I don't understand the meaning of the expression "trivial topology" or "non-trivial topology" for an electronic band structure. Does anybody have a good explanation?
[Physics] Trivial and Non-trivial topology of band structure
condensed-mattertopological-insulatorstopology
Related Solutions
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.
No. Take a free scalar theory $$ L=(d\phi)^2-m^2\phi^2 $$
If you scan $m\in(-\infty,+\infty)$, you go through the point $m=0$, where the gap closes. On either side, the theory is in the trivial topological phase.
You could do the same using free fermions, $$ L=\bar\psi d\psi-m\bar\psi\psi $$ As you scan $m\in(-\infty,+\infty)$, you again go through a gapless point. On either side, the theory may or may not be trivial. For example, in two dimensions, you get a non-trivial phase if and only if the number of fermion modes is odd. If you take an even number, then you are in the trivial phase on either side of $m=0$. In 3d you get a non-trivial phase for any number of fermions. In 4/5/6d you get a trivial phase regardless of the number of fermions. $d=7$ is similar to $d=3$. Etc.
Best Answer
One of the early triumphs of QM (through e.g. Kronig-Penney model) was the explanation of the insulating state of matter. Energy bands (and gaps) appear as the result of hybridization of many atomic orbitals, and for a specific filling you can end up with the top most pair of bands being either entirely filled (valence band) or entirely empty (conduction band). No (small) electric field can perturb them enough to cause motion, and thus you have an insulator. In this trivial insulator, although the bulk is insulating, there is possibility of for example dangling bonds introducing states that lie in an energy gap. These states are localized at the edge; however they are not robust, and as such not particularly useful.
Now if you have a material with a sufficiently strong spin-orbit interaction for example (not essential for the effect, but historically important approach), this can cause the energy bands above and below the gap to swap places. This twisting is protected by time-reversal invariance, and although still an insulator, the resulting phase is topologically different from an ordinary insulator. The twisting of the band structure is what the phrase non-trivial topology is referring to; an analogy would be the way a Mobius strip is a twisted version of an ordinary strip. This manifests itself in the fact that when you put the two in contact, the curled up band structure of the TI must unwind so that the band structure fits the one in the ordinary insulator. This unwinding will will have to close the gap near the edge, hence the topologically protected edge states. This is the interesting part of the topological insulators from the practical standpoint.
So whether the band structure is wound up or not is a topological property, and one can measure it with the topological index, also called a Chern number, defined as
$$ C=\frac{1}{2\pi}\sum_n\oint Fd\mathbf{k} $$
where the sum is over occupied bands, the integral is over the entire Brillouin zone, and the integrated quantity is the Berry curvature (analogue of the magnetic field in $\mathbf{k}$ space) $F=-i\nabla_{\mathbf{k}}\times\langle u_{n\mathbf{k}}|\nabla_{\mathbf{k}}|u_{n\mathbf{k}}\rangle$, where $u_{n\mathbf{k}}$ are the Bloch eigenvectors. If $C=0$ you have a trivial insulator, and if $C\neq0$ you have a non-trivial or topological insulator.