(i) $\mathfrak{b}_+v\subset kv$;

(ii) $\mathfrak{h}v\subset kv$ and $\mathfrak{n}_+v= 0$;

(iii) $\mathfrak{h}v\subset kv$ and $\mathfrak{g}^{\alpha}v= 0$ for all $\alpha \in B$.

(i) $=\Rightarrow$ (ii): Assume that $\mathfrak{b}_+v\subset kv$. There exists $\lambda \in \mathfrak{h}^*$ such that $v\in V^{\lambda}$. Let $\alpha \in R_+$. Then $\mathfrak{g}^{\alpha}.v\subset V^{\lambda}\cap V^{\lambda+\alpha}= 0$. Hence $\mathfrak{n}_+v= 0$.

(ii) $=\Rightarrow$ (iii): This is clear.

(iii) $=\Rightarrow$ (i): This follows from the fact that $(X_{\alpha})_{\alpha\in B}$ generates $\mathfrak{n}_+($§3, no. 3, Prop. 9 (iii)).

(i) If $U(\mathfrak{n}_-)$ denotes the enveloping algebra of $\mathfrak{n}_-$, we have $V = U(\mathfrak{n}_-).v$.

(ii) For all $\lambda \in \mathfrak{h}^*,V^{\lambda}$ is the set of $x\in V$ such that $Hx=\lambda (H)x$ for all $H\in \mathfrak{h}$. We have $V =\bigoplus_{\lambda\in\mathfrak{h}^*}V^{\lambda}$, and each $V^{\lambda}$ is finite dimensional. The space

$V^{\omega}$ is of dimension 1, and every weight of V is of the form $\omega -\sum_{\alpha\in B}n_{\alpha}.\alpha$, where the $n_{\alpha}$ are integers $\geq 0$.

(iii) V is an indecomposable $\mathfrak{g}$-module, and its commutant reduces to the scalars.

(iv) Let $U(\mathfrak{g})$ be the enveloping algebra of $\mathfrak{g}$, and $\mathscr{Z}$ the centre of $U(\mathfrak{g})$. There exists a unique homomorphism $\chi$ from $\mathscr{Z}$ to $k$ such that, for all $z\in \mathscr{Z}$, $z_V$ is the homothety with ratio $\chi (z)$.

Let $U(\mathfrak{b}_+)$ be the enveloping algebra of $\mathfrak{b}_+$. We have $U(\mathfrak{g}) = U(\mathfrak{n}_-).U(\mathfrak{b}_+)$ (Chap. I, §2, no. 7, Cor. 6 of Th. 1). Hence

Denote by $\alpha_1, . . . , \alpha_n$ the distinct elements of $R_+$. Then

is a basis of $U(\mathfrak{n}_-)$, so

V = $\sum kX_{-\alpha_1}^{p_1}. . . X_{-\alpha_n}^{p_n}v$. (1)

$(p_1,...,p_n)\in \mathbf{N}^n$

For $\lambda \in \mathfrak{h}^*$, put

$(p_1,...,p_n)\in \mathbf{N}^n, \omega -p_1\alpha_1-\cdot \cdot \cdot -p_n\alpha_n=\lambda$

By Chap. VII, §1, no. 1, Prop. 2 (ii), if $h\in \mathfrak{h},h_V|T_{\lambda}$ is the homothety with ratio $\lambda (h)$. So $T_{\lambda}\subset V^{\lambda}$. On the other hand, (1) implies that

V = $\sum T_{\lambda}$.

$\lambda \in \omega -\mathbf{N}\alpha_1-\cdot \cdot \cdot -\mathbf{N}\alpha_n$

The sum of the $V^{\lambda}$ is direct (Chap. VII, §1, no. 1, Prop. 3). From these observations it follows that $V^{\lambda}= T_{\lambda}$, that V is the direct sum of the $V^{\lambda}$, and that $V^{\lambda}$ is the set of $x\in V$ such that $hx=\lambda (h)x$ for all $h\in \mathfrak{h}$. On the other hand, dim $V^{\lambda}$ is at most the cardinal of the set of $(p_1, . . . , p_n)\in \mathbf{N}^n$ such that $p_1\alpha_1+\cdots +p_n\alpha_n=\omega -\lambda$. This proves that $V^{\lambda}= 0$ if $\omega -\lambda \notin \sum_{\alpha\in B}\mathbf{N}\alpha$,

that dim $V^{\omega}= 1$, and that the $V^{\lambda}$ are all finite dimensional.

Let $c$ be an element of the commutant of V. For all $h\in \mathfrak{h}$,

so $c(v)\in V^{\omega}$; hence there exists $t\in k$ such that $c(v) =tv$. Now, for all $(p_1, . . . , p_n)\in \mathbf{N}^n$,

so that $c=t.1$. Hence, the commutant of V reduces to the scalars. This implies (iv) and the fact that V is indecomposable.

It is clear that $\varphi (e)\in X^{\omega}_{\pi}$ for all $\varphi \in \Phi$. If $\varphi \in \Phi$ and $\varphi (e) = 0$, then $\varphi = 0$ since $e$ generates the $\mathfrak{g}$-module V. We show that, if $f$ is a non-zero element of $X^{\omega}_{\pi}$, there exists $\varphi \in \Phi$ such that $\varphi (e) =f$. Let $X'$ be the submodule of X generated by $f$. By Prop. $1, X'$ is indecomposable, hence simple since X is semi-simple. The element $(e, f)$ is primitive in the $\mathfrak{g}$-module $V\times X$. Let N be the submodule of $V\times X$ generated by $(e, f)$. Then $N\cap X\subset$ pr$_2(N) = X'$, so $N\cap X = 0$ or $X'$; if $N\cap X = X', N$ contains the linearly independent elements $(e, f)$ and $(0, f)$ which are primitive of weight $\omega$; this is absurd (Prop. 1), so $N\cap X = 0$. Thus pr$_1|N$ is an injective map $h$ from N to V; this map is surjective since its image contains $e$. Thus $\varphi =$ pr$_2\circ h^{-1}$ is a homomorphism from the $\mathfrak{g}$-module V to the $\mathfrak{g}$-module X such that $\varphi (e) =f$.

(i) every weight of V is of the form $\omega -\mu$ where $\mu$ is a radical weight $\geq 0$;

(ii) $\omega$ is the highest weight of V;

(iii) for all $\alpha \in B,\omega +\alpha$ is not a weight of V;

(iv) there exists a primitive element of weight $\omega$.

(i) $=\Rightarrow$ (ii) $=\Rightarrow$ (iii): This is clear.

(iii) $=\Rightarrow$ (iv): Assume that condition (iii) is satisfied. For all $h\in \mathfrak{h}$,

Ker($h_V-\omega (h)$)

is non-zero, contained in $V^{\omega}$, and stable under $\mathfrak{h}_V$. By induction on dim $\mathfrak{h}$, we see that there exists a non-zero $v$ in $V^{\omega}$ such that $\mathfrak{h}v\subset kv$. Condition (iii) implies that $\mathfrak{n}_+v= 0$, so $v$ is primitive.

(iv) $=\Rightarrow$ (i): Let $v$ be a primitive element of weight $\omega$. Since V is simple, V is generated by $v$ as a $\mathfrak{g}$-module. Assertion (i) now follows from Prop. 1.

Q.E.D.

Thus, for any simple $\mathfrak{g}$-module, the existence of a primitive element is equivalent to that of a highest weight, or to that of a maximal weight. There exist simple $\mathfrak{s}\mathfrak{l}(2,\mathbf{C}$)-modules V that have no weights for any Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{s}\mathfrak{l}(2,\mathbf{C}) ($§1, Exerc. $14f))$. These modules are of infinite dimension over $\mathbf{C}($§1, no. 3, Th. 1).

(i) The primitive elements of V are the non-zero elements of $V^{\omega}$.

(ii) V is semi-simple as an $\mathfrak{h}$-module.

(iii) We have $V =\bigoplus_{\lambda\in\mathfrak{h}^*}V^{\lambda}$. For all $\lambda \in \mathfrak{h}^*,V^{\lambda}$ is finite dimensional. We

have dim $V^{\omega}= 1$.

(iv) The $\mathfrak{g}$-module V is absolutely simple.

Assertions (i), (ii) and (iii) follow from Prop. 1 and Lemma 2. Assertion (iv) follows from Prop. 1 (iii) and Algebra, Chap. VIII, §7, no. 3.

This follows from (iv), cf. Algebra, Chap. VIII, §3, no. 3.

Let $X''$ be the submodule of X generated by $X^{\omega}_{\pi}$. It is clear that every submodule of X isomorphic to V is contained in $X''$. Hence $X'\subset X''$. On the other hand, let $\Phi$ be the set of homomorphisms from the $\mathfrak{g}$-module V to the $\mathfrak{g}$-module X. The length of $X'$ is dim$_k\Phi ($Algebra, Chap. VIII, §4, no. 4), that is dim$_kX^{\omega}_{\pi}$ (Prop. 2).

(ii) Let $Z^+(\lambda ) =\sum_{\lambda\not=\mu}Z(\lambda )^\mu$. Every submodule of $Z(\lambda )$ distinct from $Z(\lambda )$

is contained in $Z^+(\lambda )$.

(iii) There exists a largest submodule $F_{\lambda}$ of $Z(\lambda )$ distinct from $Z(\lambda )$. The quotient module $Z(\lambda )/F_{\lambda}$ is simple and has highest weight $\lambda$.

It is clear that $e$ generates the $\mathfrak{g}$-module $Z(\lambda )$. If $x\in \mathfrak{b}_+$,

hence (i).

The $\mathfrak{h}$-module $Z(\lambda )$ is semi-simple (Prop. 1). If G is a $\mathfrak{g}$-submodule of $Z(\lambda )$, then $G =\sum_{\mu\in\mathfrak{h}^*}(G\cap Z(\lambda )^\mu)$. The hypothesis $G\cap Z(\lambda )^{\lambda}\not= 0$ implies

that $G = Z(\lambda )$, since dim $Z(\lambda )^{\lambda}= 1$ and $e$ generates the $\mathfrak{g}$-module $Z(\lambda )$. If $G\not= Z(\lambda )$, then $G =\sum_{\mu\not=\lambda}G\cap Z(\lambda )^\mu\subset Z^+(\lambda )$.

Let $F_{\lambda}$ be the sum of the $\mathfrak{g}$-submodules of $Z(\lambda )$ distinct from $Z(\lambda )$. By (ii), $F_{\lambda}\subset Z^+(\lambda )$. Hence $F_{\lambda}$ is the largest submodule of $Z(\lambda )$ distinct from $Z(\lambda )$. It is clear that $Z(\lambda )/F_{\lambda}$ is simple and that the canonical image of $e$ in $Z(\lambda )/F_{\lambda}$ is primitive of weight $\lambda$.

In the remainder of this chapter, the $\mathfrak{g}$-module $Z(\lambda )/F_{\lambda}$ of Prop. 5 will be denoted by $E(\lambda )$.

The first assertion follows from Prop. 5 (iii). The second follows from Prop. 4.

(i) There exists a unique homomorphism of $\mathfrak{g}$-modules $\psi : Z(\lambda )\rightarrow V$ such that $\psi (e) =v$.

(ii) Assume that $v$ generates V. Then $\psi$ is surjective. Moreover, $\psi$ is bijective if and only if, for every non-zero element $u$ of $U(\mathfrak{n}_-),u_V$ is injective.

(iii) The map $u \rightarrow u\otimes 1$ from $U(\mathfrak{n}_-)$ to $Z(\lambda )$ is bijective.

Let K be the kernel of the representation of $U(\mathfrak{b}_+)$ on $L_{\lambda}$; it is of codimension 1 in $U(\mathfrak{b}_+)$. Let $J = U(\mathfrak{g})K$ be the left ideal of $U(\mathfrak{g})$ generated by K; then $L_{\lambda}$ can be identified with $U(\mathfrak{b}_+)/K$ as a left $U(\mathfrak{b}_+$)-module, and $Z(\lambda )$ can be identified with $U(\mathfrak{g})/J$ as a left $U(\mathfrak{g}$)-module. We have $K.v= 0$, so $J.v= 0$, which proves (i).

Now assume that $v$ generates V. It is clear that $\psi$ is surjective.

By the Poincaré-Birkhoff-Witt theorem (Chap. I, §2, no. 7, Cor. 6 of Th. 1), a basis of $U(\mathfrak{n}_-)$ over $k$ is also a basis of $U(\mathfrak{g})$ as a right $U(\mathfrak{b}_+)$-module. Hence the map $\varphi :u \rightarrow u\otimes 1$ from $U(\mathfrak{n}_-)$ to $U(\mathfrak{g})\otimes_{U(\mathfrak{b}_+)}L_{\lambda}$ is bijective. Let $u\in U(\mathfrak{n}_-)$. Then $\varphi^{-1}\circ u_{Z(\lambda)}\circ \varphi$ is left multiplication by $u$ on $U(\mathfrak{n}_-)$. In view of Chap. I, §2, no. 7, Cor. 7 of Th. $1,u_{Z(\lambda)}$ is injective if $u\not= 0$. Consequently, if $\psi$ is bijective, then $u_V$ is injective for non-zero $u$ in $U(\mathfrak{n}_-)$.

Assume that $\psi$ is not injective. There exists $u\in U(\mathfrak{n}_-)$ such that $u\not= 0$ and $\psi (\varphi (u)) = 0$. Then

Put $m=\lambda (H_{\alpha})$. Let $x= X^{m+1}_{-\alpha}.e\in Z(\lambda )$, and let V be the submodule of $Z(\lambda )$ generated by $x$. Then $x\not= 0$ (Prop. 6). On the other hand, $x\in Z(\lambda )^{\lambda-(m+1)\alpha}$. For $\beta \in B$ and $\beta \not=\alpha , [\mathfrak{g}^{-\alpha},\mathfrak{g}^{\beta}] = 0$ and $\mathfrak{g}^{\beta}.e= 0$, so $\mathfrak{g}^{\beta}.x= 0$. Finally, since $[X_{\alpha},X_{-\alpha}] =-H_{\alpha}$, we have

(§1, no. 1, Lemma 1 (ii)), so

Thus, $x$ is primitive of weight $\lambda -(m+ 1)\alpha$. In view of Prop. 6, the $\mathfrak{g}$-module V is isomorphic to $Z(-\alpha +\lambda -m\alpha ) = Z(-\alpha +s_{\alpha}\lambda )$.

The assertion is clear when $w= 1$. Assume that it is established whenever $w$ is of length $< q$. If $w$ is of length $q$, there exists $\alpha \in B$ such that $w=s_{\alpha}w^{'-1}$, with $l(w') =q-1$. We have $w'(\alpha )\in R_+$ (Chap. VI, §1, no. 6, Cor. 2 of Prop. 17), and hence $w^{'-1}(\lambda +\rho )(H_{\alpha}) = (\lambda +\rho )(H_{w'\alpha})$ is an integer $\geq 0$. Put

By the induction hypothesis, $Z(\mu')$ is isomorphic to a submodule of $Z(\lambda )$. On the other hand, by Chap. VI, §1, no. 10, Prop. 29 (ii),

Moreover, $\rho (H_{\alpha}) = 1$ (Chap. VI, §1, Prop. 29 (iii)), so $\mu'(H_{\alpha})+1\in \mathbf{N}$. Cor. 1 now implies that $Z(\mu)$ is isomorphic to a submodule of $Z(\mu')$, and hence also to a submodule of $Z(\lambda )$.

(i) We have $L = (U\mathfrak{n}_+)\cap U^0$, and L is a two-sided ideal of $U^0$.

(ii) We have $U^0= V\oplus L$.

It is clear that $\mathfrak{n}_-U$ (resp. $U\mathfrak{n}_+$) is the set of linear combinations of the elements $u((q_i),(m_i),(p_i))$ such that $\sum q_i>0$ (resp. $\sum p_i>0$). On the other hand

This implies that $(\mathfrak{n}_-U)\cap U^0= (U\mathfrak{n}_+)\cap U^0$. Finally, $(\mathfrak{n}_-U)\cap U^0$ (resp. $(U\mathfrak{n}_+)\cap U^0$) is a right (resp. left) ideal of $U^0$, hence (i). Further, an element $u((q_i),(m_i),(p_i))$ that is in $U^0$ belongs to V (resp. to L) if and only if $p_1=\cdots =p_n=q_1=\cdots =q_n= 0$ (resp. $p_1+\cdots +p_n+q_1+\cdots +q_n>0$), hence (ii). Q.E.D.

In view of Lemma 3, the projection of $U^0$ onto V with kernel L is a homomorphism of algebras. It is called the Harish-Chandra homomorphism from $U^0$ to V (relative to B). Recall that V can be identified with the algebra of polynomial functions on $\mathfrak{h}^*$.

Let $v$ be a primitive element of E of weight $\lambda$, and $z$ an element of the centre of U. There exist $u_1, . . . , u_p\in U$ and $n_1, . . . , n_p\in \mathfrak{n}_+$ such that $z=\varphi (z) +u_1n_1+\cdots +u_pn_p$. Then

We recall the notations of §2, no. 3, Prop. 6. We have

1 1

C = $\sum X_{\alpha}X_{-\alpha}+\sum X_{-\alpha}X_{\alpha}$

1 $'$

so

By Prop. 7,

1 $'$

Let $h_{\lambda}$ be the element of $\mathfrak{h}$ such that $\langle h_{\lambda}, h\rangle =\lambda (h)$ for all $h\in \mathfrak{h}$. By §2, no. 2, Prop. 1,

$($ 1 $)\langle$ 1 $\rangle$

$\lambda [X_{\alpha}, X_{-\alpha}]$ = $h_{\lambda},[X_{\alpha}, X_{-\alpha}]=\alpha (h_{\lambda}) =\langle \lambda , \alpha \rangle$.

Hence