The algebra of thin measurable operators is directly finite

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 ${}^{\ast }$-algebra of all $\tau$-measurable operators. Let $S_0(\mathcal{M},\tau )$ be the ${}^{\ast }$-algebra of all $\tau$-compact operators and $T(\mathcal{M},\tau )=S_0(\mathcal{M},\tau )+\mathbb{C}I$ be the ${}^{\ast }$-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.