Proposition 4

Let $(e_i)_{i \in I}$ be an orthonormal basis of the hilbertian space $E$. For every $\alpha$ in $\mathbf{N}^{(I)}$, put

Then $(z_{\alpha})_{\alpha \in \mathbf{N}^{(I)}}$ is an orthonormal basis of $\hat{\mathbf{S}}(E)$.

Let $E_0$ be the vector subspace of $E$ generated by the vectors $e_i$ for $i$ ranging over $I$. Then the $z_{\alpha}$ form a basis of the vector space $\mathbf{S}(E_0)$ (A, III, § 6, No. 6). But $E_0$ is dense in $E$, and the multilinear mapping $(x_1, \ldots, x_n) \mapsto x_1 \ldots x_n$ from $E \times \cdots \times E$ into $\mathbf{S}(E)$ is continuous for all $n \geqslant 1$; hence $\mathbf{S}(E_0)$ is dense in $\mathbf{S}(E)$. It is now enough to prove that the family of the $z_\alpha$ is orthonormal. First observe that $\hat{S}^n(E)$ and $\hat{S}^m(E)$ are orthogonal for $n \neq m$. Hence it is enough to prove the formula

when $|\alpha| = \sum_{i \in I} \alpha_i$ and $|\beta| = \sum_{i \in I} \beta_i$ are equal to the same integer $n$.

Consider a partition $(P_i)_{i \in I}$ of the set $\{ 1, 2, ..., n \}$ such that Card $P_i = \alpha_i$ for all $i \in I$. Put $x_k = e_i$ if $k$ belongs to $P_i$, then $x_1 ... x_n = \prod_{i \in I} e_i^{\alpha_i}$. Similarly we define $(Q_i)_{i \in I}$ and $y_k$ in such a way that Card $Q_i = \beta_i$ and $y_1 ... y_n = \prod_{i \in I} e_i^{\beta_i}$. Since the $e_i$ are mutually orthogonal, we have $\langle x_k | y_{\sigma(k)} \rangle = 0$ except if there exists an indice $i \in I$ such that $k \in P_i$ and $\sigma(k) \in Q_i$. By formula (15), we then have $\langle x_1 ... x_n | y_1 ... y_n \rangle = 0$ unless there exists a permutation $\sigma \in \mathfrak{S}_n$ such that $\sigma(P_i) = Q_i$ for all $i \in I$, which implies that $\alpha = \beta$. Then $\langle z_\alpha | z_\beta \rangle = 0$ for $\alpha \neq \beta$. The same argument proves that $\| x_1 ... x_n \| ^2$ is equal to the number of the $\sigma \in \mathfrak{S}_n$ such that $\sigma(P_i) = P_i$ for all $i \in I$, hence equal to $\prod_{i \in I} \alpha_i !$. We get $\| z_\alpha \| = 1$, and the proposition is proved.