Finite-Subspace CAR Inclusions Are Isometric

The generators $c_G(f)$ for $f\in F$ satisfy the defining relations of ${\mathsf{C}\mathsf{A}\mathsf{R}}^{\mathrm{a}\mathrm{l}\mathrm{g}}(F)$. Its universal property [F2] therefore defines a unique unital $\ast$-homomorphism ${\iota }_{F,G}$ sending $c_F(f)$ to $c_G(f)$ for every $f\in F$.

Let $d=\dim F$ and $m=\dim G$. Choose an orthonormal basis $e_1,\dots ,e_d$ of $F$ using [F4], and choose a finite vector-space basis $v_1,\dots ,v_m$ of $G$. Initialize the current orthonormal list by $u_j=e_j$ for $1\le j\le d$, and let $s$ denote its current length, initially $d$. Process $v_1,\dots ,v_m$ in order: set $r_k=v_k-{\sum }_{j=1}^s⟨v_k,u_j⟩u_j$, where the current list is orthonormal; then $⟨r_k,u_j⟩=0$ for every $j$. If $r_k\mathrm{\ne }0$, append $r_k\mathrm{/}\mathrm{\parallel }{r}_k\mathrm{\parallel }$; otherwise keep the list unchanged. After processing $v_k$, the list spans $F+\mathrm{span}\{v_1,\dots ,v_k\}$. Thus the resulting finite orthonormal list is a basis of $G$ and retains $e_1,\dots ,e_d$ as its initial vectors. Under these basis identifications in [F1], ${\iota }_{F,G}$ is the standard inclusion of the first $d$ modes into $m$ modes. If $d<m$, this is the composition of the inclusions from $r$ to $r+1$ for $d\le r<m$; each is isometric by [F3], so the composition is isometric. If $d=m$, the map is the identity and is isometric. Thus ${\iota }_{F,G}$ is isometric and in particular injective.