\qquad :\qquad Q^I&\equiv& \sum_x\, \psi^\dag_x \, \lambda^I \, \psi_x = 0.\label{QI}\end{aligned}$$ The $N-1$ requirements of can be written as $$\forall a=1,\dots,N\quad : \quad\sum_x \, \psi^{\dag a}_x \psi^a_x = {\rm independent \,\, of \,\,} a,$$ and since the baryon number $B$ is equal to $\frac1{N}\sum_{x,a} \psi^{\dag a}_x\psi^a_{x}-L_s$ (see ) then becomes $$\sum_x \, \psi^{\dag a}_x \psi^a_x = B+L_s, \qquad \forall a=1,2,\dots,N.\label{BLs}$$

For the rest of the generators $i\neq I$ we can indeed resolve Gauss law even for zero momentum : multiplying by $\lambda^i_{ab}$ with $a\neq b$, and defining $E^{''}_{x,ab}=\sum_{i\neq I} \, \lambda^i_{ab}\, E^{''i}_x$, and $\rho_{F,x,ab}=\sum_{i\neq I} \, \lambda^i_{ab}\, \rho^{i}_{F,x}$, one finds that the Fourier components of $E^{''}_{x,a\neq b}$ obey $$E^{''}_{p,ab} = \frac{\rho_{F,p,ab}}{1-e^{-i(p+\varphi_a - \varphi_b)}}.\label{Glaw''1}$$ (here the Fourier transformations of $E^{''}_{x,a\neq b}$ and $\rho_{x,a\neq b}$ are defined in a way similar to Eqs. (\[FTE”\]–\[FTrho\])) Note that we work in a Schröedinger picture where $\varphi_a$ is a $c$-number. Also, in we assume the absence of states with $\varphi_a-\varphi_b+p=0$. Since $p$ is quantized in units of $2\pi/L_s$, this assumption will indeed become true – see the discussion below on the importance of the Jacobian of the curvilinear coordinates $\varphi$.

We can now use the relations between $E^{''}$ and $\rho_F$, and write the electric Hamiltonian 