§ 12. Chevalley orders

1. Lattices and orders

Let V be a $\mathbf{Q}$ -vector space. A lattice in V is a free $\mathbf{Z}$ -submodule $\mathscr{V}$ of V such that the $\mathbf{Q}$ -linear map $\alpha_{\mathscr{V},V}:\mathscr{V}\otimes_{\mathbf{Z}}\mathbf{Q}\rightarrow V$ induced by the injection of $\mathscr{V}$ into V is bijective. When V is finite dimensional, this is the same as saying that $\mathscr{V}$ is a $\mathbf{Z}$ -submodule of finite type which generates the $\mathbf{Q}$ -vector space V (recall that a torsion-free $\mathbf{Z}$ -module of finite type is free by Algebra, Chap. VII, §4, no. 4, Cor. 2); moreover, in this case our definition is a special case of that of Commutative Algebra, Chap. VII, §4, no. 1, Def. 1 (loc. cit., Example 3). If W is a vector subspace of V, and $\mathscr{V}$ is a lattice in V, then $\mathscr{V}\cap W$ is a lattice in W.

If V is a $\mathbf{Q}$ -algebra, an order in V is a lattice $\mathscr{V}$ in the underlying vector space that is a $\mathbf{Z}$ -subalgebra of V; the map $\alpha_{\mathscr{V},V}$ is then an isomorphism of $\mathbf{Q}$ -algebras. If V is a unital $\mathbf{Q}$ -algebra, a unital order in V is an order in V containing the unit element.

Assume that V is a $\mathbf{Q}$ -bigebra, with coproduct $c$ and counit $\gamma$ . If $\mathscr{V}$ is a lattice in the vector space V, the canonical map $i:\mathscr{V}\otimes_{\mathbf{Z}}\mathscr{V}\rightarrow V\otimes_{\mathbf{Q}}V$ is injective; a biorder in V is a unital order $\mathscr{V}$ in the unital algebra V such that $\gamma (\mathscr{V})\subset \mathbf{Z}$ and $c(\mathscr{V})\subset i(\mathscr{V}\otimes_{\mathbf{Z}}\mathscr{V})$ ; the maps

$\gamma_{\mathscr{V}}:\mathscr{V}\rightarrow \mathbf{Z}$ and $c_{\mathscr{V}}:\mathscr{V}\rightarrow \mathscr{V}\otimes_{\mathbf{Z}}\mathscr{V}$

induced by $\gamma$ and $c$ give $\mathscr{V}$ the structure of a $\mathbf{Z}$ -bigebra, and the map $\alpha_{\mathscr{V},V}$ is then an isomorphism of $\mathbf{Q}$ -bigebras.

2. Divided powers in a bigebra

Let A be a unital $k$ -algebra, $x\in A,d\in k,n\in \mathbf{N}$ . Put

x(x-d). . .(x-d(n-1))^{n-1}

x^{(n,d)}==\prod(x-id)/(i+ 1) \tag{1}

$n$ !

i=0

In particular, $x^{(0,d)}= 1,x^{(1,d)}=x$ . We agree that $x^{(n,d)}= 0$ for $n$ an integer $<0$ . Put

$x^{(n)}=x^{(n,0)}=xn^n$ ! (2)

$(nx)=x^{(n,1)}=x(x-1). . .n($ ! $x-n+ 1)$ . (3)

Proposition 1 017V

Let A be a bigebra, with coproduct $c$ , and $x$ a primitive element (Chap. II, §1, no. 2) of A. Then

c(x^{(n,d)}) =\sum x^{(p,d)}\otimes x^{(q,d)} \tag{4}

p\in \mathbf{N},q\in \mathbf{N},p+q=n

The proposition is trivial for $n\leq 0$ . We argue by induction on $n$ . If formula (4) is true for $n$ , then

(n+ 1)c(x^{(n+1,d)}) =c(x-dn)c(x^{(n,d)})

= (x\otimes 1 + 1\otimes x-dn1\otimes 1)c(x^{(n,d)})

=\sum_{p+q=n}[xx^{(p,d)}\otimes x^{(q,d)}+x^{(p,d)}\otimes xx^{(q,d)}-(p+q)dx^{(p,d)}\otimes x^{(q,d)}]

=\sum_{p+q=n}(x-pd)x^{(p,d)}\otimes x^{(q,d)}+\sum_{p+q=n}x^{(p,d)}\otimes (x-qd)x^{(q,d)}

=\sum_{p+q=n}(p+ 1)x^{(p+1,d)}\otimes x^{(q,d)}+\sum_{p+q=n}(q+ 1)x^{(p,d)}\otimes x^{(q+1,d)}

=\sum_{r+s=n+1}rx^{(r,d)}\otimes x^{(s,d)}+\sum_{r+s=n+1}sx^{(r,d)}\otimes x^{(s,d)}

= (n+ 1)\sum_{r+s=n+1}x^{(r,d)}\otimes x^{(s,d)}

hence formula (4) for $n+ 1$ .

3. Integral variant of the poincaré-birkhoff-witt theorem

Let $\mathfrak{g}$ be a finite dimensional $\mathbf{Q}$ -Lie algebra, $U(\mathfrak{g})$ its enveloping bigebra. If I is a totally ordered set, $\mathbf{x}= (x_i)_{i\in I}$ a family of elements of $\mathfrak{g}$ , and $\mathbf{n}=$ $(n_i)_{i\in I}\in \mathbf{N}^{(I)}$ a multi-index, put

$\mathbf{x}^{(\mathbf{n})}=\prod x\overset{ni}{n_{i}}$ ! $^{^i}$ , (5)

i\in I

the product being calculated in $U(\mathfrak{g})$ in accordance with the ordered set I.

Theorem 1 017W

Let $\mathscr{U}$ be a biorder in the bigebra $U(\mathfrak{g})$ . Let $\mathscr{G}=\mathscr{U}\cap \mathfrak{g}$ , which is an order in the Lie algebra $\mathfrak{g}$ . Let $(x_i)_{i\in I}$ be a basis of $\mathscr{G}$ . Give I a total order, and assume that we are given, for all $\mathbf{n}\in \mathbf{N}^I$ , an element $[\mathbf{n}]$ of $\mathscr{U}$ such that $[\mathbf{n}]-\mathbf{x}^{(\mathbf{n})}$ has filtration $<|\mathbf{n}|$ in $U(\mathfrak{g})$ . Then, the family of the $[\mathbf{n}]$ for $\mathbf{n}\in \mathbf{N}^I$ is a basis of the $\mathbf{Z}$ -module $\mathscr{U}$ .

For $p\in \mathbf{N}$ , let $U_p(\mathfrak{g})$ be the set of elements of $U(\mathfrak{g})$ of filtration $\leq p$ ; then the images in $U_p(\mathfrak{g})/U_{p-1}(\mathfrak{g})$ of the $\mathbf{x}^{(\mathbf{n})}$ such that $|\mathbf{n}|=p$ form a basis of this $\mathbf{Q}$ -vector space (Chap. I, §2, no. 7, Th. 1); hence the $[\mathbf{n}]$ form a basis of the $\mathbf{Q}$ -vector space $U(\mathfrak{g})$ . It remains to prove the following assertion (in which we put $M =\mathbf{N}^I$ ) $:$

(*) if $u\in \mathscr{U}, (a_{\mathbf{n}})\in \mathbf{Z}^{(M)}$ , and $d\in$ N -- $\{0\}$ are such that

du=\sum_{\mathbf{n}\in M}a_{\mathbf{n}}[\mathbf{n}] \tag{6}

then $d$ divides each $a_{\mathbf{n}}$ .

For each integer $r\geq 0$ , introduce the iterated coproduct

c_i:\mathscr{U}\rightarrow \mathbf{T}^r(\mathscr{U}) =\mathscr{U}\otimes \mathscr{U}\otimes \cdots \otimes \mathscr{U}

by definition, $c_0$ is the counit of $\mathscr{U},c_1=$ Id $_{\mathscr{U}},c_2=c$ (the coproduct of $\mathscr{U}$ ), and, for $r\geq 2,c_{r+1}$ is defined as the composite $p\circ (c_r\otimes 1)\circ c:$

\mathscr{U}\longrightarrow^c\mathscr{U}\otimes_{\mathbf{Z}}\mathscr{U}^c\longrightarrow^{^r\otimes 1}\mathbf{T}^r(\mathscr{U})\otimes_{\mathbf{Z}}\mathscr{U}\longrightarrow^p\mathbf{T}^{r+1}(\mathscr{U})

where $p$ is defined by using the multiplication in the algebra $\mathbf{T}(\mathscr{U})$ . Further, consider the canonical projection $\pi$ of $\mathscr{U}$ onto $\mathscr{U}^+=$ Ker $c_0$ , and the composite

c^+_r=\mathbf{T}^r(\pi )\circ c_r:\mathscr{U}\rightarrow \mathbf{T}^r(\mathscr{U}^+)

Lemma 1 017X

Let $\mathbf{n}\in \mathbf{N}^I$ . If $|\mathbf{n}|< r$ , then $c^+_r([\mathbf{n}]) = 0$ . If $|\mathbf{n}|=r$ , then

c^+_r([\mathbf{n}]) =\sum_{\varphi}x_{\varphi(1)}\otimes x_{\varphi(2)}\otimes \cdots \otimes x_{\varphi(r)} \tag{7}

where $\varphi$ belongs to the set of maps from $\{1,2, . . . , r\}$ to I which take each value $i\in In_i$ times.

By Prop. 1,

c_r(\mathbf{x}^{(\mathbf{n})}) =\sum\mathbf{x}^{(\mathbf{p}_1)}\otimes \cdots \otimes \mathbf{x}^{(\mathbf{p}_r)}

where the summation extends over the set of sequences $(\mathbf{p}_1, . . . ,\mathbf{p}_r)$ of $r$ elements of M such that $\mathbf{p}_1+\cdots +\mathbf{p}_r=\mathbf{n}$ . In view of Chap. II, §1, no. 3, Prop. 6, the map $c^+_r$ , extended by linearity to a map from $U(\mathfrak{g})$ to $\mathbf{T}^r(U^+(\mathfrak{g}))$ , vanishes on $U_{r-1}(\mathfrak{g})$ . It follows that, for $r\geq |\mathbf{n}|$ ,

c^+_r([\mathbf{n}]) =c^+_r(\mathbf{x}^{(\mathbf{n})}) =\sum\pi (\mathbf{x}^{(\mathbf{p}_1)})\otimes \cdots \otimes \pi (\mathbf{x}^{(\mathbf{p}_r)}) \tag{8}

For $r >|\mathbf{n}|$ , the relation $\mathbf{p}_1+\cdots +\mathbf{p}_r=\mathbf{n}$ implies that at least one of the $\mathbf{p}_i$ is zero, so $c^+_r([\mathbf{n}]) = 0$ . For $r=|\mathbf{n}|$ , the only non-zero terms of the third member of (8) are those for which $|\mathbf{p}_1|=\cdots =|\mathbf{p}_r|= 1$ , hence (7).

We return to the proof of Th. 1. We retain the notations of (*) and prove, by descending induction on $|\mathbf{n}|$ , that $d$ divides $a_{\mathbf{n}}$ , which is clear when $|\mathbf{n}|$ is sufficiently large. If $d$ divides $a_{\mathbf{n}}$ for $|\mathbf{n}|> r$ then, putting

u'=u-\sum_{|\mathbf{n}|>r}(a_{\mathbf{n}}/d)[\mathbf{n}]\in \mathscr{U}

we have

du'=\sum_{|\mathbf{n}|\leq r}a_{\mathbf{n}}[\mathbf{n}] \tag{9}

For any map $\varphi$ from $\{1,2, . . . , r\}$ to I, put

e_{\varphi}=x_{\varphi(1)}\otimes \cdots \otimes x_{\varphi(r)}

and $a_{\varphi}=a_{\mathbf{n}}$ where $\mathbf{n}=$ (Card $\varphi^{-1}(i)$ ) $_{i\in I}$ . By Lemma 1, (9) implies that

dc^+_r(u') =\sum_{\varphi\in I^r}a_{\varphi}e_{\varphi} \tag{10}

so $c^+_r(u')\in \mathbf{T}^r(\mathscr{U}^+)\cap \mathbf{Q}\mathbf{T}^r(\mathscr{G})$ . But the submodule $\mathscr{G}$ of $\mathscr{U}^+$ is a direct factor (Algebra, Chap. VII, §4, no. 3, Cor. of Th. 1), so the submodule $\mathbf{T}^r(\mathscr{G})$ is a direct factor of $\mathbf{T}^r(\mathscr{U}^+)$ , and hence $c^+_r(u')\in \mathbf{T}^r(\mathscr{G})$ . On the other hand, the $x_i$ form a basis of $\mathscr{G}$ by hypothesis, so the $e_{\varphi}$ form a basis of $\mathbf{T}^r(\mathscr{G})$ . Then (10) proves that $d$ divides the $a_{\varphi}$ , that is, the $a_{\mathbf{n}}$ for $|\mathbf{n}|=r$ . This proves (*).

4. Example: polynomials with integer values

Let V be a finite dimensional $\mathbf{Q}$ -vector space, $V^*$ its dual, $\mathscr{V}$ a lattice in V, $\mathscr{V}^*$ the dual $\mathbf{Z}$ -module of $\mathscr{V}$ , which can be identified canonically with a lattice in $V^*,\mathbf{S}(V)$ the symmetric algebra of V, and

\lambda :\mathbf{S}(V)\rightarrow A(V^*)

the canonical bijection from $\mathbf{S}(V)$ to the algebra of polynomial functions on $V^*($ Algebra, Chap. IV, §5, no. 11, Remark 1). If we identify $A(V^*\times V^*)$ with $A(V^*)\otimes_{\mathbf{Q}}A(V^*)$ , then $\lambda$ transforms the coproduct of $\mathbf{S}(V)$ into the map $A(V^*)\rightarrow A(V^*\times V^*)$ which associates to the polynomial function $\varphi$ on $V^*$ the polynomial function

(x, y) \rightarrow \varphi (x+y)

on $V^*\times V^*($ Algebra, Chap. IV, §5, no. 11, Remark 2).

Denote by $(^{\mathscr{V}}_{\mathbf{Z}})$ the subset of $\mathbf{S}(V)$ consisting of the elements which correspond to polynomial maps from $V^*$ to $\mathbf{Q}$ that take integer values on $\mathscr{V}^*$ .

Proposition 2 017Y

(i) $(^{\mathscr{V}}_{\mathbf{Z}})$ is a biorder in the bigebra $\mathbf{S}(V)$ , and $(^{\mathscr{V}}_{\mathbf{Z}})\cap V =\mathscr{V}$ .

(ii) The $\mathbf{Z}$ -algebra $(^{\mathscr{V}}_{\mathbf{Z}})$ is generated by the $(^h_n)$ for $h\in \mathscr{V}, n\in \mathbf{N}$ .

(iii) If $(h_1, . . . , h_r)$ is a basis of $\mathscr{V}$ , the elements

(h)(h_1)(h_r)

= $\cdots$ ,

\mathbf{n}n_1n_r

where $\mathbf{n}= (n_1, . . . , n_r)$ belongs to $\mathbf{N}^r$ , form a basis of the $\mathbf{Z}$ -module $(^{\mathscr{V}}_{\mathbf{Z}})$ .

For $m\in \mathbf{N}$ , put $\mathbf{S}_m(V) =\sum_{i\leq m}\mathbf{S}^i$ (V), $\mathbf{S}_m(\mathscr{V}) =\sum_{i\leq m}\mathbf{S}^i(\mathscr{V})$ . By Algebra, Chap. IV, §5, no. 9, Prop. 15 and Remark,

$\mathbf{S}_m(\mathscr{V})\subset \mathbf{S}_m(V)\cap (\mathscr{V}\mathbf{Z})\subset m1$ ! $\mathbf{S}_m(\mathscr{V})$

so $(^{\mathscr{V}}_{\mathbf{Z}})\cap V =\mathscr{V}$ . Since $\mathbf{S}_m(\mathscr{V})$ is a lattice in $\mathbf{S}_m$ (V), $\mathbf{S}_m(V)\cap (_{\mathbf{Z}}^{\mathscr{V}})$ is also a lattice in $\mathbf{S}_m(V)$ . On the other hand, $\mathbf{S}_m(V)\cap (^{\mathscr{V}}_{\mathbf{Z}})$ is a direct factor of $\mathbf{S}_{m+1}(V)\cap (_{\mathbf{Z}}^{\mathscr{V}})$ (since the quotient is torsion-free), hence it admits a complement which is a free $\mathbf{Z}$ -module. It follows that $(^{\mathscr{V}}_{\mathbf{Z}})$ is a free $\mathbf{Z}$ -module. It is clear that this is a unital order in the algebra $\mathbf{S}(V)$ . Let $(u_n)_{n\in\mathbf{N}}$ be a basis of the $\mathbf{Z}$ -module $(^{\mathscr{V}}_{\mathbf{Z}})$ . This is also a basis of the $\mathbf{Q}$ -module $\mathbf{S}(V)$ and, for all

\varphi \in \mathbf{S}(V\times V) =\mathbf{S}(V)\otimes_{\mathbf{Q}}\mathbf{S}(V)

there exists a unique sequence $(v_n)$ of elements of $\mathbf{S}(V)$ such that $\varphi =\sum u_n\otimes v_n$ . As above, identify $\mathbf{S}(V)$ with $A(V^*)$ and $\mathbf{S}(V)\otimes \mathbf{S}(V)$ with $A(V^*\times V^*)$ . If $\varphi \in (^{\mathscr{V}\times\mathscr{V}}_{\mathbf{Z}})$ , the polynomial function $x \rightarrow \varphi (x, y)$ belongs to $(^{\mathscr{V}}_{\mathbf{Z}})$ for all $y\in \mathscr{V}^*$ . It follows that $v_n(y)\in \mathbf{Z}$ for all $n$ and all $y\in \mathscr{V}^*$ , in other words that $v_n\in (^{\mathscr{V}}_{\mathbf{Z}})$ . This proves that the coproduct maps $(^{\mathscr{V}}_{\mathbf{Z}})$ to $(_{\mathbf{Z}}^{\mathscr{V}})\otimes_{\mathbf{Z}}(_{\mathbf{Z}}^{\mathscr{V}})$ . If $h\in \mathscr{V}$ and $n\in \mathbf{N}$ , then $(_n^h)$ maps $u\in \mathscr{V}^*$ to the integer $(^{u(h)}_n)$ , so $(^h_n)\in (_{\mathbf{Z}}^{\mathscr{V}})$ . Assertion (iii) is now obtained by applying Th. 1 to the commutative Lie algebra V, and (ii) follows.

Corollary 017Z

Let X be an indeterminate. The polynomials $(^X_n)$ , where $n\in \mathbf{N}$ , form a basis of the $\mathbf{Z}$ -module consisting of the polynomials $P\in k[X]$ such that $P(\mathbf{Z})\subset \mathbf{Z}$ .

If $P(\mathbf{Z})\subset \mathbf{Z}$ , the Lagrange interpolation formula (Algebra, Chap. IV, §2, no. 1, Prop. 6) shows that the coefficients of P belong to $\mathbf{Q}$ ; thus, we can assume that $k=\mathbf{Q}$ and apply Prop. 2 with $V =\mathbf{Q},\mathscr{V}=\mathbf{Z}$ .

5. Some formulas

In this number, A denotes a unital associative algebra. If $x\in A$ , we write ad $x$ instead of ad $_Ax$ .

Lemma 2 0180

If $x, y\in A$ and $n\in \mathbf{N}$ ,

(ad $nx$ ! ) $^ny=\sum(-1)^qxp^p$ ! $yxq^q$ ! = $\sum(-1)^qx^{(p)}yx^{(q)}$ . (11)

p+q=np+q=n

Indeed, if we denote by $L_x$ and $R_x$ the maps $z \rightarrow xz$ and $z \rightarrow zx$ from A to A, we have, since $L_x$ and $R_x$ commute,

$n1$ !(ad $x$ ) $^n=n1$ ! $(L_x-R_x)^n=\sum_{p+q=n}(-1)^qp1$ ! $L^p_xq1$ ! $R^q_x$ .

Lemma 3 0181

Let $x, h\in A$ and $\lambda \in k$ be such that (ad $h$ ) $x=\lambda x$ . For all $n\in \mathbf{N}$ , and all $P\in k[X]$ , we have

P(h)x^{(n)}=x^{(n)}P(h+n\lambda ) \tag{12}

Since ad $h$ is a derivation of A and since (ad $h$ ) $x$ commutes with $x$ , we have

(ad $h$ ) $x^n=nx^{n-1}$ ((ad $h$ ) $x$ ) $=n\lambda x^n$ , (13) so

(ad $h$ ) $x^{(n)}=n\lambda x^{(n)}$ .

Thus, formula (12) follows from the special case

P(h)x=xP(h+\lambda ) \tag{14}

by replacing $x$ by $x^{(n)}$ and $\lambda$ by $n\lambda$ . It suffices to prove (14) when $P = X^m$ , by induction on $m$ . It is clear when $m= 0,1$ . If (14) is true for $P = X^m$ , then

h^{m+1}x=h.h^mx=hx(h+\lambda )^m=x(h+\lambda )^{m+1}

which proves (12).

Lemma 4 0182

Let $x, y, h\in A$ be such that

[y, x] =h,[h, x] = 2x,[h, y] =-2y \tag{15}

(i) For $m, n\in \mathbf{N}$ , we have

_{(n)(m)}\sum_{(m-p)}(m+n-p-1-h)_{(n-p)}

$xy$ = $yx$ . (16)

p

p\geq 0

(ii) Let $A'$ be the $\mathbf{Z}$ -subalgebra of A generated by the $x^{(m)}$ and the $y^{(m)}$ for $m\in \mathbf{N}$ . Then $(^h_n)\in A'$ for all $n\in \mathbf{N}$ .

Formula (16) can be written in the equivalent form

_{(n)(m)}\sum_{(m-p)}(m+n-p-1-h)_{(n-p)}

(ad $x$ ) $y$ = $yx.(17_m)$

p

p\geq 1

This is trivial for $m= 0$ . We argue by induction on $m$ . From $(17_m)$ , we obtain

$(m+$ 1)(ad $x^{(n)}$ ) $y^{(m+1)}=$ (ad $x^{(n)}$ ) $y^{(m)}.y+y^{(m)}$ .(ad $x^{(n)}$ ) $y$ (18)

_{(m-p)}(m+n-p-1-h)_{(n-p)(m)}-

=\sum yxy+y(n-1-h)x^{(n1)}

p

p\geq 1

(§1, no. 1, Lemma 1). Now, applying the same lemma, and then Lemma 3,

we have

(m+n-p-1-h)_{(n-p)}

xy

p

(m+n-p-1-h)_{(n-p)}-_-

= $(yx+ (n-p-1-h)x^{(np1)})$

p

(m+n-p+ 1-h)_{(n-p)}

=yx

p

(m+n-p-1-h)_{--}

+ $(n-p-1-h)x^{(np1)}$ .

p

Inserting this into (18), we obtain

$(m+$ 1)(ad $x^{(n)}$ ) $y^{(m+1)}$

_-(m+n-p+ 1-h)_-

=\sum(m-p+ 1)y^{(mp+1)}x^{(np)}

p

p\geq 1

_-(m+n-p-1-h)_{--}

+\sum y^{(mp)}(n-p-1-h)x^{(np1)}

p

p\geq 1

+y^{(m)}(n-1-h)x^{(n-1)}

_-(m+n-p+ 1-h)_-

=\sum(m-p+ 1)y^{(mp+1)}x^{(np)}

p

p\geq 1

_{(m-p)}(m+n-p-1-h)_{--}

$+\sum y(n-p-1-h)x^{(np1)}$ .

p

$p\geq 0$ Changing $p$ to $p-1$ in the second sum, and regrouping the terms, we obtain

$(m+$ 1)(ad $x^{(n)}$ ) $y^{(m+1)}=\sum_{p\geq 1}y^{(m-p+1)}A_px^{(n-p)}$ (19)

with

(m+n-p+ 1-h)(m+n-p-h)

$A_p= (m-p+ 1)+ (n-p-h)$ .

pp-1

Putting $z=m+n-p-h$ , this can also be written as

$A_p=p1$ ! $(m-p+ 1)(z+ 1)z(z-1). . .(z-p+ 2)$

+ 1 $(z-m)z(z-1). . .(z-p+ 2)$

$(p-$ 1)!

$=p1$ ! $z(z-1). . .(z-p+ 2)[(m-p+ 1)(z+ 1) +p(z-m)]$

(z)((m+ 1) +n-p-1-h)

= (m+ 1)= (m+ 1)

pp

Inserting this into (19), we obtain $(17_{m+1})$ , hence (i).

Assume that $(^h_p)\in A'$ for $p < n$ . Then, for all $P\in \mathbf{Q}[T]$ of degree $< n$ such that $P(\mathbf{Z})\subset \mathbf{Z}$ , we have $P(h)\in A'$ (no. 4, Cor. of Prop. 2). Hence, in view of (16) with $m=n$ ,

(h)(n-1-h)

$(-1)^n$ =

nn

^{n-1}_-(2n-p-1-h)_-'

$=-x^{(n)}y^{(n)}+\sum y^{(np)}x^{(np)}\in A$ ;

p

p=0

hence (ii) by induction on $n$ .

6. Biorders in the enveloping algebra of a split reductive Lie algebra

Let $\mathfrak{g}$ be a reductive Lie algebra over $\mathbf{Q},\mathfrak{h}$ a splitting Cartan subalgebra of $\mathfrak{g}$ , and $R = R(\mathfrak{g},\mathfrak{h}) ($ §2, no. 1, Remark 5).

Definition 1 0183

A lattice $\mathscr{H}$ in $\mathfrak{h}$ is said to be permissible (relative to $\mathfrak{g}$ ) if, for all $\alpha \in R$ , we have $H_{\alpha}\in \mathscr{H}$ and $\alpha (\mathscr{H})\subset \mathbf{Z}$ .

Remark 1 0184

Let B be a basis of R. A lattice $\mathscr{H}$ in $\mathfrak{h}$ is permissible if and only if $H_{\alpha}\in \mathscr{H}$ and $\alpha (\mathscr{H})\subset \mathbf{Z}$ for all $\alpha \in B$ .

Remark 2 0185

Let $\mathfrak{c}$ be the centre of $\mathfrak{g}$ . Then, a lattice $\mathscr{H}$ in $\mathfrak{h}$ is permissible if and only if $Q(R^{\vee})\subset \mathscr{H}\subset P(R^{\vee})\oplus \mathfrak{c}$ . The lattice $\mathscr{H}\cap \mathscr{D}\mathfrak{g}$ is then permissible in the Cartan subalgebra $\mathfrak{h}\cap \mathscr{D}\mathfrak{g}$ of $\mathscr{D}\mathfrak{g}$ . There may exist permissible lattices $\mathscr{H}$ such that $\mathscr{H}\not= (\mathscr{H}\cap \mathscr{D}\mathfrak{g})\oplus (\mathscr{H}\cap \mathfrak{c})$ (cf. §13, no. 1.IX).

Remark 3 0186

If $\mathfrak{g}$ is semi-simple, the permissible lattices in $\mathfrak{h}$ are the subgroups $\mathscr{H}$ of $\mathfrak{h}$ such that $Q(R^{\vee})\subset \mathscr{H}\subset P(R^{\vee})$ .

In the remainder of this number, we assume fixed a split reductive Lie algebra $(\mathfrak{g},\mathfrak{h})$ , a basis B of $R = R(\mathfrak{g},\mathfrak{h})$ and, for each $\alpha \in B$ , a pair $(x_{\alpha}, y_{\alpha})$ with

y_{\alpha}\in \mathfrak{g}^{-\alpha},x_{\alpha}\in \mathfrak{g}^{\alpha},[y_{\alpha}, x_{\alpha}] =H_{\alpha} \tag{20}

If we denote by $\mathfrak{n}_+$ (resp. $\mathfrak{n}_-$ ) the subalgebra of $\mathfrak{g}$ generated by the $x_{\alpha}$ (resp. the $y_{\alpha}$ ), we know (§3, no. 3, Prop. 9 (iii)) that

\mathfrak{g}=\mathfrak{n}_-\oplus \mathfrak{h}\oplus \mathfrak{n}_+ \tag{21}

U(\mathfrak{g}) = U(\mathfrak{n}_-)\otimes_{\mathbf{Q}}U(\mathfrak{h})\otimes_{\mathbf{Q}}U(\mathfrak{n}_+) \tag{22}

(where $U(\mathfrak{g}), . .$ . are the enveloping algebras of $\mathfrak{g}, . . .$ ).

Denote by $\mathscr{U}_+$ the $\mathbf{Z}$ -subalgebra of $U(\mathfrak{n}_+)$ generated by the $x^{(n)}_{\alpha}$ for $\alpha \in B$ and $n\in \mathbf{N}$ . Let W be the Weyl group of $R, R_+$ the set of positive roots relative to B.

Lemma 5 0187

(i) $\mathscr{U}_+$ is a lattice in the vector space $U(\mathfrak{n}_+)$ .

(ii) For all $\alpha \in B$ , we have $\mathscr{U}_+\cap U(\mathfrak{g}^{\alpha}) =\bigoplus_{n\in\mathbf{N}}\mathbf{Z}x^{(n)}_{\alpha}$ .

By definition, $\mathscr{U}_+$ is generated as a $\mathbf{Z}$ -module by the elements

$x(\varphi \mathbf{n})$ = $1\leq \prod i\leq rx(\varphi n((ii)$ ))

where $r\in \mathbf{N},\varphi = (\varphi (i))\in B^r$ , and $\mathbf{n}= (n(i))\in \mathbf{N}^r$ . Give the algebra $U(\mathfrak{n}_+)$ the graduation of type Q(R) for which each $\mathfrak{g}^{\alpha}(\alpha \in R_+)$ is homogeneous of degree $\alpha$ . A monomial $x^{(\mathbf{n})}_{\varphi}$ of the preceding type is homogeneous of degree

\sum_{1\leq i\leq r}n(i)\varphi (i)\in Q(R)

The monomials of this kind having a given degree $q$ are finite in number, and generate over $\mathbf{Q}$ the homogeneous component of $U(\mathfrak{n}_+)$ of degree $q$ . This proves (i).

If $\alpha \in B,\mathscr{U}_+\cap U(\mathfrak{g}^{\alpha})$ is contained in the sum of the homogeneous components of degrees which are multiples of $\alpha$ ; thus, by the preceding, $\mathscr{U}_+\cap U(\mathfrak{g}^{\alpha})$ is generated by the $x^{(\mathbf{n})}_{\varphi}$ such that $\sum n(i)\varphi (i)\in \mathbf{N}\alpha$ , which forces $\varphi (i) =\alpha$ for all $i$ (since B is a basis of R), so

$x^{(\mathbf{n})}_{\varphi}=x^{(n(1))}_{\alpha}. . . x^{(n(r))}_{\alpha}=(n(1) +n$ (1)! $\cdots . . . n+(nr$ )! $(r$ ))! $x^{(n(1)+\cdot \cdot \cdot+n(r))}_{\alpha}$ .

Thus, $\mathscr{U}_+\cap \mathscr{U}(\mathfrak{g}^{\alpha})\subset \bigoplus_n\mathbf{Z}x^{(n)}_{\alpha}$ , hence (ii). Q.E.D.

In the remainder of this paragraph, if E and F are $\mathbf{Z}$ -submodules of $U(\mathfrak{g})$ , we denote by $E.F$ the $\mathbf{Z}$ -submodule of $U(\mathfrak{g})$ generated by the products $ab$ , where $a\in E, b\in F$ .

Proposition 3 0188

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 $\alpha \in B$ ,

\mathscr{U}\cap \mathfrak{g}^{\alpha}=\mathbf{Z}x_{\alpha},V\cap \mathfrak{g}^{-\alpha}=\mathbf{Z}y_{\alpha}

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

(\pm h+q)

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

p

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,

\mathscr{U}_-.\mathscr{L}\subset \mathscr{L} \tag{24}

\mathscr{L}.\mathscr{U}_+\subset \mathscr{L} \tag{25}

while Lemma 3 and (23) imply that

\mathscr{U}_0.\mathscr{L}\subset \mathscr{L} \tag{26}

\mathscr{L}.\mathscr{U}_0\subset \mathscr{L} \tag{27}

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

(m(1), . . . , m(r))\in \mathbf{N}^r

We show that

x^{(n)}_{\alpha}y_{\varphi(1)}^{(m(1))}. . . y_{\varphi(r)}^{(m(r))}\in \mathscr{L} \tag{28}

or equivalently, in view of (25), that

[x^{(n)}_{\alpha}, y^{(m(1))}_{\varphi(1)}. . . y_{\varphi(r)}^{(m(r))}]\in \mathscr{L} \tag{29}

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))

y_{\varphi(1)}. . . y_{\varphi(k)}y_{\varphi(k+1)}x_{\alpha}y_{\varphi(k+2)}. . . y_{\varphi(r)} \tag{31}

p

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

\mathscr{L}.\mathscr{L}\subset \mathscr{U}_-.\mathscr{U}_0.\mathscr{L}\subset \mathscr{U}_-.\mathscr{L}\subset \mathscr{L}

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,

\mathscr{U}\cap \mathfrak{h}=\mathscr{L}\cap \mathfrak{h}=\mathscr{U}_0\cap \mathfrak{h}=\mathscr{H}

by Prop. 2 of no. 4; similarly,

\mathscr{U}\cap \mathfrak{g}^{\alpha}=\mathscr{U}_+\cap \mathfrak{g}^{\alpha}=\mathbf{Z}x_{\alpha}

by Lemma 5. This proves (ii).

Remark 4 0189

By Prop. 5 of §4, no. 4, there exists a unique automorphism $\theta$ of $\mathfrak{g}$ such that $\theta (x_{\alpha}) =y_{\alpha}$ and $\theta (y_{\alpha}) =x_{\alpha}$ for all $\alpha \in B$ , and $\theta (h) =-h$ for all $h\in \mathfrak{h}$ ; we have $\theta^2= 1$ . By construction of $\mathscr{U}$ , we see that the automorphism of $U(\mathfrak{g})$ that extends $\theta$ leaves $\mathscr{U}$ stable.

Corollary 1 018A

Put $\mathscr{G}=\mathscr{U}\cap \mathfrak{g}$ . Then $\mathscr{G}$ is an order in the Lie algebra $\mathfrak{g}$ , stable under $\theta$ . We have $\mathscr{G}=\mathscr{H}+\sum_{\alpha\in R}(\mathscr{G}\cap \mathfrak{g}^{\alpha})$ . For all $\alpha \in B$ and all $n\in \mathbf{N}$ , the maps (ad $x_{\alpha}$ ) $^n/n$ !, (ad $y_{\alpha}$ ) $^n/n$ ! leave $\mathscr{U}$ and $\mathscr{G}$ stable.

The first assertion is clear. The second follows by considering the graduation of type Q(R) on $U(\mathfrak{g})$ and $\mathscr{U}$ . The third follows from Lemma 2 of no. 5.

Corollary 2 018B

Let $w\in W$ . There exists an elementary automorphism $\varphi$ of $\mathfrak{g}$ that commutes with $\theta$ , leaves $\mathscr{G}$ and $\mathscr{U}$ stable, and extends $w$ .

It suffices to treat the case in which $w$ is of the form $s_{\alpha}(\alpha \in B)$ . Note first of all that ad $x_{\alpha}$ and ad $y_{\alpha}$ are locally nilpotent on $U(\mathfrak{g})$ , in other words that for all $u\in U(\mathfrak{g})$ there exists an integer $n$ such that (ad $x_{\alpha}$ ) $^nu=$ (ad $y_{\alpha}$ ) $^nu= 0$ .

This enables us to define the automorphisms $e^{adx_{\alpha}}=\sum_{n=0}^{\infty}\frac{1}{n!}$ (ad $x_{\alpha}$ ) $^n$ and $e^{ady_{\alpha}}$ of $U(\mathfrak{g})$ ; we verify immediately that these automorphisms of $U(\mathfrak{g})$ leave $\mathscr{U}$ stable. Put $\varphi_1=e^{adx_{\alpha}}e^{ady_{\alpha}}e^{adx_{\alpha}},\varphi_2=e^{ady_{\alpha}}e^{adx_{\alpha}}e^{ady_{\alpha}}$ . We have $\varphi_1|\mathfrak{g}=$ $\varphi_2|\mathfrak{g}($ §2, no. 2, formula (1)), so $\varphi_1=\varphi_2$ . Put $\varphi_1=\varphi_2=\varphi$ . We have $\theta \varphi \theta^{-1}=\varphi$ , so $\theta$ and $\varphi$ commute. On the other hand, $\varphi |\mathfrak{h}=w$ by §2, no. 2, Lemma 1.

Corollary 3 018C

Let $\alpha \in R$ . If $x\in \mathscr{G}\cap \mathfrak{g}^{\alpha}$ and $n\in \mathbf{N}$ , we have $x^{(n)}\in \mathscr{U}$ , and (ad $x$ ) $^n/n$ ! leaves $\mathscr{G}$ and $\mathscr{U}$ stable.

This is clear if $\alpha \in B$ , by construction of $\mathscr{U}$ and Cor. 1. In the general case, there exists $w\in W$ such that $w(\alpha )\in B$ (Chap. VI, §1, no. 5, Prop. 15). By Cor. 2, there exists an automorphism $\varphi$ of $\mathfrak{g}$ that leaves $\mathscr{G}$ and $\mathscr{U}$ stable and takes $\mathfrak{g}^{\alpha}$ to $\mathfrak{g}^{w(\alpha)}$ , hence the corollary by transport of structure.

Corollary 4 018D

There exists a Chevalley system $(X_{\alpha})_{\alpha\in R}$ in $(\mathfrak{g},\mathfrak{h}) ($ §2, no. 4, Def. 3) such that $X_{\alpha}=x_{\alpha}$ and $X_{-\alpha}=y_{\alpha}$ for $\alpha \in B$ . For every Chevalley system $(X'_{\alpha})_{\alpha\in R}$ having these properties, and for all $\alpha \in R,X'_{\alpha}$ is a basis of $\mathscr{G}\cap \mathfrak{g}^{\alpha}$ .

For $\alpha \in B$ , put $X_{\alpha}=x_{\alpha}, X_{-\alpha}=y_{\alpha}$ . For $\alpha \in R_+$ -- B, choose a $w\in W$ such that $w(\alpha )\in B$ and an automorphism $\varphi$ of $\mathfrak{g}$ such that $\theta \varphi =\varphi \theta ,\varphi (\mathscr{G}) =\mathscr{G}$ and $\varphi (h) =w^{-1}(h)$ for $h\in \mathfrak{h}$ (Cor. 2); put $X_{\alpha}=\varphi (x_{w(\alpha)}), X_{-\alpha}=\varphi (y_{w(\alpha)})$ . Then

[X_{-\alpha}, X_{\alpha}] =\varphi ([y_{w(\alpha)}, x_{w(\alpha)}]) =\varphi (H_{w(\alpha)}) =w^{-1}(H_{w(\alpha)}) =H_{\alpha}

\theta (X_{\alpha}) =\theta \varphi (x_{w(\alpha)}) =\varphi \theta (x_{w(\alpha)}) =\varphi (y_{w(\alpha)}) =X_{-\alpha}

so $(X_{\alpha})_{\alpha\in R}$ is a Chevalley system. Moreover,

\mathscr{G}\cap \mathfrak{g}^{\alpha}=\varphi (\mathscr{G}\cap \mathfrak{g}^{w(\alpha)}) =\varphi (\mathbf{Z}x_{w(\alpha)}) =\mathbf{Z}X_{\alpha} \tag{32}

\mathscr{G}\cap \mathfrak{g}^{-\alpha}=\varphi (\mathscr{G}\cap \mathfrak{g}^{-w(\alpha)}) =\varphi (\mathbf{Z}y_{w(\alpha)}) =\mathbf{Z}X_{-\alpha} \tag{33}

Let $(X'_{\alpha})_{\alpha\in R}$ be a Chevalley system such that $X'_{\alpha}=x_{\alpha}, X'_{-\alpha}=y_{\alpha}$ for $\alpha \in B$ . Let S be the set of $\alpha \in R$ such that $X'_{\alpha}=\pm X_{\alpha}$ . By §2, no. 4, Prop. 7, S is a closed set of roots. Since $S\supset B\cup (-B)$ , we have S = R (Chap. VI, §1, no. 6, Prop. 19). Thus, by (32) and (33), we have $\mathscr{G}\cap \mathfrak{g}^{\alpha}=\mathbf{Z}X'_{\alpha}$ for all $\alpha \in R$ .

Remark 5 018E

Let $(X_{\alpha})_{\alpha\in R}$ be the Chevalley system constructed above. If $\alpha , \beta , \alpha +\beta \in R$ and if we put $[X_{\alpha}, X_{\beta}] = N_{\alpha ,\beta}X_{\alpha+\beta}$ , we have $[X_{\alpha}, X_{\beta}]\in$ $\mathscr{G}\cap \mathfrak{g}^{\alpha+\beta}$ , and we recover the fact that $N_{\alpha ,\beta}\in \mathbf{Z}$ (cf. §2, no. 4, Prop. 7).

Remark 6 018F

We have obtained in passing a new proof of the existence of Chevalley systems (cf. §4, no. 4, Cor. of Prop. 5), independent of Lemma 4, §2.

7. Chevalley orders

Let $(\mathfrak{g},\mathfrak{h})$ be a split reductive Lie algebra over $\mathbf{Q}, R$ its root system. Choose:

a) a permissible lattice $\mathscr{H}$ in $\mathfrak{h}$ (no. 6, Def. 1);

b) for all $\alpha \in R$ , a lattice $\mathscr{G}^{\alpha}$ in $\mathfrak{g}^{\alpha}$ .

Put $\mathscr{G}=\mathscr{H}\oplus \sum_{\alpha\in R}\mathscr{G}^{\alpha}$ . This is a lattice in $\mathfrak{g}$ . Denote by $\mathscr{U}$ the $\mathbf{Z}$ -subalgebra

of $U(\mathfrak{g})$ generated by the $(^h_n)(h\in \mathscr{H}, n\in \mathbf{N})$ and the $x^{(n)}(x\in \mathscr{G}^{\alpha}, \alpha \in R$ , $n\in \mathbf{N})$ . Finally, for $\alpha \in R$ and $x\in \mathfrak{g}^{\alpha}$ -- $\{0\}$ , put

$w_{\alpha}(x) =$ (exp ad $x$ )(exp ad $y$ )(exp ad $x$ ),

where $y$ is the unique element of $\mathfrak{g}^{-\alpha}$ such that $[y, x] =H_{\alpha}$ . With these notations:

Theorem 2 018G

The following conditions are equivalent:

(i) There exists a Chevalley system $(X_{\alpha})_{\alpha\in R}$ of $(\mathfrak{g},\mathfrak{h})$ such that $\mathscr{G}_{\alpha}=\mathbf{Z}X_{\alpha}$ for all $\alpha \in R$ .

(ii) $\mathscr{U}\cap \mathfrak{g}=\mathscr{G}$ and $[\mathscr{G}^{\alpha},\mathscr{G}^{-\alpha}] =\mathbf{Z}H_{\alpha}$ for all $\alpha \in R$ .

(iii) For all $\alpha \in R, x\in \mathscr{G}^{\alpha}, n\in \mathbf{N}$ , the endomorphism (ad $x$ ) $^n/n$ ! of $\mathfrak{g}$ maps $\mathscr{G}$ to $\mathscr{G}$ , and $[\mathscr{G}^{\alpha},\mathscr{G}^{-\alpha}] =\mathbf{Z}H_{\alpha}$ .

(iv) For all $\alpha \in R$ and every basis $x$ of $\mathscr{G}^{\alpha},w_{\alpha}(x)$ maps $\mathscr{G}$ to $\mathscr{G}($ that is, maps $\mathscr{G}^{\beta}$ to $\mathscr{G}^{s_{\alpha}(\beta)}$ for all $\beta \in R)$ .

(i) $=\Rightarrow$ (ii): let $(X_{\alpha})_{\alpha\in R}$ be a Chevalley system in $(\mathfrak{g},\mathfrak{h})$ such that $\mathscr{G}^{\alpha}=$ $\mathbf{Z}X_{\alpha}$ for all $\alpha \in R$ , and let B be a basis of R. For $\alpha \in B$ , put $x_{\alpha}=X_{\alpha}, y_{\alpha}=$ $X_{-\alpha}$ . Let $\mathscr{U}'$ be the biorder associated by Prop. 3 of no. 6 to $\mathscr{H}$ , the $x_{\alpha}$ and the $y_{\alpha}$ . It is clear that $\mathscr{U}'\subset \mathscr{U}$ . By Cor. 3 and 4 of Prop. $3,x^{(n)}\in \mathscr{U}'$ for all $\alpha \in R,x\in \mathscr{G}^{\alpha}$ and $n\in \mathbf{N}$ . Thus $\mathscr{U}=\mathscr{U}'$ , which proves (ii).

(ii) $=\Rightarrow$ (iii): this is clear by Lemma 2 of no. 5.

(iii) $=\Rightarrow$ (iv): let $\alpha \in R$ and let $x$ be a basis of $\mathscr{G}^{\alpha}$ . Since $[\mathscr{G}^{\alpha},\mathscr{G}^{-\alpha}] =\mathbf{Z}H_{\alpha}$ , the unique $y\in \mathfrak{g}^{-\alpha}$ such that $[y, x] =H_{\alpha}$ belongs to $\mathscr{G}^{-\alpha}$ . Since exp ad $x$ and exp ad $y$ leave $\mathscr{G}$ stable by (iii), so does $w_{\alpha}(x)$ .

(iv) $=\Rightarrow$ (i): let B be a basis of R. Choose a basis $x_{\alpha}$ of $\mathscr{G}^{\alpha}$ for all $\alpha \in B$ . Let $y_{\alpha}\in \mathscr{G}^{-\alpha}$ be such that $[y_{\alpha}, x_{\alpha}] =H_{\alpha}$ . By §1, no. 5, formulas (5), we have $y_{\alpha}=w_{\alpha}(x_{\alpha}).x_{\alpha}$ so $y_{\alpha}$ is a basis of $\mathscr{G}^{-\alpha}$ by (iv). Let $\mathscr{G}'$ be the order in $\mathfrak{g}$ defined by $\mathscr{H}$ , the $x_{\alpha}$ and the $y_{\alpha}$ (no. 6, Cor. 1 of Prop. 3). Then $\mathscr{G}'$ is stable under the (ad $x_{\alpha}$ ) $^n/n$ !, (ad $y_{\alpha}$ ) $^n/n$ ! (loc. cit.), and hence under the $w_{\alpha}(x_{\alpha})$ .

Now let $\beta \in R$ . There exist $\alpha_0, \alpha_1, . . . , \alpha_r\in B$ such that

\beta =s_{\alpha_r}s_{\alpha_{r-1}}. . . s_{\alpha_1}(\alpha_0)

(Chap. VI, §1, no. 5, Prop. 15). Then $w_{\alpha_r}(x_{\alpha_r}).w_{\alpha_{r-1}}(x_{\alpha_{r-1}}). . . w_{\alpha_1}(x_{\alpha_1})$ maps $\mathscr{G}^{\alpha_0}$ to $\mathscr{G}^{\beta}$ by (iv), and maps $\mathscr{G}'\cap \mathfrak{g}^{\alpha_0}$ to $\mathscr{G}'\cap \mathfrak{g}^{\beta}$ by the preceding. Since $\mathscr{G}'\cap \mathfrak{g}^{\alpha_0}=\mathscr{G}^{\alpha_0}$ (Prop. 3 (ii)), we have $\mathscr{G}'\cap \mathfrak{g}^{\beta}=\mathscr{G}^{\beta}$ . Thus

\mathscr{G}'=\mathscr{H}\oplus \sum_{\beta\in R}(\mathscr{G}'\cap \mathfrak{g}^{\beta}) =\mathscr{H}\oplus \sum_{\beta\in R}\mathscr{G}^{\beta}=\mathscr{G}

and Cor. 4 of Prop. 3 concludes the proof.

Definition 2 018H

When conditions (i) to (iv) of Th. 2 are satisfied, $\mathscr{G}$ is said to be a Chevalley order in $(\mathfrak{g},\mathfrak{h})$ .

Remark 018I

Chevalley orders in $(\mathfrak{g},\mathfrak{h})$ always exist. Indeed, the Chevalley orders are the sets of the form $\mathscr{H}\oplus \sum_{\alpha\in R}\mathbf{Z}X_{\alpha}$ , where $(X_{\alpha})_{\alpha\in R}$ is a Chevalley system in $(\mathfrak{g},\mathfrak{h})$ and $\mathscr{H}$ is a lattice in $\mathfrak{h}$ such that

Q(R^{\vee})\subset \mathscr{H}\subset P(R^{\vee})\oplus \mathfrak{c}

$(\mathfrak{c}$ being the centre of $\mathfrak{g})$ .

Theorem 3 018J

We retain the notations at the beginning of no. 7, and assume that $\mathscr{G}$ is a Chevalley order in $(\mathfrak{g},\mathfrak{h})$ .

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

(ii) Let B be a basis of R, and $(X_{\alpha})_{\alpha\in B\cup(-B)}$ a family of elements of $\mathfrak{g}$ such that $\mathscr{G}^{\alpha}=\mathbf{Z}X_{\alpha}$ for $\alpha \in B\cup (-B)$ . The $\mathbf{Z}$ -algebra $\mathscr{U}$ is generated by the $(^h_n)$ and the $X_{\alpha}^{(n)}(h\in \mathscr{H}, \alpha \in B\cup (-B), n\in \mathbf{N})$ . If $\mathfrak{g}$ is semi-simple and $\mathscr{H}= Q(R^{\vee})$ , the $\mathbf{Z}$ -algebra $\mathscr{U}$ is generated by the $X_{\alpha}^{(n)}(\alpha \in B\cup (-B), n\in \mathbf{N})$ .

(iii) Let B be a basis of R, $R_+$ the corresponding set of positive roots, $R_-=-R_+,\mathfrak{n}_+=\sum_{\alpha\in R_+}\mathfrak{g}^{\alpha},\mathfrak{n}_-=\sum_{\alpha\in R_-}\mathfrak{g}^{\alpha}$ . Then,

\mathscr{U}= (\mathscr{U}\cap U(\mathfrak{n}_-)).(\mathscr{U}\cap U(\mathfrak{h})).(\mathscr{U}\cap U(\mathfrak{n}_+))

Let $(h_i)_{i\in I}$ be a basis of $\mathscr{H}$ . For all $\alpha \in R$ , let $X_{\alpha}$ be a basis of $\mathscr{G}^{\alpha}$ . Give the set $I\cup R$ a total order (we assume that $I\cap R =\emptyset$ ). For $\lambda \in I\cup R$ and $n\in \mathbf{N}$ , put $e^{\langle n\rangle}_{\lambda}=(^{h_{\lambda}}_n)$ if $\lambda \in I,e^{\langle n\rangle}_{\lambda}=X_{\lambda}^{(n)}$ if $\lambda \in R$ . Then the products $\prod_{\lambda\in I\cup R}e^{\langle n_{\lambda}\rangle}_{\lambda}$ , where $(n_{\lambda})$ belongs to $\mathbf{N}^{I\cup R}$ , form a basis of the $\mathbf{Z}$ -module $\mathscr{U}$ . The

products $\prod_{\lambda\in I}(^{h_{\lambda}}_{n_{\lambda}})$ , where $(n_{\lambda})$ belongs to $\mathbf{N}^I$ , form a basis of the $\mathbf{Z}$ -module $\mathscr{U}\cap U(\mathfrak{h})$ . The products $\prod_{\lambda\in R_+}X_{\lambda}^{(n_{\lambda})}$ , where $(n_{\lambda})$ belongs to $\mathbf{N}^{R_+}$ , form a basis of the $\mathbf{Z}$ -module $\mathscr{U}\cap U(\mathfrak{n}_+)$ .

Let B and $(X_{\alpha})_{\alpha\in B\cup(-B)}$ be as in (ii), and such that $[X_{-\alpha}, X_{\alpha}] =H_{\alpha}$ . Let $\mathscr{U}'$ be the $\mathbf{Z}$ -subalgebra of $U(\mathfrak{g})$ generated by the $(_n^h)$ and the $X_{\alpha}^{(n)}$ $(h\in \mathscr{H}, \alpha \in B\cup (-B), n\in \mathbf{N})$ . We have seen in the proof of Th. 2, (i) $=\Rightarrow$ (ii), that $\mathscr{U}'$ is equal to $\mathscr{U}$ and is a biorder in $U(\mathfrak{g})$ . This proves (i) and the first assertion of (ii); the second follows from Lemma 4 (ii). Assertion (iii) follows from Th. 1 (no. 3) and Prop. 3 (no. 6).

8. Admissible lattices

Generalizing the terminology adopted for vector spaces, an endomorphism $u$ of a module M is said to be diagonalizable if there exists a basis of M such that the matrix of $u$ relative to this basis is diagonal.

Lemma 6 018K

Let M be a free $\mathbf{Z}$ -module of finite type, $u$ an endomorphism of M, and $v$ the endomorphism $u\otimes 1$ of $M\otimes_{\mathbf{Z}}\mathbf{Q}$ . Assume that $(^v_n)(M)\subset M$ for all $n\in \mathbf{N}$ . Then $u$ is diagonalizable.

a) For any polynomial $P\in \mathbf{Q}[T]$ such that $P(\mathbf{Z})\subset \mathbf{Z}$ , we have $P(v)(M)\subset M$ (no. 4, Cor. of Prop. 2), so det $P(v)\in \mathbf{Z}$ .

b) Denote by $\chi_v(t) =t^d+\alpha_1t^{d-1}+\cdots$ the characteristic polynomial of $v$ . Let $k\in \mathbf{Z}, n\in \mathbf{N}$ . Applying a) to the polynomial $(^{T-k}_n)$ , we see that the number

$a_n=$ det $(v-k)=1_d$ det( $v-k$ ) det( $v-k-1$ ) $. .$ . det( $v-k-n+ 1$ )

$n(n$ !)

=\frac{(-1)^n}{4}\chi_v(k)\chi_v(k+ 1). . . \chi_v(k+n-1)

$(n$ !)

is an integer. Take $k-1<-\alpha_1/d$ . Then

\chi_v(k+n-1) =n^d+ (\alpha_1+ (k-1)d)n^{d-1}+\cdots

and

|a_n|=\frac{|\chi_v(k + n- 1)|}{d}|a_{n-1}|

n

hence, if $a_n\not= 0$ for all $n\in \mathbf{N}$ , the sequence of the $|a_n|$ is strictly decreasing for $n$ sufficiently large, which is absurd. It follows that $v$ has an integer eigenvalue $\lambda$ . Put $M'=$ Ker( $u-\lambda .1$ ) and $M''= M/M'$ . Then $M'$ is the intersection with M of a vector subspace of $M\otimes_{\mathbf{Z}}\mathbf{Q}$ , so the $\mathbf{Z}$ -module $M''$ is torsion-free of finite type, and consequently free of rank $< d$ . Arguing by induction on $d$ and applying the induction hypothesis to the endomorphism of $M''$ induced by $u$ , we conclude that all the eigenvalues of $v$ in an algebraically closed extension of $\mathbf{Q}$ are integers.

c) We show that $v$ is diagonalizable. Let $\lambda$ be an eigenvalue of $v$ and let $x\in M\otimes_{\mathbf{Z}}\mathbf{Q}$ be such that $(v-\lambda )^2x= 0$ . We have $v(vx-\lambda x) =\lambda (vx-\lambda x)$ , so

$n1$ ! $(v-\lambda -n+ 1)(v-\lambda -n+ 2). . .(v-\lambda -1)(v-\lambda )x$

=\frac{(-1)^{n-1}}{n}(vx-\lambda x)

By a), this implies that $vx-\lambda x\in nM$ for all $n\in \mathbf{N}$ , so $(v-\lambda )x= 0$ .

d) Let $\lambda$ be an eigenvalue of $v$ and let $\lambda -a, \lambda +b$ be an interval in $\mathbf{Z}$ containing all the eigenvalues of $v$ . Consider the polynomial

(T-\lambda -1)(T-\lambda -2). . .(T-\lambda -b)

P(T) = (-1)^b

$b$ !

(T-\lambda + 1)(T-\lambda + 2). . .(T-\lambda +a)

\times

$a$ !

We have $P(\mathbf{Z})\subset \mathbf{Z},P(\lambda ) = 1,P(\mu) = 0$ for $\mu\in \mathbf{Z}\cap (\lambda -a, \lambda +b)$ and $\mu\not=\lambda$ . By $a), P(v)(M)\subset M$ . By $c), P(v)$ is a projection of $M\otimes_{\mathbf{Z}}\mathbf{Q}$ onto the eigenspace corresponding to $\lambda$ . Q.E.D.

Remark 1 018L

If we only assume that $v$ is diagonalizable with integer eigenvalues, $u$ is not necessarily diagonalizable (for example, take $M =\mathbf{Z}^2$ and $u(x, y) =$ $(y, x)$ for all $(x, y)\in M$ ).

Let $\mathfrak{g},\mathfrak{h},R,\mathscr{H},\mathscr{G}^{\alpha},\mathscr{G},\mathscr{U}$ be as in no. 7, and assume that $\mathscr{G}$ is a Chevalley order in $(\mathfrak{g},\mathfrak{h})$ .

Definition 3 018M

Let E be a $\mathfrak{g}$ -module. A lattice $\mathscr{E}$ in E is said to be admissible (relative to $\mathscr{G}$ ) if the following conditions are satisfied:

(i) $\mathscr{U}$ maps $\mathscr{E}$ to $\mathscr{E}$ ;

(ii) $\mathscr{E}$ is stable under $(^h_n)$ and $x^{(n)}$ for all $\alpha \in R, x\in \mathscr{G}^{\alpha}, n\in \mathbf{N}, h\in \mathscr{H}$ .

Remark 2 018N

Let $\rho$ be the adjoint representation of $\mathfrak{g}$ on $U(\mathfrak{g})$ . Let $\alpha , x, n, h$ be as in (ii) above. We have $\rho (x^{(n)}).\mathscr{U}\subset \mathscr{U}$ by Lemma 2. On the other hand, if $p\in \mathbf{N}$ ,

$((h))_{(n)}($ ad $h)_{(n)}(n\alpha (h))_{(n)}$

$\rho x$ = $x$ = $x$

ppp

(no. 5, formula (13)), so $\rho ((^h_p)).\mathscr{U}\subset \mathscr{U}$ . This proves that $\mathscr{U}$ is an admissible lattice in $U(\mathfrak{g})$ , and it follows that $\mathscr{G}$ is an admissible lattice in $\mathfrak{g}$ (for the adjoint representation).

Remark 3 018O

Let E be a finite dimensional $\mathfrak{g}$ -module, $\mathscr{E}$ an admissible lattice in E, $\mathfrak{c}$ the centre of $\mathfrak{g}$ . By Lemma 6, every element of $\mathfrak{c}$ defines a diagonalizable endomorphism of E. Hence E is semi-simple (Chap. I, §6, no. 5, Th. 4). Thus, E is a direct sum of simple $\mathscr{D}\mathfrak{g}$ -modules on which $\mathfrak{c}$ induces homotheties. By Lemma $6,\mathscr{E}=\oplus (\mathscr{E}\cap E^{\lambda})$ and, for all weights $\lambda$ of E, we have

\lambda (\mathscr{H})\subset \mathbf{Z}

Remark 4 018P

If $\mathfrak{g}$ is semi-simple and $\mathscr{H}= Q(R^{\vee})$ , conditions (i) and (ii) of Def. 3 are equivalent, by Th. 3 (ii), to

(iii) $\mathscr{E}$ is stable under $x^{(n)}$ for all $\alpha \in R, x\in \mathscr{G}^{\alpha}, n\in \mathbf{N}$ .

Remark 5 018Q

Let B be a basis of R; in conditions (i) and (ii) above, “ $\alpha \in$ R” can be replaced by “ $\alpha \in B\cup (-$ B)” (loc. cit).

Theorem 4 018R

Let E be a finite dimensional $\mathfrak{g}$ -module. The following conditions are equivalent:

(i) E has an admissible lattice;

(ii) every element of $\mathscr{H}$ defines a diagonalizable endomorphism of E with integer eigenvalues.

(i) $=\Rightarrow$ (ii): this follows from Remark 3.

(ii) $=\Rightarrow$ (i): we assume that condition (ii) is satisfied and prove (i). By Th. 4 of Chap. I, §6, no. 5, we can assume that the elements of $\mathfrak{c}$ define homotheties of E, and that E is a simple $\mathscr{D}\mathfrak{g}$ -module. Let B be a basis of R, and $\mathfrak{g}=\mathfrak{n}_-\oplus \mathfrak{h}\oplus \mathfrak{n}_+$ the corresponding decomposition of $\mathfrak{g}$ . Let $\lambda$ be the highest weight of the $\mathscr{D}\mathfrak{g}$ -module E, and let $e\in E^{\lambda}$ -- $\{0\}$ . Put $\mathscr{E}=\mathscr{U}.e$ . It is clear that $\mathscr{U}.\mathscr{E}\subset \mathscr{E}$ . Since E is simple, $U(\mathfrak{g}).e= E$ and hence $\mathscr{E}$ generates E as a $\mathbf{Q}$ -vector space. For $h\in \mathscr{H}$ and $n\in \mathbf{N}$ , we have $(^h_n)e=(^{\lambda(h)}_n)e\in \mathbf{Z}e$ , so

(\mathscr{U}\cap U(\mathfrak{h})).e=\mathbf{Z}e

Since $U(\mathfrak{n}_+).e= 0$ , we have $\mathscr{E}= (\mathscr{U}\cap U(\mathfrak{n}_-)).e$ by Prop. 3. It now follows from Th. 3 (iii) that $\mathscr{E}$ is a $\mathbf{Z}$ -module of finite type.

Corollary 018S

If $\mathfrak{g}$ is semi-simple and $\mathscr{H}= Q(R^{\vee})$ , every finite dimensional $\mathfrak{g}$ -module has an admissible lattice.