A piecewise linear flat torus embedded in $\mathbb{R}^3$ is shown at http://www.mathcurve.com/polyedres/toreplat/toreplat.shtml. It is flat in the sense that the angle defect at the vertices is zero.

Here is a 3D printed hinged version:

enter image description here

Who came up with this construction?

I asked Robert Ferréol who maintains the mathcurve.com site. He heard of it from Guy Valette, who remembers seeing a torus like this at Oberwolfach over 30 years ago.

  • 1
    $\begingroup$ related answer: mathoverflow.net/a/34370/1345 $\endgroup$ – Ian Agol Jun 11 '15 at 14:47
  • 1
    $\begingroup$ One gets 1-parameter families of embedded flat tori by taking the middle level (which consists of a regular $n$-gon and a $2n$-star), and inserting a product region. $\endgroup$ – Ian Agol Jun 12 '15 at 8:54

I believe the originator is Victor A. Zalgaller.

Permit me to quote myself from an earlier answer:

In the paper by V. A. Zalgaller, "Some bendings of a long cylinder," Journal of Mathematical Sciences, 100(3):2228--2238, 2000 (translated from a 1997 article in the Russian journal Zapiski Nauchnykh Seminarov POMI), he proves this theorem:

"Theorem 1. A direct flat torus can be isometrically embedded in $\mathbb{R}^3$ 'in the origami style' if its development is a rectangle sufficiently large compared to its altitude."

He defines a direct flat torus as the result of identifying the opposite sides of a rectangle.

  • $\begingroup$ (I see Ian linked to the same earlier answer as I was typing this.) $\endgroup$ – Joseph O'Rourke Jun 11 '15 at 14:49
  • $\begingroup$ Thanks for this! The construction in Zalgaller's paper does seem to be closely related, but it isn't exactly the same. Presumably Zalgaller had many related constructions and only wrote up the most general? $\endgroup$ – Henry Segerman Jun 11 '15 at 15:20
  • $\begingroup$ @HenrySegerman: I am not sure. Zalgaller might respond to an email query, even at 94(!). $\endgroup$ – Joseph O'Rourke Jun 11 '15 at 15:39
  • 1
    $\begingroup$ Original is at mi.mathnet.ru/eng/znsl/v246/p66 $\endgroup$ – David Roberts Jun 12 '15 at 2:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.