# What does the Chern number physically represent?

In 2D the Chern number can be written as

$$C_m=\frac 1{2\pi}\int_{BZ}\Omega_m(\mathbf k)\cdot d^2 \mathbf k$$

where we are integrating over the Brillouin zone.

In 2D this is equivalent to finding the "flux" of the Berry curvature through the entire surface of the torus. I struggle to understand what this means. My understanding is that the Berry curvature is a "gauge invariant" (as in we just pick a different phase for the eigenfunction?) "field" which can be found by taking the curl of the Berry vector (which is not gauge invariant).

The maths seems to be rather straightforward but I can't seem to figure out what any of this means in a physical sense.

I also understand that the Hall conductance can be found to be proportional to the Chern number with integer multiples. Again, this makes no sense to me other than a mathematical trick. What is the motivation behind this and why is it important?

• I think it represents the magnetic charge of the magnetic monopole the surface encloses. – Libertarian Monarchist Bot Jun 4 '18 at 11:28

Physically, I like to view it in the following way:

You are looking at an adiabatic evolution of an electron confined to some energy band $n$. You somewhat project out the presence of other states in other bands $m\neq n$. Still these ignored bands effect the dynamics of your adiabatic state.

When other bands are approaching the energy of the adiabatic eigenstate the Berry curvature increases. If they move further away, Berry curvature decreases.

For some intuition on the QHE I recommend Laughlin's Nobel lecture! To summarize (I losely follow the Bohm's book on Geometric Phases): the basic idea is to consider a two-dimensional electron gas (2DEG) on a finite cylinder. Perpendicular to the 2DEG you have a constant magnetic field $B$. Imagine you would change the magnetic flux through the cylinder by $\delta\phi$ . This changes the energy of the system by an amount

$$\delta U = I \delta \phi,$$

where $I$ is the induced azimuthal current. If you thread this flux in an adiabatic fashion and choose $\delta\phi$ as the flux quantum $h/e$, the bulk system returns to its initial state. However, in this process the $z$-localization of the eigenstates shifts. This means if you apply an electric voltage $\delta V$ along this direction you change the energy by

$$\delta U = n e \delta V,$$

where $n$ is the number of electrons shifted from one edge to the other. Combining the two results you see that

\begin{align} \sigma &= \frac{I}{\delta V} = \frac{1}{\delta V} \frac{\delta U}{\delta \phi} = \frac{ n e}{\delta \phi} = \frac{n e^2}{\hbar}. \end{align}

A more rigorous treatment of this hand-waving argument can be found for example here.

As you know, this integer $n$ is actually a topological invariant of your electronic band structure. It might be easy to calculate the Hall effect for 2DEG's, as for example done by T. Ando et al. in 1975 (see here) predating von Klitzing's discovery of the QHE! However, this leaves many open questions. For example, why is the QHE so ridiculously precise? Von Klitzing's experiments revealed a very high accuracy.

These properties are elucidated by Berry phase physics because of its connection to mathematical theory of Chern classes.

The chern number gives you information about the wavefunction.

In the Brillouin zone, we go from spatial dependence to momentum dependence for the wavefunction.

Sometimes we can't define a wavefunction for the whole Brillouin zone.

It is just that one single function will not cover the whole area, so we have to define two parts.

The Chern number will tell you if this is the case.

• I've never heard this explanation before and it is very intriguing. Would you mind elaborating a bit? Is it possible to give any examples? – dylnan Jul 14 '18 at 4:42

First of all, the connection between the transverse conductivity and the Brillouin zone average of the Berry curvature is not a “mathematical trick”, but can be derived using linear response theory — you find that the conductivity coefficients are given in terms of the Kubo formula. This formula only involves projections and is therefore gauge invariant.

If you plug in the projection onto a single, isolated band (i. e. there are no intersections with other bands), you can easily see that this reduces to the ordinary expression for the Berry curvature. If you have families of bands, e. g. all bands up to the Fermi energy (which is assumed to lie in a gap), you obtain the trace over the multiband Berry curvature.

No mathematical tricks are being played here. The only non-obvious fact is that the Brillouin zone average of the Chern number is necessarily integer-valued. To do that one approach is to construct the Bloch vector bundle associated to the Fermi projection, and you see that the Berry curvature is in a mathematical sense the curvature of this Bloch vector bundle. For a given Berry curvature you can locally (!) choose a connection — the Berry connection. Just like vector potentials in electromagnetism you need to make a choice of gauge here.

The Chern number measures whether there is an obstruction to choosing a global gauge — this is possible if and only if the Chern number is zero. Classification theory of vector bundles tells you that the Chern number is necessarily an integer. This may be mathematically abstract, but nevertheless, no magic is involved.

Note that none of these arguments invoke “adiabatic changes in parameters”. I find this choice of words misleading and mathematically incorrect. What people describe when they invoke “adiabatic changes in parameters” is nothing but parallel transport of vectors with the help of a connection.

• For a given Berry curvature you can globally choose a connection, so long as its integrals over closed surfaces are all $2\pi$-integers. It won't be unique though. – Ryan Thorngren Sep 13 '18 at 22:34
• The Chern number over the torus (or the sphere) is always an integer. And you can only choose a global section if and only if the Chern number is non-zero. Otherwise, you can only define the connection locally and have to glue them together in overlapping regions. – Max Lein Sep 15 '18 at 9:49
• You're not using the word connection properly. A connection is only locally a differential form anyway. And a closed 2-form is not necessarily integral. It is iff it's the curvature of a global connection. Sometimes you only have local connections, which is the case when there is a nontrivial B-field. – Ryan Thorngren Sep 15 '18 at 16:12
• I think I understand what you are saying: yes, you are right, I should have written that you need to distinguish the situations where you can express the connections globally or only locally in terms of coordinates. Thank you for the clarification. – Max Lein Sep 16 '18 at 20:13