The algebra of thin measurable operators is directly finite

Yıl 2023, Cilt: 6 Sayı: 1, 1 - 5, 15.03.2023

Airat Bikchentaev

https://doi.org/10.33205/cma.1181495

Cited By: 3

Öz

Let $\mathcal{M}$ be a semifinite von Neumann algebra on a Hilbert space $\mathcal{H}$ equipped with a faithful normal semifinite trace $\tau$, $S(\mathcal{M},\tau)$ be the ${}^*$-algebra of all $\tau$-measurable operators. Let $S_0(\mathcal{M},\tau)$ be the ${}^*$-algebra of all $\tau$-compact operators and $T(\mathcal{M},\tau)=S_0(\mathcal{M},\tau)+\mathbb{C}I$ be the ${}^*$-algebra of all operators $X=A+\lambda I$
with $A\in S_0(\mathcal{M},\tau)$ and $\lambda \in \mathbb{C}$. It is proved that every operator of $T(\mathcal{M},\tau)$ that is left-invertible in $T(\mathcal{M},\tau)$ is in fact invertible in $T(\mathcal{M},\tau)$.
It is a generalization of Sterling Berberian theorem (1982) on the subalgebra of thin operators in $\mathcal{B} (\mathcal{H})$.
For the singular value function $\mu(t; Q)$ of $Q=Q^2\in S(\mathcal{M},\tau)$, the inclusion $\mu(t; Q)\in \{0\}\bigcup
[1, +\infty)$ holds for all $t>0$. It gives the positive answer to the question posed by Daniyar Mushtari in 2010.

Anahtar Kelimeler

Hilbert space, von Neumann algebra, semifinite trace, $\tau$-measurable operator, $\tau$-compact operator, singular value function, idempotent

Destekleyen Kurum

Volga Region Mathematical Center

Proje Numarası

075-02-2022-882

Kaynakça

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