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Kohmoto from TKNN(Thouless-Kohmoto-Nightingale-deNijs) who described the topology of the integer quantum hall effect always stressed the importance of the 2D Brillouin zone being a donut due to periodic boundary conditions.

--> http://www.sciencedirect.com/science/article/pii/0003491685901484

Now I don't really see why this is relevant. Shouldn't the zeros of the wavefunction always lead to a Berry phase?. What would happen if we have a sphere instead of a donut? I think I am missing a major point here, because I can't see how the Gauss-Bonnet theorem that connects topology to geometry plays a role. For a sphere with no holes the Gaussian curvature gives us 4$\pi$, for a donut if gives us 0. In both cases Stokes' theorem should still give us a nonzero value?

Charles Kane then uses this argument to compare a donut with the quantum hall state and a sphere with an insulator. He then writes down the Gauss-Bonnet theorem and immediately talks about topological insulators, and again I don't see the connection or is it just an analogy and I shouldn't waste any time on this?

If there is a connection, I would like to know if there is a simple explanation for using the Gauss-Bonnet theorem in the context of topological insulators. I'm even more confused because Xiao-Gang-Wen said in a recent post here, that a topological insulator is NOT due to topology but due to symmetries...

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I believe he is trying to bring out the fact the QHE depends on the BZ having a nontrivial first homotopy group. Otherwise it would be impossible to construct a gauge connection [$A(k,k')$ in his notation] that did not give zero $\sigma_{xy}$ by stokes theorem. –  BebopButUnsteady Jul 5 '13 at 4:40
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@BebopButUnsteady Actually it does not depend on BZ having non-trivial first homotopy group, in this particular case one can with no problem change the torus to a sphere. Despite the fact that $\pi_1(S^2)=0$, this can lead to non-zero $\sigma_{xy}$. This is because whats important is homotopy classes of maps from the base manifold $M$ to the classifying space $BG$, $[M,BG]$. For $G=U(1)$ bundles, we have $BG=\mathbb CP^\infty$. We can have non-zero $\sigma_{xy}$ with the sphere $M=S^2$, since $[S^2,BG]=\pi_2(\mathbb CP^\infty) = \pi_1(U(1)) = \mathbb Z$. –  Heidar Jul 5 '13 at 5:13
    
@Heider: You are right - I knew it didn't make sense as soon as I wrote it. –  BebopButUnsteady Jul 5 '13 at 13:03

1 Answer 1

Sorry this answer got too long. I have categorized it into three points.

(1)

I think the reason Kohmoto stresses the importance of the Brillouin zone being a torus $BZ = T^2$, is because he wants to say that BZ is compact and has no boundary. This is important because of the subtlety that makes everything work. The Hall conductance is given by $\sigma_{xy} = -\frac{e^2}h C_1$ (eq. 4.9), where the first Chern number is (eq. 4.8)

$C_1 = \frac i{2\pi}\int_{BZ} F = \frac i{2\pi}\int_{BZ} dA$.

However by naively using Stokes theorem $\int_M dA = \int_{\partial M} A$, where $\partial M$ is the boundary of $M$. Since $BZ= T^2$ and the fact that the torus has no boundary $\partial T^2$, this seem to imply that $\int_{\partial BZ} A = 0$ and thus $\sigma_{xy}=0$. There is however an important subtlety here, our use of Stokes theorem is only correct if $A$ can be constructed globally on all of $BZ$ and this cannot be done in general. One has to split the $BZ$ torus into smaller patches and construct $A$ locally on each patch, which now do have boundaries (see figure 1). The mismatch between the values of the $A$'s on the boundaries of the patches will make $\sigma_{xy}$ non-zero (see eq. 3.13).

In terms of de Rahm cohomology one can say that $F$ belongs to a non-trivial second cohomolgy class of the torus, or in other words the equation $F = dA$ is only true locally not globally. And that's why our use of Stokes theorem was wrong.

In this case, you can actually replace the torus with a sphere with no problem (why that is requires some arguments from algebraic topology, but I will shortly give a more physical picture of this). In higher dimensions and in other types of topological insulators there can be a difference between taking $BZ$ to be a torus or a sphere. The difference is that with the sphere you only get what people call strong topological insulators, while with $BZ=T^2$ you also get the so-called weak topological insulators. The difference is that, the weak topological insulators correspond to stacks of lower-dimensional systems and these exist only if there is translational symmetry, in other words they are NOT robust against impurities and disorder. People therefore usually pretend $BZ$ is a sphere, since the strong topological insulators are the most interesting anyway. For example the table for the K-theoretic classification of topological insulators people usually show (see table I here), correspond to using the sphere instead of torus, otherwise the table will be full of less interesting states.

Let me briefly give you some physical intuition about what $\sigma_{xy}$ measures by making an analogy to electromagnetism. In a less differential geometric notation, one can write (eq. 3.9)

$C_1 = \frac i{2\pi}\oint_M \mathbf B\cdot d\mathbf S$,

where $\mathbf B = \nabla_k\times \mathbf A$ can be though of as a magnetic field in k-space. This is nothing but a magnetic version of the Gauss law and it measures the total magnetic flux through the closed surface $M$. In other words, it measures the total magnetic charge enclosed by the surface $M$ (see also here). Take $M=S^2$, the sphere. If $C_1 = n$ is non-zero, that means that there are magnetic monopoles inside the sphere with total charge $n$. In conventional electromagnetism $C_1$ is always zero, since we assume there are no magnetic monopoles! This is the content of the Gauss law for magnetism, which in differential form is $\nabla\cdot\mathbf B = 0$. The analogue equation for our k-space "magnetic field" would be $\nabla\cdot\mathbf B = \rho_m$, where $\rho_m$ is the magnetic charge density (see here). If $M=BZ=T^2$ the intuition is the same, $C_1$ is the total magnetic charge inside the torus.

Another way to say the above is that the equation $\mathbf B = \nabla\times\mathbf A$ as we always use and love, is only correct globally if there are no magnetic monopoles around!

(2)

Now let me address the next point about Gauss-Bonnet theorem. Actually Gauss-Bonnet theorem does not play any role here, it is just an analogy. For a two-dimensional manifold $M$ with no boundary, the theorem says that $\int_M K dA = 2\pi (2-2g)$. Here $K$ is the Gauss curvature and $g$ is the genus. For example for the torus, $g=1$ and the integral is zero as you also mention. This is not the same as $C_1$ however. The Gauss-Bonnet theorem is about the topology of the manifold (for example the $BZ$ torus), but $\sigma_{xy}$ is related to the topology of the fiber bundle over the torus not the torus itself. Or in other words, how the Bloch wavefunctions behave globally. What plays a role for us is Chern-Weil theory, which is in a sense a generalization of Gauss-Bonnet theorem. The magnetic field $\mathbf B$, or equivalently the field strength $F$, is geometrically the curvature of a so-called $U(1)$ bundle over $BZ$. Chern-Weil theory says that the integral over the curvature

$C_1 = \frac i{2\pi}\int_{BZ} F$

is a topological invariant of the $U(1)$ bundle. This is analogous to Gauss-Bonnet, which says that the integral over the curvature is an topological invariant of the manifold. Thus this connection is mainly an analogy people use to give a little intuition about $C_1$, since it is easier to see the curvature $K$ than the curvature $F$ which is more abstract.

(3)

The comment of Xiao-Gang Wen is correct and to explain it requires going into certain deep issues about what is topological order and what is a topological insulator and what the relation between them is. The distinction between these two notions is very important and there are lots of misuse of terminology in the literature where these are mixed together. The short answer is that both notions are related to topology, but topological order is a much deeper and richter class of states of matter and topology (and quantum entanglement) plays a much bigger role there, compared to topological insulators. In other words, topological order is topological in a very strong sense while topological insulator is topological in a very weak sense.

If you are very interested, I can post another answer with more details on the comment of Xiao-Gang Wen since this one is already too big.

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Quick question: where is this comment by Xiao-Gang Wen? –  Chris White Jul 5 '13 at 7:24
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@ChrisWhite I don't know exactly, maybe Mike can give a link of where he has seen it. But the point of the comment is correct and something Xiao-Gang Wen been one of the main people to clarify in details in his papers. arxiv.org/abs/1004.3835 is a good paper on this. There topological insulators are a tiny subset of what they call "symmetry protected topological order" (SPT). See figure 3 of their paper or the wikipedia article en.wikipedia.org/wiki/… . –  Heidar Jul 5 '13 at 7:37
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@ChrisWhite Since you used to be at Caltech, you might be interested in a very recent blog post by John Preskill on SPT states and bosonic versions of topological insulators quantumfrontiers.com/2013/07/03/… . But beware that the magnetic monopoles he is talking about it not the same as those I talk about. –  Heidar Jul 5 '13 at 7:41
    
Very nice answer. I'd like to know your thoughts on the following: it seems to me that topological insulators do refer to "topology" in a very real way, but in the sense of connectedness of the single particle Hilbert space of one electron. It is easy to imagine that many-body effects could destroy this. Topological order refers to many-body systems explicitly (therefore more interesting), but the notion of "topology" seems actually less explicit. Is there an easy way to reconcile this? –  wsc Jul 5 '13 at 19:34
    
@Heidar. What would happen to the bulk-boundary correspondance if we don't have a torus but a sphere? Isn't this the crucial point, why we can consider Laughlin's argument and speak of a charge pump? Because the torus get's transformed into a cylinder due to the two contacts where we must measure the conductance with. –  Matthias Jul 6 '13 at 2:49

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