Bourbaki

Lemma 1

Tag 00V1, permanent. Cite it as [Bourbaki, Tag 00V1].

Lie Groups and Lie Algebras, chapter VII, § 3. Conjugacy theorems

Let V be a finite dimensional vector space, n\mathfrak{n} a Lie subalgebra of a=gl(V)\mathfrak{a}=\mathfrak{g}\mathfrak{l}(V) consisting of nilpotent elements.

(i) The map x→x \rightarrow exp xx is a bijection from n\mathfrak{n} to a subgroup N of GL(V)\mathbf{G}\mathbf{L}(V) consisting of unipotent elements (Chap. II, §6, no. 1, Remark 4). We have n=\mathfrak{n}= log(expn\mathfrak{n}). The map f→f∘f \rightarrow f\circ log is an isomorphism from the algebra of polynomial functions on n\mathfrak{n} to the algebra of restrictions to N of polynomial functions on End(V).

(ii) If x∈nx\in \mathfrak{n} and a∈aa\in \mathfrak{a},

(exp adax_{\mathfrak{a}}x).a=.a= (exp xx)aa(exp(−x-x)).

(iii) Let V′V' be a finite dimensional vector space, n′\mathfrak{n}' a Lie subalgebra of gl(V′)\mathfrak{g}\mathfrak{l}(V') consisting of nilpotent elements, ρ\rho a homomorphism from n\mathfrak{n} to n′\mathfrak{n}'. Let π\pi be the map exp x→x \rightarrow exp ρ(x)\rho (x) from expn\mathfrak{n} to expn′\mathfrak{n}'. Then π\pi is a group homomorphism.

By Engel’s theorem, we can identify V with knk^n in such a way that n\mathfrak{n} is a subalgebra of n(n,k)\mathfrak{n}(n, k) (the Lie subalgebra of Mn(k)\mathbf{M}_n(k) consisting of the lower triangular matrices with zeros on the diagonal). For s≥0s\geq 0, let ns(n,k)\mathfrak{n}_s(n, k) be the set of (xij)1≤i,j≤n∈Mn(k)(x_{ij})_{1\leq i,j\leq n}\in \mathbf{M}_n(k) such that xij=0x_{ij}= 0 for i−j<si-j < s. Then

[ns(n,k),ns′(n,k)]⊂ns+s′(n,k) [\mathfrak{n}_s(n, k),\mathfrak{n}_{s'}(n, k)]\subset \mathfrak{n}_{s+s'}(n, k)

(Chap. II, §4, no. 6, Remark), and the Hausdorff series defines a polynomial map (a,b)→H(a,b)(a, b) \rightarrow H(a, b) from n(n,k)×n(n,k)\mathfrak{n}(n, k)\times \mathfrak{n}(n, k) to n(n,k)\mathfrak{n}(n, k) (Chap. II, §6, no. 5, Remark 3); this map makes n(n,k)\mathfrak{n}(n, k) into a group (Chap. II, §6, no. 5, Prop. 4). By Chap. II, §6, no. 1, Remark 4, the maps x→x \rightarrow exp xx from n(n,k)\mathfrak{n}(n, k) to 1+n(n,k)1 +\mathfrak{n}(n, k) and y→y \rightarrow log yy from 1+n(n,k)1 +\mathfrak{n}(n, k) to n(n,k)\mathfrak{n}(n, k) are inverse bijections and are polynomial; by Chap. II, §6, no. 5, Prop. 3, these maps are isomorphisms of groups if n(n,k)\mathfrak{n}(n, k) is given the group law (a,b)→H(a,b)(a, b) \rightarrow H(a, b) and if 1+n(n,k)1 +\mathfrak{n}(n, k) is considered as a subgroup of GLn(k)\mathbf{G}\mathbf{L}_n(k). Assertions (i) and (iii) of the lemma now follow. Let x∈nx\in \mathfrak{n}. Denote by Lx,RxL_x,R_x the maps u→xu,u→uxu \rightarrow xu, u \rightarrow ux from a\mathfrak{a} to a\mathfrak{a}, which commute and are nilpotent. We have adax=Lx−Rx_{\mathfrak{a}}x= L_x-R_x, so, for all a∈aa\in \mathfrak{a},

(exp adax_{\mathfrak{a}}x)a=a= (exp(Lx−Rx_x-R_x))a=a= (exp LxL_x)(exp R−xR_{-x})aa (1)

=∑Liix=\sum Li^{ix}! Rj!j−xa=Rj\overset{j-x}{!}a= (exp xx)aa(exp(−x-x)).

i,j≥0i,j\geq 0

With the notation in Lemma 1,π1,\pi is called the linear representation of exp n\mathfrak{n} compatible with the given representation ρ\rho of n\mathfrak{n} on V′V'. When kk is R\mathbf{R}, C\mathbf{C}, or a non-discrete complete ultrametric field, ρ=L(π)\rho = L(\pi ) by the properties of exponential maps (Chap. III, §4, no. 4, Cor. 2 of Prop. 8).

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

Cites

  • Chap. II, §4, no. 6
  • Chap. II, §6, no. 5, Prop. 3 not in this corpus
  • Chap. II, §6, no. 5, Prop. 4 not in this corpus
  • Chap. III, §4, no. 4, Cor. 2 of Prop. 8 not in this corpus
Transcribed from the printed volume. Not translated.