Closed-subgroup theorem 2026-10-14
Every closed subgroup of a finite-dimensional Lie group is an embedded submanifold and a Lie subgroup. Closedness alone forces the compatible smooth structure; smoothness need not be assumed in advance. The proof constructs the Lie algebra of a closed subgroup from one-parameter subgroups, then uses local product coordinates for a closed subgroup.
Choose a vector subspace complementary to the Lie algebra of a closed subgroup . The differential of at is , so the inverse function theorem makes a local diffeomorphism. Sufficiently small nonzero cannot satisfy : otherwise limit directions of a closed subgroup give a nonzero element of . Consequently is locally exactly the coordinate slice , proving that it is an embedded submanifold.
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 1 5 b Solution Created 2026-10-03 Updated 2026-10-14
For , we prove closure under addition using the Lie product formula in a Lie group. In the local exponential chart, let . Smooth multiplication has differential at the identity, so Taylor's theorem givesFor each fixed real and all sufficiently large positive integers ,Every term belongs to , and is closed, proving . Thus is a real vector subspace. It is also closed: if with , continuity of the Exponential map of a Lie group gives for each .
For the Lie bracket, use the Adjoint representation of a Lie group. Conjugating a one-parameter subgroup givesConsequently . For , linearity puts in . Its limit as is . Closedness of therefore gives , proving that the Lie algebra of a closed subgroup is a Lie subalgebra of .