§ 13. Classical splittable simple Lie algebras

In this paragraph we describe explicitly, for each type of classical splittable simple Lie algebra:

(I) an algebra of this type, its dimension and its splitting Cartan

subalgebras;

(II) its coroots;

(III) its Borel subalgebras and its parabolic subalgebras;

(IV) its fundamental simple representations;

(V) those of its fundamental simple representations which are

orthogonal or symplectic;

(VI) the algebra of invariant polynomial functions;

(VII) certain properties of the groups Aut$\mathfrak{g}$, Aut$_0\mathfrak{g}$, Aut$_e\mathfrak{g}$;

(VIII) the restriction of the Killing form to a Cartan subalgebra;

(IX) the Chevalley orders.

1. Algebras of type a$_{\boldsymbol{l}}$ ($\boldsymbol{l}\geq$ 1)

(I) Let V be a vector space of dimension $l+ 1$ over $k$, and let $\mathfrak{g}$ be the algebra $\mathfrak{s}\mathfrak{l}(V)$ of endomorphisms of V of trace zero. Let $(e_i)_{1\leq i\leq l+1}$ be a basis of V; the map which associates to an element of $\mathfrak{g}$ its matrix with respect to this basis is an identification of $\mathfrak{g}$ with the algebra $\mathfrak{s}\mathfrak{l}(l+ 1, k)$ of matrices of trace zero. We know that $\mathfrak{g}$ is semi-simple (Chap. I, §6, no. 7, Prop. 8).

Recall (Algebra, Chap. II, §10, no. 3) that $E_{ij}$ denotes the matrix $(\alpha_{mp})$ such that $\alpha_{ij}= 1$ and $\alpha_{mp}= 0$ for $(m, p)\not= (i, j)$. The matrices

form a basis of $\mathfrak{g}$. Hence

dim$\mathfrak{g}=l(l+ 2)$.

Let $\widehat{\mathfrak{h}}$ be the set of diagonal elements of $\mathfrak{g}\mathfrak{l}(l+ 1, k)$; the sequence $(E_{ii})_{1\leq i\leq l+1}$ is a basis of the vector space $\widehat{\mathfrak{h}}$; let $(\widehat{\varepsilon}_i)_{1\leq i\leq l+1}$ be the basis of $\widehat{\mathfrak{h}}^*$ dual to $(E_{ii})_{1\leq i\leq l+1}$. For all $h\in \widehat{\mathfrak{h}}$,

by Chap. I, §1, no. 2, formulas (5). Let $\mathfrak{h}$ be the set of elements of $\widehat{\mathfrak{h}}$ of trace zero, and put $\varepsilon_i= \widehat{\varepsilon}_i|\mathfrak{h}$. Then $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ (Chap. VII, §2, no. 1, Example 4). Relation (1) proves that this Cartan subalgebra is splitting, and that the roots of $(\mathfrak{g},\mathfrak{h})$ are the $\varepsilon_i-\varepsilon_j(i\not=j)$. Let $\widehat{\mathfrak{h}}^*_0$ be the set of elements of $\widehat{\mathfrak{h}}^*$ the sum of whose coordinates with respect to $(\widehat{\varepsilon}_i)$ is zero. The map $\lambda \rightarrow \lambda |\mathfrak{h}$ from $\widehat{\mathfrak{h}}^*_0$ to $\mathfrak{h}^*$ is bijective. Thus, the root system R of $(\mathfrak{g},\mathfrak{h})$ is of type $A_l$ (Chap. VI, §4, no. 7). Consequently, $\mathfrak{g}$ is simple (§3, no. 2, Cor. 1 of Prop. 6). Thus, $\mathfrak{g}$ is a splittable simple Lie algebra of type $A_l$.

Every splitting Cartan subalgebra $\mathfrak{h}'$ of $\mathfrak{g}$ is a transform of $\mathfrak{h}$ under an elementary automorphism (§3, no. 3, Cor. of Prop. 10). Since Aut$_e\mathfrak{g}$ is the set of automorphisms $x \rightarrow sxs^{-1}$ of $\mathfrak{g}$ with $s\in \mathbf{S}\mathbf{L}(V)$ (Chap. VII, §3, no. 1, Remark 2; cf. also (VII)), there exists a basis $\beta$ of V such that $\mathfrak{h}'$ is the set $\mathfrak{h}_{\beta}$ of elements of $\mathfrak{g}$ whose matrix with respect to the basis $\beta$ is diagonal. Since $\mathfrak{h}_{\beta}$ contains an element with distinct eigenvalues, the only vector subspaces of V stable under the elements of $\mathfrak{h}_{\beta}$ are those generated by a subset of $\beta$. It follows that the map $\beta \rightarrow \mathfrak{h}_{\beta}$ induces by passage to the quotient a bijection from the set of decompositions of V into the direct sum of $l+ 1$ subspaces of dimension 1 to the set of splitting Cartan subalgebras of $\mathfrak{g}$. (II) Let $\alpha =\varepsilon_i-\varepsilon_j(i\not=j)$ be a root. We have $\mathfrak{g}^{\alpha}=kE_{ij}$. Since

and since $\alpha (E_{ii}-E_{jj}) = 2$, we have (§2, no. 2, Th. 1 (ii))

(III) Put $\alpha_1=\varepsilon_1-\varepsilon_2, \alpha_2=\varepsilon_2-\varepsilon_3, . . . , \alpha_l=\varepsilon_l-\varepsilon_{l+1}$. By Chap. VI, §4, no. $7.I, (\alpha_1, . . . , \alpha_l)$ is a basis B of R; the positive roots relative to B are the $\varepsilon_i-\varepsilon_j$ for $i < j$. The corresponding Borel subalgebra $\mathfrak{b}$ is the set of upper triangular matrices of trace zero.

A flag in V is a set of vector subspaces of V, distinct from $\{0\}$ and V, totally ordered by inclusion. Order the set of flags of V by inclusion. The maximal flags are the sets $\{W_1, . . . ,W_l\}$, where $W_i$ is an $i$-dimensional vector subspace and

For example, if $V_i$ denotes the subspace of V generated by $e_1, . . . , e_i$, then $\{V_1, . . . ,V_l\}$ is a maximal flag.

It is immediate that $\mathfrak{b}$ is the set of elements of $\mathfrak{g}$ leaving stable the elements of the maximal flag $\{V_1, . . . ,V_l\}$. Conversely, since $\mathfrak{b}$ contains $\mathfrak{h}$ and the matrices $E_{ij}$ for $i < j$, we see that the $V_i$ are the only non-trivial vector subspaces stable under $\mathfrak{b}$.

Now let $\delta$ be a maximal flag in V. It follows from the preceding that the set $\mathfrak{b}_{\delta}$ of elements of $\mathfrak{g}$ leaving stable all the elements of $\delta$ is a Borel subalgebra of $\mathfrak{g}$. Since every Borel subalgebra of $\mathfrak{g}$ is a transform of $\mathfrak{b}$ under an elementary automorphism, we see that the map $\delta \rightarrow \mathfrak{b}_{\delta}$ is a bijection from the set of maximal flags to the set of Borel subalgebras of $\mathfrak{g}$.

Let $\beta$ be a basis of V. By (I) and the preceding, the Borel subalgebras containing $\mathfrak{h}_{\beta}$ are those corresponding to the maximal flags each of whose elements is generated by a subset of $\beta$. These flags correspond bijectively to the total orders on $\beta$ in the following way: to a total order $\omega$ on $\beta$ is associated the flag $\{W_1, . . . ,W_l\}$, where $W_i$ is the vector subspace generated by the first $i$ elements of $\beta$ for the order $\omega$. Since there are $(l+$ 1)! total orders on $\beta$, we recover the fact that there exist $(l+$ 1)! Borel subalgebras of $(\mathfrak{s}\mathfrak{l}(V),\mathfrak{h}_{\beta}) ($§3, no. 3, Remark).

Let $\gamma$ be a flag in V. Since $\gamma$ is contained in a maximal flag, the set $\mathfrak{p}_{\gamma}$ of elements of $\mathfrak{g}$ leaving stable the elements of $\gamma$ is a parabolic subalgebra of $\mathfrak{g}$. We show that the only non-trivial vector subspaces stable under $\mathfrak{p}_{\gamma}$ are the elements of $\gamma$. For this, we can assume that $\gamma =\{V_{i_1}, . . . ,V_{i_q}\}$ with $1\leq i_1<\cdots < i_q\leq l$. Put $i_0= 0, i_{q+1}=l+ 1$. The non-empty intervals

form a partition of $\{1, . . . , l+ 1\}$, so that any square matrix of order $l+ 1$ can be written as a block matrix $(X_{ab})_{1\leq a,b\leq q+1}$. The algebra $\mathfrak{p}_{\gamma}$ is then the set $\mathfrak{p}_{i_1,...,i_q}$ of elements $(X_{ab})_{1\leq a,b\leq q+1}$ of $\mathfrak{s}\mathfrak{l}(l+ 1, k)$ such that $X_{ab}= 0$ for $a > b$. Since $\mathfrak{p}_{i_1,...,i_q}\supset \mathfrak{b}$, a non-trivial vector subspace stable under $\mathfrak{p}_{i_1,...,i_q}$ is one of the $V_i$; if $i_k< i < i_{k+1}$, the algebra $\mathfrak{p}_{i_1,...,i_q}$ contains $E_{i_{k+1},i}$ and $V_i$ is not stable, hence our assertion.

Consequently, the $2^l$ flags contained in the maximal flag $\{V_1, . . . ,V_l\}$ give rise to $2^l$ distinct parabolic subalgebras containing $\mathfrak{b}$; since there are exactly $2^l$ parabolic subalgebras containing $\mathfrak{b}($§3, no. 4, Remark), it follows that the map $\gamma \rightarrow \mathfrak{p}_{\gamma}$ is a bijection from the set of flags of V to the set of parabolic subalgebras of $\mathfrak{g}$. Moreover, $\mathfrak{p}_{\gamma}\supset \mathfrak{p}_{\gamma'}$ if and only if $\gamma \subset \gamma '$.

Recall the parabolic subalgebra $\mathfrak{p}=\mathfrak{p}_{i_1,...,i_q}(1\leq i_1<\cdots < i_q\leq l)$. Let $\mathfrak{s}$ (resp. $\mathfrak{n}$) be the set of $(X_{ab})_{1\leq a,b\leq q+1}$ in $\mathfrak{s}\mathfrak{l}(l+ 1, k)$ such that $X_{ab}= 0$ for $a\not=b$ (resp. $a\geq b$). In view of Prop. 13 of §3, no. 4, we have $\mathfrak{p}=\mathfrak{s}\oplus \mathfrak{n}$, the subalgebra $\mathfrak{s}$ is reductive in $\mathfrak{g}$ and $\mathfrak{n}$ is both the largest nilpotent ideal and the nilpotent radical of $\mathfrak{p}$.

(IV) For $r= 1,2, . . . , l$, let $\varpi_r=\varepsilon_1+\cdots +\varepsilon_r$. We have $\varpi_i(H_{\alpha_j}) =\delta_{ij}$, so $\varpi_r$ is the fundamental weight corresponding to $\alpha_r$.

Let $\sigma$ be the identity representation of $\mathfrak{g}$ on V. The exterior power $\bigwedge^r\sigma$ of $\sigma$ is a representation on $E =\bigwedge^r(V)$. Let $(e_1, . . . , e_{l+1})$ be the chosen basis of V. The $e_{i_1}\wedge \cdots \wedge e_{i_r}$, where $i_1<\cdots < i_r$, form a basis of E. If $h\in \mathfrak{h}$,

Thus, every weight is of multiplicity $1,\varpi_r$ is a weight of $\bigwedge^r\sigma$, and every other weight is of the form $\varpi_r-\mu$, where $\mu$ is a positive radical weight. Consequently, $\varpi_r$ is the highest weight of $\bigwedge^r\sigma$, and $e_1\wedge \cdots \wedge e_r$ is a primitive element. By Chap. VI, §4, no. 7.IX, the Weyl group can be identified with the symmetric group of

The orbit of $\varpi_r$ under the Weyl group thus contains all the $\varepsilon_{i_1}+\cdots +\varepsilon_{i_r}$ with $i_1<\cdots < i_r$. The simple submodule generated by the primitive element $e_1\wedge \cdots \wedge e_r$ thus admits all the $\varepsilon_{i_1}+\cdots +\varepsilon_{i_r}$ as weights and consequently is equal to E. Thus, $\bigwedge^r\sigma$ is irreducible with highest weight $\varpi_r$.

Thus, the representations $\bigwedge^r\sigma (1\leq r\leq l)$ are the fundamental representations. We have dim($\bigwedge^r\sigma$ ) $=(^{l+1}_r)$.

(V) We have $w_0(\alpha_1) =-\alpha_l, w_0(\alpha_2) =-\alpha_{l-1}, . .$. (Chap. VI, §4, no. 7, XI), so

Let

be a dominant weight. Then, the simple representation with highest weight $\omega$ is orthogonal or symplectic if and only if

(§7, no. 5, Prop. 12). In particular, if $l$ is even, none of the fundamental representations of $\mathfrak{s}\mathfrak{l}(l+ 1, k)$ is orthogonal or symplectic. If $l$ is odd, the representation $\bigwedge^i\sigma$ for $i\not= (l+ 1)/2$ is neither orthogonal nor symplectic; by Chap. VI, §4, no. 7.VI, the sum of the coordinates of $\varpi_{(l+1)/2}$ with respect to $(\alpha_1, . . . , \alpha_l)$ is

so $\bigwedge^{(l+1)/2}\sigma$ is orthogonal if $l\equiv -1$ (mod. 4) and symplectic if $l\equiv 1$ (mod. 4) (§7, no. 5, Prop. 12). This last result can be made more precise as follows. Choose a non-zero element $e$ in $\bigwedge^{l+1}(V)$. The multiplication in the exterior algebra V defines a bilinear map from

to $\bigwedge^{l+1}$(V), which can be written $(u, v) \rightarrow \Phi (u, v)e$, where $\Phi$ is a bilinear form on $\bigwedge^{(l+1)/2}(V)$. It is immediately verified that $\Phi$ is non-zero, invariant under $\mathfrak{g}$ (and hence non-degenerate), symmetric if $(l+ 1)/2$ is even, and alternating if $(l+ 1)/2$ is odd.

(VI) For all $x\in \mathfrak{g}$, the characteristic polynomial of $\sigma (x) =x$ can be written

where $f_2, . . . , f_{l+1}$ are polynomial functions invariant under $\mathfrak{g}($§8, no. 3, Lemma 2).

If $x=\xi_1E_{11}+\cdots +\xi_{l+1}E_{l+1l+1}\in \mathfrak{h}$, the $f_i(x)$ are, up to sign, the elementary symmetric functions of $\xi_1, . . . , \xi_{l+1}$ of degree $2, . . . , l+ 1$. Thus, by Chap. VI, §4, no. 7.IX, the $f_i|\mathfrak{h}$ generate the algebra of elements of $\mathbf{S}(\mathfrak{h}^*)$ invariant under the Weyl group, and are algebraically independent. Hence (§8, no. 3, Prop. $3$)$f_2, f_3, . . . , f_{l+1}$ generate the algebra of polynomial functions invariant under $\mathfrak{g}$, and are algebraically independent.

(VII) For all $g\in \mathbf{G}\mathbf{L}(l+ 1, k)$, let $\varphi_k(g) =\varphi (g)$ be the automorphism $x \rightarrow gxg^{-1}$ of $\mathfrak{g}$. Then $\varphi$ is a homomorphism from $\mathbf{G}\mathbf{L}(l+ 1, k)$ to Aut($\mathfrak{g}$). We have

$\varphi (\mathbf{S}\mathbf{L}(l+ 1, k)) =$ Aut$_e(\mathfrak{g})$

(Chap. VII, §3, no. 1, Remark 2). Let $\overline{k}$ be an algebraic closure of $k$. We have

so $\varphi_{\overline{k}}(\mathbf{G}\mathbf{L}(l+ 1,\overline{k})) =\varphi_{\overline{k}}(\mathbf{S}\mathbf{L}(l+ 1,\overline{k})) =$ Aut$_e(\mathfrak{g}\otimes_k\overline{k})$; it follows that $\varphi (\mathbf{G}\mathbf{L}(l+ 1, k))\subset$ Aut$_0(\mathfrak{g})$. On the other hand, Aut$_0(\mathfrak{g})\subset \varphi (\mathbf{G}\mathbf{L}(l+ 1, k))$, by Prop. 2 of §7, no. 1, applied to the identity representation of $\mathfrak{g}$. Hence,

Aut$_0(\mathfrak{g}) =\varphi (\mathbf{G}\mathbf{L}(l+ 1, k))$.

The kernel of $\varphi$ is the set of elements of $\mathbf{G}\mathbf{L}(l+ 1, k)$ that commute with every matrix of order $l+1$, that is the set $k^*$ of invertible scalar matrices. Thus, Aut$_0(\mathfrak{g})$ can be identified with the group $\mathbf{G}\mathbf{L}(l+ 1, k)/k^*=\mathbf{P}\mathbf{G}\mathbf{L}(l+ 1, k)$. The kernel of $\varphi '=\varphi |\mathbf{S}\mathbf{L}(l+ 1, k)$ is $\mu_{l+1}(k)$, where $\mu_{l+1}(k)$ denotes the set of $(l+$ 1)th roots of unity in $k$. Thus, Aut$_e(\mathfrak{g})$ can be identified with the group $\mathbf{S}\mathbf{L}(l+ 1, k)/\mu_{l+1}(k) =\mathbf{P}\mathbf{S}\mathbf{L}(l+ 1, k)$. On the other hand, we have the exact sequence

1 $\longrightarrow \mathbf{S}\mathbf{L}(l+ 1, k)\longrightarrow \mathbf{G}\mathbf{L}(l+ 1, k)\longrightarrow^{det}k^*\longrightarrow$ 1

and the image of $k^*$ under det is $k^{*l+1}$. It follows that there are canonical isomorphisms

Aut$_0(\mathfrak{g})/$Aut$_e(\mathfrak{g})\longrightarrow \mathbf{P}\mathbf{G}\mathbf{L}(l+ 1, k)/\mathbf{P}\mathbf{S}\mathbf{L}(l+ 1, k)$

If $k$ = $\mathbf{R}$, we see that Aut$_0(\mathfrak{g}) =$ Aut$_e(\mathfrak{g})$ if $l+ 1$ is odd, and that Aut$_0(\mathfrak{g})/$Aut$_e(\mathfrak{g})$ is isomorphic to $\mathbf{Z}/2\mathbf{Z}$ if $l+ 1$ is even.

With the notations of §$5,f(T_{\mathbf{Q}})$ is the set of automorphisms of $\mathfrak{g}$ that induce the identity on $\mathfrak{h}$, and hence is equal to $\varphi$(D), where D is the set of diagonal elements of $\mathbf{G}\mathbf{L}(l+ 1, k) ($§5, Prop. 4). Let $D'$ be the set of diagonal elements of $\mathbf{S}\mathbf{L}(l+ 1, k)$. By Prop. 3 of §5, and the determination of Aut$_e(\mathfrak{g})$, we have $f(q(T_P))\subset \varphi (D')$. We show that $f(q(T_P)) =\varphi (D')$. Let

$\lambda_1$ 0

$d=$ . ..

0 $\lambda_{l+1}$

be an element of $D'$. There exists a $\zeta \in$ Hom(Q(R)$, k^*$) $= T_Q$ such that $\zeta (\varepsilon_i-\varepsilon_j) =\lambda_i\lambda^{-1}_j$ for all $i$ and $j$. It is easy to verify that $f(\zeta ) =\varphi (d)$. By Chap. VI, §4, no. 7.VIII, P(R) is generated by Q(R) and the element $\varepsilon =\varepsilon_1$, whose image in $P(R)/Q(R)$ is of order $l+ 1$; but

so $\zeta$ extends to a homomorphism from P(R) to $k^*$. This proves that $\zeta \in q(T_P)$, so $\varphi (d)\in f(q(T_P))$.

Recall (§5, no. 3, Cor. 2 of Prop. 5) that Aut($\mathfrak{g}$) $=$ Aut$_0(\mathfrak{g})$ for $l= 1$, and that Aut($\mathfrak{g}$)$/$Aut$_0(\mathfrak{g})$ is isomorphic to $\mathbf{Z}/2\mathbf{Z}$ for $l\geq 2$. The map $\theta :x \rightarrow -^tx$ is an automorphism of $\mathfrak{s}\mathfrak{l}(l+ 1, k)$ and $a_0=\theta |\mathfrak{h}\notin W$ if $l\geq 2$ (Chap. VI, §4, no. 7.XI), so the class of $a_0$ in Aut($\mathfrak{g}$)$/$Aut$_0(\mathfrak{g})$ is the non-trivial element of this group (§5, no. 2, Prop. 4).

(VIII) The restriction to $\mathfrak{h}$ of the Killing form is

(IX) For $1\leq i < j\leq l+ 1$, put

Then, for all $\alpha \in R$, we have $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$ and $\theta (X_{\alpha}) =X_{-\alpha}$ (where $\theta$ is the automorphism $x \rightarrow -^tx$ introduced in (VII)). Consequently, $(X_{\alpha})_{\alpha\in R}$ is a Chevalley system in $(\mathfrak{g},\mathfrak{h})$.

Take $k=\mathbf{Q}$. The permissible lattices in $\mathfrak{h}($§12, no. 6, Def. 1) are those lying between the $\mathbf{Z}$-module $Q(R^{\vee})$ generated by the $E_{ii}-E_{i+1,i+1}$, that is, consisting of the diagonal matrices belonging to $\mathfrak{s}\mathfrak{l}(l+ 1,\mathbf{Z})$, and the $\mathbf{Z}$-module $P(R^{\vee})$ generated by $Q(R^{\vee})$ and $E_{11}-(l+ 1)^{-1}\sum E_{ii}$ (Chap. VI, §4, no. 7.VIII), that is, consisting of the diagonal matrices of trace zero of the form $x+ (l+ 1)^{-1}a.1$, where $x$ has integer entries and $a\in \mathbf{Z}$. It follows that $\mathfrak{s}\mathfrak{l}(l+1,\mathbf{Z})$ is the Chevalley order in $(\mathfrak{g},\mathfrak{h})$ associated to the permissible lattice $Q(R^{\vee})$ and the Chevalley system $(X_{\alpha})$. It is easy to verify that $\bigwedge^r\mathbf{Z}^{l+1}$ is an admissible lattice in $\bigwedge^r\mathbf{Q}^{l+1}$ relative to $\mathfrak{s}\mathfrak{l}(l+ 1,\mathbf{Z}) ($§12, no. 8, Def. 3).

On the other hand, $\mathfrak{g}\mathfrak{l}(l+ 1,\mathbf{Z})$ is a Chevalley order in the split reductive algebra $\mathfrak{g}\mathfrak{l}(l+ 1,\mathbf{Q})$; its projection onto $\mathfrak{s}\mathfrak{l}(l+ 1,\mathbf{Q})$ parallel to the centre $\mathbf{Q}.1$ of $\mathfrak{g}\mathfrak{l}(l+1,\mathbf{Q})$ is the Chevalley order in $(\mathfrak{g},\mathfrak{h})$ defined by the permissible lattice $P(R^{\vee})$ in $\mathfrak{h}$ and the Chevalley system $(X_{\alpha})$. We remark that $\mathfrak{g}\mathfrak{l}(l+ 1,\mathbf{Z})$ is not the direct sum of its intersections with $\mathfrak{s}\mathfrak{l}(l+ 1,\mathbf{Q})$ and the centre of $\mathfrak{g}\mathfrak{l}(l+ 1,\mathbf{Q})$.

2. Algebras of type b$_{\boldsymbol{l}}$ ($\boldsymbol{l}\geq$ 1)

(I) Let V be a finite dimensional vector space, and $\Psi$ a non-degenerate symmetric bilinear form on V. The set of endomorphisms $x$ of V such that $\Psi (xv, v') +\Psi (v, xv') = 0$ for all $v, v'\in V$ is a Lie subalgebra of $\mathfrak{s}\mathfrak{l}$(V), semi-simple for dim $V\not= 2$ (Chap. I, §6, no. 7, Prop. 9). We denote it by $\mathfrak{o}(\Psi )$ and call it the orthogonal Lie algebra associated to $\Psi$.

Assume that V is of odd dimension $2l+ 1\geq 3$ and that $\Psi$ is of maximum index $l$. Denote by Q the quadratic form such that $\Psi$ is associated to Q. We have $Q(x) =\frac{1}{2}\Psi (x, x)$ for $x\in V$. By Algebra, Chap. IX, §4, no. 2, V can be written as the direct sum of two maximal totally isotropic subspaces F and $F'$ and the orthogonal complement G of $F + F'$, which is non-isotropic and 1-dimensional. Up to multiplying $\Psi$ by a non-zero constant, we can assume that there exists $e_0\in G$ such that $\Psi (e_0, e_0) =-2$. On the other hand, F and $F'$ are in duality via $\Psi$; let $(e_i)_{1\leq i\leq l}$ be a basis of F and $(e_{-i})_{1\leq i\leq l}$ the dual basis of $F'$. Then

is a basis of V; we have

and the matrix of $\Psi$ with respect to this basis is the square matrix of order $2l+ 1$

0 0 $\cdots$ 0 1

0 0 $s$ 0 0 $\cdots$ 1 0

$S=$ 0 $-2$ 0 $,s=$ ... ... . .. ... ...,

$s$ 0 0 0 1 $\cdots$ 0 0

1 0 $\cdots$ 0 0

where $s$ is the square matrix of order $l$ all of whose entries are zero except those on the second diagonal$^6$ which are equal to 1. A basis of V with the preceding properties will be called a Witt basis of V. The algebra $\mathfrak{g}=\mathfrak{o}(\Psi )$ can then be identified with the algebra $\mathfrak{o}_S(2l+ 1, k)$ of square matrices $a$ of order $2l+ 1$ such that $a=-S^{-1}aS($Algebra, Chap. IX, §1, no. 10, formulas (50)). An easy calculation shows that $\mathfrak{g}$ is the set of matrices of the form

$y$ 0 $x$ (2)

where $x$ and $y$ are matrices with 1 row and $l$ columns and $A, B, C, D$ are square matrices of order $l$ such that $B=-s^tBs, C=-s^tCs$, and $D=-s^tAs$. Since the map $A \rightarrow s^tAs$ from $\mathbf{M}_l(k)$ to itself is the symmetry with respect to the second diagonal, it follows that

dim$\mathfrak{g}= 2l+l^2+ 2\frac{l(l-1)}{2}=l(2l+ 1)$.

Let $\mathfrak{h}$ be the set of diagonal elements of $\mathfrak{g}$. This is a commutative subalgebra of $\mathfrak{g}$, with basis the elements

$^6$ The second diagonal of a square matrix $(a_{ij})_{1\leq i,j\leq n}$ is the family of $a_{ij}$ such that

Let $(\varepsilon_i)$ be the basis of $\mathfrak{h}^*$ dual to $(H_i)$. Put

$X_{-\varepsilon_i}$ = $-2E_{-i,0}-E_{0,i}(1\leq i\leq l)$

$X_{\varepsilon_i-\varepsilon_j}$ = $E_{i,j}-E_{-j,-i}(1\leq i < j\leq l)$

$X_{\varepsilon_j-\varepsilon_i}$ = $-E_{j,i}+E_{-i,-j}(1\leq i < j\leq l)$ (3)

$X_{\varepsilon_i+\varepsilon_j}$ = $E_{i,-j}-E_{j,-i}(1\leq i < j\leq l)$

$X_{-\varepsilon_i-\varepsilon_j}$ = $-E_{-j,i}+E_{-i,j}(1\leq i < j\leq l)$.

It is easy to verify that these elements form a basis of a complement of $\mathfrak{h}$ in $\mathfrak{g}$ and that, for $h\in \mathfrak{h}$,

for all $\alpha \in R$, where R is the set of the $\pm \varepsilon_i$ and the $\pm \varepsilon_i\pm \varepsilon_j(1\leq i < j\leq l)$. It follows that $\mathfrak{h}$ is equal to its normalizer in $\mathfrak{g}$, and hence is a Cartan subalgebra of $\mathfrak{g}$, that $\mathfrak{h}$ is splitting, and that the roots of $(\mathfrak{g},\mathfrak{h})$ are the elements of R. The root system R of $(\mathfrak{g},\mathfrak{h})$ is of type $B_l$ for $l\geq 2$, and of type $A_1$ (also said to be of type $B_1$) for $l= 1$ (Chap. VI, §4, no. 5.I, extended to the case $l= 1$). Consequently, $\mathfrak{g}$ is a splittable simple Lie algebra of type $B_l$.

Every splitting Cartan subalgebra of $\mathfrak{o}(\Psi )$ is a transform of $\mathfrak{h}$ by an elementary automorphism of $\mathfrak{o}(\Psi )$, and hence by an element of $\mathbf{O}(\Psi )$ (cf. (VII)), and consequently is the set $\mathfrak{h}_{\beta}$ of elements of $\mathfrak{g}$ whose matrix with respect to a Witt basis $\beta$ of V is diagonal. We verify immediately that the only vector subspaces invariant under $\mathfrak{h}_{\beta}$ are those generated by a subset of $\beta$.

If $l= 1$, the algebras $\mathfrak{o}(\Psi )$ and $\mathfrak{s}\mathfrak{l}(2, k)$ have the same root systems, and are thus isomorphic (cf. also §1, Exerc. 16). From now on, we assume that $l\geq 2$.

(II) The root system $R^{\vee}$ is determined by means of Chap. VI, §4, no. 5.V, and we find that

(III) Put $\alpha_1=\varepsilon_1-\varepsilon_2, . . . , \alpha_{l-1}=\varepsilon_{l-1}-\varepsilon_l, \alpha_l=\varepsilon_l$. By Chap. VI, §4, no. 5.II, $(\alpha_1, . . . , \alpha_l)$ is a basis B of R; the positive roots relative to B are the $\varepsilon_i$ and the $\varepsilon_i\pm \varepsilon_j(i < j)$. The corresponding Borel subalgebra $\mathfrak{b}$ is the set of upper triangular matrices in $\mathfrak{g}$.

It is immediately verified that the only vector subspaces of V distinct from $\{0\}$ and V stable under $\mathfrak{b}$ are the elements of the maximal flag corresponding to the basis $(e_i)$, that is, the totally isotropic subspaces $V_1, . . . ,V_l$, where $V_i$ is generated by $e_1, . . . , e_i$, together with their orthogonal complements $V_{-1}, . . . ,V_{-i}:$ the orthogonal complement $V_{-i}$ of $V_i$ is generated by $e_1, . . . , e_l, e_0, e_{-l}, . . . , e_{-i-1}$ and is not totally isotropic. On the other hand, if an element of $\mathfrak{g}$ leaves stable a vector subspace, it leaves stable its orthogonal complement. Consequently, $\mathfrak{b}$ is the set of elements of $\mathfrak{g}$ leaving stable the elements of the flag $\{V_1, . . . ,V_l\}$.

A flag is said to be isotropic if each of its elements is totally isotropic. The flag $\{V_1, . . . ,V_l\}$ is a maximal isotropic flag. Since the group $\mathbf{O}(\Psi )$ operates transitively both on the Borel subalgebras of $\mathfrak{g}$ (cf. (VII)) and on the maximal isotropic flags (Algebra, Chap. IX, §4, no. 3, Th. 1), we see that, for any maximal isotropic flag $\delta$ in V, the set $\mathfrak{b}_{\delta}$ of elements of $\mathfrak{g}$ leaving stable the elements of $\delta$ is a Borel subalgebra of $\mathfrak{g}$ and that the map $\delta \rightarrow \mathfrak{b}_{\delta}$ is a bijection from the set of maximal isotropic flags to the set of Borel subalgebras of $\mathfrak{g}$.