Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 3 6E Solution Created 2026-09-24 Updated 2026-10-07
Write a member of the special linear group as with . Its Möbius transformation is , with the usual values at a pole and at infinity. For nonreal ,This follows by multiplying numerator and denominator by . Thus the sign of the imaginary part is preserved, while the extended real line is preserved as a set.
Fixing zero requires . Therefore its stabilizer subgroup and group orbit areTranslations send zero to each finite real point, and sends it to infinity.
For , comparing real and imaginary parts gives and , with . Exactly the same conditions follow from . HenceFor any and , the upper triangular matrixsends to and to . Therefore the two group orbits are the open complex upper half-plane and the open lower half-plane respectively. Along with the extended real line they exhaust the Riemann sphere, so there are exactly three orbits, the real determinant-one Möbius orbits.
Every has the form and acts by . Thus is the entire complex upper half-plane, by the same explicit matrices . Given , choose such that . Then fixes , so and . This proves the triangular-rotation factorization of real determinant-one matrices.
For uniqueness, if , then . An upper triangular rotation matrix must have , soThere are consequently exactly two matrix factorizations, and . Restricting the first diagonal entry of to be positive makes the factorization unique. For example, if and , that unique branch isAngles are understood modulo when counting matrices; allowing unrestricted real angle representatives gives infinitely many labels for those same two factorizations.