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Let $S$ be an abstract euclidean polyhedral surface. By this, I mean a orientable compact 2D topological surface obtained by gluing together some (convex) euclidean polygons (arbitrary genus).

Question: are there softwares to draw $S$ in the 3D euclidean space $\mathbb E^3$?

Formulated this way, my question is certainly too vague. What I mean by drawing $S$ in $\mathbb E^3$ is not determined. A PL isometric immersion $S\rightarrow \mathbb E^3$ would be great. But anything else leading to nice pictures would be appreciated too...

Thanks in advance!

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  • $\begingroup$ This question is related mathoverflow.net/questions/142284/embedding-of-flat-surfaces/… $\endgroup$
    – j.c.
    Sep 24, 2013 at 13:41
  • $\begingroup$ More specifically: I don't have access to the paper (the english translation is not online) but perhaps unraveling the proof of the theorem by Burago and Zalgaller described in the question I linked to above will help, as that result apparently yields an isometric immersion for arbitrary genus polyhedral surfaces. $\endgroup$
    – j.c.
    Sep 24, 2013 at 16:14
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    $\begingroup$ The proof of B&Z is quite far from algorithmic. When I studied it I could not see an upper bound. $\endgroup$ Sep 24, 2013 at 23:29

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In 2007 I wrote a "news-like article" on

"The new algorithm of Bobenko and Izmestiev for reconstructing the unique polyhedron determined by given gluings of polygons is described."

I think this partially answers your question...?
ScreenSnapshot

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    $\begingroup$ Very nice! But it concerns only the genus 0 case... $\endgroup$
    – Elbabak
    Sep 24, 2013 at 14:00
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    $\begingroup$ The link in your article is now dead but the java program is linked here www3.math.tu-berlin.de/geometrie/ps/software.shtml $\endgroup$
    – j.c.
    Sep 24, 2013 at 14:11
  • $\begingroup$ @Elbabak: That's correct. Alexandrov's theorem only applies to the genus zero case, and it is Alexandrov's theorem that underlies that algorithm. I do not believe there is anything comparable for arbitrary genus. $\endgroup$ Sep 24, 2013 at 14:28

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