0
$\begingroup$

Let $g$ be a non-identity element in a torsion-free amenable group, does there exist a finite-dimensional unitary representation $\pi$ with $\pi(g)\neq 1$?

(The word "finite-dimensional" was initially omitted: as mentioned in the comments the answer is a trivial "yes" then, by considering the left regular representation.)

Improve this question
$\endgroup$
18
  • 1
    $\begingroup$ Can I take $R(H)$ to be the set of all unitary representations (say on a Hilbert space of sufficiently small cardinality) for all $H$? $\endgroup$
    – Matthew Daws
    Commented Aug 31, 2019 at 14:55
  • 2
    $\begingroup$ So in the end, it seems this question can be reduced to "Let $G$ be a torsion-free amenable group, and let $g \neq e \in G$; then is there some unitary representation $\pi$ such that $\pi(g) \neq 1$?" $\endgroup$
    – user44191
    Commented Aug 31, 2019 at 19:03
  • 1
    $\begingroup$ So, for the new question: "yes" the left-regular representation has this property. $\endgroup$
    – Matthew Daws
    Commented Aug 31, 2019 at 19:25
  • 2
    $\begingroup$ I have rolled this back to Version 4, even though the answer then turns out to be trivial as pointed out by others in comments, because the change made after version 4 is so drastic that it invalidates the efforts and responses of the existing comments. I suggest that if the OP is specifically interested in fin-dim reps he should post this as a new question, but perhaps more thought is needed about what exactly he is looking for (i.e. not such a vague fishing expedition) $\endgroup$
    – Yemon Choi
    Commented Aug 31, 2019 at 19:29
  • 2
    $\begingroup$ @YCor I see your point, but the OP and others of similar mathematical background and training have a habit of asking ill-posed questions and subsequently moving the goalposts in seemingly arbitrary ways. (I and others in my area end up refereeing the consequences of this attitude.) It is not clear to me that "fin-dim reps" was a well-thought out condition that the OP had in mind, or just a knee-jerk response of the form "oh my question has a trivial answer, well let me just add in another condition" $\endgroup$
    – Yemon Choi
    Commented Sep 1, 2019 at 13:02

1 Answer 1

Reset to default
1
$\begingroup$

The buzzword to look for is "maximally almost periodic". This is part of the theory of Bohr Compactifications. Start with a (locally compact) group $G$. I do not believe amenability and (especially) torsion freeness has much if any baring on what following.

The Bohr Compactification is the maximal compact group $K$ for which there is a dense range homomorphism $\theta:G\rightarrow K$. Any finite-dimensional unitary rep of $G$ factors through $\theta$ (by the representation theory of compact groups). It follows that finite-dimensional unitary reps separate the points of $G$ if and only if $\theta$ is injective. By definition, this means that $G$ is "maximally almost periodic" (MAP). One can also construct $K$ using almost periodic functions, and here $G$ is MAP if and only if the almost periodic functions separate the points of $G$.

The Freudenthai-Weil theorem says that for connected locally compact groups, $G$ is MAP if and only if $G=\mathbb R^n \times L$ for some compact $L$. Chapter 16 of Dixmier's book on $C^\ast$-algebras is a good source for all of this.

Searching around MathOverFlow will find examples of groups which are not MAP.

Improve this answer
$\endgroup$
1
  • 4
    $\begingroup$ The question is asked for discrete groups. In this case, one can still answer the question in the negative. For a finitely generated group, to be MAP is equivalent to being residually finite (and there are torsion-free solvable f.g. groups for which this fails). $\endgroup$
    – YCor
    Commented Sep 1, 2019 at 12:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .