Analysis on Lie Groups - Jacques Faraut by Jacques Faraut.

By Jacques Faraut.

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4 The differential of the exponential map at A is given by ∞ (D exp) A X = exp A k=0 (−1)k (ad A)k X. (k + 1)! By putting, for z ∈ C, ∞ (z) = k=0 (−1)k k 1 − e−z z = , (k + 1)! z the statement can be written I − Exp(− ad A) , ad A where Exp T denotes the exponential of an endomorphism T of the vector space M(n, R). (D exp) A = L exp A ◦ (ad A) = L exp A ◦ Proof. (a) Let us consider the maps Fk : M(n, R) → M(n, R), X → Xk, and compute the differential of Fk at A: d (A + t X )k dt (D Fk ) A X = t=0 k−1 = Ak− j−1 X A j j=0 k−1 k− j−1 = j LA R A X.

4) can be written (D exp) A = exp A (ad A). 1 If X , Y < r = 12 log(2 − 12 2), then 1 log(exp X exp Y ) = X + Exp(ad X ) Exp(t ad Y ) Y dt. 2 If X , Y ≤ α, then exp X exp Y − I ≤ e2α − 1. Proof. 4 Campbell–Hausdorff formula and, since exp X − I ≤ e X 45 − 1 ≤ eα − 1, exp X exp Y − I ≤ (eα − 1)2 + 2(eα − 1) = e2α − 1. 3 If g − I ≤ β < 1, then 1 . 1−β log g ≤ log Proof. log g ≤ ∞ k=1 (g − I )k ≤ k ∞ k=1 βk 1 = log . 1. For X , Y < 1 2 log 2, put F(t) = log(exp X exp tY ). 2, the function F is defined for |t| ≤ 1.

If |z| < log 2, then |e z − 1| ≤ e|z| − 1 < 1, and e z z 1 − e−z = 1. −1 z Therefore, if L is an endomorphism such that L < log 2, then (e z ) (z) = ez (Exp L) (L) = I d. 4) can be written (D exp) A = exp A (ad A). 1 If X , Y < r = 12 log(2 − 12 2), then 1 log(exp X exp Y ) = X + Exp(ad X ) Exp(t ad Y ) Y dt. 2 If X , Y ≤ α, then exp X exp Y − I ≤ e2α − 1. Proof. 4 Campbell–Hausdorff formula and, since exp X − I ≤ e X 45 − 1 ≤ eα − 1, exp X exp Y − I ≤ (eα − 1)2 + 2(eα − 1) = e2α − 1. 3 If g − I ≤ β < 1, then 1 .

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