By Jacques Faraut.
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Extra info for Analysis on Lie Groups - Jacques Faraut
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 .
Proof. The terms of degree 2 and 3 are written in the following table. 5 Exercises 1. Let α be an irrational real number. (a) Show that Z + αZ is dense in R. 0 1 3 X, [X, Y ] 0 1 6 Y, [X, Y ] 48 Linear Lie groups (b) Let G be the subgroup of G L(2, C) defined by e2iπ t 0 G= 0 e2iπ αt t ∈R . Determine the closure G¯ of G in G L(2, C). (c) Show that there does not exist any closed subgroup of G L(2, C) with Lie algebra g= it 0 0 iαt t ∈R . 2. Let G be a linear Lie group and g its Lie algebra. One assumes that G is Abelian.
To a linear Lie group one associates its Lie algebra. In this way the properties of the group are translated in terms of the linear algebra properties of its Lie algebra. 3. Let us observe that G L(n, C) is a linear Lie group since it can be seen as a closed subgroup of G L(2n, R). In fact, to a matrix Z = X + iY in M(n, C) one associates the matrix Z˜ = X Y −Y X in M(2n, R), and the map Z → Z˜ is an algebra morphism which maps G L(n, C) onto a closed subgroup of G L(2n, R). 1 One parameter subgroups Let G be a topological group.