A Lie group germ over K is a system (G,e,θ,m) satisfying the following conditions: (i) G is an analytic manifold over K; (ii) e∈G; (iii) θ is an analytic mapping of G into G; (iv) m is an analytic mapping of an open subset Ω of G×G into G; (v) for all g∈G, (e,g)∈Ω, (g,e)∈Ω, m(e,g)=m(g,e)=g; (vi) for all g∈G, (g,θ(g))∈Ω, (θ(g),g)∈Ω, m(g,θ(g))=m(θ(g),g)=e; (vii) if g,h,k are elements of G such that (g,h)∈Ω, (h,k)∈Ω, (m(g,h),k)∈Ω, (g,m(h,k))∈Ω, then m(m(g,h),k)=m(g,m(h,k)).
e is called the identity element of the group germ. We often write gh instead of m(g,h) and (by an abuse of notation) g−1 instead of θ(g).
A Lie group G is a Lie group germ with the obvious choice of e,θ,m.
Let G be a Lie group germ. Then ee−1=e, that is
e−1=e.
For all g∈G,
g=eg=((g−1)−1g−1)g=(g−1)−1(g−1g)=(g−1)−1e,
that is
(g−1)−1=g.
A subset of G invariant under the mapping g↦g−1 is called symmetric.
The manifold G, with the point e, the mapping g↦g−1 and the mapping (g,h)↦hg is a Lie Group germ G∨ called the opposite of G.
The Lie group germ G is called commutative if, for all (g,h)∈G×G such that gh is defined, hg is defined and equal to gh.
Let G be a Lie group germ. The set of (g,h)∈G×G such that gh is defined is a neighbourhood of (e,e). On the other hand, the mappings (g,h)↦gh and g↦g−1 are continuous. Hence (gh)k=g(hk) for g,h,k sufficiently close to e. Similarly, (h−1g−1)(gh)=h−1(eh)=h−1h=e for g,h sufficiently close to e, whence multiplying on the right by (gh)−1,
(gh)−1=h−1g−1for g,h sufficiently close to e.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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