Theorem 2

Suppose that the dimension of $E$ is an even number $m = 2r$ and that $Q$ is neutral (§ 4, no. 2). Then the algebra $C(Q)$ is separable (chap. VIII, § 7, no. 5, def. 1) and is isomorphic to the algebra of endomorphisms of a vector space of dimension $2^r$ over $A$. Moreover, if $m > 0$, $C^+(Q)$ is separable and is direct sum of two ideals isomorphic to the algebra of endomorphisms of a vector space of dimension $2^{r-1}$ over $A$.

Indeed, since Q is neutral, E can be decomposed into the direct sum of two totally singular subspaces N and P of dimensions r (§ 4, no. 2, cor. 1 of prop. 2). The restriction of Q to N being zero, the subalgebra S of C(Q) generated by N is identified with the exterior algebra of N (no. 3, lemma 4). For $n \in \mathbf{N}$, we shall denote by $e'_n$ the mapping $t \to nt$ of S into itself.

Let $(n_1, \ldots, n_r)$ be a basis of N; we shall denote by $(p_1, \ldots, p_r)$ the basis of P such that $\Phi(n_i, p_j) = \delta_{ij}$ (§ 4, no. 2, prop. 2). For $p \in P$, we shall denote by $p'$ the linear form $n \to \Phi(n, p)$ on N, and by $i_p$ the endomorphism of S obtained by passing to the quotient from the endomorphism $i_{p'}$ of $T(N)$ associated with $p'$ as was said in lemma 1 of no. 2. By (5), we have

Put, for $x = n + p \in E$ (with $n \in \mathbf{N}$ and $p \in \mathbf{P}$), $s(x) = e'_n + i_p$. It is clear that s is a linear mapping of E into $\mathscr{L}(S)$. Since we have

by virtue of (12) and lemma 1 (no. 2), s extends to a homomorphism (which we shall still denote by s) of C(Q) into $\mathscr{L}(S)$ (no. 1, prop. 1). We shall now prove that this homomorphism is surjective, which, since C(Q) and $\mathscr{L}(S)$ are both of dimension $2^{2r}$, will imply that s is an isomorphism and will prove our first assertion.

Let us indeed denote by $I$ the interval $[1, r]$. For any subset H of I, we shall put $H' = I - H$ and denote by $n_H$ (resp. $p_H$) the product of the $n_i$ (resp. $p_i$) for $i \in H$, arranged in increasing order of the indices. Recall that the $n_H$ form a basis of S (No. 3, Theorem 1). Finally, for any two subsets H, K of I, put $x_{H, K} = n_H p_I n_K$. We shall now prove that the elements $s(x_{H, K})$ of $s(C(Q))$ generate $\mathscr{L}(S)$. Now, if $j \notin H$, we have $s(p_j)(n_H) = i_{p_j}(n_H) = 0$ by Lemma 1, since the $n_i$ for $i \in H$ belong to the kernel of the linear form $n \to \Phi(n, p_j)$ on N; on the other hand, we have

(by (12)). Since s is a homomorphism, it follows that, for any two subsets H, K of I, $s(p_K)(n_H) = 0$ if $K \not\subset H$, and that $s(p_K)(n_H) = \pm n_{H-K}$ if $K \subset H$. Since, for $M \subset I$ and $L \subset I$, by definition $s(n_M)(n_L) = n_M n_L$, and $n_M n_L$ is zero if $M \cap L \neq \emptyset$ and is equal to $\pm n_{M \cup L}$ otherwise, it follows from the above that, for any subsets $H, K, L$ of $I$, $s(x_{H,K})(n_L) = s(n_H)s(p_I)s(n_{K'}) (n_L)$ is zero if $K \neq L$ and is equal to $\pm n_H$ if $K = L$. This shows that the $s(x_{H,K})$ generate $\mathscr{L}(S)$ and completes the proof of the first assertion.

To prove the second assertion, put $S^+ = S \cap C^+$ and $S^- = S \cap C^-$; it is clear that $S^+$ (resp. $S^-$) is the subspace of S generated by the $n_H$ such that H has an even (resp. odd) number of elements, that S is the direct sum of $S^+$ and $S^-$, and that $s(C^+)$ leaves $S^+$ and $S^-$ stable. Consequently $s$ maps $C^+$ into a subalgebra of $\mathscr{L}(S)$, isomorphic to the product $\mathscr{L}(S^+) \times \mathscr{L}(S^-)$; the restriction of $s$ to $C^+$ is an isomorphism of $C^+$ onto this subalgebra, since $s$ is injective and $C^+$ and $\mathscr{L}(S^+) \times \mathscr{L}(S^-)$ both have dimension $2^{2r-1}$ (No. 2, Corollary 1 of Theorem 1). QED.