Proposition 2

Tag 0163, permanent. Cite it as [Bourbaki, Tag 0163].

Lie Groups and Lie Algebras, chapter VIII, § 9. The formula of Hermann Weyl

Let $\lambda , \mu\in P_{++}$ . In $\mathscr{R}(\mathfrak{g})$ , we have

[\lambda ].[\mu] =\sum_{\nu\in P_{++}}m(\lambda , \mu, \nu )[\nu ]

with

m(\lambda , \mu, \nu ) =\sum_{w,w'\in W}\varepsilon (ww')\mathfrak{P}(w(\lambda +\rho ) +w'(\mu+\rho )-(\nu + 2\rho ))

Let $E,F$ be finite dimensional simple $\mathfrak{g}$ -modules of highest weights $\lambda , \mu$ . Let $l_{\nu}$ be the length of the isotypical component of $E\otimes F$ of highest weight $\nu$ . It suffices to show that

l_{\nu}=\sum_{w,w'\in W}\varepsilon (ww')\mathfrak{P}(w(\lambda +\rho ) +w'(\mu+\rho )-(\nu + 2\rho )) \tag{2}

Put $c_1=$ ch(E) $=\sum_{\sigma\in P}m_{\sigma}e^{\sigma},c_2=$ ch(F), and $d= J(e^{\rho})$ , where J is defined as in no. 2. We have $\sum_{\xi\in P_{++}}l_{\xi}$ ch[ $\xi ] =$ ch(E $\otimes F$ ) $=c_1c_2$

so, after multiplying by $d$ and using Th. 1,

\sum_{\xi\in P_{++}}l_{\xi}J(e^{\xi+\rho}) =c_1J(e^{\mu+\rho}) =(\sum_{\sigma\in P}m_{\sigma}e^{\sigma})((\sum_{w\in W}\varepsilon (w)e^{w(\mu+\rho)}) \tag{3}

=\sum_{\tau\in P}(\sum_{w\in W}\varepsilon (w)m_{\tau+\rho-w(\mu+\rho)})e^{\tau+\rho}

Now, if $\xi \in P_{++},\xi +\rho$ belongs to the chamber defined by B (Chap. VI, §1, no. 10); thus, for all $w\in W$ distinct from 1, we have $w(\xi +\rho )\notin P_{++}$ . Consequently, the coefficient of $e^{\nu+\rho}$ in $\sum_{\xi\in P_{++}}l_{\xi}J(e^{\xi+\rho})$ is equal to $l_{\nu}$ . In view of (3), we obtain

l_{\nu}=\sum_{w\in W}\varepsilon (w)m_{\nu+\rho-w(\mu+\rho)}

that is, by Prop. 1,

l_{\nu}=\sum_{w,w'\in W}\varepsilon (w)\varepsilon (w')\mathfrak{P}(w'(\lambda +\rho )-(\nu +\rho -w(\mu+\rho ) +\rho ))

which proves (2).

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.

Proposition 2

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