By Jacques Faraut.

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**Extra resources for Analysis on Lie Groups - Jacques Faraut**

**Example text**

This is not a group of matrices, but it can be identified with the closed subgroup of G L(2, R) whose elements are the matrices a 0 b 1 . The matrices X1 = 1 0 0 0 , X2 = 0 0 1 0 , constitute a basis of its Lie algebra and [X 1 , X 2 ] = X 2 . Let G be the motion group of R2 , that is the group of affine linear transformations of the form (x, y) → (x cos θ − y sin θ + a, x sin θ + y cos θ + b). The group G can be identified with the subgroup of G L(3, R) whose elements are the matrices cos θ −sin θ a sin θ cos θ b .

To every closed subgroup G of G L(n, R) one associates its Lie algebra g = Lie(G) ⊂ M(n, R). However, not every Lie subalgebra of M(n, R) corresponds to a closed subgroup of G L(n, R). 4 Campbell–Hausdorff formula Let G be a linear Lie group and g = Lie(G) its Lie algebra. The Campbell– Hausdorff formula expresses log(exp X exp Y ) (X, Y ∈ g) in terms of a series, each term of which is a homogeneous polynomial in X and Y involving iterated brackets. Let us introduce the functions (z) = (z) = 1 − e−z = z ∞ (−1)k k=0 ∞ zk (k + 1)!

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). We associate to the group G the set g = Lie(G) = {X ∈ M(n, R) | ∀t ∈ R, exp(t X ) ∈ G}. 1 (i) The set g is a vector subspace of M(n, R).