Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Here's a research problem, which I think interesting. Suppose that $t$ is an invertible element in the Calkin algebra $\mathcal{Q} = \mathcal{B}(\ell_2)/\mathcal{K}(\ell_2)$ which satisfies $\sup_{n \in \mathbb{Z}} \|t^n\|<+\infty$. Does it follow $t$ is similar to a unitary element, i.e., is there an invertible element $s \in \mathcal{Q}$ such that $s t s^{-1}$ is unitary?

Béla Szőkefalvi-Nagy has proved that it is the case when one replaces $\mathcal{Q}$ with $\mathcal{B}(\ell_2)$ (or any other von Neumann algebra, e.g., $\mathcal{Q}^{\ast\ast}$). I'm asking this question, because a counterexample (which I believe exists) would provide an amenable operator algebra which is not isomorphic to a $\mathrm{C}^\ast$-algebra---The existence of such an example is an open problem.

I will explain why a counterexample provides such an example. Let $\pi\colon \mathcal{B}(\ell_2)\to \mathcal{Q}$ be the quotient map, $G$ an abelian group (which is $\mathbb{Z}$ in the above question) and $u\colon G\to \mathcal{Q}$ a uniformly bounded homomorphism. Then, the operator algebra $\mathcal{A} := \pi^{-1}( \overline{\mathrm{span}}\ u(G) )$ is amenable. If $\mathcal{A}$ is isomorphic to a $\mathrm{C}^\ast$-algebra, then by the solution of the similarity problem for amenable $\mathrm{C}^\ast$-algebras, there is $S \in \mathcal{B}(\ell_2)$ such that $S \mathcal{A} S^{-1}$ is a $\mathrm{C}^\ast$-subalgebra. Thus $s \pi(\mathcal{A}) s^{-1}$ (where $s=\pi(S)$) is an abelian $\mathrm{C}^\ast$-subalgebra of $\mathcal{Q}$ and hence it consists of normal elements. Since $u$ is uniformly bounded, $s u(\cdot)s^{-1}$ is a unitary homomorphism. (The converse is also true: if $u$ is similar to a unitary homomorphism, then $\mathcal{A}$ is isomorphic to a $\mathrm{C}^\ast$-algebra.)

The obvious thing one should try is to see whether $H^1_b(G,\mathcal{Q}(\ell_2G))\neq0$. For a starter, I looked at $H^1_b(G,\ell_\infty(G)/c_0(G))$, but it was zero for every countable exact group $G$. (Whether it is zero for every group $G$ is unclear.)

share|improve this question
1  
Great question! Is there any example of a $C^*$-algebra, where "uniformly bounded" does not imply "unitarizable" for $\mathbb Z$-actions? –  Andreas Thom May 22 '12 at 8:00
    
I agree, this is a nice question because either answer would be interesting. (My money is on every power bounded element of $Q$ being similar to a unitary.) –  Nik Weaver Jun 4 '12 at 0:07
    
(every invertible power bounded element) –  Nik Weaver Jun 4 '12 at 13:51
1  
Since this question has been bumped by edits: did you ever make any progress on this question? –  Yemon Choi Jun 30 '13 at 21:25
1  
Um, I posted the problem because I gave it up, and I didn't make any efforts. I haven't heard any progress either. –  Narutaka OZAWA Jul 11 '13 at 3:24
show 1 more comment

1 Answer 1

up vote 4 down vote accepted

Posting this purely so that Ozawa's interesting question does not stay marked as unanswered, and hence lead to unnecessary effort on the part of someone reading. (Note: could someone reading this please flag the answer for moderator attention, to make it CW.)


It turns out that every uniformly bounded representation of a discrete, countable amenable group inside the Calkin is unitarizable. This follows from techniques introduced in work of Farah and Hart, as expounded in e.g. arXiv: 1112.3898, which extend techniques of G. K. Pedersen for dealing with certain corona algebras. Details of the argument in this particular case are in Theorem 7 of the recent preprint arXiv 1309.2145 by Choi, Farah, and Ozawa.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.