Proposition 8

Assume that $\mathfrak{g}$ is semi-simple $\not= 0$. Let $\mathfrak{h}$ be a splitting Cartan subalgebra of $\mathfrak{g}$, R the root system of $(\mathfrak{g},\mathfrak{h})$, B a basis of R$,h^0$ the element of $\mathfrak{h}$ such that $\alpha (h^0) = 2$ for all $\alpha \in B$.

(i) The element $h^0$ is simple and principal.

(ii) The elements $x$ of $\mathfrak{g}$ such that there exists an $\mathfrak{s}\mathfrak{l}_2$-triplet of the form $(x, h^0, y)$ are the elements of $\sum_{\alpha\in B}\mathfrak{g}^{\alpha}$ that have a non-zero component in each $\mathfrak{g}^{\alpha}$.

The element $h^0$ is that considered in §7, no. 5, Lemma 2 (cf. loc. cit., formula (1)). It follows from this lemma that $h^0$ is simple principal and that, if $x\in \sum_{\alpha\in B}\mathfrak{g}^{\alpha}$ has a non-zero component in each $\mathfrak{g}^{\alpha}$, there exists an $\mathfrak{s}\mathfrak{l}_2$-triplet of the form $(x, h^0, y)$. Conversely, let $(x, h^0, y)$ be a $\mathfrak{s}\mathfrak{l}_2$-triplet. We have $[h^0, x] = 2x$, so $x\in \sum_{\gamma\in R,\gamma(h^0)=2}\mathfrak{g}^{\gamma}=\sum_{\alpha\in B}\mathfrak{g}^{\alpha}$. Similarly, $y\in \sum_{\alpha\in B}\mathfrak{g}^{-\alpha}$. Write

$h^0=\sum_{\alpha\in B}a_{\alpha}H_{\alpha}$ where $a_{\alpha}>0$ for all $\alpha \in B$,

$x=\sum_{\alpha\in B}X_{\alpha}$ where $X_{\alpha}\in \mathfrak{g}^{\alpha}$ for all $\alpha \in B$,

$y=\sum_{\alpha\in B}X_{-\alpha}$ where $X_{-\alpha}\in \mathfrak{g}^{-\alpha}$ for all $\alpha \in B$.

Then

so $[X_{-\alpha}, X_{\alpha}]\not= 0$ for all $\alpha \in B$.