A median space is a metric space $X$ for which for any three points $x, y , z \in X $ there exists a unique point $m$ such that $d(x,m)+ d(m, y)= d(x , y ), d(x,m)+ d(m, z)= d(x , z ), d(y,m)+ d(m, z)= d(y , z )$. $CAT(0)$-polygonal complexes are simply connected collections of glued (on their faces) polyhedra of varying dimension such that each link is flag. Does there exist a metric on $CAT(0)$-polygonal complexes such that they are median spaces (in an analogous fashion to $CAT(0)$-cube complexes)?

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    $\begingroup$ The question is unclear if you don't specify the metric on the polygons. Beware that a CAT(0) cube complex of dimension $\ge 2$, with Euclidean cubes, is not a median space; when endowed with the $\ell^1$-metric on cubes, it is median but not CAT(0)! $\endgroup$ – YCor Oct 8 '18 at 13:49
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    $\begingroup$ Another remark is that real trees are the only examples of geodesic metric spaces which are both median and CAT(0). It follows from the fact that CAT(0) spaces are uniquely geodesic. $\endgroup$ – AGenevois Oct 8 '18 at 16:25
  • $\begingroup$ @user129863: Your edited question is not clear either. What are the restrictions you impose on your new metric? Has it to induce the same topology? Has it to be Lipschitz equivalent or quasi-isometric to the previous metric? Because if you impose no restriction, just transfer a median metric through a bijection with a median space having the same cardinality. $\endgroup$ – AGenevois Oct 9 '18 at 6:50

The question is vague. The median metric defined on CAT(0) cube complexes satisfy two fundamental properties:

  1. For finite-dimensional cube complexes, it is bi-Lipschitz equivalent to the CAT(0) metric.
  2. Any automorphism of a cube complex preserves both the CAT(0) and the median metrics.

What I claim is that there exist finite-dimensional CAT(0) polyhedral complexes which do not admit a median metric satisfying the above two points. Indeed, otherwise any group acting geometrically on such a polyhedral complex would act geometrically on a median space. By combining the following observation with the fact that any group acting on a median space with an unbounded orbit does not satisfy Kazhdan's property (T), we conclude that it is far from being true.

Theorem: There exists an infinite group satisfying Kazhdan's property (T) which acts geometrically on a finite-dimensional CAT(0) polyhedral complex.

My guess is that there exist plenty of such complexes. For one example, you can see Caprace's paper A sixteen-relator presentation of an infinite hyperbolic Kazhdan group. There, the polyhedral complex is even CAT(-1). [See Yves' comment below for additional information.]

I suspect that the second point above can be removed, ie., there should exist finite-dimensional CAT(0) polyhedral complexes which do not admit a bi-Lipschitz equivalent median metric.

  • $\begingroup$ You can attribute your theorem to Bruhat-Tits (construction of the building for $SL_3(Q_p)$) and Kazhdan (Property T for lattices therein), and maybe Borel-Harish-Chandra (existence of lattice therein). All three ingredients are $\le 1970$. $\endgroup$ – YCor Oct 9 '18 at 12:45

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