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It looks just like the perfect metal action, but he has (eq 2.22) $\psi_R \propto e^{i k_F x} e^{i \phi_R}, \psi_L \propto e^{-i k_F x} e^{i \phi_L}$. That means our chiral fields are off by a minus sign:  $$ \phi_I \equiv \phi^R_I = - \phi_R^{(G))} $$  $$ \phi_{N+I} \equiv \phi^L_I =-  \phi_L^{(G))} $$ so finally I conclude that: