Holes of Toratopes (wendy's version)

Discussion of shapes with curves and holes in various dimensions.

Re: Holes of Toratopes (wendy's version)

Postby wendy » Sun Nov 29, 2009 11:25 am

If one starts off with the notion that a hole is a missing patch (in the fabric of space), it could be quite interesting to count them by working out what patches are missing. It makes some sort of sense by multitopes (vertices, edges etc, without any sort of closure).

We begin with the notion that an object without holes is piecewise construction of spherous patches. Spherous means 'topologically convex', and peicewise means that you add peices of consecutive dimensions, eg vertex+edge, edge+hedron, hedron+choron. The root form is the nulloid+vertex.

One constructs a peicewise multitope, that contains the desired figure, and removes the desired bits and peices either peicewise (which do not create a hole), or singly, which does.

t = teron =4d, c=choron = 3d. h=hedron = 2d, e = latron (edge) = 1d, v = vertex = teelon = 0d, n = wesian = nulloid = -1d.
The products in *# etc include (*) or exclude (#) the leading/trailing element. *# = prism product.

So, for example, we could demonstrate the holes in 5,5/2 (great dodecahedron), by starting with the icosahedron (which is 1c, 20h, 30e, 12v, 1n. We now add in pairs, c+h, formed by dividing a new cell from adding a pentagonal wall. (intersections are not counted here). This gives 13c.32h.30e.12v.1n. We now remove the 12 triangular faces, along with 12c, to give 1c.20h.30e.12v.n. To get the desired result, we need to remove the remaining 8 triangles, with just the 12 pentagons, but no edges etc are removed, so the result 1c.12h.30e.12.n is 8h shy of the peicewise construct, and therefore the genus is 8/2 or 4.

A circle has a hole, because the peicewise construct is n+v (vertex) e+h (fence + enclosure), while the circle is h.e.n. This means that it's missing a teelon or 1d patch. [any point pair can vanish in two different ways].

A sphere is derived from a circle-digon, by taking a circular ring, and adding a hemisphere on each side. Since the first hemisphere is the enclosure above, the second h adds h+c (second face + content). We can now remove h+e, by removing the line between the faces, and the face itself, ie a sphere's hole-less construct is 1c.1h.1v.1n, but the 1v is additional, a sphere also has a teelal (0d) hole in it. [as do all higher spheres].

If we now construct a monogon prism (basically a circle + point), we get:

1h.1e.1v.1n *# 1h.1e.1v.1n = 1t.2c.3h.2e.1v.n.

This basically is a square, wrapped into a cylinder (preserving the seam), and then hooked end to end to make a bi-cylinder, preserving that join. So we get two seams (where there are the two ends of the cylinder), one vertex (where the seeams meet), three hedra (the two circles, amd the square.

The observed result is 1h.1e.1n *# 1h.1e.1n = 1t.2c.1h.0v.n, or a difference of 2h.2e.1v. These are indeed the observed additions needed to be added to the bicylinder to prevent non-vanishing loops forming anywhere in the figure. The h fall in each of the faces (chora) to prevent loops in 3-space, while the 2e prevent loops forming in the margin (it's a torus), the v is where the 2e fall.

We now turn to the torus, constructed by a bimonogon-comb, ie 1h.1e.1v.1n ## 1h.1e.1v.1n = 1c.1h.2e.1v.1n. We see that the comb-product is a repetition of surface, and does not carry across any holes in the solid. However, we can use the 4d case, and flatten it out, by way of deleting 1t.1c (by stretching the surface out into a plane, and removing the content, and one choron (the "outside"). So we get 1c.3h.2e.1v.n. This is indeed the correct result as before.

The circle-sphere and sphere-circle combs (23 and 32 holes), can be formed from the simple 5d prism, by removing p+t on the stretch. So we get:

circle *# sphere = 1.1.1.1 * 1.1.0.1.1 = 1.2.2.2.1.1. Eliminating the 5d element, we get 1.2.2.1.1.,
The dream you dream alone is only a dream
the dream we dream together is reality.
User avatar
wendy
Pentonian
 
Posts: 1521
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: Holes of Toratopes (wendy's version)

Postby PWrong » Mon Nov 30, 2009 5:59 am

If one starts off with the notion that a hole is a missing patch (in the fabric of space)

To say that a torus is missing a patch seems to imply that you could put the patch back where it belongs, and the torus would be whole. What would that look like?

We begin with the notion that an object without holes is piecewise construction of spherous patches. Spherous means 'topologically convex', and peicewise means that you add peices of consecutive dimensions, eg vertex+edge, edge+hedron, hedron+choron. The root form is the nulloid+vertex.


As an example, do you mean the way a cube is the union of 8 points, 12 open line segments, 6 open squares and an open solid cube?

It also sounds a bit like the definition of a manifold as a topological space with a collection of "charts" (maps from a patch of the space to R^n). But this also applies to objects with holes.

I'm curious as to how something can be "topologically convex" though. A shape is convex if the line segment between any two points is contained within the shape. But a convex shape can easily be deformed into a shape that isn't convex, so convex shouldn't have any meaning in topology.
User avatar
PWrong
Pentonian
 
Posts: 1539
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Holes of Toratopes (wendy's version)

Postby wendy » Mon Nov 30, 2009 7:11 am

The notion of "topologically convex" allows for things like dints, inward curves etc, but not holes. Basically, it means that any set of lines drawn between two points, can be continuously merged into one, without crossing the surface. The term in the PG is 'spherous'.

If a figure has holes in it, than one might suppose that the holes can be enumerated, and "filled in". So we need a definition of a figure without evident holes in it. Since I have already experimented with the idea of "multitopes", one might suppose that the simplest multitope that contains all of the elements of the desired figure, and is piecewise constructible, should be the simplest figure. Removing an unmatched set of elements then provides the information of holes.

The patching of the torus would be to add a circle to stretch across the hole, a second one to stretch across the tube. The intersection of these new circles creates two edges and a point, the whole is a figure for which no space of any dimension admits non-vanishing loops of any kind.
The dream you dream alone is only a dream
the dream we dream together is reality.
User avatar
wendy
Pentonian
 
Posts: 1521
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: Holes of Toratopes (wendy's version)

Postby PWrong » Mon Nov 30, 2009 8:00 am

The notion of "topologically convex" allows for things like dints, inward curves etc, but not holes. Basically, it means that any set of lines drawn between two points, can be continuously merged into one, without crossing the surface. The term in the PG is 'spherous'.

Ah ok. Basically anything that can be deformed into a ball then.

The patching of the torus would be to add a circle to stretch across the hole, a second one to stretch across the tube. The intersection of these new circles creates two edges and a point, the whole is a figure for which no space of any dimension admits non-vanishing loops of any kind.


Hey, this is cool. This patching still leaves a "pocket". In fact the resulting shape can be deformed into a sphere. You could fill it in completely by adding a ball on the inside. This fits perfectly with the homology groups of the torus.

Torus has homology 1,2,1
Add two disks and we get 1,0,1
Add a ball and we get 1,0,0, the same homology as a point.


I'll try (211), or circle # sphere.

(211) has homology 1,1,1,1
Add a ball in the "tube", and a disk in the hole. The resulting shape has homology 1,0,0,1.
This still leaves a pocket. Add a solid glome and we get 1,0,0,0. :D
User avatar
PWrong
Pentonian
 
Posts: 1539
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Holes of Toratopes (wendy's version)

Postby Keiji » Mon Nov 30, 2009 8:39 am

Add a sphere in the "tube"


Wouldn't you be adding a ball here, not a sphere?

In any case, you're right, this is cool :P
User avatar
Keiji
Administrator
 
Posts: 1922
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Re: Holes of Toratopes (wendy's version)

Postby PWrong » Mon Nov 30, 2009 8:44 am

Yep. Just fixed it.
User avatar
PWrong
Pentonian
 
Posts: 1539
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Holes of Toratopes (wendy's version)

Postby Keiji » Mon Nov 30, 2009 10:48 am

I see a little ambiguity here, though...

Hyperballs (point, disc, ball, solid glome, etc) all have [1] as their hole-sequence. When you add a hyperball to a shape you subtract the hole-sequence of the corresponding hypersphere from the hole-sequence of the shape. But couldn't you just deform it so you only needed to add a hyperball of lesser dimension?

In any case, I can try G2: adding two discs makes it into a torus, so it's [1, 4, 1].
3-frame G2 can be reached by adding two discs and a ball, giving [1, 2]. This is equivalent to the wedge sum of two circles.
User avatar
Keiji
Administrator
 
Posts: 1922
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Re: Holes of Toratopes (wendy's version)

Postby PWrong » Mon Nov 30, 2009 12:08 pm

Hyperballs (point, disc, ball, solid glome, etc) all have [1] as their hole-sequence. When you add a hyperball to a shape you subtract the hole-sequence of the corresponding hypersphere from the hole-sequence of the shape. But couldn't you just deform it so you only needed to add a hyperball of lesser dimension?

I don't think so. Take the torus, let the larger radius be 1, and call the smaller radius r. The idea is to make r so close to 1 that instead of adding a disk you can just add a point. If r = 1, it's not really a torus anymore, because it's already connected in the middle. If r < 1, you still need a disk to patch it. There's nowhere in between where you can just add a point. You could try pinching it so that you only had to add a line, but that would have the same problem.

In any case, I can try G2: adding two discs makes it into a torus, so it's [1, 4, 1].

What's G2?
User avatar
PWrong
Pentonian
 
Posts: 1539
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Holes of Toratopes (wendy's version)

Postby Keiji » Mon Nov 30, 2009 12:55 pm

PWrong wrote:I don't think so. Take the torus, let the larger radius be 1, and call the smaller radius r. The idea is to make r so close to 1 that instead of adding a disk you can just add a point. If r = 1, it's not really a torus anymore, because it's already connected in the middle. If r < 1, you still need a disk to patch it. There's nowhere in between where you can just add a point. You could try pinching it so that you only had to add a line, but that would have the same problem.


Alright, that makes sense now.

What's G2?


Take two torii and glue them together. Then remove the face separating their insides.

Similarly, for G3 you use three torii and remove two faces. And so forth.
User avatar
Keiji
Administrator
 
Posts: 1922
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Re: Holes of Toratopes (wendy's version)

Postby PWrong » Mon Nov 30, 2009 1:14 pm

Oh, so the G stands for genus?

In any case, I can try G2: adding two discs makes it into a torus, so it's [1, 4, 1].

[1, 4, 1] is right, but doesn't this depend on where you put the disks? There are 4 places they can go, 5 if you count the part of the tube between the torii.

Filling in the hole and the tube on the same side would give you a torus.
If you fill in both the holes, you get something equivalent to a sphere with two poles.
Filling in the part of the tube in the middle separates the shape so you get a wedge sum of two torii. This has [1,4,2], so the original shape must have been [1,5,2].

There's something weird going on here.
User avatar
PWrong
Pentonian
 
Posts: 1539
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Holes of Toratopes (wendy's version)

Postby Keiji » Mon Nov 30, 2009 2:07 pm

PWrong wrote:Filling in the hole and the tube on the same side would give you a torus.


which is [1, 2, 1], so the original shape was [1, 4, 1].

If you fill in both the holes, you get something equivalent to a sphere with two poles.


which I'll call shape A. And if you fill in two parts of the pocket to make it a negated ball, you get the toratopic dual of shape A. Then if, from this dual, you fill in the two outer holes, you get a sphere, which is [1, 0, 1], so this is consistent.

Filling in the part of the tube in the middle separates the shape so you get a wedge sum of two torii. This has [1,4,2], so the original shape must have been [1,5,2].


There are actually two ways to do this. If you fill in the part of the tube in the middle with a disc running in a plane intersecting the outer holes, you get something I'll call shape B. If from shape B you fill in one of the outer holes, you get a torus [1, 2, 1]. So the original shape was [1, 4, 1]. This is consistent.

If you fill it in with a disc in a plane running between the outer holes, you get something I'll call shape C. Adding four discs to C gives... the wedge sum of two spheres. Which is [1, 0, 2]. Hmm, you're right, this would suggest the original shape was [1, 5, 2]. Where's the conflict? :\
User avatar
Keiji
Administrator
 
Posts: 1922
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Re: Holes of Toratopes (wendy's version)

Postby PWrong » Mon Nov 30, 2009 4:24 pm

Hmm, look at what happens when you put a disk inside a sphere.
User avatar
PWrong
Pentonian
 
Posts: 1539
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Holes of Toratopes (wendy's version)

Postby Keiji » Mon Nov 30, 2009 8:31 pm

...ouch.

Any idea how to resolve this? I hope the entire system isn't worthless now...
User avatar
Keiji
Administrator
 
Posts: 1922
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Re: Holes of Toratopes (wendy's version)

Postby PWrong » Tue Dec 01, 2009 8:36 am

Well this means thr we can't put disks wherever we want. We might be able to get a conjecture about the minimum number of balls it takes to reduce the shape to a point. So it doesn't matter if you put a disk in a sphere, because it's easier to fill it with a ball.
User avatar
PWrong
Pentonian
 
Posts: 1539
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Holes of Toratopes (wendy's version)

Postby PWrong » Sat Dec 05, 2009 10:29 am

Something I posted in wendy's other thread is relevant here.

As for the missing patch theory, I think I can recover it. Our problem was with things like adding a disk to the inside of a sphere. The difference between that and adding a disk to the inside of a torus is the following. You can take the disk and deform it, without moving the boundary, until it lies entirely on the sphere. You can't do that with the patches we were talking about on the torus. Thus, I claim it doesn't count. With this restriction, I think the patch method will work as a way to count homology groups, at least for toratopes.
User avatar
PWrong
Pentonian
 
Posts: 1539
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Holes of Toratopes (wendy's version)

Postby PWrong » Sat Dec 05, 2009 10:40 am

In fact I think I could eventually prove the missing patch method is equivalent to homology. If the shape is A and the patch is B, then AnB is some hypersphere and we can use Mayer-Vietoris on X = AuB.
I've just done it for the patch in the middle of a torus and I got HX = [1,1,1] as expected.
User avatar
PWrong
Pentonian
 
Posts: 1539
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Holes of Toratopes (wendy's version)

Postby PWrong » Sun Dec 06, 2009 3:20 am

Yep, it's easy to prove. Suppose B is a k+1D ball, then AnB is a k-sphere. It's easy to show that HqX = HqA except when q=k.

For q=k we have

Hk+1X --> Z --> HkA ⊕ 0 --> HkX

Now if the patch is placed just right, then the map in the middle is
a |--> (0,...,0,a,0,...,0)

The position of a depends on where the patch is. However if the patch is in the wrong place, e.g. a disk in a sphere, then the map will be the zero map (because the circle can be shrunk to a point).

The new sequence is then

Hk+1X -->> 0 --> HkA /Z >--> HkX -->> 0

HkX = HkA / Z as required.
User avatar
PWrong
Pentonian
 
Posts: 1539
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Previous

Return to Toratopes

Who is online

Users browsing this forum: No registered users and 1 guest

cron