Let us consider the surface of $(3+1)$-dimensional topological insulator, which is protected by the charge conservation $U(1)_Q$ and a time-reversal symmetry $\mathbb{Z}_2^T$. Such a surface, if not breaking $U(1)_Q\rtimes\mathbb{Z}_2^T$ explicitly, cannot be gapped with a unique ground state. Such an ``ingappability'' can be understood by either the nontrivial $U(1)_Q\rtimes\mathbb{Z}_2^T$ symmetry-protected-topological (SPT) bulk or the $U(1)_Q\rtimes\mathbb{Z}_2^T$ symmetry anomaly of itself.

If we explicitly break $\mathbb{Z}_2^T$ only (namely without breaking $U(1)_Q$) on the surface by $(2+1)$-dimensional local interaction to gap it with a unique ground state, such a $\mathbb{Z}_2^T$-explicitly-broken gapped surface state will develop a half Hall conductance.

However, purely $(2+1)$-dimensional electronic system with a unique ground state must have integer Hall conductance. In this sense, this $\mathbb{Z}_2^T$-explicitly-broken surface still needs to be attached to a bulk. My question is, does it mean that this $\mathbb{Z}_2^T$-explicitly-broken gapped surface is still "anomalous" since it cannot be consistent (e.g. non-integer Hall conductance) without a higher dimensional bulk? If so, does it contradict to the fact that $(2+1)$-dimensional Dirac fermion is actually anomaly-free if we only impose the $U(1)_Q$ symmetry without $\mathbb{Z}_2^T$? It seems such a gapped phase is a new phase since it cannot be consistent without a $U(1)_Q$-trivial SPT bulk and this bulk is a nontrivial $U(1)_Q\rtimes\mathbb{Z}_2^T$ bulk if further $\mathbb{Z}_2^T$ is imposed.


1 Answer 1


Great question. The answer is no, the surface is not anomalous if $\mathbb{Z}_2^T$ is explicitly broken. The reason is that the Hall conductance of the surface (which allegedly is 1/2) is not actually measurable (and as such, I would argue it's not actually meaningful at all). To measure a Hall conductance you have to have a sample with a boundary to which you can attach electrical contacts. Since the surface cannot have a boundary (it is already the boundary of the bulk 3-D TI, and a boundary of a boundary is always trivial), it follows that its Hall conductance is not actually defined.

Now, what you can do is the following. Since the bulk TI is in the trivial phase if you explicitly break $\mathbb{Z}_2^T$, you can now adiabatically deform it to a product state. Then the boundary really is a strictly 2-D system, and its Hall conductance becomes well-defined. However, if you do this you will find that the Hall conductance is indeed an integer. (Which integer? Actually, it depends on which adiabatic path you take in the bulk).

Why do people often say that the time-reversal broken boundary of a 3-D TI has Hall conductance $\pm 1/2$ then, if this is not actually a well-defined statement? Well, what is well-defined is the relative Hall conductance between a time-reversal broken boundary and its time-reversal conjugate. This is measurable by considering, for example, a domain wall between the two, and looking at the chiral modes that flow along this domain wall. (Actually only this quantity modulo 2 reflects bulk physics, since you can shift it by 2 just by appending an integer quantum Hall state to the symmetry-broken surfaces). For the 3-D TI one finds that this relative difference is 1 (modulo 2). If you forget everything I said about the (absolute) Hall conductance of the surface not being well-defined and try to assign a Hall conductance $\sigma_{xy}$ to the time-reversal broken surface, then its conjugate must have Hall conductance $-\sigma_{xy}$ and then $\sigma_{xy} - (-\sigma_{xy}) = 1$ implies that $\sigma_{xy} = 1/2$. But as I have emphasized, one should not take this too seriously.

  • 1
    $\begingroup$ Thanks for the great answer, which is also physically clear! But I still cannot understand the way you argued that the symmetry-broken surface of TI is strictly 2D by the adiabatic transformation of the (trivial U(1)) bulk. It is because this transformation of bulk, although not closing the bulk gap, will inevitably introduce non-local interaction on its boundary. Did you mean this is reflected in the different regulators before and after the adiabatic transformation? $\endgroup$
    – Yuan Yao
    Jul 18, 2019 at 3:49
  • $\begingroup$ @SmartYao It's not true that the bulk adiabatic transformation will necessarily induce non-local interactions on the boundary. To show this one uses the combination of two observations: (1) A gapped adiabatic path (of system without boundary) always induces a local unitary that relates the ground states [e.g. see arxiv.org/abs/1004.3835 ]; (2) any such local unitary can always be restricted (while remaining a local unitary) to act on a system with boundary, such that the action on the bulk far from the boundary is the same as it was in the absence of boundary. $\endgroup$ Jul 18, 2019 at 14:23

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