Bourbaki

Exercise 12, § 9

Lie Groups and Lie Algebras, III, 399

Let (e1,e2,e3)(e_1, e_2, e_3) be the canonical basis of K3K^3. Consider the nilpotent Lie algebra structure on K3K^3 such that [e1,e2]=e3,[e1,e3]=[e2,e3]=0[e_1, e_2] = e_3, [e_1, e_3] = [e_2, e_3] = 0. With the associated group law on K3K^3 (no. 5),

(x,y,z)(x′,y′,z′)=(x+x′,y+y′,z+z′+12(xy′−yx′)). (x, y, z)(x', y', z') = (x + x', y + y', z + z' + \frac{1}{2}(xy' - yx')).

The mapping

(λ,μ,v)↦(exp⁡λe1)(exp⁡μe2)(exp⁡v(e1+e3)) (\lambda, \mu, v) \mapsto (\exp \lambda e_1)(\exp \mu e_2)(\exp v(e_1 + e_3))

of K3K^3 into G is neither surjective (show that (0,1,1)(0, 1, 1) is not in the image) nor injective (show that (0,1,0)(0, 1, 0) and (1,1,−1)(1, 1, -1) have the same image).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.