By Mazurov V. D.
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Let D ∈ Lie Aut(g) . For every t ∈ R, Exp(t D) is an automorphism of g: for X, Y ∈ g, Exp(t D)[X, Y ] = Exp(t D)X, Exp(t D)Y . Taking derivatives of both sides at t = 0 we obtain D[X, Y ] = [D X, Y ] + [X, DY ], which means that D is a derivation: D ∈ Der(g). Conversely, let D ∈ Der(g) and put, for X, Y ∈ g, F1 (t) = Exp(t D)[X, Y ], F2 (t) = Exp(t D)X, Exp(t D)Y . We have F1 (t) = D Exp(t D)[X, Y ] = D F1 (t), F2 (t) = D Exp(t D)X, Exp(t D)Y + Exp(t D)X, D Exp(t D)Y , and, since D is a derivation of g, F2 (t) = D Exp(t D)X, Exp(t D)Y = D F2 (t).
M! The convergence of the series is uniform for t in [0, 1]. The statement is obtained by termwise integration since 1 t q1 +···+qk dt = 0 1 . 5 1 1 1 log(exp X exp Y ) = X + Y + [X, Y ] + X, [X, Y ] + Y, [Y, X ] 2 12 12 + terms of degree ≥ 4. Proof. The terms of degree 2 and 3 are written in the following table. 5 Exercises 1. Let α be an irrational real number. (a) Show that Z + αZ is dense in R. 0 1 3 X, [X, Y ] 0 1 6 Y, [X, Y ] 48 Linear Lie groups (b) Let G be the subgroup of G L(2, C) defined by e2iπ t 0 G= 0 e2iπ αt t ∈R .
X n (t) + · · · + det X 1 (t), . . , X n−1 (t), X n (t) . One can also use the fact that every matrix X ∈ M(n, C) is triangularisable: there exists an invertible matrix g, and an upper triangular matrix Y such that X = gY g −1 , If exp X = g exp Y g −1 . y 11 Y = 0 0 then e y11 exp Y = 0 0 ∗ ∗ , ∗ .. 0 ynn ∗ .. ∗ 0 e ynn ∗ , hence det(exp X ) = det(exp Y ) = etr Y = etr X . For K = R, the exponential map is a map from M(n, R) into G L(n, R)+ . For n ≥ 2 it is not injective. In fact, exp 0 −θ θ 0 exp 0 −2kπ = cos θ −sin θ sin θ cos θ , and, for every k ∈ Z, 2kπ 0 = I.