Hi.gher. Space

by **wendy** » Fri Sep 07, 2012 12:47 pm

I'm going to put the notions and notations on the various notations here, rather than to scatter them through diverse threads. This serves as a reference to the notations.

The dynkin symbol.

The underlying notion here is a kaleidoscope and motif. The kaleidoscope reflects the motif into a lot of images, to make a polytope, for example.

The node is a mirror. The branch between the nodes is the angle between the mirror. A single node makes only for a reflected line, eg x. It's when we put down mirrors at an angle that a single line can make a polygon.

We now have two mirrors at say 60 deg. If we draw a line to just one mirror, it will not close up unless the distance from the other mirror is zero. We can stand against the second mirror, and drop an edge to the first, or we can suppose an edge of zero to the second. eg x3o. means an edge of one to the first mirror and an edge of zero to the second.

As one walks around these mirrors, one effectively places an edge wherever there is an x. So while the trip is 2n long, here six, we lay down only three edges. On the other hand, x3x puts an edge at every step or six mirrors together.

A polyhedron is generated by three mirrors, forming a point at the centre. These three mirrors work like the example above, but the more common example to be found is the digon. This is a polygon of two sides that disappears as an edge. Mirrors at right angles are so common, that they are not usually indicated. In fact, they are shown as a pair of isolated nodes, like o o. We indicate these belonging to the same group by way of a 2, eg o2o. It works the same way as the triangle above. x2o drops two edges, but makes these fall on the same mirror, the edges do coincide as one. On the other hand x2x gives in general a rectangle, of which the square is an examlpe.

A polytope x3o5o is read in this manner. The faces are x3o, x.o and .o5o. We lay three edges for the first polygon. The second polygon is layed out as a digon. We step through the digon back intk another triangle, The third edge o5o laya dwon no edgea in the cycle, so comes out on the central point of the two mirrors. We get an icosahedron.

The polytope x3o5x, likewise uses the three mirrors A B C. The pair AB gives triangle. The mirrors Ac gives x2x, a square or rectangle. The third pair BC gives o5x, a pentagon. This figure gives a rhomboicosahedron.

by **wendy** » Sun Sep 09, 2012 8:04 am

Stott Expansion

One of the earliest discoveries of the modern era is due to Mrs Alice Boole Stott. This is that one can convert various polytopes into each other by expanding things.

Suppose one starts out with a square of straws, with a thread of elastic running through the straws. You can pull out the sides of the square, radially outwards, so that the elastic is stretched between the ends of the squares. You get an octagon. Now, suppose the squares are the sides of a cube (with two sets of elastic, one for each square). Pulling these outwards will reveal not just an octagon, but a truncated cube. The vertices become triangles, the edges are kept solid, and the squares become octagons. This is an edge expansion.

It is possible to imagine that starting off with a cube, with similar solid faces, and something like a rubber balloon inside, to blow up the balloon. The squares are moved radially outwards, but keep their original sizes. New rectangular faces form along the old edges, and the vertices are turned into triangles, as before. This is a face or hedron expansion.

The expansions can be kept running, as long as one tracks what the original elements of the cube become. The combination of both kinds of expansion, come by recalling that the original faces of the cube were octagons, and we need to keep the octagons from step one a constant size. If we apply step 2 before step 1, we note that the bit that stays constant are the things at the places of the cube-edges: the rectangles that form between the squares. These are expanded out radially, the new edges form at the corners, turn the squares and triangles into octagons and hexagons.

The original cube edges, which we had in straws, can also be made to grow and shrink. The shrinking of these edges to zero, makes way for three new polytopes, the octahedron, the truncated octahedron, the cuboctahedron.

If, instead of starting with a cube, we start with a molecule in the form of a cube, the expansion of vertices gives rise to the cube, and one can construct these figures by expanding the vertices, edges, and hedra (or 0, 1, 2 d) elements of the cube.

t_0 vertex x4o3o cube
t_1 edge o4x3o cuboctahedron
t_2 hedron o4o3x octahedron
t_0,1 vertex+edge x4x3o truncated cube
t_1,2 edge+hedron o4x3x truncated octahedron
t_0,2 vertex+hedron x4o3x rhombicuboctahedron
t_0,1,2 vertex+edge+hedron x4x3x trunc cuboctahedron.

This is in essence, stott's notation. The main difference is that Stott starts with t_0 = 1, and e is used to make t_0 = 0. Wythoff showed that the numbered values belong to mirrors also. It is wythoff's paper that gave Coxeter the inspiration to decorate the Dynkin or deWitt graph (which he ( C ) discovered first).

One can build these polytopes as vectors, if one has a suitable coordinate system. The t_0,1 becomes not just two operations, but the coordinate (1,1,0), whose mirror-edge reflection in the walls of an oblique coordinate system gives the truncated cube.

The vectors are for the cross polytope these. q = sqrt(2), The resulting edge is 2. The primary cell is one where all values are sorted from greatest to least, and made positive. This means, eg for (1, 5,-3), the primary coordinate is (5,3,1).
0 (q,0,0,...) = cross
1 (q,q,0,...)
2 (q,q,q,...0)
n-1 (1,1,1,1,1,1) = cube

The generalised cube-symmetry uniforms of edge 2 has the coordinates A.q + B, where all values of 1 to A occur, and B is 0 always or 1 always. So, we find, a figure 2q1, 1q1, 0q1 corresponds to q,0,0 + q,q,0 + 1,1,1 = x3x4x.

We will return to the stott arithmetic when we come to show some fancy matrix multiplications.

by **Klitzing** » Tue Sep 11, 2012 8:38 pm

A very well chosen thread, Wendy!

Let me come in on that topic too. First on those Dynkin symbols.

Dynkin himself was hunting for Lie groups. So he was putting his symbols up for those. It was Coxeter who took them over for reflection groups. He also introduced the numbers on the links, whereas Dynkin used single, double, etc. linkages.

So what is meant already was outlined by Wendy: The nodes of those graphs represent mirrors. The linkages represent the angles between those mirrors. The numbers at those links represent the sub-multiple of Pi (180 degrees). That is, a graph with just 2 nodes and those being linked with a link number 4 represents 2 mirrors at an angle of Pi/4 (45 degrees).

It should be noted here, in order that we will have a closed loop of reflections through those 2 mirrors (and their reflected mirrors) we are forced to have rational submultiples of Pi only. In fact the numerator of that rational gives the number of images of one of the mirrors (including itself), while the denominator represents the number of circuits around the intersection of those 2 connected mirrors needed in order to close again. Thus o-5/2-o represents a symmetry group where 5 mirrors of type A and 5 of type B intersect alternatingly, and only the sixth reflection of the first mirror of type A would coincide with the first one (of that type).

More generally we will have N mirrors (N being the numbers of nodes in the graph), each pair of which will have a mutual angle as specified by the numbers at the graph. Thus, a graph with 3 nodes and connections marked 2, marked 3, respectively marked 4, obviously represents the reflectional symmetry group of the cube or of the octahedron: The mirrors intersecting in the direction of the face normal of the squares would intersect at an angle of Pi/4 (45 degrees). Those intersecting in the direction of a cube vertex would intersect at an angle of Pi/3 (60 degrees). And those intersecting in the direction of the cubical edge midpoints would intersect at an angle of Pi/2 (90 degrees).

So far we note that any such Dynkin symbol represents a reflectional symmetry group. In fact, the pyrtohedral group or mere rotation groups etc. can not be described. Obviously any 2 mirrors will have some angle, even if it would become 0 degrees, i.e. the number at the link becomes infinite. So in fact the graph will always be a simplex, as any 2 nodes are connected. - This is in fact what Schläfli once describing in its dual form as an "Orthoschem", the fundamental domain of the reflection group. For example, the fundamental domain of the symmetry group of any regular polyhedron is the open cone described by the 3 mirrors defined by A) body center, face center, edge center; B) body center, face center, vertex; resp. C) body center, edge center, vertex.

The very name of Orthoschem reflects that most often a lot of those mirrors will be forced to be orthogonal. Already Dynkin himself used for those orthogonal mirrors linkages with zero lines, i.e. ones which visually seem not to be present at all. In fact, this makes it much easier to diplay those generally simplexial graphs. Reflectional polyhedral symmetry groups thus can be displayed by linear graphs: o-P-o-Q-o (were P and Q are some rationals, being subject to further restrictions), i.e. the triangular graph breaks open. In 4d the elemental groups require linear or bifurcated graphs only.

None the less it would be usefull, to represent any arbitrary (i.e. even a non-elemental) symmetry group in a linearized fashion. This is why I once introduced virtual nodes in addition: you would just begin to display the graph as long as possible in a linear way as a sequence of consecutive mirrors with pairwise angles. At some stage you would be forced to display a link back to an already displayed node. This is where virtual nodes come in. Those just represent not an additional further node, but just represent revisits of already displayed nodes. Consider for example the general pentagon

Code: Select all: oA Bo oE Co oD AB BC CD DE EA AC CE EB BD DA o----o----o----o----o---*a---*c---*e---*b---*d---*a o : (real) node --- : link *a : virtual node, representing the left-most real node *b : virtual node, representing the second real node from the left etc.

In that way virtually any graph can be displayed inline. Obviously there are several ways to do this, depending on the sequence of node selection.

Next we have the decoration of those symbols. I.e. using ringed nodes versus un-ringed nodes. Typewriter-friendly usually un-ringed nodes are displayed as "o", while ringed nodes in the past were displaed by "(o)" or by "@". Nowadays those usually are typed as "x".

These decorations represent nothing but Wythoffs kaleidoscopical construction of polytopes by means of a fundamental set of mirrors (i.e. being represented by the bare Dynkin graph) and a seed point, which will be reflected by these mirrors. This reflection will be done in kind of ray-tracing mood, i.e. the seed point and its direct mirror image do span a line segment, which becomes an edge of the polytope. The mirror images of those primal edges will yield the edge skelleton of the polytope. In fact, the seed point and its images represent the vertex set. The span of 2 adjacent edges will define a face polygon, etc. In this construction the seed point can have any choosen position with respect to the mirrors of the fundamental cone. But topologically the relevant cases are: coincident to or not coincident to any of the mirrors. This is what is represented by "o" (= coincident) and "x" (=not coincident).

So reconsider the cubical group o3o4o.
x3o4o would be defined by a seed point off to the mirror which has angles with submultiples 2 and 3 to the other ones, but on the other 2 types. So this is the octahedron.
o3o4x on the other hand is the cube.
o3x4o would be the cuboctahedron.
x3x4o = truncated octahedron.
x3o4x = rhombicuboctahedron.
o3x4x = truncated cube.
x3x4x = great rhombicuboctahedron (omnitruncated cube).

It was Wendy, who introduced a notational re-distinction of those topological classes, simply by introducing different letters for different edge lengths. That is,

"x" represents an edge of unit length (the seed point is off to a single mirror; that seed point together with its mirror image represents an edge).
"q" represents an edge of length sqrt(2).
"f" represents an edge of length tau.
"v" has length 1/tau.
"h" has length sqrt(3).
"u" has length 2.
"w" has length 1+sqrt(2).
(Others may be defined ad hoc - i.e. should be explained nearby.)

So you could consider deformed variants: u3x3o, i.e. a polyhedron similar to the truncated tetrahedron, where the edges connecting the triangles to the hexagons have unit length, while the ones connecting 2 hexagons have the double length. I.e. the triangles are still regular ones, while the hexagons are semiregular only.

There is an easy way to interprete those Coxeter-Dynkin symbols in the sense of facet elements. This is done just by removing any arbitrary mirror from the construction. Accordingly we have:

Code: Select all: x3o5x = rhombicosidodecahedron . o5x = pentagons x . x = squares x3o . = triangles

--- rk

by **wendy** » Wed Sep 12, 2012 7:38 am

Groups and Counting things

The undecorated dynkin graph represents a group, particularly a class of Lie groups. We are not going into this mathematics, but it needless to say that i bought a book on the subject, gracefully adorned with the 2_21, 3_21 and 4_21 on the front cover. Still, 'twasn't dear.

What we are going to do is some simple walks to set the notion of 'rooms'. We then use the room idea to catch a single surtope, and the size of said room, divided into the size of total space, gives the count.

The mirrors divide the surface into cells (in PG, a cell is a bubble in a foam: cells are solid in the space they exist). We imagine now that a cell has doors to adjacent cells (which it shares a wall), numbered A, B, C, ... Going through a door gives a complete mirror image, the names match the mirror image. In the dynkin graph, these doors or walls correspond to the nodes of the graph, viz A---B---C-5-D .

A walk is then a trip from cell I to some other cell J. The name of the cell J is the path from I. There are some obvious identies, like the path AA or BB amounts to walking through a door, and back through it. A walk around an angle, like ABABAB.. must bring you back to the room, after an even number of steps. One can divide this path into halves, and consider ABA.. = BAB.. for n steps. In fact, these two kind of relation suffices to define the complete coxeter reflective group that the graph represents.

If two sets of mirrors are related by the relation pA = Ap for all mirrors A,B,C, and p,q,r..., then the groups are a direct product, and one can express any combination of upper and lower case symbols by moving all of the lower case symbols to the left, eg pABqCrD = pqr . ABCD. This means that the order of unconnected node-sets, is the product of the individual node-sets. So a group ( o o-5-o ) is the product of ( o ) and ( o-5-o )

One can do all sorts of fancy walks. The fancy one is [ABC..], which brings one back to the start in h moves. This is the petrie polygon. It crosses each mirror exactly twice, so 2m = nh is the relation of the mirror-count to the petrie polygon. In three dimensions, the petrie polygon is the thing that girths the o3xPo.

Rooms

A subgroup of the general group arises by removing some of the nodes, and their branches, from the symbol. So, the mirrors AB make a subgroup of ABC. What this means is that if one uses doors A and B exclusively, one can only walk around a range of cells, and the outer walls are made of C. The rooms are identical by reflection in C, so the number of rooms is the total order g(ABC) divided by the order of the room g(AB).

For example, where C represents the mirror opposite the face-centre of the dodecahedron xA--5-oB--oC , the room is laid out in mirrors A and B only. The edges fall perpendicular to A, run against the wall of the mirror. The entire interior of the pentagon is in the room, and there is just one. We find the room size g(AB) is 10, the total size g(ABC) is 120, so there are 12 pentagons.

Catching the Surtope

This exercise is meant to make a room, that the interior of the surtope is inside the room. What this means is keeping just those mirrors (or doors), that leave the surtope (vertex, edge, hedron, choron, etc..), in the same place.

There are two kinds of mirror to count. The first one is the one that moves it onto itself, but in a different orientation. These are s or surround-mirrors. These nodes are directly connected to the vertex-node.

The second kind of mirror to count is the ones that leave the thing unmoved. The thing is drawn on the surface of the mirror itself. These mirrors change things around the surtope, but leave the surtope undisturbed.

A simple example of this, is to consider the line from z=1 to z=-1, at x=0, y=0. The mirror z -> -z is a surround-mirror. It still gives the same line, but the points are reflected end to end. The mirrors x=> -x and y =>-y are arround-mirrors. The whole line is undisturbed, although other things are counted. If we mean to capture interior on interior, we need to count all of these mirrors in the S and A sets.

Mirrors that are connected to the sub-group by a branch larger than 2, reflect the surtope onto a different copy of it, ie they reflect between rooms.

So, for example, the number of edges of the x---o---o--5-o is found by finding the room it is in. The edge's S mirrors amount to just the first one, A. The mirror B is at an angle to it, is a 'room-wall' mirror. The mirrors C and D are at right-angles to A, ie AC, AD are '2' branches. So we need to count these in the 'a' mirrors. So the graph looks like S---W---A-5-A we drop the W node to get the room is S × A-5-A. The total order is 14400, divided by 2 and 10, gives the 720 edges of this polytope.

by **Klitzing** » Wed Sep 12, 2012 2:14 pm

Today I want to show some concepts of snubbing. An other name somehow related is hemiation. But I will show that there is something what N. Johnson called holosnubbing, which does not divide the vertex count by 2. I also will show, that snub nodes, as applied to Dynkin diagrams, in fact display alternated facetings, and not describe uniform snub polytopes, as they are usually understood. They just describe a topological variant of those. - This better understanding of the true coded content of snub nodes in Dynkin diagrams makes the way free for more general alternations too: not only of vertices alone.

Consider any Dynkin diagram with all nodes being ringed. I'll use x3x4x for an example, i.e. the omnitruncated cube or alternatively being called the great rhombicuboctahedron, girco. Next consider as alternation rule the following: locally alternate the vertices. One being kept, the next being omitted, etc. (In the case of omnitruncates we have no troubles here, we allways will return to a former spot in the correct parity.) Omition would mean to remove the vertex pyramid so that not only the vertex, but also all the incident former edges would be removed completely. Instead a new facet would emerge, here the vertex figure, the base of that pyramid.

In our example the octagons will be reduced to vertex inscribed squares (every second vertex), the hexagons are reduced to vertex inscribed triangles, and the squares will be reduced to one of its diagonals. As new facets underneath the omitted vertices some irregular triangles would emerge. This is what s3s4s would describe in the first run. - Yes, I admit, s3s4s usually is used to describe the snub cube. An uniform polyhedron. But what I described here is effectively not uniform, it would have edges of 3 different lengths (the secants of a square, a hexagon, and an octogon). None the same, it is topologically equivalent to the snub cube, it is a variation of the uniform figure.

This is much more general. The alternated faceting, just being described for that special case here, would apply in general, for many other applications, as to be described below. In some of those resulting figures it might be possible to do a resizement of edges afterwards, so that the total figure would become uniform. But sadly this latter part is not general. It highly depends on the degree of freedom, which in turn depends on the dimension of space. On the other hand there are the number of different to be resized edges, which in general also depend on the dimension, but most often in an higher degree. So in result, the higher the dimension, the less cases will be "uniformable". - In the past this was the reason to omit any application from consideration. But this is not fair. Alternated faceting is a valuable operation to be considered on its own. And this is what truely is coded by the application of snub nodes to Dynkin symbols!

In the example above we considered omnitruncated figures only, i.e. Dynkin symbols with all nodes being ringed, for snubbing. Now we will go on. We not only allow polytopes, the Dynkin symbols of which do use unringed nodes as well, for starting figures. We even would allow to alternate not only vertices, but even higher dimensioanl elements as well, e.g. edges, faces, cells, etc.

The procedure would be as follows. Take any Dynkin diagram of any reflectional symmetry, and decorate the nodes by any choice with ringed nodes, with un-ringed nodes, and with snub nodes. I would use as an example x3o4s, but it would apply generally.

1st step: replace any "s" node by an "x" node (here: x3o4x = sirco). This would be the starting figure.
2nd step: omit any "s" node together with all incident links (here: x3o . = triangle). This is the element to be alternated.
Alternation again takes place as above, just that we dont have pyramids any longer, but more general cupolaic elements, which are to be removed locally in an alternating way. Still there are new facets underneath the omitted elements, which come in as new facets, while the other ones are to be diminished somehow.

In our example the cut off elements are x3o || x3q (triangles atop semiregular hexagons). Thus we omit any second triangle, and replace those by those hexagons. The other triangles would be kept in place. The squares in cubical positions are to be reduced to one of their diagonals, the squares in rhombical positions are completely withdrawn together with those alternated triangles. So we would result with a figure, which is a topological variant of the truncated tetrahedron, in fact it is q3x3o.

One last thing is to be mentioned here. We pointed out that the application has to be done locally. So far a local and a global application would not make any difference. But what about s3o4x? In here the squares in cubical positions are to be alternated from the sirco. But there are 3 such squares around the cubical body diagonal. Thus the parity does not close after a single circuit. But our rule was alternation. We have only 2 states. Therefore clearly a second circuit would close again. This is what N. Johnson once was introducing (with respect to vertex alternation in those days) by the concept of holosnubbing. As alternation will omit a figure in the one circuit, but retain it in the other, the number of vertices would not be halved, as for the other alternations, but would them maintain. But the mere process still applies, independant what so ever!

Could we tell a priori, whether a normal snubbing (alternation) or an holosnubbing would have to apply? Yes we can! Consider the set of those links between a snub node "s" and an other node "o" or "x". If all those links carry even numbers, the normal snubbing (alternation) would apply, i.e. there is no difference between local and global application. If at least one of those link marks is odd, we need to apply our procedure locally, holosnubbing will be the result.

Finally, having spoken of s3o4x. So what would be the outcome? You guess what? - Yes, that one again is not only topologically equivalent to a known figure, it even will be exactly what is known as socco (Bowers acronym).

by **Klitzing** » Sat Sep 15, 2012 1:59 pm

Would like to outline within this thread Wendys Zoo of Lace thingies, as the lace prisms became rather freequent in this archive lately.

The lacing notion as such was named after the childrens or military drum: the lateral thread zick-zacking up and down, connecting the upper and lower base. Meanwhile she has come up with a rich zoo of such terms. But mainly they are all derived from the lace prisms. Here is what all those terms are meant for:

Lace Prism

Lace prisms use 2 polytopes for bases, both of which belong to the same (undecorated) Dynkin graph symmetry. E.g. consider the cubical antiprism within 4D, i.e. cube || oct: the Dynkin symbol of the cube is x4o3o, that of the oct is o4o3x. So both belong to the symmetry group o4o3o. Accordingly that 4D fellow could be denoted as x4o3o || o4o3x. More economically the mentioning of the symmetry group could be given once only. But still, we have 2 decorations of it for the top and the bottom base. Those will be condensed into a single node position of the group, but given as a sequence on every such node instead. We would have xo4oo3ox&#x for its symbol: The first decoration on every node corresponds to the top base, while the second decoration on every node corresponds to the bottom base. The suffix "&#x" just is to be read: "and additionally" (&) "there is a lacing" (#) "of edge size: x".

The sub-elements of those base polytopes immediately can be derived from those (decorated) normal Dynkin symbols. The lacing edges clearly are given by the suffix itself. The lacing 2D elements are given by all the single node pairs (provided at least one is ringed: In our example we have xo .. ..&#x (triangles pointing down) and .. .. ox&#x (triangles pointing up). In other diagrams xx&#x (squares) might occure as well. 3D lacing elements would be derived similar, using any 2 node positions. Etc.

Compounds

You could also consider "prisms" of zero height, omitting the lacings, i.e. compounds of 2 polytopes with the same symmetry. Those could be denoted by omition of the former suffix "&#x". That is, a compound of a cube and an oct in according relative orientation just would be xo4oo3ox. - But be aware, Wendy sometimes uses her own symbols inconsistently. So it often can be found that she denotes lace prisms without suffix, as if it would be implied. But this usage then clashes with that of compounds!

Lace towers

You could consider stacks of several lace prisms, one atop of the other in a linear sequence. It would become obvious how to denote these, provided the notation of lace prisms is given. None the less Wendy invented an additional mark here: the suffix will be extended by an additional "t" (becoming &#xt) for towers. So the ico could be given such: oct || co || oct, i.e. xox3oxo4ooo&#xt.

It should be mentioned here that not any 2 adjacent layers need to be connected. Edges might connect, say, layer 2 and 4 also. Consider e.g. sirco = x3o4x in its triangle first representation. Then it will be represented as lace tower xxwoqo3oqowxx&#xt. (Here q = sqrt(2)-edge, w = (1+sqrt(2))-edge.) Her there are lacing edges between 1st and 2nd layer, between 2nd and 3rd layer, but also ones between 2nd and 4th layer. But ones between 3rd and 4th layer wont occur! Putting bottom up we have here likewise lacings between 3rd and 5th layer, ones between 4th and 5th layer, and ones between 5th and 6th layer. Moreover it can be seen from this example already as well, that the "edges", used to display the geometry of the non-extremal layers, not necessarily belong to the final structure: x3o4x surely has no edges of size q nor w! In fact those q-edges are the diagonals of some outer squares, but those w would even be inner distances!

Lace simplices

Instead of considering linear stackings you could extend lace prisms to more than just node pairs by considering that any 2 pairs out of the multituple would provide an prism. Here the tower suffix will be not usable, but still we would get more than 2 decorations per node position. So ooo&#x denotes a triangle, oox&#x denotes a tetrahedron, etc.

Lace simplices are a very essential concept. In fact, any vertex figure of a Wythoffian polytope (i.e. one which can be described by a Dynkin diagram) is nothing but a lace simplex! (Sure, not all edges then would have the same lengths, and accordingly a closed form might become somehow difficult to display, but essentially, the concept of several diagrams, pairwise being connected by some lacings, would still hold. - Perhaps Wendy or I would outline that concept in a different post.)

Lace rings

Similar to the linear stacks of towers you could consider stacks bending back to the first position. The suffix will be extended by an "r" here. So a xxx&#xr represents a triangular prism, for example. Or you could denote a square as oooo&#xr.

Lace cities

Lace cities surely are one of the most genial findings of Wendy. Lace towers kind of display higher dimensional polytopes along a linear sequence. So they display them with a common axial symmetry, or kind of projectioning out D-1 space. Lace cities OTOH are similarily kind of projections which reduce only by D-2!

The term itself is derived easily: Towers are stacks of symbols on different floors. So we align those atop each other. A city OTOH is similarily a composition of towers. Thus we get a 2D display from that: in one direction the stacks of symbols on top of each other, and in an orthogonal 2nd dimension, the alignment of those towers. This derives an ASCII art 2D representation of higher dimensional polytopes.

To provide some first, rather simple cases, we would commence with lace cities for 3D objects:

Code: Select all: x x : this is the alignment of the 4 lateral edges of a cube; in fact x x it is the stack of 2 towers, a square atop a square; the square in turn is a stack of 2 edges. x x : this is the alignment 8 edges of size x=1 x w w x and 4 edges of size w=1+sqrt(2), or, as towers, a square (x || x), a regular octagon x w w x (x || w || w || x), an other octagon, and an x x other square, i.e. the description of sirco.

For 3D figures D-2 clearly is 1 only, so we get edges within the orthogonal direction of the display. Thus those lace cities are kind of rather trivial. But lace cities can be provided for higher dimensional figures too! The following displays a lace city of ex = 600-cell = hydrochoron:

Code: Select all: o5o o5o o5x x5o x5o o5o f5o o5o o5f o5f o5x o5x f5o f5o o5o x5x o5o o5f o5f x5o x5o f5o f5o o5o o5f o5o o5x o5x x5o o5o o5o

Reading this city in vertical columns, we get for free: pt || ike || doe || f-ike || id || f-ike || doe || ike || pt (f = edge of size tau).

It was esp. with resp. to the Gosset figures, that Wendy exposed the might of that display, depicting some D-2 dimensional sub-symmetries:

Code: Select all: naq = 3_1,2: h h* h =x3o3o3o3o (hix) h* d h h*=o3o3o3o3x h h* d =o3o3x3o3o (dot) laq = 2_1,3: o o + - o + = o3x3o3o3o (rix) - = o3o3o3x3o - # + # = x3o3o3o3x (scad) o + - o o lin = 1_2,3: = = = x3o3o3o3o (hix) 1 1 # = o3o3o3o3x (hix) # + # + = x3o3o3x3o (spix) - - - = o3x3o3o3x (spix) 1 0 1 1 = o3o3x3o3o (dot) + + 0 = xo3xo3oo3ox3ox ((tix, inv tix)-compound) = - = 1 1 #

A further advantage of lace cities (in general) is, that once you have derived the right display, you might rotate that display into a slightly different orientation, and then you would be able to deduce a different alignment of different towers therefrom, completely for free!

Best regards,
--- rk

by **quickfur** » Sat Sep 15, 2012 8:41 pm

wendy wrote:Stott Expansion

One of the earliest discoveries of the modern era is due to Mrs Alice Boole Stott. This is that one can convert various polytopes into each other by expanding things.
[...]
One can build these polytopes as vectors, if one has a suitable coordinate system. The t_0,1 becomes not just two operations, but the coordinate (1,1,0), whose mirror-edge reflection in the walls of an oblique coordinate system gives the truncated cube.

The vectors are for the cross polytope these. q = sqrt(2), The resulting edge is 2. The primary cell is one where all values are sorted from greatest to least, and made positive. This means, eg for (1, 5,-3), the primary coordinate is (5,3,1).
0 (q,0,0,...) = cross
1 (q,q,0,...)
2 (q,q,q,...0)
n-1 (1,1,1,1,1,1) = cube

The generalised cube-symmetry uniforms of edge 2 has the coordinates A.q + B, where all values of 1 to A occur, and B is 0 always or 1 always. So, we find, a figure 2q1, 1q1, 0q1 corresponds to q,0,0 + q,q,0 + 1,1,1 = x3x4x.
[...]

I find this very interesting, because I independently discovered the same thing while hunting for a way to generate coordinates of uniform polychora. My first (re)discovery was that the Cartesian coordinates of the n-cube family of uniform polytopes can be trivially read off their CD diagrams. I see that you have placed the edge labelled 4 on the right; my custom is to orient the diagram such that this edge is on the left. So given some CD diagram .4.3.3... one reads from left to right, generating the coordinates of a "reference point" as follows:
1) If the first node is ringed, the first coordinate is 1; otherwise it is 0.
2) For all subsequent nodes, if the node is unringed, then its corresponding coordinate is a repeat of the previous coordinate; otherwise it is the previous coordinate plus sqrt(2).

After the reference point is constructed, we simply take all permutations of coordinate and sign, and that gives all the vertices of the uniform polytope.

In seeking to generalize this scheme to polytopes of other families, I eventually (re)discovered that uniform n-polytopes of the n-simplex family occur as facets in (n+1)-polytopes of the n-cube family. Specifically, the facet perpendicular to (1,1,1,...). So this means that by just taking all permutations of coordinate (without changes of sign) in the above algorithm, one obtains the coordinates of the corresponding uniform n-simplex truncate in (n+1)-space.

At the time, I was interested in coordinates in n-space, so I considered how project the coordinates into n-space. My thought was to find a rotation R that will rotate (1,1,1,...) to (k,0,0,0...), then I can just rotate the polytope by R then drop the first coordinate. Now, there are two ways to make R: the most obvious way is to rotate (1,1,1,...) to (k,0,0,0,...) directly, that is, in the plane defined by these two vectors. I discovered, however, that this does not yield "nice" coordinates in n-space; the polytope ends up in some arbitrary-looking orientation that makes it hard to manipulate.

The other way to make R is to rotate (1,...,1,1,1) to (1,...,1,p,0), then rotate the latter to (1,..,q,0,0), then to (1,...,r,0,0,0), etc., until we get (k,0,0,0). I discovered that if these rotations are combined together, they can be represented by a matrix with some nice patterns. But more importantly, the n-space coordinates produced by this scheme has the property that (1) the apex of the n-simplex thus projected (or the corresponding element of the uniform truncate) is lined up with one of the coordinate axes, (2) the base of the n-simplex has bilateral symmetry across a coordinate hyperplane, and (3) the base of the n-simplex itself recursively has these same 3 properties.

The coordinates of the n-simplex itself thus obtained exhibits a fascinating pattern: Let A(i) be the square root of the inverse of the i'th triangular number, and let B(i) be the square root of the i'th square number divided by the i'th triangular number. Then the coordinates of the n-simplex are:
(B(1), B(2), B(3), ... B(n))
(A(1), B(2), B(3), .. B(n))
(0, A(2), B(3), ... B(n))
(0, 0, A(3), ... B(n))
...
(0,0,... 0, A(n))

These coordinates are origin-centered, in a "nice" orientation as described above, and always have edge length 2. The involvement of the square and triangular numbers is particularly fascinating; the inverse triangular numbers in particular have an interesting telescoping property when summed consecutively, which ultimately leads to the consistent edge length. Interestingly enough, one obtains the same coordinates by applying the following recursive procedure:
1) The 0-simplex has coordinates (), that is, the empty vector.
2) Given the coordinates of the i-simplex, construct a pyramid of it in (i+1) dimensions, that is, find a point exactly 2 units away from all points in the i-simplex.
3) Shift the newly-added i'th coordinate such that the resulting vertices are origin-centered.

In any case, this covers the coordinates of the n-cube and n-simplex families. Later on, I considered how to generalize the scheme to work with other symmetries. The 120-cell family, for example, eluded me for a while, because given Coxeter's construction of the 600-cell from the 24-cell via the snub 24-cell, it was unclear how one may derive a uniform truncate of the 600-cell using a similar scheme. Eventually, I wrote a program that used trigonometry to derive coordinates using Stott expansion. However, it was not very satisfying, since I felt that I still lacked an intrinsic handle on the problem.

Eventually, I realized that every symmetry group (of a polytope) in n-dimensions can be represented by n sets of vectors, or n stars, each corresponding with a node in the CD diagram. To generate coordinates, one starts with a single point on the origin, then for each ringed node in the CD diagram, take the Minkowski sum of the current coordinates with the corresponding star, or equivalently, add the vectors of each star to the vertices of the corresponding elements of the polytope. This is essentially what is being done in the reading of the n-cube truncate coordinates described above. These stars can be derived by measuring the radius of the n elements of a regular polytope of that given symmetry. For example, in 3D, there are 3 sets of vectors, one corresponding with the vertices, one with the midpoints of the edges, and one with the faces. The Minkowski sum of various subsets of these stars generate the Archimedean solids with cubic symmetry. The star corresponding with the midpoints of the edges, for example, gives the quasiregular cuboctahedron. If each star is scaled so that edge lengths are always equal, then the Minkowski sums will also yield uniform polytopes.

by **wendy** » Mon Sep 17, 2012 7:54 am

You can find the vertices of the various rectates of the simplex-faced regulars, by this rule:
Imagine that the face of the x3o3o3... is set by vectors radiating from the centre to the various vertices.

Then x3o3o3..Po = one vector.
o3x3o3..Po = two vectors
o3o3x3..Po = three vectors
o3o3o3..Pp = N vectors.

The last vector, by adding all of the vectors to the face, gives a xPo3o3o3o... of size q for P=4, f for p=5, h for p=6, etc.

So, where the vertices of a face of a cross=polytope of edge r2, is 1,0,0,0,0 and 0,1,0,0,0 and 0,0,1,0,0 etc.
for o3x3o4o 1,1,0,0
for o3o3x4o 1,1,1,0
for o3o3o4q 1,1,1,1 (the edge itself is 2, or q × sqrt(2)).

by **wendy** » Wed Sep 19, 2012 7:10 am

There are more lace thingies than r klitzing mentions.

lace tegums

The dual of a lace prism is a lace tegum. Just as a lace prism is a progression from base to base, the lace tegum is the intersection of lace cones.

lace cones

A lace cone is the region in the apex of a pyramid, the base being the desired figure. When the pyramid is not solid, it is multiplied by the across-space, that is, if the thing is N-d dimensions, then it is made solid, by the cartesian product of the orthogonal d dimensions. A lace tegum is the intersection of lace cones.

Consider the tetrahedron. You can hold it with a thumb and finger from each of two hands, so that each face has just one digit applied.

Now, suppose the left hand touches 'blue' faces and the right hand touches 'red' faces. The blue cone corresponds to the planes of the blue faces, as far as the line of crossing. The actual 'cone' is a point between the blue faces, and the line between the red faces. This line serves to hold open the blue planes. Because a triangle is not solid in 3d, it is multiplied (extended into) the across-space. This means that instead of a point with lines spanning the strut, it becomes a line with planes so held.

antitegum

The dual of an antitegum is an antiprism. It's the lace-tegum of a figure and its dual. All surtopes of the antitegum are antitegums. The prismatotope is a simplex-antitegum, for example. The Hass antitegum is the diagram of connections of the surtopes of a polytope. It is in essence, the antitegum of the polytope itself. The "antitegmal sequence" is the sequence of 'decent of the faces of the dual', where one considers at various points the intersection of a cube of size x, and a octahedron of size a-x. This corresponds to parallel sections of the corresponding antitegum. In the case of the regulars, the centres of surtopes of every order occur at the same point, so the antitegmal sequence passes through the truncates and rectates in order.

by **Klitzing** » Wed Sep 19, 2012 7:25 am

Hy Wendy,
do I get your lace cone concept correctly, if I would translate it into what elsewhere is called wedges?
I.e. the top base (say) has to be degenerate?
So, the tetrahedron, positioned as point || triangle, is a lace cone with respect to its top base (vertex only), thus it is a pyramid in fact. But it is not a lace cone with respect to its bottom base, as the triangle already is full dimensional.
The tetrahedron, positioned as line || gyro line, is a lace cone with respect to either base (both are not full dimensional).
Correct?
--- rk

by **Klitzing** » Wed Sep 19, 2012 7:49 am

Klitzing wrote:Hy Wendy,
do I get your lace cone concept correctly, if I would translate it into what elsewhere is called wedges?
I.e. the top base (say) has to be degenerate?
So, the tetrahedron, positioned as point || triangle, is a lace cone with respect to its top base (vertex only), thus it is a pyramid in fact. But it is not a lace cone with respect to its bottom base, as the triangle already is full dimensional.
The tetrahedron, positioned as line || gyro line, is a lace cone with respect to either base (both are not full dimensional).
Correct?
--- rk

... or would you ask for a further condition, e.g.
(*) the 2 bases have to be co-dimensional?

Then, the tetrahedron surely would be a lace cone in every of the above given orientations.
The triangular prism surely is no lace cone in its orientation with triangle || triangle.
But it would be a lace cone in its orientation with line || square, if (*) would be neglected, but not if (*) would be considered.
And the square pyramid clearly is a lace cone in its orientation with point || square.
But in its orientation with line || triangle it would be a lace cone, if (*) would be neglected, and none if (*) would be considered.

So please shed some light onto this.
--- rk

by **Klitzing** » Wed Sep 19, 2012 8:25 am

wendy wrote:[...]

lace tegums

The dual of a lace prism is a lace tegum. Just as a lace prism is a progression from base to base, the lace tegum is the intersection of lace cones.
[...]

antitegum

The dual of an antitegum is an antiprism. It's the lace-tegum of a figure and its dual. All surtopes of the antitegum are antitegums. The prismatotope is a simplex-antitegum, for example. The Hass antitegum is the diagram of connections of the surtopes of a polytope. It is in essence, the antitegum of the polytope itself. The "antitegmal sequence" is the sequence of 'decent of the faces of the dual', where one considers at various points the intersection of a cube of size x, and a octahedron of size a-x. This corresponds to parallel sections of the corresponding antitegum. In the case of the regulars, the centres of surtopes of every order occur at the same point, so the antitegmal sequence passes through the truncates and rectates in order.

So you get the followings:

A prism (of whatever dimension) would be the special case of a lace prism, where both bases are congruent (and, as general for lace prisms, are aligned accordingly)
An antiprism (of whatever dimension) would be the special case of a lace prism, where both bases are vice versas duals
A tegum would be the dual of such a prism (of whatever dimension)
An antitegum would be the dual of such an antiprism (of whatever dimension)
A lace tegum, as you wrote, would be the dual of (any individual) lace prism (of whatever dimension)

Esp. in 3D we get as tegum corresponding to the uniform n-gonal prism just an n-gonal dipyramid, the faces being 2n triangles. The equatorial section just is that n-gon. The antitegum derived from the uniform n-gonal antiprism would likewise have 2n faces, but those would be kites instead. The equatorial circuit is a zick-zack of 2n edges.

The tegum should be related here in addition to your tegum-product as well. The latter is defined by using any 2 (convex) shapes, positioning those both with their centers at the origin, but spanning co-dimensional subspaces. Then the tegum product of those 2 shapes would be the figure which is the convex hull (coat, latin: tegum) of that arangement. Esp. what you above defind tobe a (mere) tegum, obviously is the tegum product of a figure and a line segment. In fact, that "figure" itself is nothing but the dual of either base of the associated prism.

--- rk

by **wendy** » Wed Sep 19, 2012 9:02 am

Hi Richard

You got the lace prism thing correct.

Lace prisms generally include lace simplexes. The dual of a lace prism/simplex is a lace tegum, the intersection of lace cones.

The pyramid product is a lace tegum or lace prism, where each base of the product is taken in prism-product with points, ie they correspond to the solid triangle (x,0,0), (0,y,0) and (0,0,z), where x, y, and z are the polytope bases.

So, although we usually represent a lace prism with some sort of height, like a line in x, a line in y, and a height in z, it is more correct to visualise the lace product happening in a plane like an octahedron-face, ie the line w at (1,0), and the line x at (0,1). The line between (1,0) and (0,1) represents the progression of rectangles as w grows and x shrinks. This is how it is with all pyramid products.

When one has a different sections on the same symmetry, such as do cupolae, the line does not run from (1,0) to (0,1), but say (1,1) to (0,2) or something, In general, one might suppose the triangular cupolae runs from (1) ie xo to (3) ie xx. In general, a lace prism might run from something like (xxo, ox) to (oxo, xo), where these are coordinates in a polyhedral and polygonal group.

But, in general, the rule is, the span of the several bases, and the altitude (a simplex between the several bases), must span all-space without overlap. This is usually implemented by specifying that all of the bases reside in the same product-symmetry.

All the other lace-towers and lace-city stuff, are generalisations of the coordinate by section of polytopes. Instead of giving coordinates in right-angle forms, they are given as dynkin graphs. One can further generalise it by writing a figure as a compound of prisms, and then expanding out one of the bases.

So, the tetrahedron, positioned as point || triangle, is a lace cone with respect to its top base (vertex only), thus it is a pyramid in fact. But it is not a lace cone with respect to its bottom base, as the triangle already is full dimensional.
The tetrahedron, positioned as line || gyro line, is a lace cone with respect to either base (both are not full dimensional)

When you look at the tetrahedron as point // triangle, in a lace cone, the top 3 faces are parts of an open pyramid (ie a cone), stretching from the point downwards. The bottom is actually a second cone, being represented by a point, the cone stretches upwards. Because this second cone is not solid in three dimensions, it is multiplied by the across-space (say the xy plane), so it represents the plane of the face, and everything stretching upwards: ie a half-space. The tetrahedron is then represented by the intersection: the open pyramid above the plane.

Since we have the tetrahedron as point // triangle or point-triangle lace-prism. In the dual, it is the intersection of a point-cone and a triangle-cone, over the measure of the height. The total space is 3d, the height is 1d, so we have a 2d around-space. This means that the triangle-cone is already solid. The point-cone radiates along the altitude, but is not solid. It is made solid by filling in the missing dimensions, into half-space.

When the simplex is regarded as a product of points, the entire interior of the simplex is taken as altitude, and the individual points are then extended into planes, by way of a plane or n-1 space. The simplex is then formed by the intersection of planes. An example is to consider the classical coordinates of the orthotope, where the points are in the plane (0,1,1), (1,0,1), (1,1,0). The corresponding point cones are then these points, extended in all other spaces, ie x=0, y=0, and z=0. The simplex is then bound in the first octant by these three point lace-cones.

Looking at line // gyro line. Suppose we have height in z, the line runs at z=-1, y=0 for some length X. The gyro-line runs at z=+1, x=0, for some length Y.

The actual 'cone' for the top line lies entirely in the yz plane, consists of the point z=-1, y=0, and two lines spanning upwards, so that the length Y is set at height z=+1, ie there's a strut between z=1, y=Y/2, and z=1, y=-Y/2. These hold the rays of the cone open. Because these are not solid in xyz, the cone is then stretched into all values of x. This creates a solid section to intersect with.

The second 'cone' is similarly placed at z=1, x=0, lying in the xz plane. This one points down, the rays run from z=1,x=0, around the length of the line X, ie z=-1, x=X/2 to z=-1, x=-X/2). and continue beyond. Because this is not solid in xyz space, it is expanded over all values of y (ie multiplied by the y axis). This gives a solid region.

The tetrahedron is then the intersection of these two regions.

The tegum-product of antiprisms, is itself the antiprism of the pyramid product of the antiprism-bases, and the prism-product of antitegums, is the antitegum of the same product. This is the mystery of quickfur's question on the square pyramid // dual square pyramid.

by **wendy** » Wed Sep 19, 2012 9:25 am

In general, lace prisms and lace tegums have very little to do with the prism and tegum product, unless you can involve the height in the product. Instead, they're more kindred to the pyramid product. It's what happens when you push the historical 'antiprism' to its logical conclusions.

In general, where the prism gives rise to the bi-pyramid, the two pyramids of that are the two lace cones of the x.Po. and .xP.o . The figure is the corresponding intersection. There was some considerable debate over what 'antiprism' might mean, since these can be represented as s2sPs. In my mind, it is more useful to regard s2sPs as a half-diminishing of x2xPx, and the antiprism as a relation of duals. Lace prisms derive from antiprisms by general extension.

The dual of a figure might be constructed by placing tangential points at the vertices (which works wonderfully with uniforms). If one does this where the thing is a face of a larger figure, one can tilt the planes along where they intersect with the figures' plane, and make them cross at a point. This point represents the dual of the face of the figure. But at this point, we have nothing to restrict them radiating outwards, so it's a 'cone' rather than a based pyramid.

Just like we have two or more bases in the lace-prism, we have two or more lace cones, all of the same construction.

If one has two lace cones, in the shape of a pentagon-peak, one can intersect so that the rays through the vertex of the pentagons meet. This would create a pentagon bipyramid (or tegum). When we rotate one pentagon relative to the other, we get something like a dodecahedron, with pyramids on opposite faces. The lines radiating away from the poles no longer intersect, and instead of a coplanar equator, we get a reverse-version of the zigzag seen on the antiprism.

When you do the same sort of thing for the cupolae xoPxx&x, you get some the 2P triangles from one pole, and from the other pole, the faces come out like the startrek logo (a strombus).

Apart from the fact that kepler used the term 'antiprism' for the class-leader, there is very little connection between the two. In practice, it is more the other way: lace prisms are more tegmic than prismatic, and lace tegums are more prismic than tegmic. For example, lace-prisms and tegums quickly multiply the number of faces, while the other two just add faces. The anti-tegum has only antitegmic elements, in the manner of the cube (which is a triangular antitegum). But like prisms, lace-prisms are more likely to be uniform or at least have edges of equal length, ie a C-UC. Tegums and Lace tegums do so by exception

by **wendy** » Thu Sep 20, 2012 10:01 am

The class of spheres, ellipsoids, and kindred figures, are handled by the overloaded dynkin symbol, but using the special rune 'O' for the circle.

/O = circle, /OO = sphere, /OOO = glome usw.

The placing of / states that there is always an increase, so

/OO/ is a prolate ellipsoid (ie x=y < z) = xOoOx
/O/O is an oblate ellipsoid (oe x < y = z) = xOxOo
/O/O/ is a general ellipsoid = xOxOx

One then has by placing things in product, the likes of oxOoo a cone

In four dimensions, one has more elliposids, including /OO/O w=x < y=z , a possible stable shape for a double-rotation.

The sphere-product gives rise to the radiant product rss(). This means the final surface, is set by unit rays r² = x² + y² + ..., where each ray x, goes from the centre (0) to the surface (1). It defines a coherent unit, where the sphere in N dimensions is a unit crind-solid volume.

Names of various products. Expressions like P1C2 are varieties of what's shown after. This symbols like [x(xx)] state a product of [ prism ] over two elements x and a crind-product of (x,x).

/O circle C2 (xx)
/OO sphere C3 (xxx)
/O&/ cylinder P1C2 = [x(xx)]
(intersection of cylinders) C1P2 = (x[xx]).
xoOoo cone (a lace -prism)
/O&/O duocylinder = bi-circular prism PC2C2 [(xx)(xx)]
\O&\O bi-cylinder tegum TC2C2 <(oo)(oo)>
/OO&/ spherinder P1C3 [x(ooo)]
\OO&\ sphere-tegum T1C3 <x(ooo)>
/OOO glome C4 (xxxx)
/O&/4 cubinder = P2C2 [xx(xx)]

by **Klitzing** » Thu Sep 20, 2012 2:51 pm

Found some threads in this archive dealing with specific vertex figures.

So it looks apropriate to publish Wendy's findings about vertex figure derivations right in this very topic of notions and notations. Not only 'cause it is kind a central topic for techniques etc. It even would fit in here, as her technique starts with Dynkin symbols. That is, it just acts on those symbols only!

1st step: Vertex figures of regular polytopes

Any regular polytope can be given as some linear Dynkin diagram, one end of which is ringed, while all other nodes being un-ringed. I.e. something like x-P-o-Q-o-...-o-R-o. In fact that symbol is a mere transliteration of the according Schläfli symbol, which then would read {P, Q, ..., R}.

For those it is well-known that the vertex figure is nothing but the (also regular) polytope of 1 dimension less, which has the Schläfli sub-symbol {Q, ..., R}.

We could leave off here, and procede to the next. But stop! In order to be armed for what comes next, we should not only derive the shape itself, but also the correct scaling. So, the dodecahedron, the cube, and the tetrahedron all would have a regular triangle for vertex figure. But the base of those vertex pyramids, which is nothing but the requested the vertex figure, differ in scale! That of the tetrahedron clearly has size 1 (same as the tetrahedral edges). That of the cube would have sides of the size of the squares diametral, i.e. sqrt(2) times the former edge length. And that of the dodecahedron would have those sizes which correspond to the secant of the pentagon, which omits a single vertex, i.e. runs from one vertex to the 2nd. - As is known from the geometry of the pentagon, this would be tau times the side length, the golden ratio.

Accordingly the correct answer to the vertex figure ("verf" for short) of x-P-o-Q-o-...-o-R-o would be: x(P)-Q-o-...-o-R-o, here x(P) would be an edge again, but its size is exactly the length of that secant of the regular (unit sized) P-gon (or x-P-o).

Wendy is here rather pragmatic. She just defines the most often to be used sizes by specific letters:

Code: Select all: x = edge of size 1 (corresponding to the verf. of a regular triangle) q = edge of size sqrt(2) (corresponding to the verf. of a reg. square) f = edge of size tau (corresp. to the verf. of a reg. pentagon) v = edge of size 1/tau (corresp. to the verf. of a reg. pentagram) h = edge of size sqrt(3) (corresp. to the verf. of the reg. hexagon)

For the more arithmetically scilled ones, here would be the general values:

x(p) : x = sin(2π/p)/sin(π/p) for any rational p>1
2nd step: Vertex figures of quasiregular polytopes

Now we are considering polytopes, the Dynkin diagram of which needs no longer be linear only, but still just a single node is ringed only. That one further more does no longer need to be positioned at any end of the diagram, but could be positioned anywhere.

Again we might go back into the times of Schläfli symbols. We just bend the Dynkin diagram at the ringed node and comb the rest of the symbol from there into a single direction. (Sure, this would not apply to any closed loops in a Dynkin diagram, but at least for all linear ones or even for the bifurcated ones.) This is what is known as the Coxeter extension to Schläfli symbols.

So, what is the vertex figure of such figures? And why did we all this cosmetics to the Dynkin symbol? - Well, Coxeter already gave the solution to this in a more topological sense. We just have to split that (extended) Schläfli symbol according to the leading numbers into subsymbols, either of which start with one of these, and contains the therefrom emanating rest. Then that leading entry has to be dropped in eiter of those.

This kind of sounds a bit dubious, as for all this Schläfli symbol stuff. So we would translate it - before going into details - into the Dynkin symbol context: Coxeter would advice just to delete that single ringed node of the Dynkin diagram, likewise all therefrom emanating links, and finally ring all those nodes, which formerly where connected to that now deleted node.

Again, this result of Coxeter is only true in a topological sense. The sizes of the vertex figure edges are in general derived wrong, esp. as now the verf. would have more than a single edge class (ringed node). But again, we can get it metrically correct, by applying edge lengths x(P) to those nodes, which were formerly connected by a link marked P.

Note, that this only gets out of the problem of knowing that a linear dynkin symbol (for example) in fact is a simplectical graph, just that most links are marked 2 and therefore are not shown. - Our rule just would allow to reconsider these links marked 2, and apply accordingly a node symbol x(2). As x(2) is the secant of a digon, that secant obviously has size zero, and it comes out correct, i.e. equates to an un-ringed node, as would be the result if those links marked 2 where omitted before the application of the rule.

3rd step: Vertex figure of arbitrary Wythoffian polytopes

First of all, what is a Wythoffian polytope? - That's the most easiest question! Dynkin diagrams just encode the kaleidoscopical construction b yuse of mirror-cone and seed-point, as was devised by Wythoff. Thus a Wythoffian polytope is just anything what can be given as a dynkin symbol, hehe. I.e. some reflectional symmetry group graph, decorated by rings (classes of unit edges) or un-ringed nodes.

With respect to the vertex figures, one of Wendy's findings comes in here.

We just have to multiply that given Dynkin symbol: once for any ringed node.
We delete all ringed nodes but that specific one, together with the emanating links.
Then we apply to that sub-graph the vertex figure derivation according to step 2. (Be sure to do it metrically correct.)
Now consider any 2 of those formerly ringed nodes, and, if necessary, the connecting link. This small sub-graph represents a polygon. The vertex figure of that polygon is geometrically obvious (the according secant). that is we would derive specific edge lengths here, corresponding to pairs of formerly ringed nodes.
Finally we would have to set up a lace simplex, which has for its (graphical) vertices those single-ring vertex figures, and whoes (graphical) edges represent the lacing edges, which have the appropriate size, corresponding to those 2 nodes under consideration!

Just to provide an example. Consider x3x4x. We would have 3 subgraphes: x . ., . x ., and . . x - so the single-node vertex figures all become rather easy here: the verf of an edge is just a point (vertex). More relevant are the pairings: x3x ., x . x, and . x4x - accordingly representing a 6-gon, a 4-gon, and a 8-gon. Thus the relevant link sizes are x(6), x(4), and x(8). Finally, when putting all together, we would get a triangle with 3 different sides, one being x(6), one being x(4), one being x(8). - And, in fact, this is what is the vertex figure of the omnitruncated cube, i.e. the base polygon of the relevant vertex pyramid, cutting down to the nearest vertices. (Kind of a facet underneath.)

--- rk

by **wendy** » Fri Sep 21, 2012 7:53 am

One should recall that it took many deep insights over the course of a century, to bring the mathematics up to the level it is to-day. Even by me, it took nearly a year to decode the decorations on the graph, based on the limited zoo in 'regular complex polytopes'. Still, this has been done.

The Vertex-Node

The vertex node was my attempt to read the decorated dynkin graphs. Instead of marked nodes, one connects these nodes to a decorative vertex-node. Because this is read straight as a vertex-figure, the branches that connect to the vertex-node were marked with polygons whose shortchords become lengths. 3 started off as unmarked, ie '/'. while the other numbers used modifying lower-case letters /q etc. These become x, q, etc.

A symbol was eventually added for '3' to allow the use of a coordinate system, and for '2' to allow one to write decorated schwarz triangles (what are mis-called Wythoff symbols). R and r can be used to represent a miniscule figure, a quasi-zero. For example, one might imagine r4o3o to be a mini-cube, where the 8 vertices stand sholder to sholder, while an r4o3r is a rhombocuboctahedron, so done. It's useful to use r to show where rectangles etc grow. U specifically applies to the horogon, not the straight line, or general infinite lines in hyperbolic space.

The use of letters is one of the contributions of twelfty. This pushed the notation sufficiently far enough (for example, we don't want 335 to become 2.95), that letters were used for the branches. In turn, the lower-case letters become the short-chords, and the letters can be used with numbers to form a coordinate system, eg 1 S 0 F 0 is the vertex of an icosahedron.

Code: Select all: f----x v x q f h u x r | | V| | Q| F| H| U| S| R| 5| | 5|2 | 4| 5| 6| oo| 3| 2| | | | | | | | | | | o----o o o o o o o o o

One should imagine that the open branches all connect to a single vertex or $ node. In the earliest representation, the lines were crossed with a double-line, like an equal sign.

When one has the vertex-node separate to the other nodes, rather than having marked nodes on the symbol, it is easier to determine the surtopes, and their adjoining around-symmetries. This is because a surtope has as many nodes as the simplex of the same dimension, and at least one vertex-node. Moreover, if there are nodes not connected to a vertex-node, a zero-height prism comes.

the stott matrix

In order to make the vectors from above useful, one has to have a process to do the dot matrix with them. It took a bit of fancy stuff, but the basic theory was laid down in the study of the heptagon. One does a 'matrix-dot'. This involves creating the matrix a_ij, so that a_ij. v_i . w_j does the dot product of the vectors v and w.

The actual construction of the matrix consists of considering every single, and then double-marked nodes. The single nodes give a_ii and a_jj. The double-nodes give a_ii + a_jj + 2_ij. One would then first draw out the vertex figure, and then calculate its over-diameter. From this, it is a pretty straight forward process to filling a_ij.

lace-prisms and klitzing

Richard Klitzing was at this time fiddling around with segmentotopes. The particular entry that was worrying me at the time was o3x3o4o3x3z. This consists of a pair of rotated cubes, gyrated around a face. The idea hit me that if one regarded all of the r2 lines as a connection, the outcome then become in rk's notation as x2x4o || x2o4x. The idea came that instead of a point-wedge, one can use a line-wedge, and put a line there to reflect around. The example of type is the xPo // oPx = antiprism. Several names, like 'exotic prism' were tried out. Exotic is heavily overused, so the next choice comes from the lacing of a drum, giving rise to the 'lace prism'.

Richard's segmentotopes are generally different to the vertex-figures that lace prisms are supposed to represent. But none the same, it's because the segmentotopes have unit lacing, where our lacings are x2x, x3x, x4x and x5x, and x6x. the shortchords of these polygons. The process is none the less the same.

Because the ultimate diagram of note is to have several vertex-nodes in the graph, the symmetry group is similarly reduced, ie 2 vertex nodes gives a line-height, 3-vertex-nodes gives a hedral-height, and so forth.

Still, using the vertex-node, it is pretty straight-forward to evaluate the vertex-figure. Here is the vertex-figure of x3o4x.

Code: Select all: $A vert $A vert 4/ A 1 4 / A / B 1 / B B$-----$C C 1 1o-----$B edges edges x4x3x AB x4x = r(3.414) x4o3x AB = r(2,000) AC x2x = r(2.000) A1 = 1.41421 BC x3x = r(3.000) B1 = 1.00000

Every figure requires at least one vertex, but can have more. If there is only one vertex, the figure falls in a space vertical to the rut or join of mirrors. So in the second example, A1 and B1 are parallel, and AB connects the ends (as a trapezium). In the first case, there is no symmetry, so the whole thing is singular.

The polygons that appear are either something like $N$ (vertex figure of a 2N), or $No (vertex figure of an N-gon), the latter gives the general symbol 'n'.

One can even generate the count of elements: in the first case, the overall symmetry is 1. So each element is unique. In the second, the symmetry is of order 2, so if there is an element that has a single-symmetry, such as A, B, and AB, they occur twice. The elements A1 and B1 have a symmetry of 2, so they occur just once.

by **wendy** » Sat Oct 06, 2012 11:15 am

The transport of symmetry and number

The dynkin graph also shows the transport of number and symmetry.

An odd branch connects mirrors of the same type. That is, were one to put a red spot on one mirror, it would jump across odd branches, but not even ones. The size of such a mirror group, relative to a symmetry, can be found by removing all of the mirrors without the red spot. This contains a room whose walls all come from mirrors with the red spot.

The group o----o-4-e contains two kinds of mirror, shown as o and e. The size of the mirrors 'e' is six times that of the (3,4), which makes its order 8 = 48/8. The mirrors of the type 'o' belong to a group whose order is twice that of 'o', or 24.

While groups in euclidean and spheric geometry only fall to one set, in the hyperbolic space, one can divide a group by two different mirrors, which can not happen together. The group *9,3,3 can be divided by mirrors into four cells giving * 9,9/3,3, or into two cells giving *18/2, 3, 2, but these mirrors can not be applied together.

The use of transport to determine the size of different groups, shows that the vast number of the crystalographic hyperbolic groups, are the intersection of a much smaller set of mirrors.

The transport of number

More remarkably, one can determine what kind of numbers to expect, from the inspection of the graph. Each branch is a bridge between two structures of the form Ab, where A is a number-system, and b is a location-variable number, being a square-root of a number in A.

The number system in a group is the simplest structure that contains the product of all of the A's, and any circle-constants. The values of the b's is purely relative, but crossing the bridges can change the sizes of 'b'. If one returns to a mirror with a value of b different to what is started off with, the ratio becomes a circle-constant.

The number-system for a odd polygon p, or for an even polygon 2p, is Zp. This corresponds to the span of chords of a polygon (p). For Z1, Z2, and Z3, the ordinary integers apply, since these polygons all have integer chords. For Z4, Z5 and Z6 the span of chords are numbers of the form a + b sqrt(n). These are the most common numbers met in polytopes.

The bridge constant is 1 for all odd branches, and something different for even ones. The most common are sqrt(2) for '4', and sqrt(3) for '6'. For polygons with a divisor 4n+3, then one can use any of the sqrt(4n+3) as the bridge constant. For example, in the polygon {21}, one can use sqrt(3) or sqrt(7), because these are directly connected to the integer system Z21, by the common value (sqrt(7)+sqrt(3))/2, which can be constructed from chords of the {21}.

The group o----o--4--o has branches of orders Z3 and Z2 (which are the ordinary integers), so the general coordinates are integer. However, if one looks at the three nodes, the relative values are 1, 1, r2 (because the branch '4' has a bridge constant of r2), or r2, r2, 1. If one constructs a cube with integer coordinates, it has an integer edge. However, an octahedron x3o4o, would have vertices which involve r2 to have an integer edge, or involve edges of r2 (ie q3o4o), to have integer coordinates.

If one has the same value on both sides, eg x4o3x, the relative values are 1, q, q, and so one would expect the coordinates to involve numbers of the type x+y.r2.

The group o---o-4-o---o-6-z (a loop 3,4,3,6:), has again a through-integer sytsem of Z2 Z3, which is the integers. The nodes can be evaluated by supposing the first node is 1. Then one goes 1, 1, r2, r2, and returns with r6 to the first node. It then continues with r6, r6, r3, r3 and back to 1. This means that the integer system for this group is (1,r6), and the two bits are held at a ratio of 1:r2 or 1:r3, the constant r3+r2 will make both of these into the integer system.

The group o---o---o-5-o comprises of odd branches, but the through system is Z3, Z3, Z5 or Z5 (since the integers are a subset of any larger set). This means that polytopes will characteristically comprise of numbers of the form x+y.sqrt(5).

One does not have to study every possible number-set to see what its consist is. One can create a case where one has two nodes, and connect them with branches of the type 2p (p odd). For example, with Z21 mentioned above, we use branches of 6 and 14, connecting two nodes. The through-system here is Z3.Z7 = Z7. The branch constants are sqrt(3) and sqrt(7). One has an odd number on one side and an even number of these on the other side: ie 1, sqrt(21) on the left hand side, and sqrt(3) and sqrt(7) on the other side. It is indeed possible to construct sqrt(21) from the chords of the {21}, but not sqrt{3} or sqrt{7}. One needs a {42} to get that result.

The integer systen for powers of 2, are made by placing branches representing 3, 4, 8, etc into the point pair, and using the next power to make the bridge constant. All twice-odd systems have a bridge-constant of r2. All four-odd systems have the same bridge constant as Z4, and so forth.

For all hyperbolic groups, it is almost sufficient to apply the laws of symmetry and of transport, to divide the groups into small pens where subgroups might be found. The groups associated with Z5 are the most difficult, since there are many entries, but few subgroups. The group Z4Z5 has a unique member (o---o-4-o---o-5-z), whose through-system is Z5, but the looping transport 1 and q to every node, so Z4 is also part of it.

It's only of euclidean polytopes where it is possible for a group say Z5 or Z4, to contain a subgroup like Z1.

by **wendy** » Tue Oct 09, 2012 7:14 am

A product is a process, where f(a×b) = f(a)*f(b).

There are four shape-making products, which make the four families of infinite regular polytopes.

1. pyramid product, which makes simplexes
2. tegum-product, which makes the cross-polytopes
3. prism product, which makes the measure polytopes
4. comb products, which makes the cubic honeycombs.

One element of the product is f(), applied over the Euler sum (eg for a cube, 6h + 12e + 8v). The product of the elements has a f(ab..) = f(a) f(b)...

There are two different elements that might be added, which leaves four different parts.

n. Adding an element at N=-1 (nulloid, essence), makes the product into a draught, rather than a repetition.
c. Adding the element at N=n, (content, body), makes the product into one of content, rather than content.

If E represents the euler sum (eg 6h+12e+8v), then

pyramid = c + E + n = draught or drawing of content -> simplex polytopes = point ^(N+1)
tegum = E + n = draught or drawing of surface -> cross polytopes = diagonal ^ N
prism = c + E = repetition of content -> measure content = edge ^ N
comb = E = repetiton of surface -> cubic product = girthing polygon ^ (N-1).

The products of surface only make sense when there is a surface

The etymologies

pyramid = fire
tegum = cover, cognates include 'toga', 'deck' (roof one can walk on) and 'thatch'. The surfaces of the elements hold up the surface of the figure as does a tent.
prism = offcut (of timber), a section of a longer board, cut from the end.
comb = tunnel (as in honey-comb, cata-combs).

The products of volume define coherent units. These are powers of length. The radiant model of a polytope is to hold a 'centre' at some point inside. The surface is set at unity. This makes the polytope appear as a spherical function, where the surface is one. Values outside are larger than one.

In a product of several peices, the coordinates say, x1 x2 x3 gives a radiant value of r1, and y1, y2, gives r2. The value of the polytope at x1 x2 x3 y1 y2 is then f(y1, y2), is the radiant value of the product-polytope.

There are three radiant products accepted. (rss = root-sum-square = square of the sum of squares)

1. prism product: r0 = max(r1, r2, ...) 1d = edge, 2d = square L, 3d = cubic L, 4d = tesseractic L
2. tegum product: r0 = sum(r1, r2, ...) 1d = diagonal, 2d = rhombic L, 3d = octahedral L 4d = biquadrate L
3. crind product: r0 = rss(r1, r2, ...) 1d = diameter, 2d = circular L, 3d = spheric, 4d = glomic,

These products are coherent, that is, when measures are in the units of the crind powers, then the volume of the crind product is the product of the crind volumes.

O/T post 1000 = 8.40. But i believe to get to pentonian one has to go to 1024 = 8.64 = 4^5.

by **Klitzing** » Wed Feb 06, 2013 12:17 pm

In here, I'd like just to refer to that post, where I've outlined an introduction to additions of Schwarz triangles respectively Goursat tetrahedra. (These are the fundamental domains of 3D resp. 4D reflection groups.)

(Even so it was ment as a reply onto a question within that thread, its very explanatory nature would recommend its positioning within this thread likewise...)

--- rk

by **wendy** » Sun Oct 06, 2013 10:29 am

KEPLER OPERATORS

These are things you can do to any polytope, not just the regular ones. The general application depends pretty much on how 'regular' the solid is. It should be noted that kepler operators do not equate to stott-operators (the marks on the dynkin symbol, eg), but there is a generous set of equalities which have prompted a naming pattern

n-TRUNCATED, n-RECTIFIED, etc

The truncated/rectified come from sections of the antitegmal series. One imagines that A is increasing in size, and the dual B is decreasing in size in dual position. The antitegmal series is the intersection of these, because if the sections are strung out in series, they form an A antitegum.

At a point where B has cut off A's vertices, comes 'truncated A'. This continues until the whole edge disappears, when it becomes the 'rectified' A. The process continues as the intersection of A and B now head from the edge-centre to the hedron-centre (2d element), this becomes the 'bi-truncated' while there are bits of the A-hedrons there, and bi-rectified, when the A-hedra disappear. The process continues to tri-X as the vertex of AB continues to the mid-points of the chora of A, etc.

DIMINSHED x

A diminished X removes with the vertex, the whole of the edges etc connected to it, and creates a new face formed by adjacent vertices.

n-APICULATED, n-SURTEGMATED etc

One imagines here that you have a vertex-frame of A (at fixed size), and a vertex-frame of B which increases. A vertex-frame is just a mob of vertices of a polytope kept at the correct size (like the four corners of a square). Covering A is some kind of flexiable skin.

As B strikes the face-centres of A, it starts to push the skin outwards. At first, this leads to pyramids on the faces, or raising 'peaks' (ie apiculation). At some point, the pyramids are high enough that the angle between the bases of one and that in the next pyramid become 180 degrees, What happens here is that the edge of B is now part of the surface, and the face of the hull become tegums of the margins of A, and the edges of B. This is a 'sur(face )tegum' of A.

A B rises, the edges of B are now part of the surface, and one gets pyramids formed between the edges of B, and the margin (face-faces) of A, ie a pyramid of an 1, and N-3 element. This is the 'bi-apiculate', which cumulates in the 'bi-surtegmate' At this point, the hedra (2d elements) of B now become parts of the surface, and are in tegum-product with the N-3 elements of A. (This gives an N-1 element, a face). This happens until A is completely swallowed by B.

n-CANTETRUNCATE. n-CANTELATE

Are the truncates and rectate of the n-rectate. For example, one has the tCO as "truncated rectified cube", is a cantelate.

RUNCINATE, STROMBIATE

A runcinate is fromed by moving the faces of A away from the centre of A, without changing size. A skin over the surface of A will cause the N-x elements of A to form a prism with an x-1 element of B.

The strombiate is formed by putting the surtopes of A and B together on the same surface, rather like putting a tesseract and 16-choron togehter on the same sphere. The faces of the new figure is formed by anti-tegums of the vertex-figures of the faces of A (or B), which like A and B themselves, are duals. You then a figure whose faces are tangent to the sphere at the centre of lines connecting the face-centres and vertices, and is hence dual to the runcinate.

The sequence of runcinates form an 'antiprism sequence'. That is, progressive slices through the truncinate will give an antiprism, if one supposes the edges of A and B add together to a constant sum.

OMNITRUNCATE and VANIATED

These are related to 'pennent theory'. In essence, a flag is a simplex formed by v_0, v_1, v_2, &c, where v_n is the centre of an n-dimensional surtope. Vaniated X means that the simplex thus formed is the face, while omnitruncate X means that one puts a point in the centre of x, and drop perpendiculars to each wall.

You can apply these operators to any wythoff or conway-hart constructions, even when the flag does not correspond to coxeter's flags. All that is necessary is that there is a tiling of simplexes, where it is possible to identify p_0, p_1, etc such that at p_n, one only has v_n of the simplexes. The flags of polyhedra and wythoff simplexes are examples of pennents.

Making a wythoff symbol from these

These do not exactly correspond to stott-operators, but it is possible to devise a set of rules which allow one to create a dynkin symbol from it.

The name consists of (m-)operator - polytope

m- is a depth operator, as stated above, as in 'bi' in 'bi-truncated'

bi = o, tri = oo, tetra = ooo, etc. It's one less circle than the number.

The operator consists of a central token, with various attachments. The attatchments are replaced by various combinations of o or whatever is adjacent. The result is to produce a figure with the same number of x, o, m etc as the dimension.

* = depth attachment, as listed previously (bi, tri, tetra, etc)
~ = zero-fill operator, one justifies left and right of this, and fills the rest with 'o' nodes
# = block-fill, one justifies, and fills the missing places with what is adjacent to #, eg #xo might give xxxo in 4d.

The third bit is the operator,

truncate = *xx~ apiculate = ~mm*
rectate = *ox~ surtegmate = ~mo*
cantelate = *xox~
cantetruncate = *xxx~
runcinate = x~x strombiate = m~m
omnitruncate = x#x vaniated = m#m snub = s#s.

The symbol is made by expanding ~ and # until there are n elements. All *, ~, #, should have been removed

The schlafli symbol of the regular is then interspaced between the x, o, m etc.

So the 'bi-truncated pentaact (a 5d figure), gives
truncate = *xx~
bi * -> o : bitruncate = oxx~
(to 5d) ~ => oo : in 5d oxxoo
Pentaract = 4333 -> o4x3x3o3o

To read o3m3m4o -> ommo 334 = ~ mm (o) 334 = (bi) apiculated 334=16choron.

by **student91** » Wed Dec 11, 2013 11:12 pm

One thing I'd like to be highlighted are the incidence matrices. As far as I foud out from klitzing's site, these list the incidences of parts of polytopes. But still there are two things unclear to me:
1 what is exactly ment by incidense, is there a clear definiton for this, and how is it quantified in the matrix?
2 what do the coloums stand for? Are these the same as the rows, but then why is a_ij not equal to a_ji?

student91

by **Klitzing** » Thu Dec 12, 2013 7:04 am

student91 wrote:One thing I'd like to be highlighted are the incidence matrices. As far as I foud out from klitzing's site, these list the incidences of parts of polytopes. But still there are two things unclear to me:
1 what is exactly ment by incidense, is there a clear definiton for this, and how is it quantified in the matrix?
2 what do the coloums stand for? Are these the same as the rows, but then why is a_ij not equal to a_ji?

student91

So you most probably never read my subpage explaining all that in detail?
Cf. http://bendwavy.org/klitzing/explain/incmat.htm

--- rk

by **wendy** » Thu Dec 12, 2013 7:44 am

Incidence and Incidence Matrices.

Incidence is a relation between surtopes (surface polytopes, ie vertex, edge, ..., face) of a polytope. It is notionally a line connecting the centres of surtopes, where the S end represents a surtope of the surtope at the A end. We imagine the line being sent to S from A.

If we look at a dodecahedron, there are pentagonal faces, and edges. Where an edge belongs to a pentagon, then there is an insidence where the edge is S and the pentagon is A. Lines leaving the point for A, would lie completely in the A (the pentagon), ending at the surface (edges) of the pentagon. A line arriving at S would come perpendicular to the space that S is in, and is around the surtope "S".

In the case of a dodecahedron, there are twelve pentagons, each producing 5 S ends. There are 30 lines, each being the arrival of 2 A ends.

So we have 12*5 = 30*2 = 60, the count is of actial "incidences".

Incidence Matrices

Incidence matricies come in the much less common "specific case" and the very common "type cases". The matrices that Richard Klitzing produces are of the second type, for good reason. In the specific case, one lists every vertex separately (ie 20 for the dodecahedron), and every face (12), and produces a 12*20 matrix just for those. In it, one puts a one or zero, as the vertex is part of the face or not. There are 240 positions in the matrix, of which 60 become 1, and the remaining 180 are zero.

The type case groups all similar vertices together, and all similar faces, and has a single entry for each. So there are 20 vertices, amd 12 faces. Each vertex has 3 faces, and each face has 5 vertices. We see that 20*3 = 12*5 = 60, gives the count of incidence lines.

The actual matrix is laid out with the surtopes by kind, listed across and down, so that P_i is both the i row and i column.

The entries M_ii represent the count, either absolute or proportional, of the surtope in question. It works on infinite tilings like lattices too.

The entry M_ij represents the number of times Pj is incident on Pi. Because if Pj is incident on Pi, then Pi is incident on Pj, one then has the same incidences on two rows, and M_ii * M_ij = Mjj * Mji everywhere.

If P_i and P_j are of the same dimension, they can not be incident on each other. Richard normally lines out the surtopes by dimension, and indicates a common dimensionality by writing M_ij = '-' at these instances. Numerically, it's another kind of 0.

I refer to these matrices as S\A matrices, the bit below the diagonals represent the sub-surtopes, (that is, the surtopes that belong to the named one), and the A as ortho-surtopes (from where the incidence-line arives at right-angles to the surtope). The line through the value P, then consists of (1) the subsurtopes of P, (2) At M_ii, the count of P, and then (3) the super-surtopes containing P.

Here are some incidence matrices.

Code: Select all: v e1 e2 h1 h2 v e h c v 24 2 1 1 2 v 600 4 6 4 e 2 1200 3 3 e1 38 2 24 - 1 1 h 5 5 720 2 e2 88 2 - 12 0 2 c 20 30 12 120 h1 3 3 3 0 8 - h2 8 8 4 4 - 6

In the first case (truncated cube), there are two kinds of edge, one between 3 and 8, and the second between two octagons. These have different rows in the A bit. Likewise, the two hedra (2-d patches), are shown separately. Note that these are stull 'types', so we have the corner relation 24*2 = 8*6.

In the second case, (twelftychoron or "120cell"), this is by type, so each of the 600 vertices connect to 4 dodecahedra, and the 120 dodecahedra connect to 20 vertices, which is the 2400 incidence lines of this kind.

by **student91** » Thu Dec 12, 2013 4:37 pm

Thank you both, I understand it now

by **wendy** » Mon Dec 16, 2013 11:01 am

Another place where one might use "incidence lines" is when one is trying to work out how the cells of an apeirotope (tiling) come together.

For example, the tiling of rhombo-cuboctahedron, cubes, and tetrahedra, such as might be represented by "x4o3oxAo", might look like this:

Code: Select all: 4 8 6 6 tetra ---(/3)----- rCO -----(/4)----- cube | 12 (/2/) rectangle | =

The lines represent the walls shared between the cells, /3 = triangle, /4 = square, /2/ = rectangle.

The /3 line says that there are eight triangles in the rCO, but only four on the triangles. This means there are twice as many tetra as rCO.

The /4 line tells us that there are six squares of the rCO shared with cubes, and the cube has six squares shared with rCO, there are equal numbers of each.

The /2/ line runs into an equal sign. The /2/ are rectangles, whose sides are adjacent to the triangles and squares. The fact they run to an equal sign means they find another cell of the same nature there. These of course do not contribute to the count, but show that all faces of the rCO are accounted for.

Code: Select all: 4B1/ = ||---(72)----/4B =====(27)===== 4/B ----72--||

The above shows the tiling, in six dimensions of gosset's polytope 2_21. One can divide these into two sets, which gives different colours (black, white), which have a self connection of 72 walls (simplexes), and an 'other' connection of the 27 walls (orthotopes). Because the 27 is at both ends, we see the tiling of 4B1/ has equal numbers of /4B and 4/B.

Stellating the 72 simplexes into the middle of the faces of the other 2_21's, one gets a tiling of the Duals of the 2_21 (ie \4B), which is yet another instance where X and its dual tile space. (another example is in 4d, of the tesseract and 16choron.

by **wendy** » Mon Feb 17, 2014 9:03 am

(this is horribly confusing).

The hyper-complex plane is a plane of the form X+jY, where j<sup>2</sup>=+1. The conjecate is found by putting -j for j, ie (X+Yj)* = X-Yj. The modulus of this number, is found by AA*, ie X²-Y².

Just as the complex plane has circles, and R. cis(th), the hyperbolic plane has R cish(th), and hyperbolae, eg unit hyperbolae.

There is a 'zero' and an 'alt-zero' lines. These are where X+jY=0, and X-jY = 0. When one sets J as j.sqrt(N), one has x+Jy for the real magnitude, and x-Jy for the alt-magnitude. One finds some of the most beautiful patterns when the alt-magnitude is heavily restricted in size.

One overlays both the real and hyperbolic planes with integer lattices, usually of the form X+aY, where a² is in Z, and if a = 1, mod 4, then X and Y can be integer-halfs as well. So the eisenstein integers (a=-3), the integers in sqrt(-7), the pentagonal integers (+5), the real integers (+1), are all expressable in 'integer-half' notation (eg 2.5+0.5 sqrt(5)).

One difference between the complex and the hypercomplex numbers is that the latter has two real axies (real and alt-real). What we normally see is the real axis. However, if you consider the projection into the alt-real axis as well, much of life is made easier. This is because one can take the real and alt-real axies as different shadows of linkages that happen on the hyper-complex number. We can generally only see these linkages, for example.

One implication is that in polytopes, 8 <--> 8/3, 5 <--> 5/2, 10 <--> 10/3 and 12 <--> 12/5 as j goes between +1 and -1.

For example, there are tilings like s4s4o, whose vertices lie on various vertices of a dodecagon 12. Putting s4/3s4o makes the isomorph, which causes the same sequence of vertices but on the dodecagram 12/5. Any of the uniform figures which have marks on both sides of a '4', as in x4o3x, or x4x3o, have a corresponding isomorph with the octagons replaced by octagrams (x4/3o3o, x4/3o3o, both d7).

The evaluation of the rings of the {3,3,3,5} was made possible, because it is the same linkage as {3,3,3,5/2}, a real, all be it dense, polytope. This allows us to draw a limiting line on the hypercomplex plane where vertices would not occur. It is then a case of enumerating the densities of the rings corresponding to the numbers spread that far apart.

Penrose Tilings

The projections of penrose tilings onto the hypercomplex plane, reveals in one axis, the tiling itself, and in the other axis, a very tight sphere as the conjucates of the tiles fold back on themselves. Because it *is* a small sphere, one never gets a pattern that repeats for ever. Instead, if a pattern were to repeat, it would bave to produce an isomorph that was extremely tiny, so that many tiny steps can cross the disk.

Penrose tilings can be made by "rescaling". R Klitzing wrote a rather interesting account on this, it is in german, though. In rescaling, one is effectively using a cish(N) relation, replacing eg A => AB, B => A (gives the pentagonal integers). One can as easily use an engine which generates the next number, from that the alt-shadow of A -> AB, B -> A gives A=1, B=v, the alt-form gives A=1, B=-f.

The idea here is that in a sequence like ABAAB ABA ABAAB the reals will lurch forward in steps of 1 or 1/f, while the alt series will step between -1 and f. You add 1 if the number is less than 0.618033, and subtract f if it is over this.

Penrose tiles do much the same thing. In one case, the tiles are restricted to some order on the plane. But on the other side, one has a series of pleat-folds, which keep all the alt-tiles in the same space.

Hyperbolic Symmetries

Hyperbolic symmetries work on the same 'real-altreal' linkage thing. In one case, the projection might give {5,3,4}, which is hyperbolic, but its alt-form is {5/2,3,4}, which is real. There are other ones like {5/2,3,6} and {5,3,6} which are both hyperbolic. The symmetries, and even the encountered distances, can never co-incide. In fact, the statement above is sufficient to demonstrate that these contain no more than one vertex in common.

The reason is that the chordal lengths representing the arcs in {5,3,4}, are NP, while in {5,3,6}, NN. That means that all of the chords of the first have negative squares (which indicate a hyperbolic tiling), while the alt-squares are all positive. In the latter, both the squares and the alt-squares are negative (a different field).

Knowing that the alt-real part is a real polytope, allows us to construct the size of this, and this in turn tells us that the alt-shadow of the linkages can not go past that line.

by **Marek14** » Tue Feb 18, 2014 7:20 am

I explored hypercomplex plane and related number system once. One fun result was that multiplying numbers in hypercomplex plane corresponds to adding of velocities in Special Theory of Relativity (not that surprising since it's based on a similar metric).

by **wendy** » Mon Mar 31, 2014 9:31 am

The Conway-Thurston Notation is a generic notation for two-dimensional symmetries, in much the same way that the Coxeter-Dynkin notation serves for the mirror symmetries, particularly the hyperbolic groups.

William Thurston proved, that all possible 2d symmetries can be resolved into four possible elements. Conway devised a notation to show these.

These things are called orbifolds.

Cone. A rotation around a point. It is imagined that you cut out one cyclet, and make it into a cone, around the point of rotation.
Mirror While on the sphere, you only have a polygon, in the other geometries, you can have alleyways and yickles bounded by mirrors. The mirrors are designated by a '*', followed by a cycle of the angles between the mirrors. A pair of parallel mirrors is written as ** (ie two separate chains of mirrors, wih no angles in either cycle.
Wonders (o) It helps to recall this sounds like wander (walkabout). Basically, a non-reflecting process not produced by a cone. A simple example is a glide or translation.
Miricales (×) A mirrorless reflection. A rotary glide is an example of this.
"Quotes" are used to separate the orbifold from the surrounding text, if needed.

A group like 3 * 2 "Pyritohedral", is parsed as an order-3 cone, followed by a cycle of mirrors, at right-angles. The symmetry is a third of the face of an octahedron: the mirrors are the edges of the octahedron, and the 3-cone is at the centre of the face. You then extend a 120-degree searchlight out, and you see not all three sides of the octahedron-face, but bits of two sides, and a right-angle. That means, there is one chain, with a right-angle.

The general form of a CD symbol, like {p,q}, is " * 2 P Q ". The rotation-group sPsQs is "2 P Q"

Conway Archifold Notation

Archifold is a name i suggested to JHC when he brought up the idea. It does not really follow from the notions of the thing above, but is seen in its company.

Basically, what it does is track edges departing a vertex, and gives an indication on how it arrives at the next. There are three kinds of edge.

<> Rhombic edge. The edge lies completely in the mirror, as a rhombus-diagonal lies in the rhombus-mirrors.
() Rotation-edge. The ends are of the same parity (as by rotation).
[] Reflection-edge The ends are of different parity (like a mirror reflection).

The vertices are all identical, up to reflection. One supposes that mirrors and rotations might be active, and so one counts off just one sector. The sector might run clockwise or anticlockwise. If there is a parity change, you need to connect these with a reflection-edge. For the same kind, use a rotation-edge. The arms are then numbered from 1 to n in the sector.

A rotation-edge with a single number (like (2) ), is a digonal rotation. A rotation-edge with two next numbers (2,3), form a polygon-angle, the centre of which is a cone. A rotation-edge with non-next numbers, (like 2,4), make for a wander.

A reflection-edge with a single number, like [2], makes for a mirror. With 2 numbers, next [2,3] or otherwise [2,4], is a kind of miracle.

A rhombus-edge with one number, like <1>, is an edge which has full digonal symmetry, of order 4. A rhombus-edge with two numbers, next or otherwise, represents a set of edges, which form a polygon which closes the symmetry region, eg <1,3>

Some examples.

(1,2) (3,4) (5) This is a snub P,Q The (1,2) forms a P, the (3,4) forms a Q, and (5) forms a digon.

[1] [2] ... A wythoff polygon with n marked nodes appears thus.

<1> A regular figure. This is repeated around the vertex.

(1,3) (2,4) A group in the square lattice, where one has vertical blue lines, and horizontal red lines. The blue line leaves at 1 (top), and arrives at 3 (bottom). The horizontal leaves to the right (2), and arrives at the left (4).

This notation is not intended to be a full-blown description of the polyhedron, but when used with the orbifold, one can build a polytope out of it. Note that it is completely capable of showing symmetric grain (eg if the faces have colour or pattern, on the symmetry, then this reduces the allowed symmetry.)

Decorated Orbifolds

One of my pet projects is to find a way to decorate the orbifold, in the same way that one decorates the Coxeter-Dynkin graph to get the Wythoff figure. That is, can one meaningfully set a mob of extra symbols, in a way to make a polyhedron? And are there applicable 'laws of symmetry', which by one can find a most symmetric form.

In order to see the subtitlities, one must understand some notions.

Wrap: A digon is a polygon. A polygon or series can be freely extended. So, if {3,7} exists, then {3,8}. Likewise, 2 3 7, suggests the existance of 2 3 7 4.
Swallowed Edge: If a polygon is entirely inside a cell, it can be 'swallowed' by an adjacent cell, such that the edge of the cell becomes replaced by n-1 edges of the swallowed polygon. For example, it is possible to draw on a dodecahedron, one of the inscribed pentagons, such that full rotational symmetry is kept on the faces, edges, and vertices. The edges of this pentagon are swallowed edges.
Any polygon can be opened up into a miricale, and any two vertex-figures can be joined over a miricale. For example, the (1,3) (2,4) figure, we might suppose is (1,3)(2) ø4 and (1,3)(4) ø2 , this gives a pair of polygonal prisms, joined together by renting the digonal into a polygon, and then joining two of these together.

What happens with the orbifold, is that there is an 'active region' or more. This means that the symmetry cell can be crossed by an asortment of edges and cells. which have no symmetry of its own, or might be annexed to an wall-cell (which has a symmetry). For example, in the 2 3 5 group, (snub dodecahedron), there is a cell (a triangle), which has no apparent symmetry of its own, because each wall is adjacent to a different polygon (digon, triangle, pentagon). It lies entirely in the active region. Where the edge against the triange set to 'h' = sqrt(3), the snub-faces fall in the same plane as a triangle, and being 1:1;r3, fill around the r3 triangle, to make a hexagon, and the figure to an x3x5o.

The current misgivings about the conway orbifold notation, is that while it correctly describes the symmetry as a group, it might need heavy modifications to make it work as a decoration-frame for polyhedra. Unless you do things like insist on colours and decorations on faces and even edges, you need some fairly heavy hyperbolic figures, well beyond the scope of the poincare projection's limits, to resolve these critters.

by **mr_e_man** » Tue Dec 28, 2021 2:29 am

wendy wrote:A product is a process, where f(a×b) = f(a)*f(b).

There are four shape-making products, which make the four families of infinite regular polytopes.

1. pyramid product, which makes simplexes
2. tegum-product, which makes the cross-polytopes
3. prism product, which makes the measure polytopes
4. comb products, which makes the cubic honeycombs.

One element of the product is f(), applied over the Euler sum (eg for a cube, 6h + 12e + 8v). The product of the elements has a f(ab..) = f(a) f(b)...

There are two different elements that might be added, which leaves four different parts.

n. Adding an element at N=-1 (nulloid, essence), makes the product into a draught, rather than a repetition.
c. Adding the element at N=n, (content, body), makes the product into one of content, rather than content.

If E represents the euler sum (eg 6h+12e+8v), then

pyramid = c + E + n = draught or drawing of content -> simplex polytopes = point ^(N+1)
tegum = E + n = draught or drawing of surface -> cross polytopes = diagonal ^ N
prism = c + E = repetition of content -> measure content = edge ^ N
comb = E = repetiton of surface -> cubic product = girthing polygon ^ (N-1).

The products of surface only make sense when there is a surface

For the product A×B, couldn't the nulloid of A, the body of A, the nulloid of B, and the body of B, each independently be omitted or included? It seems that there are in fact 16 different products!

One nice thing about the prism product is that it's always uniquely defined, for any arbitrary asymmetric polytopes, at least in Euclidean space. No centre is needed, in contrast with the tegum or pyramid product.

At least the products are defined for abstract polytopes A and B. Let A'=A, or A'=A\{nulloid(A)}, or A\{body(A)}, or A\{nulloid(A),body(A)}, and similarly for B'. (The backslash means "except", or "difference of sets".) Take the Cartesian product C'=A'×B', and define a partial order on C' in terms of the partial orders on A and B, thus: (a₁,b₁)≤(a₂,b₂) if and only if a₁≤a₂ and b₁≤b₂. Now C' has a nulloid (a least element) if A' and B' have nulloids; otherwise it must be added explicitly. (There's a special case when A is a 0-polytope: then A\{nulloid(A)} still has a least element, equal to its greatest element.) So, let C=C'∪{nulloid(C),body(C)}, as the final result of the product. This is a partially ordered set, which may or may not be another abstract polytope.

For the prism, tegum, and pyramid products, the result C is always a polytope. For the comb product, C is a polytope if A and B are greater than 1-polytopes.

For the other 12 products, C is not a polytope (except in trivial cases where C=A or B or a 0-polytope). For example, suppose A' has a nulloid and B' doesn't. Then C' doesn't have a nulloid, so it must be added to get C. Between this new nulloid and something 2 ranks higher (rank 0 in A, rank 0 in B), there is only 1 element (rank -1 in A, rank 0 in B), which violates the dyadic rule. So you can forget about these other products....

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