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 ?

seethe ISP. If $T$ is an ISP counter-example, then $T$ is irreducible, noncompact-commuting, nonnormal and with spectrum strictly continuous. We could deduce some properties of $C^{*}(T)$. Unfortunately, this list here is not sufficient, because there are weight shifts checking this listandthe ISP. It's the purpose of my post about banded operators, if they all check the ISP, we could then addnon-bandedto the list, and there is still not a candidate checking this completed list. $\endgroup$ – Sebastien Palcoux Aug 2 '13 at 21:35seethe ISP, and so it would be more relevant to investigate $C(T)$ or $W(T)$ which is known to bereductive: it's an important property, because RAD implies ISP (see Mike answer below). $\endgroup$ – Sebastien Palcoux Aug 2 '13 at 21:412more comments