[x_0,x+2\pi]$, that is, $c_{t,x}=e_{t,x}^{-1}\circ e'_{t,x}$. Hence we have by basic properties of the holonomy \[3.20\] h\_[t,x]{}(A\^):=(h\_[e\_[t,x]{}]{}(A\^))\^[-1]{} h\_[e’\_[t,x]{}]{}(A\^) By the very definition of the holonomy of a connection along a path (\[3.18a\]) we find with the parameterization $c_{t,x}(s)=x+s,\;s\in 
[0,2\pi)$ \[3.21\] \_x h\_[t,x]{}(A\^)=\[h\_[t,x]{}(A\^),A\_x(t,x)\] On the other hand we have \[3.22\] \_t h\_[t,x]{}(A\^) &=& \_[0]{} (h\_[t+,x]{}(A\^)-h\_[t,x]{}(A\^))\
&=& \_[0]{} (h\_[e\_[,t,x]{}]{}(A\^)\^[-1]{}\[h\_[e\_[,t,x]{}]{}(A\^) h\_[t+,x]{}(A\^)\]-h\_[t,x]{}(A\^))\
&=& \_[0]{} (h\_[e\_[,t,x]{}]{}(A\^)\^[-1]{}\[h\_[t,x]{}(A\^) h\_[e\_[,t,x]{}]{}(A\^)\] -h\_[t,x]{}(A\^))\
&=& \[h\_[t,x]{},A\^\_t(t,x)\] where $e_{\epsilon,t,x}(s)=(t+s,x),\;s\in[0,\epsilon]$. Here we have used that the loop $c_{t,x}\circ e_{\epsilon,t,x} \circ c_{t+\epsilon,x}^{-1}
\circ e_{\epsilon,t,x}^{-1}$ is contractible and the zero curvature condition in the third step.

Conversely, postulating equations (\[3.21\]) and (\[3.22\]) we discover, using the Jacobi identity \[3.23\] 2\_[\[]{} \_[\]]{} h\_[t,x]{}(A\^) =\[h\_[t,x]{}(A\^),F\^\_[,]{}\] where $F^\pm$ is the curvature of $A^\pm$. Hence the zero curvature condition is the integrability condition for the equations $\partial_\alpha h_{t,x}(A^\pm)=[h_{t,x}(A^\pm),A^\pm_\alpha(t,x)]$.

We now claim that for any $n\le N$ the functions \[3.24\] T\^n\_[t,x]{}(A\^):=\_N(\[h\_[t,x]{}(A\^)\]\^n) are independent of both $x,t$. This follows immediately from (\[3.21\]) and (\[3.22\]) as well as the cyclicity of the trace. What is, however, even more remarkable is that (\[3.24\]) has vanishing Poisson brackets with $H(u,v)$ [*for all $u,v$*]{} even though $A^\pm_t=(v\pm u) A^\pm_x$ depends explicitly on $u,v$. The reason for this is that $A_x^\pm$ is actually independent of $u,v$ and 