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Which means that   \begin{align}  [\rho_e, \left[ \rho_e,  e^{i m_J \phi_J}] \phi_J} \right]  &=& \sum_I [ \frac{\partial_x \phi_I}{2\pi}, e^{i m_J \phi_J}] \\ &=& - \sum_{I,L} K^{-1}_{IL} [ \Pi_L, e^{i m_J \phi_J} ] \\  &=& -\sum_{I,L,J} K^{-1}_{IL} \delta_{L,J} m_J e^{i m_J \phi_J} \delta(x-x') \\  &=& -\sum_{I,J} K^{-1}_{IJ} m_J e^{i m_J \phi_J} \delta(x-x') \\  &=& \sum_{I,J} t_I K^{-1}_{IJ} m_J e^{i m_J \phi_J} \delta(x-x')  \end{align}  which allows us to immediate conclude that the correct charge vector in our conventions is   $$ t_I = -1 $$  {\bf Current operators.}  We agree that $\rho_e = \rho_R + \rho_L = e \sum_I \left( \psi^\dagger_{R,I} \psi_{R,I} + \psi^\dagger_{L,I} \psi_{L,I} \right) = \frac{e}{2\pi} \sum_{I=1}^N \left( \partial_x \phi_I + \partial_x \phi_{N+I} \right)$.