Product of operators and numerical range preserving maps
Tom 174 / 2006
Streszczenie
Let ${\bf V}$ be the $C^*$-algebra $B(H)$ of bounded linear operators acting on the Hilbert space $H$, or the Jordan algebra $S(H)$ of self-adjoint operators in $B(H)$. For a fixed sequence $(i_1, \dots, i_m)$ with $i_1, \dots, i_m \in \{1, \dots, k\}$, define a product of $A_1, \dots, A_k \in {\bf V}$ by $A_1* \cdots * A_k = A_{i_1} \cdots A_{i_m}$. This includes the usual product $A_1* \cdots * A_k = A_1 \cdots A_k$ and the Jordan triple product $A*B = ABA$ as special cases. Denote the numerical range of $A \in {\bf V}$ by $W(A) = \{ (Ax,x): x \in H,\, (x,x) = 1\}.$ If there is a unitary operator $U$ and a scalar $\mu$ satisfying $\mu^m = 1$ such that $\phi:{\bf V} \rightarrow {\bf V}$ has the form $$A \mapsto \mu U^*AU \quad \hbox{or} \quad A \mapsto \mu U^*A^tU,$$ then $\phi$ is surjective and satisfies $$W(A_1 *\cdots *A_k) = W(\phi(A_1)* \cdots *\phi(A_k)) \quad\ \hbox{for all } A_1, \dots, A_k \in {\bf V}.$$ It is shown that the converse is true under the assumption that one of the terms in $(i_1, \dots, i_m)$ is different from all other terms. In the finite-dimensional case, the converse can be proved without the surjectivity assumption on $\phi$. An example is given to show that the assumption on $(i_1, \dots, i_m)$ is necessary.
