By Koornwinder T.H.
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Also observe that ξA (st) = ξsA (t). Now define the exponential map exp: g → G by exp(A) := ξA (1). Then exp(tA) = ξA (t). So exp: g → G is C ∞ and t → exp(tA): R → G is a C∞ homomorphism with derivative at 0 equal to A. 17 Let G, g be as above. Then d exp0 : T0 g → Te G. But T0 g can be identified with g and Te G = g. Hence d exp0 linearly maps g to g. Proposition d exp0 = id. There is an open neighbourhood U of 0 in g and an open neighbourhood V of e in G such that exp: U → V is a C ∞ diffeomorphism.
Km ! + O(|t|n+1 ) as t → 0. 22 have Proposition Let G and g be as above. f ). Proof Let x ∈ G. We will expand f (x exp(tA) exp(tB) exp(−tA) exp(−tB)) in two different ways as a Taylor series in t up to degree 2, where t → 0 in R. Then we obtain the result by equality of second degree terms in both expansions. f ))(x) + O(|t|3 ). f )(x) + O(|t|3 ). 23 Corollary Let G be a Lie group with g := Te G and Lie(G) the Lie algebra of left invariant vector fields on G. Put [A, B] := b(A, B) (A, B ∈ g).
28 Let G be a Lie group. 7 for the case of right invariant vector fields on G. Let A, B ∈ Te G. Let v, w be left invariant vector fields on G such that ve = A, we = B and let v, w be right invariant vector fields on G such that ve = A, we = B. What is the relationship between [v, w]e and [v, w]e ?