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Let $(X, d)$ be a geodesic space which is also a closed manifold. Suppose $d$ is a local $CAT(0)$ metric. Here is the question:

Is it true that $X$ admits a triangulation? (No requiring the triangulation to be PL).

An example of Davis and Januszkiewicz showed that there is a non-triangulable aspherical 4-manifold. Let $P^3$ be the Poincare Homology Sphere. They firstly constructed an aspherical 4 dimensional homology manifold via Gromov's hyperbolization, then replace the neighborhood around non-manifold point (the cone over $P^3$) by Freedman's 4-dimensional contractible manifold $W$ with the boundary $P^3$. It is the second step that the $CAT(0)$ condition is lost and I don't know anything about possible geometric properties of the mysterious $W$.

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  • $\begingroup$ What does a "closed geodesic space" mean? Do you mean a compact geodesic space? $\endgroup$ Commented May 31, 2022 at 22:31
  • $\begingroup$ I mean it is a closed manifold, i.e. compact without boundary. $\endgroup$
    – J. GE
    Commented May 31, 2022 at 23:11
  • $\begingroup$ I see. Then, most likely, this is unknown. $\endgroup$ Commented May 31, 2022 at 23:17
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    $\begingroup$ 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. $\endgroup$ Commented Jun 1, 2022 at 0:14
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    $\begingroup$ I take back my comment. I communicated with Lafont, and they definitely don't prove what I claimed. $\endgroup$ Commented Jun 8, 2022 at 13:59

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