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Preservers of $λ$-Aluthge transforms

Published 20 Dec 2017 in math.OA | (1712.07499v2)

Abstract: Let $M$ and $N$ be arbitrary von Neumann algebras. For any $a$ in $M$ or in $N$, let $\Delta_{\lambda}(a)$ denote the $\lambda$-Aluthge transform of $a$. Suppose that $M$ has no abelian direct summand. We prove that every bijective map $\Phi:M\to N$ satisfying $$\Phi(\Delta_{\lambda}(a\circ b*))=\Delta_{\lambda}(\Phi(a) \circ \Phi(b)*), \hbox{ for all } a,\;b\in M,$$ (for a fixed $\lambda\in [0,1]$), maps the hermitian part of $M$ onto the hermitian part of $N$ (i.e. $\Phi (M_{sa}) = N_{sa}$) and its restriction $\Phi|{M{sa}} : M_{sa}\to N_{sa}$ is a Jordan isomorphism. If we also assume that $\Phi (x +i y ) = \Phi (x) +\Phi (i y)$ for all $x,y\in M_{sa}$, then there exists a central projection $p_c$ in $M$ such that $\Phi|{p_c M}$ is a complex linear Jordan $*$-isomorphism and $\Phi|{(\textbf{1}-p_c) M}$ is a conjugate linear Jordan $*$-isomorphism. Given two complex Hilbert spaces $H$ and $K$ with dim$(H)\geq 2$, we also establish that every bijection $\Phi: \mathcal{B}(H)\to \mathcal{B}(K)$ satisfying $$\Phi(\Delta_{\lambda}(a b*))=\Delta_{\lambda}(\Phi(a) \Phi(b)*), \hbox{ for all } a,\;b\in \mathcal{B}(H),$$ must be a complex linear or a conjugate linear $*$-isomorphism.

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