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Are there some examples of CAT(-1) spaces which are not trees which have disconnected Gromov boundary?

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  • $\begingroup$ Yes: let $T$ be (the 2-skeleton of) an equilateral triangle in the hyperbolic plane $H^2$. Consider two copies of $T$ glued on their vertices, and take the universal covering, with the length metric. Then it is obviously QI to a tree and CAT($-1$), but not isometric to a tree. $\endgroup$ – YCor Mar 14 '15 at 15:05
  • $\begingroup$ Second example (if you don't want something quasi-isometric to a tree): consider a horodisc in the hyperbolic plane. Then it's CAT($-1$) and its Gromov boundary is reduced to a point. $\endgroup$ – YCor Mar 14 '15 at 15:08
  • $\begingroup$ @YCor How is a point disconnected? $\endgroup$ – Igor Rivin Mar 14 '15 at 16:59
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    $\begingroup$ oh, I saw "totally disconnected". Otherwise it's even much easier, just take a wedge of two copies of $H^2$, then the Gromov boundary is a disjoint union of 2 circles. $\endgroup$ – YCor Mar 14 '15 at 23:31
  • $\begingroup$ This is homework; the question should be moved. As a hint - you should think of examples of CAT(-1) spaces and their boundaries. Then think about how you can cut spaces into pieces (or glue spaces together) and how the boundary changes under those operations. $\endgroup$ – Sam Nead Mar 15 '15 at 2:21
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Yes, the free product of any two word hyperbolic groups has disconnected Gromov boundary. For proof see the nice survey of Kapovich-Benakli., section 7.

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  • $\begingroup$ word hyperbolic groups are not exactly CAT(-1). (But there are many examples.) $\endgroup$ – Anton Petrunin Mar 14 '15 at 17:19
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    $\begingroup$ You need at least one example for which you know that the free product is $CAT(-1)$; it's OK for the free product of a surface group with $\mathbf{Z}$. $\endgroup$ – YCor Mar 14 '15 at 23:32
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For sure you can not make it to be manifold without boundary.

You can start with a tree and glue to it many compact pieces applying Reshetnyak gluing theorem.

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