Counting a toratope's degrees of freedom

Discussion of shapes with curves and holes in various dimensions.

Counting a toratope's degrees of freedom

Postby mr_e_man » Thu Sep 20, 2018 6:47 pm

Any toratope is finite and symmetric, so it has a definite centre point. In n-space, this point has n translational degrees of freedom. From now on, we can assume that the toratope is centred at the origin.

Each radius of a toratope is a free parameter, so each pair of parentheses () adds one DOF. Simple enough.

What about orientation?

In 2D, all circles (II) have the same orientation (0 DOF). Same with any (n-1)-sphere (III), (IIII), etc.

In 3D, the orientation of a torus ((II)I) is determined by its core circle (II)I, or by the plane of that circle. This can be described with a bivector. Bivectors in 3D have 3 DOF, but the magnitude will be ignored, leaving 2 DOF.

In 4D, the orientation of a ((II)II) is again determined by the core circle (II)II or its plane. General bivectors in 4D have 6 DOF, but one representing a plane must be simple; its wedge product with itself must be B^B=0. This is a quadvector equation, which in 4D is equivalent to a scalar equation, only removing 1 DOF. Thus, simple bivectors have 5 DOF. Again, we don't care about the magnitude of B, so a plane's orientation has 4 DOF.

The orientation of a ((III)I) depends on the 3D subspace of its core sphere (III)I. This is represented by a trivector T, which in 4D is equivalent to a vector, having 4 DOF. Ignoring the magnitude of T, this leaves 3 DOF for the core's orientation.

The orientation of a (((II)I)I) must be studied in two parts, the circle (II)II, and the torus ((II)I)I. We'll leave this for later.

In general, a k-dimensional subspace of n-space is represented by a k-blade B, a special type of grade k multivector. It's special in that its square BB must be a scalar. Its magnitude has no effect on orientation, so we can require its square to be exactly 1 or -1. The expression BB has components of grade 0,2,4,...,2k-2,2k, each of which can be further decomposed into (n choose j) scalar components. So the multivector equation BB=1 is decomposed into this many different scalar equations:

sum_{0<=j<=2k, j even} (n choose j) = 1 + n(n-1)/2! + n(n-1)(n-2)(n-3)/4! + ... + (n choose 2k)

The problem is, some of these equations are redundant, and don't remove a DOF. For example, with n=3 and k=1 (ordinary vector v), the above expression is 1+3; the first equation would say v is a unit vector (|vv|=1), but the last three would say the bivector part of vv is 0, which was already true for any vector.
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Re: Counting a toratope's degrees of freedom

Postby mr_e_man » Tue Sep 25, 2018 6:29 am

Okay, I looked it up (after several failed solution attempts :sweatdrop: ):

https://mathoverflow.net/questions/1030 ... pace-of-rd

A k-dimensional subspace has (n-k)k degrees of freedom. Here's a table:

Code: Select all
n  k ||  0  |  1  |  2  |  3  |  4  |  5  |  6  |  7  |  8
=====||=====================================================
0    ||  0  |  -  |  -  |  -  |  -  |  -  |  -  |  -  |  -
1    ||  0  |  0  |  -  |  -  |  -  |  -  |  -  |  -  |  -
2    ||  0  |  1  |  0  |  -  |  -  |  -  |  -  |  -  |  -
3    ||  0  |  2  |  2  |  0  |  -  |  -  |  -  |  -  |  -
4    ||  0  |  3  |  4  |  3  |  0  |  -  |  -  |  -  |  -
5    ||  0  |  4  |  6  |  6  |  4  |  0  |  -  |  -  |  -
6    ||  0  |  5  |  8  |  9  |  8  |  5  |  0  |  -  |  -
7    ||  0  |  6  |  10 |  12 |  12 |  10 |  6  |  0  |  -
8    ||  0  |  7  |  12 |  15 |  16 |  15 |  12 |  7  |  0


Now, the (((II)I)I) has a core torus ((II)I)I, which is represented as ((II)I) in its own 3D subspace. This subspace has 3 DOF for orientation. Within that 3-space, the torus' core circle (II)I has 2 DOF. Thus the total is 5.

(I don't want to call these toratopes by name. :\ The "ditorus" is actually a 3-torus. The "spheritorus" would better be called a "sphericircle" or something. "Toraglominder" should be simply "toraglome". "Tiger" is good; it's short, and there's no potential for confusion. :) )

The tiger ((II)(II)) has a Clifford torus as its core. How many DOF does that have? Obviously, the first circle uniquely determines the other one (by orthogonality), so a first guess is 4 DOF. Another way to obtain 4 is to consider the torus' stereographic projection into 3D; it has 2 DOF for orientation of the core circle, and 2 DOF for translation within the 3-sphere. (The 3rd translation direction, the torus' axis, would cycle the torus along itself, not moving the object as a whole.)

:?: There ought to be a more reliable way to calculate these. Any suggestions?

Anyway, here's what we have so far (with a few obvious additions):

Code: Select all
n  || toratope's orientation DOF
===||============================
0  || :0
1  || I:0
2  || II:0, (II):0
3  || III:0, (III):0, (II)I:2, ((II)I):2
4  || IIII:0, (IIII):0, (III)I:3, ((III)I):3, (II)II:4, ((II)II):4, ((II)I)I:5, (((II)I)I):5, (II)(II):4, ((II)(II)):4
5  || IIIII:0, (IIIII):0, (IIII)I:4, ((IIII)I):4, (III)II:6, ((III)II):6, ((III)I)I:7, (((III)I)I):7, (II)III:6, ((II)III):6, ((II)II)I:8, (((II)II)I):8, ((II)I)II:8, (((II)I)II):8, (((II)I)I)I:9, ((((II)I)I)I):9, (II)(II)I:8, ((II)(II)I):8, ((II)(II))I:8, (((II)(II))I):8, (III)(II):6, ((III)(II)):6, ((II)I)(II):8, (((II)I)(II)):8


(See my first post on the site for the intended meaning of the notation; II is not a square.)

Actually, I've found the pattern in these calculations, and could continue indefinitely. But I'm still not confident that it's correct.

ICN5D, are you sure about your name? ;) Can you verify these?
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Re: Counting a toratope's degrees of freedom

Postby ICN5D » Sun Sep 30, 2018 8:57 pm

Well, I think you're a step ahead of me on seeing the general pattern here. I'm still not completely sure what a degree of freedom is. Can you provide some basic visual aids? Multivectors and blades and wedge products are a bit out of my comfort zone. Are you talking about the number of unique orientations an n-D object has in n-D space? I think so, but some of your findings aren't clear in the mind to me (like (((II)I)I) having 5 DOF). How does that work?
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Re: Counting a toratope's degrees of freedom

Postby wendy » Mon Oct 01, 2018 12:16 pm

It really does matter what is implied by DOF. The symbol ((II)(II)) is a general class that has seven degrees of freedom. But this includes three of size and aspect, which means that there are three degrees of freedom creating a particular example from the symbol.

The 'tiger' is a member of a class that has seven degrees of freedom. Specifically, it's a 'C-spherated A-glomolatric B-glomolatric prism', or to use an RPN style of name, a A-glomohedrix B-glomohedrix duoprism C-spherate. The tiger has A=B>C. So its seven degrees of freedom are three aspect and four of orientation.

I'm still not exactly sure if the tiger is a "tri=latric comb". Topologically, these are identical, but i really can't see a construction of the tiger as a tri-comb. In four dimensions, the topological equivalence of the surface does not imply topological equivalance of the contained solid. A sphere-circle comb and a circle-sphere comb are topologically different creatures, with a topologically identical surface.

The class of tri-circular combs might well have more degrees of freedom than the root class of which the tiger is a member.
mr_e_man wrote:It's special in that its square BB must be a scalar. Its magnitude has no effect on orientation, so we can require its square to be exactly 1 or -1.


The main difference between what we do here and the field of mathematics is that magnitude does count. It's a separate degree of freedom. The aspects also count, this means that the general rectangle x2x has two degrees of freedom, which might be length and bredth, or size and aspect. The DOF is enumerated against the symbol, and spent against the figure. x3x5x has three degrees of freedom, of which the seven uniform polyhedra that come from it are from x=0 or 1 freely.

It is less than apparent about the modes of rotation if you do not enumerate magnitude as a free variable. So the tiger is governed by a circle, for which there are six degrees of freedom. This is because a circle maps onto a teelix on a bi-glomohedric prism, which is a 4d fabric (terix) in six dimensions (ectix). Note here that the teelix is a point-pair when the line is drawn through the centre, because the prism is actually representing great arrows in six dimensions, not great circles.

(I don't want to call these toratopes by name. :\ The "ditorus" is actually a 3-torus. The "spheritorus" would better be called a "sphericircle" or something. "Toraglominder" should be simply "toraglome". "Tiger" is good; it's short, and there's no potential for confusion. :) )


3-torus is actually confusing. Of course, the tiger is an example of the general swirl-torus. These are in essence where one follows a series of clifford-parallels, the projection on the 'lattitude-sphere' represent different points, say the vertices of an icosahedron or dodecahedron. The tiger is a developed bi-polar swirl-prism. It is only in this sense, where the C-circle is pondered against the A and B circles, that the tiger is a tri-glomolatral comb.

Really, too, you should not go vandalising ICN5D's notation, without running it past him first. This causes a lot of confusion also.
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Re: Counting a toratope's degrees of freedom

Postby mr_e_man » Mon Oct 01, 2018 5:43 pm

ICN5D wrote: Multivectors and blades and wedge products are a bit out of my comfort zone.


wendy wrote:
mr_e_man wrote:It's special in that its square BB must be a scalar. Its magnitude has no effect on orientation, so we can require its square to be exactly 1 or -1.


The main difference between what we do here and the field of mathematics is that magnitude does count. It's a separate degree of freedom.


The bivectors e1^e2 and -5e1^e2 both represent the xy-plane. That's what I meant by "magnitude doesn't matter".

But that was part of a "failed attempt" at counting DOF for orientation of a subspace; it's not necessary for understanding my second post on this page, so you can ignore it.

ICN5D wrote: I'm still not completely sure what a degree of freedom is.


It's basically a free parameter / variable, or a "generalized dimension".

A circle in 2D (II) has a total of 3 degrees of freedom. Its centre has 2 coordinates xc,yc, and it has a radius a.

A circle in 3D (II)I has 6 degrees of freedom. Its centre has 3 coordinates xc,yc,zc, and it has a radius a. The circle's axis points in a certain direction, which can be described with 2 angles (like latitude and longitude).

A torus in 3D ((II)I) has 7 degrees of freedom. In addition to the 3 position coordinates and 2 orientation angles, it has 2 radii a and b.

wendy wrote: It really does matter what is implied by DOF. The symbol ((II)(II)) is a general class that has seven degrees of freedom. But this includes three of size and aspect, which means that there are three degrees of freedom creating a particular example from the symbol.

The 'tiger' is a member of a class that has seven degrees of freedom. Specifically, it's a 'C-spherated A-glomolatric B-glomolatric prism', or to use an RPN style of name, a A-glomohedrix B-glomohedrix duoprism C-spherate. The tiger has A=B>C. So its seven degrees of freedom are three aspect and four of orientation.


Yes. The size and aspect are determined by its 3 different radii (one for each pair of parentheses). I'm also counting position DOF, which is obviously 4 in 4D, so a general tiger has 7+4 = 11 DOF.

...But you said A = B. Would that mean it has only 2 aspect/size DOF? I don't want to restrict tigers to equal radii.

ICN5D wrote: Well, I think you're a step ahead of me on seeing the general pattern here.


In addition to my MO link, you can understand the DOF for a subspace in this way: A k-dimensional subspace of an n-dimensional vector space is spanned by k different vectors, each of which has n components. So this set of vectors has nk degrees of freedom. But a different set of vectors would determine the same subspace. Any of the vectors can be rotated or scaled within the k-dimensional subspace (as long as they remain linearly independent), so we subtract k DOF for each of the k vectors: nk - kk = (n - k)k.

As for the pattern in the toratopes, I think it's exemplified below my (n-k)k table. If you still don't understand it, I could show more examples of how I derived the 5D toratopes' DOF.

wendy wrote:
(I don't want to call these toratopes by name. :\ The "ditorus" is actually a 3-torus. The "spheritorus" would better be called a "sphericircle" or something. "Toraglominder" should be simply "toraglome". "Tiger" is good; it's short, and there's no potential for confusion. :) )


3-torus is actually confusing.


It's called a 3-torus because it's topologically (but not geometrically) a Cartesian product of 3 circles. "Ditorus" to me suggests a product of 2 circles. (Or maybe a double torus, a connected sum of 2 tori.)

Really, too, you should not go vandalising ICN5D's notation, without running it past him first. This causes a lot of confusion also.


ICN5D, what do you think of this? viewtopic.php?f=24&t=1994&start=30#p26262
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Re: Counting a toratope's degrees of freedom

Postby wendy » Tue Oct 02, 2018 7:34 am

A number in front of a name is usually read as a figure in that dimension, so a 3-cube is a cube in 3d, and an 8-cube is a cube in eight dimensions. So it could as easily be seen as a spherated circle in 4d (ie sphere-circle comb).

On the other hand, we already have a term representing the 'cartesian product of the surface' (ie comb). The comb product is not communitive, that is AB <> BA. The surface none the same retains the same topology. In essence, one can form tube of 1d, into a circle either by connecting the ends of a hose, or by rolling the fabric down as one might take off a sock. If the base of a cylinder is a sphere, then the hose makes a sphere-circle comb, which would contain a circle but not a sphere, while the sock move makes a circle-sphere comb, which would cover the hollow 3-sphere in 4d.

Whereas the mathematicians have only one product, we have five. There are five different ways to topologically multiply two solids. These represent the five regular solids in all dimensions. A good deal of the terminology i wrote has to deal with the short-comings of Coxeter's "Regular Polytopes" and "Regular Complex Polytopes".

If you are supposing that a torus can be represented by a series of nested subspaces, then your formula ought give some clue on where to start looking. For example, the ((III)(III)) consists of two orthogonal spheres in 6-space, so you have 9 degrees of freedom, and because there are three sets of brackects, three of aspect, all together, 12 DOF apart from position. One can in part, regard the () as a kind of spheration around the shell of the previous, or an axial array.
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Re: Counting a toratope's degrees of freedom

Postby ICN5D » Mon Nov 05, 2018 6:02 pm

Ah, I see how it works now, mostly. The position and radius DOF are easy to count, but the rotational DOF seems a bit tricky to see.

For instance, does the clifford torus in R^4 have any rotational DOF? A circle on a 2D plane has no rotational DOF. A clifford torus is a product of two circles embedded in a product of two 2D planes, so it seems like there are also none. At least that's my current reasoning.
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Re: Counting a toratope's degrees of freedom

Postby mr_e_man » Mon Nov 26, 2018 10:35 am

The Clifford torus, a product of two circles, does have rotational freedom. Each circle is fixed in its plane, but the planes can rotate together. A 2-plane in 4D has (n-k)k = (4-2)*2 = 2*2 = 4 rotational DOF. I did try to explain this in my second post.

The 5D toratope (((III)I)I) has the same rotational freedom as its core ((III)I)I, which sits in a 4D subspace. The subspace itself has (5-4)*4 = 4 DOF, and a ((III)I) rotating within that subspace has 3 DOF (determined previously; or, (4-3)*3 = 3). Add these to get 7 DOF for (((III)I)I).

Any toratope can be broken down by this pattern. As wendy noted, I'm considering toratopes as series of nested subspaces.

(((III)(II)II)II): 7*2 + 5*2 + 3*2 = 30.
((II)(II)(II)): 4*2 + 2*2 = 12.
((IIII)III): 4*3 = 12.
((((II)I)II)I): 5*1 + 3*2 + 2*1 = 13.
((II)I): 2*1 = 2.
(III): 3*0 = 0.
(((II)I)(II)): 3*2 + 2*1 = 8.

:nod:
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