As a result, $W_{TEK}({\cal C})$ is zero if $\cal C$ is an open contour in the weak coupling limit.

A key point is that the solution for this problem is obvious only at the classical level. That is to say, there is no guarantee to maintain the ${\mathbb Z}_L^4$ symmetry if we take into account the quantum fluctuation. Going away from the weak coupling limit, the configurations fluctuate around the twist-eater. The situation can be displayed in the eigenvalue distribution of the link variables. In the weak coupling limit the $N$ eigenvalues distribute regularly and uniformly on the unit circle in the complex plane, and then they are ${\mathbb Z}_L$ symmetric. If we decrease $\beta$, the eigenvalues begin to fluctuate around the location of the twist-eater. If the fluctuation is not too large, the ${\mathbb Z}_L$ symmetric distribution is maintained. However, large fluctuation can make the uniform distribution shrink to a point, which corresponds to $U_\mu=\mathbbm{1}_N$ configuration. In the strong coupling region the distribution is randomly uniform, and then the symmetry is restored.

Although there is no guarantee to maintain the ${\mathbb Z}_N^4$ symmetry in the intermediate coupling region, the 1980s numerical simulations suggested that the symmetry was unbroken. And 