Does the concept of a Möbius torus make sense: taking a cylinder (instead of a rectangle as in the case of the Möbius strip) and twisting it before joining its ends? Or will the resulting twisted torus be indistinguishable from the normal torus in any relevant respect?

[This equivalent to the well-known Möbius strip should be called Möbius cylinder but it would have so much in common with a torus that I preferred to call it a Möbius torus.]

Embedded in Euclidean space the twisted and untwisted torus "look" the same - opposed to Möbius strip and cylinder -, the difference would be only in their intrinsic properties. But can there be such differences? And how do I specify them?

PS: I posted a follow-up question here.

  • 11
    $\begingroup$ This sounds suspiciously like the Klein bottle. $\endgroup$ Oct 20, 2012 at 17:49
  • $\begingroup$ Are you looking for the Klein Bottle, which joins the ends of a cylinder in the opposite orientation to a torus? $\endgroup$ Oct 20, 2012 at 17:51
  • 4
    $\begingroup$ A mere twist before identifying the ends of the "hose" again produces an ordinary torus with a locally euclidean metric, but its global conformal type has changed. If you identify the ends reversing the orientation you get a Klein bottle, which is a nonorientable surface. $\endgroup$ Oct 20, 2012 at 18:21
  • 1
    $\begingroup$ Just to add to Christian's comment, a twist will be isotopic to the identity map, and so the resulting space will be homeomorphic to a regular torus. Think about it: if you can do it in R^3, it's not a Klein bottle. You need a degree -1 map on the circle boundaries to get the Klein bottle. $\endgroup$
    – user641
    Oct 21, 2012 at 4:00
  • $\begingroup$ @Christian Blatter: Can you please help me to understand better: the simply twisted torus (which I tried to ask for) has a locally euclidean metric (just like the Klein bottle), is homeomorphic to the regular torus, but has another conformal type. What exactly is the conformal type? Does it make sense to think of (closed) geodesics in this context? $\endgroup$ Oct 23, 2012 at 14:14

3 Answers 3


As remarked above: What you get is the Klein bottle. Put differently: The result is what you get when you take two Möbius strips (which both have one boundary) and glue both boundaries together (which does not work when embedded in 3d space but works in theory). See this image from http://im-possible.info/english/articles/klein-bottle/: enter image description here

  • 2
    $\begingroup$ They might not meet all the technical details of the definition; but as long as you're willing to overlook the defect where it punches through its side, you can buy glass klein bottles as conversation pieces even if you're trapped in 3-space. $\endgroup$ Oct 20, 2012 at 20:51

In the course of experimenting with 3-D models, I developed a method for twisting a toroid of arbitrary cross section. I called these forms Möbioids. Of course, if the cross-section is circular the result will be indistinguishable. What I did was to consider cylinders of non-circular cross-section. However, then the twist angles are quantized. Depending on the particular cross-section and number of twists, you can get forms with one or more surfaces. The figures below show an astroid cross-section with one twist of $\pi/4$ (left) and a pentacuspid cross-section with six twists of $2\pi/5$ (right). Each has a single surface. Again, not all twists lead to a single surface. You can find more images and some animations at A New Twist on Möbius1.

Some rendered images of Möbioids as taken from the link in the text: http://old.nationalcurvebank.org////moebius2/moebius2.htm

1 Archived version in case the link above ever dies


You twist it by $\pi$ and you get that "Möbius torus" Möbioid.

Twist it by $\frac{2\pi}3$ and you get a nice impossible triangle 😊

Penrose triangle https://en.wikipedia.org/wiki/Penrose_triangle

I just noticed that it regularly tiles the torus with three pairs of same colored toric (concave) rectangular hexagons, all those six faces connected exactly like a cube. Cubic contentedness (although different topology) but with hexagons, instead of squares...

You twist it by $e$, or any other irrational number, not necessarily transcendental, and any underlying astroid, rectangular other cross section will get smoothed out into a blurry thick-toroidal Möbioid ("surface" Hausdorff dimension of three, instead of normal two).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .