Talk:Clifford torus

Duoprism relation?

 * It would appear this torus is approximated by an n-m-gonal duoprism, or a 3D subset of a duoprism? Tom Ruen (talk) 05:43, 7 February 2011 (UTC)
 * 23,29-duoprism_stereographic_closeup.jpg

Like the image above is very similar, just an edge-framework, so apparently the square are faces on the torus surface, and the 23,29-gonal polygons are ignored? Tom Ruen (talk) 05:45, 7 February 2011 (UTC)
 * That seems accurate to me... may be worth adding a section describing the relation. The important bit is that the m-gon and n-gon must each have the same radius so that the resulting shape lies entirely on the 3-sphere.  JasonHise (talk) 06:00, 7 February 2011 (UTC)

Duocylinder relation?
And is a Clifford torus the same as a duocylinder?? Tom Ruen (talk) 05:51, 7 February 2011 (UTC)
 * Very possible - the animation on the duocylinder page is exactly what my model looks like when I do a more distant projection (rather than stereographic)


 * NOT, but Clifford torus is the ridge (edge) of duocylinder. Actually duocylinder points here. 90.180.192.165 (talk) 09:31, 2 December 2012 (UTC)


 * I'm not sure if what I'm about to say is right, or will cause many topologists to break down in tears, but I think that a duocylinder is to a Clifford/flat/square torus as a sphere is to a ball, or a circle is to a disk. That is, the Duocylinder would be the surface, and the Clifford Torus would be the internal volume bounded by it. Can someone tell me if I'm right or wrong in my interpretation? — Preceding unsigned comment added by 146.90.88.202 (talk) 01:06, 9 May 2014 (UTC)


 * Nearer to the other way ’round. Consider how an ordinary cylinder can be inscribed in a 2-sphere, touching it on two circles.  If a duocylinder (a hypersolid body) is inscribed in a 3-sphere (a hypersurface), they meet in a Clifford torus. —Tamfang (talk) 23:10, 4 February 2017 (UTC)

Assessment comment
Substituted at 01:53, 5 May 2016 (UTC)

eponym
And why is it named Clifford? —Tamfang (talk) 23:10, 4 February 2017 (UTC)

Euclidean properties
The description of the Clifford torus as an example of "Euclidean geometry" is surely wrong. The Euclidean plane is equivalent to $$\mathbb R^2$$, but the Clifford torus is equivalent only to a finite subset. If you set up a Cartesian coordinate system with the origin at the centre of the fundamental square, it will bump into itself across the joining zippers, where finite coordinate values must suddenly change sign. If you adjust the metric to be infinite at the zippers (known to some geometers as the absolute line) then Euclidean geometry will never reach them and the square is not even joined up to become a torus. Sure it is locally Euclidean, but then so is any smooth coordinate manifold. &mdash; Cheers, Steelpillow (Talk) 11:57, 12 November 2019 (UTC)