Does there exist a geometric action of a non-finitely generated group on a non-proper geodesic metric space?

Question

I am studying Svarc-Milnor lemma :

If a group $G$ acts geometrically on a proper geodesic metric space $(X,d)$, then $G$ is finitely generated and any of its Cayley graphs (with some finite generating subset) is quasi-isometric to $(X,d)$.

I would like to check with examples the necessity of each hypothesis in the lemma. In particular, I'm having trouble finding a geometric action of a non-finitely generated group on a non-proper geodesic metric space.

I was thinking about the space $l_2$ (of sequences) which is not proper, but I don't find any geometric action with it (for example with reflexions, it is not cocompact).

Note : geometrically means that the action is by isometry, properly discontinuous and cocompact (the set of orbits is compact). The properly discontinuity means here that for every compact $K \subset X$, there exists only a finite number of $g \in G$ such that $ g \cdot K$ intersects $K$.

Thank you.

What examples have you checked? What research have you done into your question? — amWhy, Jan 26 '23 at 18:33
You should spell out the definition of a geometric action, otherwise people will misunderstand your question. — Moishe Kohan, Jan 26 '23 at 19:25
And your definition of proper discontinuity needs a correction. What book are you reading? — Moishe Kohan, Jan 26 '23 at 19:49
The answer to the question is negative, but a proof is a bit long. I will write a sketch once the question is reopened. Moreover, one can even prove that a geodesic metric space $X$ is proper, provided it admits a geometric group action. — Moishe Kohan, Jan 27 '23 at 16:04
@mr_e_man: The linked example is irrelevant since the action there is not cocompact. — Moishe Kohan, Jan 27 '23 at 16:05
@Moishe Kohan Thank you! Is there some paper that explains this? — johnny658, Jan 27 '23 at 18:17

Moishe Kohan · Accepted Answer · 2023-01-28T03:58:09.470

1

Yes, one can drop the properness (of $X$) assumption in the M-S lemma. You can find a proof (a bit sketchy but correct) in

Roe, John, Lectures on coarse geometry, University Lecture Series 31. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3332-4/pbk). vii, 175 p. (2003). ZBL1042.53027.

Specifically, see pages 8-10.

edited Jan 28 '23 at 03:58

answered Jan 28 '23 at 01:51

Moishe Kohan

111,854

Thanks, that's what I was looking for. – johnny658 Jan 28 '23 at 07:44

Does there exist a geometric action of a non-finitely generated group on a non-proper geodesic metric space?

1 Answers1