Rings in 4d

[anderscolingustafson]

One interesting thing about rings in 4d is that it is possible to have two rings that are pointed in different directions that are basically perpendicular were each ring has all its points the same distance from the other ring.

[PWrong]

Hey you're right. The circles are
(cos t, sin t, 0, 0)
(0, 0, cos s, sin s)

the distance between any two points is

sqrt(cos²t + sin²t + cos²s + sin²s) = sqrt(2)

[mindspace]

I know this may seem a silly question, but I'm a 'bear of little brain'.

How can we guarantee that the two individual rings in their own planes are connected and therefore form a single object?

[PWrong]

They won't be connected. The two rings are separate, in fact every point in one ring is a fixed distance from every point in the other.

[mindspace]

Hm... I'm still feeling like a bear of very little brain here and would appreciate further help to sort my befuddled thoughts out...

Let's call the four lanes x, y, z and w (the first three following a conventional 3D system). For the ring defined by the parameter t, you have z=0 and w=0, whilst for the ring defined by parameter s, you have x=0 and y=0. [I know this is obvious, but it is scene setting so to speak.]

When the t-parameter ring keeps z=0 and w=0, and the s=parameter plane keeps to its x=0 and y=0, then the distance between any point on one ring and any point on another is the sqrt(2). Now what happens if we move the s parameter ring to x=1 and y=0? The distance between any point on one ring and any point on the other ring becomes sqrt(3). This all goes to demonstrate, quite correctly, that an instant in time, your statement is correct. The important point here is AT AN INSTANT IN TIME.

We can only guarantee that the distance between the two rings will remain unchanged if one is not moving with respect to the other. The question then becomes how do we know they are or are not moving with respect to each other? I would be interested in your thoughts on this.

I hope all is well with you and yours.

[wendy]

This in four dimensions, is the representation of 'clifford parallels' of the space S3, being represented as the surface of a sphere in 4d (ie a glome or glomohedron).

Clifford parallels come in 'left' and 'right' handed versions, but for a given (say left-hand) set, and a given circle+direction [great-arrow], there is exactly one circle through any given point, and that circle has a direction too. Every such circle does not intersect with any other circle, and has a common centre.

The rotation is rigid, and because in physics, the nature of rotations is that different but linked rotations of the same object will adjust so that they both have the same energy, the natural fate of any rotating object is to become a clifford rotation.

For those familiar with complex numbers, the point x,y (complex), traces into x.cis(wt), y.cis(wt), where w is a constant speed, and t is time.

[PWrong]

The question then becomes how do we know they are or are not moving with respect to each other?

Well we're not worried about that at all. We're just doing geometry here, not talking about actual rings where we have to check if they're moving around. If we say the rings are centred on (0,0,0,0) then that's where they are.

Now what happens if we move the s parameter ring to x=1 and y=0?

You then have
(cos t, sin t, 0, 0)
(0, 0, cos s + 1, sin s)

The distance is then
sqrt(cos²t + sin²t + (cos s + 1)² + sin²s)
= sqrt(cos²t + sin²t + cos² s + 2 cos s + 1 + sin²s)
= sqrt(2) + 2 cos s + 1

So if one of the circles moves slightly off centre, they're no longer equidistant.

[quickfur]

Congratulations! You've just rediscovered a property of 4D that's quite ubiquitous in a certain class of shapes. The duocylinder is one such shape, consisting of two toroidal surfaces that interlock each other: each toroid is a 3-manifold that corresponds to one of your rings. The duoprisms consist of n-gonal and m-gonal prisms that form two interlocking rings, in exactly the same manner. The Grand Antiprism consists of two such rings of pentagonal antiprisms, connected by a layer of tetrahedra. The Grand Antiprism can be derived from the 600-cell by removing two rings of linked vertices -- the very same rings you've discovered -- which means the 600-cell itself contains many such pairs of two rings.

This shouldn't be surprising, since, in fact, the Hopf fibration of the 3-sphere (what non-mathematicians would call a 4D sphere) is based on the possibility of such interlocking 2 rings; the fibres basically run parallel to both rings, which are themselves in the fibre set. Polytopes like the 600-cell approximate the 3-sphere (or rather, they tile the 3-sphere, if you allow the cells to be curvy), so it's no surprise that the 2 rings structure should be reflected in them too.

In fact, almost all 4D uniform polytopes have the 2-rings structure as a sub-symmetry. The tesseract, for example, can be decomposed into two rings of 4 cells each. Equivalently, the vertices of the 16-cell lie on two such rings.

[anderscolingustafson]

In a 4d world one place this could be seen would be a 4d planet as a 4d planet would not have two poles but one single pole that would form a ring around the planet. All of the points on the pole would be equally distant from the equator and vice versa so to a tetronian this would have a practical effect. This would also meant that the directions of Marp and Garp would be more like extra kinds of North and South than East and West as they would point to the polar ring so unless the polar ring was to have a magnetic North and South Pole telling the difference between Marp and Garp and North and South would be difficult.

[quickfur]

That may not necessarily be true. A 4D planet could easily rotate in both planes in a Clifford double rotation, so there is no stationary point on the surface, only at the center of the planet.

But of course, to decide one way or another really requires a workable 4D physics, which so far hasn't been forthcoming.

[Keiji]

How would a 4D compass work anyway? (Assuming physics, of course)

[Eric B]

Marp and Garp? Never heard of that one! I imagine that is ana and kata changed into global circles. A second kind of lattitude or longitude?

Still, is a way of visualizing these rings, by looking at how the poles are equidistant from the equator, and then letting the poles generate new rings in 4D, and those woulgs would still be equidistant at every point to the equator?
(I also recognized the two "rings" dynamic in the duoprisms).

[quickfur]

It depends on what you mean by "equator". My definition of equator is the set of points midway between two antipodal points (i.e., the north pole and south pole). In the case of a 4D sphere (a 3-sphere), the equator is a spherical surface, not merely a ring. The fact that in 3D the equator is also a great circle is a coincidence. In 2D, the only great circle is the circle itself, and the equator is two disjoint points. 3D is the only case where the equator is also a great circle. In 4D, the equator is a sphere, a 2D surface, whereas the great circles are still... circles. In 5D, the equator is a 3-sphere, and the great circles are... well, still great circles.

But of course, if you adopt a different definition of equator, then you will draw different conclusions. But in the end, they aren't contradictory; they are just the same concepts expressed with different terminology.

[Eric B]

OK, that's right. I was thinking of the "equator" as the original 2D ring, and imagining the antipodal "poles" generated in 4D into new rings. I was not dealing with the equatorial ring generating a sphere, as we were only discussing rings in 4D. (Of course, I'm sure not all points on that spherical equator would be equidistant to the polar rings).

[Eric B]

Then there is a 2D toroidal sheet wrapped around the planet's surface, which experiences the greatest total velocity. This sheet corresponds with the ridge of the duocylinder. By analogy with a 3D planet's equator, which constitutes the points with greatest velocity, wouldn't it be reasonable to call this sheet the "equator"?

I take it this "sheet" would be the so-called "true 2-torus" made by joining opposite edges of a square (with the second pair of edges joined together in 4D so you don't stretch/crunch the cylinder formed by the joining of the first pair of edges). We had discussed this object before, and it was said to be the "ridge" or maybe it was "frame" of the whole duocylinder.
(Sounds like such an interesting object. Wish we could visualize true 4D shapes).

[quickfur]

[...](Of course, I'm sure not all points on that spherical equator would be equidistant to the polar rings).

Yes, they are exactly equidistant. Welcome to the 3-sphere. Well actually, they are equidistant to the polar POINTS. The polar rings are something else altogether.

[...] I take it this "sheet" would be the so-called "true 2-torus" made by joining opposite edges of a square (with the second pair of edges joined together in 4D so you don't stretch/crunch the cylinder formed by the joining of the first pair of edges).

Correct. Although it is deformed in the sense that the cylinder is bent into 4D and wrapped around back to itself.

We had discussed this object before, and it was said to be the "ridge" or maybe it was "frame" of the whole duocylinder.
(Sounds like such an interesting object. Wish we could visualize true 4D shapes).

It's only the ridge. A 2D surface does not bound a 4D volume. You need a 3D surface to bound a 4D volume. In the case of the duocylinder, the two bounding 3-manifolds are two circular toruses (tori) which meet each other at this ridge.

[Eric B]

Yes, they are exactly equidistant. Welcome to the 3-sphere. Well actually, they are equidistant to the polar POINTS. The polar rings are something else altogether.

Well, the OP did say "two rings that are pointed in different directions that are basically perpendicular were each ring has all its points the same distance from the other ring". That's what I was asking about.
If the poles sweeping out rings will have their every point equidistant from the original ring of the equator.

[...] I take it this "sheet" would be the so-called "true 2-torus" made by joining opposite edges of a square (with the second pair of edges joined together in 4D so you don't stretch/crunch the cylinder formed by the joining of the first pair of edges).

Correct. Although it is deformed in the sense that the cylinder is bent into 4D and wrapped around back to itself.

Yes, though the perimeter of the square (length and orthogonality of sides) will still be the same, though in the familiar "donut" torus, they are stretched.

It's only the ridge. A 2D surface does not bound a 4D volume. You need a 3D surface to bound a 4D volume.

Yes, I remember that was said the last time. Just making sure it was the same object.

In the case of the duocylinder, the two bounding 3-manifolds are two circular toruses (tori) which meet each other at this ridge.

Are both of these the same shape (congruent to each other); just aligned differently?

[quickfur]

[...]Well, the OP did say "two rings that are pointed in different directions that are basically perpendicular were each ring has all its points the same distance from the other ring". That's what I was asking about.
If the poles sweeping out rings will have their every point equidistant from the original ring of the equator.

The distance between the two polar rings is constant, and the distance from each ring to the equatorial torus is also constant.

[...]

In the case of the duocylinder, the two bounding 3-manifolds are two circular toruses (tori) which meet each other at this ridge.

Are both of these the same shape (congruent to each other); just aligned differently?

Yes.