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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 ?

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  • $\begingroup$ If $T \in B(H)$ is an ISP counter-example then $W^{*}(T) = B(H)$, but what's the list of all the properties of $C^{*}(T)$ ? Conversely, if $C^{*}(T)$ verifies this list, is $T$ an ISP counter-example ? (idem with $W(T) = \overline{\langle T \rangle }^{wot} $). $\endgroup$
    – Sebastien Palcoux
    Commented Jul 31, 2013 at 11:26
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    $\begingroup$ It seems that you are looking for properties of C^*(T) which are, in particular, invariant under replacement of T by any S^{-1}TS. Perhaps you could obtain a more focused question by adding some specific properties you think might be relevant? (E.g. commutativity is not such a property, and I suspect nuclearity is not.) $\endgroup$
    – Yemon Choi
    Commented Aug 2, 2013 at 20:56
  • $\begingroup$ @YemonChoi : intuitively, the question is to know if the $C^{*}$-algebras see the 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 list and the ISP. It's the purpose of my post about banded operators, if they all check the ISP, we could then add non-banded to the list, and there is still not a candidate checking this completed list. $\endgroup$
    – Sebastien Palcoux
    Commented Aug 2, 2013 at 21:35
  • $\begingroup$ Now, perhaps, the $C^{*}$-algebras do not see the ISP, and so it would be more relevant to investigate $C(T)$ or $W(T)$ which is known to be reductive : it's an important property, because RAD implies ISP (see Mike answer below). $\endgroup$
    – Sebastien Palcoux
    Commented Aug 2, 2013 at 21:41
  • $\begingroup$ I think your last comment is more likely to be the correct reading. All these vague and hopeful questions about the STAR-algebras generated by T seem far less convincing than approaches via non-self-adjoint op alg theory, for the reasons hinted at in my previous comment $\endgroup$
    – Yemon Choi
    Commented Aug 2, 2013 at 21:44

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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.

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  • $\begingroup$ Thank you @MikeJury. A counter-example $T \in B(H) $ of the ISP is of course non-normal. If $R\in B(H)$ is normal, there is a theorem due to Sarason : $\overline{\mathbb{C}[R]}^{wot}$ is a $∗$-algebra if and only if $R$ is reductive, i.e. every closed invariant subspace $K \subset H$ of $R$, is a reducing subspace. If I'm not mistaken, the RAP generalizes this theorem to every operator. $\endgroup$
    – Sebastien Palcoux
    Commented Jul 29, 2013 at 15:15
  • $\begingroup$ In particular, if $T$ is a counter-example of the ISP, then $\mathcal{A} = \overline{\mathbb{C}[T]}^{wot}$ is reductive because its invariant subspace are $\{0 \}$ and $H$ (obviously reducing the algebra). So if the RAP is true, $\mathcal{A}$ is self-adjoint, in particular, $T^{*} \in \mathcal{A}$, but $\mathcal{A} \subset \{ T\}'$ and $T$ is non-normal, so contradiction, and ISP is true. $\endgroup$
    – Sebastien Palcoux
    Commented Jul 29, 2013 at 15:21
  • $\begingroup$ So if $T$ is a counter-example of the ISP, $\overline{\mathbb{C}[T]}^{wot}$ is a reductive and non-selfadjoint algebra. For the converse, if $\mathcal{A} = \overline{\mathbb{C}[T]}^{wot}$ has no non-trivial invariant subspace, then it's reductive and also non-selfadjoint algebra (because if $\mathcal{A}$ selfadjoint, then it's irreducible and so $\mathcal{A} = B(H)$, contradiction with the commutativity of $\mathcal{A}$). So the question is : does $\mathcal{A} = \overline{\mathbb{C}[T]}^{wot}$ has no non-trivial invariant subspace, implies $T$ a counter-example of the ISP ? $\endgroup$
    – Sebastien Palcoux
    Commented Jul 29, 2013 at 15:47
  • $\begingroup$ Radjavi and Rosenthal - Invariant subspaces. $\endgroup$
    – LSpice
    Commented May 22, 2020 at 18:52

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