One-mode factorization of symmetric Fock space

By [F4], $K$ is a Hilbert space, so its symmetric tensor sectors and Fock space in the Statement are defined. Fix $k,m\ge 0$, write $n=k+m$, and let ${\psi }_m\in K_s^{(m)}$. Expanding the permutation average and using symmetry of ${\psi }_m$ gives $P_{s,n}(e^{\otimes k}\otimes {\psi }_m)={\left(\genfrac{}{}{0px}{}{n}{k}\right)}^{-1}{\sum }_{\begin{array}{c}I\subseteq \{1,\dots ,n\}\\ \mathrm{\mid }I\mathrm{\mid }=k\end{array}}{v}_I$, where $v_I$ is the tensor with $e$ in the slots in $I$ and the $m$ factors of ${\psi }_m$ in the remaining slots, symmetrized there. Indeed, for each fixed placement $I$, the $k!m!$ permutations of the equal $e$ factors and the symmetric ${\psi }_m$ give the same tensor, and $k!m!\mathrm{/}n!={\left(\genfrac{}{}{0px}{}{n}{k}\right)}^{-1}$.

Distinct placements $v_I$ are orthogonal: if $I\mathrm{\ne }{I}^{\mathrm{\prime }}$, some slot contains $e$ in one tensor and a vector of $K$ in the other, and $e\perp K$. Each $v_I$ has norm $\mathrm{\parallel }{\psi }_m\mathrm{\parallel }$, so step 1.1 gives $\mathrm{\parallel }{P}_{s,n}(e^{\otimes k}\otimes {\psi }_m){\mathrm{\parallel }}^2={\left(\genfrac{}{}{0px}{}{n}{k}\right)}^{-1}\mathrm{\parallel }{\psi }_m{\mathrm{\parallel }}^2$. The same orthogonality shows that the images for different $k$ in a fixed $n$ sector are orthogonal. Thus the factor $\sqrt{\left(\genfrac{}{}{0px}{}{n}{k}\right)}$ makes $J_{k,m}$ isometric and their sector ranges are orthogonal.

Algebraic elementary tensors in $H^{\widehat{\otimes }n}$ are dense by [F1]. Expand each factor using $H=\mathbb{C}e\oplus K$, distribute the finite tensor product, and apply $P_{s,n}$. Every resulting term has exactly $k$ factors $e$ and $m=n-k$ factors in $K$, so it lies in the range described in step 2.1. Since $P_{s,n}$ is bounded and algebraic elementary tensors are dense, their symmetrizations are dense in $H_s^{(n)}$; the range of the isometry is closed, hence it equals $H_s^{(n)}$. Taking the orthogonal direct sum over $n=k+m$ proves that $J$ is onto and unitary.

On a vector $e^{\otimes k}\otimes {\psi }_m$ with $n=k+m$, creation by $e$ acts by $\sqrt{n+1}{P}_{s,n+1}(e\otimes J_{k,m}(e^{\otimes k}\otimes {\psi }_m))$. Since $P_{s,n+1}$ averages over all permutations, symmetrizing the last $n$ slots first does not change the result. This equals $\sqrt{n+1}\sqrt{\left(\genfrac{}{}{0px}{}{n}{k}\right)}{P}_{s,n+1}(e^{\otimes (k+1)}\otimes {\psi }_m)$. The binomial identity $\left(\genfrac{}{}{0px}{}{n+1}{k+1}\right)=\frac{n+1}{k+1}\left(\genfrac{}{}{0px}{}{n}{k}\right)$ makes this exactly $J_{k+1,m}(\sqrt{k+1}{e}^{\otimes (k+1)}\otimes {\psi }_m)$. Thus creation becomes the stated one-mode raising map. Annihilation by $e$ is zero for $k=0$, because every first tensor factor lies in $K$. For $k\ge 1$, the placement formula of step 1.1 shows that contraction keeps exactly the $\left(\genfrac{}{}{0px}{}{n-1}{k-1}\right)$ placements with $e$ in the contracted slot. Therefore $a(e)J_{k,m}{\psi }_m=\sqrt{n}\sqrt{\left(\genfrac{}{}{0px}{}{n}{k}\right)}\frac{\left(\genfrac{}{}{0px}{}{n-1}{k-1}\right)}{\left(\genfrac{}{}{0px}{}{n}{k}\right)}{P}_{s,n-1}(e^{\otimes (k-1)}\otimes {\psi }_m)=\sqrt{k}{J}_{k-1,m}{\psi }_m$, using $\left(\genfrac{}{}{0px}{}{n}{k}\right)=\frac{n}{k}\left(\genfrac{}{}{0px}{}{n-1}{k-1}\right)$. Thus annihilation becomes the stated one-mode lowering map and leaves the $K$ factor unchanged.