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Let $G$ be a locally compact (second countable) group and let $$ G_0 = \cap \{ \ker\pi : \pi \text{ is a continuous finite-dimensional unitary representation of } G \}. $$ This is the kernel of the Bohr compactification $G \to bG$, and in particular a closed normal subgroup. Suppose that $G/G_0$ is compact (so that $bG = G/G_0$).

Q: Is it true that $G_0$ is minimally almost periodic, that is, admits no non-trivial finite-dimensional unitary representations?

If $G/G_0$ is finite, then the answer is positive, because in this case a finite-dimensional representation of $G_0$ induces a finite-dimensional representation of $G$.

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  • $\begingroup$ Do you know the answer for say Nilpotent Lie groups? $\endgroup$
    – Asaf
    Commented Dec 14, 2013 at 16:44
  • $\begingroup$ Asaf: I have convinced myself that the answer to Q2 is positive for connected nilpotent Lie groups. Indeed, $G_0\leq [G,G]$, so $G/[G,G]$ is compact. This seems to imply that $G$ is compact via the Baker-Campbell-Hausdorff formula, so one can take $H$ trivial. $\endgroup$
    – pavel
    Commented Dec 14, 2013 at 18:34
  • $\begingroup$ How is that $G/[G,G]$ is compact? this is the abelinization of $G$, hence might be an abelian non-compact grp. I thought about this class of groups because they are amenable, non-compact and it is easy to write down their representations. For abelian groups, $G_{0}={e}$, as you have a proper embedding into the Bohr compactification. Notice that say for semi-simple groups (say $SL{2}$), $G_{0}={e}$ by definition. but those groups are non-amenable (by the Tits' alternative). $\endgroup$
    – Asaf
    Commented Dec 14, 2013 at 19:11
  • $\begingroup$ By the assumption the quotient $G/G_0$ is compact, so the quotient by any larger subgroup is also compact. $\endgroup$
    – pavel
    Commented Dec 14, 2013 at 19:18
  • $\begingroup$ I still don't get it. What is your assumption? Even for $\mathbb{Z}$, $Bohr(G)$ is not a surjective image of $G$ (but indeed, $G$ embeds as a dense subgrp). Can you do the computation for say the Heisenberg group (take even the integral points inside if you insist to be countable). $\endgroup$
    – Asaf
    Commented Dec 14, 2013 at 19:45

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No, a counterexample—courtesy of Klaus Schmidt—is the semidirect product $SO(2) \ltimes \mathbb{R}^2$ taken with respect to the defining action of $SO(2)$ on $\mathbb{R}^2$. The von Neumann kernel of this group is $\mathbb{R}^2$, which follows from the results in “The structure of homomorphisms of connected locally compact groups into compact groups” by A. I. Shtern, Izv. Ross. Akad. Nauk Ser. Mat. 75.6 (2011), pp. 195–222.

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