Bourbaki

Lemma 1

Tag 017X, permanent. Cite it as [Bourbaki, Tag 017X].

Lie Groups and Lie Algebras, chapter VIII, § 12. Chevalley orders

Let n∈NI\mathbf{n}\in \mathbf{N}^I. If ∣n∣<r|\mathbf{n}|< r, then cr+([n])=0c^+_r([\mathbf{n}]) = 0. If ∣n∣=r|\mathbf{n}|=r, then

cr+([n])=∑φxφ(1)⊗xφ(2)⊗⋯⊗xφ(r)(7) c^+_r([\mathbf{n}]) =\sum_{\varphi}x_{\varphi(1)}\otimes x_{\varphi(2)}\otimes \cdots \otimes x_{\varphi(r)} \tag{7}

where φ\varphi belongs to the set of maps from {1,2,...,r}\{1,2, . . . , r\} to I which take each value i∈Inii\in In_i times.

By Prop. 1,

cr(x(n))=∑x(p1)⊗⋯⊗x(pr) c_r(\mathbf{x}^{(\mathbf{n})}) =\sum\mathbf{x}^{(\mathbf{p}_1)}\otimes \cdots \otimes \mathbf{x}^{(\mathbf{p}_r)}

where the summation extends over the set of sequences (p1,...,pr)(\mathbf{p}_1, . . . ,\mathbf{p}_r) of rr elements of M such that p1+⋯+pr=n\mathbf{p}_1+\cdots +\mathbf{p}_r=\mathbf{n}. In view of Chap. II, §1, no. 3, Prop. 6, the map cr+c^+_r, extended by linearity to a map from U(g)U(\mathfrak{g}) to Tr(U+(g))\mathbf{T}^r(U^+(\mathfrak{g})), vanishes on Ur−1(g)U_{r-1}(\mathfrak{g}). It follows that, for r≥∣n∣r\geq |\mathbf{n}|,

cr+([n])=cr+(x(n))=∑π(x(p1))⊗⋯⊗π(x(pr))(8) c^+_r([\mathbf{n}]) =c^+_r(\mathbf{x}^{(\mathbf{n})}) =\sum\pi (\mathbf{x}^{(\mathbf{p}_1)})\otimes \cdots \otimes \pi (\mathbf{x}^{(\mathbf{p}_r)}) \tag{8}

For r>∣n∣r >|\mathbf{n}|, the relation p1+⋯+pr=n\mathbf{p}_1+\cdots +\mathbf{p}_r=\mathbf{n} implies that at least one of the pi\mathbf{p}_i is zero, so cr+([n])=0c^+_r([\mathbf{n}]) = 0. For r=∣n∣r=|\mathbf{n}|, the only non-zero terms of the third member of (8) are those for which ∣p1∣=⋯=∣pr∣=1|\mathbf{p}_1|=\cdots =|\mathbf{p}_r|= 1, hence (7).

We return to the proof of Th. 1. We retain the notations of (*) and prove, by descending induction on ∣n∣|\mathbf{n}|, that dd divides ana_{\mathbf{n}}, which is clear when ∣n∣|\mathbf{n}| is sufficiently large. If dd divides ana_{\mathbf{n}} for ∣n∣>r|\mathbf{n}|> r then, putting

u′=u−∑∣n∣>r(an/d)[n]∈U u'=u-\sum_{|\mathbf{n}|>r}(a_{\mathbf{n}}/d)[\mathbf{n}]\in \mathscr{U}

we have

du′=∑∣n∣≤ran[n](9) du'=\sum_{|\mathbf{n}|\leq r}a_{\mathbf{n}}[\mathbf{n}] \tag{9}

For any map φ\varphi from {1,2,...,r}\{1,2, . . . , r\} to I, put

eφ=xφ(1)⊗⋯⊗xφ(r) e_{\varphi}=x_{\varphi(1)}\otimes \cdots \otimes x_{\varphi(r)}

and aφ=ana_{\varphi}=a_{\mathbf{n}} where n=\mathbf{n}= (Cardφ−1(i)\varphi^{-1}(i))i∈I_{i\in I}. By Lemma 1, (9) implies that

dcr+(u′)=∑φ∈Iraφeφ(10) dc^+_r(u') =\sum_{\varphi\in I^r}a_{\varphi}e_{\varphi} \tag{10}

so cr+(u′)∈Tr(U+)∩QTr(G)c^+_r(u')\in \mathbf{T}^r(\mathscr{U}^+)\cap \mathbf{Q}\mathbf{T}^r(\mathscr{G}). But the submodule G\mathscr{G} of U+\mathscr{U}^+ is a direct factor (Algebra, Chap. VII, §4, no. 3, Cor. of Th. 1), so the submodule Tr(G)\mathbf{T}^r(\mathscr{G}) is a direct factor of Tr(U+)\mathbf{T}^r(\mathscr{U}^+), and hence cr+(u′)∈Tr(G)c^+_r(u')\in \mathbf{T}^r(\mathscr{G}). On the other hand, the xix_i form a basis of G\mathscr{G} by hypothesis, so the eφe_{\varphi} form a basis of Tr(G)\mathbf{T}^r(\mathscr{G}). Then (10) proves that dd divides the aφa_{\varphi}, that is, the ana_{\mathbf{n}} for ∣n∣=r|\mathbf{n}|=r. This proves (*).

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

Nothing.

Cites

  • Algebra, Chap. VII, §4, no. 3, Cor. of Th. 1 not in this corpus
  • Chap. II, §1, no. 3, Prop. 6 not in this corpus
  • Prop. 1
  • Th. 1
Transcribed from the printed volume. Not translated.