Why is the flux quantized in 4D quantum Hall effect? I am reading "Topological Field Theory of Time-Reversal Invariant Insulators" by Qi, Hughes, and Zhang (https://arxiv.org/abs/0802.3537). It argues that time reversal invariant (TRI) insulators in 2+1 and 3+1 dimensions are descendants of the fundamental TRI insulator in 4+1 dimensions. When it discusses quantum hall effect in 4D, the current is Eq. (58)
$$ \int dx dy \,j_w = \frac{C_2 N_{xy}}{2\pi}E_z$$
where $N_{xy}=\int dx dy \, B_z / 2 \pi$. It says that the flux quanta $N_{xy}$ is always quantized to be an integer. It may be a trivial question. Why is it quantized? Is this related the periodic boundary conditions in $x,y$ directions?
 A: In the integer quantum Hall effect  geometry, the magnetic flux through a two dimensional surface, on which the  electrons are confined to the lowest Landau level (LLL) should be quantized, even if the surface is noncompact. The reason is that in very strong magnetic fields, in the projected dynamics to the lowest Landau level, the coordinates become noncommutative:
$$[x, y] = i \frac{\hbar c}{e B_z}$$
Please see for example the following work by Richard Szabo (equation 14). The density of states per unit area of this system is just $(2 \pi)^{-1}$ times the reciprocal of the right hand side:
$$\rho =(2 \pi)^{-1} \frac{ e B_z }{ \hbar c}$$
Thus the number of states on the surface is given by:
$$N = \int \rho dxdy$$
Since this number counts states in quantum theory, it should be quantized. (Qi, Hughes and Zhang very briefly mention this argument on page 15 after equation 70).
(This argument is completely analogous to the quantization of the number of states in the case of a free particle; Here we have $[x, p] = i \hbar$ and the number of states equal to the phase space volume in units of the Planck's constant: $N = \frac{\int dx dp}{2 \pi \hbar}$).
