied by both invariants becoming trivial.