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  Topological insulators: why K-theory classification rather than homotopy classification?

+ 6 like - 0 dislike

I am reading a 2009 paper by Kitaev on K-theory classification of topological insulators. In the 4th page, 1st paragraph in the section "Classification principles", he says,

Continuous deformation, or homotopy is part of the equivalence definition, but it is not sufficient for a nice classification.

Why is not homotopy sufficient? The only shortcoming of homotopy classification I can think of is that, a general homotopic deformation of the Hamiltonian may close some band gaps, so the satisfactory classification in terms of homotopy must be "equivalence class (of maps from Brillouin zone to Hamiltonians)classified up to homotopies that respect band gap", clearly this ought to give a finer classification than just "classified up to homotopy". Is this the reason for introducing the K-theory classification?

I know a bit algebraic topology but K-theory is quite over my head, and Kitaev's paper is too terse for me to figure out if I got it right. Could someone more familiar with the paper explain it to me? Or is there some more expository paper on the subject?

EDIT: Although it is already clear from Heidar's answer, let me stress here in the main post that the homotopy classification scheme does take into account the band gap nonclosing condition, I misunderstood this fact before seeing Heidar's answer and some other materials. I hope this edit makes the point clearer and eliminate the possiblity of misleading new learners of the subject who read this post.

This post imported from StackExchange Physics at 2014-06-25 20:59 (UCT), posted by SE-user Jia Yiyang
asked Mar 19, 2014 in Theoretical Physics by Jia Yiyang (2,640 points) [ no revision ]

1 Answer

+ 6 like - 0 dislike

Very loosely speaking the reasoning is this. Imagine a two band system in which the fermi sea has one filled band with Chern number $n$ and another system with $N$ filled bands but also with Chern number $n$. Physically they have the same topological properties (for example the same Hall conductance, edge states and so on), but cannot be deformed homotopically into each other since the Hamiltonians are of different size. Physically you would not consider those as two different phases and your classification should know that.

In general, consider two Hamiltonians $H_1(\bf k)$ and $H_2(\bf k)$ of the same size. It might be true they do not belong to the same homotopy class, and thus cannot be deformed into each other. However, by adding a few trivial bands (and thus trivially enlarging the Hamiltonians) one might be able to homotopically deform them into each other. Physically these trivial bands always exist, but we usually ignore them and consider finite dimensional Hamiltonians that describe the low energy bands. But since they are in the same homotopy class after adding a few trivial bands (which does not change the, say, Chern number), they must physically describe the same phase.

So a more physical equivalence relation is to not only consider homotopy classes of Hamiltonians, but also allow the addition of trivial bands. Topological $K$-theory is essentially the classification of vector bundles, not up to homotopy equivalence, but up to stable equivalence which essentially means that you are allowed to add (direct sum) trivial bundles. In this more relax equivalence relation, for example bundles of different rank can be in the same equivalence class. This is physically more sensible than considering vector bundles up to homotopy equivalence.

You can also think of it as homotopy classificaion of very very large matrices in order to get rid of the small dimension exceptions that usually exist. See for example how chaotic homotopy groups of spheres are for low dimenions: wikipedia.

As a simple example, take a two band system in $3$ dimensions and for simplicity lets assume the Brillouin zone is a sphere $S^3$ rather than a torus for simplicity (it doesn't change much). We can in general write this as

$$ H(\bf k) = \epsilon(\bf k) I + \bf d(\bf k)\cdot\bf{\sigma}, $$ with the spectrum $E(\bf k) = \epsilon(\bf k)\pm |\bf d(\bf k)|$. We can thus continuously deform (homotopy) this into the Hamiltonian

$$ \tilde H(\bf k) = \hat{\bf d}(\bf k)\cdot\bf{\sigma}, $$ where $\hat{\bf d}(\bf k)$ is just $\bf d(\bf k)$ normalized. Thus we have flattened the bands into $\tilde E(\bf k) = \pm 1$, without closing the gap. Now we see that the space of gapped two-band Hamiltonians are topologically classified by the homotopy classes of maps $\hat{\bf d}:S^3\rightarrow S^2$, and thus $\pi_3(S^2)$. From the famous Hopf map it is known that $\pi_3(S^2) = \mathbb Z$, and there are thus many different non-trivial phases for two-band gapped Hamiltonians. However from the general classification of topological insulators (based on $K$-theory) it is known that there are no non-trivial topological insulators in three-dimensions for charge conserving systems with no symmetry. This is because by adding a few trivial bands, one can show that no topological phase survives. Therefore, $K$-theory is a physically more robust classification that remove the strange behavior of systems with small Hamiltonians.

This post imported from StackExchange Physics at 2014-06-25 20:59 (UCT), posted by SE-user Heidar
answered Mar 20, 2014 by Heidar (855 points) [ no revision ]
Most voted comments show all comments
+1, very clear exposition. But one more follow-up question: is the band gap closing also an issue? Suppose a Hamiltonian A can be continuously deformed to another Hamiltonian B, what if all such homotopies close some band gaps, even say in the Kitaev's K-theory frame work? Shouldn't we consider A and B as topologically distinct?

This post imported from StackExchange Physics at 2014-06-25 20:59 (UCT), posted by SE-user Jia Yiyang
@JiaYiyang Your reasoning is completely correct, but it is taken into account in both the homotopy and K-theory classifications. One way to think about the problem is that we are not interested in the space of all Hamiltonian $\mathcal A$, but only the sub-manifold containing gapped Hamiltonians only $\mathcal M\subset\mathcal A$ (insulators are gapped by definition). Any gapped Hamiltonian is just a point of this manifold $H\in\mathcal M$. The interesting thing is now is the topology of $\mathcal M$, in particular is this space connected or not (as measured by $\pi_0(\mathcal M)$). (cont.)

This post imported from StackExchange Physics at 2014-06-25 20:59 (UCT), posted by SE-user Heidar
(continued) If $\mathcal M$ is path-wise connected, then we can deform (by local perturbations) any gapped Hamiltonian into any other without closing the gap and there is thus only one phase. If the space is not connected, then each connected component of $\mathcal M$ corresponds to a distinct topological phase, because you cannot deform a Hamiltonian from one connected component into one from another component without closing the gap! In other words, on the manifold $\mathcal M$ any perturbation that closes the gap is NOT a continues deformation and thus not homotopy! (continued)

This post imported from StackExchange Physics at 2014-06-25 20:59 (UCT), posted by SE-user Heidar
Homotopies are deformations of the Hamiltonian that do NOT close the gap, because we are interested in the manifold of gapped Hamiltonians. Whenever the gap closes, it can be a critical point/phase transition between two topologically distinct phases. So your reasoning is correct.

This post imported from StackExchange Physics at 2014-06-25 20:59 (UCT), posted by SE-user Heidar
Note however that what I describe is in general very hard to do. It can be done if you restrict $\mathcal M$ to be only the space of gapped fermionic Hamiltonians, which are non-interacting and possibly have certain symmetry (time-reversal, charge conservation, particle-hole). This will lead to homotopy classification of topological insulators. But weakening the equivalence notion as described in the answer, you end up with the K-theory classification. The general classification for interacting theories are under active research, but by very different methods.

This post imported from StackExchange Physics at 2014-06-25 20:59 (UCT), posted by SE-user Heidar
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Hi @Heidar, after reading this nice answer again, I am puzzled by your statement "...and for simplicity lets assume the Brillouin zone is a sphere $S^3$ rather than a torus for simplicity (it doesn't change much)", I don't get how we can take any Brillouin zone to be a sphere, doesn't the periodicity in k-space force us to have a torus?

This post imported from StackExchange Physics at 2014-06-25 20:59 (UCT), posted by SE-user Jia Yiyang
Dear Heidar, I've rewritten my above question in the comment into a question, I'd appreciate it if you can take a look:physics.stackexchange.com/questions/111440/…

This post imported from StackExchange Physics at 2014-06-25 20:59 (UCT), posted by SE-user Jia Yiyang

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