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Given a geodesic metric space $X$ together with a choice of midpoints $m:X\times X\rightarrow X$ (i.e. $d(m(x,y),x)=d(m(x,y),y)=d(x,y)/2$). Assume furthermore, that the following nonpositive curvature condition is satisfied:

$d(m(x,y),m(x,z))\le \frac{d(y,z)}{2}$ for all $x,y,z\in X$ . This is just a special case of the CAT(0) inequality for the "triangle" $x,y,z$. Lets call such a space a M-space.

Such a space needn't be CAT(0), as the example $(\mathbb{R}^n,d^1)$ shows, where $d^1$ is the $l^1$ metric. The choice of midpoints is given by $m(x,y)=\frac{x+y}{2}$. It also needn't be unique geodesic.

But this space can be equipped with another metric, that makes it a CAT(0) space.

So my question is: Is every group, that acts properly, isometrically and cocompactly on a M-space already a CAT(0)-group?

Improve this question
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    $\begingroup$ You might find this paper and references therein helpful: front.math.ucdavis.edu/0704.3749 See also: ams.org/mathscinet-getitem?mr=2106770 ams.org/mathscinet-getitem?mr=2197811 $\endgroup$
    – Ian Agol
    Sep 7, 2010 at 15:47
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    $\begingroup$ Tiny nitpick: I would call this a 'non-positive curvature' condition, rather than a 'negative curvature' condition. $\endgroup$
    – HJRW
    Sep 7, 2010 at 20:25
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    $\begingroup$ Ian, median spaces are not the same as M-spaces. Otherwise the answer would be positive (median spaces admitting co-compact proper isometric actions should be closely related to CAT(0)-cubical complexes). I think that an M-space does not necessarily have quadratic "isoperimetric" function, hence the answer should be in fact negative. $\endgroup$
    – user6976
    Sep 8, 2010 at 1:25
  • $\begingroup$ The first link in Ian's comment is broken, here's a replacement: arxiv.org/abs/0704.3749 $\endgroup$
    – David Roberts ♦
    Mar 29, 2022 at 8:33

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