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Suppose $G$ is a word hyperbolic group i.e. every geodesic triangle in a cayley graph with respect to a finite generating set of $G$ is $\delta$-thin, for some $\delta>0$. There are various analytical properties these groups admits. Like some hyperbolic groups have Haagerup property, e.g. free groups with $n$-generators, fundamental groups of genus $g(\geq 2)$ surfaces etc. But there are some other hyperbolic groups which satisfy Property (T), which is a strong negation of Haagerup property. One example of this kind of groups are uniform lattices of $Sp(n,1)$. Recently there is a paper https://arxiv.org/abs/2011.09276, which gave other examples of hyperbolic groups with property (T).

My question: For a word hyperbolic group $G$, is it true that $G$ either has Haagerup property or has Property (T)? I am familiar with the work of Y. Cornulier, where he showed that if a group does not have Haagerup property then it has relative property (T), but I don't know any such examples in hyperbolic groups.

Haagerup property: $G$ has haagerup property if it has a metrically proper affine action on a Hilbert space.

Property (T): $G$ has Property (T), if for every affine action on a Hilbert space there is a fixed point under the action.

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No, there is no such dichotomy. If $G$ is an infinite group with Property (T) and $H$ is any non-trivial group, then $G*H$ has neither of the two properties. This is because groups with Property (T) have Property FA (any action on a tree has a fixpoint) and thus cannot be a non-trivial free product and the Haagerup property passes to subgroups. The free product of two hyperbolic groups is however hyperbolic.

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  • $\begingroup$ Ok, so in general there is no dichotomy. Then taking this fact in mind can one say that: if $G$ is hyperbolic relative to $\{H_1,H_2,...,H_K\}$, and all the $H_i$'s have Haagerup property then $G$ has Haagerup property? $\endgroup$
    – ggt001
    Commented May 11, 2023 at 11:00

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