Existence of a non-split quotient with a maximal lifting ideal

Determine whether there exists an infinite-dimensional Banach space E and a positive integer n for which a surjective operator T from E^n onto E lacks a bounded linear right inverse, while the associated lifting ideal Lift(T) is maximal; equivalently, require that the row operator [T S] from E^n ⊕ E onto E be right invertible for every operator S on E that does not belong to Lift(T).

Background

The paper studies maximal right ideals in the Banach algebra B(E) of bounded operators on a complex Banach space E. Every finitely generated right ideal can be represented as a lifting ideal Lift(T) = {TU : U ∈ B(E, E^n)} for an operator T : E^n → E. A lifting ideal contains all finite-rank operators precisely when T is surjective, and it is the whole algebra precisely when T has a bounded linear right inverse.

The unresolved issue concerns the possible existence of a surjective but non-right-invertible operator T whose lifting ideal is nevertheless maximal. By the paper’s exact obstruction result, this occurs precisely when every row operator [T S] is right invertible for operators S outside Lift(T). A positive answer would yield a non-fixed finitely generated maximal right ideal of B(E), whereas a negative answer for a specified Banach space would imply that all finitely generated maximal right ideals of B(E) are fixed. The authors explicitly note that they have no general argument converting the row-completion property into a right inverse for T.