as in (\[path1\]). However, these symmetries are preserved if we average over the $N$ cyclic permutations, as well as the $N$ permutations that are obtained from them by reversing the order, [*i.e.*]{} by combining them with the permutation $\rho: j \mapsto N+1-j$. We thus define $$\begin{aligned}
\label{path2}
{\cal O}_{\rm reg}^{(N)}(R) =
{\rm Tr}_R\; (t^{a_1}&&
\hskip -0.5cm \cdots t^{a_N}) \left( \prod_{i=1}^{N} \int_0^{2\pi} 
d\sigma_i \right) \;\theta_{\sigma_1>\cdots >\sigma_N}\times\\ 
\nonumber
&& \qquad \qquad
\times {1\over 2N}  \left( J_{\rm reg}^{a_1}(\sigma_1)\cdots 
J_{\rm reg}^{a_N}(\sigma_N)\;
+ {\rm cyclic}\ + {\rm reversal} \right)\ .\end{aligned}$$ Note that since the bare currents at non-coincident points commute, the choice of ordering is part of the regularisation prescription for the loop operator. Note also that our prescription (which is not unique) guarantees that $ {\cal O}_{\rm reg}^{(N)}(R)$ commutes with the generator of $\sigma$-translations. As explained in the previous section, such chiral operators can be transported freely to the boundary of the half-infinite cylinder.

Plugging the mode decomposition (\[jreg\]) in (\[path2\]), and performing explicitly the integrals leads to the following expressions for the first few values of $N$ : $$\label{R2}
{\cal O}_{\rm reg}^{(2)}(R)  = 2\pi^2\;
{\rm Tr}_R(t^a t^b)\;  J_0^a J_0^b\ ,$$ $$\label{R3}
{\cal O}_{\rm reg}^{(3)}(R) =   {2\pi^2\over 3} \; {\rm Tr}_R(t^a
t^b t^c) \left[ 