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.)

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    $\begingroup$ Great question! Is there any example of a $C^*$-algebra, where "uniformly bounded" does not imply "unitarizable" for $\mathbb Z$-actions? $\endgroup$ May 22, 2012 at 8:00
  • $\begingroup$ 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.) $\endgroup$
    – Nik Weaver
    Jun 4, 2012 at 0:07
  • $\begingroup$ (every invertible power bounded element) $\endgroup$
    – Nik Weaver
    Jun 4, 2012 at 13:51
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    $\begingroup$ Since this question has been bumped by edits: did you ever make any progress on this question? $\endgroup$
    – Yemon Choi
    Jun 30, 2013 at 21:25
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    $\begingroup$ Um, I posted the problem because I gave it up, and I didn't make any efforts. I haven't heard any progress either. $\endgroup$ Jul 11, 2013 at 3:24

1 Answer 1


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.


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