Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 115 2 ii Solution Created 2026-10-03 Updated 2026-10-05
Using the left-invariant frame of the real Heisenberg group, horizontality givesFor an arbitrary starting point ,This is the control equation for the Heisenberg horizontal distribution. The final coordinate records a signed area-like integral, so changing the path in the plane can change the endpoint in the central direction.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 115 2 i Solution Created 2026-10-03 Updated 2026-10-05
A smooth rank- smooth distribution is involutive distribution if the Lie bracket of vector fields of two local sections remains a section. It is an integrable distribution if every point lies on an immersed -dimensional integral manifold with tangent spaces equal to . The Frobenius theorem says these conditions are equivalent, and gives local coordinates with .
For necessity, fields tangent to an integral manifold have brackets tangent to it: they annihilate functions vanishing on the manifold, and so does their commutator. For sufficiency, induct on . The rank-zero case is immediate. Straighten a nonvanishing local section using the flow-box theorem to obtain . Choose a frame with the having no component. Involutivity gives for the column of these fields and a smooth matrix . Solve the matrix ordinary differential equation , with on . It stays invertible, and satisfies . On the transverse slice, the span an involutive rank- distribution. The induction hypothesis supplies adapted slice coordinates; extend them independently of . These give the required rank- coordinate distribution and its integral manifolds.
For the real Heisenberg group, multiplication isDifferentiating left translation at the identity gives its left-invariant frame of the real Heisenberg group:Then and the other basis brackets vanish. Since , the Heisenberg horizontal distribution is not involutive, hence not integrable by the Frobenius theorem.