# Group action on quasi-isometric geodesic metric space [closed]

If a group $G$ acts on a geodesic metric space $X$, then does $G$ act on a geodesic metric space $Y$ which is quasi-isometric to $X$?

• Why should it? There is a notion of a quasi-action, see arxiv.org/pdf/math/0010136.pdf Commented Aug 22, 2018 at 3:22
• This question is senseless if you don't assume anything on the action of $G$ on $X$ (so the assumption is empty), and also the conclusion is empty, since $Y$ is quasi-isometric to $X$.
– YCor
Commented Aug 22, 2018 at 15:44
• @YCor is emphasizing the fact that, without specifying further assumptions, any group acts on any space, by taking the trivial action. You seem to have an implicit assumption which I am guessing is that your actions are by isometries and with no fixed points or something alike. Maybe further, you want a given qi from $X$ to $Y$ to intertwine your actions. Whatever is your assumption, you maybe want to consider the following space $Y$, which is qi to $X=\mathbb{R}$ but has no non-trivial isometries: take $\mathbb{R}$ and attach a $[0,1]$ segment at each prime. Commented Aug 23, 2018 at 6:50
• @UriBader even assuming that actions are geometric, taking $Y=X$ and the same action yields a tautological statement, so I don't know what you have in mind.
– YCor
Commented Aug 23, 2018 at 16:42
• I had the meaning "If $G$ acts on $X$, does there exist $Y$ that is QI to $X$..." which is tautological (although it could make sense with some more adjectives).From Anthony's complement to his answer I at least understand that some of you have the meaning "if $G$ acts on $X$ and $Y$ is QI to $X$, does...". Of course some more imagination is still necessary to make it a relevant question.
– YCor
Commented Aug 23, 2018 at 18:04

Edit. I understood OP's question as follows: If $G$ acts "nicely" by isometries on a geodesic metric space $X$, and if $Y$ is a geodesic metric space quasi-isometric to $X$, does it imply that $G$ acts non-trivially on $Y$?

Some general results:

• Any hyperbolic group is quasi-isometric to a CAT(0) cube complex, so if you take any hyperbolic group $G$ satisfying Kazhdan's property (T), for instance a cocompact lattice in the quaternionic hyperbolic space or a random group (taking a good density), then it acts geometrically on a space which is quasi-isometric to a CAT(0) cube complex (namely itself) but any isometric action of $G$ on a CAT(0) cube complex has a fixed point.
• Any acylindrically hyperbolic group admits a non-trivial (more precisely, acylindrical) action on a quasi-tree. However, plenty of acylindrically hyperbolic groups satisfy Serre's property (FA), so that they cannot act on a tree without fixing a point.

For a more elementary example, consider the triangle group $$T= \langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(ac)^3=(bc)^3=1 \rangle.$$ It is the symmetry group of the tesselation of the plane by equilateral triangles. Such a triangle complex is quasi-isometric to the square complex $X$ obtained by tiling the plane with squares. However, as noticed in one of my previous answers, $\mathrm{Isom}(X)= (D_\infty \times D_\infty) \rtimes \mathbb{Z}_2$ does not contain ordre-three elements, so that any action of $T$ on $X$ must fix a point.

• Since the OP's question is senseless (as meaningful as "if there exists a group $G$, does there exist a group $H$?"), it would be helpful to write down which question you claim to answer.
– YCor
Commented Aug 23, 2018 at 16:40
• @YCor: I added my interpretation of the question. Commented Aug 23, 2018 at 16:49
• What do you assume of the action on $X$? you always have the trivial action.
– YCor
Commented Aug 23, 2018 at 16:55
• Right. The action on $X$ has to be "sufficiently rich", otherwise the question (at least as I interpretated it) is not interesting. Commented Aug 23, 2018 at 18:41