Bourbaki

Exercise 20, § 4

Lie Groups and Lie Algebras, I, 96

Bourbaki marks this one as harder than the rest.

Let g be a Lie algebra and U its enveloping algebra.

(a) Let g′g' be an ideal of g, U′⊂UU' \subset U its enveloping algebra, x∈gx \in g and a1,…,ana_1, \ldots, a_n in U. Suppose that ai=xa_i = x for the indices j1,…,jpj_1, \ldots, j_p and ai∈U′a_i \in U' for the other indices (let k1,…,kqk_1, \ldots, k_q be these indices, with k1<k2<⋯<kqk_1 < k_2 < \cdots < k_q). Then a1a2…an−xpak1ak2…akqa_1a_2 \ldots a_n - x^pa_{k_1}a_{k_2} \ldots a_{k_q} is a sum of terms of the form xp′bx^{p'}b, where b∈U′b \in U' and p′<pp' < p. (Argue by induction on pp.)

(b) Suppose henceforth that g is nilpotent and K is of characteristic 0. Let g′g' be an ideal of codimension 1 in g, U′U' its enveloping algebra, x an element of g not belonging to g′g' and U0U_0 (resp. U0′U'_0) the subalgebra of U (resp. U′U') annihilated by a set d\mathfrak{d} of derivations of g (which extend to derivations of U) mapping g into g′g'. Suppose that U0U_0 is contained in the centre of U and U0⊄U′U_0 \not\subset U'. Show that there exist a1∈U0′, a2∈U′a_1 \in U'_0,\ a_2 \in U' such that a1≠0a_1 \neq 0 and a=xa1+a2∈U0a = xa_1 + a_2 \in U_0. (Let xmbm+xm−1bm−1+⋯+b0∈U0x^mb_m + x^{m-1}b_{m-1} + \cdots + b_0 \in U_0 with bm,…,b0b_m, \ldots, b_0 in U′U', m>0, bm≠0m > 0,\ b_m \neq 0. Show, using (a), that, for every derivation D∈dD \in \mathfrak{d} of g, Dbm=0, m(Dx)bm+Dbm−1=0Db_m = 0,\ m(Dx)b_m + Db_{m-1} = 0 and hence D(mxbm+bm−1)=0D(mxb_m + b_{m-1}) = 0.) Show that U0U_0 is contained in the algebra K[a,a1−1,U0′]K[a, a_1^{-1}, U'_0] generated by a, a1−1, U0′a,\ a_1^{-1},\ U'_0 in the field of fractions of U0U_0 (which can be formed by Corollary 7 to Theorem 1 of § 2). Deduce that the field of fractions of U0U_0 is the field generated by aa and U0′U'_0. Show that aa is transcendental over K(U0′)K(U'_0).

(c) Let {0}=g0⊂g1⊂…⊂gn=g\{0\} = g_0 \subset g_1 \subset \ldots \subset g_n = g be a sequence of ideals of gg of dimensions 0,1,…,n0, 1, \ldots, n. Let xjx_j be an element of gjg_j not belonging to gj−1g_{j-1}. Let j1<j2<⋯<jqj_1 < j_2 < \cdots < j_q be the indices jj such that there exists in UjU^j (enveloping algebra of gjg_j) an element of the centre U0U_0 of UU not belonging to Uj−1U^{j-1}. By (b) there exists aj1∈U0∩Uja_{j1} \in U_0 \cap U^j and aj2∈Uj−1a_{j2} \in U^{j-1} such that aj1≠0a_{j1} \neq 0 and aj=xjaj1+aj2∈U0∩Uja_j = x_j a_{j1} + a_{j2} \in U_0 \cap U^j. Show by induction on nn that

U0⊂K[aj1,…,ajq,aj1−1,…,ajq−1] U_0 \subset K[a_{j_1}, \ldots, a_{j_q}, a_{j_1}^{-1}, \ldots, a_{j_q}^{-1}]
and that the field of fractions of U0U_0 is generated by the algebraically independent elements aj1,…,ajqa_{j_1}, \ldots, a_{j_q}. In particular, the field of fractions of the centre of UU is a pure transcendental extension of KK.

Solution

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Transcribed from the printed volume. Not translated.