Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.

Invariant subspace problem: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant subspace?

Remark: This problem is known for the Banach spaces in general, but is still open for an Hilbert space.

The ISP is an operator theoretic problem, and I ask here about an operator algebraic reformulation.

A counter-example $T \in B(H)$ of the ISP is a fortiori an irreducible operator, i.e. $W^{*}(T) = B(H)$. But of course the converse is false : there are many irreducible operators which are not counter-examples of the ISP (for example the unilateral shift).

Conclusion : the von Neumann algebras don't see the ISP, because the property to be a counter-exemple of the ISP can't be encode in $W^{*}(T)$.

Is there a property $P$ of $C^{*}$-algebras verifying : $T$ is a counter-example of the ISP iff $C^{*}(T)$ is $P$ ?

Perhaps the $C^{*}$-algebras are also not relevant for the ISP, I don't know...
In this case, is there a class of operator $*$-algebras which is relevant for the ISP ?

Perhaps the operator $*$-algebras are also not relevant for the ISP...
Is there a class of operator algebras (non necessarily self-adjoint), which is relevant for the ISP ?

Is there an operator algebraic reformulation of the invariant subspace problem ?

To be more precise, is there a class $\mathcal{C}$ of operator algebras and a property $\mathcal{P}$, such that the algebra $\mathcal{C}(T)$ of class $\mathcal{C}$ generated by $T \in B(H)$, is $\mathcal{P}$ iff $T$ is a counter-example of the ISP ?