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QUESTION. Let $G$ be a group of bounded operators on $\ell^2$, satisfying $\sup_{x\in G} \lVert x\rVert <\infty$, whose elements are all of the form "identity+compact" (sometimes called "thin operators"). Is $G$ always similar, inside $B(\ell^2)$, to a group of unitary operators on $\ell^2$?

There are well-documented examples of bounded subgroups of ${\rm Inv}(B(\ell^2))$ that are not similar to unitary subgroups, but they don't seem to be contained in the space of thin operators. For example: I had a look at the examples in Pytlik–Swarc (Acta Math., 1986) but if I have understood the constructions correctly, they have the form $\pi(F)$ where $F$ is a free group and the representation $\pi: F \to {\rm Inv}(B(\ell^2(F)))$ satisfies $\pi(x)-\lambda(x)\in K(\ell^2(F))$ for all $x\in F$, so $\pi(x)-I \notin K(\ell^2(F))$ for all $x\in F \setminus \{e\}$. Nevertheless, perhaps one can modify these representations to get a negative answer to the question above?

In the other direction: if one is looking for a positive answer to the question, the closest result I could find is in work of Miglioli+Schlicht, where the following result is obtained as Corollary 4.13.

THEOREM (Miglioli–Schlicht). Let $G$ be a subgroup of ${\rm Inv}(B(\ell^2))$ which satisfies $\sup_{x\in G} {\lVert I-x\rVert}_{\sf HS} <\infty$, where ${\sf HS}$ denotes the Hilbert–Schmidt norm. Then there exists $s\in {\rm Inv}(B(\ell^2))$ such that $sGs^{-1}\subseteq {\mathcal U}(\ell^2)$ (and I think that the proof shows we can get ${\lVert I-s\Vert}_{\sf HS}<\infty$).

One might hope to reduce the original question to this result by a perturbation argument, but since the Miglioli–Schlicht result requires the group to be bounded in the norm of ${\mathbb C}I + {\rm HS}(\ell^2)$, I can't see how to make this approach work.

REMARK. I have tagged this post "fixed-point problems" because one can reformulate unitarizability questions for groups as a fixed point property. Namely: given a unital ${\rm C}^\ast$-algebra $D$ and a subgroup $G \subset {\rm Inv}(D)$, $G$ is similar inside $D$ to a unitary subgroup if and only if there exists $h\in {\rm Inv}(D)\cap D_+$ such that $xhx^\ast=h$ for all $x\in G$. The Miglioli–Schlicht result is proved by applying the Bruhat–Tits fixed point theorem to a cone of positive operators equipped with a suitable metric.

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    $\begingroup$ Yemon, I assume you want your $G$ to be bounded in norm? You mention it later, but not in the question itself. $\endgroup$
    – Mateusz Wasilewski
    Commented Sep 22, 2023 at 9:04
  • $\begingroup$ @MateuszWasilewski Oops - you are quite right. I will edit the question. $\endgroup$
    – Yemon Choi
    Commented Sep 22, 2023 at 15:52
  • $\begingroup$ What is the motivation, since the corresponding result is false for B(H) where H is finite-dimensional? For example take any nilpotent group; this cannot be symmetric to a group of unitaries...are you saying this is the obstruction, that I+K = I in Calkin algebra? Is there a reason why groups that are U+K for another U beside I must not be similar to unitary groups, or is it just that more is known about the thin case? $\endgroup$
    – Derek
    Commented Sep 22, 2023 at 22:06
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    $\begingroup$ @Derek: There is also the result of Day / Dixmier / Nakamura-Takeda that every bounded representation of an amenable group on Hilbert space is unitarizable. $\endgroup$
    – Yemon Choi
    Commented Sep 22, 2023 at 22:38
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    $\begingroup$ Does this problem have a reformulation in terms of derivations like Kadison's similarity problem? I am asking because Popa (S. Popa: The commutant modulo the set of compact operators of a von Neumann algebra) has shown that derivations with values in the compact operators are necessarily inner. $\endgroup$
    – MaoWao
    Commented Sep 25, 2023 at 9:53

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