By Jacques Faraut.

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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. Furthermore F(0) = log(exp X ) = X , and 1 F(1) = F(0) + F (t)dt, 0 hence 1 log(exp X exp Y ) = X + Exp(ad X ) Exp(t ad Y ) Y dt. 4 (Campbell–Hausdorff formula) If √ 1 2), then 2 log(exp X exp Y ) = X + ∞ k=0 (−1)k k+1 E(k) X , Y < 1 2 log(2 − 1 (q1 + · · · + qk + 1) (ad X ) p1 (ad Y )q1 . . q1 ! . m! where, for k ≥ 1, E(k) = { p1 , q1 , .

A) Show that every matrix in S L(2, R) is conjugate to one of the following matrices a 0 0 1 a 1 0 , t 1 , −1 t 0 −1 , cos θ −sin θ sin θ cos θ (a ∈ R, a = 0, t ∈ R, θ ∈ R). (b) Show that the range of the exponential map exp : x z y −x x, y, z ∈ R → S L(2, R), is equal to {g ∈ S L(2, R) | tr(g) > −2} ∪ {−I }. 3. Polar decomposition of complex matrices. Show that every matrix in g ∈ G L(n, C) can be written g = k exp X with k ∈ U (n) and X ∈ Herm(n, C). 30 The exponential map Show that the decomposition is unique, and that the map (k, X ) → g = k exp X, U (n) × Herm(n, C) → G L(n, C), is a homeomorphism.

N belong to [α, β], then Y ≤ sup | f (λ)| X . α≤λ≤β (ii) Let p be a polynomial. The differential of p˜ at A is defined by d p˜ (A + t X ) t=0 . dt One assumes here that the matrix A = is diagonal. Show that (D p˜ ) A (X ) = (D p˜ ) (X ) = M p, (X ). Hint. Consider first the case of p(λ) = λm . Recall that d (A + t X )m dt m−1 t=0 = Am−k−1 X Ak . k=0 (iii) Show that, if A = k k , where k is an orthogonal matrix and diagonal, then T (D p˜ ) A (X ) = k M p, (k T X k)k T . (iv) Show that, if f ∈ C 1 (R), then the map f˜ is differentiable, and that, if is diagonal, (D f˜ ) (X ) = M f, (X ).