I'm currently trying to understand this paper [1], in which the authors give a rigorous proof of the Laughlin's charge pump argument for the quantum hall effect.
In Lemma 4.4, it proves that if $u:\mathbb{C}\backslash 0\to\mathbb{S}^1\subseteq \mathbb{C}$ has winding number $N(u)$, e.g., $re^{i\theta} \mapsto e^{i\theta}$, then $$ \int_{\mathbb{R}^2}dx \left(1-\frac{u(x-a)}{u(x-b)} \right) \left(1-\frac{u(x-b)}{u(x-c)} \right)\left(1-\frac{u(x-c)}{u(x-a)} \right) = 2\pi i N(U) \text{Area}(a,b,c) $$ where $a,b,c\in \mathbb{R}^2$ and $\text{Area}(a,b,c)=a\wedge b +b\wedge c +c\wedge a$ (e.g., $a\wedge b = a_1b_2-a_2 b_1$).
For the moment, let us not care about the details of convergence and consider a sufficiently large disc $D$ which contains $a,b,c$ so that the integral over $\mathbb{R}^2$ is replaced by over $D$. Then the author shows that $$ (\partial_{a_1} \partial_{b_2} - \partial_{a_2} \partial_{b_1})C(a,b,c)= 2i \Im\int_{\partial D} \bar{u}(x-a)du(x-b) = 4\pi iN(u) $$ Up to this point, I understand the calculations. The author then claims that these are the only non-vanishing second derivative (except for cyclic permutations) and $C(a,b,c)$ can be reconstructed if we integrate the above equation twice.
There are 2 things that I don't quite understand.
- Why do the other second derivatives (e.g. $\partial_{a_1} \partial_{a_2} C =0$) vanish?
- Accepting that the other second derivatives indeed vanish, how can $C(a,b,c)$ be reconstructed by integrating twice?
[1] Avron, J.E., Seiler, R. & Simon, B. Charge deficiency, charge transport and comparison of dimensions. Commun.Math. Phys. 159, 399–422 (1994).