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For translationally invariant systems, we can define some topological invariant based on the translational symmetry, which is referred to "weak" topological invariant. For example, according to Kitaev's K-theory classification (https://arxiv.org/abs/0901.2686), the 3D T-invariant insulators are classified by $\tilde{K}^{-1}_\mathbb{R}(T^3) \cong \mathbb{Z} \oplus 3\mathbb{Z}_2$, where $T^3$ is the momentum space which forms a 3-torus. The $\mathbb{Z}_2$ part is the "weak" invariant.

For general systems, the result is given by $\tilde{K}^{-q}_\mathbb{R}(\bar{S}^d)$. My question is: what's the meaning of this manifold $\bar{S}^d$? Kitaev said that it's compactified momentum space (The asymptotic of Hamiltonian is fixed for $|p|\rightarrow \infty$). However, if the translational symmetry is broken, we can't even define "momentum space" by Fourier transform. So, why can we claim that the classification by K-theory over the compactified momentum space $\bar{S}^d$ gives the "strong" topological invariant, which is robust in the presence of disorder breaking the translational symmetry?

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When we compute the reduced KO theory of $\bar S^d$, a sphere, by the Atiyah-Hirzebruch spectral sequence it's $$\tilde H^d(\bar S^d, KO_{-q-d}(pt)) = KO_{-q-d}(pt).$$ Note that $\tilde H^0(\bar S^d, KO_{-q}(pt)) = 0$ since $\bar S^d$ is connected. The K theory of the point classifies indices of real Dirac operators. Not all of these are fully interacting (cobordism) invariants. Some cobordisms cause the indices to shift by certain amounts, and some are not the index of any Dirac operator. The ones that are are the "free fermion" phases, and their classification forms a quotient of the above.

Under the map $T^d \to \bar S^d$, which collapses the boundary of the Brillouin zone to a point, the K theory classes pullback to form the subgroup of "strong" invariants, which don't depend on translation symmetry.

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  • $\begingroup$ Do you mean that strong part $\tilde{K}O_{-q}(pt)$ classify the Dirac operators in position space and the weak part of $\tilde{K}O_{-q}(\bar{S}^d)$ needs the presence of translational symmetry? Why do we use $\bar{S}^d$ instead of $T^d$? $\endgroup$
    – Yu-An Chen
    Jul 21, 2018 at 6:34
  • $\begingroup$ Oh sorry I was confused by the bar and thought $\bar S^d$ was the Brillouin zone (a torus). Taking the reduced KO theory of $S^d$ just amounts to a shift of the KO-groups of the point. The weak invariants are in the KO-theory of the Brillouin zone. I have edited my answer. $\endgroup$
    – Ryan Thorngren
    Jul 22, 2018 at 19:24

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