Local behavior of smooth triangulations

Question

If $M$ is a smooth $n$- manifold, a smooth triangulation is defined to be a homeomorphism from a simplicial complex $K$ to $M$ whose restriction to each simplex is a smooth embedding. It's a well-known theorem of Whitehead that such triangulations always exist.

Given such a triangulation, I'm wondering if the following is true: For each point $p \in M$, we may find smooth coordinates on a neighborhood $U$ of $p$ such that the intersection of each $k$-simplex with $U$ is contained in a linear $k$-plane in $\mathbb{R}^n$. In other words, the triangulation is smoothly modeled on a linear triangulation of $\mathbb{R}^n$. It's easy to do this individually for each simplex (i.e., find a chart for which the inclusion of that simplex is linear) using the inverse function theorem, but that doesn't address the issue of finding a single such chart for all the simplices simultaneously.

I think I see how to do it using some of the approximation results in Munkres' Elementary Differential Topology, but I'm just wondering if this appears in the literature anywhere. I haven't managed to find it stated in that form.

Answer 1

It's one of these standard mistakes on triangulations (found even in some research papers by well-known people). The answer is generically negative, provided that the combinatorial link of some cell in $K$ is complicated enough --- here, ``generically'' refers to the simplexwise smooth embedding. Indeed, the linear representations of quivers come into play.

For a simple counterexample, consider a smooth triangulation $K$ of any manifold $M$ of dimension $n\ge 3$, and a codimension-$2$ cell $\alpha$ of $K$ lying in the boundary of at least four codimension-$1$ cells $\eta_i$ of $K$ ($1\le i\le 4$). At every point $x\in\alpha$, the four lines $\tau_x\eta_i/\tau_x\alpha$ have in the $2$-plane $\tau_xM/\tau_x\alpha$ a cross-ratio $c(x)\in R\setminus\{0,1\}$; and the function $c$ on $\alpha$ is generically not constant on any neighborhood of a given point $x_0\in\alpha$. Then, $K$ is not linearizable in any neighborhood of $x_0$.