matrix for the operator, as well as the projector: $$\label{eq:23}
P_{T^{2}}|k,q_{0}+\frac{ln}{A}>=\sum_{n^{\prime
}}M(k,q_{0})_{n^{\prime }n}|k,q_{0}+\frac{ln^{\prime }}{A}>,$$It is easy to find that the sufficient and necessary condition for $P^{2}=P$ is [@9] $$M(k,q_{0})^{2}=M(k,q_{0}).$$When $T^{2}$ satisfies $Z_{N}$ symmetry, since after $R_{N}$ rotation $%
U_{i}^{\prime }$ can be expressed by monomial of $\{U_{i}\}$ and their inverses, the state vector $R_{N}|k,q_{0}+\frac{ln}{A}>$ is still the common eigenstate of the operators $U_{1}$ and $U_{2}$. With the completeness of $%
\{|k,q+\frac{ls}{A}>\}$ and the A-fold degeneracy eigenvalues of $U_{i}$ in the $kq$ representation, the state can be expanded in the basis $\{|k^{\prime },q^{\prime }+\frac{ls^{\prime }}{A}>\}$ $$\label{eq:17}
R_{N}|k,q_{0}+\frac{ln}{A}>=\sum_{n^{\prime
}}A(k,q_{0})_{n^{\prime }n}|k,q_{0}+\frac{ln^{\prime }}{A}>$$[^4]where $k^{\prime }\in \lbrack 0,2\pi /l),q^{\prime }\in \lbrack
0,l/A)$ are definite and $$\label{eq:18}
R_{N}^{-1}|k^{\prime },q_{0}^{\prime }+\frac{ln^{\prime }}{A}%
>=\sum_{n"}A^{-1}(k,q_{0})_{n"n^{\prime }}|k^{\prime },q^{\prime }_{0}+\frac{ln"}{A}>.$$We can get the expression for the relation between $k^{\prime }$,$%
q_{0}^{\prime }$ and $k$, $q_{0}$, The mapping $W:(k,q_{0})\longrightarrow (k^{\prime },q_{0}^{\prime
}),W^{N}=id$, is essentially a linear relation, and area-preserving. By this fact and since $R_{N}$ is unitary, we conclude that the matrix $A$ is a unitary matrix, that is to say $$A^{\ast }(k,q_{0})_{nn^{\prime }}=A^{-1}(k,q_{0})_{n^{\prime }n}.
\label{eq:19}$$The projector on the noncommutative orbifold $T^{2}/Z_{N}$ satisfies $%
R_{N}^{-1}PR_{N}=P$, then from (\[eq:23\])(\[eq:17\])(\[eq:18\]) one obtains $$\label{eq:35}
R_{N}^{-1}PR_{N}|k,q_{0}+\frac{ln}{A}>=\sum_n^{'}[A^{-1}(k,q_{0})M(k^{\prime
},q_{0}^{\prime })A(k,q_{0})]_{n^{'}n}|k,q_{0}+\frac{ln^{'}}{A}>,$$ which should be equal to : $$\label{eq:36}
P|k,q_{0}+\frac{ln}{A}>=\sum_{n"}M(k,q_{0})_{n"n}|k,q_{0}+\frac{ln"}{A}>.$$So, 