Proposition 14

(i) The orthogonal of $M_z$ in $E^*$ is the set of $x^* \in E^*$ such that $x^* \perp z = 0$.

(ii) The subspace $M_z$ associated with z is the image of $\Lambda^{p-1}(E^*)$ under the mapping $\lambda_z : u^* \mapsto u^* \perp z$ of $\Lambda^{p-1}(E^*)$ into E.

Let N denote the image of $\lambda_z$. For $x^* \in E^*$ and $u^* \in \Lambda^{p-1}(E^*)$,

Therefore, for $x^*$ to be orthogonal to N, it is necessary and sufficient that $x^* \perp z$ be orthogonal to $\theta_{\Lambda}(\Lambda(E^*))$. Now, the latter condition is equivalent to saying that $x^* \perp z = 0$; for let $(e_{\lambda})_{\lambda \in L}$ be a basis of E; giving L a total ordering, it has been seen (§ 7, no. 8, Theorem 1) that the $e_J$, for J running through the set $\mathscr{F}(L)$ of finite subsets of L, form a basis of $\Lambda(E)$; it then follows from formula (30) of no. 5 that the elements $\theta_{\wedge}(e_J^*)$ are, to within a sign, the coordinate forms on $\wedge(E)$ relative to the basis $(e_J)$; whence our assertion.

The orthogonal of N therefore consists of the $x^* \in E^*$ such that $x^* \perp z = 0$ and the conclusion of (i) will therefore follow from (ii).

We show first that $N \subset M_z$. Let M be a vector subspace of E such that $z \in \wedge(M)$ and let $j : M \to E$ be the canonical injection; let $\mu_z$ denote the mapping $v^* \mapsto v^* \perp z$ of $\wedge^{p-1}(M^*)$ into M; it follows from formula (60) of no. 7 that there is a canonical factorization

which proves that $N \subset M$ and hence $N \subset M_z$ by definition of $M_z$. It remains to verify that $N = M_z$. Suppose the converse: there would then exist a basis $(e_i)_{1 \leq i \leq n}$ of $M_z$ and an element $x^* \in E^*$ such that $\langle x^*, e_1 \rangle = 1, \langle x^*, e_j \rangle = 0$ for $2 \leq j \leq n$ and such that $x^*$ is orthogonal to N and hence $x^* \perp z = 0$. We write $z = \sum_H a_H e_H$, where the sum is taken over the subsets of $\{1, n\}$ with $p$ elements. By (68) (no. 9),

which shows that the relation $x^* \perp z = 0$ implies $a_H = 0$ for $1 \in H$. But this is impossible, for $z$ would then belong to $\wedge^p(M')$, where $M'$ is the subspace of M generated by $e_2, \ldots, e_n$.