2
$\begingroup$

I was reading this article: https://arxiv.org/abs/1512.08882 on the 10 fold way which gives a nice explanation of the possible topological phases for each of the symmetry classes. The example explanation(in section III.B.) for symmetry class A (T=S=C=0) seems to be band dependent:

Topological order requires a gap, but allows for continuous deformations. This allows one to break the Hamiltonian (dimension N+M) at a momentum $k$ into two sets of eigen states(of dimention N and M) such that the first set had energy +1 and the second had energy -1. For symmetry class A, a valid transformation is part of the lie group U(N+M), while a valid transformation which doesn't mix bands is part of $[U(N)\otimes U(M)]$. Since mixing states within a band doesn't change the topology, topological order was classified by the homotopy groups of $U(N+M)/[U(N)\otimes U(M)]$.

If I generalize this to 3 bands, topological order would be classified by the homotopy groups of $U(N+M+L)/[U(N)\otimes U(M)\otimes U(L)]$. I'm then left at the following questions:

  • Are these homotopy groups different?
  • Does this generalize to the other symmetry classes and give new homotopy groups?

I know that for ground state physics, the third band is irrelevant because mixing the the second and third band doesn't change the topology of the first band, but maybe in a dynamical situation the restriction for not mixing the bands would matter. Anyways, we could save the discussion on the physical relevance of this restriction to a separate question.

$\endgroup$
  • $\begingroup$ Good question. This recent pre-print explores this: arxiv.org/abs/1808.07469 (see Eq.(8)!). It turns out that now one needs non-abelian Wilson loops! $\endgroup$ – Ruben Verresen Nov 13 '18 at 0:00
2
$\begingroup$

That is a good question. Yes, the classification theory will be different, but the answer will be much, much more complicated than the Ten-Fold Way. There are mathematical frameworks (e. g. the K-theoretic framework by Freed and Moore) which are powerful enough to investigate the question you have, but keep in mind that you are asking a difficult question.

$\endgroup$
  • $\begingroup$ Why will it be so dramatically different? Can't you still look at the homotopy groups of quotient groups? $\endgroup$ – Shane P Kelly Nov 2 '18 at 19:35
  • $\begingroup$ The problem is not the homotopy definition here. The reason why you have to use such heavy machinery as various (e. g. equivariantly twisted) K-theories and such is that you want to be able to compute whether two specific hamiltonians are in the same phase or not. That is why you want to be able to compute topological invariants, for example, because they allow you to decide whether two hamiltonians can or cannot be in the same topological phase. With the homotopy definition that is hopeless, unless you restrict yourself further to models which depend only on finitely many parameters or so. $\endgroup$ – Max Lein Nov 5 '18 at 2:52
1
$\begingroup$

Following Ruben's urging (he commented on your question above) and the fact that I recently successfully applied similar ideas to study band-structure degeneracies in crystalline solids (see here https://arxiv.org/abs/1808.07469), let me perhaps make some thoughts/comments on your questions (even though I don't answer your questions per se).

$${\color{white}.}$$ $${\color{white}.}$$

First, the homotopy groups that you ask for, are relatively easy to find. All you need is that for a coset space (or fiber bundle) $B = E/F$ there is a long exact sequence of homotopy groups $$\cdots \to \pi_p(E) \to \pi_p(B) \to \pi_{p-1}(F) \to \pi_{p-1}(E) \to \cdots .$$ In this sequence, each arrow is a group homomorphism. The "exactness" means that the image of any arrow is exactly equal to the kernel of the next arrow. The "long" simply states that the sequence is infinite (or 'semi-infinite' since zeroth homotopy groups terminate the sequence to the right).

As for your example, let's say that you want to know the first and the second homotopy group $$\pi_1[U(N_1+N_2+\ldots + N_\ell)/U(N_1)\times U(N_2)\times \ldots \times U(N_\ell)] \equiv H.$$ $$\pi_2[U(N_1+N_2+\ldots + N_\ell)/U(N_1)\times U(N_2)\times \ldots \times U(N_\ell)] \equiv G.$$ Of course, it is assumed that each $N_i \geq 1$.

To find it, first not that the second homotopy group $$\pi_2[U(N_1+N_2+\ldots + N_\ell)] = \mathbb{0}$$ is trivial (as is true for all Lie groups). Furthermore, we find $$\pi_1[U(N_1)\times U(N_2) \times \ldots \times U(N_\ell) ] = \pi_1[U(N_1)]\oplus \pi_1[U(N_2)] \oplus \ldots \oplus \pi_1[U(N_\ell)]= \mathbb{Z}^\ell$$ where each $\mathbb{Z}$ counts the winding of the $U(1)$ phase of the determinant of the corresponding $U(N_i)$. We also need to use that $\pi_1[U(N_1+N_2 + \ldots + N_\ell)] = \mathbb{Z}$, too. Finally, since unitary group as a manifold is path-connected, its zeroth homotopy group is trivial.

The relevant segment of the long exact sequence, starting with $\pi_2(E)$ and ending with $\pi_0(E)$, leads to $$\ldots \stackrel{\varphi_a}{\longrightarrow} \mathbb{0} \stackrel{\varphi_b}{\longrightarrow} G \stackrel{\varphi_c}{\longrightarrow} \mathbb{Z}^{\ell} \stackrel{\varphi_d}{\longrightarrow} \mathbb{Z} \stackrel{\varphi_e}{\longrightarrow} H \stackrel{\varphi_f}{\longrightarrow} \mathbb{0} \stackrel{\varphi_g}{\longrightarrow} \ldots$$ where I gave the group homomorphsms some names. Clearly, the image $\textrm{im}\, \varphi_b = \mathbb{0}$ is trivial, such that by exactness also the kernel $\textrm{ker}\,\varphi_c = \mathbb{0}$. Furthermore, for any group homomorphism $\varphi:G_1\to G_2$ there is an isomoprhism $\textrm{im}\,\varphi \cong G_1/\textrm{ker}\,\varphi$, so we deduce $\textrm{im}\,\varphi_c = G = \textrm{ker}\,\varphi_d$, where at the last equation mark we used the exactness again. One arrow further we find $\textrm{ker}\,\varphi_e = \mathbb{Z}^\ell/G$. On the other hand, moving from the right we can similarly find $\textrm{im}\,\varphi_e = H$. By complementarity of the image and the kernel of $\varphi_e$, we obtain $$(\mathbb{Z}^\ell / G) \oplus H = \mathbb{Z}.$$ There is more than one solution to this equation. But with some physics insight, we can guess the correct one: $$H = \mathbb{0}\qquad\textrm{and}\qquad G = \mathbb{Z}^{\ell-1}.$$

What is the physics insight? Each of the $N_i$-plets of bands carries an integer Chern number. However, the sum of all Chern numbers must be zero, therefore only $\ell-1$ of the Chern numbers are independent.

The implication for band-structure nodes in 3D Briollouin zone (BZ) would be as follows. The $\ell$ $N_i$-plets of bands are separated by $\ell-1$ band gaps. The $\ell-1$ integer charges on some sphere $S^2 \subset \textrm{BZ}$, corresponding to the second homotopy group $\pi_2(...) = \mathbb{Z}^{\ell-1}$, simply tell the total chirality of Weyl points formed inside each of these band gaps, which are contained inside the $S^2$.

$${\color{white}.}$$ $${\color{white}.}$$

Second, let me comment on the meaning of such topological charges. In the ten-fold way classification, one looks at so-called stable limit of homotopy groups. There is a good physics motivation for considering the stable limit, yet such limit is physically meaningless for the topological charges that you ask about.

Let me first explain the "conventional" case of stable limit on a simple example. Let's say that there is a symmetry (such us composition of spatial inversion + time reversal, and no spin-orbit coupling, see more here https://arxiv.org/abs/1705.07126), which forces Hamiltonian to be a real symmetric matrix. Then the space of Hamiltonians is the real Grassmanian $O(N+M)/O(N)\times O(M)$, rather than the complex one. It can be easily shown that for two occupied and one unoccupied band $$\pi_2[O(3)/O(2)\times O(1)] = \mathbb{Z},$$ while for sufficiently many bands (i.e. the stable limit) $$\lim_{N,M\to\infty}\pi_2[O(N+M)/O(N)\times O(M)] = \mathbb{Z}_2$$ i.e. it becomes smaller.

This disparity implies that three-band models can exhibit some topological obstruction, which can be trivially disentangled if one allows hybridization with additional bands. Since in solids there are countless bands available at sufficiently large energies, each of which could, at least in principle, be brought to the three bands and hybridize with them, the three-band topological obstruction is fragile (i.e. opposite of stable). This is why it is of often convenient to use $K$-theoretic methods to find the topological obstruction, since those methods are naturally susceptible to the stable limit.

However, the notion of stable limit of a three-component partition of bands, $N_1 + N_2 + N_3$, does not seem very meaningful in solid state physics. You could conjure up extra bands at very large positive or very large negative energy (i.e. inside the two outer components of the partition), but you cannot just snap your fingers and magically create an extra band in the middle (the central partition). Therefore, it seems to me, the stable limit is physically not available here.

On the other hand, as long as the three degrees of freedom are weakly coupled to all the other degrees of freedom, the fragile $\mathbb{Z}$ charge/obstruction is unremovable (for example, it would imply the absence of atomic limit consisting of exponentially localized Wannier orbitals). So under special circumstances it has a real impact on the band structure topology.

$${\color{white}.}$$ $${\color{white}.}$$

Finally, let me briefly comment on the real case, which we actually analyzed along similar ways in work https://arxiv.org/pdf/1808.07469 (supplement at https://arxiv.org/src/1808.07469v1/anc/supplement.pdf). We considered the extreme case $O(N)/O(1)\times O(1)\times\ldots \equiv M_N$. It turns out that the first homotopy group $\pi_1(M_N)$ is non-Abelian (even in non-interacting systems!), and equals the quaternion group $$Q = \{\pm 1,\pm\mathrm{i},\pm\mathrm{j},\pm\mathrm{k}\}$$ for $N=3$ bands. The noncommutative character of the first homotopy group implies nontrivial exchange (or 'braiding') rules for band-structure degeneracies in momentum space. For example it poses strict constraints on the admissible nodal line compositions in 3D. In our work, we argue that these non-trivial exchange rules survive even in the limit $N\to\infty$, and they are very easy to formulate.

So one can definitely learn something useful using the consideratiosn that you suggested in your question -- even though the notion of stable limit is rather obscure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.