of $(1,y)$ to any lattice point generating the lattice with $(1,y)$). Therefore there will be a point in the lattice in the interior of $V_0$ unless this other point is $(0,1)$. Therefore (considering the other case by interchanging $x$ and $y$) there will be integers $K$ and $L$ such that either $$(Kc-La=1~~\mbox{ and}~~Kd-Lb=0)~~~\mbox{ or }~~~
(Kc-La=0~~\mbox{and}~~Kd-Lb=1).$$ Thus, in the invertible case the matrix $M$ is determined by which of these cases occurs. One can easily solve to show that either of these cases can occur. If $k=1$ or $l=-1$ then the other (horocyclic) cases occur. [\
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It follows from this result that most parabolic maps are noninvertible in the following sense:

The set of noninvertible maps is open and dense within the set of parabolic maps.

Because the $(\bmod ~1)$ map $g$ commutes with integer matrix and $g^2 = g$, all conjugations of parabolic maps with integer coefficients by automorphisms of the torus ${{\bf T}}^2$ remain integer parabolic maps. For an integer parabolic map with rational $\alpha = r/s$, where $(r, ~s) = 1$, since there exist $m,~n \in {{\bf Z}}$ such that $mr+ns=1$, there exists an automorphism whose $GL(2,{{\bf Z}})$ matrix has bottom row $(r,~s)$. Conjugation by this 