Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-50/3/solution
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 50 3 Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
Acting on the column gives the faithful Matrix Lie group representationMultiplication and inversion areThus is the real affine group. Differentiating the matrices at the identity gives its Lie algebraWiththe matrix commutator is , and .
The left Maurer-Cartan form and its right analogue areFor a constant , left translation leaves unchanged, and right translation leaves unchanged. Taking their dual vector fields givesThe first pair consists of left-invariant vector fields; the second consists of right-invariant vector fields. These are the invariant frames of the real affine group. Computing their Lie brackets of vector fields yieldsAll brackets of a field with itself vanish. The sign difference is necessary: evaluation at the identity identifies left-invariant vector fields with the matrix Lie algebra as a homomorphism, whereas right-invariant vector fields realize the opposite Lie algebra. Equivalently, is a homomorphism for the same matrix commutator. There is no convention in which these particular coordinate fields both have the positive structure constant while retaining the usual Lie bracket of vector fields.
The Maurer-Cartan equation gives the same check. If , thenFor , insteadThus the left equation is and the right equation is , explaining the opposite bracket signs directly.
Finally, the two given matrix curves are and . Their left action on a general group element isDifferentiating at gives and respectively. Hence Left translations on a Lie group are generated by right-invariant vector fields. More generally, the velocity of is , which is right-invariant because left and right multiplication commute. In the other direction, the flows generate left-invariant vector fields and are right translations on a Lie group.
New to topics? Read the docs here!