This example is confusing me.

Is $\mathbb{Z}_3 = \langle a\vert a^3\rangle$ $\operatorname{CAT}(0)$ and/or (Gromov) $\delta$-hyperbolic?

The Cayley graph clearly has bounded diameter, therfore it is (Gromov) $\delta$-hyperbolic.

But triangles are more "fat" than comparison triangles in $\mathbb{R}^2$. Which would mean that it is not CAT(0).

I also read that it is an open question if all Gromov hyperbolic groups are also CAT(0). Therefore for this simple example I think it is more likely that the group is in fact CAT(0).

I can't see where I make a mistake.


Edit: As pointed out by Lee Mosher in comments, it is sufficient to say that finite groups act geometrically on a point: of course, a point is hyperbolic and CAT(0)...

Finite groups are clearly hyperbolic, therefore $\mathbb{Z}_3$ so is.

Then, whether a group is CAT(0) is not readable on its Cayley graph: we want to know if it acts geometrically on a CAT(0) space. In the case of $\mathbb{Z}_3$, you can make it act on a disk $D^2$ thanks to a rotation of order 3: $D^2$ is CAT(0) and the action is geometric, so $\mathbb{Z}_3$ is a CAT(0) group.

More generally, it is not difficult to prove that any finite group $F$ is CAT(0): let $T$ be the graph whose vertices are $F \cup \{ o \}$ and where $o$ is linked by an edge with any vertex of $F$ - therefore, $T$ looks like a star. Now, let $F$ act on $T$ fixing the vertex $o$. Clearly, $T$ is CAT(0) - it is a tree - and the action $F \curvearrowright T$ is geometric.

  • $\begingroup$ What rotation of order 3 acts on the square $[0,1]^2$? $\endgroup$ – Lee Mosher Feb 11 '15 at 15:37
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    $\begingroup$ Also, every finite group acts geometrically on a point. $\endgroup$ – Lee Mosher Feb 11 '15 at 15:38
  • $\begingroup$ @LeeMosher: You are completely right, I searched something really too complicated! I added a remark in my answer, thank you for your comment. $\endgroup$ – Seirios Feb 11 '15 at 18:55
  • $\begingroup$ Also, every finite graph quasi-isometrizes to a point, so they are automatically $\delta$-hyperbolic. $\endgroup$ – Balarka Sen Feb 11 '15 at 19:12

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