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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.

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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. –  Yves Cornulier 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
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1 Answer

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.

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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"! –  Yves Cornulier 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
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