by wendy » Sun Nov 29, 2009 11:25 am
If one starts off with the notion that a hole is a missing patch (in the fabric of space), it could be quite interesting to count them by working out what patches are missing. It makes some sort of sense by multitopes (vertices, edges etc, without any sort of closure).
We begin with the notion that an object without holes is piecewise construction of spherous patches. Spherous means 'topologically convex', and peicewise means that you add peices of consecutive dimensions, eg vertex+edge, edge+hedron, hedron+choron. The root form is the nulloid+vertex.
One constructs a peicewise multitope, that contains the desired figure, and removes the desired bits and peices either peicewise (which do not create a hole), or singly, which does.
t = teron =4d, c=choron = 3d. h=hedron = 2d, e = latron (edge) = 1d, v = vertex = teelon = 0d, n = wesian = nulloid = -1d.
The products in *# etc include (*) or exclude (#) the leading/trailing element. *# = prism product.
So, for example, we could demonstrate the holes in 5,5/2 (great dodecahedron), by starting with the icosahedron (which is 1c, 20h, 30e, 12v, 1n. We now add in pairs, c+h, formed by dividing a new cell from adding a pentagonal wall. (intersections are not counted here). This gives 13c.32h.30e.12v.1n. We now remove the 12 triangular faces, along with 12c, to give 1c.20h.30e.12v.n. To get the desired result, we need to remove the remaining 8 triangles, with just the 12 pentagons, but no edges etc are removed, so the result 1c.12h.30e.12.n is 8h shy of the peicewise construct, and therefore the genus is 8/2 or 4.
A circle has a hole, because the peicewise construct is n+v (vertex) e+h (fence + enclosure), while the circle is h.e.n. This means that it's missing a teelon or 1d patch. [any point pair can vanish in two different ways].
A sphere is derived from a circle-digon, by taking a circular ring, and adding a hemisphere on each side. Since the first hemisphere is the enclosure above, the second h adds h+c (second face + content). We can now remove h+e, by removing the line between the faces, and the face itself, ie a sphere's hole-less construct is 1c.1h.1v.1n, but the 1v is additional, a sphere also has a teelal (0d) hole in it. [as do all higher spheres].
If we now construct a monogon prism (basically a circle + point), we get:
1h.1e.1v.1n *# 1h.1e.1v.1n = 1t.2c.3h.2e.1v.n.
This basically is a square, wrapped into a cylinder (preserving the seam), and then hooked end to end to make a bi-cylinder, preserving that join. So we get two seams (where there are the two ends of the cylinder), one vertex (where the seeams meet), three hedra (the two circles, amd the square.
The observed result is 1h.1e.1n *# 1h.1e.1n = 1t.2c.1h.0v.n, or a difference of 2h.2e.1v. These are indeed the observed additions needed to be added to the bicylinder to prevent non-vanishing loops forming anywhere in the figure. The h fall in each of the faces (chora) to prevent loops in 3-space, while the 2e prevent loops forming in the margin (it's a torus), the v is where the 2e fall.
We now turn to the torus, constructed by a bimonogon-comb, ie 1h.1e.1v.1n ## 1h.1e.1v.1n = 1c.1h.2e.1v.1n. We see that the comb-product is a repetition of surface, and does not carry across any holes in the solid. However, we can use the 4d case, and flatten it out, by way of deleting 1t.1c (by stretching the surface out into a plane, and removing the content, and one choron (the "outside"). So we get 1c.3h.2e.1v.n. This is indeed the correct result as before.
The circle-sphere and sphere-circle combs (23 and 32 holes), can be formed from the simple 5d prism, by removing p+t on the stretch. So we get:
circle *# sphere = 1.1.1.1 * 1.1.0.1.1 = 1.2.2.2.1.1. Eliminating the 5d element, we get 1.2.2.1.1.,
The dream you dream alone is only a dream
the dream we dream together is reality.