Which traces are spectral? (Q2445298)

Let \(\mathcal I\subset \mathcal B(H)\) be a proper ideal and \(\varphi\) a trace on \(\mathcal I\). A natural question posed by \textit{A. Pietsch} in [Math. Nachr. 145, 7--43 (1990; Zbl 0709.47021)] asks which traces are spectral that is: if \(\lambda(T)\in \mathcal B(H)\) denotes the vector of eigenvalues (counting multiplicity) of an operator \(T\in\mathcal I\) arranged as a diagonal operator for a fixed orthonormal basis, one is looking for \(\varphi\) such that \(\varphi(T)=\varphi(\lambda(T))\). The main obstacle here are non-normal operators, and the key notion introduced by the authors is that of being closed with respect of logarithmic submajorization: \(B\prec\prec_{\mathrm{log}} A\) if \(\prod_{k=0}^n \mu_k(A)\leq \prod_{k=0}^n\mu_k(B)\) for all \(n\geq 0\) (here, \(\mu_k\) denotes the singular values of \(A\), that is, the eigenvalues of \(|A|\) counted with multiplicity and arranged in decreasing order); an ideal \(\mathcal I\) is closed with respect to this preorder if \(B\prec\prec_{\mathrm{log}} A\in\mathcal I\) implies \(B\in\mathcal I\). The following are the main results of the paper: Theorem 8. if \(\mathcal I\) is closed with respect to log submajorization, then any trace \(\varphi\) is spectral; and if \(\mathcal I\) is not closed, then there exists \(T\in\mathcal I\) such that \(\lambda(T)\notin \mathcal I\). Theorem 10. If \(T\in \mathcal I\) is any ideal and \(\varphi\) is a positive trace on \(\mathcal I\), then \(\varphi\) is monotone with respect to log submajorization if and only if \(\varphi\) is spectral.