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 see the ISP ?

  • $\begingroup$ Is $\mathcal{S}(T)$ the operator algebra generated by $T$? $\endgroup$ – Ulrich Pennig Aug 4 '13 at 12:57
  • $\begingroup$ Yes, $\mathcal{S}(T)$ is the operator algebra (of category $\mathcal{S}$) generated by $T \in B(H)$. Just a precision, the $C^{∗}$-algebras and von Neumann algebras are here 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. $\endgroup$ – Sebastien Palcoux Aug 4 '13 at 14:00

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.

  • $\begingroup$ Thank you Leonel ! This map is continuous for the main topologies of operators algebras (norm-topology, weak-topology, strong-topology...), and this argument runs also without a $\star$-structure (on the algebra), so that it shows that no category of operator algebras see the ISP. Is it right ? $\endgroup$ – Sebastien Palcoux Sep 26 '13 at 13:09
  • $\begingroup$ @SébastienPalcoux: I guess so. Though I'm not sure that the question cannot be made more interesting by tweaking it a little. $\endgroup$ – Leonel Robert Sep 26 '13 at 20:11
  • $\begingroup$ A generic way for tweaking a question, is to improve it by excluding the counter-examples. Here we can improve the definition of "see the ISP" by : $\forall T, T' \in B(H)$ with $\mathcal{S}(T) \simeq \mathcal{S}(T') $ and $T' \ne v_{1} T v^{*}_{1} + v_{2} T v^{*}_{2}$ (with $v_{1}$, $v_{2}$ as in your answer), then : "$T$ is an ISP counter-example" $\Leftrightarrow$ "$T'$ is an ISP counter-example". Is this what you thought? Else what do you suggest ? $\endgroup$ – Sebastien Palcoux Sep 27 '13 at 8:09

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