Exercise 5, § 9

Tag 0169, permanent. Cite it as [Bourbaki, Tag 0169].

Lie Groups and Lie Algebras, VIII, 254

Bourbaki marks this one as harder than the rest.

a) Let $\alpha \in R, x\in \mathfrak{g}^{\alpha}, y\in \mathfrak{g}^{-\alpha}$ , and let E be a $\mathfrak{g}$ -module. Show that, for all $\lambda \in P$ ,

Tr(( $xy$ ) $_E|E^{\lambda}$ ) $=$ Tr(( $xy$ ) $_E|E^{\lambda+\alpha}$ ) $+\lambda ([x, y])$ dim $E^{\lambda}$ .

Deduce that

$\sum_{\lambda\in P}\lambda ([x, y])$ dim $E^{\lambda}. e^{\lambda}= (1-e^{-\alpha})\sum_{\lambda\in P}$ Tr(( $xy$ ) $_E|E^{\lambda}$ ) $. e^{\lambda}$ .

b) Give $\mathfrak{h}^*$ a non-degenerate W-invariant symmetric bilinear form $\langle ,\rangle$ . Let $\Delta$ be the endomorphism of the vector space $k[P]$ such that $\Delta (e^\mu) =\langle \mu, \mu\rangle e^\mu$ for all $\mu\in P$ ; if $a, b\in k[P]$ , put

\Delta '(a, b) =\Delta (ab)-a\Delta (b)-b\Delta (a)

Prove that

$\Delta (J(e^\mu)) =\langle \mu, \mu\rangle J(e^\mu)$ for $\mu\in P$ ,

$\Delta '(e^{\lambda}, e^\mu) = 2\langle \lambda , \mu\rangle e^{\lambda+\mu}$ for $\lambda , \mu\in P$ ,

$\Delta '(ab, c) =a\Delta '(b, c) +b\Delta '(a, c)$ for $a, b, c\in k[P]$ .

c) Let $\lambda \in P_{++},c_{\lambda}=$ ch(E( $\lambda$ )) and $d= J(e^{\rho})$ . Prove that

\Delta (c_{\lambda}d) =\langle \lambda +\rho , \lambda +\rho \rangle c_{\lambda}d

(Use $a$ ) $,b)$ , §6, Cor. of Prop. 7, and Chap. VI, §3, no. 3, formula (3).)

d) Deduce another proof of the formula of H. Weyl from the above and §7, no. 2, Prop. 5 (iii).

e) For all $\lambda \in \mathfrak{h}^*$ , put dim $E^{\lambda}=m(\lambda )$ . Deduce from a) that

Tr(( $xy$ ) $_E|E^{\lambda}$ ) $=\sum^{+\infty}_{i=0}(\lambda +i\alpha )([x, y])m(\lambda +i\alpha )$

\sum_{i=-\infty}^{+\infty}(\lambda +i\alpha )([x, y])m(\lambda +i\alpha ) = 0

f) Let $\langle \cdot ,\cdot \rangle$ be a non-degenerate invariant symmetric bilinear form on $\mathfrak{g}$ whose restriction to $\mathfrak{h}$ is the inverse of the form chosen above. Let $\Gamma$ be the corresponding Casimir element. Assume that E is simple; put $\Gamma_E=\gamma .1$ , where $\gamma \in k$ . By using e) and §2, no. 3, Prop. 6, show that

\gamma m(\lambda ) =\langle \lambda , \lambda \rangle m(\lambda ) +\sum_{\alpha\in R}\sum^{+\infty}_{i=0}\langle \lambda +i\alpha , \alpha \rangle m(\lambda +i\alpha )

for all $\lambda \in \mathfrak{h}^*$ , and then that

\gamma m(\lambda ) =\langle \lambda , \lambda \rangle m(\lambda ) +\sum_{\alpha\in R_+}m(\lambda )\langle \lambda , \alpha \rangle + 2\sum_{\alpha\in R_+}\sum^{+\infty}_{i=1}m(\lambda +i\alpha )\langle \lambda +i\alpha , \alpha \rangle

=\langle \lambda , \lambda + 2\rho \rangle m(\lambda ) + 2\sum_{\alpha\in R_+}\sum^{+\infty}_{i=1}m(\lambda +i\alpha )\langle \lambda +i\alpha , \alpha \rangle

g) Continue to assume that E is simple; let $\omega$ be its highest weight. Deduce from f) that, for all $\lambda \in \mathfrak{h}^*$ ,

(\langle \omega +\rho , \omega +\rho \rangle - \langle \lambda +\rho , \lambda +\rho \rangle )m(\lambda ) = 2\sum_{\alpha\in R_+}\sum^{+\infty}_{i=1}m(\lambda +i\alpha )\langle \lambda +i\alpha , \alpha \rangle

(Recall that, by Prop. 5 of § $7,\langle \omega +\rho , \omega +\rho \rangle >\langle \lambda +\rho , \lambda +\rho \rangle$ if $\lambda$ is a weight of E distinct from $\omega$ . The preceding formula thus gives a procedure for calculating $m(\lambda )$ step by step.) $^{18}$

Solution

No solution has been written yet.

Cited by

Nothing.

Exercise 5, § 9

Cited by

Cites