F} \,  \left( E^i_x-\left( U^{{\rm}{Adj.}}_{x-1}\right)_{ji} E^j_{x-1} \right) {\cal F}^\dag = {\cal F} \rho^i_{F,x} \, {\cal F}^\dag.$$ This equation, in terms of $E^{'i}_x$, reads $$E^{'i}_x-\left( U^{{\rm}{Adj.}}_{x-1}\right)_{ji} E^{'j}_{x-1} = {\cal F} \rho^i_{F,x} \, {\cal F}^\dag = \sum_{ab}\, \lambda^i_{ab}\, {\cal F} \, \psi^{\dag a}_x \psi^b_x \, {\cal F}^\dag =  \sum_{ab \atop a'b'}\, \lambda^i_{ab}\,\, \psi^{\dag a'}_x \, \psi^{b'}_x \, V^\dag_{x,a'a} V_{x,bb'}.$$ Multiplying this equation by $\lambda^i_{cd}$, summing over $i$, and writing it in terms of the adjoint representation of the operator $V$, we get $$\sum_{i}\, \left( V^{\rm Adj.}_x \right)_{ik} \,  \left( E^{'i}_{x}-\left( U^{ {\rm}{Adj.}}_{x-1}\right)_{ji} E^{'j}_{x-1} \right) = \rho^k_{F,x}.$$ Finally, if we define $$E^{''k}_x=\left(V^{\rm Adj.}_x\right)_{ik} \, E^{'i}_x, \label{E''def}$$ and $$E^{'j}_x=\left(V^{\rm Adj.}_x\right)_{jk}\, E^{''k}_x,$$ then we get $$E^{''i}_{x}-\left( V^\dag_x \, U^\dag_{x-1}\, V_{x-1}\right)^{\rm Adj.}_{ij} E^{''j}_{x-1}  = \rho^i_{F,x}.$$ Using this gives $$E^{''i}_{x}-\left(e^{-i\varphi}\right)^{\rm Adj.}_{ij} E^{''j}_{x-1}  = \rho^i_{F,x}. \label{Glaw''}$$ is the starting point to the discussions in Sections \[GLresolve\]–\[0mode\] : first, in Section \[GLresolve\], we solve and write the fields $E^{''i}_x$ in terms of the color charge densities $\rho^i_{F,x}$. This cannot be done to all of the components of $E^{''}$, and a subset of these remains independent of the fermionic charge densities. The way we treat these remaining components is explained in Section \[0mode\].

Resolution of the Gauss law constraint {#GLresolve}
