On the automorphisms of the spectral unit ball

Abstract:

Let $\mathcal {A}$ be a (complex, unital) semisimple Banach algebra and denote by $\Omega _{\mathcal {A}}$ its open spectral unit ball, that is, the set \begin{equation*} \Omega _{\mathcal {A}}=\{a\in \mathcal {A}:\sigma \left ( a\right ) \subseteq \mathbf {D}\}, \end{equation*} where $\sigma \left ( a\right )$ denotes the spectrum of $a$ in $\mathcal {A}$ and $\mathbf {D}$ is the open unit disc in the complex plane. We prove that if $F:\Omega _{\mathcal {A}}\rightarrow \Omega _{\mathcal {A}}$ is a holomorphic map satisfying $F\left ( 0\right ) =0$ and $F^{\prime }\left ( 0\right ) =I$ (the identity of $\mathcal {A}$), then for $a$ in $\Omega _{ \mathcal {A}}$ the intersection of all closed discs lying inside $\mathbf {D}$ and containing $\sigma \left ( a\right )$ equals the intersection of all closed discs lying inside $\mathbf {D}$ and containing $\sigma \left ( F\left ( a\right ) \right )$. When all the elements of $\mathcal {A}$ have an at most countable spectrum and $F$ is biholomorphic, this implies that $F$ preserves the convex hull of the spectrum. As an application of the same equality, we prove that if $\mathcal {B}$ is a semisimple Banach algebra and $T: \mathcal {A } \rightarrow \mathcal {B}$ is a unital surjective spectral isometry, then $\sigma \left ( T\left ( a\right ) \right ) =\sigma \left ( a\right )$ in the case when $\sigma \left ( a\right )$ has exactly two elements.