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Let $G$ be a finitely generated group. Consider the Floyd boundary as defined in https://www.unige.ch/math/folks/karlsson/free.pdf by A. Karlsson. For a Floyd function f, we denote the Floyd boundary of G by $\partial_f(G)$.

We know that the action $G \curvearrowright \partial_f(G)$ is convergence action. Can we say anything about the point stabilizers under the action? i.e. are all of them amenable or else?

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    $\begingroup$ Are you really assuming that the group is hyperbolic ? In such case, the Floyd boundary is the Gromov boundary. Also in Karlsson's paper, the action of G on its Floyd boundary is a convergence action, whether G is hyperbolic or not, you just need to assume that the Floyd boundary is non-trivial. $\endgroup$
    – M. Dus
    Commented May 17 at 9:25
  • $\begingroup$ @M.Dus sorry for the confusion, I don't assume the group to be hyperbolic in my mind, but don't know why I typed that. The answer you provide is the thing I needed. Thanks $\endgroup$
    – ggt001
    Commented May 18 at 13:12
  • $\begingroup$ Further do you know whether there is any way to classify which points are conical, which are not under the action? $\endgroup$
    – ggt001
    Commented May 18 at 13:14
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    $\begingroup$ Okay then. Maybe you should edit the question accordingly :) And about your new question, I do not know any general statement, but with additional assumptions, you can say a bit more. If the group is relatively hyperbolic, then the Floyd boundary covers the Bowditch boundary and the pre-image of a conical limit point (in the sense of relatively hyperbolic groups )is reduced to a point and is still conical (in the sense of the convergence action of the group on the Floyd boundary). $\endgroup$
    – M. Dus
    Commented May 18 at 15:08

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No, this is certainly not true that the stabilizer of a point on the Floyd boundary is amenable. And more generally, stabilizers can be pretty wild. Typically, mapping class groups have trivial Floyd boundary, this is because "too many elements comutes at infinity". More generally, see the paper Thick groups have trivial Floyd boundary by Levcovitz, to appear in Proceedings of the American Mathematical Society.

Now, if you want an example with infinite Floyd boundary, you just need to take the free product of a group $G$ with trivial Floyd boundary and any group. Then, the free factor $G$ will be the stabilizer of a point in the Floyd boundary.

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  • $\begingroup$ The question assumes the group is hyperbolic. $\endgroup$
    – HJRW
    Commented May 16 at 16:42
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    $\begingroup$ @HJRW Thank you very much, I had misread the question. But for hyperbolic groups, the Floyd boundary is the Gromov boundary.... I asked the OP in comments if they really assume that the group is hyperbolic and if so I will delete my answer. $\endgroup$
    – M. Dus
    Commented May 17 at 9:26
  • $\begingroup$ I think your answer is still interesting! But you might like to edit it to make it clear that it doesn't quite answer the question as stated. $\endgroup$
    – HJRW
    Commented May 17 at 10:04

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