Maps preserving the maximal numerical range of the triple product of operators

Let \(\mathscr {L}(\mathscr {H})\) be the algebra of all bounded linear operators on a complex Hilbert space \(\mathscr {H}\). For an operator \(T\in \mathscr {L}(\mathscr {H})\), let \(W_0(T)\) be the maximal numerical range of T. We show that a map \(\varphi \) from \(\mathscr {L}(\mathscr {H})\) onto itself satisfies

$$\begin{aligned} W_0\left( \varphi (S)\varphi (T)\varphi (S)\right) ~=~W_0(STS), \qquad (T,~S\in \mathscr {L}(\mathscr {H})), \end{aligned}$$

if and only if there are a unitary operator \(U\in \mathscr {L}(\mathscr {H})\) and \(\lambda \in \mathbb {C}\) such that \(\lambda ^3=1\) and either \(\varphi (T)= \lambda UTU^*\) for all \(T\in \mathscr {L}(\mathscr {H})\), or \(\varphi (T)= \lambda UT^\top U^*\) for all \(T\in \mathscr {L}(\mathscr {H})\). Here, \(T^\top \) denotes the transpose of any operator \(T\in \mathscr {L}(\mathscr {H})\) relative to a fixed but arbitrary orthonormal base of \(\mathscr {H}\). When the triple product “STS” is replaced by the skew-triple product “\(TS^*T\)”, we arrive at the same conclusion but with \(\lambda =1\).