Visualizing 3-hyperplanes

Ideas about how a world with more than three spatial dimensions would work - what laws of physics would be needed, how things would be built, how people would do things and so on.

Re: Visualizing 3-hyperplanes

Postby quickfur » Wed Jan 29, 2014 11:47 pm

ICN5D wrote:Yes, this would work. An interesting algorithm would be one that computes a degenerate height based on the combination of operators in unit edge lacing.

What is the general effect on n-simplices, as n grows? What happens to the clustering of vertices? Where is the volume concentrated?

I haven't studied the n-simplex enough to say for sure, but from the Wikipedia volume formula, it would appear that as n increases the total bulk enclosed inside an n-simplex decreases, and tends towards 0 as n approaches infinity. That is, the higher the dimension, the less volume the n-simplex occupies. Also, its dihedral angle approaches 90° as n approaches infinity, so the simplex becomes more and more "square-like" in very high dimensions.

The topological method is the original rubric in my notation, in a tesselationist method. The unit edges are a new thing, and interesting as well.

I think I'm going to include a subscript for the axis in motion with the spin operator. So, the |O>|Ow means to hold plane XYZ stationary, allowing the cone endcaps stretched along W to rotate in a non-bisecting manner into V. This will clear up all of the "prismic, triangular, toric" ambiguity. Does that work better?

If you're going to be using subscripts, I highly recommend using numerical subscripts instead of letters, because (1) our unfortunate convention of starting at X, Y, Z means that in 4D and beyond we have to do an ugly jump back to W then count backwards, which is very jarring. And (2) we run out of letters at 26D. What if we wanted to describe a 27D object?

I vote for using numerical subscripts. It also makes the notation more self-consistent, since ||||O4 tells you exactly which symbol corresponds with the axis that will participate in the rotation, as opposed to ||||Ow, which requires mental effort to determine which symbol it might correspond with. (In this case it doesn't matter, but in some cases it does.)

This should allow the cone atop cone and line atop cylinder projections to work as one, and create the duocylinder-circle convex hull.

:nod: Excellent idea! :nod:
quickfur
Pentonian
 
Posts: 2935
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Visualizing 3-hyperplanes

Postby quickfur » Thu Jan 30, 2014 12:21 am

@ICN5D: btw, I don't know if you're aware, but some years ago (2008) I invented a notation that generates n-dimensional shapes just by extrusion and tapering alone. This somewhat overlaps with your notation, but it also has its own quirks, in that it doesn't actually include rotation, but allows cones and such to be constructed by using non-linear tapering, and also includes bipyramids. Because of this "fake rotation", it generates a bunch of crind-like shapes instead of "true" rotated/lathed shapes like the duocylinder. For example, applying the O operator to a triangular prism produces a 4D shape with 5 curved surfaces with lune-like shapes: 2 triangular lunes and 3 square lunes.

This may represent another direction for exploration. ;)
quickfur
Pentonian
 
Posts: 2935
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Visualizing 3-hyperplanes

Postby Polyhedron Dude » Thu Jan 30, 2014 5:58 am

I did this on Stella 4D - it is the 11-cell unfolded, the cells themselves are also "unfolded" into true ikes - so each ike needs to be imagined as though two opposite faces are one and two opposite vertices are one, same with edges.

Image
Whale Kumtu Dedge Ungol.
Polyhedron Dude
Trionian
 
Posts: 196
Joined: Sat Nov 08, 2003 7:02 am
Location: Texas

Re: Visualizing 3-hyperplanes

Postby ICN5D » Thu Jan 30, 2014 6:16 am

Oh, wow! It has some wild symmetry. It's got four branching off behind it, and three rows emanating towards. So, the 57-cell is just a 57 room version of the strange effects? My question is, is the inverted matter turned into antimatter? Would you really want to take the inverted dollar? Kaboom, perhaps?
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby ICN5D » Thu Jan 30, 2014 6:48 am

quickfur wrote:@ICN5D: btw, I don't know if you're aware, but some years ago (2008) I invented a notation that generates n-dimensional shapes just by extrusion and tapering alone. This somewhat overlaps with your notation, but it also has its own quirks, in that it doesn't actually include rotation, but allows cones and such to be constructed by using non-linear tapering, and also includes bipyramids. Because of this "fake rotation", it generates a bunch of crind-like shapes instead of "true" rotated/lathed shapes like the duocylinder. For example, applying the O operator to a triangular prism produces a 4D shape with 5 curved surfaces with lune-like shapes: 2 triangular lunes and 3 square lunes.

This may represent another direction for exploration. ;)



No, I didn't know that. That's actually pretty cool. I started working on my notation around the same time. In something like this, it's probably rare to come across a like-minded individual. If you're interested in them, I put a whole bunch of those computation tables up, in the "Cool Tricks ...." thread. Also, I just read about some of your contributions to this website, by discovering new unit edge shapes. That looks like some pretty math heavy stuff. None of which I am familiar with. But, I happen to have some good pattern recognition skills, and I really just documented what I saw happening in the projections. I sort of created a way to describe what I was seeing.

As you can tell, I like to use a surtope pairing method, and I build up shapes or lace them together with those tables. They're more like matrices, and you augment the operator matrix to the surtopes. With lathing, it's easier to tell which surtopes are bisected by the stationary plane, in the pairing method. I'm still discovering things about the commuting spin, and working on a toratope cut algorithm. I'd have to say that probably the coolest part was adapting cartesian products to the system, because it provides an additional way to derive the surtopes, like that crazy Conindric Trianglinder. It's a good method for double checking the math. Kind of like a triad relationship, the three algorithms support each other, and offer new insights. I'm still learning the real vocabulary, I've created words for just about everything, and am still trying to get away from it.

Well, that's good that you're interested in the lathing method. Like I said, I've been working on that one for a while! I'd be happy to help in any way I can :)


--Philip
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby wendy » Thu Jan 30, 2014 8:16 am

The 11-cell by my reading belongs to a large collection of figures called 'cyclic maps'. The simplest one of these is a digon, representing the black and white squares of a chess-board. One has similar sorts of things like x5o5oA5p, where p is a petrie-node. This is a section of a {5,5}, where every petrie polygon on it is 5 edges long. It has 66 hedra, 165 (dec) edges, and 66 vertices.

The 11-cell is an 11-fold colouring of the {3,5,3}, such that opposite each cell is a cell of the same colour as the one you're in. It sounds interesting, and is worth the look. In any case, the are known to be many hundreds of these sorts of things, some going into the hundred-thousands of cells.

In some way, it's rather like mapping the gaussian integers onto {4,4}, and then making different colourings as the remainder divided by 2+i or 3-2i or whatever. You get 5 or 13 {4,4}.

Unlike the euclidean groups, the 11-cell tells us that you can't sequence the 11 colours. In 10 dimensions, the group A_10 can be presented as a tiling of the omnitruncated simplex, (ie all nodes marked), and there the numbering of cells produces lines that step in multiples of x modulo 11, eg 0, 5, 10, 4, 9, 3, ... The {3,5,3} tells us that if you step through opposite faces of the icosahedron, you end up at the place you start from, eg 0,5,0,5, or 0,8,0,8 or 5,8,5,8 or etc.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: Visualizing 3-hyperplanes

Postby ac2000 » Thu Jan 30, 2014 4:34 pm

Klitzing wrote:Just to provide you some feeling on that mentioned Clifford stuff:
[...]
So far that might look kind of arbitrary. But there is one specific type of polytopal figures within 4D, which just bow to that special Clifford symmetry! Consider some n-gon, centered at the origin within the x-/y-plane. Consider then some m-gon, centered at the origin within the z-/w-plane. Then build the direct sum (cartesian product) of those polygons, and you will get the n,m-duoprism, a figure being bounded by m cells, being n-gonal prisms, plus n cells, being m-gonal prisms. Btw. the tesseract here is just the special case, where n=m=4. And obviously, that 4D polytope has an isometry which applies any n-fold rotation within the one direction (2D subspace) and simultanuously any m-fold rotation within the other!
--- rk


Oh dear, I was a bit slow to answer and keep track of all the posts here.

Thank you Mr Klitzing for you helpful explanations on the rotations!
I also didn't know that these duoprisms could be so easily constructed, so that just the number of sides becomes the number of cells of the other one, and vice versa. I had a look at the Wiki-Page about those duoprisms and also on quickfur's homepage there's a section on duoprisms, some of them rotating. Very interesting , indeed!
ac2000
Dionian
 
Posts: 28
Joined: Sat Mar 12, 2011 7:15 am
Location: Berlin, Germany

Re: Visualizing 3-hyperplanes

Postby Polyhedron Dude » Fri Jan 31, 2014 4:45 am

ICN5D wrote:Oh, wow! It has some wild symmetry. It's got four branching off behind it, and three rows emanating towards. So, the 57-cell is just a 57 room version of the strange effects? My question is, is the inverted matter turned into antimatter? Would you really want to take the inverted dollar? Kaboom, perhaps?


Yes the 57-cell will appear to look like 57 dodecahedral rooms with that weird mirror-you in the opposite location. Very likely that the mirror-you would be made of antimatter, so in reality this would be a very dangerous place, for even the air itself would cause matter-antimatter collisions! Although for geometric purposes, we could suppose there is some way to mirror-image the effects without changing matter to antimatter to avoid complete annihilation of the entire structure.
Whale Kumtu Dedge Ungol.
Polyhedron Dude
Trionian
 
Posts: 196
Joined: Sat Nov 08, 2003 7:02 am
Location: Texas

Re: Visualizing 3-hyperplanes

Postby ICN5D » Fri Jan 31, 2014 6:12 am

That would be a good idea!

Which reminds me of an old idea I've had about creating an antimatter converter by use of a Hypermobius Conveyor Belt. If one were able to snip a strip of space and give it a half-twist, then sew it back in, it would create a structure of space that inverts the matter-wave. This could be used to generate lots of antimatter very easily. Of course, this thing would be an indestructible structure of space, not really made out of anything, and potentially the worst threat to the entire universe. It would travel the universe and land on planets or whatever and blow everything up. So, good thing it's probably impossible. At least, if it required a destructible machine that created the twist, we'd be okay.

Another idea I've always had was an actual perpetual motion machine using wormholes. Assuming these wormholes don't require gravity to exist, they could be used to power an infinite energy generator. Let's say one was placed above the other, in a gravity field, and we place the bottom wormhole in a container of water. There would be a continuous flow of water going in the bottom and coming out of the top, only to be caught by the bucket again. Then, enclose this flowing system to prevent evaporation, and place a hydroelectric turbine in the middle. Voila, there's your power generator, assuming that the wormholes don't require the power of a 1000 tesla magnet, in order to bend space :)
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby quickfur » Fri Jan 31, 2014 6:20 am

Polyhedron Dude wrote:
ICN5D wrote:Oh, wow! It has some wild symmetry. It's got four branching off behind it, and three rows emanating towards. So, the 57-cell is just a 57 room version of the strange effects? My question is, is the inverted matter turned into antimatter? Would you really want to take the inverted dollar? Kaboom, perhaps?


Yes the 57-cell will appear to look like 57 dodecahedral rooms with that weird mirror-you in the opposite location. [...]

How do the rooms connect to each other, though? The 11-cell is kinda special in the sense that it is a kind of "simplex": each room connects to every other room. But here, obviously there's more to the structure than that, since otherwise there'd only be 7 rooms. So where do the other 50 fit in?
quickfur
Pentonian
 
Posts: 2935
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Visualizing 3-hyperplanes

Postby Polyhedron Dude » Sat Feb 01, 2014 12:18 am

quickfur wrote:How do the rooms connect to each other, though? The 11-cell is kinda special in the sense that it is a kind of "simplex": each room connects to every other room. But here, obviously there's more to the structure than that, since otherwise there'd only be 7 rooms. So where do the other 50 fit in?


I haven't completely solved that one yet. So far I got to this point: There are 6 hemi-does surrounding the original one, then there are 30 more that connect to the 15 edges in pairs. This gives us 37, so there are 20 more to fill in the gaps - perhaps they form a double 10 set up at the vertices, not sure here yet.
Whale Kumtu Dedge Ungol.
Polyhedron Dude
Trionian
 
Posts: 196
Joined: Sat Nov 08, 2003 7:02 am
Location: Texas

Re: Visualizing 3-hyperplanes

Postby wendy » Sat Feb 01, 2014 7:29 am

It's probably not hard to see how the 57 cells of {5,3,5} fit together.

The cell itself makes one, each of the six faces add six, and fifteen edges give another 30. So there's room for another 20. The thirty-seven cells given here are those that share an incidence with the central one, so we can leave that one alone.

For the six dodecahedra added in step 1, all of its faces are taken, and only the petrie polygons-edges need to be filled.

For the thirty dodecahedra, added edge first, there are four sets of faces to be allocated, all together, some 1.00 sets. This is exactly what 20 further dodecahedra would allow, but this glue is not yet cristal clear how it might work. One might need to 'walk' the central dodecahedron around a bit to see what adds in, but 57 is certainly allowable. Supposing that the 20 fall into 3 edge-on configurations across the repeating boundary, then we would be looking for 60 or 100 points, relative to the centre, which divide into two sets of self-images, and then into sixty edges, each dodecahedron represented three times in the region from the centre.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: Visualizing 3-hyperplanes

Postby ICN5D » Mon Feb 10, 2014 4:31 am

So, what's really the concept behind the 11-cell? It exists in some really bizarre space, but why does it happen with 11 rooms? Why couldn't it be just one room? Is it all about unit edge lacing, and this is one of the things that are possible with a special requirement? Curious how the're prime numbers, 11 and 57.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby quickfur » Mon Feb 10, 2014 6:07 am

ICN5D wrote:So, what's really the concept behind the 11-cell? It exists in some really bizarre space, but why does it happen with 11 rooms? Why couldn't it be just one room? Is it all about unit edge lacing, and this is one of the things that are possible with a special requirement? Curious how the're prime numbers, 11 and 57.

It's an abstract polytope. Abstract polytopes are a kind of idealized generalization where you dispense with physical constructibility or even having any underlying space at all, and consider only the combinatorial structure of the polytope, that is, the containment hierarchy of which k-dimensional faces contain which j-dimensional faces. Of course, to prevent it reducing to just pure set containment hierarchies (it becomes just set theory), some restrictions are imposed on what kinds of hierarchies are considered to be "polytopes" (as opposed to just some random partially-ordered set). I won't get into the details here, unless you want to get into the subject, but basically, by not requiring the existence of any underlying space (even though in some cases such a space might actually exist), a far wider set of objects can be treated as polytopes, and we can apply the results of polytope theory (at least the non-geometric parts thereof) to it.

The challenge, of course, is in finding whether a given abstract polytope has what is called a faithful representation, that is, an embedding into some actual space (you can think of this as building a model of the abstract polytope in some chosen space, such that it accurately represents the underlying abstract structure). While the 11-cell's cells, the hemi-icosahedra, are constructible in projective space, it is not clear (to me, anyway) what kind of space the 11-cell itself would require in order to have a faithful representation. That was what I was originally alluding to when I said that the 11-cell would seriously challenge your visualization ability, because any such space that can contain a faithful representation of the 11-cell would have to be so strange that it is questionable whether we can even understand it "visually", rather than only abstractly in mathematical symbols.

As for why it's 11 rooms and nothing else, that's not an easy question to answer. It's a similar question to why there are exactly 5 Platonic solids in 3D and exactly 4 non-convex regular polyhedra, or why there are exactly 6 convex regular polychora in 4D and exactly 10 non-convex regular polychora, or why in 5D and onwards, there are only 3 convex regular polytopes and no regular non-convex polytopes. Why the dodecahedron has exactly 12 regular pentagons, and not some other number? It's a combination of different things -- the angles of regular pentagons, the topological properties of the same, combine together in such a way that the only regular solution is a 12-faced polyhedron. In the case of the 11-cell, there are no angles to speak of, since it is treated abstractly, but the hemi-icosahedra retain their topological properties, which ultimately dictate that there must be exactly 11 cells, otherwise some parts of the polytope would not close up in the required fashion.
quickfur
Pentonian
 
Posts: 2935
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Visualizing 3-hyperplanes

Postby ICN5D » Mon Feb 10, 2014 6:34 am

Interesting. It makes sense when you mention the hemi-icosahedra. This is an 11-sided shape, and it will certainly fold up into 11 cells in the method you are describing. I have no idea how to fully visualize the true nature of it, yet. The strange number patterns remind me of when I was playing around with the n-cells of the n-cubes. I was dividing the sequence of cells from one row over another. The end result sequence was pascal's triangle, and thus binomial expansion. It was very cool to see that come out of the number of n-cube cells. There must be some similar pattern at work with the 12 pentagon faces. It goes back to number theory, and perhaps makes more sense with a certain counting base. Who knows.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby quickfur » Mon Feb 10, 2014 7:15 am

ICN5D wrote:Interesting. It makes sense when you mention the hemi-icosahedra. This is an 11-sided shape, and it will certainly fold up into 11 cells in the method you are describing. I have no idea how to fully visualize the true nature of it, yet. The strange number patterns remind me of when I was playing around with the n-cells of the n-cubes. I was dividing the sequence of cells from one row over another. The end result sequence was pascal's triangle, and thus binomial expansion. It was very cool to see that come out of the number of n-cube cells. There must be some similar pattern at work with the 12 pentagon faces. It goes back to number theory, and perhaps makes more sense with a certain counting base. Who knows.

Hemi-icosahedra have 10 faces, not 11. :)
quickfur
Pentonian
 
Posts: 2935
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Visualizing 3-hyperplanes

Postby Klitzing » Mon Feb 10, 2014 10:08 am

Once again, for Grünbaum-Coxeter polytopes confer e.g. to my corresponding webpage.

For these figures you usually have 2 ways of construction:

Either consider the hemicosahedron = elike (elliptical space version of icosahedron) and the hemidodecahedron = eldoe (elliptical space version of the dodecahedron) first. Those are mappings of the usual ike resp. doe onto a circumscribing sphere (spherical space) and then identifying pairs of opposite points each (elliptical space).

Then
Code: Select all
elike

hemi(. . .) | 6 |  5 |  5
------------+---+----+---
hemi(x . .) | 2 | 15 |  2
------------+---+----+---
hemi(x3o .) | 3 |  3 | 10

Code: Select all
eldoe

hemi(. . .) | 10 |  3 | 3
------------+----+----+--
hemi(x . .) |  2 | 15 | 2
------------+----+----+--
hemi(x5o .) |  5 |  5 | 6


Then you build up a (still dyadic) polychoron which has elikes for cells and eldoes for vertex figures. That one comes out to have 11 cells and 11 vertices too. Or you could reverse the usage, using elikes for vertex figures and eldoes for cells. This asks for closure for 57 such cells and also 57 vertices.

Code: Select all
11-cell = GC(x3o5o3o)

mod(e(. . . .)) | 11 | 10 | 15 |  6   verf = eldoe
----------------+----+----+----+---
mod(  x . . . ) |  2 | 55 |  3 |  3
----------------+----+----+----+---
mod(  x3o . . ) |  3 |  3 | 55 |  2
----------------+----+----+----+---
mod(e(x3o5o .)) |  6 | 15 | 10 | 11   elike

Code: Select all
57-cell = GC(x5o3o5o)

mod(e(. . . .)) | 57 |   6 |  15 | 10   verf = elike
----------------+----+-----+-----+---
mod(  x . . . ) |  2 | 171 |   5 |  5
----------------+----+-----+-----+---
mod(  x5o . . ) |  5 |   5 | 171 |  2
----------------+----+-----+-----+---
mod(e(x5o3o .)) | 10 |  15 |   6 | 57   eldoe



OTOH, you could consider the opposite application of operations too. First build a quite normal polychoral structure, and do apply identifications of "opposite" points afterwards. - Sure you would Need to take refuge to hyperbolical space structures, in order to have large enough structures for the required mod-wrapping. And yes, "opposite" here truely has to be in quotes. You will have to look for some subsymmetry then which describes all those elemnts to be identified. (Sure, it has to be a true subsymmetry. Else you just result in mod-wrapping in nothing but a single element, instead of some still interesting remaining structure.)

And it is that hyperbolic {3,5,3} = x3o5o3o allows for a consistent color-subsymmetry, which uses exactly 11 different colors for cells, and simultanuously 11 different colors for vertices. That one is applied as follows: Apply a first color to the first ike. Then use further colors to all 20 incident ikes, subject to the requirement that opposite ones shall have the same color. This coloring has to be continued in the same sense throughout all of x3o5o3o. It turns out, that you Need no more than the Initial 11 cell-colors. And furthermore that you derived exactly 11 different vertex configurations (different color incidences): 10 for either pair of opposite vertices of the starting icosahedron plus one more color-configuration elsewhere. Then finally apply a corresponding mod-wrap to this infinte, colored structure. That one would simply identify all likewise colored elements. This mod-wrap of the such colored x3o5o3o reproduces nothing but the affore mentioned 11-cell.

You well could consider to apply the same procedure onto the hyperbolic {5,3,5} = x5o3o5o. Again start coloring the first doe with the first color, then further 6 colors for each pair of opposite adjacent does, etc. on. Thus you would need for 57 colors in total. Wrt. to vertex color-configurations this then amounts in the same number. Finally considering the mod-wrap according to the identification of all like-colored elements, you'd result in the 57-cell.

That's all. - In fact it was Branko Grünbaum who once found that 11-cell first, and it was Harold Scott MacDonald Coxeter, who applied Grünbaum's techniques to that other structure, resulting then in the 57-cell.

--- rk
Klitzing
Pentonian
 
Posts: 1637
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Visualizing 3-hyperplanes

Postby ICN5D » Mon Feb 10, 2014 7:18 pm

Doesn't an icosahedron have 20 sides? Cutting in half will leave 10 sides, plus the face that comes from the slicing?
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby quickfur » Mon Feb 10, 2014 8:33 pm

ICN5D wrote:Doesn't an icosahedron have 20 sides? Cutting in half will leave 10 sides, plus the face that comes from the slicing?

The hemi-icosahedron is not produced by cutting an icosahedron. Rather, it is by identifying opposite faces to be the same face. Or, if you like, to warp it around in projective space so that opposite faces touch and become one and the same.
quickfur
Pentonian
 
Posts: 2935
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Visualizing 3-hyperplanes

Postby ICN5D » Mon Feb 10, 2014 9:46 pm

Ah, I see now. It's all about that crazy curved space where everything loops back into itself. The hemi-icosahedron, in this respect, is only possible in this curvature. It is interesting to see them having the same number of faces as vertices.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby Klitzing » Mon Feb 10, 2014 11:44 pm

Those not only have the same number of cells and vertices, but likewise the same number of edges and faces.
In fact, in an abstract sense those are self-dual just like the pentachoron and as the icositetrachoron!

Code: Select all
 V   E   F  C
 5  10  10  5
11  55  55 11
24  96  96 24
57 171 171 57

--- rk
Klitzing
Pentonian
 
Posts: 1637
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Visualizing 3-hyperplanes

Postby ICN5D » Tue Feb 11, 2014 12:51 am

Okay, I see. You know, I think that this is the first time I understood what the matrices meant. It's very similar to mine, but much better, as it's more descriptive with the inclusion of all other cell types. I can do this for the rotopes, for the wiki. I also just learned what self dual actually means, too. The definition in the wiki is really cryptic, in my opinion. But, I'm sure it makes sense to those who have taken the formal math studies.

So, really, the icosahedron has a unique symmetry with its faces. It is this special arrangement that allows for the faces to equally curve back onto themselves. Now two become one, existing in two places at once in elliptical (hyperbolic?) space. I see how that works with the mirror transformation of everything in the room. That's some wacky stuff, right there. I see what all the hype is about with the 24-cell. You're right, it just suddenly appears out of nowhere, and has no analog that shares all of its properties. It's made of 24 octahedra surcells, right? Unit edge and everything?

You know, I'll bet that there's tons of crazy unit edge shapes that can be made that use rotopes as tilings. Imagine some wild looking thing with cylconinders, tesserinders, cylhemoctahedrinders, and anything else. I see how lots of them have square pyramids and such for tiling pieces. These could also be analogous to a cartesian product with a circle, making ||>|O in place. They would be CRF rotopes, and capable of extending into higher dimensions than 19, due to the nature of the spin. Some would have tons of crazy flat rotope faces, all connected by a single colossal toratope face.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby Keiji » Tue Feb 11, 2014 6:47 am

ICN5D wrote:So, really, the icosahedron has a unique symmetry with its faces. It is this special arrangement that allows for the faces to equally curve back onto themselves.


Well, no, not really. It's nothing special about the icosahedron. You could do it with any polytope that has point symmetry - which is basically all uniform polytopes except the simplex families.

ICN5D wrote:You know, I'll bet that there's tons of crazy unit edge shapes that can be made that use rotopes as tilings...


Sure, but the study of polytopes generally excludes anything with curved elements. If nothing else, incidence matrices become ambiguous when you try to use them for curved tapertopes! I myself tried to do this, and found two tapertopes with the same imat (forget which two they were, though).
User avatar
Keiji
Administrator
 
Posts: 1984
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Re: Visualizing 3-hyperplanes

Postby wendy » Tue Feb 11, 2014 7:05 am

There is of course, another self-dual polytope in 4D. It's 3(3)3(3)3(3)3, in Coxeter's "regular complex polytopes" of 1971.

Its order is of course, the product of the first column (as does all regular solids), gives 200 * 27 * 8 * 3 = 10.96.00. The matrix is in decimal, but the long diagonal translates 200, 1800, 1800, 200 twelfty. (I'm pretty sure i got them right!)

Code: Select all
   240    27    72    27     v
     3  2160     8     8       e   3
     8     8  2160     3       h   3(3)3
     27   72    27   240      c   3(3)3(3)3


It is self dual.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: Visualizing 3-hyperplanes

Postby ICN5D » Fri Feb 14, 2014 1:48 am

Keiji wrote:Sure, but the study of polytopes generally excludes anything with curved elements. If nothing else, incidence matrices become ambiguous when you try to use them for curved tapertopes! I myself tried to do this, and found two tapertopes with the same imat (forget which two they were, though).


Really? That's interesting. I'd like to better understand the incidence matrices and what they really display. Do you have any of them for the rotopes left over? Hopefully you didn't delete them all, I'd be interested to see them. You know my style, I'll always try something out, no matter how crazy wild of an idea. Within reason.....
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby Keiji » Fri Feb 14, 2014 7:01 am

This is what I came up with a while ago. It doesn't include the two that ended up the same though - I'm not sure what happened to them (probably didn't save them). Some may be wrong. The toracubinder diagram is certainly wrong (since when did it have four distinct "tubular 2-surfaces"?) but it's been too long for me to remember what I'd intended with that.

Image
User avatar
Keiji
Administrator
 
Posts: 1984
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Re: Visualizing 3-hyperplanes

Postby wendy » Fri Feb 14, 2014 7:25 am

Being solids, the round figures have surtopes as surfaces. But because the faces are not flat, there is no need for a chain of incidences to run through the dimensions.

The cone has four surtopes (apex point, circle edge, sloping face, disk face). The indicence diagram is then "apex--sloping--circle---disk" with all incident above and below. The figures even obey the surtope-product rules. For example, the cone is an point-circle-pyramid.

Point = 1, 1 Circle = 1, 1, 0, 1 Product = 1, 2, 1, 1, 1.

Because the pyramid product adds a term at each end, and multiplies so that the simplex of N vertices is the N'th power, we get

Point *has* no surface.
Circle = 2d = edge + 0 vertices
Cone = 2 hedra + 1 latron (edge) + 1 teelon (vertex)

However, the incidence matrices for these are:

I had to fix them up. They're S\A matrices. The bit below the diagonal tells us how many of the named columns appear on the surtope on the row. The bit above the diagonal tells us how many times the surtope in the column is part pf the row. So, in the cylinder, the hw (hedron-wall), has two edges, but each edge has only hedron-wall, and one hedron-disk on it, because it divides these. The rest is correct.

Code: Select all
              Cone                  Cylinder                  Sphere
         v   e    hs    hd            e    hd     hw            h
   v     1   0     1     0       e    2     1      1      h     1
   e     0   1     1     1       hd   1     2      -
   hs    1   1     1     -       hw   2     -      1
   hd    0   1     -     1

         hs = sloping edge        hd = disk-face         h = sphere-face
         hd = disk-face           hw = wall (curved)


These also have an incidence diagram (or 'hass antitegum') as well. This is the complete incidence diagrams for the cone, cylinder and sphere. You can see that the cone is a product, there is a surtope for each of (n,v), by (n,e,hd). The pyramid product has a Hass antitegum that is the prism-product of the hass antitegums of the bases.

Code: Select all
          Cone             Cylinder          Sphere       D

           c                  c             c             3
          / \                /|\            |
        hs  hd             d  w  d          h              2
         |\ /              | / \ |          |
         | e               e    e           |              1
         | |                \  /            |
         v |                \  /            |              0
          \|                \ /             |
           n                 n              n             -1


You can sort of see the two products in the cone and cylinder.

In the cone, one has a rectangle where the height is v-n, and the other base is n-e-hd. These represent a point (v-n), multiplied by a circle (n-e-hd), by way of draught (think 'chewing-gum'). The n is drawn to create a vertex. The edge on the circle is drawn out (ie by drawing lines between pairs of points of), between the apex v, and the points on the circle e. This creates the sloping edge. The content is made by a solid measure of gum from the interior of the circle to the (interior of) the point.

In the prism, the bases are e-w-e (which represents a line of height, as vertex-body-vertex of the line), and e-d ( a circle edge and disk). The product of v.x gives x, so v.e gives the two edges, (for the two vertices), and v.d gives the two disk-faces. The product of eh (the height of the prism, as an edge-body), gives e.h * e. gives the curved face, (ie a curved prism), and e.h * d gives the full content c.

And indeed, there =are= figures in four dimensions, which have these as their sections.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: Visualizing 3-hyperplanes

Postby ICN5D » Fri Feb 14, 2014 7:47 am

Wow, those are super cool, Wendy and Keiji! I understand the rotopes and tapertopes really well. So, seeing these diagrams let me understand the patterns of them. It's those really wild CRF shapes that most of you are working on that are blowing my mind. I've been studying those incmats of them and it all looks so complicated and amazing. I've been trying to figure out those toratope tree notations that you posted, Keiji. But, I don't recognize anything in them.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby ICN5D » Fri Feb 14, 2014 4:12 pm

The toracubinder diagram is certainly wrong (since when did it have four distinct "tubular 2-surfaces"?) but it's been too long for me to remember what I'd intended with that.


Keiji, I'll bet you were applying the sock rolling method to the cubinder. That would snip out the cylinders and roll back the square torus into the shape you described. But, the spheration happens only to the square product in (II)II, not the actual square torus.
It is by will alone, I set my donuts in motion
ICN5D
Pentonian
 
Posts: 1135
Joined: Mon Jul 28, 2008 4:25 am
Location: the Land of Flowers

Re: Visualizing 3-hyperplanes

Postby wendy » Wed Feb 19, 2014 10:45 am

I've fixed up some errors in my post, and made some new observations on how cone = point ** circle and cylinder = line *# circle can be seen in the hass antitegums. Note Hass was a person who described the incidence diagram, but i don't think he was aware that it gives an antitegum.

The errors were in that the S\A or incmat for the cylinder was transposed. The surtopes under a figure fall under the diagonal, while those where the diagonal surtope is part of a larger one are above the table. The incidence comes from a_ii a_ij = a_ji a_jj in the matrix. That is, the number of J at an I, multiplied by the number of I's, is the same as the number of I's in a J, multiplied by the number of J's. For example, of the dodecahedron, at each of the 20 vertex, there are 3 pentagons, and there are 5 vertices on each of the 12 pentagons, , so we have 20 * 3 = 5 * 12. (= 60, but this number does not matter save to be equal on both sides.)

Here is some suggestions for the left half of Keiji's picture. I normally descend all the way down to the namon, (nulloid), but where there are no surtopes in the path, the lines are drawn through to where there is a direct connection. The idea is that there is an incidence if there is a path, and no incidence if there is no path.

The crind product is not a surtope product, so one does not find surtope arrays in this table. The comb product (eg torus = circle - circle comb), is, though, but you really can't make a one-by-one array into something fancy - or can you?

Code: Select all
           point     edge       triangle           pt   edg  tri  tet

     3d                                                             1

     2d                             h                          1    4
                                   /|\
     1d               e           e e e                   1    3    6
                     / \          |X X|
     0d      v      v   v         v v v             1     2    3    4
              \      \ /           \|/
    -1d        n      n             n          1    1     1    1    1


   The simplexes as powers of a point: the hass diagram is an N-cube.

     4d

     3d               c                           c            c        c
                    / |                        / || \          |        |
     2d     h     hd  hs       h             hd  hw  hd        h        h#
            |      | /|       ||             || /   \||        |        |
     1d     e      e  |        e    eh       e       e         |        |
            |      |  |        :    / \       :     :          |        |
     0d     |      |  a        :   v   v       :   :           |        |
            |      |/          :    : :         : :            |        |
    -1d     n      n           n     n           v             n        n

          circle  cone      circle  line       cylinder      sphere   torus




The torus HA actually hides an array, formed by the product of the circle (less h,n), by a curcle (less h,n), but e*e = h#, is all the array contains. You need something fancier to see a real array. But even the triangle-triangle is to give 9, 18, 9 points, which is a little overbearing. Anyway, if one holds the circle to have a vertex and an edge, by turning it into a 'monogon', one can see that the circle forms powers through the comb product, in the same way as the simplex is a point-power.

The incidences shown with colons are there, but not part of the product. Likewise, some incidences are shown as double-lines, are likewise the same sort, but help highlight the array.

Code: Select all

         circle     torus      tiger

     4d                           t
                                  :
     3d              c            c
                     :           /|\
     2d     h        h          h h h
            :       / \         |X X|
     1d     e      e   e        e e e
            |       \ /          \|/
     0d     v        v            v
            :        :            :
    -1d     n        n            n



In each case, we make the surface by wrapping a line, rectangle, and prism into a torus-shape. The ends of the line meet, so the circle has one of each. The hedron would normally have four edges, but these have met in pairs to make the two edges (around and along the tyre), while we can fold a three-dimensional block into a tiger, which joins the opposite faces by pairs, the edges by fours, and the vertices by eights.

As before, the colons do not take part in the product, but are part of the incidence diagram. The comb-product is a repetition of surface.

The di-torus has the same incidence matrix as the tiger, but the folds happen differently. In the tiger, suppose you have x=4, y=4, z=1. The first roll is in wx. The second roll is in yz. The erstwhile z is now wrapped in a circle. In a ditorus, one supposes x=4, y=2, z=1. The first roll connects w,x. The second roll connects w,y, and the third role connects w,z. Topologically they're equal shapes, because the solid holes are identical, but this is not shown here.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

PreviousNext

Return to Higher Spatial Dimensions

Who is online

Users browsing this forum: Vector_Graphics and 7 guests