Essentially Invertible Measurable Operators Affiliated to a Semifinite von Neumann Algebra and Commutators

Abstract

Suppose that a von Neumann operator algebra ${\mathcal{M}}$ acts on a Hilbert space ${\mathcal{H}}$ and $\tau$ is a faithful normal semifinite trace on ${\mathcal{M}}$. If Hermitian operators $X,Y\in S({\mathcal{M}},\tau)$ are such that $-X\leq Y\leq X$ and $Y$ is $\tau$-essentially invertible then so is $X$. Let $0<p\leq 1$. If a $p$-hyponormal operator $A\in S({\mathcal{M}},\tau)$ is right $\tau$-essentially invertible then $A$ is $\tau$-essentially invertible. If a $p$-hyponormal operator $A\in{\mathcal{B}}({\mathcal{H}})$ is right invertible then $A$ is invertible in ${\mathcal{B}}({\mathcal{H}})$. If a hyponormal operator $A\in S({\mathcal{M}},\tau)$ has a right inverse in $S({\mathcal{M}},\tau)$ then $A$ is invertible in $S({\mathcal{M}},\tau)$. If $A,T\in{\mathcal{M}}$ and $\mu_{t}(A^{n})^{\frac{1}{n}}\to 0$ as $n\to\infty$ for every $t>0$ then $AT$ ($TA$) has no right (left) $\tau$-essential inverse in $S({\mathcal{M}},\tau)$. Suppose that ${\mathcal{H}}$ is separable and $\dim{\mathcal{H}}=\infty$. A right (left) essentially invertible operator $A\in{\mathcal{B}}({\mathcal{H}})$ is a commutator if and only if the right (left) essential inverse of $A$ is a commutator.