Proposition 10

Let $\lambda , \mu\in P_{++}$. Let $E,F,G$ be simple $\mathfrak{g}$-modules with highest weights $\lambda , \mu, \lambda +\mu$. Let $\mathscr{X}$ (resp. $\mathscr{X}',\mathscr{X}''$) be the set of weights of E (resp. $F,G$). Then $\mathscr{X}''=\mathscr{X}+\mathscr{X}'$.

We have $E =\bigoplus_{\nu\in P}E^{\nu},F =\bigoplus_{\sigma\in P}F^{\sigma}$, so $E\otimes F$ is the direct sum of the

By Prop. 9, G can be identified with a $\mathfrak{g}$-submodule of $E\otimes F$, so $\mathscr{X}''\subset \mathscr{X}+\mathscr{X}'$. We have $G^{\tau}= G\cap (E\otimes F)^{\tau}$, and it is enough to show that, for $\nu \in \mathscr{X}$ and $\sigma \in \mathscr{X}'$, we have $G\cap (E\otimes F)^{\nu+\sigma}\not= 0$. Let $(e_1, . . . , e_n)$ (resp. $(f_1, . . . , f_p)$) be a basis of E (resp. F) consisting of elements each of which belong to some $E^{\nu}$ (resp. $F^{\sigma}$), and such that $e_1\in E^{\lambda}$ (resp. $f_1\in F^\mu$). The $e_i\otimes f_j$ form a basis of $E\otimes F$. Suppose that the result to be proved is false. Then there exists a pair $(i, j)$ such that the coordinate of index $(i, j)$ of every element of G is zero. Let U be the enveloping algebra of $\mathfrak{g}, U'$ the dual of $U,c$ the coproduct of U. For all $u\in U$, let $x_i(u)$ (resp. $y_j(u)$) be the coordinate of $u(e_1)$ (resp. $u(f_1)$) of index $i$ (resp. $j$); let $z_{ij}(u)$ be the coordinate of index $(i, j)$ of $u(e_1\otimes f_1)$. Then $x_i, y_j, z_{ij}\in U'$. Now $e_1$ generates the $\mathfrak{g}$-module E, so $x_i\not= 0$, and similarly $y_j\not= 0$. By the definition of the $\mathfrak{g}$-module $E\otimes F$ (Chap. I, §3, no. 2), if $c(u) =\sum u_s\otimes u'_s$, we have

In other words, $z_{ij}$ is the product of $x_i$ and $y_j$ in the algebra $U'$. But this algebra is an integral domain (Chap. II, §1, no. 5, Prop. 10), so $z_{ij}\not= 0$. Since $u(e_1\otimes f_1)\in G$ for all $u\in U$, this is a contradiction.