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10

Assume that $S$ is a finite 2-dimensional simplicial complex equipped with a metric $d$ such that each triangle is isometric to a plane triangle (so $(S,d)$ is a polyhedral space).

Is it possible to subdivide the triangulation of $S$ so that it will have acute triangles only?

Comments:

It is well known that any triangle admits a triangulation with only acute triangels; one can see the proof in the following picture. But I do not see a way to fit such triangulations together.
Such a triangulation would be useful in the proof of Zalgaller's theorem: any n-dimensional polyhedral space admits a length-preserving piecewise linear map to $\mathbb R^n$. Well, it would help only if $n=2$, for larger $n$, we should say that a simplex is acute if it contains its circumcenter.

flag	asked May 15 2012 at 10:57 Anton Petrunin 16.4k●41●110

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Related though you are probably aware of it: It is shown in [Kopczynski, Pak, Przytycki, Acute triangulations of polyhedra and $\mathbb{R}^n$][mimuw.edu.pl/~pprzytyc/combinatorica.pdf] that the 4-cube and $\mathbb{R}^n$ for $n\ge 5$ do not admit acute triangulations. Where acute means in their sense that all dihedral angles are smaller than 90°. – HenrikRüping May 15 2012 at 14:11

Gjergji Zaimi · Accepted Answer · 2012-05-15 11:31:50Z UTC

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There is a theorem in Y. D. Burago, V. A. Zallgaller, "Polyhedral embedding of a net" (Russian), Vestnik Leningrad. Univ., 15 (1960), 66–80, which says that any 2-dimensional polyhedral surface has an acute triangulation.

Another proof for this theorem can be found in "Acute and nonobtuse triangulations of polyhedral surfaces" by S. Saraf. Moreover they show that any polyhedral subdivision can be further refined into a non-obtuse sub-triangulation.

answered May 15 2012 at 11:31

Gjergji Zaimi
37.5k●2●72●149

Thank you so much. I was surprised that acute triangulations are so popular. Is anything known in the higher dimensional case? (acute simplex = simplex which contains its circumcenter.) – Anton Petrunin May 15 2012 at 14:16

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I have no idea. I think people in discrete geometry call this a "well centered mesh", but I haven't been able to find anything relevant. I know of the reference above from Igor Pak's book and as it was mentioned in the comments, Pak has investigated acute triangulations in the sense of acute dihedral angles. – Gjergji Zaimi May 15 2012 at 15:14

Saraf's paper for free: web.mit.edu/shibs/www/acute.pdf – Anton Petrunin Aug 5 at 14:10

Igor Rivin · Answer 2 · 2012-05-15 13:50:05Z UTC

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Just to add to @Gjergji's answer: there are also algorithms to compute such things. See, for example,Erten and Ungor, CCCG 2007 and references therein.

answered May 15 2012 at 13:50

Igor Rivin
31.1k●3●37●71

Acute triangulation

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