Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 2 / 14G (Geometry)

The characteristic polynomial of $A$ isWhen $|\mathrm{tr}A|>2$, its discriminant is positive, so $A$ has two distinct real eigenvalues $\lambda ,{\lambda }^{-1}$ and two independent real eigenvectors. Let $P$ have those eigenvectors as columns. Rescaling one column makes $\det P=1$ without changing the diagonalization. Hence $P\in S{L}_2(\mathbb{R})$ andThis is the normal form of a hyperbolic element of PSL2(R).

Write $B=\mathrm{diag}(\lambda ,{\lambda }^{-1})$ and $q={\lambda }^2>0$, so its Möbius transformation is $z\mapsto qz$, with $q\ne 1$. For $z=u+iv$, the Hyperbolic distance in the Poincare half-plane satisfiesAt $z=i$, the final parenthesis is $1$. Since $\cosh$ is strictly increasing on nonnegative arguments, every $z$ off the imaginary axis has

A fixed point of $A$ satisfiesIts discriminant iswhere $ad-bc=1$ was used. Thus the two roots are nonreal complex conjugates, and exactly one lies in the upper half-plane. Consequently every elliptic element of PSL2(R) fixes a point of $\mathbb{H}$.

Take the parabolic element of PSL2(R)It fixes no point in $\mathbb{H}$. A Geodesic in the Poincare half-plane model is determined by its unordered pair of ideal endpoints: either two real numbers or one real number together with $\infty$. Translation adds one to every finite endpoint, so no such pair is invariant. Therefore this $A$ preserves neither a point nor a hyperbolic line in $\mathbb{H}$.