Now, consider any pair of the images - for example, the one that is seen through the yellow window (call it Y) and the one that is seen through the red window (call it R). Suppose a person standing in Y looks out through the window that faces towards R - i.e. through the pentagon which adjoins his red and blue windows but which isn't yellow. Furthermore, suppose that, looking out of this window, he sees an image of himself and his cell, rotated by 36°, in just the same way that I see images of myself outside each of my windows.
Remarkably, the image of his own cell which he sees will then be just as if he could see just the same image R which I see through my red window. In other words, R is related to Y by just the same sort of translation-and-twist by which R is related to my own cell.
It is a bit like this. My friend is sitting at his table looking at a plate of food in front of him. I catch sight of my friend, and his plate of food, in a mirror. What is the image of my friend looking at? He is of course looking directly at the image of his plate of food. If I weren't so used to this property of mirrors, I might be rather surprised!
However, the keen-witted will have noticed that the three images Y, R and B can't be in quite the correct geometrical relationship with one another. The whole configuration would only work exactly if the dihedral angle between adjacent faces of a regular dodecahedron were exactly 120°. But it isn't! It's more like 116.565° (the angle whose tangent is -2). So there's a bit of rattling around, or 'angle deficit' (meaning the angles aren't quite big enough for everything to fit together snuggly).
We won't let that get us down, however. For the moment let's pretend that we are topologists - who allow themselves to stretch their objects and pull them around as though they were made of rubber. So an angle deficit of 3.435° is easily dealt with by moulding the rubbery dodecahedron.
Now Poincaré Dodecahedral Space (PDS) was originally conceived by Poincaré as a purely topological object. In fact when first conceived by Poincaré it had nothing to do with the dodecahedron at all! That connection came quite a few years later. But we will stick with the dodecahedral way of getting at it.
Think of PDS as a three-dimensional analogue of the construction of a toroidal surface in two dimensions. Start with a rectangular piece of paper. It is decreed that a closed surface is to be constructed by identifying the top edge of the paper with its bottom edge, and its left-hand edge with its right-hand edge. In other words, the paper is to be conceived as a sort of map of the surface, which is quite faithful in most places, but falls down at the edges. For the paper comes to an end at any of its edges, but the actual surface has no edges. The top and bottom edges of the map correspond to a single line drawn on the surface. If you are wandering around on the surface and happen to cross this line, then your path on the map will go off the edge (the top edge, say) and reappear at the corresponding point on the bottom edge.
If you cross back over the line, your representative point will go back off the bottom edge and reappear at the top edge.
In this way the surface is perfectly well defined, in an 'intrinsic' sort of way. But most of us are never quite satisfied with the 'intrinsic' point-of-view, and want to know what the surface is really like. Fortunately, in this case, this desire can be fulfilled, by going into three dimensions. We can 'indentify' the top and bottom edges of the paper simply by bending it into a tube. Then the left and right-hand edges can be 'identified' by bending the tube round into a torus. A paper tube wouldn't bend like that, of course. But we're doing topology, remember? So this is rubbery paper.
So the surface we were told to construct turns out to be topologically identical ('homeomorphic' in topology-jargon) with the surface of a torus - which is topologically distinct from the surface of a sphere, by the way, because one can draw on it closed paths which cannot be continuously shrunk to a point (they are condemned to go round the torus, one way or another).
Next, suppose the instructions were like this. Take a hexagonal piece of paper; identify its opposite edges. What do we 'really get' this time? This is a bit more tricky. Roll up the hexagon into a tube, bend the tube round - and the ends of the tube don't fit together properly. What we have to do is to stretch the tube out lengthways, twist one end of it through 180°, and then join the ends. We're back at the torus - a bit twisted, but that makes no difference from the toplogical point-of-view (though it might from others).
Ready to move on to three dimensions? You could start by taking a cuboid, and identifying its opposite faces. This defines a new space, but can we visualize what it's 'really like'? Working in the usual three dimensions, we can only get part of the way. Stretch out the cuboid in one direction, bend it round and join one pair of opposite faces. What we've got is a torus with a rectangular cross-section. Stretch the solid torus out into a long cylinder, bend it round and join a second pair of opposite faces. Now we've got a hollow torus, made out of a certain thickness of material. This solid has an outer surface and an inner one. Unfortunately, these two surfaces (inner and outer) are the descendants of the third pair of faces that are to be joined - and there is clearly no way of joining them without going outside three-dimensional space. All we can say is that the constructed space is a 'three-dimensional torus' - like a two-dimensional one, only more so.
OK, we're ready for the PDS. Take a rubbery dodecahedron, stretch it out in one direction, bend round and join opposite faces, giving them a twist of 36° (in the anti-clockwise sense) just before joining them.
Then do the same with the remaining 5 pairs of opposite faces.
Well, that sounds easy enough! Not so easy to visualize, though...
Our original, solid dodecahedron is the 'map' of the PDS. At each pentagonal face, one goes off the boundary of the map - but the actual space (of which it is a map) has no boundaries. Instead it has six pentagonal surfaces drawn in it (each one corresponding to two opposite faces of the dodecahedron). As one wanders back and to across one of these surfaces, ones representative point jumps from one face of the dodecahedron to the opposite one.
To be sure that the constructed space is well-defined, we need to make sure of what happens when you wander about near one of the lines in the PDS which correspond to the edges of the dodecahedral map - for example E in the figure below (click to enlarge).
Now E is common to two of the pentagonal faces, each of which is supposed to be joined up with its opposite number (with a twist) in the construction of the PDS. This means that E will be identified not just with the edge e1 (when the top face is identified with the bottom face) but also with the edge e2 (when the front face is identified with the back face).
But e1 is also the boundary of another face, the one at bottom-front-left, which needs to be identified with the face at top-back-right. But that's no problem: in the identification-with-twist, e1 will be identified directly with e2. So far so good: which ever way you look at it, E, e1 and e2 will all correspond to the same line in the PDS.
In fact, this line will be the common boundary of three pentagonal surfaces (corresponding to the pairs top-bottom, front-back and bottom-front-left, top-back-right). In the figure, we also see how the wedge-shaped bits of dodecahedron just near to the three edges fit together when the edges are identified to make a nice bit of three-dimensional space. Note that the wedges each have to be given a relative twist (anti-clockwise) as they are fitted together.
Having satisfied ourselves about the edges, it only remains to check what happens at the vertices of the dodecahedron. A vertex - such as V, below - belongs to three faces, and so is party to three identifications, leading to V being identified with three partners - marked v1, v2, v3.
Note that given one of the four partners, the other three three may be determined as follows. The given vertex belongs to three pentagonal faces. Choose one of them. Go across it following the left-hand diagonal (as explained in the figure) and from there go down the edge that is not in the plane of the pentagon. You arrive at one of the partner vertices. Get the other two by starting across different pentagons.
The partners are mutually equidistant - in other words, they are the vertices of a regular tetrahedron. In fact, all 20 vertices of the dodecahedron are grouped into 5 sets of four, yielding 5 tetrahedra, all symmetrically placed with respect to one another, the so-called 'compound of five tetrahedra'. (Note that there is a second compound, the mirror-image of the first, that would be obtained by traversing the faces on the right-hand diagonal instead of the left.)
Our picture also shows how the solid angles at the vertices V, v1, v2 and v3 come together at one point in the PDS. Once again, the pyramidal pieces need to be twisted as they are brought together.
One of the compounds of five tetrahedra
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