The Countable CAR Algebra Has the Universal Inductive-Limit Norm

Restrict the given family to its first $n$ modes. The quotient universal property of Finite-Mode Complex CAR Algebra gives the finite algebraic $\ast$-homomorphism on generators, and [F2] identifies it with a contractive unital $\ast$-representation ${\rho }_n:{\mathsf{C}\mathsf{A}\mathsf{R}}_n\to \mathcal{B}(H)$. If $m\le n$, the maps ${\rho }_n\circ {\iota }_{m,n}$ and ${\rho }_m$ agree on all generators of ${\mathsf{C}\mathsf{A}\mathsf{R}}_m$, so uniqueness of the finite quotient map gives ${\rho }_n\circ {\iota }_{m,n}={\rho }_m$.

Let ${\mathcal{P}}_{\mathrm{f}\mathrm{i}\mathrm{n}}(\mathbb{N})$ be all finite subsets of $\mathbb{N}$, and let ${\mathcal{F}}_{\mathrm{\infty }}={\mathrm{\ell }}^2({\mathcal{P}}_{\mathrm{f}\mathrm{i}\mathrm{n}}(\mathbb{N}))$ with coordinate vectors $e_S$. By [F8], this is a complex Hilbert space, the vectors $e_S$ are an orthonormal basis, and their finite span is dense. Define $c_j{e}_S=0$ if $j\mathrm{\notin }S$, and $c_j{e}_S=(-1{)}^{\mathrm{\mid }\{s\in S:s<j\}\mathrm{\mid }}{e}_{S\setminus \{j\}}$ if $j\in S$; define $c_j^{†}$ by the reverse toggle with the same sign. Each map is a partial isometry on the basis and extends boundedly, and the coefficient of each reverse basis transition shows $c_j^{†}$ is its adjoint. For any finite set of indices, choose $n$ containing them; under the decomposition of the basis into first-$n$ and tail subsets, the first $n$ operators act as the finite occupation representation tensored with the identity. Hence the CAR hold for every pair of indices. The finite universal property gives compatible contractive representations of all ${\mathsf{C}\mathsf{A}\mathsf{R}}_n$ on ${\mathcal{F}}_{\mathrm{\infty }}$, which extend to a representation ${\pi }_{\mathrm{\infty }}$ of the inductive limit.

Define ${\rho }_0$ on ${\mathsf{C}\mathsf{A}\mathsf{R}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ by ${\rho }_0(x)={\rho }_n(x)$ whenever $x\in {\mathsf{C}\mathsf{A}\mathsf{R}}_n$. Compatibility in step 1.1 makes this well defined, and contractivity at each stage gives $\mathrm{\parallel }{\rho }_0(x)\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }$. The space $\mathcal{B}(H)$ is complete: for an operator-norm Cauchy sequence $(S_k)$, each $(S_{k}x)$ converges in $H$; its pointwise limit $Sx$ is linear and bounded because $(\mathrm{\parallel }{S}_k\mathrm{\parallel })$ is bounded, and taking the pointwise limit in the uniform Cauchy bound shows $\mathrm{\parallel }{S}_k-S\mathrm{\parallel }\to 0$. Thus ${\rho }_0$ extends uniquely and contractively to the completion. Products pass to limits by $\mathrm{\parallel }AB\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }$. For any bounded operator $A$ with a specified adjoint, [F6] and [F7] give $\mathrm{\parallel }{A}^{\ast }y\mathrm{\parallel }={\sup }_{\mathrm{\parallel }x\mathrm{\parallel }\le 1}\mathrm{\mid }⟨A^{\ast }y,x⟩\mathrm{\mid }={\sup }_{\mathrm{\parallel }x\mathrm{\parallel }\le 1}\mathrm{\mid }⟨Ax,y⟩\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, hence $\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }$. Conjugating [F6] shows $A$ is the adjoint of $A^{\ast }$; uniqueness gives $(A^{\ast }{)}^{\ast }=A$, so applying the same inequality to $A^{\ast }$ yields $\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }A\mathrm{\parallel }$. Therefore if $S_k\to S$, the adjoints $S_k^{\ast }$ are Cauchy and converge to some $R$; passing to the limit in [F6] gives $⟨Sx,y⟩=⟨x,Ry⟩$, so $R$ is the adjoint of $S$. Thus the extension takes values in ${\mathcal{B}}_{\mathrm{a}\mathrm{d}}(H)$, preserves $\ast$, and is a unital $\ast$-homomorphism into this subalgebra.

Under the basis bijection ${\mathcal{P}}_{\mathrm{f}\mathrm{i}\mathrm{n}}(\mathbb{N})\cong {\mathcal{P}}_n\times {\mathcal{P}}_{\mathrm{f}\mathrm{i}\mathrm{n}}(\{n+1,n+2,\dots \})$, the first $n$ toggle signs depend only on the first subset coordinate because every tail index is larger than every mode $j\le n$. Thus the restriction of ${\pi }_{\mathrm{\infty }}$ to ${\mathsf{C}\mathsf{A}\mathsf{R}}_n$ is the finite matrix representation acting identically on each tail coordinate. The tail contains the empty subset, so it is nonzero; by the isometric amplification statement in [F2], $\mathrm{\parallel }{\pi }_{\mathrm{\infty }}(x)\mathrm{\parallel }=\mathrm{\parallel }x{\mathrm{\parallel }}_{{\mathsf{C}\mathsf{A}\mathsf{R}}_n}$ for every $x\in {\mathsf{C}\mathsf{A}\mathsf{R}}_n$.

For any bounded CAR representation $\sigma$ and $x\in {\mathsf{C}\mathsf{A}\mathsf{R}}_n$, its restriction is contractive by [F2], so $\mathrm{\parallel }\sigma (x)\mathrm{\parallel }\le \mathrm{\parallel }x{\mathrm{\parallel }}_{{\mathsf{C}\mathsf{A}\mathsf{R}}_n}$. Step 2.2 supplies a representation attaining this norm. Therefore the supremum over all representations equals the inductive-limit norm on each finite stage. Every element of ${\mathsf{C}\mathsf{A}\mathsf{R}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ lies in some finite stage, so the universal norm is finite and equals the inductive-limit norm on the entire union. Step 1.2 proves the representing family is nonempty. The extension in step 2.1 is unique because ${\mathsf{C}\mathsf{A}\mathsf{R}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ is dense.