Does this group act geometrically on a Median space?

Let $G$ be the semidirect product of $\mathbb{Z}^2$ with $\mathbb{Z}/6$ where $\mathbb{Z}/6$ acts by the order 6 element of $SL_2(\mathbb{Z})$. We can think of this group as the group of order preserving isometries of the tesselation of $\mathbb{R^2}$ with regular triangles.

Does this group acts properly, isometrically and cocompactly on a median space??

Let for two points in a metric space $[x,y]=\{z|d(x,z)+d(z,y)=d(x,y)\}$. If $X$ is a geodesic metric space than this is just the set of all points lying on some geodesic from $x$ to $y$. $X$ is called a median space if for every triple of points $x,y,z$ we have that $[x,y]\cap[x,z]\cap[y,z]$ consists of exactly one point - the median of $x,y,z$. Examples for median spaces are trees and $\mathbb{R}^n$ with the $l^1$- metric.

The motivation is that the one skeleton of a CAT(0) cube complex is a median graph. If a group acts geometrically on this CAT(0)-cube complex it also acts that way on that graph. For example this group acts properly and isometrically on $\mathbb{R}^3$. This gives a proper and isometric action on a median space, but this action is not cocompact. So I was wondering whether there is a better action. The problem seems to be that the automorphism of $\mathbb{Z}^2$ does not extend to a cube-complex automorphism of $\mathbb{R}^2$, but I could not make this precise.

• Perhaps you could remind us of the definition of a median space? – HJRW Dec 10 '12 at 13:07
• @Joseph: you just gave the definition of a geodesic median space . A median space (as given by Henrik) does not require geodesics. He just mentions what is $[x,y]$ is in case the metric space is geodesic. – YCor Dec 10 '12 at 16:34
• @Yves: I've deleted my comment now that Henrik has defined "median space." – Joseph O'Rourke Dec 10 '12 at 22:29

It is known that the group you mention does not act geometrically on a CAT(0) cube complex. See for instance this answer for a possible argument, based on Lemma 16.12 in Wise's monograph The structure of groups with a quasiconvex hierarchy. I think you can adapt the proof of this lemma in order to prove the following statement:

Proposition: Let $$G$$ be a group acting properly and cocompactly on a median metric space $$(M,d)$$ of finite rank. Assume that $$G$$ contains a finite-index subgroup $$H \simeq \mathbb{Z}^n$$, $$n\geq 2$$. Then $$G$$ acts properly and cocompactly on $$(\mathbb{R}^n, \ell^1)$$.

Sketch of proof. As shown by Bowditch in Some properties of median metric spaces, there exists a CAT(0) metric $$\sigma$$ on $$M$$, and, if I understand the construction correctly, this metric satisfies the following properties: (1) any isometry of $$(M,d)$$ induces an isometry of $$(M,\sigma)$$; (2) the metrics $$d$$ and $$\sigma$$ are biLipschitz equivalent; (3) halfspaces of $$M$$ are $$\sigma$$-convex.

As a consequence, the flat torus theorem can be applied, and we find a $$G$$-invariant and $$\sigma$$-convex subspace $$\Sigma \subset M$$ which is $$\sigma$$-isometric to $$\mathbb{R}^n$$.

Now, we consider the structure of measured wallspace of $$\Sigma$$ induced by the walls of $$M$$. By $$\sigma$$-convexity of $$\Sigma$$ and of the walls, we must have $$m$$ families of parallel hyperplanes $$\mathbb{R}^{n-1}$$ in $$\Sigma$$. (Here, we ignore the collections of hyperplanes which lie in the neighborhood of a single hyperplane. As a consequence, $$m \leq n$$ since otherwise it would be possible to embed coarsely $$\mathbb{R}^{n+1}$$ into $$\Sigma \simeq \mathbb{R}^n$$.)

Let $$F$$ denote the median space associated to the previous wallspace. Then $$F$$ decomposes as the $$\ell^1$$-product of $$m$$ (discrete or continuous) unbounded lines. Up to replacing discrete lines with continuous lines, we may suppose that $$F$$ is $$(\mathbb{R}^m,\ell^1)$$.

Notice that, because $$G$$ acts properly on $$\Sigma$$, necessarily it also acts properly on $$F$$. So we must have $$m \geq n$$. But we already know that $$m \leq n$$, so $$m=n$$.

So far, we have proved that $$G$$ acts properly on $$(\mathbb{R}^n,\ell^1)$$. As $$G$$ is virtually $$\mathbb{Z}^n$$, we conclude that $$G$$ acts geometrically on $$(\mathbb{R}^n,\ell^1)$$. $$\square$$

In your specific example, the question is now: does $$\mathbb{Z}^2 \rtimes \mathbb{Z}_6$$ act geometrically on $$(\mathbb{R}^2,\ell^1)$$? The same argument as the one followed here works. A presentation of the group is $$T=\langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(bc)^3=(ac)^3=1 \rangle.$$ Notice that $$\mathrm{Isom}(\mathbb{R}^2, \ell^1)= (\mathbb{R}\rtimes \mathbb{Z}_2)^2 \rtimes \mathbb{Z}_2$$ does not contain elements of order three, so, for any action $$T \curvearrowright (\mathbb{R}^2, \ell^1)$$ by isometries, the elements $$ab$$, $$bc$$ and $$ac$$ must be trivial. Consequently, such an action must factorise through the quotient $$T \twoheadrightarrow \mathbb{Z}_2$$ sending all the generators to $$1$$. A fortiori, the action cannot be geometric (and even proper).

Remark: The argument above does not completely answer the question as the median metric space is supposed to have finite rank. I do not know what happens if the rank is infinite, but it seems reasonable to think that the same conclusion holds.