Do two triangulations of a smooth manifold have a common subdivision?

Question

The Hauptvermutung (ie. the question in the title) is known to be false for PL manifolds and topological manifolds, but I can't find a result for smooth manifolds (with boundary), though I recall reading it is true. If not, is something close to this true?

Answer 1

No, this is false.

For example, there exists a triangulation of the smooth manifold $S^5$ which does not have a common subdivision (not even up to isotopy) with the standard triangulation of $S^5$ as the boundary of a $6$-simplex. This triangulation of $S^5$ was discovered by Edwards, who proved that if $M$ is certain homology $3$-sphere not homeomorphic to $S^3$ then the double suspension of $M$ is homeomorphic to $S^5$ (this was generalized by Cannon to all homology spheres not homeomorphic to spheres). If you then take any triangulation of $M$, and if you carry out the double suspension construction on the triangulation itself, the resulting triangulation of $S^5$ is the desired counterexample.

Now the thing one must be careful about here is that this triangulation of $S^5$ is not smoothable, i.e. it is not smooth and it cannot be isotoped to a smooth triangulation.

I am guessing that what you read was something like this: for any smooth manifold and for any two smooth triangulations of that manifold, those two triangulations have common subdivisions up to isotopy.