Remark

Retain the notations of Prop. 9. Let C be the isotypical component of $E\otimes F$ of highest weight $\lambda +\mu$. Then C depends only on E and F and not on the choice of $\mathfrak{h}$ and the basis B. In other words, let $\mathfrak{h}'$ be a splitting Cartan subalgebra of $\mathfrak{g}, R'$ the root system of $(\mathfrak{g},\mathfrak{h}')$, and $B'$ a basis of $R'$; let $\lambda ', \mu'$ be the highest weights of $E,F$ relative to $\mathfrak{h}'$ and $B'$; let $C'$ be the isotypical component of $E\otimes F$ of highest weight $\lambda '+\mu'$; then $C'= C$. Indeed, to prove this we can assume, by extension of the base field, that $k$ is algebraically closed. Then there exists $s\in$ Aut$_e(\mathfrak{g})$ that takes $\mathfrak{h}$ to $\mathfrak{h}', R$ to $R', B$ to $B'$. Let $S\in \mathbf{S}\mathbf{L}(E\otimes F)$ have the properties in Prop. 2 of no. 1. Then $S((E\otimes F)^{\lambda+\mu}) = (E\otimes F)^{\lambda'+\mu'}$ and S(C) = C. Hence $(E\otimes F)^{\lambda'+\mu'}\subset C'\cap S(C) = C'\cap C$, so $C'= C$. Thus, to 2 classes of finite dimensional simple $\mathfrak{g}$-modules we can associate canonically a third; in other words, we have defined on the set $\mathfrak{S}_{\mathfrak{g}}$ of classes of finite dimensional simple $\mathfrak{g}$-modules a composition law. With this structure, $\mathfrak{S}_{\mathfrak{g}}$ is canonically isomorphic to the additive monoid $P_{++}$.