)G_{00}(u,v)$$is unequal to zero for any real variables $u$ and $v$. It is easy to prove that when $f_{1}$ is equal to $f_{2}^{\ast }$, the related denominator $$D=A\sum_{n}<k,q+\frac{ln}{A}|\phi ><\phi |k,q+\frac{ln}{A}>=A\sum_{n}|<k,q+%
\frac{ln}{A}|\phi >|^{2}.$$Thus if $D=0$,then $$<k,q+\frac{ln}{A}|\phi >=0~~~~~~~~n=0,1,\cdot \cdot \cdot ,A-1.
\label{eq:11}$$The zero points of the state vector $|\phi >$ in $|k,q>$ representation should be points equally spaced along $q$ with interval of $%
\frac{l}{A}$. The mapping from $k$ and $q$ to $<k,q+\frac{ln}{A}|\phi >\in C$ is a mapping from plane to plane. In general, $<k,q+\frac{ln}{A}|\phi >=0$ are some discrete points, and thus it is casual that $D$ is equal to zero. So in this sense, for most of $f_{1}=f_{2}^{\ast }$, this still not happen (in some sense, the measure of $D=0$ event is zero.) Specially, when the state $|\phi _{1}>=|\phi _{2}>=|0>$, It can be proved [@16] that $D$ is not equal to zero everywhere. Thus set $$|\phi _{1}>=|0>+\epsilon |\psi _{1}>,~~~~~~~<\phi
_{2}|=<0|+\epsilon <\psi _{2}|$$$D$ is also not equal to zero everywhere for small enough $\epsilon $. But we don’t know the situation for general $f_{1}\neq f_{2}^{\ast }.$

[99]{} A. Connes, Non-commutative Geometry, Academic Press, 1994.

G. Landi,“ An introduction to non-commutative space and their geometry”, hep-th/9701078; J. Varilly, “An introduction to non-commutative Geometry”, 