Hyperbolic Tilings &c

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

Hyperbolic Tilings &c

Postby wendy » Wed Apr 01, 2015 11:39 am

I've decided that trying to decorate the orbifold is not the way to go, because this notation is not really the correct answer. Instead, I've decided to attack the problems with the archifolds, or Conway's 'arm' notation. When you add a few features into it, and change a few things, it really is quite good.

The first really big change is that we expect the numbers to be in order, and where this is not possible, we use separate 'arm' notations. So Conway's edge of the type (5, 8), would be written as (A) at position 5, and #A at position 8. An edge (5,6) is written (i,i) or ultimately, (u), and an edge (5) as (i). Because the wrap operator can replace an edge like (6) with (6,7), the edge (5,8) would become (5,9).

Mirror edges

A mirror edge is an edge whose vertices are images of each other: the edge is perpendicular to the mirror. The representation of this is [x]. As with the dynkin symbol, the symmetry lies between the edges. So x3x5x = [x]3[x]5[x]2: First to note it's a full cycle, which is what the colon represents. Second, one might note that we preserve Conway's [] notation, which means that the vertices are mirror images of each other on the line.

The [o] node makes a zero-length edge. The vertex then falls on the mirror that that edge is perpendicular to. So [x]3[o]5[x]2: is a polyhedron with triangles [x]3[o], pentagons [o]5[x], and squares (rectangles) [x]2[x]. The vertex can be moved into a corner, so that the two [o] are adjacent. Because this is a loop, the first and last [] are adjacent.

You can have any number of [] edges, giving in general, a polygon with angles 1/2x. So [x]2[x]3[x]2[x]5: works. You can only make a pair of adjacent x's or a single x into a o, so [x]2[x]3[x]2[o]5: is valid, as is [x]2[x]3[o]2[o]5:, but not [x]2[o]3[x]2[o]5: .

Rotary edges

The edge-type () connects verticies of the same parity, as might be got by rotating an edge. In the simple form, the edge types are (i) = conway's (5), and (ii) or (u) = conway's (5,6). The first designates a digon rotation, the second designates a polygon in rotation, the edges 5 and 6 surround the tip of the polygon. In both cases, the symmetry is on the edge, rather than between the edge, so we write the symmetry inside the brackets, so (u3) is a triangle, and (u5) is a pentagon.

Unlike the mirror edges, the rotary edges lie between a polygon surrounding a 'cone' (centre of the polygon u), and an 'active region'. The active region here is a polygon with an edge facing each polygon.

(i)(i)(i) produces a tetrahedron, the faces of which can be scalene.
(i)(i)(u5) produces a pentagonal antiprism.
(i)(u3)(u5) produces a snub dodecahedron.

Note here that even though the active region can be scalene, the faces enclosed in brackets are always regular, produced by rotation of the edge in the cone.

As with the ME figures, we can set individual edges to varying lengths, the two that generate further uniforms are (o) and (%). Neither of these are in Conway's notation: instead, one has to figure these out by joining an orbifold to it.

Setting an edge to % makes the edge into the chord of a large number. An active region in (i)(u3)(u5) is a triangle that joins the digon, triangle and pentagon. Turning one of the edges into a % turns that edge into an internal edge, so (i)(u3)(%5) will join the triangles to the pentagon, so the erstwhile boundary between these become chords of a decagon. The (i) and (u3) still remain.

Clearly you can't have two % in the same active region.

The other activity is that you can shrink the polygon to zero size, ie (o) or (o5). So (i)(u3)(o5) makes the pentagon zero-edge, and the triangle turns into a digon between the digon and triangle. It has two degrees of freedom, but (i) >= (u3) >= (i), is solved by making them equal. One has then an icosahedron.

Note that (%)(u3)(u5) still preserves the triangle and pentagon, but because the digon has absorbed two triangles, it becomes a square. In (o)(u3)(u5), the snub triangle has turned into a 0:1:1 triangle, and that the triangle and pentagon still remain, but are equal sized.

In (i)(i)(i)(i) tells us that any quadralateral you can draw, will tile space, the quadralateral is the active region, and the edges surround the cones. We can set one of these to zero, which makes (i)(i)(i)(o), which makes a quadralateral of x,y,z,0, edges (ie triangle), and this is repeated twice around the vertex. On the other hand (i)(i)(i)(%) makes a hexagon, because removing one edge of the quadralateral joins it to the next, and the three remaining edges form a parallel-edge hexagon. You can remove both (in any position), to produce a parallelopied (i)(o)(i)(%) or (i)(i)(o)(%), each of which gives a triangle with digonal rotation at the apex, and a joining chord (%) which turns the triangle into a parallelopied.

So, you can have as many () edges as you like, but only one (o) and only one (%).

A chain of Mirrors

Things get more interesting with the chain of mirrors, such as the orbifold 3 * 2 or 2 * 3 3 represent. When there's only one cone, the mirrors form a closed box, being the chain repeated by the number in front, so 3 * 2 = * 2 2 2 [this is the pyritohedral group containing the rectanuloid group {2,2}], and the second repeats 3 3 twice to give * 3 3 3 3.

In both cases, the cone splits out the same (end) mirror twice, and it is possible to have two mirror-edges to the same mirror. We then imagine then that in 3*2, that (i3) [x]2[x] is possible, and that for the second 2*33 gives (i)[x]3[x]3[x] is possible.

Imagine you have something like 3*2 or 3*5, the symmetry here is a subgroup of *2 2 2 or * 5 5 5 (faces of {3,4} or {3,10}. The nature of (u3) is to place a triangle in the centre, but this triangle can freely spin. One can then construct two mirror-edges to each mirrors, because the vertex is copied three times inside the cell, and the edges of the mirror-edge polygons touch only the vertices of the triangle. The active region is now the edge of the triangle (u3), and two perpendiculars to the edges, which are the outer mirror-edges.

If we set (i) to (o), the vertex falls in the centre of the cell, and none of the [x]'s can be zero. More over, the first and last mirror merge, and one of these must become [%]. But this is no harm, because the produced polygon has become zero-height.

We can no longer set (i) to (%), because the active region is not a complete polygon, having a mirror through it. Joining a polygon like this would make an endless polygon, not a larger polygon. We can only join the active region over a mirror-edge.

(3)[]2[] is the pyritohedral group. edges are (a)[ b ][c]

(o3)[x]2[%] C The triangle is a point in the centre, the calculated edge is the same as the free edge, gives an equalateral rectangle or square.
(u3)[o]2[o] O The rectangle is reduced to a point, the vertex is in the corner of the reflective region, and the triangle dominates. The active region is an edge.
(u3)[x]2[o] I The rectangle is a line-edge, ie b*0, the triangle is edge a, and the active region is a trapezium 0 a b a = triangle.
(u3)[x]2[x] rCO The rectangle is b*c, there are adjacent, trapezia b a c a, and a triangle c. The uniform case converts quadralaterals to squares.
(u3)[x]2[%] tC The rectangle B% swallows the adjacent trapezia %aba to give an octagon BabaBaba. The edge is a digon with B on one side and b on the other.
(u3)[%]2[o]. CO. It's the same as I above, except that the trapezia is 0 a % a, are joined on % to make a a a a (a square).

In the case of the 3 9 9 9 by Marek, we find it's

(2)[]3[]3[] = 2 * 3 3 is the ruling symmetry. Here the digon (2) strikes the quadralateral []3[]3[]3[]3: obliquely, so it offers up to three mirror-edges per vertex. Because the digon does not cut the middle mirror 3[]3, only one vertex can drop an edge to it.

We can still set (o), but this leads to (o)[x]3[x]3[%]. This gives as one expects {6,4}, but the edges of the hexagons can freely alternate without destroying the symmetry.

When the vertex falls on the edge, the active region is now ]()[ which is one sloping side between two edges perpendicular to a mirror: ie a trapezium. We can not set () to zero, because the active region would exit to two symmetries. So we can only set either of the ends to %.

So, we can only set one of ()[x]3[]3[x] to %, and one or two adjacents of ()[x]3[x]3[x] to zero. Note here that the ends are no longer adjacent, since they don't intersect in the symmetry. Note that appending the active region to a polygon will replace every % edge by the active region - %.

So (i)[x]3[o]3[%] gives an active region as i x i %, which is merged onto o3% (ie triangles), to give a nine-gon. The triangles arise from [x]3[o]. Were the middle edge an [x], we would get (i)[x]3[x]3[%]. The active region is aba%, which is merged onto alternating faces of a hexagon c%c%c%. This leads to a twelve-gon abacabacabac and hexagons bcbcbc.

Something like (i)[o]3[o]3[%] the active region is bada = o a % a. This is a triangle joined onto the faces of o3% to give a hexagon. The triangle o3o disappears to give a reflection in both edges at the vertex, so we get {6,6}.

Multiple Cones

A symmetry group like ()()[] has chains of unconnected mirrors. Such a group corresponds to "2 2 *", where you have parallel mirrors, and digonal rotations between the mirrors. The mirrors form long corridors. When there are several mirrors ()()[]2[], it produces a kind of hyperbolic chain, such as might be found in {6,4}, by removing sets of 'vertical' edges.

In either case, much of the above applies. There are still two verticals to the mirror-chain, but the trapezium is replaced by a hexagon or octagon since [] are once in the polygon, and () are twice in the mirror. These are indeed 'marek snubs'.

The zero-edge can of course fall in one object () or []2[], and in the latter, at one adjacent corner. The % can fall as before at the end of an active region, ie where [%] at the end of a chain.

Of wanders and Miracles

Tomorrow, perhaps.
The dream you dream alone is only a dream
the dream we dream together is reality.

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Re: Hyperbolic Tilings &c

Postby wendy » Thu Apr 02, 2015 9:59 am

Of Wanders and Miracles.

This is simply Conway-Thurston jargon for rotationless glides and mirrorless reflections. Things like rotary inversion and glide reflections all belong here. But Conway's solution is horridly simple for this. It is also a mine trap that makes the western front look like hyde park. Still, we can tackle it.

An apeirogon is a polygon.

Something that follows a straight line (such as a euclidean straight line, or a hyperbolic equidistant), is an apeirogon for this purpose.

The seven polygon-prism groups repeat into seven frieze groups. It's a one for one substitution of P for U in *PP, PP, *22P, 22P, P*, 2*P and P×. by seven frieze groups. Frieze groups are not among the wallpaper groups, since the symmetry is purely equatorial. The difference between P and U is that P generates rotations, and U generates a glide. The same seven appear in hyperbolic symmetry as well, by replacing U with W.

You can make 2d groups by alternating freeze-groups in euclidean geometry U, but not with P or with W. What happens is that in hyperbolic geometry, frieze-groups are not restricted to stripes between a pair of U's, but something that might be bounded inside W, three at a corner. In other words, what we're saying is that a 'layer' could be something as complex as the hexagons and squares in x3x6x, where the dodecagons are 'opened out'. We call branching layers bounded by straight lines 'yickles', so people don't get confused by the added meanings of layer.

Laminate polytopes

The polytope x2xUo, might be represented as a stripe of squares, with a half-plane above and below it. One can equally have an antiprism made from s2sUs, which consists of a single band of triangles between two apeirogons.

The edges of the apeirogon are 'etches' of the 'etching' (since the surface is assumed flat), and the lines inside the band of squares or triangles are 'struts'. It is important that the strut runs from side to side, there can be no internal vertices in a layer. If we want to make the tiling uniform, we can put any second layer that matches the first, so we can have prism/prism (ie 4,4) or antiprism/antiprism (ie 3,6) or prism/antiprism (ie LPC1). What does not work is parallel stripes formed by rows of hexagons and triangles extracted from o3x6o. There are vertices between the etchings, which greatly restricts what can be layered with it.

So we can write for example, a prism as [o]P[x]2[x]2: and an antiprism as (sP)(i)(i) and somehow 'add' them to get a new figure, by slipping the entire figure 1 into (sP) of figure 2. Conway's solution, which we follow here in a different form, is to have an edge with its two ends at different parts of the same vertex figure. We shall use the notation of a capital letter L with the full symmetry at the start, and a #L at the second end.

That is, [o]P[x]2[x]2: + (sP)(i)(i) = [P]2[x] 2#P (i)(i). It kind of looks messy, but this is the start of it. It reduces to [x]2[x]2[x](i)(i).

Some examples of laminate edges

A simple example of a laminate polytope might be found where consecutive polygons in a hatch mirror group are identical, for example

[x]2[x]2[x]6[x]6: consists of alternating bands of squares and hexagons, and one might be tempted to do oWx3x + oWx2x, which achieves exactly this result. On the other hand, one can join two xWx3x together, these having a structure of alternating squares and hexagons. We see here our first worry: the addition of xWx3x, can on one hand lead to alternating bands of squares and hexagons, and on the other hand, it could lead to x4x6x, if the join is set as a glide inversion (ie reflect and move one edge).

A second example is to take some snub like (u3) (u3) (u4) (u4) (u4). We can draw a chord across this, and turn it into (uW) (u3)(u3) + (uW) (u4)(u4)(u4). The original vertex figure is 3 5 3 5 4 5 4 5 4 5, where there are 3,3,4,4,4 around the pentagon, and pentagons around the triangle and square, into something like

W s 3 s 3 s + W q 4 q 4 q 4 q where s and q are snub faces, giving s 3 s 3 s q 4 q 4 q 4 q, which means that the s join two triangles and a q, and the q join three squares and an s.

Worse still, the triangle and square can individually swallow the adjacent s and q, since the triangles are entirely adjacent to s, and squares to q, giving

6* 3 6 8 4 12* 4 12 4 12 = (A) (%3) (u3) #A (u4) (%4) (u4) (u4). The asterix note the swallowing polygon, is there as an assist.

We can still set any of the u edges to o, which would result in a simpler section, but repeated three or four times.

Next

The laws of moot and of symmetry
The dream you dream alone is only a dream
the dream we dream together is reality.

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