In the classification of free fermion systems in condensed matter, physicists usually divide the systems into ten symmetry classes, first discovered by Altland and Zirnbauer. In their classification, they use the presence/absence of three symmetries, time reversal, particle hole, and chiral symmetry, to classify their systems.

I was reading a paper by Ryu et al http://arxiv.org/pdf/0912.2157.pdf and they explain the justification for this. Unfortunately, I still don't understand.

So here's my question: why do we only use these three symmetries to classify Hamiltonians in the tenfold way? Why not others? What's special about these three?

  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/0912.2157 $\endgroup$
    – Qmechanic
    Commented Aug 24, 2018 at 6:02

1 Answer 1


The papers by Ryu et al are not very well written and contain quite a few substantive mathematical errors (e. g. the classification in terms of strong topological invariants is not the same as the homotopy classification, unless you are in very low dimension).

Commuting symmetries

Let us start with Wigner's theorem: it states that symmetries of quantum systems are those that preserve transition probabilities and are therefore implemented via unitaries or antiunitaries which commute with the Hamiltonian.

All commuting, unitary symmetries need to be reduced out before applying the Altland-Zirnbauer scheme. This step is usually not mentioned explicitly in the literature, but essential. Otherwise, you have way more than 10 distinct classes, because of the interaction between unitary, commuting and other symmetries. The most prominent example here are crystallographic symmetries, for which a classification exists, but that is much, much more complicated (the state of the art is to my knowledge the work by Shiozaki, Sato and Gomi).

Antiunitary, commuting symmetries are the other option admitted by Wigner's theorem. If you assume in addition that they square to $\pm \mathbf{1}$, then these are even ($+$) and odd ($-$) time-reversal symmetries.

Anticommuting symmetries

Now if you are in a periodic system, then commuting symmetries give you additional information of Bloch functions associated to a fixed band at potentially different $k$ vectors. Time-reversal symmetries typically flip $k \mapsto -k$, and therefore give you a relation between $\varphi_n(k)$ and $\varphi_n(-k)$.

The next level of complication are symmetries which relate two bands or, more generally, an even number of bands with one another. Chiral and particle-hole type symmetries relate Bloch functions of one band $E_n(k)$ with that of its symmetric partner $E_{-n}(k) = - E_n(\pm k)$, i. e. they exchange bands. Mathematically this means they need to anticommute with the Hamiltonian. (Since these fall outside of the purview of Wigner's theorem, some authors such as Zirnbauer call those pseudosymmetries, but I will not make that distinction.) These types of symmetries arise naturally in the context of fundamental equations (e. g. the Dirac equation sports a particle-hole symmetry) or when you consider effective tight-binding models after renormalizing the zero energy level to fall in the middle of a band crossing or avoided band crossing.

As before, the antiunitary, anticommuting symmetries, particle-hole-type symmetries, come in two flavors, even and odd depending on whether they square to $\pm 1$. Chiral symmetries come in a single flavor for if $U^2 = \mathrm{e}^{\mathrm{i} \vartheta} \, \mathbf{1}$, then $U' = \mathrm{e}^{-\mathrm{i} \frac{\vartheta}{2}} U$ squares to $+ \mathbf{1}$.

The Ten-Fold Way

To summarize, you have four types of symmetries, two types of commuting and two types of anticommuting symmetries. However, you have reduced out all unitary, commuting symmetries, which leaves you with three types of symmetries. The antiunitary symmetries come in two flavors, even and odd. So you have 1 case of no symmetries (class A), 5 cases of one symmetry (chiral, 2x TR, 2x PH) and $2 \times 2 = 4$ cases of 3 symmetries (if you have two symmetries, then the product is automatically the third). Overall you have $1 + 5 + 4 = 10$ cases. That is the Ten-Fold Way.

Higher-order symmetries

You could ask yourself whether you could consider more complicated symmetries which “exchange bands” (think of an operator that implements the permutation of a three-band system that maps 1—>2, 2—>3, 3—>1). Absolutely, and I know some people who are toying with this idea. But AFAIK there is no real application for this at the moment and the existing scheme is already complicated enough — especially once you add crystallographic symmetries into the mix.

Interaction with crystallographic symmetries

To finish, let me give you an example of how the “interaction” between symmetries can give rise to new phenomena. Consider a two-dimensional periodic system that possesses an even time-reversal symmetry and parity. If you disregard parity, this system is topologically trivial (it is of class AI, and in $d = 2$ there is only a single phase). However, because of the presence of parity you can split your system into even and odd contributions (eigenstates to $\pm 1$ of the parity operator). There are cases where the time-reversal symmetry is block-offdiagonal with respect to this decomposition, i. e. time-reversal maps even to odd functions and vice versa. This is the case in quasi-two-dimensional photonic crystals with honeycomb structure (investigated by Wu and Hu). So that means that time-reversal symmetry is broken on the even and odd subspace! Put another way, you can split class AI system into two class A subsystem which are “conjugate to another”. Here, edge states appear in pairs, but the edge states have a definite parity (even or odd). So if you are able to selectively excite, say, states with even parity and only introduce perturbations which preserve parity symmetry, then your edge modes will be topologically protected. The even parity Chern number gives you the net number of edge states with even parity; the odd parity Chern number necessarily has to be equal in magnitude but have opposite sign when compared to the even parity Chern number. This way the total Chern number of the system adds up to 0, as it should be for a system of class AI.


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