Timeline for Are negatively curved $2$-complexes homeomorphic to quotients of the form $\mathbb{H}^{2}/G$, where $G$ is some group?
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| Mar 26, 2017 at 19:34 | history | edited | Jean Raimbault | CC BY-SA 3.0 |
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| Mar 26, 2017 at 15:49 | comment | added | YCor | Actually gluing infinitely copies of $H^2$ in this way, in an arborescent way, with singularities at all points with integer Busemann function, yields "natural" spaces with cocompact (nondiscrete) isometry groups, and called millefeuille spaces. Groups such as the semidirect product $(\mathbf{R}\times\mathbf{Q}_p)\rtimes\mathbf{Z}$ with contracting action, e.g., by multiplication by $(1/2,p)$, admit isometric cocompact proper action on such spaces (branched $(1+p)$ times at every point of integral Busemann function). | |
| Mar 26, 2017 at 15:28 | history | answered | Jean Raimbault | CC BY-SA 3.0 |