Bourbaki

Exercise 13, § 2

Lie Groups and Lie Algebras, II, 184

Bourbaki marks this one as harder than the rest.

(In this exercise we assume that every projective module over KK is free; this is so for example if KK is a principal ideal domain.)

Let l=∑n=1∞lnl = \sum_{n=1}^\infty l_n be a graded Lie algebra admitting a basic family BB consisting of homogeneous elements and let bb be a graded Lie subalgebra of ll, which is a direct factor of ll considered as a module.

(a) For all i≥0i \geq 0, let l(i)l^{(i)} be the graded subalgebra of ll such that

lj(i)=bjif j≤i.lj(i)=ljif j>i. l_j^{(i)} = b_j \quad \text{if} \ j \leq i. \\ l_j^{(i)} = l_j \quad \text{if} \ j > i.
Then l=l(0)⊃l(1)⊃⋯⊃bl = l^{(0)} \supset l^{(1)} \supset \cdots \supset b. Show, arguing by induction on ii, the existence of a basic family B(i)B^{(i)} of l(i)l^{(i)}, consisting of homogeneous elements, such that the elements of B(i−1)B^{(i-1)} and of B(i)B^{(i)} of degree <i−1< i - 1 are the same. (Suppose that B(i−1)B^{(i-1)} has been constructed. Let mim_i be the intersection of li=li(i−1)l_i = l_i^{(i-1)} with the subalgebra generated by the bjb_j for j<ij < i and let bib_i be the submodule of lil_i generated by the elements of B(i−1)B^{(i-1)} of degree ii. The induction hypothesis implies that li=mi⊕bil_i = m_i \oplus b_i. As mi⊂bim_i \subset b_i, we can decompose bib_i as bi=ηi⊕δib_i = \eta_i \oplus \delta_i, so that bi=mi⊕δib_i = m_i \oplus \delta_i; by the hypothesis on KK the modules ηi\eta_i and δi\delta_i are free;

changing the elements of B(i−1)B^{(i-1)} of degree ii if necessary, we can assume that zi\mathfrak{z}_i is generated by a subset of B(i−1)B^{(i-1)}. Applying Exercise 11 to l(i−1)\mathfrak{l}^{(i-1)} and its ideal l(i)\mathfrak{l}^{(i)}, we obtain a basic family B(i)B^{(i)} of l(i)\mathfrak{l}^{(i)} with the desired properties.)

(b) Deduce from (a) the fact that h\mathfrak{h} has a basic family consisting of homogeneous elements.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.