Spherical knots - Cuboctahedral bend(s)

[Ruby]

maybe you can take those button knots as Spherical knots.

like diamond bend.
or extended diamond bend. some over-2-under-2 pattern.

or torus knot?
http://www.mi.sanu.ac.rs/vismath/pennock/index.html

[xarax]

   I am sure that knot tyers who are interested in the boundless space of Decorative knots will have many more things to say about "spherical" or "polyhedral" knots - but my intention was different. I had tried to figure out a "third" way to represent a Practical knot - which way, on the one hand, will reveal its "topological" properties better than a photographic image or a 2D sketch of the "final" tight/compact knot ( where the "over"/"under" crossings of some segments of the rope are oftentimes hidden deep inside the "closed" nub, or covered under other overlying crossings of some other segments ), and, on the other, will reveal its "geometrical" properties better than a 2D tying diagram ( where the particular 3D geometry of the knot - which, regarding the way a practical knot "works" under tension, is what counts more - is destroyed by the 2D image of the "loosened", "spread out" and "flattened" knot ). I imagined a model of a knot made of a somewhat elastic material which will first be "inflated"/ "exploded", so that all its segments will move apart from each other and will settle being arranged on the surfaces of one or more concentric spherical "shells", and then this "spheroid" will be projected on a plane, without been "spread out" and deformed any more.
   Clear as mud ?   
   In other words, when we want to represent a knot, in order to reveal its topology and its geometry better, it may be useful to project its lines on the surface of a "surrounding" sphere, from a point at the centre of the knot, and not on a flat plane, from a point outside the knot.

[rrmmff2000@gmail.com]

Thanks for this inspiration -- here is my tripled version on a Rubik cube

[struktor]

Cube and two trefoil knots

[Tex]

xarax:

First, the crossing points of a spherical knot can be arranged in such a way that their number is minimal. On the contrary, most of the times, on a loose and flattened knot this will not happen.

This was in the context of talking about knot topology.  This is highly misleading.  Ok, maybe most of the time it "won't" happen, but the fact is:

If a knot can be represented on a spherical surface with N crossings then it can also be represented on a flat surface with N crossings.

It's very simply to understand.  Take a point on the back side of the sphere as you look at it.  Choose a point where there is not rope.  Prick a hole.  Expand the hole out and out and out until it is much larger than the sphere, stretching all surfaces as needed to make it all flatten out.  You're done.  No crossings were added or removed.

[Tex]

I'm going to add one caveat to this and that is regarding the actual ends/tails of the ropes. 

In a flat representation the ends/tails as they come into the knot from infinitely far away might cross over two ropes before passing under one.  On a sphere, I'm not sure what we're supposed to do with the tails.  There is no infinite flat plane for them to approach from. I guess we make them turn out 90 degrees to the surface and just go away.  Ok, in that case the two "over" crossings would disappear.   This though isn't really a result of using a sphere but is a result that the sphere forces you to cheat with the ends.  We can also cheat that way on a plane, have the ends turn 90 degrees up off the plane.  That would also avoid those two crossings on the plane and get the same number of crossings.

We can instead "fix" the sphere by drawing a small circle somewhere on the sphere.  No part of the knot should be inside that small circle and four rope ends start there.  If you prefer, they turn out at 90 degrees there.  So long as no rope crosses this small circle then this will be equivalent to the normal rules for flat representation and will require the same number of crossings.

[xarax]

  If a knot can be represented on a spherical surface with N crossings then it can also be represented on a flat surface with N crossings.

  Who argued about this tautology ? Not me...
  I was talking about the number of crossing points of "the loose and flattened knot"(sic). Read what you quote !
  Take a loose knot. Project it on a plane / "flatten" it. You get a 2D representation / diagram of this knot.
  In general, and if this knot is not a very simple one ( an overhand knot, for example ), the diagram which you will get will not have the minimum number of crossing points ! You will need to "explode" the "under" side of the loose knot a lot, so the "over" side will not be projected on top of it, but inside it. On the contrary, in a spherical representation of the same knot, you will not have to distort the one side more than the other.
   Draw the plane representation of the Cuboctahedral bend with the minimum number of crossings, and compare it to the spherical representation.
   My main point was that, if we do not project on a plane, from a point outside the knot, the "over" and the "under" sides of a knot the one on top of the other, but we project them on a sphere, from a point inside the knot, the "over" side on the one hemisphere and the "under" side on the other hemisphere, we get a spherical diagram which, geometrically, will not differ much from the shape of the loose knot itself.

[Tex]

I guess it all comes down to understanding what you meant (or didn't mean) to imply by "loose and flattened". 

[xarax]

   Start from the "final", compact form of the knot, and imagine that the path of the rope in 3D remains the same, but the size of the rope shrinks by, say, 50%. What you get is a loose form of the same knot.

[struktor]

Triangular bipyramid.
Compact form 4 trefoil knots.
Hint : geodesic loop.

[struktor]

Another method on a sphere

http://www.littleimpact.de/permanent/math/sphere_filling/