This post is an appendix of this one.

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?

**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 seethe ISP ?

separable(the categories $C^{∗}$ and $W^{∗}$). If we can prove that $C^{∗}(T)$ is a Cuntz algebra (with $T\in B(H)$ an ISP counter-example), the result should follow. – Sebastien Palcoux Aug 4 '13 at 14:00