MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

Very nice question! My feeling is that the answer is no. I only have a rough sketch of proof. By Chepoi, conversely a median graph is the 1-skeleton of a CAT(0) cube complex. So your question is equivalent to whether there is a proper cocompact action of this group on a CAT(0) cube complex (necessarily finite-dimensional and locally finite). First subdivide your cube complex to ensure that no element fixes a cube without fixing its boundary. Then consider the action of $\mathbf{Z}^2$ and let $M$ be the set of elements that have minimal displacement for all elements of $\mathbf{Z}^2$: the CAT(0) flat torus theorem (Bridson-Haefliger, Theorem 7.1) implies that $M$ splits as a product (in the CAT(0)-metric) $Y\times\mathbf{R}^2$. On the other hand, it follows from the assumption that $M$ is a subcomplex. Now some further efforts should be made to ensure that the product decomposition of $M$ is compatible with the combinatorial structure (this has maybe already be done somewhere). If done, then we can expect that the action of the whole group factors "modulo $Y$" and thus we get an action on a 2-dimensional CAT(0) square complex homeomorphic to $\mathbf{R}^2$ and a contradiction as expected.

Edit: I misread the question and answered with "median graph". It's unclear if this approach can work in the case of median spaces, nor whether the answer is the same.

share|cite|improve this answer
But a median space need not be a median graph. So conceivably there could be a median space action which does not contain, as an invariant subspace, a CAT(0) cube complex. – Lee Mosher Dec 10 '12 at 15:37
@Lee thanks, I would have read the question 10 times seeing each time "median graph"! – YCor Dec 10 '12 at 16:28
That is also one of my own downfalls with MO questions :-/ – Lee Mosher Dec 10 '12 at 16:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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