the commutation relations of $E^{'i}_x$ and the operator $$\left(P_y\right)_{ab} = \left( U_1 \, U_2 \cdots U_{y} \right)_{ab}.$$ The result is easy to obtain and can be written in terms of the operators $S$ and $\varphi$ (see ), and the operator $V$ (), $$\left[E^{'i}_x,\left(P_y\right)_{ab}\right] = \theta(y-x) \left( S^\dag \, e^{i\varphi \, x} V^\dag_x \lambda^i \, V_x e^{-i\varphi \, x } S P_y  \right)_{ab}.$$ Here $\theta(x)= 1$ if $x\ge 0$ and zero otherwise. Next we use the definition of $E^{''I}$ in and after some algebra find that (here note that we focus only on the Cartan sub-algebra) $$\left[E^{''I}_x,\left(P_y\right)_{ab}\right] = \theta(y-x) \left( S^\dag \, \lambda^I \, S P_y  \right)_{ab}.\label{comm1}$$ Since the dependence on the site index $x$ in the r.h.s. of is trivial, we can easily write the commutation relation of the zero mode $E^{''I}_{p=0}=\frac1{\sqrt{L_s}} \, \sum_x E^{''I}_x$, and we find $$\left[E^{''I}_{p=0},\left(P_y\right)_{ab}\right] = \frac{y}{\sqrt{L_s}} \, \left( S^\dag \, \lambda^I \, S P_y  \right)_{ab}.\label{comm2}$$ In particular, for $y=L_s$, we use and get $$\left[E^{''I}_{p=0},P_{ab}\right] = \sqrt{L_s} \, \left( S^\dag \, \lambda^I \, e^{i\varphi L_s} S  \right)_{ab}.\label{comm3}$$ If we now define ${\cal E}_a\equiv \sum_I \, \lambda^I_{aa}E^{''I}_{p=0}$, we get $$\begin{aligned}
\left[{\cal E}_c,\left(S^\dag e^{i\varphi L_s}S\right)_{ab}\right]&=&\frac12 \, \sqrt{L_s} \left[\, S^{\dag}_{ac} e^{i\varphi_c L_s} S_{cb} -\frac1{N} \, P_{ab}\right].\label{comm4_1}\end{aligned}$$ In Appendix \[gauge\_fixing\] 