Proposition 2

Tag 0156, permanent. Cite it as [Bourbaki, Tag 0156].

Lie Groups and Lie Algebras, chapter VIII, § 8. Symmetric invariants

Let E be a finite dimensional $\mathfrak{g}$ -module. Let $U(\mathfrak{h}) =\mathbf{S}(\mathfrak{h})$ be the enveloping algebra of $\mathfrak{h}$ . If $u\in U(\mathfrak{h})$ , then

Tr( $u_E$ ) $=\langle \psi$ (ch $E$ ) $, u\rangle$ .

It suffices to treat the case in which $u=h_1. . . h_m$ with $h_1, . . . , h_m\in \mathfrak{h}$ . For all $\lambda \in P$ , let $d_{\lambda}=$ dim $E^{\lambda}$ . Then ch $E =\sum_{\lambda}d_{\lambda}e^{\lambda}$ , so $\psi$ (ch $E$ ) $=\sum_{\lambda}d_{\lambda}$ exp( $\lambda$ ) and hence

\langle \psi(\mathrm{ch}\,E),u\rangle = \sum_{\lambda} d_\lambda\langle\exp\lambda,h_1\ldots h_m\rangle

= \sum_{\lambda} d_\lambda\lambda(h_1)\ldots\lambda(h_m)\qquad\text{(no. 1)}

= \mathrm{Tr}\,u_E.

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

Corollary 1, § 8

Cites

Nothing.