Bourbaki

Exercise 13, § 9

Lie Groups and Lie Algebras, III, 399

(a) A multiplication is defined on R3\mathbf{R}^3 as follows:

(x,y,z).(x′,y′,z′)=(x+x′cos⁡z−y′sin⁡z,y+x′sin⁡z+y′cos⁡z,z+z′). (x, y, z).(x', y', z') = (x + x' \cos z - y' \sin z, y + x' \sin z + y' \cos z, z + z').

Show that a solvable real Lie group G is thus obtained such that DG=R2×{0}DG = \mathbf{R}^2 \times \{0\}. The centre Z of G is {0}×{0}×2πZ\{0\} \times \{0\} \times 2\pi \mathbf{Z}.

(b) For (x,y,z)∈G(x, y, z) \in G, let π(x,y,z)\pi(x, y, z) be the mapping

(λ,μ)↦(λcos⁡z−μsin⁡z+x,λsin⁡z+μcos⁡z+y) (\lambda, \mu) \mapsto (\lambda \cos z - \mu \sin z + x, \lambda \sin z + \mu \cos z + y)

of R2\mathbf{R}^2 into R2\mathbf{R}^2. Show that π\pi is a morphism of G onto a Lie subgroup G′G' of the affine group of R2\mathbf{R}^2, generated by the translations and rotations of R2\mathbf{R}^2. Show that Ker π=Z\mathrm{Ker}\, \pi = \mathbf{Z}.

(c) We canonically identify L(G)=T(0,0,0)GL(G) = T_{(0, 0, 0)}G with R3\mathbf{R}^3. Let (e1,e2,e3)(e_1, e_2, e_3) be the canonical basis of R3\mathbf{R}^3. Show that [e1,e2]=0,[e3,e1]=e2,[e3,e2]=−e1[e_1, e_2] = 0, [e_3, e_1] = e_2, [e_3, e_2] = -e_1.

(d) Show that, for c≠0c \neq 0,

exp⁡G(a,b,c)=(1c(asin⁡c+bcos⁡c−b),1c(−acos⁡c+bsin⁡c+a),c). \exp_G(a, b, c) = \left( \frac{1}{c} (a \sin c + b \cos c - b), \frac{1}{c} (-a \cos c + b \sin c + a), c \right).

Deduce that, for all u∈L(G)u \in L(G), Z⊂exp⁡(Ru)Z \subset \exp(\mathbf{R}u). Show that exp⁡G:L(G)→G\exp_G : L(G) \to G is neither injective nor surjective.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.