Proposition 3

Let $\mathscr{H}$ be a permissible lattice in $\mathfrak{h}$. Let $\mathscr{U}_+,\mathscr{U}_-,\mathscr{U}_0$ be the $\mathbf{Z}$-subalgebras of $U(\mathfrak{g})$ generated respectively by the elements $x_{\alpha}^{(n)}$ $(\alpha \in B,n\in \mathbf{N}),y_{\alpha}^{(n)}(\alpha \in B, n\in \mathbf{N}),(^h_n)(h\in \mathscr{H}, n\in \mathbf{N})$. Let $\mathscr{U}$ be the $\mathbf{Z}$-subalgebra of $U(\mathfrak{g})$ generated by $\mathscr{U}_+,\mathscr{U}_-,\mathscr{U}_0$.

(i) $\mathscr{U}$ is a biorder in the bigebra $U(\mathfrak{g})$.

(ii) We have $\mathscr{U}=\mathscr{U}_-.\mathscr{U}_0.\mathscr{U}_+,\mathscr{U}\cap \mathfrak{h}=\mathscr{H}$ and, for all $\alpha \in B$,

By Lemma 5 and Prop. $2,\mathscr{U}_+,\mathscr{U}_-,\mathscr{U}_0$ are orders in the $\mathbf{Q}$-algebras $U(\mathfrak{n}_+),U(\mathfrak{n}_-),U(\mathfrak{h})$, respectively, and

$\in \mathscr{U}_0$ for $h\in \mathscr{H}, q\in \mathbf{Z}, p\in \mathbf{N}$. (23)

Put $\mathscr{L}=\mathscr{U}_-.\mathscr{U}_0.\mathscr{U}_+\subset U(\mathfrak{g})$. By (22), $\mathscr{L}$ is a lattice in $U(\mathfrak{g})$. By construction,

while Lemma 3 and (23) imply that

Let $\alpha \in B, n\in \mathbf{N}, r\in \mathbf{N}, \varphi = (\varphi (i))\in B^r$, and

We show that

or equivalently, in view of (25), that

We argue by induction on $r$. The bracket to be studied is the sum of the terms

$(m(1))(m(k))(n)(m(k+1))(m(k+2))(m(r))$

$y_{\varphi(1)}. . . y_{\varphi(k)}[x_{\alpha}, y_{\varphi(k+1)}]y_{\varphi(k+2)}. . . y_{\varphi(r)}$. (30) For $\alpha \not=\varphi (k+ 1),x_{\alpha}$ and $y_{\varphi(k+1)}$ commute, so $[x^{(n)}_{\alpha}, y^{(m(k+1))}_{\varphi(k+1)}] = 0$. If $\alpha =\varphi (k+ 1)$, the expression (30) is, by (17), the sum of expressions of the form

$(m(1))(m(k))(m(k+1)-p)(q-h)(n-p)(m(k+2))(m(r))$

where $q\in \mathbf{Z}, p\in$ N -- $\{0\}, h\in \mathscr{H}$. The induction hypothesis, together with (24) and (26), proves that the expression (31) belongs to $\mathscr{L}$. We have thus proved (28).

By (28), $x^{(n)}_{\alpha}\mathscr{U}_-\subset \mathscr{L}$; thus, by (25) and (27), $x^{(n)}_{\alpha}\mathscr{L}\subset \mathscr{L}$, so $\mathscr{U}_+.\mathscr{L}\subset \mathscr{L}$ and

Thus, $\mathscr{L}$ is a $\mathbf{Z}$-subalgebra of $U(\mathfrak{g})$, so $\mathscr{U}=\mathscr{L}$. If $c$ is the coproduct of $U(\mathfrak{g}),c(\mathscr{U})\subset \mathscr{U}\otimes_{\mathbf{Z}}\mathscr{U}$ (no. 2, Prop. 1). Let $\gamma$ be the counit of $U(\mathfrak{g})$. Since $\gamma (x^{(n)}_{\alpha}) =\gamma (y^{(n)}_{\alpha}) =\gamma ((^h_n))= 0$ for $n >0$, we have $\gamma (\mathscr{U})\subset \mathbf{Z}$. This proves (i). On the other hand,

by Prop. 2 of no. 4; similarly,

by Lemma 5. This proves (ii).