@20] $$|k,q>=\sqrt{\frac{l}{2\pi }}e^{-i\tau _{1}\hat{y_{2}}^{2}/2\tau
_{2}}\sum_{j}e^{ijkl}|q+jl>,$$where the ket on the right is a $\hat{y_{1}}$ eigenstate. We have $$\begin{aligned}
U_{1}|k,q>= &&e^{-ilk}|k,q>,~~~~~~~U_{2}|k,q>=e^{il\tau
_{2}q}|k,q>=e^{2\pi
iqA/l }|k,q>,  \nonumber  \label{eq:7} \\
id &=&\int_{0}^{\frac{2\pi }{l}}dk\int_{0}^{l}dq|k,q><k,q|.\end{aligned}$$It also satisfies $$\label{eq:46}
|k,q>=|k+\frac{2\pi }{l},q>=e^{ilk}|k,q+l>.$$Consider the equation (\[eq:6\]), namely the standard expansion of operators on $T^{2}$ we have $$\label{eq:15}
\Psi
_{st}(u_{1}^{A},u_{2}^{A})|k,q>=\Psi _{st}(e^{-ilk},e^{-2\pi
iqA/l})|k,q>\equiv \psi _{st}(k,q)|k,q>,$$where $\psi _{st}$ is a function of the independent variables $k$ and $q$, called the eigen value function of $\Psi _{st}(u_{1}^{A},u_{2}^{A})$. From (\[eq:15\]), we see that the function $\psi _{st}$ is invariant when $q\rightarrow q+l/A$, $$\psi _{st}(k,q+\frac{ln}{A})=\psi _{st}(k,q).  \label{eq:16}$$As long as the eigen value function is obtained, the operator on the noncommutative torus can be completely determined. Introducing new basis $%
|k,q_{0};n>\equiv |k,q_{0}+\frac{ln}{A}>, k\in \lbrack 0,\frac{2\pi }{l}%
), q_{0}\in \lbrack 0,\frac{l}{A})$, we have from (\[eq:7\]) $$\sum_{n=0}^{A-1}\int_{0}^{\frac{2\pi }{l}}dk\int_{0}^{\frac{l}{A}%
}dq_{0}|k,q_{0}+\frac{ln}{A}><k,q_{0}+\frac{ln}{A}|=id,$$$$u_{1}|k,q>=|k,q+\frac{l}{A}>,  \label{eq:21}$$$$u_{2}|k,q>=e^{-2\pi i\frac{q}{l}}|k,q>.  \label{eq:22}$$From the above equation and (\[eq:46\]), we see that when any power of the operators $u_{1}$ and $u_{2}$ act on the $|k,q_{0}+\frac{ln}{A}>$, the result can be expanded in the basis $|k,q_{0}+\frac{ln^{\prime }}{A}>$ with the same $k,q_{0}$. So the operators on the noncommutative torus have the same property, namely don’t change $k$ and $q_{0}$. Thus, for every $k$ and $%
q_{0}$ we get a $A\times A$ matrix, called reduced 