Bourbaki

Exercise 5, § 6

Lie Groups and Lie Algebras, II, 203

Let gg be a nilpotent Lie Q\mathbf{Q}-algebra which we give the Hausdorff group law, denoted by (x,y)↦x.y(x, y) \mapsto x.y.

(a) Show that, if x∈g,y∈gx \in g, y \in g, then

x.y.x−1=∑n=0∞1n!(ad x)n(y). x.y.x^{-1} = \sum_{n=0}^{\infty} \frac{1}{n!} (\mathrm{ad}\ x)^n(y).
(Use Exercise 2.)

(b) Let h\mathfrak{h} be a subset of gg. Show that h\mathfrak{h} is a Lie subalgebra (resp. an ideal) of gg if and only if it is a saturated subgroup (resp. a saturated normal subgroup) of the group gg. (Use the formulae of Exercise 4 (d) to pass from the group law to the Lie algebra law.)

(c) Let h\mathfrak{h} be a subgroup of the group gg. Show that the P\mathbf{P}-saturation of h\mathfrak{h} is Q.h\mathbf{Q}.\mathfrak{h}.

(d) Let h\mathfrak{h} be a Lie subalgebra of gg. Show that the centralizer (resp. normalizer) of h\mathfrak{h} in the group gg is the set of x∈gx \in g such that (ad x)(h)=0(\mathrm{ad}\, x)(\mathfrak{h}) = 0 (resp. (ad x)(h)⊂h(\mathrm{ad}\, x)(\mathfrak{h}) \subset \mathfrak{h}).

(e) Show that the lower central series of the Lie algebra gg coincides with that of the group gg and that the associated graded Lie algebra gr(g)\mathrm{gr}(g) is the same from the "group" point of view as from the "Lie algebra" point of view.

Solution

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Transcribed from the printed volume. Not translated.