Given an $n$-dimensional star-convex polyhedron $P\subset \mathbb{R}^n$ with simplicial facets, is it always possible to construct a regular triangulation $K$ of $P$ which does not subdivide the boundary of $P$, i.e, $\partial K = \partial P$?
3
-
$\begingroup$ Could you please define "star-convex"? $\endgroup$– Joseph O'RourkeCommented Jul 19, 2019 at 11:01
-
$\begingroup$ There exists a point $p\in int(P)$ such that for any $x\in P$, the interior of the line segment $[x,p]$ joining $x$ and $p$, lies in the interior of $P$. $\endgroup$– user136604Commented Jul 19, 2019 at 12:04
-
$\begingroup$ Also, I should mention that I am interested in full dimensional polyhedra $P \subset \mathbb{R}^n$, i.e, $dim(P)=n$... so existence of a function that is strictly convex across codimension one simplexes makes sense. $\endgroup$– user136604Commented Jul 19, 2019 at 12:07
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Schoenhardt polyhedron (wikipedia) is a star-shaped polyhedron in $\mathbb{R}^3$ with triangular faces that cannot be triangulated without subdividing its faces. So the answer is no even without requiring the regularity of the subdivision.
EDIT: As the OP remarks in the comment, this does not answer the question because vertices in the interior are permitted (and then every star-shaped polyhedron has a triangulation, by starring the boundary from a point which sees the whole boundary).
-
$\begingroup$ Thanks for your reply! By 'triangulation', I allow new vertices. So if it's star convex a cone over the boundary is a triangulation for me. $\endgroup$ Commented Jul 19, 2019 at 16:56
-
$\begingroup$ For the required regular triangulation as well, any number of new vertices in the interior are permitted, but none on the boundary. $\endgroup$ Commented Jul 19, 2019 at 17:08