Bourbaki

Proposition 30

Tag 0AAM, permanent. Cite it as [Bourbaki, Tag 0AAM].

Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra

Let G be a finite-dimensional Lie group with a countable base and

H and H' Lie subgroups of G. Suppose that K is of characteristic 0 and that HH' is locally closed in G.

(i) HH' is a submanifold of G and Te(HH′)=L(H)+L(H′)T_e(HH') = L(H) + L(H').

(ii) Suppose that every element of H commutes with every element of H'. Then HH' is a Lie subgroup of G. Let ϕ\phi be the mapping (h,h′)↦hh′(h, h') \mapsto hh' of H×H′H \times H' onto HH'. The kernel of ϕ\phi is the set of (m,m−1)(m, m^{-1}) where m∈H∩H′m \in H \cap H' and the morphism of (H×H′)/Ker ϕ(H \times H')/\mathrm{Ker}\ \phi onto HH' derived from ϕ\phi by passing to the quotient is a Lie group isomorphism.

Let H×H′H \times H' operate on G on the right by the mapping ((h,h′),g)↦hgh′−1((h, h'), g) \mapsto hg{h'}^{-1}. The orbital mapping ρ\rho of e is (h,h′)↦hh′−1(h, h') \mapsto h{h'}^{-1}. By Proposition 14 (iii) of § 1, no. 7, HH' is a submanifold of G and Te(HH′)=Im Te(ρ)T_e(HH') = \mathrm{Im}\ T_e(\rho). Now

Te(ρ)(L(H)×{0})=L(H)andTe(ρ)({0}×L(H′))=L(H′) T_e(\rho)(L(H) \times \{0\}) = L(H) \quad \text{and} \quad T_e(\rho)(\{0\} \times L(H')) = L(H')

and hence Te(HH′)=L(H)+L(H′)T_e(HH') = L(H) + L(H'). Suppose that every element of H commutes with every element of H'. Then HH' is a subgroup of G. By (i), it is a Lie subgroup of G. The rest of the statement follows from Proposition 28.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Transcribed from the printed volume. Not translated.