We have now explained why there are ten rows and the meaning of the symmetry classes, but the table as a whole still does not have a coherent structure. Is there a way that we can connect systems between different symmetry classes and different dimensions?
Given a gapped Bloch Hamiltonian \(H_d\) in a certain symmetry class and dimension \(d\), there is a systematic procedure by which we can increase its dimensionality. This changes its symmetry class, but makes sure that the topological classification is unvaried. This procedure thus allows us to connect different rows of the table.
The basic idea is to add a new momentum \(k_{d+1}\) to the Bloch Hamiltonian in such a way that the gap does not close for any value of \(k_{d+1}\) in the Brillouin zone. This means that the new Hamiltonian must have the same value of the topological invariant as the initial one. The tricky (but also beautiful) part is that adding a momentum can change the symmetries of the Hamiltonian, and one has to keep track of that carefully.
This procedure is slightly different depending on whether the initial \(H_d\) has chiral symmetry or not. In one case the procedure removes symmetries, in the other it adds them. Let’s start with the first.
Let’s first suppose that \(H_d\) has chiral symmetry \(\mathcal{C}\). The Hamiltonian then has \(n\) pairs of energy bands symmetric around zero, which we can denote as \(\pm \epsilon_{n,d}\), hiding their dependence on the different momenta appearing in \(H_d\).
Then consider the following higher-dimensional Hamiltonian:
\[ H_{d+1} = H_d\,\cos k_{d+1} + \mathcal{C} \sin k_{d+1}. \]
This Hamiltonian has the same number of bands as \(H_d\), even though the bands are higher-dimensional. Given its simple form, every band \(\epsilon_{n,d+1}^n\) is directly related to a band \(\epsilon_{n,d}\) of \(H_d\),
\[ \epsilon_{n,d+1} = \pm \sqrt{\epsilon_{n, d}^2\,\cos^2 k_{d+1} + \sin^2 k_{d+1}}. \]
This expression guarantees that the gap of \(H_{d+1}\) only closes if that of \(H_d\) closes. In other words, whatever value of the topological invariant \(H_d\) has, \(H_{d+1}\) will have the same value.
What are the discrete symmetries of \(H_{d+1}\)? First of all, note that \(H_{d+1}\) has no chiral symmetry since the second term by construction commutes with \(\mathcal{C}\), while the chiral symmetry would require anti-commutation. So \(H_{d+1}\) and \(H_d\) belong to a different symmetry classes. Which symmetry class does \(H_d\) belong to?
If \(\mathcal{C}\) is the only discrete symmetry of \(H_d\), i.e. if \(H_d\) belongs to class AIII, then \(H_{d+1}\) has no symmetries at all, so it is in class A. We can express this by writing AIII \(\to\) A, meaning that the topological classification is the same to the left of the arrow in \(d\) dimensions, and to the right of the arrow in \(d+1\) dimensions.
If instead \(\mathcal{C}\) is not the only discrete symmetry of \(H_d\), then \(H_d\) must have both \(\mathcal{P}\) and \(\mathcal{T}\), because we know that two of the symmetries imply the third. Hence \(H_d\) must be in one of the symmetry classes BDI, CI, CII, DIII. Because \(H_{d+1}\) has no chiral symmetry, it can only have one remaining symmetry, either \(\mathcal{P}\) or \(\mathcal{T}\). It is therefore in one of the four classes AI, AII, C, D. With some patience, it is possible to work out exactly the symmetry class of \(H_{d+1}\) given that of \(H_d\) in all four cases.
We won’t do that, but state that the result is that by removing chiral symmetry and adding one dimension, one obtains that BDI \(\to\) D, DIII \(\to\) AII, CII \(\to\) C and CI \(\to\) AI.
Let’s now start from a Hamiltonian without chiral symmetry. Our procedure this time involves a doubling of the number of bands of \(H_d\). That is, we introduce a new set of Pauli matrices \(\tau\), not present in \(H_d\), and write \(H_{d+1}\) as
\[ H_{d+1} = H_d\,\cos k_{d+1}\,\tau_x + \sin k_{d+1}\,\tau_y. \]
Note that just like in our previous argument, the topological invariant of \(H_{d+1}\) must be the same as that of \(H_d\). Also, by construction, \(H_{d+1}\) has a chiral symmetry given by \(\mathcal{C}=\tau_z\), which anticommutes with all the terms in the Hamiltonian.
What are the symmetries of \(H_{d+1}\) in this case? To begin with, \(H_d\) cannot have both anti-unitary symmetries, because then it would have \(\mathcal{C}\) as well. It either has none (class A) or just one of them (classes AI, AII, C, or D). We thus have two cases:
Now that we have learned how to extend the topological classification by adding one dimension to the Hamiltonian and changing its symmetry in an appropriate way, nothing forbids us to repeat the procedure, by going two dimensions higher, three dimensions higher, etc…
So we have made the procedure of finding topological phases systematic.