Order preserving maps on Hermitian matrices (Q2866895)

The space \(H_n\) of all \(n \times n\) Hermitian matrices is a partially ordered set with the partial order defined by \(A \leq B\) if \(\langle Ax, x \rangle \leq \langle Bx, x \rangle\) for every \(x \in {\mathbb C}^n\). A (not necessarily linear) map \(\phi: H_n \longrightarrow H_n\) is said to be order preserving if for every pair \((A,B) \in H_n^2\) we have \(A \leq B \Leftrightarrow \phi(A) \leq \phi(B)\). Quite a number of earlier studies deal with linear maps preserving order, most of them in the infinite-dimensional case. Recently, \textit{L. Molnár} [J. Math. Phys. 42, No. 12, 5904--5909 (2001; Zbl 1019.81005)] obtained a result under weaker assumptions. He showed that every bijective map \(\phi\) preserving order on \(H_n\) in both directions and satisfying \(\phi(0) = 0\) must be a congruence transformation, possibly composed with transposition. Motivated by Molnár's paper, the authors prove the following theorem that replaces bijectivity by continuity:NEWLINENEWLINELet \(\phi\) a continuous order preserving map on \(H_n\) satisfying \(\phi(0) = 0\). Then there exists an invertible \(n \times n\) complex matrix \(T\) such that either \(\phi(A) = T A T^*\) or \(\phi(A) = T A^t T^*\) for every \(A \in H_n\), \(n \geq 2\).NEWLINENEWLINEIt is noticeable that this result cannot be extended to the infinite-dimensional case. Moreover, the assumption that \(\phi\) preserves order in both directions cannot be weakened. The linearity of \(\phi\) is a mere consequence of the theorem.