Products of irreducible operators in factors

TL;DR

This paper proves that in separable factors, every non-zero operator can be expressed as a product of two irreducible operators, extending the understanding of operator factorizations within von Neumann algebras.

Contribution

It establishes that all operators in a separable factor are products of two irreducible operators, except for zero in certain type I factors, providing a multiplicative analogue to Radjavi's sum result.

Findings

Every operator in a separable factor is a product of two irreducible operators.

Zero operator cannot be expressed as such in type I_{2n+1} factors.

Extends the theory of operator decompositions in von Neumann algebras.

Abstract

Let $\mathcal M$ be a separable factor. An operator $T$ in $\mathcal{M}$ is said to be irreducible in $\mathcal{M}$ if the von Neumann algebra $W^*(T)$ generated by $T$ is an irreducible subfactor of $\mathcal{M}$, i.e., $W^*(T)'\cap\mathcal{M}=\mathbb{C}I$. In this paper, we show that every operator in a separable factor $\mathcal{M}$ is the product of two irreducible operators in $\mathcal{M}$, except the zero operator in factors of type $\mathrm{I}_{2n+1}$ for $n\geqslant 1$. This may be viewed as a multiplicative analogue of Radjavi's result which asserts that every operator on a separable Hilbert space is the sum of two irreducible operators.