Analysis on Lie Groups - Jacques Faraut - download pdf or read online

By Jacques Faraut.

ISBN-10: 0511422962

ISBN-13: 9780511422966

ISBN-10: 0521719305

ISBN-13: 9780521719308

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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). 52 Lie algebras Thus F1 and F2 are solutions of the same differential equation with the same initial data: F1 (0) = F2 (0) = [X, Y ]. Hence, for every t ∈ R, F1 (t) = F2 (t). This means that, for every t, Exp(t D) is an automorphism of g, and that D ∈ Lie Aut(g) . An ideal J of a Lie algebra g is a subalgebra which furthermore satisfies ∀X ∈ J, ∀Y ∈ g, [X, Y ] ∈ J.

2, the function F is defined for |t| ≤ 1. 3, F(t) < 1 2 log 2. From the inequality XY − Y X ≤ 2 X Y it follows that ad X ≤ 2 X , hence ad F(t) < log 2. Let us prove that the function F satisfies the differential equation F (t) = Exp(ad F(t) Y. One can write exp F(t) = exp X exp tY. Taking the derivative at t: (D exp) F(t) F (t) = (exp X exp tY )Y. 4, we obtain ad F(t) F (t) = Y. Since ad F(t) < log 2 this can be written F (t) = Exp(ad F(t)) Y. 46 Linear Lie groups We can also write F (t) = Ad(exp F(t) Y = Ad(exp X ) Ad(exp tY ) Y = Exp(ad X ) Exp(ad tY ) Y.

N Put  log λ  1 X =k ..  k −1 . log λn Then exp X = p. (b) Injectivity. Let X and Y ∈ Sym(n, R) be such that exp X = exp Y . Let us diagonalise X and Y : λ  1 .. X = k   k −1 , . λn e λ1 .. exp X = k   k −1 , e λn 1 .. Y = h  exp Y = h   . µ   h −1 , . µn e µ1 .. k ∈ O(n), h ∈ O(n),   h −1 . e µn Let us show that X and Y commute. If f is a polynomial in one variable such that f (eµi ) = µi , i = 1, . . , n, then f (exp Y ) = Y , hence Y X = f (exp Y )X = f (exp X )X = X f (exp X ) = X f (exp Y ) = X Y.

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