Lemma 6

Let $\alpha , \beta \in B$ with $\alpha \not=\beta$. Then

The second formula follows from the first by using the automorphism $\theta$. To prove the first, it suffices to show that $[x_{\gamma}, y_{\alpha \beta}] = 0$ for all $\gamma \in B$. We distinguish three cases.

Case 1$:\gamma \not=\alpha$ and $\gamma \not=\beta$. In this case, $x_{\gamma}$ commutes with $x_{-\alpha}$ and $x_{-\beta}$, and hence with $y_{\alpha \beta}$.

Case 2$:\gamma =\beta$. In this case, $x_{\gamma}$ commutes with $x_{-\alpha}$, so

$[x_{\gamma}, y_{\alpha \beta}] =$ (ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}[x_{\gamma}, x_{-\beta}]$

$=-$(ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}h_{\beta}=-n(\alpha , \beta$)(ad $x_{-\alpha}$)$^{-n(\beta ,\alpha)}x_{-\alpha}$.

If $n(\beta , \alpha )<0$, this expression is zero since (ad $x_{-\alpha}$)$.x_{-\alpha}= 0$. If $n(\beta , \alpha ) = 0$, then $n(\alpha , \beta ) = 0$. In both cases, $[x_{\gamma}, y_{\alpha \beta}] = 0$.

Case 3$:\gamma =\alpha$. In the algebra of endomorphisms of $\mathfrak{a}$, $[$-ad $h_{\alpha}$, ad $x_{-\alpha}] = 2$ ad $x_{-\alpha}$

and [ad $x_{\alpha}$, ad $x_{-\alpha}] =$-ad $h_{\alpha}$; thus, by §1, Lemma 1,

[ad $x_{\alpha}$, (ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}] = (1-n(\beta , \alpha$))(ad $x_{-\alpha}$)$^{-n(\beta ,\alpha)}($-ad $h_{\alpha}-n(\beta , \alpha ))$.

Consequently,

$[x_{\gamma}, y_{\alpha \beta}] =$ [ad $x_{\alpha}$, (ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}]x_{-\beta}+$ (ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}$(ad $x_{\alpha}$)$x_{-\beta}$

$=-(1-n(\beta , \alpha$))(ad $x_{-\alpha}$)$^{-n(\beta ,\alpha)}$(ad $h_{\alpha}+n(\beta , \alpha )$)$x_{-\beta}$

+ (ad $x_{-\alpha}$)$^{1-n(\beta ,\alpha)}$(ad $x_{\alpha}$)$x_{-\beta}$.

Now $[h_{\alpha}, x_{-\beta}] +n(\beta , \alpha )x_{-\beta}= 0$ and $[x_{\alpha}, x_{-\beta}] = 0$, so $[x_{\gamma}, y_{\alpha \beta}] = 0$.