Finite-Dimensional CAR over an Inner-Product Space Is a Matrix Algebra

A finite orthonormal basis exists by [F4]; fix the basis in Given. For $f={\sum }_{j=1}^n{\lambda }_j{e}_j$, set $C(f)={\sum }_j{\lambda }_j{a}_j^{†}$ in the standard finite-mode algebra. Since the inner product is linear in the first variable, $⟨g,f⟩={\sum }_j{\mu }_j\overline{{\lambda }_j}$ when $g={\sum }_j{\mu }_j{e}_j$. Bilinearity of the anticommutator and the finite-mode relations give $\{C(f),C(g)\}=0$, $\{C(f{)}^{\ast },C(g{)}^{\ast }\}=0$, and $\{C(f{)}^{\ast },C(g)\}={\sum }_j\overline{{\lambda }_j}{\mu }_{j}1=⟨g,f⟩1$. Thus [F1] gives a unital $\ast$-homomorphism $\mathrm{\Phi }:{\mathsf{C}\mathsf{A}\mathsf{R}}^{\mathrm{a}\mathrm{l}\mathrm{g}}(F)\to {\mathsf{C}\mathsf{A}\mathsf{R}}_n^{\mathrm{a}\mathrm{l}\mathrm{g}}$.

Conversely, the elements $c(e_j{)}^{\ast }$ and $c(e_j)$ in ${\mathsf{C}\mathsf{A}\mathsf{R}}^{\mathrm{a}\mathrm{l}\mathrm{g}}(F)$ satisfy the standard finite-mode relations by [F1] and orthonormality. The universal property [F2] gives $\mathrm{\Psi }:{\mathsf{C}\mathsf{A}\mathsf{R}}_n^{\mathrm{a}\mathrm{l}\mathrm{g}}\to {\mathsf{C}\mathsf{A}\mathsf{R}}^{\mathrm{a}\mathrm{l}\mathrm{g}}(F)$. The maps $\mathrm{\Phi }\mathrm{\Psi }$ and $\mathrm{\Psi }\mathrm{\Phi }$ fix, respectively, every standard generator and every $c(e_j)$ and its adjoint. Since the $e_j$ span $F$, linearity makes them fix all generators; hence the maps are inverse $\ast$-isomorphisms.

By [F3] and step 2.1, the finite-mode universal C*-norm transports to ${\mathsf{C}\mathsf{A}\mathsf{R}}^{\mathrm{a}\mathrm{l}\mathrm{g}}(F)$ and its completion is $M_{2^n}(\mathbb{C})$. The same transport identifies any unital $\ast$-representation of this completion with a unital representation of the standard finite-mode algebra. Such a representation is isometric by [F3]. If $n=0$, both algebras are $\mathbb{C}$ and the identity representation is isometric as well.