Past exam of the mathematics course of the University of Cambridge 2019 ib Paper 4 15E Created 2026-09-24 Updated 2026-09-29
The groupacts on the Poincare half-plane model byA direct calculation givesThereforeso every element of PSL2(R) is an isometry of the hyperbolic metric.
Past exam of the mathematics course of the University of Cambridge 2021 ib Paper 2 11E Solution Created 2026-09-24 Updated 2026-09-25
The Poincare half-plane model iswith the orientation inherited from the complex plane. Forone hasThese identities show directly that . The map is holomorphic with nonzero derivative, so it preserves orientation. The matrices and induce the same map, giving the action of PSL2(R).
Conversely, let be an orientation-preserving isometry. The PSL2(R) action is transitive on , so compose with an element taking back to . The resulting isometry fixes and acts on by an orientation-preserving orthogonal map, hence a rotation. The stabilizer of in PSL2(R),realizes every such tangent rotation. An isometry is determined by its value and differential at one point because it preserves geodesics and the exponential map. Thus the composed isometry belongs to PSL2(R), and so does .
The map is an orientation-reversing isometry. Composing any orientation-reversing isometry with gives an orientation-preserving one, so
A hyperbolic line is a vertical Euclidean line or a semicircle orthogonal to the real axis. Its hyperbolic reflection is the unique orientation-reversing isometry that fixes every point of . If meet at angle , thenis the hyperbolic rotation about through angle . The generators satisfyIf in lowest terms, then has order and the generated group is the finite dihedral group of order . If is irrational, has infinite order and the generated group is the infinite dihedral group. In the degenerate case , the group has order two.