10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Perhaps you could remind us of the definition of a median space? $\endgroup$ – HJRW Dec 10 '12 at 13:07
  • $\begingroup$ @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. $\endgroup$ – YCor Dec 10 '12 at 16:34
  • $\begingroup$ @Yves: I've deleted my comment now that Henrik has defined "median space." $\endgroup$ – Joseph O'Rourke Dec 10 '12 at 22:29
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.