Hilbert exterior powers and induced operators

Fix $n\ge 1$. In any finite tensor sum, all factor vectors lie in a finite-dimensional subspace $E\subseteq H$. Choose a finite orthonormal basis $(f_j)$ of $E$ by [A4]. Multilinearity expands the sum in the elementary tensors $f_{j_1}\otimes \cdots \otimes f_{j_n}$; the product form makes these tensors orthonormal. Their linear independence follows by applying the multilinear coordinate functionals $(x_1,\dots ,x_n)\mapsto {\prod }_{r=1}^n⟨x_r,f_{j_r}⟩$, which descend to the quotient and extract their coefficients. Thus the form is positive definite on every finite tensor span. Its completion is the Hilbert tensor power $H^{\otimes n}$.

Each $U_{\pi }$ is unitary and $U_{\pi }{U}_{\tau }=U_{\pi \tau }$. Replacing $\pi$ by ${\pi }^{-1}$ in the adjoint sum gives $A_n^{\ast }=A_n$; grouping the $n!$ pairs with product $\rho$ gives $A_n^2=A_n$. Thus $A_n$ is an orthogonal projection and is bounded with norm at most $1$ by the orthogonal decomposition into its range and kernel. Its range is closed. It is exactly the alternating subspace because the signed average is alternating and fixes every alternating tensor.

Let $C$ be any bound for $T$ from [A3]. After permuting tensor factors, a finite tensor sum can be written ${\sum }_{j=1}^m{x}_j\otimes y_j$ with $(y_j)$ an orthonormal basis of the finite-dimensional span of its remaining-factor tensors. Its squared norm is ${\sum }_j\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2$, while applying $T$ in that factor gives squared norm ${\sum }_j\mathrm{\parallel }T{x}_j{\mathrm{\parallel }}^2\le C^2{\sum }_j\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2$. Since factor permutations are unitary, this proves the bound $C$ for applying $T$ in any one slot. Composing over the $n$ slots extends $T^{\otimes n}$ to the completion with bound $C^n$.

By step 2.1 the normalized wedges are vectors in the alternating range. For pure tensors, self-adjointness and idempotence give $⟨\sqrt{n!}{A}_{n}x,\sqrt{n!}{A}_{n}y⟩=n!⟨x,A_{n}y⟩$. Expanding the signed average yields ${\sum }_{\pi \in S_n}\mathrm{sgn}(\pi ){\prod }_i⟨x_i,y_{\pi (i)}⟩$, which is the determinant in [A2]. This proves the stated Gram formula and its positivity from the Hilbert-space norm.

On elementary tensors $T^{\otimes n}$ commutes with every permutation, hence preserves $\mathrm{ran}{A}_n$ and restricts to a bounded ${\mathrm{\Lambda }}^{n}T$ with bound $C^n$. Its action on wedges is the displayed formula. Applying that formula twice proves ${\mathrm{\Lambda }}^n(ST)={\mathrm{\Lambda }}^{n}S{\mathrm{\Lambda }}^{n}T$ on a dense span and therefore everywhere. For $n=0$ the space is $\mathbb{C}$ and the induced map is its identity; for $n=1$, $A_1=I$ and ${\mathrm{\Lambda }}^{1}T=T$. If $T=0$ and $n\ge 1$, its induced map is zero.

If $(e_j)$ is a supplied orthonormal basis of $H$, its finite span is dense by [A5]. Approximate the factors of each elementary tensor by finite linear combinations of the $e_j$. The telescoping tensor identity and $\mathrm{\parallel }{x}_1\otimes \cdots \otimes x_n\mathrm{\parallel }={\prod }_r\mathrm{\parallel }{x}_r\mathrm{\parallel }$ show that elementary tensors in those finite spans are dense in $H^{\otimes n}$. Since $A_n$ is bounded, their images are dense in $\mathrm{ran}{A}_n$. Applying $A_n$ to a basis tensor gives zero if indices repeat and otherwise a signed multiple of the wedge with increasing indices. These wedges are orthonormal by step 3.1 and span a dense subspace, so they form an orthonormal basis of ${\mathrm{\Lambda }}^{n}H$. If $H$ is finite-dimensional with $n>\dim H$, there are no increasing $n$-tuples and ${\mathrm{\Lambda }}^{n}H=\{0\}$.