# Singularity of piecewise linear (PL-) manifold with non-positive curvature

Let $X^3$ be a simply connected piecewise linear(PL-) manifold with non-positive curvature. I am curious about that can we know about the singular set of $X^3$ explicitly. For example, what is the singular set of the hyperbolic space $H^3$? Is it depend on triangulation of $H^3$?

In my best knowledge, the definition of singular set(Cao and Escobar's paper, Definition 1.10) is as follows.

Let $\tau$ be a given triangulation of $X^3$ and if $k$-dimensional simplex $\sigma^k \subset X^3$ is said to be singular if $Link(\sigma^k, X^3)$ is not isometric to the unit sphere $S^{n-k-1}(1)$.

• You question requires some work. First of all, what do you mean by a PL manifold of nonpositive curvature? Do you mean that each simplex is equipped with a flat metric? Then what do you mean by the hyperbolic space $H^3$ in this context? Or maybe you mean that the metric has constant curvature on each simplex; then $H^3$ makes sense, but the singular locus is empty. – Misha Mar 18 '17 at 14:47
• I think Cao and Escobar's paper considers PL manifold of nonpositive curvature as simplicial complex whose simplices are equipped with flat metric and the metric on the complex is given as Bridson's paper. Because, in Theorem 4.1, they used Gauss Bonnet theorem and they calculated total Gaussian curvature of the interior of $\hat{\Omega}$ is zero. – Donghwi Seo Mar 19 '17 at 14:07
• Thus my question about hyperbolic space seems not appropriate to understand PL manifold with non-positive curvature. – Donghwi Seo Mar 19 '17 at 14:09

• If you remove singular locus, then the remaining part has to be two-convex, as defined "Sweeping out sectional curvature" by Dima Panov and me. In other words, for any direction $\xi$ at a singular point there will be a singular direction $\zeta$ such that $\measuredangle(\xi,\zeta)\le\tfrac\pi2$. This is true in all dimensions.