## Spheration.

Discussion of shapes with curves and holes in various dimensions.

### Spheration.

I want to make sure I understand spheration properly, and then I have a few questions about it.

First, I'll enumerate the 4D toratopes in product notation (at least, the product notation that's on the list of rotopes page, which could be nonsense for all I know ):

Tesseract - 1x1x1x1
Glome - 4
Cubinder - 2x1x1
Toracubinder - 2#3
Duocylinder - 2x2
Tiger - (2x2)#2
Spherinder - 3x1
Toraspherinder - 3#2
Torinder - (2#2)x1
Ditorus - (2#2)#2

Now, the first thing I notice is that those numbers don't all add up to 4. If # counted as -1, then everything apart from the tiger would - with the tiger being 5.

So, I'm imagining that the Cartesian product of two circles used in the tiger is really referring to the 2-frame duocylinder, and that 2D surface is being spherated by a circle to produce the 3-net, 4-bounding-space tiger.

Apart from hyperspheres, all closed toratopes must have a spheration operator at the lowest level of precedence (i.e. the "outermost" spheration operator). This is simple to show as all closed toratopes have a group around the entire shape and only hyperspheres have only 1s inside that group. So the dual of a toratope (i.e. turning it inside out) is given by reversing the operands of the outermost spheration operator:

dual toracubinder = 3#2 = toraspherinder
dual toraspherinder = 2#3 = toracubinder
dual tiger = 2#(2x2) = ?
dual ditorus = 2#(2#2) = ?

Now, what is the dual tiger and the dual ditorus? They must exist, since they are given by taking the tiger or ditorus and turning them inside-out (a la this, except in 4D). For the ditorus, if spheration is associative (it certainly isn't commutative), it could be self-dual, since associativity would mean (2#2)#2 = 2#(2#2).

However, I have no idea how to represent the dual tiger, 2#(2x2), in toratopic notation - this shape would be "putting a duocylinder at every point in a circle, oriented perpendicular to the circle". To me, this sounds like the shape would be 5D, but of course we are referring to the 2-frame duocylinder, so the resulting shape would be 3-frame. Then the question is, can the resulting shape be embedded in four dimensions without self intersections?

So, my questions:
1. Is everything I've said above correct?
2. Is the spheration operator associative (and thus the ditorus self-dual)?
3. How do you find A#B where B is not a hypersphere?
4. What is the dual of the tiger? If the tiger is self-dual, then why is 2#(2x2) equivalent to (2x2)#2?
5. If the dual of the tiger isn't a toratope, does this mean the set of toratopes is too restrictive?

Keiji

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### Re: Spheration.

So, I'm imagining that the Cartesian product of two circles used in the tiger is really referring to the 2-frame duocylinder, and that 2D surface is being spherated by a circle to produce the 3-net, 4-bounding-space tiger.

This is right. We always use the minimal frame for spheration.

I'm not sure about the idea of every object having a dual. The torus can be turned inside out by poking a hole in it, but you don't need a hole to deform (31) into (211). If a deformation is the only requirement, then duel isn't a good word for it. This would mean (22) and ((21)1) are duel, but also (221), (32) and ((21)11) are all duals of each other. Having a pair of duals is unique to 4D.

Now as for reversing the order of #, it really can't be done. Spheration uses the normal space, the hodge dual of the tangent space. For a k-frame object in nD, the tangent space is kD and the normal space is n-k D. Now although the tangent space is isomorphic to R^k, there is no particular unique isomorphism. So if we want to spherate by anything that isn't perfectly symmetric, there's no "proper" way to do it. You can only spherate by a sphere or a ball, and it's dimension has to be at least n-k.

2#(2x2) is like 2#11. There are as many ways to do it as the power set of the reals (aleph 2? Not really sure how that works).

PWrong
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### Re: Spheration.

Hmm maybe there's some cases where you can do it. A#B might work if B is the cartesian product of a sufficiently high dimensional sphere with something else. You can put the sphere in the normal space and the rest goes into other dimensions, which have a basis.

For example, (2x2)#(2x1) might work. The circle fits in the normal space of the duocylinder, and the line goes into an extra axis. So this would be a tigerinder.

Actually 2#(1x1) could exist uniquely as well. Maybe. The normal space is 1D, so you can put a line in there. The other line goes into the z axis. So this is a pair of concentric cylinders. The wireframe is four circles, so this shape is similar to a cubinder.

2#(2x1) could work in two ways. You could fit the 1 in the normal space and put the circle in z-w, giving two concentric cylinders cross a circle. Or you could fit the circle in the n-z plane and the line in w, giving a torus cross a circle. Both of these are can be deformed into 221.

(2x2)#(1x1) would never work. You have a normal plane with no standard basis, so no hints to where the square can go. So there are still some truly invalid spherations.

2#(2x2) does work. The circle has a normal axis. You can put part of a circle in there and the rest in the z axis. Then the other circle goes in the w-t plane. So it's torus x circle, or (21)2.

It seems that all valid spherations are homeomorphic to an existing toratope. That makes sense, because spheration is just the cartesian product with some axes moved around.

PWrong
Pentonian

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### Re: Spheration.

Basically, we can define A#(B x C) = (A#B) x C,
where B is a sphere whose dimension is at least n-k.

PWrong
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### Re: Spheration.

PWrong wrote:I'm not sure about the idea of every object having a dual. The torus can be turned inside out by poking a hole in it, but you don't need a hole to deform (31) into (211). If a deformation is the only requirement, then duel isn't a good word for it. This would mean (22) and ((21)1) are duel, but also (221), (32) and ((21)11) are all duals of each other. Having a pair of duals is unique to 4D.

The dual is the shape obtained by performing the "turning inside-out", which is done inside its own bounding space. So hyperspheres are self-dual and so is the torus. The toraspherinder and toracubinder are duals of each other. The dual of the tiger is what you get when you perform this on a tiger and likewise for the ditorus. What I'm asking is what these shapes are. "Dual" is not the same as "homologous". And it may not be always accomplished by reversing the outer spheration; this was a mere observation from some of the toratopes.

If I think about it, 2#(2#2) should be doable. I'm not sure what it is though. I think if you align the torus with its polar axis tangent to the circle you get a ditorus and if you align it perpendicular to the curve you get a tiger.

Keiji

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Joined: Mon Nov 10, 2003 6:33 pm
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### Re: Spheration.

What does it mean to turn something inside out though? Do you have to puncture the object first, like with a torus or a sphere? I'm not entirely convinced that (31) is the dual of (211), although it's plausible.

Maybe we can look at what happens to the [url=fundamental polygon]http://en.wikipedia.org/wiki/Fundamental_polygon[/url] when a shape is turned inside out. That might give some hints as to what happens to the ditorus. Shapes that are homeomorphic have the same fundamental polygon, so tiger and ditorus have the same fundamental polyhedron. I suspect that tiger and ditorus will turn out to be dual to each other.

If I think about it, 2#(2#2) should be doable.

Surely not. At least not with the current definition. You only have a normal axis. What would you put in there?

I think if you align the torus with its polar axis tangent to the circle you get a ditorus

You can't put stuff in the tangent space, points will overlap.

PWrong
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### Re: Spheration.

PWrong wrote:
I think if you align the torus with its polar axis tangent to the circle you get a ditorus

You can't put stuff in the tangent space, points will overlap.

Errm... yeah. What I meant was putting a line perpendicular to the polar axis normal to the circle. In any case, were my two results right?

Keiji

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### Re: Spheration.

Ok, it took me a while to visualise it but I think you're right. Kind of. That is, if there is an answer at all it will be your answer.

I can't work out if the lack of basis causes a problem the way it does with 2#(1x1). Perhaps the torus has sufficient symmetry that you don't need a basis.

PWrong
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