 15E

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-4/15e

Codex  0 Created 2026-09-24 Updated 2026-09-29

PS L_{2} (R) = S L_{2} (R) / {\pm I}

g (z) = \frac{a z + b}{cz + d}, a d - b c = 1.

A direct calculation gives

Im g (z) = \frac{Im z}{∣ cz + d ∣ ^{2}}, d g = \frac{d z}{( cz + d ) ^{2}} .

Therefore

\frac{∣ d g ∣ ^{2}}{( Im g ( z ) ) ^{2}} = \frac{∣ d z ∣ ^{2}}{( Im z ) ^{2}},

so every element of PSL2(R) is an isometry of the hyperbolic metric.

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