Can local dissipation prepare a pure chiral topological state?
In plain words
Engineered loss can pump atoms into a chosen state, like water settling at the bottom of a bowl. It is unknown whether this works for states with a built-in handedness, such as the quantum Hall states, using only short-range processes.
Precise statement
Fermions or spins on a 2D lattice evolving under a Lindblad master equation with finite-range jump operators and finite-range Hamiltonian. Determine whether a pure state with nonzero Chern number (or nonzero Hall conductance) can be the unique steady state with a Liouvillian gap that stays finite as $N \to \infty$. Known obstructions: quadratic (Gaussian) local Lindbladians cannot do it (Budich, Zoller, Diehl, PRA 2015), and local commuting-projector Hamiltonians have zero Hall conductance (Kapustin, Fidkowski, CMP 2019). Answer: a general no-go proof covering interacting Lindbladians, or an explicit local Lindbladian that works.
What would settle it
A theorem extending the Gaussian obstruction to interacting finite-range Lindbladians, or a constructed counterexample with a proven size-independent Liouvillian gap.
Status in the literature
Approximate dissipative schemes for Chern and fractional Chern insulators exist (Liu, Bergholtz, Budich, Phys. Rev. Research 2021), but none gives an exact pure chiral steady state with a finite gap.