Lemma 1

Let V be a finite dimensional vector space, Q a non-degenerate quadratic form on V$,\Psi$ the symmetric bilinear form associated to Q, C(Q) the Clifford algebra of V relative to Q$,f_0$ the composite of the canonical maps

(the 1st is the canonical injection, the 3rd is defined by the canonical isomorphism from $V^*$ to V corresponding to $\Psi$, the 4th is defined by the multiplication in C(Q), cf. Algebra, Chap. IX, § 9, no. 1). Put $f=\frac{1}{2}f_0$.

(i) If $(e_r),(e'_r)$ are bases of V such that $\Psi (e_r, e'_s) =\delta_{rs}$, we have $f_0(a) =$ $\sum_r(ae_r)e'_r$ for all $a\in \mathfrak{o}(\Psi )$.

(ii) If $a, b\in \mathfrak{o}(\Psi )$, we have $\sum_r(ae_r)(be'_r) =-\sum_r(abe_r)e'_r$.

(iii) If $a\in \mathfrak{o}(\Psi )$ and $v\in V$, we have $[f(a), v] =av$.

(iv) If $a, b\in \mathfrak{o}(\Psi )$, we have $[f(a), f(b)] =f([a, b])$.

(v) $f(\mathfrak{o}(\Psi ))$ generates the associative algebra $C^+(Q)$.

(vi) Let N be a left $C^+(Q)$-module and $\rho$ the corresponding homomorphism from $C^+(Q)$ to End$_k(N)$. Then $\rho \circ f$ is a representation of $\mathfrak{o}(\Psi )$ on N. If N is simple, $\rho \circ f$ is irreducible.

Assertion (i) is clear. If $a, b\in \mathfrak{o}(\Psi )$, we have (putting $\Psi (x, y) =\langle x, y\rangle$ )$:$

which proves (ii). Next, for all $v\in V$, we have by (i),

1 $''$

2

= $1\sum^r((ae_r)e'_rv+ (ae_r)ve'_r-(ae_r)ve'_r-v(ae_r)e'_r)$

2

$r$

1 $''$

= $\sum((ae_r)\langle e_r, v\rangle - \langle ae_r, v\rangle e_r)$

2

= $1a(\sum^r\langle e'_r, v\rangle e_r)+1\sum\langle e_r, av\rangle e'_r=1av+1av=av$,

2 $_r$ 2 $_r$ 2 2

which proves (iii). Then

$[f(a), f(b)] =[f(a),\frac{1}{2}\sum(be_r)e'_r]$ by (i)

$r$

$=\frac{1}{2}\sum([f(a), be_r]e'_r+ (be_r)[f(a), e'_r])$ $=\frac{1}{2}\sum^r((abe_r)e'_r+ (be_r)(ae'_r))$ by (iii)

$r$

$=\frac{1}{2}\sum((abe_r)e'_r-(bae_r)e'_r)$ by (ii)

$r$

$=f([a, b])$ by (i)

which proves (iv). To prove (v), we can, by extending scalars, assume that $k$ is algebraically closed. Choose then a basis $(e_r)$ of V such that $\Psi (e_r, e_s) =\delta_{rs}$, so $e'_r=e_r$. If $i\not=j$, then $E_{ij}-E_{ji}\in \mathfrak{o}(\Psi )$ and

but the $e_ie_j$ generate $C^+(Q)$.

Assertion (vi) follows from (iv) and (v). Q.E.D.

Recall now the notations used at the beginning of this number. Put $\widetilde{V} = F + F'$ and let $\widetilde{Q}$ (resp. $\widetilde{\Psi}$) be the restriction of Q (resp. $\Psi$ ) to $\widetilde{V}$. Then $\widetilde{Q}$ is a non-degenerate quadratic form of maximum index $l$ on the space $\widetilde{V}$ of dimension $2l$ and the Clifford algebra $C( \widetilde{Q})$ is a central simple algebra of dimension $2^{2l}($Algebra, Chap. IX, §9, no. 4, Th. 2). Let N be the exterior algebra of the maximal isotropic subspace $F'$ generated by $e_{-1}, . . . , e_{-l}$. Identify F with the dual of $F'$ by means of $\Psi$ and for $x\in F'$ (resp. $y\in F$) denote by $\lambda (x)$ (resp. $\lambda (y)$) the left exterior product with $x$ (resp. the left interior product with $y$) in N; if $a_1, . . . , a_k\in F'$, then

It is easily verified that $\lambda (x)^2=\lambda (y)^2= 0$ and that

It follows (Algebra, Chap. IX, §9, no. 1) that there exists a unique homomorphism (again denoted by $\lambda$ ) from $C( \widetilde{Q})$ to End(N) extending the map $\lambda : F\cup F'\rightarrow$ End(N). Since dim $N = 2^l$ and since $C( \widetilde{Q})$ has a unique class of simple modules, of dimension $2^l($Algebra, Chap. IX, §9, no. 4, Th. 2), the representation of $C( \widetilde{Q})$ on N defined by $\lambda$ is irreducible and is a spinor representation of $C( \widetilde{Q}) ($loc. cit.).

Consider now the map $\mu:v \rightarrow e_0v$ from $\widetilde{V}$ to $C^+(Q)$. For $v\in \widetilde{V}$, we have

and $\mu$ extends uniquely to a homomorphism, again denoted by $\mu$, from $C( \widetilde{Q})$ to $C^+(Q)$. Since $C( \widetilde{Q})$ is simple and since

dim $C^+(Q) =$ dim $C( \widetilde{Q}) = 2^{2l}$,

we see that $\mu$ is an isomorphism. Consequently, $\lambda \circ \mu^{-1}$ defines a simple $C^+$(Q)-module structure on N and $\rho =\lambda \circ \mu^{-1}\circ f$ is an irreducible representation of $\mathfrak{g}$ on N (Lemma 1 (vi)).

On the other hand, in view of Lemma 1 (i), we have

Since $e_ie_{-i}=-e^2_0e_ie_{-i}=e_0e_ie_0e_{-i}$ and $e_ie_{-i}+e_{-i}e_i= 1$, we have

$_{-1}$ 1 1

2 2 We deduce that, for $1\leq i_1<\cdots < i_k\leq l:$

$\rho (H)(e\wedge \cdots \wedge e) =\{-\frac{1}{2}e_{-i_1}\wedge \cdots \wedge e_{-i_k}$ if $i\in \{i_1, . . . , i_k\}$

$^{i-i_1}-_{i_k}\frac{1}{2}e_{-i_1}\wedge \cdots \wedge e_{-i_k}$ if $i \notin \{i_1, . . . i_k\}$

and for $h\in \mathfrak{h}$

This shows that the highest weight of $\rho$ is $\varpi_l$. We call $\rho$ the spinor representation of $\mathfrak{g}$. Note that its weights are all simple (moreover, $\varpi_l$ is a minuscule weight).

(V) We have $w_0=-1$, so every finite dimensional simple representation of $\mathfrak{g}$ is orthogonal or symplectic. By Chap. VI, §4, no. 5.VI, the sum of the coordinates of $\varpi_r$ with respect to $(\alpha_1, . . . , \alpha_l)$ is integral for $1\leq r\leq l-1:$ thus, the representation $\bigwedge^r\sigma$ is orthogonal. Moreover, it leaves invariant the extension $\Psi_{(r)}$ of $\Psi$ to $\bigwedge^rV$.

For the spinor representation, the sum of the coordinates of $\varpi_l$ with respect to $(\alpha_1, . . . , \alpha_l)$ is $\frac{1}{2}(1 +\cdots +l) =\frac{l(l+1)}{4}($loc. cit.). Thus, it is orthogonal for $l\equiv 0$ or $-1$ (mod. 4) and symplectic for $l\equiv 1$ or 2 (mod. 4). In fact, consider the bilinear form $\Phi$ on $N =\bigwedge F'$ defined as follows: if $x\in \bigwedge^pF'$ and $y\in \bigwedge^qF'$, put $\Phi (x, y) = 0$ if $p+q\not=l$ and

if $p+q=l$. It is easily verified that $\Phi$ is non-degenerate and is orthogonal for $l\equiv 0,-1$ (mod. 4) and alternating for $l\equiv 1,2$ (mod. 4). On the other hand, in view of Lemma 1 (i),

for $1\leq i\leq l$ and

for $1\leq i < j\leq l$, and similarly

hence

and

$\mu^{-1}\circ f(X_{\pm\varepsilon_i\pm\varepsilon_j}) =ce_{\pm i}e_{\pm j}$ for $1\leq i, j\leq l, i\not=j$ with $c\in \{1,-1\}$. It is now painless to verify that $\Phi$ is actually $\mathfrak{g}$-invariant (cf. Exerc. 18).

(VI) For $x\in \mathfrak{g}$, the characteristic polynomial of $\sigma (x)$ takes the form

where $f_1, . . . , f_{2l+1}$ are invariant polynomial functions on $\mathfrak{g}$.

If $x=\xi_1H_1+\cdots +\xi_lH_l\in \mathfrak{h}$, the $f_i(x)$ are, up to sign, the elementary symmetric functions of $\xi_1, . . . , \xi_l,-\xi_1, . . . ,-\xi_l$; these symmetric functions are zero in odd degrees, and

so that $f_2, . . . , f_{2l}$ are, up to sign, the elementary symmetric functions of $\xi_1^2, . . . , \xi_l^2$, which are algebraically independent generators of $\mathbf{S}(\mathfrak{h}^*)^W$ (Chap. VI, §4, no. 5.IX). In view of §8, no. 3, Th. 1 (i), we see that $f_1=f_3=f_5=\cdots = 0$ and that $(f_2, f_4, . . . , f_{2l})$ is an algebraically free family generating the algebra of invariant polynomial functions on $\mathfrak{g}$.

(VII) Since the only automorphism of the Dynkin graph is the identity, we have Aut($\mathfrak{g}$) $=$ Aut$_0(\mathfrak{g})$.

Let $\Sigma$ be the group of similarities of V relative to $\Psi$. For all $g\in \Sigma$, let $\varphi (g)$ be the automorphism $x \rightarrow gxg^{-1}$ of $\mathfrak{g}$. Then $\varphi$ is a homomorphism from $\Sigma$ to Aut($\mathfrak{g}$). We show that it is surjective. Let $\alpha \in$ Aut($\mathfrak{g}$) $=$ Aut$_0(\mathfrak{g})$. By Prop. 2 of §7, no. 1, there exists $s\in \mathbf{G}\mathbf{L}(V)$ such that $\alpha (x) =sxs^{-1}$ for all $x\in \mathfrak{g}$. Then $s$ transforms $\Psi$ into a bilinear form $\Psi '$ on V that is invariant under $\mathfrak{g}$, and hence proportional to $\Psi ($§7, no. 5, Prop. 12). This proves that $s\in \Sigma$.

Since the identity representation of $\mathfrak{g}$ is irreducible, its commutant reduces to the scalars (§6, no. 1, Prop. 1), so the kernel of $\varphi$ is $k^*$. Thus, the group Aut($\mathfrak{g}$) $=$ Aut$_0(\mathfrak{g})$ can be identified with $\Sigma /k^*$. But, it follows from Algebra, Chap. IX, §6, no. 5, that the group $\Sigma$ is the product of the groups $k^*$ and $\mathbf{S}\mathbf{O}(\Psi )$; hence Aut($\mathfrak{g}$) $=$ Aut$_0(\mathfrak{g})$ can be identified with $\mathbf{S}\mathbf{O}(\Psi )$.

Let $\mathbf{O}^+_0(\Psi )$ be the reduced orthogonal group of $\Psi ($Algebra, Chap. IX, §9, no. 5). Since $\mathbf{S}\mathbf{O}(\Psi )/\mathbf{O}^+_0(\Psi )$ is commutative (loc. cit.), the group Aut$_e(\mathfrak{g})$ is contained in $\mathbf{O}^+_0(\Psi ) ($§11, no. 2, Prop. 3); in fact, it is equal to it (Exerc. 7).

(VIII) The canonical bilinear form $\Phi_R$ on $\mathfrak{h}^*$ is given by

$\Phi_R(\xi_1\varepsilon_1+\cdots +\xi_l\varepsilon_l, \xi '\varepsilon_1+\cdots +\xi '\varepsilon_l) =$ 1 $(\xi_1\xi '+\cdots +\xi_l\xi ')$

(Chap. VI, §4, no. 5.V). The isomorphism from $\mathfrak{h}$ to $\mathfrak{h}^*$ defined by $\Phi_R$ takes $H_i$ to $(4l-2).\varepsilon_i$. Thus, the inverse form of $\Phi_R$, that is the restriction to $\mathfrak{h}$ of the Killing form, is

(IX) Recall the $X_{\alpha}(\alpha \in R)$ defined by the formulas (3). It is easy to verify that $[X_{\alpha}, X_{-\alpha}] =-H_{\alpha}$ for $\alpha \in R$. On the other hand, let M be the matrix $I+E_{0,0}$; since $M=S^tM^{-1}S$, the map

is an automorphism of $\mathfrak{g}$ and $\theta (X_{\alpha}) =X_{-\alpha}$ for all $\alpha \in R$. Consequently, $(X_{\alpha})$ is a Chevalley system in $(\mathfrak{g},\mathfrak{h})$.

Assume that $k=\mathbf{Q}$. The Cartan subalgebra $\mathfrak{h}$ has two permissible lattices: the lattice $Q(R^{\vee})$ generated by the $H_{\alpha}$ and the lattice $P(R^{\vee})$ that is generated by the $H_i$ and consists of the diagonal matrices in $\mathfrak{h}$ with integer entries. It follows that $\mathfrak{o}_S(2l+ 1,\mathbf{Z})$ (the set of matrices in $\mathfrak{g}$ with integer entries) is the Chevalley order $P(R^{\vee}) +\sum\mathbf{Z}.X_{\alpha}$ in $\mathfrak{g}$. Since $(X_{\pm\varepsilon_i})^2= 2E_{\pm i,\mp i},(X_{\pm\varepsilon_i})^3= 0$ and $(X_{\pm\varepsilon_i\pm\varepsilon_j})^2= 0$, we see that the lattice $\mathscr{V}$ generated by the Witt basis $(e_i)_{-l\leq i\leq l}$ is an admissible lattice for $\mathfrak{o}_S(2l+ 1,\mathbf{Z})$ in V. The same is true for $\bigwedge^r\mathscr{V}$ in $\bigwedge^rV$.

Now consider the spinor representation $\rho$ of $\mathfrak{g}$ on $N =\bigwedge F'$. As its weights do not map $P(R^{\vee})$ to $\mathbf{Z}$, it has no admissible lattice for $\mathfrak{o}_S(2l+ 1,\mathbf{Z})$. On the other hand, the lattice $\mathscr{N}$ generated by the canonical basis $(e_{-i_1}\wedge \cdots \wedge e_{-i_k})$ of N (for $1\leq i_1<\cdots < i_k\leq l$) is an admissible lattice for the Chevalley order $\mathscr{G}= Q(R^{\vee}) +\sum_{\alpha\in R}\mathbf{Z}.X_{\alpha}$. Indeed, it is immediate that $\mathscr{N}$ is stable under

the exterior product with the $e_{-i}$ and the interior product with the $e_i$ (for $1\leq i\leq l$). The formulas of (V) then show that $\mathscr{N}$ is stable under $\rho (\mathscr{G})$. Moreover, since $\rho (X_{\alpha})^2= 0$ for all $\alpha \in R$, it follows that $\mathscr{N}$ is admissible.