Canonical Finite-Mode CAR Inclusions Are Isometric

The first $n$ generators in ${\mathsf{C}\mathsf{A}\mathsf{R}}_{n+1}$ satisfy the $n$-mode CAR relations. The quotient universal property in [F3], followed by the finite-mode C*-completion in [F1], therefore gives the unique unital $\ast$-homomorphism ${\iota }_n$ with the stated generator values.

Identify $e_S\otimes e_{\epsilon }$ with $e_{S\cup (\{n+1\}\text{ if }\epsilon =1)}$, where $S\subseteq \{1,\dots ,n\}$ and $\epsilon \in \{0,1\}$. For $j\le n$, the number of occupied indices below $j$ is unchanged by the last occupation bit. The formulas in [F2] therefore give $c_j^{(n+1)}=c_j^{(n)}\otimes I_2$ and $(c_j^{†}{)}^{(n+1)}=(c_j^{†}{)}^{(n)}\otimes I_2$. By multiplication and linearity, the representation of ${\iota }_n(x)$ is ${\pi }_n(x)\otimes I_2$ for every $x\in {\mathsf{C}\mathsf{A}\mathsf{R}}_n$.

For any $A\in M_{2^n}(\mathbb{C})$ and $(x_0,x_1)\in {\mathbb{C}}^{2^n}\oplus {\mathbb{C}}^{2^n}$, $\mathrm{\parallel }(A\otimes I_2)(x_0,x_1){\mathrm{\parallel }}_2^2=\mathrm{\parallel }A{x}_0{\mathrm{\parallel }}_2^2+\mathrm{\parallel }A{x}_1{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }A{\mathrm{\parallel }}^2(\mathrm{\parallel }{x}_0{\mathrm{\parallel }}_2^2+\mathrm{\parallel }{x}_1{\mathrm{\parallel }}_2^2)$, so $\mathrm{\parallel }A\otimes I_2\mathrm{\parallel }\le \mathrm{\parallel }A\mathrm{\parallel }$. Conversely, vectors $(x,0)$ with $\mathrm{\parallel }x{\mathrm{\parallel }}_2\le 1$ give $\mathrm{\parallel }A\otimes I_2\mathrm{\parallel }\ge \mathrm{\parallel }A\mathrm{\parallel }$. By [F1] and step 1.2, $\mathrm{\parallel }{\iota }_n(x)\mathrm{\parallel }=\mathrm{\parallel }{\pi }_n(x)\otimes I_2\mathrm{\parallel }=\mathrm{\parallel }{\pi }_n(x)\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$. Thus ${\iota }_n$ is isometric and therefore injective; the argument includes $n=0$.