This post is an appendix of this [one](https://mathoverflow.net/questions/138070/is-there-an-operator-algebraic-reformulation-of-the-invariant-subspace-problem). 


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? 


**Hypothesis** : The ISP admits a negative answer, i.e., there are ISP counter-examples. 


**Definition** : A category $\mathcal{S}$ of operator algebras **see the ISP** if $ \forall T, T' \in B(H)$ with $\mathcal{S}(T) \simeq \mathcal{S}(T')$:
$$
T \text{ is an ISP counter-example} \Leftrightarrow T' \text{ is an ISP counter-example }
$$


**Proposition**: The category $W^{\*}$ of von Neumann algebras, **doesn't see** the ISP.  

**proof**: Under the previous *hypothesis*, let $T \in B(H)$ be an ISP counter-example. Then $T$ is irreducible, i.e., $W^{\*}(T) = B(H)$. But there are many irreducible operators checking the ISP, for example, the **unilateral shift** $S$. So $W^{\*}(T) \simeq W^{\*}(S)$, $S$ checks the ISP and $T$ not. $\square$


This post asks about an equivalent result for the category of $C^{\*}$-algebras : 



> 
> Is there a proof that the category of $C^{\*}$-algebras *doesn't see* the
>  ISP ?
> 
> 
> 



Is there a proof that the $C^{*}$-algebras don't see the invariant subspace problem?
C\*-algebras don't see the ISP. The operators $T\in B(H)$ and $T\oplus T\in B(H\oplus H)$
generate isomorphic C\*-algebras, but the latter clearly has non-trivial invariant subspaces.
To have both operators in the same Hilbert space, pick isometries $v\_1,v\_2\in B(H)$ with orthogonal ranges that add up to $H$. Then
$$
T\mapsto v\_1Tv\_1^\*+v\_2Tv\_2^\*
$$ 
is an injective \*-endomorphism of $B(H)$ that maps $T$ to an operator with non-trivial invariant subspaces.


