satisfying both inequalities.

Using the previous Lemma we obtain the main result of this section.

\[thmnoninvert\] The only invertible parabolic maps $f$ have $M$ equal to one of $${\left(\begin{array}{cc} {k} & {-(k-1)^2/l} \\ {l} & {2-k}
\end{array}\right)},~~
{\left(\begin{array}{cc} {2+l} & {-k} \\ {(l+1)^2/k} & {-l}
\end{array}\right)},~~
{\left(\begin{array}{cc} {1} & {b} \\ {0} & {1}
\end{array}\right)}~~\mbox{ or }~~
{\left(\begin{array}{cc} {1} & {0} \\ {b} & {1}
\end{array}\right)},$$ where $(k,l)$ are integers and $b$ is real.

[[**Proof**]{} ]{}The square $V_\epsilon=(-1-\epsilon,1+\epsilon)^2$ is a convex region symmetrical about the origin with area $4(1+\epsilon)^2$. A variant of Minkowski’s theorem says that for any $\epsilon>0$ there will be a point on the lattice $\Lambda=\{(Kc-La,Kd-Lb)~:~(K,L)\in{{\bf Z}}^2\}$ other than zero inside $V_\epsilon$, if the lattice has determinant $\Delta=ad-bc\geq 1$. Taking the limit $\epsilon\rightarrow 0$ and using compactness we see that there must be a point $(\kappa,\lambda)\in \Lambda$ other than zero in $\overline{V_0}$.

If there is a lattice point $(\kappa,\lambda)$ in the interior of $V_0$, we are done. If not, assume without loss of generality that $(\kappa,\lambda)=(1,y)$ with $|y|\leq 1$. Thus exactly one point in the line segment from (but not including) $(-1,1-y)$ to $(0,1)$ that is in the lattice (one can generate the same lattice by adding multiples 