Bourbaki

Proposition 9

Tag 0101, permanent. Cite it as [Bourbaki, Tag 0101].

Lie Groups and Lie Algebras, chapter VIII, § 3. Subalgebras of split semi-simple Lie algebras

Let b\mathfrak{b} be the Borel subalgebra of (g,h)(\mathfrak{g},\mathfrak{h}) defined by a chamber C of R. Let n=gR+(C)=∑α∈R,α>0gα\mathfrak{n}=\mathfrak{g}^{R_+(C)}=\sum_{\alpha\in R,\alpha >0}\mathfrak{g}^{\alpha}. Let l=l= dimh\mathfrak{h}.

(i) If h∈hh\in \mathfrak{h} and x∈nx\in \mathfrak{n}, the characteristic polynomial of adg(h+x)_{\mathfrak{g}}(h+x) is Tl∏α∈R(T−α(h))T^l\prod_{\alpha\in R}(T-\alpha (h)).

(ii) The largest nilpotent ideal of b\mathfrak{b} is equal to n\mathfrak{n} and to [b,b][\mathfrak{b},\mathfrak{b}]. This is also the set of elements of b\mathfrak{b} nilpotent in g\mathfrak{g}.

(iii) Let B be the basis of R associated to C. For all α∈B\alpha \in B, let XαX_{\alpha} be a non-zero element of gα\mathfrak{g}^{\alpha}. Then (Xα)α∈B(X_{\alpha})_{\alpha\in B} generates the Lie algebra n\mathfrak{n}. We have [n,n]=∑α∈R,α>0,α∉Bgα[\mathfrak{n},\mathfrak{n}] =\sum_{\alpha\in R,\alpha >0,\alpha \notin B}\mathfrak{g}^{\alpha}.

There exists a total order on hQ∗\mathfrak{h}^*_{\mathbf{Q}} compatible with its vector space structure and such that the elements of R+(C)R_+(C) are >0>0 (Chap. VI, §1, no. 7). Let h,xh, x be as in (i) and y∈gαy\in \mathfrak{g}^{\alpha}. Then [h+x,y]=α(h)y+z[h+x, y] =\alpha (h)y+z where z∈∑β>αgβz\in \sum_{\beta >\alpha}\mathfrak{g}^{\beta}. Then,

with respect to a suitable basis of g\mathfrak{g}, the matrix of adg(h+x)_{\mathfrak{g}}(h+x) has the following properties:

1) it is lower triangular;

2) the diagonal entries of the matrix are the number 0(l0 (l times) and the α(h)\alpha (h) for α∈R\alpha \in R.

This proves (i). It also shows that the characteristic polynomial of adb(h+x)_{\mathfrak{b}}(h+x) is T∏α∈R+(C)l(T−α(h))T\prod_{\alpha\in R_+(C)}^l(T-\alpha (h)). It follows from the preceding that the

set of elements of b\mathfrak{b} nilpotent in g\mathfrak{g}, as well as the largest nilpotent ideal of b\mathfrak{b}, are equal to n\mathfrak{n}. We have n=[b,b]\mathfrak{n}= [\mathfrak{b},\mathfrak{b}] by Prop. 5 (i). Finally, assertion (iii) follows from §2, Prop. 4 (ii) and Chap. VI, §1, no. 6, Prop. 19.

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.

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