The Finite-Mode CAR Algebra Is a Full Matrix Algebra

By [F2] the vector-space dimension of ${\mathsf{C}\mathsf{A}\mathsf{R}}_n^{\mathrm{a}\mathrm{l}\mathrm{g}}$ is the dimension of ${\mathrm{C}\mathrm{l}}^{+}({\mathbb{C}}^{2n},Q)$. Applying [F3] with $2n$ basis vectors gives ${\dim }_{\mathbb{C}}{\mathsf{C}\mathsf{A}\mathsf{R}}_n^{\mathrm{a}\mathrm{l}\mathrm{g}}=2^{2n}=4^n$, including $n=0$.

The occupation representation in [F4] is a surjective $\ast$-homomorphism onto $\mathrm{End}({\mathcal{F}}_n)$. Since ${\mathcal{P}}_n$ has $2^n$ elements, ${\dim }_{\mathbb{C}}\mathrm{End}({\mathcal{F}}_n)=(2^n{)}^2=4^n$. By step 1.1, the source also has dimension $4^n$, so the kernel has dimension zero. Thus the occupation representation is a $\ast$-isomorphism.

Put $m=2^n$ and identify ${\mathsf{C}\mathsf{A}\mathsf{R}}_n^{\mathrm{a}\mathrm{l}\mathrm{g}}$ with $M_m(\mathbb{C})$ by step 2.1. In any unital $\ast$-representation $\rho$ on a nonzero Hilbert space $K$, the projections $P_S=\rho (E_{S,S})$ are pairwise orthogonal and sum to $I$. Set $K_0=P_{\mathrm{\varnothing }}K$. The map $U_S=\rho (E_{S,\mathrm{\varnothing }}){\mathrm{\mid }}_{K_0}:K_0\to P_{S}K$ is unitary because the matrix-unit relations give $U_S^{\ast }{U}_S=I_{K_0}$ and $U_S{U}_S^{\ast }=P_S$. Hence $U:{⨁}_{S\in {\mathcal{P}}_n}{K}_0\to K$, $U(({\xi }_S{)}_S)={\sum }_S{U}_S{\xi }_S$, is unitary and carries each $\rho (E_{S,T})$ to the standard matrix unit on the subset coordinate with identity multiplicity on $K_0$.

The space $K_0$ is nonzero: if $y\in P_{S}K$, then $y=\rho (E_{S,\mathrm{\varnothing }})\rho (E_{\mathrm{\varnothing },S})y$, and $\rho (E_{\mathrm{\varnothing },S})y\in K_0$; if $K_0=0$, this forces every $P_{S}K=0$, contradicting ${\sum }_S{P}_S=I$ and $K\mathrm{\ne }\{0\}$. Under the unitary in step 3.1, $A=(A_{S,T})$ acts on ${⨁}_S{K}_0$ as $A\otimes I_{K_0}$. For each vector $({\xi }_T{)}_T$, its finitely many coordinates span a finite-dimensional subspace of $K_0$; expand them in a finite orthonormal basis and apply the scalar matrix norm bound to obtain $\mathrm{\parallel }A\otimes I_{K_0}\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }$. Conversely, for a unit vector $\eta \in K_0$, vectors $(x_S\eta {)}_S$ give $\mathrm{\parallel }A\otimes I_{K_0}\mathrm{\parallel }\ge \mathrm{\parallel }A\mathrm{\parallel }$ after taking the supremum over unit $x\in {\mathbb{C}}^m$. Thus every unital representation is isometric.

The identity representation on ${\mathbb{C}}^m$ is one of the representations in the universal supremum, and step 4.1 shows every unital representation has the same matrix operator norm. The universal norm is therefore finite and equals the matrix operator norm. By [F5] and the C*-axioms in [F6], this is a complete C*-norm, so the algebraic isomorphism of step 2.1 already identifies the universal C*-completion with $M_m(\mathbb{C})$. For $n=0$, $m=1$ and the same argument gives $\mathbb{C}$.