Analysis on Lie Groups - Jacques Faraut by Jacques Faraut.

By Jacques Faraut.

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For g ∈ G, X ∈ g, t ∈ R, g exp(t X )g −1 = exp(tg Xg −1 ). Hence g Xg −1 ∈ g. The map Ad(g) : X → Ad(g)X = g Xg −1 is an automorphism of the Lie algebra g, Ad(g)[X, Y ] = [Ad(g)X, Ad(g)Y ] (X, Y ∈ g). Furthermore Ad(g1 g2 ) = Ad(g1 ) ◦ Ad(g2 ), and this means that the map Ad : G → Aut(g) is a group morphism. 2 (i) For X ∈ g, d Ad(exp t X ) dt = ad X. t=0 (ii) Let us denote by Exp the exponential map from End(g) into G L(g). Then Exp(ad X ) = Ad(exp X ) (X ∈ g). 3 Linear Lie groups are submanifolds 41 Proof.

11. Let N p denote the set of nilpotent matrices of order p, N p = {X ∈ M(n, C) | X p = 0}, and U p the set of unipotent matrices of order p, U p = {g ∈ G L(n, C) | (g − I ) p = 0}. Show that the exponential map is a bijection from N p onto U p , whose inverse is the logarithm map. Hint. For X ∈ N p , log(exp t X ) − t X is a polynomial in t, vanishing on a neighbourhood of 0, hence identically zero. 12. Let A ∈ M(n, C) be a complex matrix for which there exists a constant C such that ∀t ∈ R, exp(t A) ≤ C.

Then f : R → M(n, R) is C ∞ , and f (t) = = ∞ α(s)γ (t − s)ds −∞ ∞ −∞ α(s)γ (−s)ds · γ (t). We will choose the function α in such a way that the matrix B= ∞ −∞ α(s)γ (−s)ds is invertible. It will follow that γ is C ∞ . If B − I < 1 then it holds. Let α ≥ 0, with integral equal to one. Then B−I ≤ ∞ −∞ α(s) γ (−s) − I ds. Since γ is continuous at 0, for every > 0 there exists η > 0 such that, if |s| ≤ η, then γ (s) − I ≤ . If the support of α is contained in [−η, η], then B−I ≤ . 2 Lie algebra of a linear Lie group Let G be a linear Lie group, that is a closed subgroup of G L(n, R).

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