Bourbaki

Exercise 5, § 4

Lie Groups and Lie Algebras, III, 377

Let (e1,e2,e3)(e_1, e_2, e_3) be the canonical basis of R3\mathbf{R}^3. We give R3\mathbf{R}^3 the nilpotent Lie algebra structure gg 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. Let cijkc_{ijk} denote the constants of structure of gg relative to the basis (e1,e2,e3)(e_1, e_2, e_3).

(a) Show that on R3\mathbf{R}^3 the differential forms

ω1=dx1,ω2=dx2,ω3=−x1dx2+dx3 \omega_1 = dx_1,\quad \omega_2 = dx_2,\quad \omega_3 = -x_1 dx_2 + dx_3
satisfy the relations
dωk=−∑i<jcijkωi∧ωj(k=1,2,3) d\omega_k = -\sum_{i < j} c_{ijk} \omega_i \wedge \omega_j \qquad (k = 1, 2, 3)
and are linearly independent at each point of R3\mathbf{R}^3.

(b) Deduce that there exists, on an open neighbourhood of 0 in R3\mathbf{R}^3, a Lie group germ structure GG such that L(G)=gL(G) = g and

(a1,a2,a3)(x1,x2,x3)=(x1+a1,x2+a2,x3+a3+a1x2) (a_1, a_2, a_3)(x_1, x_2, x_3) = (x_1 + a_1, x_2 + a_2, x_3 + a_3 + a_1 x_2)
for (a1,a2,a3), (x1,x2,x3)(a_1, a_2, a_3),\ (x_1, x_2, x_3) sufficiently close to 0. Show that in fact formulae (1) define a nilpotent Lie group structure on R3\mathbf{R}^3.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.