Algebra isomorphisms between standard operator algebras
Volume 191 / 2009
Abstract
If $X$ and $Y$ are Banach spaces, then subalgebras ${\mathfrak A}\subset B(X)$ and ${\mathfrak B}\subset B(Y)$, not necessarily unital nor complete, are called standard operator algebras if they contain all finite rank operators on $X$ and $Y$ respectively. The peripheral spectrum of $A\in \mathfrak A$ is the set $\sigma_\pi(A)=\{\lambda\in\sigma(A) : |\lambda|=\max_{z\in\sigma(A)}|z|\}$ of spectral values of $A$ of maximum modulus, and a map $\varphi\colon{\mathfrak A}\to\mathfrak B$ is called peripherally-multiplicative if it satisfies the equation $\sigma_\pi(\varphi(A)\circ\varphi(B))=\sigma_\pi(A B)$ for all $A,B\in\mathfrak A$. We show that any peripherally-multiplicative and surjective map $\varphi\colon{\mathfrak A}\to\mathfrak B$, neither assumed to be linear nor continuous, is a bijective bounded linear operator such that either $\varphi$ or $-\varphi$ is multiplicative or anti-multiplicative. This holds in particular for the algebras of finite rank operators or of compact operators on $X$ and $Y$ and extends earlier results of Molnár. If, in addition, $\sigma_\pi(\varphi(A_0))\neq-\sigma_\pi(A_0)$ for some $A_0\in\mathfrak A$ then $\varphi$ is either multiplicative, in which case $X$ is isomorphic to $Y$, or anti-multiplicative, in which case $X$ is isomorphic to $Y^\ast$. Therefore, if $X\not\cong Y^*$ then $\varphi$ is multiplicative, hence an algebra isomorphism, while if $X\not\cong Y$, then $\varphi$ is anti-multiplicative, hence an algebra anti-isomorphism.
