Short version: Is it possible to arrange the fluxes for the Kagomé lattice with triangle flux $\phi_\triangle=\frac{\pi}2$ and hexagon flux $\phi_{hex}=0$ using a single unit cell?
Longer version: I am looking at fermionic mean field theories on the Kagomé lattice that describe a chiral spin liquid state for spin-1/2. Skipping the derivation (in 1 and 2), the mean field Hamiltonian is
$H = -\sum\limits_{\langle i, j\rangle,\sigma}~\rho~e^{iA_{ij}}~f^\dagger_{i\sigma}~f_{j\sigma} + H.c.$
This basically says that $f$-fermions hop along nearest-neighbor links, picking up a phase $A_{ij}$ one way along the bond, and $-A_{ij}$ hopping in the reverse direction. Phases (aka gauge field) are assigned to each lattice link in a unit cell, often depicted with arrows. The flux $\phi$ is defined as the sum of phases on a closed lattice plaquette along, say, the counter-clockwise direction.
In Marston's paper (Ref. 1), and Ran's paper (Ref. 2), they introduce a state with triangle flux $\frac{\pi}2$ and hexagon flux 0 (equivalent to Chua's (Ref. 5) (d)), and SL-[$\frac{\pi}2$,0] in Kim's diagram (Ref. 6).
My problem is with that with Kim's (Ref. 6) phase assignment for the triangles, I don't see how it is possible to achieve zero (or 2$\pi$) hexagon flux, unless one uses a doubled magnetic unit cell, as is claimed in Chua's work (Ref. 5). Ran (Ref. 2) doesn't explicitly state whether the unit cell has to double or not. Am I missing something here?
