Bourbaki

Exercise 11, § 2

Lie Groups and Lie Algebras, II, 184

Bourbaki marks this one as harder than the rest.

Let YY and ZZ be two disjoint sets (with YY totally ordered), let X=Y∪ZX = Y \cup Z and l=L(X)l = L(X). Let aYa_Y be the submodule of ll generated by ZZ and [l,l][l, l]; it is an ideal of the Lie algebra ll. We propose to find explicitly a basic family of aYa_Y.

Let MM be the set of positive integral functions on YY of finite support. If m∈Mm \in M, we denote by θm\theta^m the endomorphism of ll given by

θm=∏y∈Y(ad⁡y)m(y), \theta^m = \prod_{y \in Y} (\operatorname{ad} y)^{m(y)},
where the product is relative to the order relation given on YY. Let S\mathscr{S} denote the subset of Y×MY \times M consisting of the ordered pairs (u,m)(u, m) such that there exists y∈Yy \in Y for which y>uy > u and m(y)≥1m(y) \geq 1. Show that the elements
θm(u)for(u,m)∈Sandθm(z)for(z,m)∈Z×M \theta^m(u) \quad \text{for} \quad (u, m) \in \mathscr{S} \quad \text{and} \quad \theta^m(z) \quad \text{for} \quad (z, m) \in Z \times M
form a basic family of the Lie algebra aYa_Y.

(Reduce the problem to the case where YY is finite and argue by induction on Card⁡(Y)\operatorname{Card}(Y). Use the Corollary to Proposition 10 applied to the greatest element yy of YY and apply the induction hypothesis to Y−{y}Y - \{y\}.)

Solution

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Transcribed from the printed volume. Not translated.