Slices of a Penteract in 4D?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Slices of a Penteract in 4D?

Postby ICN5D » Mon Jun 15, 2015 6:46 am

I was imagining what these would look like, when passing a 5D cube through a 4D hyperplane. So far, I came to this:

Tesseract first: is an unchanging tesseract.

Cube first: a cube appears, expands into rectangular tesseract, collapses to cube.

Square first: a square morphs into triangular diprism, hexagon diprism, then reverse all the way back into a square.

Line first: a line expanding into a tetrahedral prism, morph into octahedral prism then reverse all the way back into a line.

Vertex first: a pentachoron, that truncates into something in between that plays the role of the octahedron. I still haven't figured out what this thing would be in 4D.

So, what is the 4D midsection shape of a vertex-first scan of a 5D cube? A cube has a hexagon, and tesseract is an octahedron. What does a pentachoron truncate into, as the center shape?

After further thought, the 5 vertices of the pentachoron would truncate into 5 tetrahedra, that grow larger. But I still can't figure out what they would evenly morph into as the midsection.

Well, after even more thought, I believe it would be a bipyramid shape with 10 tetrahedral faces. This follows the same pattern we see with a hexagon, having six sides corresponding to a cube, and an octahedron, having 8 triangle faces, corresponding to a tesseract. This is the next logical step, since a penteract has 10 faces, where we get 10 tetrahedra built around a shape with bipyramid symmetry. Would this be correct?
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Re: Slices of a Penteract in 4D?

Postby wendy » Mon Jun 15, 2015 8:53 am

If one wants to find out the 'slice' of a simplex or pyramid product, then it goes like this.

The plane x_1 + x_2 + ... = 1, crosses in the positive region (ie x_1 , x_2, &c ... > 0), in a simplex. For the axies, ye can put the shapes that are making the pyramid, so that at 1, you have a size-1 figure, at 2, a size-2 copy, and so forth.

The section at the point x_1, x_2, &c, is then a prism of x_1 B_1 * x_2 B_2 ...

So your slice of the simplex goes from eg 2 points (line) to 4 points (tetra) as

100% line o--------------------(50 % line * 50% tetra) ---------( 30 % line * 70 % tetra) ----------- o (0 % line * 100 % tetra).

Ye can replace the two ends with any combo that adds up to N+1, eg 6.


CUBE-LIKE STRUCTURES.

The measure polytope, in any dimension, is the simplex antitegum, and the orthotope is a simplex antiprism. One talks of the 'antitegmal sequence' as a kind of sequenced truncates and rectates. Here too, one can provide a numbered series.

The anti-tegmal series, works by the intersection of A and its dual B, when the sizes of A+B are constant.

You start with b=0, and A at 5-b, and B = b. The intersection is a point. As b increases to 1, it is entirely contained inside A, and so you have a B only of size b. When b=1, its vertices are touching the face-centres of A. As b goes past 1, parts of B are outside of a, and you get a truncated B. The truncated series continues until the edges of B hit the face-edges of A, in other words, a rectate.

So, b = 0 rectate = point o3o3o3o
0 < b < 1 truncate = miniture B x3o3o3o
b = 1 rectate = full B x3o3o3o
1 < b < 2 truncate = truncate B x3x3o3o
b = 2 rectate = rectified B o3x3o3o
2 < b < 3 truncate = bi-truncate B = o3x3x3o
b = 3 bi-rectate = o3o3x3o

until one reaches exhaustion.

If you start a cube from a lesser cube, then it's a 'simplex-antitegum' * 'lesser cube', and the sections follow the above rule, except that the extra dimensions of the lesser cube are not distorted, so the pentachoron, beginning with a square, gives the (cube, beginning with a point, * square) prism.
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Re: Slices of a Penteract in 4D?

Postby Klitzing » Mon Jun 15, 2015 11:55 am

If you start with a unit square x4o, and consider its sections vertex first, you'll start with that vertex, i.e. a point, which itself is nothing but an edge of size zero: o. Cutting deeper, that zero sized edge becomes larger and larger until it reaches the size of the diagonal of that square, i.e. sqrt(2) times the original edge of that square. This is what Wendy usually labels q. After that stage the sectioning line decreases to zero again. Thus our square is something like the sequence o => q => o. Therefore the square also could be described as a lace tower oqo&#xt : 3 layers, described by the individual node positions, those layers furthermore additionally (&) being laced (#), and that lacing edge would have metrically the size x. t then just marks that the sequence is ment as a tower. Or as a stack of segmentotopes it also could be described as o || q || o.

Next take a cube. The vertex-first axial subsymmetry here is triangular. thus it ought be described by symmetry o3o, where those nodes might get edges of varying length, depending on the depth of sectioning. In the first step we start at the vertex itself. Thus the edge size of our sectioning triangle will be zero. Accordingly we have o3o here. (Note that this symbol now is ment metrically exact, not just describing the mere symmetry group as such.) Thereafter the triangle increases in size, again until we reach the diagonals of the vertex incident squares. Thus our final shape at this stage then would be q3o. After that stage the size of the sectioning line of the still intersected squares becomes smaller again, while now inbetween in the now opening gaps, the opposite squares become interseced as well, and that sectioning line size here increases by the same amount as the former would decrease. Thus the general section in that realm would be (q-a)3a, for o < a < q. Esp. for a=q we obviously get o3q, the dual triangle. Thereafter the former squares are no longer intersected, and the sectioning line size of the latter squares now decreases again. Thus our sectioning sequence here is o3o => q3o => o3q => o3o. I.e. we could rewrite the cube as the lace tower oqoo3ooqo&#xt. Or as a stack of segmentotopes you'd have o3o || q3o || o3q || o3o.

Note furthermore, that this sectioning sequence would mark the vertex layers perpendicular to the axial direction of movement of our sectioning plane. Here we have exactly 4 such layers. The medial stage, i.e. cutting the cube right in the middle of this axial orientation, would not hit any vertices of the cube. Rather it would just intersect the edges only. From what was mentioned before that medial stage would be described by a regular hexagon y3y, where this edge size here would be derived from y = q-a = a, i.e. y = q/2.

Next consider the tesseract, vertex first. Then our axial subsymmetry would be o3o3o (tetrahedral). Our sections would be o3o3o (metrically exact), then something like a3o3o for o < a < q, then q3o3o (exact, next vertex layer), then (q-a)3a3o, then o3q3o (exact, next vertex layer), then o3(q-a)3a, then o3o3q (exact, next vertex layer), then o3o3(q-a), and finally o3o3o (exact, opposite vertex). Thus our sectioning sequence here is o3o3o => q3o3o => o3q3o => o3o3q => o3o3o. Lace tower here is oqooo3ooqoo3oooqo&#xt, segmentotopal description is o3o3o || q3o3o || o3q3o || o3o3q || o3o3o. - Btw, we could count the vertices in any of those layers individually. Then we get 1 (vertex) + 4 (tetrahedron) + 6 (octahedron) + 4 (dual tetrahedron) + 1 (opposite vertex) = 16. Thus we get the total count of the tesseract's vertices again.

Going still one dimension up, where we finally would reach your question, i.e. the vertex first sectioning sequence of the penteract, then we'd find accordingly: Subsymmetry is o3o3o3o (pentachoral). Thus o3o3o3o (metrically exact), then something described by a3o3o3o for o < a < q, then q3o3o3o (exact), then (q-a)3a3o3o, then o3q3o3o (exact), then o3(q-a)3a3o, then o3o3q3o (exact), then o3o3(q-a)3a, then o3o3o3q (exact), then o3o3o3(q-a), and finally o3o3o3o (exact: opposite vertex). Thus our section of vertex layers is o3o3o3o => q3o3o3o => o3q3o3o => o3o3q3o => o3o3o3q => o3o3o3o. The lace tower thus is oqoooo3ooqooo3oooqoo3ooooqo&#xt or the segmentopal stack description is o3o3o3o || q3o3o3o || o3q3o3o || o3o3q3o || o3o3o3q || o3o3o3o.

So what then would be the medial section here? Again this one here is not a true vertex layer. Rather it is again midway on some of the edges of the penteract. In fact it is that stage of o3(q-a)3a3o, where again y = q-a = a, i.e. y = q/2. Thus we also could write here o3y3y3o. This (up to scaling) is nothing but the bitruncated pentachoron, which also is known as decachoron. The incidence matrix of this figure is
Code: Select all
o3x3x3o

. . . . | 30 |  2  2 |  1  4  1 | 2 2
--------+----+-------+----------+----
. x . . |  2 | 30  * |  1  2  0 | 2 1
. . x . |  2 |  * 30 |  0  2  1 | 1 2
--------+----+-------+----------+----
o3x . . |  3 |  3  0 | 10  *  * | 2 0
. x3x . |  6 |  3  3 |  * 20  * | 1 1
. . x3o |  3 |  0  3 |  *  * 10 | 0 2
--------+----+-------+----------+----
o3x3x . | 12 | 12  6 |  4  4  0 | 5 *
. x3x3o | 12 |  6 12 |  0  4  4 | * 5
which shows that all its facets (cells) would be 10 identical truncated tetrahedra. Metrically exact we should have replaced each occurance of x by that afore defined y. The triangles here stem from sectioning a cube midway of the 3 vertex incident edges, and the hexagons stem from cubes being cut exactly halfway in their axial direction.

--- rk
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Re: Slices of a Penteract in 4D?

Postby ICN5D » Mon Jun 15, 2015 5:06 pm

Wow, excellent! So, I guess I was spot-on with the 10-sided object with tetrahedral faces. I was seeing it as a 4D version of a hexagon-like shape. The real trick would be to animate those scans, as projections of the 4D slices. When I master projection rendering, that will be on the bucket list.
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Re: Slices of a Penteract in 4D?

Postby ICN5D » Mon Sep 14, 2015 6:29 pm

My mind had a revisit to the 5-cube passing through a 4-plane, corner-first. I was wondering, what would be the A || B configuration of the decachoron? I understand it's a bipyramid of two groups of 5 tetrahedral cells, reflected on either side of a bisecting 3-plane. But, is there another distinct orientation of it, by lacing shapes across 4-space?
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Re: Slices of a Penteract in 4D?

Postby Klitzing » Tue Sep 15, 2015 12:21 pm

Well, the only facets of deca are tuts. Thus assumed you would like to have some A||B structure, you well could have tuts for A and/or B. But then the lacing cells too ought be tuts. And because a tut at least has 3 vertex layers (depending on orientation), you'll never come out with a monostratic description only!

But you can get o3x3x3o = x3x3o || (pseudo) o3u3o || o3x3x = oox3xux3xoo&#xt. Having then the incidence description:
Code: Select all
oox3xux3xoo&#xt

o..3o..3o..      & | 24 * |  2  1  1 | 1 2  1  2 | 1 3
.o.3.o.3.o.        |  * 6 |  0  0  4 | 0 0  2  4 | 0 4
-------------------+------+----------+-----------+----
... x.. ...      & |  2 0 | 24  *  * | 1 1  0  1 | 1 2
... ... x..      & |  2 0 |  * 12  * | 0 2  1  0 | 1 2
oo.3oo.3oo.&#x   & |  1 1 |  *  * 24 | 0 0  1  2 | 0 3
-------------------+------+----------+-----------+----
o..3x.. ...      & |  3 0 |  3  0  0 | 8 *  *  * | 1 1
... x..3x..      & |  6 0 |  3  3  0 | * 8  *  * | 1 1
... ... xo.&#x   & |  2 1 |  0  1  2 | * * 12  * | 0 2
... xux ...&#xt    |  4 2 |  2  0  4 | * *  * 12 | 0 2
-------------------+------+----------+-----------+----
o..3x..3x..      & | 12 0 | 12  6  0 | 4 4  0  0 | 2 *
oox3xux ...&#xt  & |  9 3 |  6  3  9 | 1 1  3  3 | * 8


You well can orient deca in axial o2o3o subsymmetry too. Then you'd get o3x3x3o = o2x3o || (pseudo) x2u3o || (pseudo) u2x3x || (pseudo) x2o3u || o2o3x = oxuxo-2-xuxoo-3-ooxux-&#xt. Again using the additional top-down symmetry (here in contrast being antiprismatical) you then can derive:
Code: Select all
oxuxo xuxoo3ooxux&#xt
o.... o....3o....      & | 6  *  * | 2  2 0  0  0 | 1 1  4  0 0 0 | 2 2
.o... .o...3.o...      & | * 12  * | 0  1 1  2  0 | 0 1  2  1 2 0 | 1 3
..o.. ..o..3..o..        | *  * 12 | 0  0 0  2  2 | 0 0  2  2 1 1 | 2 2
-------------------------+---------+--------------+---------------+----
..... x.... .....      & | 2  0  0 | 6  * *  *  * | 1 0  2  0 0 0 | 2 1
oo... oo...3oo...&#x   & | 1  1  0 | * 12 *  *  * | 0 1  2  0 0 0 | 1 2
.x... ..... .....      & | 0  2  0 | *  * 6  *  * | 0 1  0  0 2 0 | 0 3
.oo.. .oo..3.oo..&#x   & | 0  1  1 | *  * * 24  * | 0 0  1  1 1 0 | 1 2
..... ..x.. .....      & | 0  0  2 | *  * *  * 12 | 0 0  1  1 0 1 | 2 1
-------------------------+---------+--------------+---------------+----
..... x....3o....      & | 3  0  0 | 3  0 0  0  0 | 2 *  *  * * * | 2 0
ox... ..... .....&#x   & | 1  2  0 | 0  2 1  0  0 | * 6  *  * * * | 0 2
..... xux.. .....&#xt  & | 2  2  2 | 1  2 0  2  1 | * * 12  * * * | 1 1
..... ..... .ox..&#x   & | 0  1  2 | 0  0 0  2  1 | * *  * 12 * * | 1 1
.xux. ..... .....&#xt    | 0  4  2 | 0  0 2  4  0 | * *  *  * 6 * | 0 2
..... ..x..3..x..        | 0  0  6 | 0  0 0  0  6 | * *  *  * * 2 | 2 0
-------------------------+---------+--------------+---------------+----
..... xux..3oox..&#xt  & | 3  3  6 | 3  3 0  6  6 | 1 0  3  3 0 1 | 4 *
oxux. xuxo. .....&#xt  & | 2  6  4 | 1  4 3  8  2 | 0 2  2  2 2 0 | * 6


The idea on how to derive the latter can be obtained from the lace city investigation. The first description here just outlines the tower of towers: tut = o3x || (pseudo) o3u || x3x, and u-oct = o3u || u3o. Therefore:
Code: Select all
o3x  o3u  x3x 
              
              
 o3u       u3o
              
              
  x3x  u3o  x3o


But you could write down the same stacking not with resp. to o3o symmetry, but wrt. o2o symmetry. Then you'll have: tut = x2o || (pseudo) u2x || (pseudo) x2u || o2x, and u-oct = o2o || (pseudo) u2u || o2o. Therefore:
Code: Select all
   x o   u x   x u   o x   
                          
                          
                          
o o         u u         o o
                          
                          
                          
   o x   x u   u x   x o   


The clue then finally is to rotate this latter lace city within its representation plane resp. read new towers not as horizontal lines but as highly slanted ones. Then you get (o2o || x2o) || (o2x || u2x) || (x2u || u2u || x2u) || (u2x || o2x) || (x2o || o2o) - which happens to be nothing but that stacking, which was described above within the second incidence matrix...

Hehe, the same surely can also been read off from the first lace city as well, when using such highly slanted towers: o3x || (o3u || o3u) || (x3x || x3x) || (u3o || u3o) || x3o.

It even ought be possible to read off some stack of antiprisms: x3x || (o3u || u3o) || (o3x || x3o) || (o3u || u3o) || x3x. - I have not elaborated that one so far. Not even the required lacing edge lengths for the there being used 2 types of non-uniform antiprisms within those vertex layers.

--- rk
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