# Boundary triangulation induces triangulation

In $R^n$ (the real space) we have an open connected set $D$, such that $\partial D$ is triangulable. Can we prove the closure $\bar{D}$ is triangulable or any counterexample?

Furthermore, the $\partial D$ are piecewise algebraic in the question I am considering, I do not know whether this would be helpful for the above statement.

Thanks for any help.

• If $\partial D$ is piecewise algebraic, then there seem to be an obvious way to proceed: find a homeomorphism of $\mathbb R^n$ taking $\partial D$ to a locally finite union of affine simplices, and extend the latter to a triangulation of $\mathbb R^n$, which in particular, should triangulate $\bar D$. Oct 6 '16 at 1:55
• Is there any standard way to build the homeomorphism, or how to find this kind of homeomorphism? Is it by gluing? Thanks a lot. Oct 6 '16 at 15:37
• If I had all the references I would post it as an answer. Even the case when $\partial D$ is smooth is nontrivial (but course classical). I would search online for "extending a triangulation real algebraic". Once this is done take a regular neigborhood of the subcomplex. Its boundary would be a PL submanifold and you can then extend a triangulation as in Igor Rivin's answer. Oct 6 '16 at 20:41
• Here is a reference: see theorem 4.1 in deepblue.lib.umich.edu/handle/2027.42/63851, "Triangulation of Locally Semi-Algebraic Spaces", by Kyle Hofmann (The result is not proved there; rather it points original references). Oct 6 '16 at 21:30

In many categories, the answer is known to be yes, see

Emil Saucan, MR 2184196 Note on a theorem of Munkres, Mediterr. J. Math. 2 (2005), no. 2, 215--229.

• I have read the note, and find they are all applied to $C^r$ manifold with boundary. But in my question, the boundary is only piecewise smooth (algebraic), i.e. no $C^r$ structure on the boundary. Will those results still be true (by approximation)? Oct 6 '16 at 19:27
• If the boundary is algebraic, the whole set must be semi-algebraic, and Hironaka's result should apply. Oct 6 '16 at 19:47
• That would be cool for algebraic part. But which Hironaka's result are you pointing to? Can you kindly offer some reference? Thanks a lot. Oct 6 '16 at 21:07
• @lunzhang this one? evernote.com/shard/s24/sh/3a549739-3274-4640-901b-40d9e58e9a71/… Oct 6 '16 at 21:51
• Good. That seems to be what I try to find. Oct 7 '16 at 1:35

Although you do not seem to require that the triangulation of the closure to be compatible with the triangulation of the boundary, it is true in $\mathbb{R}^3$ that a triangulated polyhedron $P$ has a compatible interior tetrahedralization. Bern proved that, if $P$ has $n$ vertices, such a tetrahedralization by $O(n^2)$ tetrahedra exists (and can be found quickly):

Bern, Marshall. "Compatible tetrahedralizations." Fundamenta Informaticae 22.4 (1995): 371-384. (ACM link.)

In fact, he proved all of $\mathbb{R}^3$ can be tetrahedralized compatible with $P$'s surface triangulation (with some tetrahedra having a vertex at $\infty$).

It is interesting that if you change "triangulable" to "hexahedral-able," and ask if the surface mesh can be extended compatibly to an interior mesh, the answer is unknown:

"No algorithm is known to construct hexahedral meshes compatible with an arbitrary given quadrilateral mesh, or even to determine when a compatible hex mesh exists, even for the simple examples shown in Figure 1" Erickson, Jeff. "Efficiently hex-meshing things with topology." Discrete & Computational Geometry 52.3 (2014): 427-449. (PDF download.)

• So all the above statement for the $\mathbb{R}^3$ space, right? Indeed, I do not require compatibility of triangulation, but want result in $\mathbb{R}^n$ in general Oct 6 '16 at 15:40
• @lunzhang: Yes, my references are only to $\mathbb{R}^3$. Oct 6 '16 at 16:53