Preservers of Operator Commutativity

Abstract

Let $\mathfrak{M}$ and $\mathfrak{J}$ be JBW${}^{\ast }$-algebras admitting no central summands of type $I_1$ and $I_2,$ and let $\hat{I}¦:\mathfrak{M}\to \mathfrak{J}$ be a linear bijection preserving operator commutativity in both directions, that is, $$[x,\mathfrak{M},y]=0\iff [\hat{I}¦(x),\mathfrak{J},\hat{I}¦(y)]=0,$$ for all $x,y\in \mathfrak{M}$, where the associator of three elements $a,b,c$ in $\mathfrak{M}$ is defined by $[a,b,c]:=(a\circ b)\circ c-(c\circ b)\circ a$. We prove that under these conditions there exist a unique invertible central element $z_0$ in $\mathfrak{J}$, a unique Jordan isomorphism $J:\mathfrak{M}\to \mathfrak{J}$, and a unique linear mapping ${\hat{I}}^2$ from $\mathfrak{M}$ to the centre of $\mathfrak{J}$ satisfying $$\hat{I}¦(x)=z_0\circ J(x)+{\hat{I}}^2(x),$$ for all $x\in \mathfrak{M}.$ Furthermore, if $\hat{I}¦$ is a symmetric mapping (i.e., $\hat{I}¦(x^{\ast })=\hat{I}¦(x{)}^{\ast }$ for all $x\in \mathfrak{M}$), the element $z_0$ is self-adjoint, $J$ is a Jordan ${}^{\ast }$-isomorphism, and ${\hat{I}}^2$ is a symmetric mapping too. In case that $\mathfrak{J}$ is a JBW${}^{\ast }$-algebra admitting no central summands of type $I_1$, we also address the problem of describing the form of all symmetric bilinear mappings $B:\mathfrak{J}\times \mathfrak{J}\to \mathfrak{J}$ whose trace is associating (i.e., $[B(a,a),b,a]=0,$ for all $a,b\in \mathfrak{J})$ providing a complete solution to it. We also determine the form of all associating linear maps on $\mathfrak{J}$.