\\
& 0 &  &  \\
&  & 0 &  \\
&  &  & \ddots%
\end{array}
\right) s(k,q_0),  \nonumber$$

$$\label{eq:8}
M(k,q_0)_{nn^{\prime}}=s^{-1}(k,q_0)_{n0}
s(k,q_0)_{0n^{\prime}}\equiv a(k,q_0)_{n}b(k,q_0)_{n^{\prime}}.$$

Explicit calculation about $R_{N}$ acting on $|k,q;n>$ shows that we can divide the complete area $\Sigma :\{k\in \lbrack 0,2\pi
/l),q_{0}\in \lbrack 0,l/A)\}$ into $N$ subarea $\sigma
_{0},\cdots ,\sigma _{N-1}$,making $W:\sigma _{i}\rightarrow
\sigma _{i+1},(i=0,1,\cdots ,N-2),\sigma _{N-1}\rightarrow
\sigma _{0}.$ If we construct a reduced matrix $M(k,q_{0})$ to satisfy ([eq:8]{}) in the area $\sigma _{0}$, then the projector corresponding to continuous $\psi _{st}$ with minimal trace is completely determined. In area $\sigma _{0}$, set $$a_{n}=<k,q_{0}+\frac{ln}{A}|\phi _{1}>,~~~~~~~~b_{n}=<\phi _{2}|k,q_{0}+%
\frac{ln}{A}>,  \label{eq:9}$$where $$\sum_{n}a_{n}b_{n}=trM(k,q_{0})=1.$$In the other areas $\sigma_{j}$ with $(k,q_{0})\rightarrow
(k_{j},q_{0j})$ by mapping $W^{j}$, we demand $$\begin{aligned}
a_{n}(k_{j},q_{0j}) &=&<k_{j},q_{0j}+\frac{ln}{A}|\phi _{1}>  \nonumber \\
&=&A^{j}(k,q_{0})_{nn^{\prime }}a_{n^{\prime }}(k,q_{0}), \\
b_{n}(k_{j},q_{0j}) &=&<\phi _{2}|k_{j},q_{0j}+\frac{ln}{A}>  \nonumber \\
&=&b_{n^{\prime }}(k,q_{0})A^{-j}(k,q_{0})_{n^{\prime }n}.\end{aligned}$$We thus have all coefficients of $|\phi_{1}>, <\phi_{2}|$ in $\sigma_{0},\cdots,\sigma_{N-1}$. Owing to the completeness of $|k,q_{0}+\frac{ln}{A}>$ in the area $%
\Sigma $, $|\phi _{1}>$ and$<\phi _{2}|$ can be determined by the coefficient (\[eq:9\]) of $|\phi _{1}>$ and$<\phi _{2}|$. Meanwhile, in the area $\sigma _{j}$, we have $$\begin{aligned}
M(k_{j},q_{0j})_{nn^{\prime }} &=&a_{n}(k_{j},q_{0j})b_{n^{\prime}}(k_{j},q_{0j})  \nonumber \\
&=&[A^{j}(k,q_{0})M(k,q_{0})A^{-j}(k,q_{0})]_{nn^{\prime }}.\end{aligned}$$The matrix $M(k,q_{0})$ really satisfies the equation (\[eq:10\]). Consider the state vector $$\begin{aligned}
|\phi _{1}>&= &\int dkdq_{0}\sum_{n}|k,q_{0}+\frac{ln}{A}><k,q_{0}+\frac{ln}{%
A}|\phi 