The transitive algebra problem

Corollary 1.2 of the principal manuscript: the commutant of the constructed nonzero quasinilpotent operator $S$ is a proper, strongly closed, unital, transitive complex algebra, so it is not strongly dense in $B(H)$. This answers the question negatively on every infinite-dimensional separable Hilbert space. It is an immediate consequence of the hyperinvariant counterexample, since the commutant of an operator without hyperinvariant subspaces is transitive. The algebra is a commutant, so nothing is said about singly generated transitive algebras, which is the invariant subspace problem.

Corollary 1.2 of the principal manuscript: the commutant of the constructed nonzero quasinilpotent operator $S$ is a proper, strongly closed, unital, transitive complex algebra, so it is not strongly dense in $B(H)$. This answers the question negatively on every infinite-dimensional separable Hilbert space. It is an immediate consequence of the hyperinvariant counterexample, since the commutant of an operator without hyperinvariant subspaces is transitive. The algebra is a commutant, so nothing is said about singly generated transitive algebras, which is the invariant subspace problem.