Particle-Hole Exchange Is an Automorphism of Countable-Mode CAR

On a finite stage, exchange $a_j$ with $a_j^{†}$. The first two CAR families are exchanged, while $\{a_i^{†},a_j\}-{\delta }_{ij}1$ is the same relation as $\{a_j,a_i^{†}\}-{\delta }_{ji}1$ because ${\delta }_{ij}={\delta }_{ji}$. The exchange also respects $\ast$, since $(a_j^{†}{)}^{\ast }=a_j$. The quotient universal property in [F1] therefore gives a unital $\ast$-homomorphism ${\alpha }_n:{\mathsf{C}\mathsf{A}\mathsf{R}}_n\to {\mathsf{C}\mathsf{A}\mathsf{R}}_n$. Applying it twice fixes every generator, so ${\alpha }_n^2=I$ and ${\alpha }_n$ is an automorphism.

For $m\le n$, the maps ${\alpha }_n\circ {\iota }_{m,n}$ and ${\iota }_{m,n}\circ {\alpha }_m$ agree on every generator of ${\mathsf{C}\mathsf{A}\mathsf{R}}_m$, hence agree on that stage by its universal property. The compatible automorphisms therefore define an automorphism ${\alpha }_0$ of the algebraic union. For $x$ in a finite stage, the universal norm there is the supremum of $\mathrm{\parallel }\rho (x)\mathrm{\parallel }$ over its unital representations; precomposition by the bijection ${\alpha }_n$ permutes those representations, so $\mathrm{\parallel }{\alpha }_n(x)\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$. Thus ${\alpha }_0$ is isometric and extends continuously to the completion in [F2]. Its inverse extends from the compatible maps ${\alpha }_n^{-1}={\alpha }_n$, and ${\alpha }^2=I$ because this holds on the dense union.