Proposition 5

Let $(e_i)_{i \in I}$ be an orthonormal basis of the hilbertian space E. Endow I with a totally ordered structure. Then the set of all elements $e_{i_1} \wedge ... \wedge e_{i_n}$ for $i_1 < ... < i_n$ is an orthonormal basis of $\hat{\Lambda}^n(E)$.

We know (A, III, § 7, No. 8) that the elements in question form a basis of the vector space $\Lambda^n(E_0)$ where $E_0$ is the vector subspace of E generated by the vectors $e_i$. Further, for $i_1 < ... < i_n$, the matrix of scalar products $\langle e_{i_k}|e_{i_l} \rangle$ is the unit matrix of order n; by (26), we have $\|e_{i_1} \wedge ... \wedge e_{i_n}\| = 1$. Finally, if $(i_1, ..., i_n)$ and $(j_1, ..., j_n)$ are two distinct, strictly increasing sequences of elements of I, then there exists an element $j_\ell$ distinct from $i_1, ..., i_n$ and so we have $\langle e_{i_k}|e_{j_\ell} \rangle = 0$ for $1 \leq k \leq n$, and by (26), $\langle e_{i_1} \wedge ... \wedge e_{i_n}|e_{j_1} \wedge ... \wedge e_{j_n} \rangle = 0$. In other words, the family of elements $e_{i_1} \wedge ... \wedge e_{i_n}$, for $i_1 < ... < i_n$, is orthonormal.

But $E_0$ is dense in E, and the mapping $(x_1, ..., x_n) \mapsto x_1 \wedge ... \wedge x_n$ from $E \times \cdots \times E$ into $\Lambda^n(E)$ is continuous. Consequently, $\Lambda^n(E_0)$ is dense in $\Lambda^n(E)$, and proposition 5 follows.