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
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