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


**[Invariant subspace problem](http://en.wikipedia.org/wiki/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 ? 



Is there an operator algebraic reformulation of the invariant subspace problem?
To the weakest form of your question ("Is there a class of operator algebras relevant to the ISP?"), there is the "reductive algebra problem," which is not quite of the form you're asking for but is, I think, in the same spirit. An algebra of bounded operators on Hilbert space is called *reductive* if it is WOT-closed, contains 1, and every invariant subspace for the algebra is reducing for the algebra. The *reductive algebra problem* asks if every reductive algebra is self-adjoint. It is known that a "yes" answer to RAP implies a "yes" answer to ISP. (In fact, if I am not mistaken, a "yes" to RAP actually implies a "yes" to the hyperinvariant subspace problem.) I am not sure about the status of (ISP implies RAP). If you haven't already, you might want to look at the book "Invariant Subspaces" by Radjavi and Rosenthal.


