to investigate other phases, one has to solve the third order equation with respect to $\theta_1$, $$\left(1+{{t\alpha}\over{24\pi^2}}\right)\theta_1^3
-\left({{t\alpha}\over{8\pi}} +6\pi\right)\theta_1^2
+\left({{t\beta}\over{24}}-\mbar^2+12\pi^2\right)\theta_1
+{{t\gamma}\over{24}}\pi +2\pi\mbar^2 -8\pi^3=0,
\label{eqevendt}$$ where $\alpha, \beta$ and $\gamma$ are constants and any solution to Eq.(\[eqevendt\]) has to lie in the range $$\begin{aligned}
\pi \le \theta_1 \le 2\pi ,
\label{restricteven}\end{aligned}$$ as shown in the appendix. This equation has the very different structure from that for $N=$ odd (See Eq.(\[cohiggsu\]) in the appendix). It does not have the solution of $\theta_1=2\pi$ for any values of $t$ and $\mbar$. This implies that the Higgs phase $${\rm type~~II}~~\cdots~~\left\{\begin{array}{l}
gL\vev{A_y}={\rm diag}\left(0, {{N-2}\over{N-1}}\pi,\cdots,
{{N-2}\over{N-1}}\pi, -{{(N-2)^2}\over{N-1}}\pi\right),\\[0.3cm]
\vev{\Phi}={1\over\sqrt{2}}(\sqrt{2\over\lambda}m,0,\cdots, 0)^T,
\end{array}\right.
\label{higgsd}$$ does not exist unlike the case $N=$ odd. The Higgs phase can be realized only in the limit $L \rightarrow \infty$ (or $\bar{m}\rightarrow \infty$), as we will see later. The $SU(N)$ gauge symmetry is broken to $SU(N-1)\times U(1)$ by the Wilson line and the Higgs VEV breaks the $U(1)$, so that the residual gauge symmetry is $SU(N-1)$ for the configuration (\[higgsd\]).

In order to see that the vacuum configuration is expected to approach the Higgs phase in the limit $\bar{m}\rightarrow \infty$, let us first note that the classical Higgs potential dominates the effective potential in the 