Bourbaki

Definition 5

Tag 0A8E, permanent. Cite it as [Bourbaki, Tag 0A8E].

Lie Groups and Lie Algebras, chapter III, § 1. Lie groups

A Lie group germ over K is a system (G,e,θ,m)(G, e, \theta, m) satisfying the following conditions: (i) GG is an analytic manifold over KK; (ii) e∈Ge \in G; (iii) θ\theta is an analytic mapping of GG into GG; (iv) mm is an analytic mapping of an open subset Ω\Omega of G×GG \times G into GG; (v) for all g∈Gg \in G, (e,g)∈Ω(e, g) \in \Omega, (g,e)∈Ω(g, e) \in \Omega, m(e,g)=m(g,e)=gm(e, g) = m(g, e) = g; (vi) for all g∈Gg \in G, (g,θ(g))∈Ω(g, \theta(g)) \in \Omega, (θ(g),g)∈Ω(\theta(g), g) \in \Omega, m(g,θ(g))=m(θ(g),g)=em(g, \theta(g)) = m(\theta(g), g) = e; (vii) if g,h,kg, h, k are elements of GG such that (g,h)∈Ω(g, h) \in \Omega, (h,k)∈Ω(h, k) \in \Omega, (m(g,h),k)∈Ω(m(g, h), k) \in \Omega, (g,m(h,k))∈Ω(g, m(h, k)) \in \Omega, then m(m(g,h),k)=m(g,m(h,k))m(m(g, h), k) = m(g, m(h, k)).

ee is called the identity element of the group germ. We often write ghgh instead of m(g,h)m(g, h) and (by an abuse of notation) g−1g^{-1} instead of θ(g)\theta(g).

A Lie group GG is a Lie group germ with the obvious choice of e,θ,me, \theta, m.

Let GG be a Lie group germ. Then ee−1=eee^{-1} = e, that is

e−1=e. e^{-1} = e.

For all g∈Gg \in G,

g=eg=((g−1)−1g−1)g=(g−1)−1(g−1g)=(g−1)−1e, g = eg = ((g^{-1})^{-1}g^{-1})g = (g^{-1})^{-1}(g^{-1}g) = (g^{-1})^{-1}e,

that is

(g−1)−1=g. (g^{-1})^{-1} = g.

A subset of GG invariant under the mapping g↦g−1g \mapsto g^{-1} is called symmetric.

The manifold GG, with the point ee, the mapping g↦g−1g \mapsto g^{-1} and the mapping (g,h)↦hg(g, h) \mapsto hg is a Lie Group germ G∨G^\vee called the opposite of GG.

The Lie group germ GG is called commutative if, for all (g,h)∈G×G(g, h) \in G \times G such that ghgh is defined, hghg is defined and equal to ghgh.

Let GG be a Lie group germ. The set of (g,h)∈G×G(g, h) \in G \times G such that ghgh is defined is a neighbourhood of (e,e)(e, e). On the other hand, the mappings (g,h)↦gh(g, h) \mapsto gh and g↦g−1g \mapsto g^{-1} are continuous. Hence (gh)k=g(hk)(gh)k = g(hk) for g,h,kg, h, k sufficiently close to ee. Similarly, (h−1g−1)(gh)=h−1(eh)=h−1h=e(h^{-1}g^{-1})(gh) = h^{-1}(eh) = h^{-1}h = e for g,hg, h sufficiently close to ee, whence multiplying on the right by (gh)−1(gh)^{-1},

(gh)−1=h−1g−1for g,h sufficiently close to e. (gh)^{-1} = h^{-1}g^{-1} \quad \text{for } g, h \text{ sufficiently close to } e.

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

Cited by

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Cites

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Transcribed from the printed volume. Not translated.