Timeline for Is every geodesic space with non-positively curved metric triangulable?

8 events

when toggle format	what		by	license	comment
Jun 8, 2022 at 13:59	comment	added	Igor Belegradek		I take back my comment. I communicated with Lafont, and they definitely don't prove what I claimed.
Jun 1, 2022 at 0:14	comment	added	Igor Belegradek		It seems the answer follows from [Aspherical manifolds that cannot be triangulated, M. Davis, J. Fowler, and J-F. Lafont, Algebr. Geom. Topol. (2014)], people.math.osu.edu/lafont.1/no_triangulation.pdf. Namely, on p.7 in the subsection "Word hyperbolicity" they build an example by gluing two locally CAT(-1) spaces along locally convex subspaces, and the result is a closed non-triangulable manifold of any dimension $\ge 6$. By gluing theorem this closed manifold is locally CAT(-1). What bothers me is that the authors only argue (in a different way) that $\pi_1$ is word hyperbolic.
May 31, 2022 at 23:17	comment	added	Moishe Kohan		I see. Then, most likely, this is unknown.
May 31, 2022 at 23:11	history	edited	J. GE	CC BY-SA 4.0	edited body
May 31, 2022 at 23:11	comment	added	J. GE		I mean it is a closed manifold, i.e. compact without boundary.
May 31, 2022 at 22:31	comment	added	Moishe Kohan		What does a "closed geodesic space" mean? Do you mean a compact geodesic space?
May 31, 2022 at 14:42	history	edited	J. GE		edited tags
May 31, 2022 at 9:15	history	asked	J. GE	CC BY-SA 4.0