Curvature and edge state

Question

If the boundary of quantum hall fluid has non-constant curvature, how will it affect the edge state which is usually described in chiral Luttinger fluid?

Answer 1

A naive guess would be that there isn't any real difference.

The theoretical logic behind my guess is that at low-energies FQH states are described by 2+1D Chern-Simons theories, which are topological gauge theories. Although the bulk does not have any local degrees of freedom, the boundary does. This is because in the presence of a boundary $\partial M$ one has to impose boundary conditions and reduce the set of gauge transformations to those that respect this BC, therefore there will be an infinite number of states which are not gauge equivalent anymore and therefore correspond to physical degrees for freedom. More formally, the boundary dynamics are described by a Wess-Zumino-Witten theory which I think is nothing but a chiral Luttinger liquid in the simplest case. Now this is a conformal field theory and only depend on the conformal class of the boundary metric, not the metric itself. 2D manifolds, like the boundary $\partial M$, are however all conformally flat and therefore the boundary dynamics are insensitive to the curvature.

This robustness against boundary curvature, impurity scattering and so on is a general feature of quantum Hall states. If you have access to Nature, see (here and here) the simulations done for photonic crystal analogs of IQHE boundary states. It is here seen that the light wave goes around any boundary defect, curvature or impurity without any reflections at all. It is quite non-intuitive that light can go around a mirror without any reflection!