The algebra of thin measurable operators is directly finite

TL;DR

This paper proves that in a semifinite von Neumann algebra, every left-invertible operator in a certain algebra of operators is actually invertible, extending a classical theorem to a broader non-commutative setting.

Contribution

It generalizes Sterling Berberian's theorem on thin operators to the algebra of all operators of the form A plus a scalar multiple of identity in a semifinite von Neumann algebra.

Findings

Left-invertible operators in T(M,τ) are invertible.

The singular value function μ(t;Q) is either zero or at least one.

The result affirms a question posed in 2010 about the structure of these operators.

Abstract

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 $\lambda \in \mathbb{C}$. We prove 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 $Q=Q^2\in S(\mathcal{M},\tau)$ we have $\mu(t; Q)\in \{0\}\bigcup [1, +\infty)$ for all $t>0$. It gives the positive answer to the question posed by Daniyar Mushtari in 2010.