Tegum polytopes

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

Tegum polytopes

Postby Klitzing » Wed Aug 24, 2016 3:30 pm

Wendy once coined the notion of tegums. That one derives from the latin verb "tegere" (to cover, to cloth) resp. noun "tegumentum" (the cover, the hull, the armour, the shell). Herself often refers the connotion to "tent" (even so that one derives from latin "tenta" and therefore has a different etymology).

In its application to polytopes it is quite closely related to the hull operator.

Infact, we meanwhile know of 2 such applications, the tegum sum and the tegum product. While the tegum sum refers to the "cover" of 2 or more elements within the same space, the tegum product refers to such a "cover" of 2 or more elements within orthogonal spaces. More rigorously and formally, if P and Q would be any 2 polytopes those tegum terms could be defined as
  • tegum sum(P, Q) = hull(P + Q) = hull( P ∪ Q )  -  thus we need dim(P) = dim(Q) here.
  • tegum product(P, Q) = hull(P ⊕ Q) = hull( (P,0) + (0,Q) )
Esp. the last term shows, that any tegum product also can be considered a tegum sum too: tegum product(P, Q) = tegum sum((P,0), (0,Q)).

The easiest example here for sure is the tegum product of N equivalent line segments. That one then is nothing but the N-dimensional crosspolytope. - Right in that example it becomes apparent, that the size of these line elements and the size of the lacing edges differs by a factor of sqrt(2). And so in general, the size of the lacing edges highly depends on the overall size of the respective factors or addends.


Student91 once asked for a new lace term, which he, in his first posts, termed "&#U" in the reading of "lace union". That one later settled in "&#zx", kind a lace prism with zero height. But, infact, in more formal rigourness, it is rather the following
  • Let P and Q be the components of some compound, e.g. P = q3o3o and Q = o3o3q, thus comp(P, Q) = qo3oo3oq (a stella octangula);
  • Provided that compound gives rise to a hull polytope hull(P, Q) = tegum sum(P, Q), where all lacing edges have some specific (say unit) size,
  • Then we are allowed to write comp(P, Q)&#zx, e.g. qo3oo3oq&#zx (which one is nothing but the unit cube).

It should be noted that there is not a restriction here to all vertex pairings. Only those will be considered, which provide an outside visible lacing edge of the hull of the compound components, i.e. the true edges. Whereas those false lacing "edges", which would become internal, do not contribute to that restriction!

From the hull operation it becomes apparent as well, that way not all subelements of the compound components would contribute as subelements of the resulting polytope. Esp. the component polytopes themselves usually are "pseudo elements" only. (Only when one compound component is completely internal to the other, then the larger component "survives" as hull.)


For completeness it should be added, that we usually consider Dynkin describable polytopes, i.e. the Wythoffians. Those thus are all uniform and therefore circumscribable. Thus the compounding procedure here becomes much more definite: both the circumcenters will have to coincide and moreover the describing symmetry ought be the same for such a compounding description like in qo3oo3oq.


From a more abstract point of view, here we just have given a constructive definition for new sets of polytopes, the tegum product and the tegum sum, on the base of some formerly known polytopes P and Q. Esp. the restricting construction asking additionally for equal sized lacings only ("&#zx") is, from a mere mathematical point of view, rather "wild".

Therefore here a task, open to be solved:
  • Could those "Student91-ian lace tegums" ("&#zx") could be classified outside from their construction device?
  • Esp. could one start (and luckily also end) an exhaustive enlisting of those?
Any opinions to that?

Beside the more theoretical aspect of this post (and its supposed answers) I further hope to get in here - in this new thread - a broard collection of already found such polytopes (resp. the polytopal descriptions thereof).

--- rk
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Re: Tegum polytopes

Postby wendy » Thu Aug 25, 2016 8:53 am

The names of the tegum product, and coverings in general, are only late additions to a well-established process. By me, 'tegum' refers only to the product, and 'thatch' to the sum. But it is prehaps necessary to return to the radiant-space to understand what is going on here. This is where all of the products were established.

The radiant model of a polytope, is that the surface represents a radial function at 1, the centre at 0, and this defines in every direction the value of space. So the surface at 2 represents a copy twice the size, and the same centre, etc. For this process, we discard the angular part, and note that this equates to an axis x, where a coordinate x represents a figure x.X, in the space where X is solid.

What we are interested is the coordinate space (x,y,z,..) where each axis is a different polytope. Each point then represents a position-product xX * yY * zZ * ... in prism product. Teragon does a good picture here. It's not a good way of visualising a figure, since we don't construct 3d figures here, and there is 'no box to stand on to peer over the fence'.

When no altitude is given, the figure represents overlays. For example, a circular arc represents a circle (r.cos(þ), r.sin(þ)). One often sees non-rectangular figures constructed in rectangular blocks, and this model gives pay to that. The tegum product, then is the octahedron quadrant drawn on this, the prism is then simply the prism (ie 1,1,1,1,1 decended to each axis). The analasys of the surface of the tegum leads to the pyramid product.

When altitude is added, then the points say, in a line, is not an overlay, but rather the progress on line defines the height, and figure describes the section. One of johnson's uniform polytopes xxo2oxx&#xt gives the line going from (1,,0) to(1,,1) and then (1,,1) to (0,,1), as y is let vary. The line is bent, but that's because a change of shape, rather than dimension.

Note now that when we spread a figure across from the points (1,0,0) to (0,1,0) to (0,0,1), the figure formed is a triangle here, is the 'altitude' of the figure. It represents ultimately the lace prism ( xooX oxoY ooxZ &#x), But you can equally consider (xxoX oxxY xoxZ &#x), is still a lace-prism, but not a simple product.

"Tegum" is set opposite "Prism", and used only where they are interchangable. The simple process of cutting is 'truncation', and the simple process of covering is 'thatching'. These words can be translated to the local language. The need for a term like 'thatch' is because Bower's figures spread across a given frame, can not be described as a 'hull', because that term supposes convexity. Moreover, it is possible thatch a given frame (vertices + edges of ike) in more than one way (icosahedron, great dodecahedron), and that the vertices and edges of the {5/2,5} sisid, the intent here is not to find the hull, but thatches of the frame giving sisid and great icosahedron, resp. The hull is the icosahedron here.

Tegum is a specific covering made by the 'drawing or draught of surface'. Specifically, for two orthogonal figures, where p is a point on one, and q on the other, there exists a unique line pq, for every pair of points.

Student91's lace-union

This is a specific case of a thatch over one or more figures. The series of appiculates (the thatch of x.X and (1-x).dual(X)), as x increases, is another lace-union. In practice, when one divides a figure into parts, and cover the results, the result is a lace-union, eg qo3oo4xo&z is a cover of two figures x-cube and q-octahedron, so the edges cross.

It is less obvious with the cube qo3oo3oq&z, but it is still perfectly legitimate if one wishes to show the devolution of symmetry.

A lace-union here would show all of the lacing-edges here, even if the base edges disappear. There are 12 lace-edges in total, these are on the edge of the symmetry, which is why the lacings are in pairs of cells. This is normal.

What makes the lace-unions different to the lace-towers, is that some of the lace-tower surtopes are internal, and hence disappear. It does not devalue the lace-union though.

Of course, it makes no sense to restrict to unit edges. There are equally interesting flat lace-towers formed on the {3,3,5} when the lacing is set to 'f', and even the ZomeTools, give each kind of edge in the ratio of v:1:f,

Student91's lace-unions, in their simplest forms, are symmetrically placed overlays, the projections of higher figures onto the same plane. One might suppose that quickfur's projections of figures are examplers of this.

The lace-unions would be hard to enumerate completely, since any representation of a polytope as a thatched union is thus. One does not need the elements of the union to be the same dimension as the figure, since oxf2xfo2fox&#xz is a lace-union of golden rectangles, none the less makes an icosahedron. Similarly, the ex is a thatch of the five separate 24ch that are generate it lefthanded-wise (or ten in both directions). The lace union is then here a statement that the vertices of a figure is simply a compound of lesser figures.
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Re: Tegum polytopes

Postby student91 » Thu Aug 25, 2016 9:16 am

Hello,

I agree that the construction that the suffix &#zx proposes, shoud be more explicitly defined. Once, i already defined it in a less rigour way, Cf. this post.

I have been thinking about how the CD's should work formally, and I've written some ideas down in this draft. (I wrote this, because the other draft became quite unwieldy and lacked the rigour I was after). As you can see, I define a polytope e.g. o5o3x to be a set of (n-1)-dimensional elements (Which are themselves sets of (n-2)-dimensional elements, etc.). Thus this polytope starts with a vertex {o5o3x}, then you make the edge {o5o3x,o5x3(-x)}, then you make the face {{o5o3x,o5x3(-x)},{o5o3x,f5(-x)3o},{f5(-x)3o,o5x3(-x)}}, and then your polytope consists of all 20 faces that can be produced by quirking this face.

So far, this definition does look very rigorous. Only when edges coincide, e.g. when describing (-x)3x, this approach doesn't work, but some more math can fix that by saying that one should map a x3x to anything without an o, and so on for higher-dimensional stuff.

Note how this approach does not care for convexity, it works just as fine on convex polytopes as on non-convex ones (Given that nothing coincides, which can be solved as stated.)

I am looking for a similarly rigorous approach on things like xofox2ofxfo&#xt or fox2xfo2oxf&#zx, or fo(-x)2xfo2oxf&#zx or (-x)ofo(-x)2ofxfo&#xt.

The first, xofox2ofxfo&#xt, gives an icosahedron. However, when the diagram is read literally, we would get x2o||o2f+o2f||f2x+f2x||o2f+o2f||x2o, thus we would be missing the lacings from x2o to f2x, which can be solved by inserting xfo&#x-patches. However, this is not implied by the notation.

Alternatively, one could approach this in a &#zx-manner. Then one makes the complex \xofox2\ofxfo&#xt, in the "most-convex" way. I will describe the procedure I think one should follow when a &#zx is encountered.
The patch \xofox2\ofxfo&#xt can also be interpreted as \xofox2\ofxfo2\fxo(-x)(-f)&#zx, where the last node gives the heights according to some well-chosen mirror plane.
Making it in "the most convex" way then means that whenever you have a distance of x between two points, you should make an edge. Then, when you see a set of edges that make a polygon (i.e. the edges can be put vertex-to-vertex in a circular manner, and the distance between two points is equal to the corresponding diagonal of an n-gon), you should do so, and in general when there is a set of (n-m)-dimensional things that make a (n-m+1)-dimensional thing, you should do so. Thus we make the edges {\x2\o2\f,\o2\f2\x}, {\x2\o2\f,\f2\x2\o}, {\o2\f2\x,\f2\x2\o}, {\f2\x2\o,\o2\f2\(-x)}, {\f2\x2\o,\x2\o2\(-f)}, {\o2\f2\(-x),\x2\o2\(-f)} and {\o2\f2\x,\o2\f2\(-x)}, and the faces {{\x2\o2\f,\o2\f2\x}, {\x2\o2\f,\f2\x2\o}, {\o2\f2\x,\f2\x2\o}}, {{\f2\x2\o,\o2\f2\(-x)}, {\f2\x2\o,\x2\o2\(-f)}, {\o2\f2\(-x),\x2\o2\(-f)}} and {{\f2\x2\o,\o2\f2\x}, {\f2\x2\o,\o2\f2\(-x)}, {\o2\f2\x,\o2\f2\(-x)}}.
Then, we will start mirroring in the mirrors, one by one.
So we will have to think what xofox2\ofxfo2\fxo(-x)(-f)&#zx means (note the omitted backslash). Of course, our previous patches will be mirrored. Furthermore, any edge created by allowing the mirror (i.e. when a node on the active mirror is x, as in &#xz) must be made, so we make the additional edges {\x2\o2\f,\(-x)2\o2\f} and {\x2\o2\(-f),\(-x)2\o2\(-f)}. Then yet again, we will look for any hereby created surtopes, going one dimension up at a time. We see that these edges are part of one new triangle each, {{\x2\o2\f,\(-x)2\o2\f}, {\x2\o2\f,\o2\f2\x}, {\(-x)2\o2\f,\o2\f2\x}} and {{\x2\o2\(-f),\(-x)2\o2\(-f)}, {\x2\o2\(-f),\o2\f2\(-x)}, {\(-x)2\o2\(-f),\o2\f2\(-x)}}.

Similarly, when the second mirror is activated (So we get \xofox2ofxfo2\fxo(-x)(-f)&#zx), we get the new edge {\f2\x2\o,\f2\(-x)2\o}, which gives rise to the new triangles {{\x2\o2\f,\f2\x2\o},{\x2\o2\f,\f2\(-x)2\o},{\f2\x2\o,\f2\(-x)2\o}} and {{\x2\o2\(-f),\f2\x2\o},{\x2\o2\(-f),\f2\(-x)2\o},{\f2\x2\o,\f2\(-x)2\o}}.

Thus, when we activate both mirrors (and thus get the full "patch" xofox2ofxfo2\fxo(-x)(-f)&#zx), we get 3×4 triangles from \xofox2\ofxfo2\fxo(-x)(-f)&#zx, 2×2 triangles from xofox2\ofxfo2\fxo(-x)(-f)&#zx and 2×2 triangles from \xofox2ofxfo2\fxo(-x)(-f)&#zx for a total of 20 triangles. Then we look wether these 20 triangles make up a complete poluhedron, which it does, so we're happy. (A polyhedron thus is a set of faces, where every edge of every face is shared with exactly 1 other face of the polyhedron).

We see that this approach again does not allow duplicate-edges as in xofx3ofo(-x)&#xt.
Furthermore, the construction as such can (and will most of the time) fail. Whenever the final set of polygons (So after activating all mirrors) does not give all faces for a polyhedron, it does not work.
However, this approach does make not-very-convex polytopes like fox2ofx2(-x)fo&#zx look the way we want them to look.

I hope this part has been clear, I apologise for the huge amount of code-like text.

For the classification outside the construction device, any unit-edged polytope with a given symmetry, where the "fundamental patch" under that symmetry is "convex" in some specific sense, can be described using this construction device.
As such, a lot of polytopes will be describable in this manner, and thus exhaustingly enlisting it could be quite undoable. However, when one would add some more restrictions, e.g. allow only a small number of vertices, I guess one could be able to exhaustively enlist the possibilities. This is, though, beyond the scope of the CRF-project.
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Re: Tegum polytopes

Postby Klitzing » Wed Jan 18, 2017 7:05 pm

Chamferd polytopes, i.e. regular polytopes with rasped down edges (solely), could be mentioned here.

I found that chamfered polyhedra exactly can be given as
Code: Select all
axPooQoc&#zx
with some pseudo edge sizes 'a' and 'c'. Obviously 'a' then is nothing but the tip-2-tip length of the (just axial) hexagons, and 'c' is the width of those hexagons between the 2 parallel sides, which in turn are parallel to the rasped off edge. In fact, the hexagons themselves are described by
Code: Select all
ax .. oc&#zx
while the only other faces are
Code: Select all
.xP.o ..
In this case it is immediate that the new parallel edge 'x' (first node) can generally be made the same size as the lacing one ('&#x'), thus making the hexagon at least all unit edged.

But on how to continue that edges-only-rasping-down process within 4D, I'm still struggling. So far it still succeeds to elude my mind on the one or other end... :angry:

--- rk
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Re: Tegum polytopes

Postby Klitzing » Wed Jan 18, 2017 11:17 pm

Okay, solved it on my own. :)

The hard part, which tricked me so far, was, that I assumed the chamfered polychora ought likewise have 2 vertex types only. But rather they would ask for 3!
In fact, here is the general form they have:
Code: Select all
abxPoooQoocRodo&#zx =

a..Po..Qo..Ro..     +
.b.P.o.Q.o.R.d.     +
..xP..oQ..cR..o


There are (beside the false edges 'a', 'b', 'c', and 'd') only the following 3 types of true edges:
Code: Select all
oo.Poo.Qoo.Roo.&#x
.ooP.ooQ.ooR.oo&#x
..x ... ... ...
Note that those once more all are sized 'x'!

The faces will thus be:
Code: Select all
... ... ... odo&#xt  (rhombs)
.bx ... .oc ...&#zx  (axial hexagons)
..xP..o ... ...      (regular P-gons)


The cells then are:
Code: Select all
abx ... oocRodo&#zx  (elongated R-gonal rhombohedra)
.bxP.ooQ.oc ...&#zx  (chamfered xPoQo)
where those elongated rhombohedra represent the facets underneath the rasped off former edges, and the chamfered polyhedra represent the remainders of the former cells. (Those elongated rhombohedra consist from R axial hexagons as the equatorial girdle, and then R rhombs on either end.)

--- rk

Edit: cf. some corrections here.
Last edited by Klitzing on Sun Jan 29, 2017 8:32 am, edited 1 time in total.
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Re: Tegum polytopes

Postby Klitzing » Fri Jan 20, 2017 10:40 am

Here now the explicite combinatorical information of these chamfered polychora will be provided, i.e. their respective incidence matrices. :]

The chamfered pentachoron:
Code: Select all
abx3ooo3ooc3odo&#zx = a-pen + (b,d)-spid + (x,c)-srip

o..3o..3o..3o..     | 5  *  * |  4  0  0 |  6  0  0 |  4 0  verf: tetrahedron
.o.3.o.3.o.3.o.     | * 20  * |  1  3  0 |  3  3  0 |  3 1
..o3..o3..o3..o     | *  * 30 |  0  2  2 |  1  4  1 |  2 2
--------------------+---------+----------+----------+-----
oo.3oo.3oo.3oo.&#x  | 1  1  0 | 20  *  * |  3  0  0 |  3 0
.oo3.oo3.oo3.oo&#x  | 0  1  1 |  * 60  * |  1  2  0 |  2 1
..x ... ... ...     | 0  0  2 |  *  * 30 |  0  2  1 |  1 2
--------------------+---------+----------+----------+-----
... ... ... odo&#xt | 1  2  1 |  2  2  0 | 30  *  * |  2 0  rhomb {(r,R)^2}
.bx ... .oc ...&#zx | 0  2  4 |  0  4  2 |  * 30  * |  1 1  axial hexagon {(h,H,H)^2}
..x3..o ... ...     | 0  0  3 |  0  0  3 |  *  * 10 |  0 2  regular triangle {3}
--------------------+---------+----------+----------+-----
abx ... ooc3odo&#zx | 2  6  6 |  6 12  3 |  6  3  0 | 10 *  elongated trigonal rhombohedron
.bx3.oo3.oc ...&#zx | 0  4 12 |  0 12 12 |  0  6  4 |  * 5  chamfered tetrahedron


The chamfered hexadecachoron:
Code: Select all
abx3ooo3ooc4odo&#zx = a-hex + (b,d)-sidpith + (x,c)-rico

o..3o..3o..4o..     | 8  *  * |  8   0  0 | 12  0  0 |  6  0  verf: cube
.o.3.o.3.o.4.o.     | * 64  * |  1   3  0 |  3  3  0 |  3  1
..o3..o3..o4..o     | *  * 96 |  0   2  2 |  1  4  1 |  2  2
--------------------+---------+-----------+----------+------
oo.3oo.3oo.4oo.&#x  | 1  1  0 | 64   *  * |  3  0  0 |  3  0
.oo3.oo3.oo4.oo&#x  | 0  1  1 |  * 192  * |  1  2  0 |  2  1
..x ... ... ...     | 0  0  2 |  *   * 96 |  0  2  1 |  1  2
--------------------+---------+-----------+----------+------
... ... ... odo&#xt | 1  2  1 |  2   2  0 | 96  *  * |  2  0  rhomb {(r,R)^2}
.bx ... .oc ...&#zx | 0  2  4 |  0   4  2 |  * 96  * |  1  1  axial hexagon {(h,H,H)^2}
..x3..o ... ...     | 0  0  3 |  0   0  3 |  *  * 32 |  0  2  regular triangle {3}
--------------------+---------+-----------+----------+------
abx ... ooc4odo&#zx | 2  8  8 |  8  16  4 |  8  4  0 | 24  *  elongated square rhombohedron
.bx3.oo3.oc ...&#zx | 0  4 12 |  0  12 12 |  0  6  4 |  * 16  chamfered tetrahedron


The chamfered hexacosachoron:
Code: Select all
abx3ooo3ooc5odo&#zx = a-ex + (b,d)-sidpixhi + (x,c)-srix

o..3o..3o..5o..     | 120    *    * |   20    0    0 |   30    0    0 |  12   0  verf: dodecahedron
.o.3.o.3.o.5.o.     |   * 2400    * |    1    3    0 |    3    3    0 |   3   1
..o3..o3..o5..o     |   *    * 3600 |    0    2    2 |    1    4    1 |   2   2
--------------------+---------------+----------------+----------------+--------
oo.3oo.3oo.4oo.&#x  |   1    1    0 | 2400    *    * |    3    0    0 |   3   0
.oo3.oo3.oo4.oo&#x  |   0    1    1 |    * 7200    * |    1    2    0 |   2   1
..x ... ... ...     |   0    0    2 |    *    * 3600 |    0    2    1 |   1   2
--------------------+---------------+----------------+----------------+--------
... ... ... odo&#xt |   1    2    1 |    2    2    0 | 3600    *    * |   2   0  rhomb {(r,R)^2}
.bx ... .oc ...&#zx |   0    2    4 |    0    4    2 |    * 3600    * |   1   1  axial hexagon {(h,H,H)^2}
..x3..o ... ...     |   0    0    3 |    0    0    3 |    *    * 1200 |   0   2  regular triangle {3}
--------------------+---------------+----------------+----------------+--------
abx ... ooc5odo&#zx |   2   10   10 |   10   20    5 |   10    5    0 | 720   *  elongated pentagonal rhombohedron
.bx3.oo3.oc ...&#zx |   0    4   12 |    0   12   12 |    0    6    4 |   * 600  chamfered tetrahedron


The chamfered icositetrachoron:
Code: Select all
abx3ooo4ooc3odo&#zx = a-ico + (b,d)-spic + (x,c)-srico

o..3o..4o..3o..     | 24   *   * |   6   0   0 |  12   0  0 |  8  0  verf: octahedron
.o.3.o.4.o.3.o.     |  * 144   * |   1   4   0 |   4   4  0 |  4  1
..o3..o4..o3..o     |  *   * 288 |   0   2   2 |   1   4  1 |  2  2
--------------------+------------+-------------+------------+------
oo.3oo.4oo.3oo.&#x  |  1   1   0 | 144   *   * |   4   0  0 |  4  0
.oo3.oo4.oo3.oo&#x  |  0   1   1 |   * 576   * |   1   2  0 |  2  1
..x ... ... ...     |  0   0   2 |   *   * 288 |   0   2  1 |  1  2
--------------------+------------+-------------+------------+------
... ... ... odo&#xt |  1   2   1 |   2   2   0 | 288   *  * |  2  0  rhomb {(r,R)^2}
.bx ... .oc ...&#zx |  0   2   4 |   0   4   2 |   * 288  * |  1  1  axial hexagon {(h,H,H)^2}
..x3..o ... ...     |  0   0   3 |   0   0   3 |   *   * 96 |  0  2  regular triangle {3}
--------------------+------------+-------------+------------+------
abx ... ooc3odo&#zx |  2   6   6 |   6  12   3 |   6   3  0 | 96  *  elongated trigonal rhombohedron
.bx3.oo4.oc ...&#zx |  0   6  24 |   0  24  24 |   0  12  8 |  * 24  chamfered octahedron


The chamfered tesseract:
Code: Select all
abx4ooo3ooc3odo&#zx = a-tes + (b,d)-sidpith + (x,c)-srit

o..4o..3o..3o..     | 16  *  * |  4   0  0 |  6  0  0 |  4 0  verf: tetrahedron
.o.4.o.3.o.3.o.     |  * 64  * |  1   3  0 |  3  3  0 |  3 1
..o4..o3..o3..o     |  *  * 96 |  0   2  2 |  1  4  1 |  2 2
--------------------+----------+-----------+----------+-----
oo.4oo.3oo.3oo.&#x  |  1  1  0 | 64   *  * |  3  0  0 |  3 0
.oo4.oo3.oo3.oo&#x  |  0  1  1 |  * 192  * |  1  2  0 |  2 1
..x ... ... ...     |  0  0  2 |  *   * 96 |  0  2  1 |  1 2
--------------------+----------+-----------+----------+-----
... ... ... odo&#xt |  1  2  1 |  2   2  0 | 96  *  * |  2 0  rhomb {(r,R)^2}
.bx ... .oc ...&#zx |  0  2  4 |  0   4  2 |  * 96  * |  1 1  axial hexagon {(h,H,H)^2}
..x4..o ... ...     |  0  0  4 |  0   0  4 |  *  * 24 |  0 2  square {4}
--------------------+----------+-----------+----------+-----
abx ... ooc3odo&#zx |  2  6  6 |  6  12  3 |  6  3  0 | 32 *  elongated trigonal rhombohedron
.bx4.oo3.oc ...&#zx |  0  8 24 |  0  24 24 |  0 12  6 |  * 8  chamfered cube


The chamfered hecatonicosachoron:
Code: Select all
abx5ooo3ooc3odo&#zx = a-hi + (b,d)-sidpixhi + (x,c)-srahi

o..5o..3o..3o..     | 600    *    * |    4    0    0 |    6    0   0 |    4   0  verf: tetrahedron
.o.5.o.3.o.3.o.     |   * 2400    * |    1    3    0 |    3    3   0 |    3   1
..o5..o3..o3..o     |   *    * 3600 |    0    2    2 |    1    4   1 |    2   2
--------------------+---------------+----------------+---------------+---------
oo.5oo.3oo.3oo.&#x  |   1    1    0 | 2400    *    * |    3    0   0 |    3   0
.oo5.oo3.oo3.oo&#x  |   0    1    1 |    * 7200    * |    1    2   0 |    2   1
..x ... ... ...     |   0    0    2 |    *    * 3600 |    0    2   1 |    1   2
--------------------+---------------+----------------+---------------+---------
... ... ... odo&#xt |   1    2    1 |    2    2    0 | 3600    *   * |    2   0  rhomb {(r,R)^2}
.bx ... .oc ...&#zx |   0    2    4 |    0    4    2 |    * 3600   * |    1   1  axial hexagon {(h,H,H)^2}
..x5..o ... ...     |   0    0    5 |    0    0    5 |    *    * 720 |    0   2  regular pentagon {5}
--------------------+---------------+----------------+---------------+---------
abx ... ooc3odo&#zx |   2    6    6 |    6   12    3 |    6    3   0 | 1200   *  elongated trigonal rhombohedron
.bx5.oo3.oc ...&#zx |   0   20   60 |    0   60   60 |    0   30  12 |    * 120  chamfered dodecahedron


What still has to be evaluated here, are the corresponding metrical values, that is the lengths of all these pseudo edges, used within those descriptions, i.e. the respective values of 'a', 'b', 'c', and 'd' (wrt. the unit edge size 'x'), as well as the respective values of those smaller and larger rhomb angles 'r' and 'R' respectively the smaller and larger axial hexagon angles 'h' and 'H'. - But then 'b', 'c', 'h', and 'H' fully belong to the 3D geometry of the chamfered polyhedra only, i.e. here can be assumed to be known (just ought to be provided herewith as well, in order to be self-contained). :P

--- rk
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Re: Tegum polytopes

Postby wendy » Sat Jan 21, 2017 7:49 am

I use 'bevel' as a general form of the keplerish rhombo-truncate in N dimensions.

The rectates are the family of wythoff-polytopes with one node. The extreme cases (like the cube), have the marked node at the end, are taken as the zero-rectate.

The dual of the rectates are the surtegmates, which correspond to a tegumsum of the two dual figures (b.s. dodecahedron, icosahedron), at various sizes. At different sizes, the kD surtopes of ine cross the (n-k-1) surtopes of the other, and the faces of the tegumsum are tegums of the crossing elements.

The face planes of the various surtegums correspond to the possible face-planes of the wythoff-figures.

The rhombo- icosadodecahedron, then corresponds to the intersection of a rhombo-tricontahedron and some sort of truncated dodecahedron. It's not a perfect match, but the cut is taken like this.

The generalised form is to allow the intersection of any surtegum (bs o3m4o) and some wythoff figure. This corresponds to beveling the parts that point in that direction.

Every wythoff-mirror-edge figure is then an intersection of surtegums.

If one takes a cube, and bevel its edges, one starts off with long hexagons, but when the bevel is so severe that the faces of the cube are lost, the edges of the cube have become rhombuses.

The general form of the hexagon, say in 4d, is a peak-truncated polygonal tegum xyx3ooo&#xt, or a rim-truncated polygonal tegum (a prism with peaked ends, eg oxxo3oooo&#xt.

It now comes to find the coordinates of these points.

Since the plane is perpendicular to the stott-vector like (0,1,0,0) or (0,0,0,1) (with just one '1' and the rest are zero), the relevant calculations are then that the matrix-dot of this vector and any other vector will give the parallel plane the vertex falls on, and the second vector can be proportionally lengthened to match.

The spreadsheet already calculates these terms for every vector, and simply exposing the vector S*r (where r is the coordinate of the vertex), will give the plane where r crosses the parallels in each of the nodal directions,

The vertices of the polytope formed above will be variously, the vector (1,0,0,0...), or (0,0,0...,1), or some (x,0,0,0,y) in between [where the uncut part of the faces cross the rhombus]
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Re: Tegum polytopes

Postby Klitzing » Sun Jan 29, 2017 12:18 am

Klitzing wrote:Okay, solved it on my own. :)

The hard part, which tricked me so far, was, that I assumed the chamfered polychora ought likewise have 2 vertex types only. But rather they would ask for 3!
In fact, here is the general form they have:
Code: Select all
abxPoooQoocRodo&#zx =

a..Po..Qo..Ro..     +
.b.P.o.Q.o.R.d.     +
..xP..oQ..cR..o


Sadly I have to withdraw from that post again a bit. This is because this 'general form' does work out metrically correct only for R=4. The main problem here are not the 3 superimposed layers (which are correct), rather it is the final &#zx part.
There are (beside the false edges 'a', 'b', 'c', and 'd') only the following 3 types of true edges:
Code: Select all
oo.Poo.Qoo.Roo.&#x
.ooP.ooQ.ooR.oo&#x
..x ... ... ...
Note that those once more all are sized 'x'!

This now is the point where the problems happen: The latter 2 edge sizes clearly can be made the same, essentially this is nothing but the existance of the chamfered versions of the regular polyhedra. But the first line then rather in general ought read
Code: Select all
oo.Poo.Qoo.Roo.&#y

and thus no longer all would be sized 'x'!
The faces will thus be:
Code: Select all
... ... ... odo&#xt  (rhombs)
.bx ... .oc ...&#zx  (axial hexagons)
..xP..o ... ...      (regular P-gons)


And in consequence what I just meant to be rhombs, rather are kites in general only. Solely in the case of R=4 these kites happen to be truely rhombs (i.e. y=x).
The cells then are:
Code: Select all
abx ... oocRodo&#zx  (elongated R-gonal rhombohedra)
.bxP.ooQ.oc ...&#zx  (chamfered xPoQo)
where those elongated rhombohedra represent the facets underneath the rasped off former edges, and the chamfered polyhedra represent the remainders of the former cells. (Those elongated rhombohedra consist from R axial hexagons as the equatorial girdle, and then R rhombs on either end.)

In fact here too I'd to speak of elongated kitohedra as well.

And here now is why:
Consider to look at those additional cells, which occur underneath the rasped down edges in their tip-2-tip orientation. Then we would get e.g. for R=3 the following picture:
Code: Select all
      C     
     / \   
    /   \   
   B_   _B 
  /   A   \
 /    |    \
C-----B-----C

Here A represents the vertices of the first layer, B those of the second, and C those of the third. All 3 types occur as top-and-bottom pairs. The hexagons from the chamfered polyhedra then would be represented by any such 'C---B---C' side views. The projections of A-B and B-C respectively into the picture plane, then would be orthogonal (for any R). Esp. the cycle (ABCB) generally is no rhomb. Rather it is a kite. For R<4 (as in the picture) the sharper corner is at C and the flatter one at A. For R>4 it would be reversed. Accordingly we get y<x for R<4 and y>x for R>4.

 
This withdrawel does not mean that chamfered polychora would not exist. Rather it will just state that in general we cannot get all edges to be made of the same size. Solely in the special case of R=4 this can be achieved. Therefore the according incidence matrix (provided in the other post) remains fully valid.
The chamfered hexadecachoron:
Code: Select all
abx3ooo3ooc4odo&#zx = a-hex + (b,d)-sidpith + (x,c)-rico

o..3o..3o..4o..     | 8  *  * |  8   0  0 | 12  0  0 |  6  0  verf: cube
.o.3.o.3.o.4.o.     | * 64  * |  1   3  0 |  3  3  0 |  3  1
..o3..o3..o4..o     | *  * 96 |  0   2  2 |  1  4  1 |  2  2
--------------------+---------+-----------+----------+------
oo.3oo.3oo.4oo.&#x  | 1  1  0 | 64   *  * |  3  0  0 |  3  0
.oo3.oo3.oo4.oo&#x  | 0  1  1 |  * 192  * |  1  2  0 |  2  1
..x ... ... ...     | 0  0  2 |  *   * 96 |  0  2  1 |  1  2
--------------------+---------+-----------+----------+------
... ... ... odo&#xt | 1  2  1 |  2   2  0 | 96  *  * |  2  0  rhomb {(r,R)^2}
.bx ... .oc ...&#zx | 0  2  4 |  0   4  2 |  * 96  * |  1  1  axial hexagon {(h,H,H)^2}
..x3..o ... ...     | 0  0  3 |  0   0  3 |  *  * 32 |  0  2  regular triangle {3}
--------------------+---------+-----------+----------+------
abx ... ooc4odo&#zx | 2  8  8 |  8  16  4 |  8  4  0 | 24  *  elongated square rhombohedron
.bx3.oo3.oc ...&#zx | 0  4 12 |  0  12 12 |  0  6  4 |  * 16  chamfered tetrahedron


The incidence matrices for the other provided cases clearly would provide the correct topology, i.e. the numbers all would be correct. Only the assumption of having y=x there too was wrong.

In that specific case of the chamfered hexadecachoron we then would have
  • a,b,c,d all pseudo edges,
  • d = q*q/2 = x (shorter diagonal of rhombs),
  • c = q (width of hexagons),
  • b = w = x+q (total tip-2-tip diagonal of hexagons),
  • a = 2(b-x)+x = 2b-x = 2q+x (total tip-2-tip diagonal of elong. rhombohedra),

 
Therefore, in the stronger sense of asking for all equal edge sizes as well, there is only a single finite chamfered version of regular polychora, the chamfered {3,3,4}.
WRT infinite polychora the above restriction would imply the existance of chamfered {P,Q,4} alike (providing euclidean or hyperbolic honeycombs).

--- rk
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Re: Tegum polytopes

Postby Klitzing » Fri Mar 31, 2017 8:45 am

Dear all,

the last days I re-thought the topic of partial Stott expansions / contractions. Original Stott expansions / contractions are meant to use some elements under the full symmetry of the object, say vertices, edges, some face type etc., and pull them appart while introducing new faces inbetween, respectively its reverse process of deleting such faces inbetween. Partial Stott expansions in contrast use some subsymmetry only. This then might imply as well that larger patches of the starting figure are to be pulled apart as a single unit.

For e.g. we have a Stott expansion from the octahedron (oct, x3o4o) which pulls the triangles radially apart and results in the small rhombicuboctahedron (sirco, x3o4x). Or we have a further Stott expansion which starts with the cube (cube, o3o4x) and pulls the edges radially apart and results in the truncated cube (tic, o3x4x). In fact, all the Wythoff constructed polytopes could be derived from the regular ones by these Stott expansions. And indeed it was both this influence of Stott and of Dynkin on Coxeter, which finally lead to our nowadays decorated Coxeter-Dynkin symbols (whereof the above examples are just Krieger type typewriter friendly representations). Therefore all Stott expansions and contractions of Wythoffian polytopes again result in Wythoffians. And thus all are describable with such single layered Coxeter-Dynkin symbols.

Partial Stott expansions in contrast take refuge in some subsymmetry only. For e.g. we could expand the octahedron along all 3 cartesian directions independently. For sure, when doing so within all 3 directions at the same time, we would be back to the normal Stott expansion. But when doing so in 1 direction only, we would result in the elongated square dipyramid (esquidpy, J15), when doing so in 2 such directions, we would result in the square orthobicupola (squobcu, J28). - So far I derived that already way back in 2013, exactly 100 years after Alicia Boole-Stotts own according publication.

But then it occured to me, that it kind of depends of the former orientation of edges, faces etc. of the starting figure wrt. the being applied subsymmetry, whether or not the final result would still use regular polygonal faces only. All my so far considered cases did for sure. But, for instance, when starting with a simple square, we could pull the pairs of adjoining edges apart, to result within an irregular hexagon with corner angles being 90, 135, 135, 90, 135, 135 degrees (in that sequence). - By use of the layered Krieger-Dynkin symbols (of lace towers) that one could be described as xwx&#xt or as oqqo&#xt (where q represents a (pseudo) edge of size sqrt(2) and w=x+q is a similar of size 1+sqrt(2)). By use of the Gevaert-Krieger symbols deviced for tegum sums (i.e. hulls of compounds) it well could be also described as wx oq&#zx. - A second such expansion (of adjoined edge triples) into the orthogonal direction then would result in the regular octagon again (x8o = x4x = xwwx&#xt = wx xw&#zx). And indeed, it is right this diametrally expanded square, which occurs very often in such partial Stott expansion cases. In fact in every application, where the starting figure uses 45 degrees rotated squares wrt. to the chosen subsymmetry application.

When allowing for that irregular hexagon as an additional to be used polygonal face, we can enlarge the already be known partial Stott expansions be quite a lot further figures.

E.g. within 3D we have:

  • pacop = partially contracted octagonal prism = xxx xwx&#xt = xxxx oqqo&#xt, i.e. the prism of that irregular hexagon. - Faces are 2 irregular hexagons and 6 squares. - It occurs within the sequence: cube <-> pacop <-> op.
  • pexco = partially expanded cuboctahedron = wx oq4xo&#zx = oqqo4xoox&#xt. - Faces are 4 irregular hexagons, 2 squares, 8 triangles. - It occurs within the sequence: co <-> pexco <-> pactic <-> tic.
  • pactic = partially contracted truncated cube = qo xw4xo&#zx = xwx4xox&#xt. - Faces are 4 irregular hexagons, 2 octagons, 8 triangles. - It occurs within the sequence: co <-> pexco <-> pactic <-> tic.
  • pextoe = partially expanded truncated octahedron = xux4ooq Xwx&#zxt = xuxux4ooqoo&#xt (where X=x+2q=w+q). - Faces are 4 irregular hexagons, 6 squares, 8 (regular) hexagons. - It occurs within the sequence: toe <-> pextoe <-> pac girco <-> girco.
  • pac girco = partially contracted great rhombicuboctahedron = xuxux4xxwxx&#xt = Qqo xux4xxw&#zxt (where Q=2q). - Faces are 4 irregular hexagons, 8 squares, 8 (regular) hexagons, 2 octagons. - It occurs within the sequence: toe <-> pextoe <-> pac girco <-> girco.
  • patex cube = partially tetrahedrally-expanded cube = chamfered tetrahedron = wx3oo3oq&#zx. - Faces are 6 irregular hexagons, 4 triangles. - It occurs within the sequence: cube <-> patex cube <-> tic.
  • patex sirco = partially tetrahedrally-expanded small rhombicuboctahedron = wx3xx3oq&#zx. - Faces are 6 irregular hexagons, 12 squares, 4 triangles, 4 (regular) hexagons. - It occurs within the sequence: sirco <-> patex sirco <-> girco.

Note, that the regular hexagons in the last 2 cases occur as Stott expanded triangles: x3o <-> x3x = x6o.

(Up to my knowledge, the here provided Bowers-style acronyms do not provide any clashes to other OBSAs. - Jonathan, would these get your ok as well?)

Stay tuned for the according applications within 4D!

--- rk
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Re: Tegum polytopes

Postby Klitzing » Sat Apr 01, 2017 8:12 am

Here I continue on the lately investigated set of the partially Stott expanded polytopes which include that irregular 90, 135, 135, 90, 135, 135 degrees hexagon. (Mere CRF ones were already provided at https://bendwavy.org/klitzing/explain/stott.htm since 2013.) Now I will consider such cases of 4D. In this post I will describe from that realm more precisely those partial expansions, which would be derived solely by independently chosen cartesian axial directions. Within that subset we have:

  • pexrit = partially expanded rectified tesseract = oo3xo4oq wx&#zx = oooo3xoox4oqqo&#xt. - It has a total of 40 vertices. Cells are 2 coes, 16 tets, and 6 pexcoes. - It occurs within the sequence: rit = o3o3x4o <-> pexrit <-> pabexrit <-> pactat <-> tat = o3o3x4x.
  • pabexrit = partially bi-expanded rectified tesseract = xo4oq ox4wx&#zx. - It has a total of 48 vertices. Cells are 16 tets, 4 pexcoes, and 4 pactics. - It occurs within the sequence: rit = o3o3x4o <-> pexrit <-> pabexrit <-> pactat <-> tat = o3o3x4x.
  • pactat = partially contracted truncated tesseract = oo3xo4xw qo&#zx = ooo3xox4xwx&#xt. - It has a total of 56 vertices. Cells are 2 tics, 16 tets, and 6 pactics. - It occurs within the sequence: rit = o3o3x4o <-> pexrit <-> pabexrit <-> pactat <-> tat = o3o3x4x.
  • pexrico = partially expanded rectified icostetrachoron = oxx3xxo4ooq Xwx&#zxt = oxxxxo3xxooxx4ooqqoo&#xt. - It has a total of 120 vertices. Cells are 18 coes, 12 cubes, 8 trips, 12 pacops, and 6 pexcoes. - It occurs within the sequence: rico = x3o3x4o <-> pexrico <-> pabexrico <-> pacproh <-> proh = x3o3x4x.
  • pabexrico = partially bi-expanded rectified icostetrachoron = xuxo4qooq xoxu4xwwx&#zx. - It has a total of 144 vertices. Cells are 4 cubes, 16 coes, 16 trips, 4 ops, 16 pacops, 4 pexcoes, and 4 pactics. - It occurs within the sequence: rico = x3o3x4o <-> pexrico <-> pabexrico <-> pacproh <-> proh = x3o3x4x.
  • pacproh = partially contracted prismatorhombated hexadecachoron = oxx3xxo4xxw Qqo&#zxt = oxxxo3xxoxx4xxwxx&#xt. - It has a total of 168 vertices. Cells are 16 coes, 2 tics, 24 trips, 12 ops, 12 pacops, and 6 pactics. - It occurs within the sequence: rico = x3o3x4o <-> pexrico <-> pabexrico <-> pacproh <-> proh = x3o3x4x.
  • pextah = partially expanded tesseractihexadecachoron = xoo3xux4ooq Xwx&#zxt = xoooox3xuxxux4ooqqoo&#xt. - It has a total of 120 vertices. Cells are 2 toes, 16 tuts, 8 trips, and 6 pextoes. - It occurs within the sequence: tah = o3x3x4o <-> pextah <-> pabextah <-> pac grit <-> grit = o3x3x4x.
  • pabextah = partially bi-expanded tesseractihexadecachoron = xuxo4ooqQ oxux4Xwxx&#zx. - It has a total of 144 vertices. Cells are 16 tuts, 16 trips, 4 pextoes, and 4 pac gircoes. - It occurs within the sequence: tah = o3x3x4o <-> pextah <-> pabextah <-> pac grit <-> grit = o3x3x4x.
  • pac grit = partially contracted great rhombated tesseract = xoo3xux4xxw Qqo&#zxt = xooox3xuxux4xxwxx&#xt. - It has a total of 168 vertices. Cells are 2 gircoes, 16 tuts, 24 trips, and 6 pac gircoes. - It occurs within the sequence: tah = o3x3x4o <-> pextah <-> pabextah <-> pac grit <-> grit = o3x3x4x.
  • pextico = partially expanded truncated icositetrachoron = xuxx3xxux4oooq AXwx&#zxt (where A=X+q=w+Q) = xuxxxxux3xxuxxuxx4oooqqooo&#xt. - It has a total of 240 vertices. Cells are 12 cubes, 18 toes, 8 hips, 12 pacops, and 6 pextoes. - It occurs within the sequence: tico = x3x3x4o <-> pextico <-> pabextico <-> pac gidpith <-> gidpith = x3x3x4x.
  • pabextico = partially bi-expanded truncated icositetrachoron = xuxDux4ooqoqQ xuDxux4Xwxwxx&#zx (where D=3x). - It has a total of 288 vertices. Cells are 4 cubes, 16 toes, 16 hips, 4 ops, 16 pacops, 4 pextoes, and 4 pac gircoes. - It occurs within the sequence: tico = x3x3x4o <-> pextico <-> pabextico <-> pac gidpith <-> gidpith = x3x3x4x.
  • pac gidpith = partially contracted great disprismated tesseractihexadecachoron = xuxx3xxux4xxxw ZQqo&#zxt (where Z=3q) = xuxxxux3xxuxuxx4xxxwxxx&#xt. - It has a total of 336 vertices. Cells are 2 gircoes, 16 toes, 24 hips, 12 ops, 12 pacops, and 6 pac gircoes. - It occurs within the sequence: tico = x3x3x4o <-> pextico <-> pabextico <-> pac gidpith <-> gidpith = x3x3x4x.

Stay tuned for the next post with the partially Stott expanded polychora wrt. hexadecachoral subsymmetry which include that diagonally elongated square!

--- rk
Last edited by Klitzing on Sun Apr 02, 2017 11:44 am, edited 1 time in total.
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Re: Tegum polytopes

Postby Klitzing » Sun Apr 02, 2017 10:55 am

This time I want to elaborate on the partial Stott expansions of 4D polychora, having that diagonally elongated square as irregular polygonal face. Esp. on those which would emerge while applying hexadecachorally subsymmetric expansions. (The mere CRF ones of this subsymmetry where already contained within https://bendwavy.org/klitzing/explain/stott.htm - since 2013.)

  • pox rico = partially octachorally-expanded rectified icositetrachoron = wx3oo3oq4xo&#zx. - It has a total of 160 vertices. Its cells are 8 cubes, 24 pexcoes, and 16 patex cubes. - It occurs within the sequence: rico = o3x4o3o <-> pox rico <-> poccont <-> cont = o3x4x3o.
  • poccont = partially octachorally-contracted tetrachontoctachoron = qo3oo3xw4xo&#zx. - It has a total of 224 vertices. Its cells are 8 tics, 24 pactics, and 16 patex cubes. - It occurs within the sequence: rico = o3x4o3o <-> pox rico <-> poccont <-> cont = o3x4x3o.
  • (As an aside, rico itself would be describable within Gevart-Krieger tegum-sum notation as qo3oo3oq4xo&#zx. And cont there would be wx3oo3xw4xo&#zx.)
  • pox srico = partially octachorally-expanded small rhombated icositetrachoron = wx3xx3oq4xo&#zx. - It has a total of 384 vertices. Its cells are 8 sircoes, 96 trips, 24 pexcoes, and 16 patex sircoes. - It occurs within the sequence: srico = o3x4o3x <-> pox srico <-> pibox srico <-> grico = o3x4x3x.
  • pibox srico = partially bi-octachorally-expanded small rhombated icositetrachoron = qo3xx3xw4xo&#zx. - It has a total of 480 vertices. Its cells are 8 gircoes, 96 trips, 24 pactics, and 16 patex sircoes. - It occurs within the sequence: srico = o3x4o3x <-> pox srico <-> pibox srico <-> grico = o3x4x3x.
  • (As an aside, srico itself would be describable within Gevart-Krieger tegum-sum notation as qo3xx3oq4xo&#zx. And grico there would be wx3xx3xw4xo&#zx.)
  • pox tico = partially octachorally-expanded truncated icositetrachoron = Xwx3ooo3ooq4xux&#zxt (where here and below X=w+q=x+Q). - It has a total of 320 vertices. Its cells are 8 cubes, 32 trips, 24 pextoes, and 16 patex cubes. - It occurs within the sequence: tico = x3x4o3o <-> pox tico <-> pibox tico <-> grico = x3x4x3o.
  • pibox tico = partially bi-octachorally-expanded truncated icositetrachoron = Qqo3ooo3xxw4xux&#zxt (where here and below Q=2q). - It has a total of 448 vertices. Its cells are 8 tics, 64 trips, 24 pac gircoes, and 16 patex cubes. - It occurs within the sequence: tico = x3x4o3o <-> pox tico <-> pibox tico <-> grico = x3x4x3o.
  • (As an aside, tico itself would be describable within Gevart-Krieger tegum-sum notation as Qqo3ooo3ooq4xux&#zxt. And grico there would be Xwx3ooo3xxw4xux&#zxt.)
  • pox prico = partially octachorally-expanded prismatorhombated icositetrachoron = Xwx3xxx3ooq4xux&#zxt. - It has a total of 768 vertices. Its cells are 8 sircoes, 64 trips, 128 hips, 24 pextoes, and 16 patex sircoes. - It occurs within the sequence: prico = x3x4o3x <-> pox prico <-> poc gippic <-> gippic = x3x4x3x.
  • poc gippic = partially octachorally-contracted great prismatotetracontoctachoron = Qqo3xxx3xxw4xux&#zxt. - It has a total of 960 vertices. Its cells are 8 gircoes, 32 trips, 160 hips, 24 pac gircoes, and 16 patex sircoes. - It occurs within the sequence: prico = x3x4o3x <-> pox prico <-> poc gippic <-> gippic = x3x4x3x.
  • (As an aside, prico itself would be describable within Gevart-Krieger tegum-sum notation as Qqo3xxx3ooq4xux&#zxt. And gippic there would be Xwx3xxx3xxq4xux&#zxt.)
It should be noted that grico here occurs in both its orientations as x3x4x3o and as o3x4x3x. Therefore a name like “poc grico” would not be unique. This is why I took refuge to these 2 “pibox”-names instead.

I additionally just want to remind you on the usage of “&#zxt”. “&” means “and”, “additionally”. “#” means “lacing”. The leading “z” means we have a zero-height lacing, i.e. the layer polytopes are within the same hyperspace, the lace prism (or lace tower, lace simplex) becomes degenerate. This is what is required for tegum sums, which are nothing but hulls of compounds. Here the compound components would be right these layers. And finally, for tegum sums of more than 2 “layers” usually not all layers would be connected to any of the others by direct lacings. This is why I usually write something like “all existing heights = 0”. Whenever only pairwise lacings exist, which make up a mere sequence, I therefore also use, to highlight this fact, the notation “&#zxt”, i.e. lacings occur only in a lace tower like fashion.

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Re: Tegum polytopes

Postby username5243 » Sun Jun 04, 2017 6:05 pm

While I was looking at those lists, I thought of another possible expansion, this time in demi-tessic subsymmetry. These will still use that irregular hexagon, as yours do.

Tes (x4o3o3o, tesseract) can be written as qo3oo3oq *b3oo&#zx, while tat (x4x3o3o, truncated tesseract) can be written as wx3oo3xw *b3oo&#zx. So I thought that there should be a qo3oo3xw &#zx too, that uses that hexagon among its faces (I'm fairly sure it's still all unit-edged. Further, similar things should work with some other tessics - I'm fairly sure there is this kind of demi-tessic intermediate between srit (x4o3x3o) and grit (x4x3x3o), sidpith (x4o3o3ox) and proh (x4o3o3x), as well as prit (x4o3x3x) and gidpith (x4x3x3x.

I haven't worked these out in full as to what cells they should have, but they would use patex cube or patex sirco as some of its cells. I guess we could name them something like this:

Code: Select all
qo3oo3xw *b3oo&#zx = pahtex tes (partially demitesseractically expanded tesseract)
qo3xx3xw *b3oo&#zx = pahtex srit (partially demitesseractically expanded small rhombated tesseract)
qo3oo3xw *b3xx&#z = pahtex sidpith (partially demitesseractically expanded small disprismatotesseractihexadecachoron)
qo3xx3xw *b3xx&#zx = pahtex prit (partially demitesseractically expanded prismatorhombated tesseract)


Are these four valid polychora? After this, I can't think of too many more undiscovered ones (not counting the prisms of the 3D partial expansions)...
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Re: Tegum polytopes

Postby Klitzing » Sun Jun 04, 2017 10:39 pm

Great idea, username5243!

and here is the first of those:
Code: Select all
wx3oo3oq *b3oo&#zx -> height = 0
(tegum sum of w-hex and (x,q)-rit)

o.3o.3o. *b3o.     | 8  * |  4  0 |  6  0 | 4 0 verf: tet
.o3.o3.o *b3.o     | * 32 |  1  3 |  3  3 | 3 1
-------------------+------+-------+-------+----
oo3oo3oo *b3oo&#x  | 1  1 | 32  * |  3  0 | 3 0
.x .. ..    ..     | 0  2 |  * 48 |  1  2 | 2 1
-------------------+------+-------+-------+----
wx .. oq    ..&#zx | 2  4 |  4  2 | 24  * | 2 0
.x3.o ..    ..     | 0  3 |  0  3 |  * 32 | 1 1
-------------------+------+-------+-------+----
wx3oo3oq    ..&#zx | 4 12 | 12 12 |  6  4 | 8 * patex cube
.x3.o .. *b3.o     | 0  4 |  0  6 |  0  4 | * 8 tet
= demitessically truncated tes

Others tomorrow ;-)
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Re: Tegum polytopes

Postby Klitzing » Mon Jun 05, 2017 6:50 am

next one: pahtex srit
Code: Select all
wx3xx3oq *b3oo&#zx -> height = 0
(tegum sum of (w,x)-thex and (x,x,q)-tah)

o.3o.3o. *b3o.     | 48  * |  4  2  0  0 |  2  2  1  4  0  0 | 1 2  2 0
.o3.o3.o *b3.o     |  * 96 |  0  1  1  2 |  0  0  1  2  2  1 | 0 2  1 1
-------------------+-------+-------------+-------------------+---------
.. x. ..    ..     |  2  0 | 96  *  *  * |  1  1  0  1  0  0 | 1 1  1 0
oo3oo3oo *b3oo&#x  |  1  1 |  * 96  *  * |  0  0  1  2  0  0 | 0 2  1 0
.x .. ..    ..     |  0  2 |  *  * 48  * |  0  0  1  0  2  0 | 0 2  0 1
.. .x ..    ..     |  0  2 |  *  *  * 96 |  0  0  0  1  1  1 | 0 1  1 1
-------------------+-------+-------------+-------------------+---------
.. x.3o.    ..     |  3  0 |  3  0  0  0 | 32  *  *  *  *  * | 1 1  0 0
.. x. .. *b3o.     |  3  0 |  3  0  0  0 |  * 32  *  *  *  * | 1 0  1 0
wx .. oq    ..&#zx |  2  4 |  0  4  2  0 |  *  * 24  *  *  * | 0 2  0 0
.. xx ..    ..&#x  |  2  2 |  1  2  0  1 |  *  *  * 96  *  * | 0 1  1 0
.x3.x ..    ..     |  0  6 |  0  0  3  3 |  *  *  *  * 32  * | 0 1  0 1
.. .x .. *b3.o     |  0  3 |  0  0  0  3 |  *  *  *  *  * 32 | 0 0  1 1
-------------------+-------+-------------+-------------------+---------
.. x.3o. *b3o.     |  6  0 | 12  0  0  0 |  4  4  0  0  0  0 | 8 *  * * oct
wx3xx3oq    ..&#zx | 12 24 | 12 24 12 12 |  4  0  6 12  4  0 | * 8  * * patex sirco
.. xx .. *b3oo&#x  |  3  3 |  3  3  0  3 |  0  1  0  3  0  1 | * * 32 * trip
.x3.x .. *b3.o     |  0 12 |  0  0  6 12 |  0  0  0  0  4  4 | * *  * 8 tut

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Re: Tegum polytopes

Postby Klitzing » Mon Jun 05, 2017 8:50 am

next one: pahtex sidpith
Code: Select all
wx3oo3oq *b3xx&#zx -> height = 0
(tegum sum of (w,x)-rit and (x,x,q)-rico)

o.3o.3o. *b3o.     | 32  * |  3  3  0  0 |  3  3  6  0  0  0 | 1 1  3  3 0
.o3.o3.o *b3.o     |  * 96 |  0  1  2  2 |  0  2  2  1  2  1 | 0 1  2  1 1
-------------------+-------+-------------+-------------------+------------
.. .. ..    x.     |  2  0 | 48  *  *  * |  2  0  2  0  0  0 | 1 0  1  2 0
oo3oo3oo *b3oo&#x  |  1  1 |  * 96  *  * |  0  2  2  0  0  0 | 0 1  2  1 0
.x .. ..    ..     |  0  2 |  *  * 96  * |  0  1  0  1  1  0 | 0 1  1  0 1
.. .. ..    .x     |  0  2 |  *  *  * 96 |  0  0  1  0  1  1 | 0 0  1  1 1
-------------------+-------+-------------+-------------------+------------
.. o. .. *b3x.     |  3  0 |  3  0  0  0 | 32  *  *  *  *  * | 1 0  0  1 0
wx .. oq    ..&#zx |  2  4 |  0  4  2  0 |  * 48  *  *  *  * | 0 1  1  0 0
.. .. ..    xx&#x  |  2  2 |  1  2  0  1 |  *  * 96  *  *  * | 0 0  1  1 0
.x3.o ..    ..     |  0  3 |  0  0  3  0 |  *  *  * 32  *  * | 0 1  0  0 1
.x .. ..    .x     |  0  4 |  0  0  2  2 |  *  *  *  * 48  * | 0 0  1  0 1
.. .o .. *b3.x     |  0  3 |  0  0  0  3 |  *  *  *  *  * 32 | 0 0  0  1 1
-------------------+-------+-------------+-------------------+------------
.. o.3o. *b3x.     |  4  0 |  6  0  0  0 |  4  0  0  0  0  0 | 8 *  *  * * tet
wx3oo3oq    ..&#zx |  4 12 |  0 12 12  0 |  0  6  0  4  0  0 | * 8  *  * * patex cube
wx .. oq    xx&#zx |  4  8 |  2  8  4  4 |  0  2  4  0  2  0 | * * 24  * * pacop
.. oo .. *b3xx&#x  |  3  3 |  3  3  0  3 |  1  0  3  0  0  1 | * *  * 32 * trip
.x3.o .. *b3.x     |  0 12 |  0  0 12 12 |  0  0  0  4  6  4 | * *  *  * 8 co

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Re: Tegum polytopes

Postby Klitzing » Mon Jun 05, 2017 11:15 am

And here now is the final one of this series:
pahtex prit
Code: Select all
wx3xx3oq *b3xx&#zx -> height = 0
(tegum sum of (w,x,x)-tah and (x,x,x,q)-tico)

o.3o.3o. *b3o.     | 96   * |  2  1   2  0  0  0 |  1  2  1  2  2  0  0  0 | 1 1  1  2 0
.o3.o3.o *b3.o     |  * 192 |  0  0   1  1  1  1 |  0  0  1  1  1  1  1  1 | 0 1  1  1 1
-------------------+--------+--------------------+-------------------------+------------
.. x. ..    ..     |  2   0 | 96  *   *  *  *  * |  1  1  0  1  0  0  0  0 | 1 1  0  1 0
.. .. ..    x.     |  2   0 |  * 48   *  *  *  * |  0  2  0  0  2  0  0  0 | 1 0  1  2 0
oo3oo3oo *b3oo&#x  |  1   1 |  *  * 192  *  *  * |  0  0  1  1  1  0  0  0 | 0 1  1  1 0
.x .. ..    ..     |  0   2 |  *  *   * 96  *  * |  0  0  1  0  0  1  1  0 | 0 1  1  0 1
.. .x ..    ..     |  0   2 |  *  *   *  * 96  * |  0  0  0  1  0  1  0  1 | 0 1  0  1 1
.. .. ..    .x     |  0   2 |  *  *   *  *  * 96 |  0  0  0  0  1  0  1  1 | 0 0  1  1 1
-------------------+--------+--------------------+-------------------------+------------
.. x.3o.    ..     |  3   0 |  3  0   0  0  0  0 | 32  *  *  *  *  *  *  * | 1 1  0  0 0
.. x. .. *b3x.     |  6   0 |  3  3   0  0  0  0 |  * 32  *  *  *  *  *  * | 1 0  0  1 0
wx .. oq    ..&#zx |  2   4 |  0  0   4  2  0  0 |  *  * 48  *  *  *  *  * | 0 1  1  0 0
.. xx ..    ..&#x  |  2   2 |  1  0   2  0  1  0 |  *  *  * 96  *  *  *  * | 0 1  0  1 0
.. .. ..    xx&#x  |  2   2 |  0  1   2  0  0  1 |  *  *  *  * 96  *  *  * | 0 0  1  1 0
.x3.x ..    ..     |  0   6 |  0  0   0  3  3  0 |  *  *  *  *  * 32  *  * | 0 1  0  0 1
.x .. ..    .x     |  0   4 |  0  0   0  2  0  2 |  *  *  *  *  *  * 48  * | 0 0  1  0 1
.. .x .. *b3.x     |  0   6 |  0  0   0  0  3  3 |  *  *  *  *  *  *  * 32 | 0 0  0  1 1
-------------------+--------+--------------------+-------------------------+------------
.. x.3o. *b3x.     | 12   0 | 12  6   0  0  0  0 |  4  4  0  0  0  0  0  0 | 8 *  *  * * tut
wx3xx3oq    ..&#zx | 12  24 | 12  0  24 12 12  0 |  4  0  6 12  0  4  0  0 | * 8  *  * * patex sirco
wx .. oq    xx&#zx |  4   8 |  0  2   8  4  0  4 |  0  0  2  0  4  0  2  0 | * * 24  * * pacop
.. xx .. *b3xx&#x  |  6   6 |  3  3   6  0  3  3 |  0  1  0  3  3  0  0  1 | * *  * 32 * hip
.x3.x .. *b3.x     |  0  24 |  0  0   0 12 12 12 |  0  0  0  0  0  4  6  4 | * *  *  * 8 toe

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Re: Tegum polytopes

Postby Klitzing » Tue Jun 13, 2017 3:58 pm

I revisited the grand antiprism lately.
This one is the exeptional convex uniform polychoron being found only in 1965 by Conway and Guy.

It occured to me that it could be given in tegum sum notation as xofo-5-oxof-2-ofxo-5-foox-&#zx, describing its vertex set as that of a compound of 4 differently oriented x5o2f5o, i.e. non-uniform variations of the dipentagonal duoprisms. It then happens that all layer lacings (of the tegum sum) exist as edges, the f-edges of those duoprisms would become false internal "edges". All true edges would thus be of size "x".


Then I considered the dual of the grand antiprism.

It then occured to me that it too can be given as tegum sum within the same symmetry. In fact it could then be described as of|foxfv-5-of|ofxvf-2-fo|fvxof-5-fo|vfxfo-&#z(f,v). There the nodes before the pipe represent the vertices, which occur as duals of the pentagonal antiprism cells, while the nodes after the pipe represent the vertices, which occur as duals of the tetrahedra. The lacing symbol "&#z(f,v)" here shall denote that we still have a zero height layering of the individual components, that is, we again consider the hull of the respective compound, and that the lacing edges are either of size f = 1.618 or v = 0.618. In fact, the formers would occur between layer pairings 1-3, 1-4, 2-6, 2-7, while the latter occurs between pairings 3-5, 4-5, 5-6, 5-7. In fact, all true edges of this polychoron are either the in-layer f-edges of the layers before the pipe, the just described f-lacings, the in-layer v-edges of the layers after the pipe, and the just described v-lacings.

The cells of the dual of the grand antiprism could best be described as a subsymmetrical stellation of the dodecahedron. In fact onto which 2 small stellated dodecahedron tips would be attached at 2 neighbouring pentagons, and inbetween these peaks a wedge type simplex would be spanned (in order to get it convex again). That one would look like the edge support of the great dodecahedron. Edges are of 2 sizes, again those of the dual of the grand antiprism, for sure. Then the remaining ones of the dodecahedron kernel are of size v, while those of the attaching pyramids, which are collinear with original dodecahedral ones, thus become of size f. And the spanning edge between those tips too will be of size f.

This cell polyhedron could be given as ofoxv-2-fvfxo-&#(f,v)t too. Here the trailing "&#(f,v)t" denotes that we have a true lace prism representation. The respective heights would be (within that being used edge scaling) height(1,2) = (3+sqrt(5))/4 = 1.309017, height(2,3) = height(4,5) = (sqrt(5)-1)/4 = 0.309017, height(3,4) = (3-sqrt(5))/4 = 0.190983.

Faces of these cells are v-scaled pentagons, kites with edge cycle vvff (with corner angle of 36 degrees between the 2 Long edges, while all others are 108 degrees), and trapezia with edge cycle vfff (corner angles being 72 degrees resp. 108 degrees).

The vertex figures of the dual of the grand antiprism also would fall into 2 types. Those for the vertices, which are represented by the node layers after the pipe, clearly are tetrahedra. This is because the grand antiprism also has some tetrahedra for faces, and their duals are tetrahedra again. The other vertex figure type will be a pentagonal antitegum or antibipyramid, as those are the duals of the pentagonal antiprisms of the grand antiprism. The faces of those antitegums then are exactly the same kites as before. In fact, just as the antiprism itself could be obtained as the parabidiminishing of the icosahedron, this antitegum could be obtained a a parabiapiculation of a dodecahedron. Accordingly this antitegum can be given as oxoo5ooxo&#(f,v,f)t. Here the polar edges are the long ones, while the tropal zick-zack uses the short ones. The in-layer x-edges would feature only as diagonals of the kites and thus in fact are false ones. The respective heights between the layers of this tower representation then are height(1,2) = height(3,4) = sqrt[(5+2 sqrt(5))/5] = 1.376382, resp. height(2,3) = sqrt[(5-2 sqrt(5))/5] = 0.324920.


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Re: Tegum polytopes

Postby username5243 » Tue Jun 13, 2017 11:56 pm

I'm fairly sure most of the duals of the convex uniform polytopes could be represented in that tegum-sum notation. I think I know whta their symbols are like, though I'm not sure what the lengths of the edges represented by the nodes are. For example, the dual of srid would be given as aoo3obo5ooc&#z(y,z) for some pseudo-edge lengths a, b, and c, and some (real) edge lengths y and z, though I don't know what those lengths are. Most of the other duals of uniforms have similar representations.
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Re: Tegum polytopes

Postby Klitzing » Wed Jun 14, 2017 6:54 am

Ha, srip (small rhombated pentachoron, x3o3x3o) has indeed for dual m3o3m3o = aoo3obo3ooo3oox&#z(c,d,e), where a = 2/3 = 0.666667 (pseudo), b = 4/7 = 0.571429 (pseudo), c = lacing(1,2) = 8/21 = 0.380952, d = lacing(2,3) = 3 sqrt(2)/7 = 0.606092, e = lacing(1,3) = a = 2/3 = 0.666667. Cells then are bipyramids on the pseudo triangle with edge cycle (xee).


Well, I even know the structure of sadi (snub disicositetrachoron) = idex (icositetra-diminished hexacosachoron) as well of its dual quidex (quadri-icositetra-diminished hexacosachoron). In fact sadi has incidence matrix
Code: Select all
demi( . . . . ) | 96 |   3   6 |  3   9  3 |  3  1  4  verf = teddi
----------------+----+---------+-----------+---------
      . s4o .   |  2 | 144   * |  0   2  2 |  1  1  2
sefa( s3s . . ) |  2 |   * 288 |  1   2  0 |  2  0  1
----------------+----+---------+-----------+---------
      s3s . .   |  3 |   0   3 | 96   *  * |  2  0  0
sefa( s3s4o . ) |  3 |   1   2 |  * 288  * |  1  0  1
sefa( . s4o3o ) |  3 |   3   0 |  *   * 96 |  0  1  1
----------------+----+---------+-----------+---------
      s3s4o .   | 12 |   6  24 |  8  12  0 | 24  *  *  ike
      . s4o3o   |  4 |   6   0 |  0   0  4 |  * 24  *  tet
sefa( s3s4o3o ) |  4 |   3   3 |  0   3  1 |  *  * 96  tet

while quidex has
Code: Select all
24  *  * |  8  12  0 |   6  24 | 12  : verf = doe
 * 24  * |  0   0  4 |   6   0 |  4  : verf = tet
 *  * 96 |  0   3  1 |   3   3 |  4  : verf = tet
---------+-----------+---------+---
 2  0  0 | 96   *  * |   0   3 |  3  : F-edges
 1  0  1 |  * 288  * |   1   2 |  3  : f-edges
 0  1  1 |  *   * 96 |   3   0 |  3  : v-edges
---------+-----------+---------+---
 1  1  2 |  0   2  2 | 144   * |  2  : oxo&#(v,f)t kites
 2  0  1 |  1   2  0 |   * 288 |  2  : oF&#f
---------+-----------+---------+---
 3  1  4 |  3   9  3 |   3   6 | 96  : titdi

where teddi = trigonally-diminished icosahedron and titdi = tri-trigonally-diminished icosahedron. The latter also could be described as tri-apiculated dodecahedron. That latter figure then is
Image
and can be represented as oxoo3ooFo&#(v,f,f)t.

But I neither have a tegum sum representation for sadi nor for quidex. - Could you come up with those?

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Re: Tegum polytopes

Postby wendy » Wed Jun 14, 2017 7:52 am

Username5243 is indeed correct on the duals. It's actually straignt-forward. If 'S_ij is the stott matrix for the symmetry, and v_i is the vector representing the marked nodes, then S_ij.V_i = W_j represents the equation of the plane of the faces. One then creates a second vector K/W_j, where K is an arbitary constant. This gives the intercepts on each of the axies, and in turn the edges of the single-node polytopes that the figure is the tegum-sum of.

Most of these are interior, it's only the ones where the original figures have face-nodes that we get vertices.

For example, the polytope o3x4o gives a vector (0,1,0). When this is dotted with the S[3,4] matrix, we get the vector (2, 4, 2q). We see then that the axis-intercepts are then (2, 1, q), when these are divided into 4. This means the dual of o3x4o, corresonds to a tegumsum of uoo3oxo4ooq&#z but since the middle of these figures produce no proud vertices, we can drop it from the compound, to give qo3oo4ox&#z which makes the rhombic dodecahedron the lace-union of a cube o3o4x and an octahedron q3o4o.
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Re: Tegum polytopes

Postby Klitzing » Wed Jun 14, 2017 9:03 am

wendy wrote:but since the middle of these figures produce no proud vertices, we can drop it from the compound

Could you elaborate on this?
It is clear from the outcome here that it ought be dropped. But how could one decide on that in general?
(The remainder of your mail was a greatly apreciated input, Wendy!)
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Re: Tegum polytopes

Postby wendy » Wed Jun 14, 2017 9:48 am

The reason that they can be dropped, is because they do not contribute to the forming of the tegumsum, because their vertex is interior to the faces of what is already there.

Suppose you look at the example given, the dual of o3x4o. The symmetry region forms in the cell, as the vertices of u3o4o, o3x4o, and o3o4q. But the vertices of o3x4o fall in the centre of the rhombs of o3m4o, and the o3m4o only has faces at the nodes marked $ in $3o4o$.

If you were calculating the volume, you do need all three of these measurements (as g K^n/Prod(v_i) n! where g is the order of the group, K is a constant, and n the dimension. But the actual tegum-sum is only where the faces (N-1 surtopes) fall in the figure.
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Re: Tegum polytopes

Postby Klitzing » Wed Jun 14, 2017 11:03 am

okay, I'm with you so far.
But how to decide in general, without laborously calculating the actual vertex coordinates and facet equations, whether some layer does contribute or not?
I think, that ought be already somewhere intrisically coded and thus quite readily deducible from the so far obtained numbers, ain't it?
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Re: Tegum polytopes

Postby username5243 » Wed Jun 14, 2017 11:15 am

I think that just the ones that actually have facets in the uniform versions will have those layers in the dual. Using wendy's example of the dual of co, take the symbol o3x4o. Remove the second node gives o . o (a mere vertex), as opposed to any actual faces. So the dual won't use the second layer vertices.

On another note, I'm not sure this works for the duals of snubs and the like...
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Re: Tegum polytopes

Postby wendy » Wed Jun 14, 2017 11:25 am

The tegum-sum compound is a compound of polytopes whose vertices correspond to the faces of the dual.

"Connected to a vertex" means that the node is either marked or is connected by a chain of non-zero branches to a marked node. It supposes that the marked nodes are connected to a special vertex-node, and the unmarked ones are not.

If you remove a given node, and the remainder of nodes are connected to the vertex-node, then a face forms there. Specifically, it means that a non-zero height prism has formed there. Since the vertices of the catalan correspond to the faces of the uniform, then it is sufficient to find the nodes connected to the vertex-node.

So in a figure like x3o4x3o, you get connections at the first, second and fourth nodes, but the third node isolates the fourth one. In o3x3o5o, the nodes a and d give rise to faces, but removal of b or c would leave at least one unconnected node, and thus no face. In the x3o3o3o3*a, the removal of any node except the first, leaves the rest connected to the vertex, and thus faces would form there.

All of the figures would have their vertices on the same plane as i given them, but if there is no face on the uniform figure, the vertex does not stand proud (that is above the surface), and thus is not a vertex of the face of the catalan.
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Re: Tegum polytopes

Postby wendy » Wed Jun 14, 2017 11:50 am

The dual of the snubs would be a tegum-sum of a snub (for the triangular faces), and the two dual figures (eg cube and octahedron).

It's do-able but a bit tricky, since it requires a few reflections. In part, you construct s3s4s as a3c4e (with general values of a,c,e), so that a²+ac+c² = c²+(q)ce+e² = a²+e² (which gives cubic equation), q=sqrt(2), then You would reflect the point a,c,e in the three mirrors. The first mirror reflects a,c,e into (-a, a+c, e), this is formed by subtracting the columns of the dynkin matrix (here (a,c,e)-a(2,-1,0) , (a,c,e)-c(-1,2,-q), and (a,c,e)-e(0,-q,2)) gives the images of the point (a,c,e) in the three closest mirrors. These are the vertices of a face here. If we suppose that the triangle is equalateral, it is possible to find the vector of the face as [ 3(a,c,e)+a(-2,1,0)+c(1,-2,q)+e(0,q,-2) ]/3, gives (a+c, a+c+qe, qc+e)/3.

You then reflect this point back through the mirrors, to get the three faces relative to a vertex.
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Re: Tegum polytopes

Postby Klitzing » Wed Jun 14, 2017 1:19 pm

Ah, yes, for sure! :idea:
I already got the feeling, that it ought be somehow deducible from the Dynkin diagram of the uniform starting figure.
But I missed the simple connection to the facet derivation thereof, which clearly is responsible for the vertices of the dual.
Thus, thank you both for that missing link. :D



As to the snubs you'd say
* dual( sPsQs ) = asoPosoQosb&#z(c,d,e) ?
* dual( sPsQo ) = asoPosoQoob&#z(c,d,e) ?
* dual( sPoQo ) = asoPoooQoob&#z(c,d,e) ?
each with individually to be derived a,b,c=lacing(1,2),d=lacing(2,3),e=lacing(1,3)

And thus then esp.
* quidex = dual( sadi ) = dual( s3s4o3o ) = aso3oso4ooo3oob&#z(c,d,e) ?

And how has the snub layer usage to be understood there?
Is it meant to be restricted to the single layer only,
or would it be meant as the tegum sum with the un-snubbed layer, which only thereafter will have to be snubbed as a total?

How could the "extremal layers" be considered duals, when applying for the antiprism s2sNs ?
In fact there we should get antitegums, which topologically indeed is something like as2osNos&#zb,
but metrically the therein used tropical antiprism no longer is uniform, rather it is somehow flattened for N>3.
Therefore we don't re-use the edges of s2sNs (together with the lacing b's).

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Re: Tegum polytopes

Postby Klitzing » Sun Jun 18, 2017 8:17 am

Thought of the representation of o3o4o3o by means of o3o3o *b3o, provided the 3 legs are to be identified.

Using this symmetry ex = 600-cell becomes fox|o-3-ooo|f-3-xfo|o *b3-oxf|o-&#zx.
Accordingly sadi = s3s4o3o = f-ico-diminished ex = idex then becomes fox-3-ooo-3-xfo *b3-oxf-&#zx.

The dual of sadi is known to be quidex = 4x f-ico-diminished ex,
resp., as hi = 120-cell = dual of ex = 5x-f-ico-diminished ex, we likewise could say quidex = ico-stellated hi!

Thus we'd need to know about some o3o3o *b3o representation of hi.
Then, by means of that stellation one could derive a tegum sum representation of quidex too!



Just as a minor aside:
is the symmetry equivalence given in the first line not using the full trigonal symmetry of the symbol?
It striked me, that the given sadi representation just uses the cyclical subsymmetry thereof.
Wouldn't that mean, that sadi won't have full ico symmetry, but rather the mere rotational subgroup thereof?
That is, sadi then would be chiral, just like the 3D snubs. - Or did I get something wrong here?

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Re: Tegum polytopes

Postby student91 » Sun Jun 18, 2017 8:49 am

Klitzing wrote:Just as a minor aside:
is the symmetry equivalence given in the first line not using the full trigonal symmetry of the symbol?
It striked me, that the given sadi representation just uses the cyclical subsymmetry thereof.
Wouldn't that mean, that sadi won't have full ico symmetry, but rather the mere rotational subgroup thereof?
That is, sadi then would be chiral, just like the 3D snubs. - Or did I get something wrong here?

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here we also doiscussed the correspondence of ex and ico.
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Re: Tegum polytopes

Postby wendy » Sun Jun 18, 2017 8:55 am

The relation of sadi to o3o3o3 *b3o, is that the shared symmetry is not this group, but an order-3 subgroup by rotating the arms, viz a-b-c-a. This corresponds to replacing the octahedra by icosahedra, Given that the coordinates correspond to (0,1,f,v) + (2,0,0,0) + (1,1,1,1) EPAC. then the coordinates correspond to the first (as vxo3ooo3ovx *b3xov), + a second figure I imagine as o3x3o *b3o.

So your coordinates are right, scaling mine by f.

The full symmetry of ico is 1152 (dec). The shared symmetry between [3,4,3] and [3,3,5] is 576, gives by Coxeter [3,4,3+]. The o3o3o *3o is order 192, a subgroup of 3, by removing the rotation of the three arms from [3,4,3+]. The icosahedral faces are actually pyritohedral, being s3s4o = s3s3s. Just as you can create s3s3s with two different size triangles, and a different third side (its orbifold is "3a 3e 2i"), you can construct the sadi out of three different length edges a,e,i, by rotating these through the symbol. The faces are all tetrahedra where the opposite edges are equal.
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