The Countable-Mode CAR Fock Representation Is Irreducible

Write $\xi ={\sum }_S{\xi }_S{e}_S\in {\mathcal{F}}_{\mathrm{\infty }}$ and suppose $c_j\xi =0$ for every $j$. For any nonempty finite $S$, take its least element $j$. The coefficient of $e_{S\setminus \{j\}}$ in $c_j\xi$ is $(-1{)}^{\mathrm{\mid }\{s\in S:s<j\}\mathrm{\mid }}{\xi }_S={\xi }_S$, and no other input subset maps to this output under $c_j$. Hence ${\xi }_S=0$ for every nonempty $S$, so $\xi ={\xi }_{\mathrm{\varnothing }}{e}_{\mathrm{\varnothing }}$. Conversely every annihilator kills $e_{\mathrm{\varnothing }}$.

If $\xi \mathrm{\ne }0$, choose a finite subset $S$ with ${\xi }_S\mathrm{\ne }0$ and let $j$ be the least positive integer not in $S$. The coefficient of $e_{S\cup \{j\}}$ in $c_j^{†}\xi$ is $(-1{)}^{\mathrm{\mid }\{s\in S:s<j\}\mathrm{\mid }}{\xi }_S\mathrm{\ne }0$, and this output has the unique input $S$. Thus $c_j^{†}\xi \mathrm{\ne }0$, proving that no nonzero vector is killed by every creator.

Let $K\subseteq {\mathcal{F}}_{\mathrm{\infty }}$ be a nonzero closed subspace invariant under ${\pi }_F({\mathsf{C}\mathsf{A}\mathsf{R}}_{\mathbb{N}})$. Choose $0\mathrm{\ne }\xi \in K$ and, by [F2], a finite subset $S=\{j_1<\cdots <j_q\}$ with ${\xi }_S\mathrm{\ne }0$. If $q=0$, put $\eta =\xi$; otherwise put $\eta =c_{j_q}\cdots c_{j_1}\xi$. In either case $\eta \in K$. The operator used is the identity when $q=0$ and the displayed annihilator product otherwise. On basis vectors it vanishes unless $S\subseteq U$, and then gives a sign times $e_{U\setminus S}$; hence only $U=S$ contributes to its empty-set coefficient, so ${\eta }_{\mathrm{\varnothing }}=\pm {\xi }_S\mathrm{\ne }0$.

Given $\epsilon >0$, use [F2] to choose a finite-support vector $\zeta$ with $\mathrm{\parallel }\eta -\zeta \mathrm{\parallel }<\epsilon \mathrm{/}2$. For each $n$, put $N_j=c_j^{†}{c}_j$ and $Q_n={\prod }_{j=1}^n(I-N_j)$. By [F1], $N_j{e}_T={\mathbf{1}}_{j\in T}{e}_T$, so the factors commute and $Q_n$ is the diagonal projection that kills $e_T$ when $T\cap \{1,\dots ,n\}\mathrm{\ne }\mathrm{\varnothing }$; it belongs to ${\pi }_F({\mathsf{C}\mathsf{A}\mathsf{R}}_{\mathbb{N}})$ and has norm at most one. Choose $n$ at least every element of every nonempty set in the finite support of $\zeta$ (take $n=1$ if there are none), so $Q_n\zeta ={\zeta }_{\mathrm{\varnothing }}{e}_{\mathrm{\varnothing }}$. Then $\mathrm{\parallel }{Q}_n\eta -{\eta }_{\mathrm{\varnothing }}{e}_{\mathrm{\varnothing }}\mathrm{\parallel }\le \mathrm{\parallel }{Q}_n(\eta -\zeta )\mathrm{\parallel }+\mathrm{\mid }{\zeta }_{\mathrm{\varnothing }}-{\eta }_{\mathrm{\varnothing }}\mathrm{\mid }<\epsilon$: the first term is bounded by $\mathrm{\parallel }\eta -\zeta \mathrm{\parallel }$ since $Q_n$ is a contraction, and the second by the same norm since a coordinate functional on an orthonormal basis has norm one. Thus for every $\epsilon >0$ there is $Q_n\eta \in K$ within $\epsilon$ of ${\eta }_{\mathrm{\varnothing }}{e}_{\mathrm{\varnothing }}$. Closedness and ${\eta }_{\mathrm{\varnothing }}\mathrm{\ne }0$ give $e_{\mathrm{\varnothing }}\in K$.

For every finite $S$, a finite creator product sends $e_{\mathrm{\varnothing }}$ to $e_S$ up to a sign by [F1]. Invariance gives every basis vector in $K$, and their finite span is dense by [F2]; hence $K={\mathcal{F}}_{\mathrm{\infty }}$. This proves there is no nonzero proper closed invariant subspace, so ${\pi }_F$ is irreducible.