Lemma 2

The endomorphisms $X_{\alpha}^0,H_{\beta}^0,X_{-\gamma}^0$, where $\alpha , \beta , \gamma \in B$, are linearly independent.

Since $X_{-\alpha}^0(\emptyset ) =\alpha$, it is clear that the $X_{-\alpha}^0$ are linearly independent.

Assume that $\sum_{\alpha}a_{\alpha}H_{\alpha}^0= 0$; then, for all $\beta \in B$,

since det($n(\beta , \alpha )$)$\not= 0$, it follows that $a_{\alpha}$ = 0 for all $\alpha$. Assume that $\sum_{\alpha}a_{\alpha}X_{\alpha}^0= 0$. In view of formulas (7), (8), (9),

for all $\beta \in B$. It follows that $a_{\beta}= 0$ for all $\beta$. Since $X_{\alpha}^0, H_{\alpha}^0, X_{-\alpha}^0$ are of degree $-1,0,1$, respectively, the lemma follows from what has gone before.

Let I be the set $B\times \{-1,0,1\}$. Put $x_{\alpha}= (\alpha ,-1),h_{\alpha}= (\alpha ,0)$, and $x_{-\alpha}= (\alpha ,1)$. Let $\mathfrak{a}$ be the Lie algebra defined by the generating family I and the following set $\mathscr{R}$ of relators:

$[x_{\alpha}, x_{-\beta}]$ if $\alpha \not=\beta$

(cf. Chap. II, §2, no. 3). By Lemma 1, there exists a unique linear representation $\rho$ of $\mathfrak{a}$ on E such that

In view of Lemma 2, this proves the following result: