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It is mentioned on the Wikipedia article for Hadamard spaces that the Cayley graphs of a word-hyperbolic (f.g.) group are CAT(0) metric spaces. Is it so? My question comes from the fact that the Cayley graph for S$L(2,\mathbb{Z})$ with the two usual generators $z\mapsto z+1$ and $z\mapsto -z^{-1}$ is very tree-like but still has 1-sided equilateral triangles (whose inner distances are obviously larger than those on Euclidean space).

At the same time, if it were true, then any word-hyperbolic group would act properly and cocompactly on its Cayley graph, and we would have an affirmative answer to this very similar (and as far as it seems, open) question. Am I missing something? (I tend to assume that general math articles on the English Wikipedia are correct, yes)

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RickyGervais
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  • But a Hadamard space is a complete CAT(0) space (at least that's the definition I have from Ballmann's lectures, Wikipedia has the midpoint inequality definition and then says they are equivalent)... – RickyGervais Apr 08 '16 at 19:19
  • It looks like a mistake, I looked at the history and it used to say cayley graph of discrete groups, which is certainly not true, and I guess the person who corrected that ended up inserting a more subtle error, although it is a well known open problem if hyperbolic groups are CAT(0) (note there is no reference to the claim). –  Apr 08 '16 at 20:46
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    A graph which is not a tree, equipped with a geodesic metric putting length 1 on each edge as Cayley graphs are, is certainly is not CAT(0), for example any minimal length embedded circle in the graph can be used to disprove the CAT(0) inequality. So the only groups that can have a CAT(0) Cayley graph are free groups. – Lee Mosher Apr 09 '16 at 00:42
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    Alternatively, CAT(0) spaces are simply connected (in fact, contractible), so trees are the only CAT(0) graphs. – Seirios Apr 09 '16 at 09:59
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    I edited the Wikipedia page, and edited the link on this question to go to the old version to preserve the question (just letting future people who see this question know). –  Apr 09 '16 at 16:57
  • Perhaps the word "graph" should have been "complex". Refuting this is harder, although I am sure it is still false. (Even if all hyperbolic groups are $CAT(0)$, I would be surprised if it was because their Cayley Complexes' are $CAT(0)$.) – user1729 Apr 10 '16 at 15:22
  • @user1729: This is false for Cayley complexes as well, just take $G=Z_2$. – Moishe Kohan Apr 20 '16 at 05:13
  • @studiosus what if we assume $G$ is torsion-free? – user1729 Apr 20 '16 at 15:56
  • @user1729: The most interesting examples are in Theorem 2 in http://www2.math.ou.edu/~nbrady/papers/cathyp2.pdf, since, in addition, these groups have geometric dimension 2. If you do not require geometric dimension 2, just take the fundamental group of a closed hyperbolic 3-manifold. Its cohomological dimension is 3, hence, it cannot have a contractible Cayley complex. – Moishe Kohan Apr 20 '16 at 16:13
  • The correct statement is that the Cayley graph of every hyperbolic group is quasiisometric to a CAT(-1) space. – Moishe Kohan May 09 '16 at 22:49

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This question has been answered by Lee Mosher and Seirios in the comments. The only graphs that are CAT(0) are trees, so the groups whose Cayley graphs are CAT(0) are free groups.

Jim Belk
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