I am following notes on Berry phase and Chern number here. In section 3.2 they make the argument that the Chern number

$$Q=-\frac{1}{2\pi}\oint_S \mathbf{B} \cdot d\mathbf{S}$$

which is an integral over a closed surface $S$ in $k$-space, can be split up into two surface integrals $(S_1,S_2)$ joined by a closed contour $C$. They use a surface element with opposite orientation in each case so that this reads

$$Q=\oint_S \mathbf{B} \cdot d\mathbf{S} = \int_{S1} \mathbf{B} \cdot d\mathbf{S} - \int_{S2} \mathbf{B} \cdot d\mathbf{S}$$

Because the two surface integrals are just the Berry phase of the closed loop $C$ and a Berry phase is defined modulo 2$\pi$, they conlude that the Chern number is

$$Q = n_1 - n_2$$

where $n_1$ and $n_2$ are the integers that come from the two Berry phases calculated from the corresponding surfaces. Ok so I just about understand the reasoning that the Chern number is an integer in this case, but what I don't understand is how it is an invariant. Surely these two integers come from an arbitrary Gauge choice of the basis functions so they can be chosen how I like, meaning the Chern number can be any integer. I know this is not the case but I don't see how to deduce it from the logic followed in these notes. Any clarifications gratefully receieved!

  • $\begingroup$ not all kinds of gauge choice is allowed. you need to choose the gauge which makes the vector potential is well defined everywhere in your integral and then you can use Stokes' theorem. That's why people say that there is an obstruction to finding a smooth gauge for topological nontrivial bundles. $\endgroup$
    – FangXie
    Dec 14 '18 at 22:53

Okay so here's my interpretation:

$\vec B$ is defined in terms of the vector potential $\vec A$. Changing Gauge takes $\vec A$ to $\vec A -\nabla\alpha$. (look on page 22 of the link you sent) Then $\vec B$ is defined as $\nabla\times\vec A$, and is invariant under the transformation because the curl of a gradient is always zero. So the B-field is an invariant. $Q$ is defined in terms of $\vec B$, so $Q$ is an invariant.

  • $\begingroup$ Welcome to the site! Good username ;) $\endgroup$
    – Kyle Oman
    Feb 23 '18 at 7:48

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