Clifford parallels in position-space

Distance

There is a very natural notion of distance in position-space. Given any two points in the space, A and B, we already know about the line joining A to B. The distance between A and B is simply the angle of the rotation which is required to transform position A into position B. The largest possible distance between two points is 180° or π radians. Such points are polar to one another.

Clifford parallels

As we have seen, there are no coplanar parallel lines. However, given any line in the space there are many other non-coplanar (or skew) lines which are parallel to it in the sense of Clifford (see wiki on W.K.Clifford). Suppose one is given two lines, L and M. Given a point X on L, there will be a certain minimum distance between X and all the points of M. If this distance is the same, wherever X is chosen along the line L, then M is said to be parallel to L in Clifford's sense.

Using an arrow on the sphere to represent positions, a line L is represented by a circle with arrows all along it, all making a constant angle with the circle. There are two sets of Clifford parallels to L, as follows.

1) Concentric Lines
Suppose a second line, M, is based on a trajectory that is concentric with L's. Note that the arrows of M may not make the same angle with their trajectory as the arrows of L do with theirs.Choose a point X belonging to L: the symmetry of the situation makes it clear that the minimum distance from X to the points of M will be the same wherever X is chosen.


2) Rotated copies of the line L
Let A be the centre of the trajectory of L. Any point B of the sphere may be chosen as the centre of a copy of L. Let the copy be called M. Without loss of generality, we may regard M as arising out of L by means of a rotation through 180° (as the centre of rotation take C, the mid-point of AB). Now let X be a point of L, Y a point of M. One way of getting from X to Y is the following. Rotate through 180° about C, then through an angle θ about B. In a formula:

X C[180] B[θ] = Y.

But the combination of two rotations is a third, which we may find by the half-angle, spherical triangle method. Let CB be the base of the triangle. Draw a side through C, making an angle of 90° with CB. Draw a second side through B, making an angle θ/2 with BC. Where the sides meet, is the centre of the third rotation; its angle will be twice the angle between the sides.


The question is, what choice of Y makes the distance XY least; or makes the angle of rotation least; or what value of θ makes the angle between the sides least? The answer is that θ must be 180°; then the angle of rotation will be twice the spherical distance CB, or, simply, the spherical distance AB. This is the minimum possible distance from X to Y - and it is clearly independent of the choice of X. Thus the lines L and M are parallel in Clifford's sense.

The geometry of position-space (continued)

So far, we know that position-space is three-dimensional, and that between any two points a 'line' can be drawn. The lines are topologically circles, and the angles of a triangle add up to more than 180°; so, wherever we are, we aren't in plain old Euclidean space. Another question that might arise, is whether position-space is sufficiently like Euclidean space to possess planes. By 'plane' we mean a two-dimensional subset such that, if any two points in the plane are given, then the line joining the two points lies wholly in the plane. Once we know whether there are any planes, we can then go on to ask whether any three points determine a plane, and so on. First of all, we need some new definitions.

Polars and polar planes

Any two positions A and B determine a rotation, which takes A into B. If the angle of the rotation is maximal, i.e. 180° we say that A is polar to B, and vice versa (think of the expression 'poles apart'). In terms of arrows drawn on the circumsphere (surrounding our teapot or other rotating object), here are some positions that are polar to position A.

Now consider the set of positions that are polar to A. The set is two-dimensional. At each point of the sphere, just one arrow may be drawn that is polar to A. We will now show that this set of positions is in fact a plane, i.e. that it satisfies the plane-condition.

To this end, suppose that B and C are polar to A. We need to show that any point on the line BC is polar to A. B and C are both reached from A by means of rotations of 180°, suppose the centres of the rotations are p and q. (In fact, p and q must be the mid-points of the lines ab and ac, respectively - where, once again, a stands for the base-point of the arrow A, and so on. But we won't be using that fact, just yet.) Suppose that r is the centre of rotation for getting from B to C.

Thus p,q,r are the centres of rotation associated with the triangle ABC: and the angles of rotation are twice the angles of the spherical triangle pqr. But the rotations about p and q are through 180°. Therefore the angles of pqr at p and q must be 90°. The spherical triangle pqr has two of its angles right angles. Knowing where p and q are, r is determined, as follows. Draw the line pq; draw perpendiculars at p and q; the perpendiculars must meet at one or other of the poles associated with the great circle pq. (The poles are the points in which the axis of symmetry of the great circle meets the sphere.) Thus r is one or other of these poles.

Now let D be a point on the line BC: so D (like C) is reached from B by means of a rotation about r, but through a different angle. Consider the triangle ABD. p and r are the centres of rotation associated with the sides AB and BD. Suppose the centre of the third rotation, associated with DA, is s. The angle at p of the triangle prs is known to be 90°. Therefore ps must be an arc of the great circle through p and q: in other words, s must lie on the line pq. Therefore the angle of the trangle at s must also be 90°, and D must indeed be polar to A.

Thus the set of points polar to A is indeed a plane, and we may call it the polar plane of A.

Any three points lie on a plane

Let A, B and C be any three distinct points. We will show how to find a point X such that all three given points are polar to X. Then A, B and C will all lie in a plane, namely, the polar plane of X.

Suppose that α, β and γ are the centres of rotation associated with the lines BC, CA and AB, respectively. Suppose that A, B and C are in fact polar to X, and that p, q and r are the centres of rotation for moving the three points to X (by means of a half-turn).

From the last sub-section we know that p and q must lie on C_γ, the great circle whose pole is γ. Similarly, q and r must lie on C_α, and r and p must lie on C_β. Consequently, p must coincide with one or other of the intersections of C_γ and C_β (and similar statements apply to q and r).

Therefore, X, if it exists, must be the point obtained by rotating A about the centre

p=C_γ ∩ C_β

through a half-turn.

Conversely, suppose that a point X is obtained in this way. Then X is certainly polar to A, but what about B and C?

Suppose that the angle of rotation (about γ) for getting from A to B is θ. Then we have a way of getting from X to B: first rotate about p, through 180°; then rotate about γ through θ. But two rotations in succession are equivalent to a third: and we know a recipe for finding it. In this case we need to draw a spherical triangle pγq, on the line pγ as base, with angle 90° at p, and angle θ/2 at γ. But because p lies on C_γ, the side of the triangle going up from p will be an arc of the great circle C_γ; and the angle at q will be 90°, whatever θ is. Therefore the rotation for getting from X to B is a rotation of 180°, and X is polar to B as well as A.

By an exactly similar argument, X is also polar to C, so A, B and C all lie in the polar plane of X.

A property of all the lines in a given plane

We already know that any point D on the line AB can be reached from X by a half-turn, about a centre, s, that lies in C_γ (where γ is the 'centre' of AB, in an obvious sense). Since the rotation is a half-turn, this implies that s is the mid-point of the line xd (where x and d are the root-points of the vectors X and D, as usual). Or, to put it the other way round, d could be obtained by producing the line segment xs to d, such that sd=xs.

Let perpendiculars be dropped from x and d to C_γ. Then the lengths of the perpendiculars must be equal (a case of congruent triangles). But the length of the perpendicular from x is a constant; therefore, so must be the perpendicular from d - which implies that as s moves along C_γ, the distance from γ of the moving, derived point d stays constant. In other words, d moves in a circle about γ - a fact that we already knew (it follows directly from the definition of γ).

Suppose x is at the bottom of the sphere, and that the great circle C_x is the equator. Then any great circle, such as C_γ or C_α intersects the equator in two antipodal points. When X is rotated about these points it moves up to the top of the sphere, to the antipode of x, call it x'. But this implies that the trajectory of the line AB (or BC) passes through x'. The same applies to any line in the plane: its trajectory (the locus of the root-points of the arrows belonging to the line) must pass through x'. Think back to the last post. The trajectories of the sides of a triangle ABC all seemed to intersect in a point. Now we can see that this was no accident: they had no choice but to pass through the point x', the antipode of x. It's just that, at that stage, we had no idea about X and its root-point x.

Two coplanar lines always meet

Any line in the plane can be obtained by 'doubling' a great circle, in the way described in the last subsection (xs doubled to xd). Any two lines are generated by two great circles. But two great circles always meet in two antipodal points, which define an axis of rotation. When X is rotated through 180° about this axis, its image is a point which belongs to both lines: this is the point in which the lines meet.

There are no parallel lines in the planes of position-space: Euclid's fifth postulate fails to be satisfied. Once again, whatever geometry we're in, it isn't the same as Euclid's.

The geometry of position-space

Triangles

Here is a picture of a triangle in position-space. The vertices are the arrows (not just their base points) labelled A, B, C. It is the (relative) orientation of the arrows A, B that determines the 'trajectory' of the line AB - by which we mean the path traced by the reference-point (or arrow-base) across the sphere. This path is always an arc of a circle (not necessarily a great circle, of course).


It appears likely from the picture that the trajectories meet in a point (call it X - near the bottom right of the figure). Can you think of any reason why this should be the case? (Note that the lines in position-space AB, BC, and CA do not meet when produced, even though the trajectories do: the arrow-directions will not agree at X.)

Angles

Consider the angle between the lines AB and AC. To move away from A in the direction of B, the teapot/circumsphere has to be rotated, clockwise, about a certain point of the sphere (call it p). To move away from A in the direction of C necessitates rotation (clockwise) about a different point, q. We define the angle between AB and AC to be the spherical distance between p and q.

Let's have a look at the sum of the angles in the triangle ABC.

To move from B towards A we need to rotate clockwise about p', the antipode of p. To move from B towards C we rotate about r. To move from C towards A we rotate about q' (antipode of q); to move from C towards B we rotate about r'. So the sum of the angles is the sum of the spherical distances

pq + p'r + q'r' = pq + qr + rp'.

This is the distance from p to its antipode p' going via q and r, which will be greater than the shortest distance from p to p', which is π (in radian measure) or 180°, unless q and r happen to lie on a spherical line (or great circle) going from p to p'.

Thus the sum of the angles of the triangle will almost always be greater than 180°. Is it at all likely that p, q and r lie in a straight line? We will get to the bottom of that in the next section.

Combining rotations

So far, we have that rotation about p (through a certain angle) gets us from position A to position B, rotation about r gets us from B to C, and rotation about q gets us from A to C. So we have two ways of getting from A to C: directly, in one rotation about q; or indirectly, via B, rotating first about p, then about r. Remarkably, knowing the locations of the centres p, q and r is enough to tell us the angles of the rotations, as follows.

Rotation about the point p, through an angle θ, say, is equivalent to two reflections in lines through p, separated by an angle θ/2. Similarly, rotation about r through angle ψ is equivalent to reflection in two lines through r, separated by angle ψ/2.

There is nothing to stop us choosing as one of the lines, in both cases, the line pr.


We see that reflection in l1, followed by reflection in pr (taking a to a' and then a'') amounts to rotation about p through angle θ; and that reflection in pr then l2 amounts to rotation about r through angle ψ.

Therefore rotation about p followed by rotation about r is equivalent to reflection in l1 followed by reflection in l2 (since the reflections in pr cancel out).

But the lines l1 and l2 have to meet somewhere (these are spherical lines, remember?), in a point s, say, so the two reflections are equivalent to a rotation about s.

But the rotations about p and then r get us from position A to B and then C, which can be done in one step by rotation about q. So a rotation about s is equivalent to a rotation about q: therefore, either s and q must be the same point, or else antipodal points. In the latter case, we can replace s by its antipode, by producing the lines l1 and l2 in the opposite direction.

So the angles of rotation can be found by drawing lines between the centres p, q and r to form the spherical triangle pqr. The required angles of rotation will then be just twice the angles of the triangle pqr.

As a corollary, we note that if p, q and r are in a straight line, the triangle will collapse. Two of its angles will be zero, the third will be 180°. Doubling up the angles, we find that the rotations will be null in each case: therefore A, B and C were all the same point. If ABC is any non-trivial triangle in position-space, its angles will add to more than 180°.

The space of positions of a teapot - and of a regular dodecahedron

A 'natural' definition of the Poincaré Dodecahedral Space runs as follows: Poincaré Dodecahedral Space is the space of positions of a regular dodecahedron. We're not concerned here with the translational position of the dodecahedron (as you move it from that side of the table to this), just its rotational position (as you turn it round in your hand).

It's easier to think about the space of positions of an unsymmetrical object (we're talking rotational symmetry here) such as a teapot, to start with. The space of positions of a dodecahedron is a bit smaller, because some positions of the dodecahedron, which would be different if the dodecahedron was less symmetrical (for example, if one or more of its faces were visibly scratched), count as indistinguishable if the dodecahedron is absolutely, ideally regular - if it is a truly Platonic dodecahedron, in other words. The ideal dodecahedron has fewer distinguishable positions than one that is scratched or otherwise imperfect.

Indeed, we can readily see that the space of positions of an ideal dodecahedron has some of features that agree with what we already know of the PDS. Imagine your dodecahedron sitting on your desk, and that you turn it slowly about a vertical axis. All its positions are different, until you have turned it through 72°, at which point its position becomes indistinguishable from its starting position. Your journey through the 'space of positions' has come back to its starting-point.

However, this is not the only way to return to the starting-point. Suppose you turn the dodecahedron slowly about a horizontal axis, parallel to one edge of the top pentagonal face. After you have turned it through 63.43495° (= inverse tan of 2) the centre of the top face will have taken the position that was originally occupied by the centre of a next-to-top face. If you now further rotate the dodecahedron through 36° about the axis which passes through this centre, and through the centre of the opposite face (the one that started off at the bottom), the dodecahedron will again be in a position that is indistinguishable from its starting position.

In fact, what we have achieved by means of two rotations (through 63.43495° and 36°) can actually be achieved by a single rotation of 72°. Just let the axis pass through the centre of another one of the next-to-top faces (a next-door one). Then the topmost face will move directly into the place of the next-to-top one.

There are five possible choices of axis - six, including the vertical one - which gives us six directions in the space of positions which lead back to the starting-point. Or 12 directions, counting clockwise and anticlockwise turns as distinct. The similarity - to put it no stronger - with the PDS (as defined earlier in this blog) is striking. What is not clear, as yet, is whether it makes dense to regard the 12 privileged directions as 'equably distributed' over the 'celestial sphere' of directions at a point within the 'space of positions'. This is something we need to investigate.

* * *

Returning to the space of positions of a teapot, how big is the space? How much of it is there? For a start, how many dimensions does it have?

It will often be helpful to imagine the teapot as fixed within a transparent sphere, with a reference point and arrow (based at that point) marked on the sphere. When the teapot is rotated into a new position, so is the sphere. We can tell what has happened to the sphere by noting 1) where the reference point now is and 2) in which direction the arrow is now pointing. The position of the point is fixed by two coordinates; the direction of the arrow by one more coordinate. Thus three coordinates suffice to fix the position of the sphere, and hence the position of the teapot: its space of positions is three-dimensional.

Having got a bit of a handle on the space of positions, our next move is to turn it into a sort of geometry, by saying what we mean by a line in the space of positions.

Given two positions, A and B, say, of the teapot, one can always get from one position to the other by rotating the teapot through a certain angle, about a certain axis (this is not obvious, but will be shown below). The 'line' joining A and B consists of all the intermediate positions reached by rotating about the same axis, but through different angles. It is clear that the 'line' is not the sort of line that goes on forever. Really it is more like a circle: if you go far enough along it, you come back to your starting-point. But so long as we are speaking of 'geometry', it will be useful to continue to call it a line.

As promised, we will show that one can always get from A to B by means of a rotation. Think in terms of the circumsphere with its reference point and arrow. How do we get the arrow from position A to position B? Suppose the reference point needs to move from point a to point b. The locus of points equidistant from a and b is a certain plane, passing through the mid-point of the line ab, and also through O, the centre of the sphere. (The plane is the perpendicular bisector of ab.) We can get the arrow from a to b by reflecting in this plane. It is of course a plane of symmetry of the sphere, so the whole sphere is mapped to itself.

The trouble is, that although the arrow has been moved to the right place, it will most likely be pointing in the wrong direction. However, this can be corrected by a second reflection, this time in a plane that passes through b, and through O. So, we can get the arrow into the right position by means of two reflections.

But the effect of two reflections, in two planes, is identical with the effect of a rotation, about the line which is the intersection of the two planes, through an angle which is twice the dihedral angle between the planes.

In this case, the line of intersection necessarily passes through O and is therefore a diameter of the sphere, intersecting the sphere in two antipodal points.

Each of the planes intersects the sphere in a great circle. Taking a more 'intrinsic' view of what is going on, we may regard these great circles as 'lines' in the geometry of the sphere, and the reflections as reflections in these 'lines'. The lines meet in two points, at a certain angle (which is the same at both points). The two reflections are equivalent to rotation about either of these points, through twice the angle between the lines.

It may be the case that the second line of reflection is actually the (spherical) line ab, in which case the meet of the two lines is the mid-point of ab, and the angle of intersection is 90°. The equivalent rotation is rotation about the mid-point, through 180°. This is of course the largest possible angle of rotation.

The smallest possible angle of rotation occurs when the second line of reflection (passing through b) is perpendicular to ab. Then the axis of the rotation is the pole of the great circle ab, and the angle is the length of the arc ab (times 360 divided by the circumference of the sphere, assuming you want it in degrees).

The two extreme cases, together with an intermediate case, are illustrated in the following figure. In each case the line in arrow-position-space, joining A to B is also shown.

(To get better resolution, click on the image)

From the 120-cell to E8 (Part 4)

A more explicit definition of the E8 lattice runs as follows. The point with coordinates (x1,x2,...x8) belongs to the lattice iff

either x1..x8 are all integers, and their sum is even
or x1..x8 are all half-integral (i.e. an integer plus ½) and their sum is an even integer

In this case, the lattice-points closest to the origin, such as

(½, ½, -½, ½, ½, -½, ½, ½)

or

(0, 1, 0, 0, 0, -1, 0, 0)

are all at distance √2. But once again, there are 240 of them (128 of the half-integral kind, 28×4=112 of the integral kind). And once again we can arrange them in 'alphabetical order', and find a set of simple roots.

The first positive root is

A = (0, 0, 0, 0, 0, 0, 1,-1);

the second (linearly independent of the first) is

B = (0, 0, 0, 0, 0, 0, 1, 1).

After that we have a run of five:

C = (0, 0, 0, 0, 0, 1,-1, 0);
D = (0, 0, 0, 0, 1,-1, 0, 0);
E = (0, 0, 0, 1,-1, 0, 0, 0);
F = (0, 0, 1,-1, 0, 0, 0, 0);
G = (0, 1,-1, 0, 0, 0, 0, 0);

and finally

H = (½, -½, -½, -½, -½, -½, -½, ½)

(note: the number of minus signs must be even). The Dynkin diagram is


Recalling the diagram generated by our first lattice,


we see that the simple roots correspond as follows:

a → B
b → D
c → F
d → A
e → C
f → E
g → H
h → G.

This correspondence of the simple roots determines a linear transformation from the first space to the second, as follows. The coordinate basis vectors e5 ( = (0, 0, 0, 0, 1, 0, 0, 0)), e6, e7, e8 are none other than d, c, b, a and map to A, F, D, B, respectively. Thus

e5 maps to A = (0, 0, 0, 0, 0, 0, 1,-1);
e6 maps to F = (0, 0, 1,-1, 0, 0, 0, 0);
e7 maps to D = (0, 0, 0, 0, 1,-1, 0, 0);
e8 maps to B = (0, 0, 0, 0, 0, 0, 1, 1);

e4 is 2e + a + b + d and maps to 2C + B + D + A = (0, 0, 0, 0, 1, 1, 0, 0);

e3 is 2f + e4 + b + c and maps to 2E + 3C + 2B + D + A = (0, 0, 1, 1, 0, 0, 0, 0);

e2 is 2g + e3 + e4 + d and maps to 2H + (0, 0, 1, 1, 1, 1, 0, 0) + A
= (1, -1, 0, 0, 0, 0, 0, 0);

e1 is 2h + e2 + e3 + c and maps to 2G + (1, -1, 1, 1, 0, 0, 0, 0) + F
= (1, 1, 0, 0, 0, 0, 0, 0).

So the matrix for mapping a vector in the first space to a vector in the second (both written as column vectors) takes the remarkably simple form:

[ 1  1  0  0  0  0  0  0  ]
[ 1 -1  0  0  0  0  0  0  ]
[ 0  0  1  0  0  1  0  0  ]
[ 0  0  1  0  0 -1  0  0  ]
[ 0  0  0  1  0  0  1  0  ]
[ 0  0  0  1  0  0 -1  0  ]
[ 0  0  0  0  1  0  0  1  ]
[ 0  0  0  0 -1  0  0  1  ]

The transformation could be expressed verbally as follows. First permute the coordinates of your vector, from

(x1,x2,x3,x4,x5,x6,x7,x8)

to

(x1,x2,x3,x6,x4,x7,x8,x5),

then take the coordinates in pairs and act on them with the matrix

[ 1  1 ]
[ 1 -1 ].

This last step takes the pair

(0,0) (which we have called 0) to (0,0)
(0,½) or 1 to (½,-½)
(½,0) or 2 to (½,½) and
(½,½) or 3 to (1,0).

To arrive at a vector of the desired type (i.e. a root of the E8 system in the conventional coordinates), the product of the first stage of the transformation must either be a mixture of the pairs 0 and 3, or else a mixture of 1 and 2 (giving rise to integral vectors and half-integral vectors, respectively).

Further, for the sum of the ccordinates to be even, the number of occurrences of 3 (in the integral case) or 2 (in the half-integral case) must be even, that is, 0, 2 or 4. The possibilities are

0033, 1111, 1122, 2222 (and permutations thereof).

These are 6+1+6+1=14 in number. Multiplying by 16, for all the sign combinations, that is 224 possibilities. Once more there is an opportunity for mathematical pottering. It is amusing to verify that the even permutations of 0123, together with 1111 and 2222, are transformed by the first stage (permutation of the last five coordinates) into patterns of the required form. For example

0123 = (0, 0, 0, 1, 1, 0, 1, 1) is permuted into (0, 0, 0, 0, 1, 1, 1, 1) = 0033,

and so on. Odd permutations of the pairs give rise, of course, to sequences that are illegal in one way or another.

This concludes our exploration of the way that leads from the 120-cell to E8. It has taken quite a bit of explaining, but with hindsight we can see that it is really only a little bit of a tweak that turns one into the other.

I recommend David Richter's web site for more on the 600-cell and the E8 root system (especially this page). The splendid photograph on the front page shows Richter holding a model which is a projection into 3 dimensions of the 600-cell. The sun shines through the model and casts a shadow on the ground. The beauty of the shadow is that it corresponds to something known as the Van Oss projection of the 600-cell. It bears a more than passing resemblance to a well-known diagram of the E8 root-system (see for example here).

From the 120-cell to E8 (Part 3)

In the last post, we saw how the cell-centres of the 120-cell in 4 dimensions generate a lattice in 8 dimensions. So far, just about the only thing we know about the lattice is that each lattice-point is surrounded by 240 nearest neighbours, each at distance 1.

The 240 nearest neighbours of the origin constitute what is known as a root system. (See the 'Root System' wiki to learn what is involved in this, in general.) We will not attmept to prove it. We only note that the dot-product of any two members of the system is either ½, 0, -½ or -1 (+1 only occurs when a root is dotted with itself). Given a root, one root (itself) has dot product 1 with it, 56 have dot-product ½, 126 have dot-product 0, 56 have dot-product -½, and one (its negative) has dot -1. If you enjoy a bit of mathematical pottering (like me) you may like to pick a root and verify that the system is closed under reflection in the hyperplane through the origin, orthogonal to your chosen root. Thus the roots with dot-product ½ are in 1-to-1 correspondence with the roots whose dot-product is -½.

Given a root system, we can find a set of simple roots, out of which the other roots may be formed as linear combinations with integer coefficients, as follows. The first step is to list all the roots in 'alphabetical order'. In this case, the alphabet consists of the five 'letters' -1, -½, 0, ½, 1, and we regard each root as an 8-letter 'word' (each coordinate being interpreted as a letter). So in the list, all the words beginning with the letter -1 (actually, there is only one of them) come first, then all the words beginning with -½, then 0, and so on. Then the words beginning with -½ are ordered using the second letter, and so on. The origin is not a root, and so does not appear in the list; but if it were, it would appear precisely half-way through the list. The roots that occur after this half-way point we denote positive roots.

Now, what is the first positive root in the list? Its 8-letter 'word' will begin with as many 0's as possible; then the first non-zero 'letter' or coordinate will be as small as possible, but positive.

The very first positive root is in fact

a = (0, 0, 0, 0, 0, 0, 0, 1).

The half-integral roots are ruled out, at this stage, because they cannot have as many 0's on the left. In fact, the first half-integral root to appear in the list will be

e = (0, 0, 0, ½, -½, 0, -½, -½).

Before that, we'll find the simple roots

b = (0, 0, 0, 0, 0, 0, 1, 0)
c = (0, 0, 0, 0, 0, 1, 0, 0)
d = (0, 0, 0, 0, 1, 0, 0, 0).

Note that the -½'s in e can be turned into +½'s by adding a, b or d. So the next root in the list that cannot be obtained as a linear combination of the simple roots already discovered, is

f = (0, 0, ½, -½, 0, -½, -½, 0).

After that we find

g = (0, ½, -½, -½, -½, 0, 0, 0)

and

h = (½, -½, -½, 0, 0, -½, 0, 0).

(Note that the pairs occuring in e, f, g, h have the patterns 0123, 0312, 1320, and 3210, respectively - all even permutations of 0123.)

We now have in our possession eight simple roots, out of which all positive roots may be formed by positive superpositions.

We now appeal to the fact that root systems can be classified by the geometrical configuration of their simple roots. We form the Cartan matrix which records the dot-products between the simple roots:

 | a  b  c  d  e  f  g  h
-------------------------
a| 1  0  0  0 -½  0  0  0
b| 0  1  0  0 -½ -½  0  0
c| 0  0  1  0  0  0 -½ -½
d| 0  0  0  1 -½  0 -½  0
e|-½ -½  0 -½  1  0  0  0
f| 0 -½ -½  0  0  1  0  0
g| 0  0  0 -½  0  0  1  0
h| 0  0 -½  0  0  0  0  1

An alternative way to encode these geometrical relationships is the Dynkin diagram. Here each simple root is represented by a dot; each non-zero dot-product by a line between two dots. Starting with e, the simple root that is most connected to other simple roots, it is straightforward to draw the diagram corresponding to the above matrix:



The connectedness of the diagram means that the simple roots cannot be split into 2 camps, belonging in mutually orthogonal subspaces of 8-dimensional space. The root system is said to be 'irreducible'. This particular diagram is the signature of the root system called E8; it gives rise to the largest of the exceptional Lie groups, also called E8.

This suffices to establish that the lattice we have obtained from the cell-centres of the 120-cell is indeed the E8 lattice. In the next post (Part 4) we will be a bit more explicit about this, and show how our lattice can be transformed into a more familiar representation of E8.

From the 120-cell to E8 (Part 2)

So far we have 120 points on the unit sphere in 4 dimensions, with coordinates as follows (since the last section, I've swopped two axes):

(±1, 0, 0, 0) and permutations thereof (8 points)
(0, ±1/2φ, ±1/2, ±φ/2) and even permutations thereof (96 points)
(±1/2, ±1/2, ±1/2, ±1/2) (16 points).

Each point has 12 nearest neighbours, all at distance 1/φ. By joining up all the nearest neighbours we obtain the skeleton of a regular polytope in 4-dimensions, having 600 tetrahedral-cells. The 120 points may also be interpreted as the cell-centres of the 120-celled polytope, whose cells are dodecahedra.

Now consider the lattice generated by the position vectors of the 120 points. Each lattice-point is a linear combination of the 120 vectors, with integer coefficients.

Adding a point to an almost opposite one we get the difference vector between nearby points, which, as we've seen, will be one of the original points, divided by φ (=1.61803...). So the lattice will contain a scaled-down copy of the original 120 points. But it will also contain the difference vectors between points of the scaled-down copy which will lead to a second copy, scaled down still further - and so on. A tiny sphere, centred on the origin, will contain an infinity of lattice-points; and the same goes for any other point of the lattice.

In a manner of speaking, there are too many lattice-points to fit into 4 dimensions. The result - the lattice-points pile up almost on top of one another. The remedy - to move into more than 4 dimensions!

The coordinates of the 120 points (and hence of all the lattice-points) are a mixture of rational and irrational parts. Suppose we separate them out and assign the irrational parts to 4 new dimensions? There are many ways one could do this, but in the present case a very simple method presents itself. All the coordinates arising in the lattice consist of a rational number plus a rational multiple of φ (1/φ is no problem because it can be written φ-1). In other words, they can be written as a+bφ, with a and b rational. In fact, things will work out a bit neater if we use as our 'fundamental' irrational not φ, but ψ = φ-1. Each coordinate can be written a+bψ, and to it we can assign the pair of rational numbers (a,b). Thus

φ = 1+ψ goes to (1,1)
1 goes to (1,0)
1/φ = ψ goes to (0,1) and
1/φ² = 1-ψ goes to (1,-1).

To the point (0, 1/2φ, 1/2, φ/2), for example, we assign the rational 8-tuple

(0, 0, 0, 1/2, 1/2, 0, 1/2, 1/2).

The even permutations of the coordinates then give rise to even permutations of the pairs

(0, 0), (0, 1/2), (1/2, 0), (1/2, 1/2).

If any pair is replaced by its negative, we get another point belonging to the set of 120. The point (1/2, 1/2, 1/2, 1/2) translates as

(1/2, 0, 1/2, 0, 1/2, 0, 1/2, 0).

Subtracting this from the point already considered we get

(-1/2, 0, -1/2, 1/2, 0, 0, 0, 1/2),

which does not belong to the original set because of the pair (-1/2, 1/2) in second place (this pair corresponds to the real number -1/2φ²). Similarly, we obtain all even permutations of the pairs

(0, 0), (0, 1/2), (1/2, 0), (1/2, -1/2),

and any of these pairs can be replaced by its negative. Taking the difference of the points

(0, 0, 0, 1/2, 1/2, 0, 1/2, 1/2) and (0, 0, 0, -1/2, 1/2, 0, 1/2, 1/2)

we obtain

(0,0,0,1,0,0,0,0).

Clearly that 1 could be in any of the 8 places, which implies that any 8-tuple of integers belongs to the lattice. Finally, we note that the sum of

(0, 0, 0, 1/2, 1/2, 0, 1/2, 1/2)

and its even permutation

(1/2, 1/2, 1/2, 0, 0, 1/2, 0, 0)

is

(1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2),

from which we could if we wanted subtract

(1/2, 0, 1/2, 0, 1/2, 0, 1/2, 0)

to obtain

(0, 1/2, 0, 1/2, 0, 1/2, 0, 1/2).

From all of this it follows that any 8-tuple of the following form belongs to the lattice: start with an arbitrary 8-tuple of integers (possibly negative), then add a half-integral part taking one of the following forms:

0000, 1111, 2222, 3333, 0123 or any even permutation of 0123.

Here 0 stands for the pair (0,0), 1 for (0, 1/2), 2 for (1/2, 0), 3 for (1/2, 1/2).

However, this set of 8-tuples turns out to be closed under addition, which implies that the lattice consists exactly of this set of 8-tuples. Why it should be closed under addition is not obvious to me - perhaps it is to you? It's all a question of those half-integer parts. When we add them, we make take 1/2+1/2 = 1 to be equivalent to 0, since we are not worried about the integer parts. So we have the addition table for pairs

 | 0 1 2 3
---------
0| 0 1 2 3
1| 1 0 3 2
2| 2 3 0 1
3| 3 2 1 0

(Klein's four-group). Note that the rows are all even permutations of one another, which implies that when 1111 (or 2222 or 3333) is added to 0123 we get an even permutation of 0123. We don't need the complete addition table for pair-quadruples, but note that if the column under 0000 is

0000
1111
2222
3333
0123
0312
0231
1032
1203
1320
2301
2130
2032
3210
3021
3102

then the column under 0123 (obtained by adding 0123 to the previous column) will be

0123
1032
2301
3210
0000
0231
0312
1111
1320
1203
2222
2013
2130
3333
3102
3021.

Note that the origin has 240 nearest neighbours in the lattice, all at (8-dimensional) distance 1. These are

(±1, 0, 0, 0, 0, 0, 0, 0) and permutations (16 points)
(±1/2, 0, ±1/2, 0, ±1/2, 0, ±1/2, 0) (16 points)
(0, ±1/2, 0, ±1/2, 0, ±1/2, 0, ±1/2) (16 points)
(0, 0, 0, ±1/2, ±1/2, 0, ±1/2, ±1/2) and all even permutations of the four pairs (192 points).

From the 120-cell to E8 (Part 1)

Take three golden rectangles made of cardboard. Make a slot down the centre of each one, parallel to the longer side, and equal in length to the shorter side. Insert the second rectangle into the slot in the first one, thus:

Finally, insert no. 3 into the slot in no.2, at the same time contriving that no.1 goes through the slot in no.3 (!).

Then the 12 vertices of the 3 rectangles mark out the 12 vertices of a regular icosahedron.

To see this, let the short side of the rectangles be 1, the long side φ (the golden number, such that φ² = φ+1). We shall show that the length of the vector joining a vertex of rectangle no.1 to a nearby vertex of no.2 is 1. In a suitable coordinate system, the coordinates of the first point are (φ/2,0,1/2), and of the second, (1/2, φ/2,0); so the coordinates of the vector are (-(φ-1)/2, φ/2, -1/2) or (-1/2φ, φ/2, -1/2). So the square of the length of the vector is

(1/φ² + φ² + 1)/4 = (φ² - 2φ + 1 + φ² + 1)/4 = 1,

as was to be shown. Thus the length of the vector is the same as the short side of a rectangle. Consequently, any vertex is surrounded by five nearby vertices all at distance 1. Moreover, each of the five is separated from its neighbours by the same distance. Thus the vertices enclose 20 equilateral triangles, and thus mark out the vertices of a regular icosahedron.

Note that the coordinates of the 12 vertices are (±φ/2,0,±1/2) , and all even permutations of that form (i.e. permutations obtained by an even number of exchanges: the even permutations of abc are bca - obtained by swopping the first two elements, then the second two - and cab).

Ready for four dimensions? Consider the set of points

(±1, 0, 0, 0) and permutations thereof
(±φ/2, ±1/2, 0, ±1/2φ) and even permutations thereof
(±1/2, ±1/2, ±1/2, ±1/2).

There are 8 of the first sort, 8 times 12 of the second, 16 of the third: altogether, 8 times 15 or 120. From the calculation we already did for the icosahedron, we see that each of these 120 points lies at distance 1 from the origin: in other words, they all lie on a (hyper)sphere of radius 1.

Now consider the nearest neighbours of (1,0,0,0). These are (φ/2, 1/2, 0, 1/2φ) and 11 other points obtained by evenly permuting the last three coordinates, and plus-or-minus-ing them. Components of the vector:

(-(2-φ)/2, 1/2, 0 , (φ-1)/2).

Observe that 2-φ is the square of φ-1 and so equals 1/φ²; so the vector is equal to

(-1/2φ, φ/2, 0, 1/2)

divided by φ. So its length is 1/φ. (In fact, we note that the vector times φ is actually the position vector of one of the original 120 points.)

It is clear that the distance between neighbours among the 12 points at height φ/2 is also 1/φ.

We won't go through the calculations, but it is straightforward to show that each and every one of the 120 points has 12 nearest neighbours at distance 1/φ. The points are thus disposed in a highly symmetrical manner on the (3-dimensional) 'surface' of the hypersphere, and are in fact the vertices of 600 tetrahedra, which fit together to form a 600-celled polytope, which is the 4-dimensional analogue of the 3-dimensional icosahedron.

The topmost point, (1,0,0,0), lies 'inside' the 12 points at level φ/2, which mark out the vertices of an icosahedron. Joining up vertices, we get 20 tetrahedra all sharing the topmost point as vertex. 120 times 20 makes 2400, but each tetrahedron has 4 vertices so we divide by 4 to obtain 600 as the number of tetrahedra.

The 'dual' of the 600-cell (each of whose cells is a tetrahedron) is the 120-cell (each of whose cells is a dodecahedron). To see this, consider the 20 tetrahedra which meet at a given vertex of the 600-cell. The centres of the tetrahedra are symmetrically disposed, and mark out the vertices of a dodecahedron. The geometrical centre of the dodecahedron does not quite coincide with the given vertex, but lies on the radius joining the origin to the vertex.

A neighbouring vertex will give rise to a second dodecahedron. But its cluster of 20 tetrahedra will have 5 in common with the cluster of the first vertex. This implies that the two dodecahedra have a pentagon in common: their pentagonal faces will touch. The dodecahedra will be 'glued' together.

In the same way, each vertex of the 600-cell gives rise to a dodecahedron; and the 120 dodecahedra are glued together, thus forming the cells of a second regular polytope, the 120-cell.

What we like about the Cosmic PDS Hypothesis

We must recognize that this whole PDS business may turn out to be moonshine - without basis in fact. If that's the case, then, with heavy heart, we will have to turn away from it, and look to pastures new. Meanwhile, there's no reason why we shouldn't take a look at some of the reasons we would like it to be true.

Firstly, there is the long-standing 'mysticism of the dodecahedron'.

***Plato, in his Timaeus, says of it (after describing the other four so-called 'Platonic' solids):

There still remained a fifth construction, which the God used for embroidering the constellations on the whole heaven.

The other four solids were assigned to the four terrestrial elements (fire, air, water and earth). The dodecahedron was set apart and associated with the heaven, i.e. with the sphere of the fixed stars, thought of as an outermost skin or shell of the (physical) world.

***Proposition 17 of book 13 of Euclid's Elements, the second last proposition of the whole book, explains how to construct a dodecahedron inside a sphere (touching it in 20 points). Its position in the book makes it hard not to think of this construction as a sort of grand finale of the whole work.

***Johannes Kepler was extremely susceptible to this sort of mysticism. He said that the greatest moment of his life was when he realized, in a moment of dazzling illumination , that the orbits of the planets were determined by celestial polyhedra, nested inside one another, as detailed in Kepler's famous diagram:

***Maybe Salvador Dali had the last word on dodecahedron-mysticism, in the shape of his painting of 1955, entitled 'Sacrament of the Last Supper'. The painter described it as an "arithmetic and philosophical cosmogony based on the paranoiac sublimity of the number twelve".

Secondly, there is the curious coincidence that if Roukema is right, our sphere of visibility, here and now (limited by our 'cosmological horizon') has almost the same diameter as the cosmic dodecahedron. If we had occurred much earlier in the history of the cosmos, our visibility sphere would have been much smaller than the dodecahedron; much later, and it would have been much greater than the dodecahedron - big enough, in fact, to contain many copies of the dodecahedron.

To some minds, this coincidence is so absurd that it seems to be a reason to reject the hypothesis. Myself, I relish the absurdity, the unlikeliness of it. It's as good as the fact that we live in an era when the moon is just big enough in the sky to blot out the sun when she stands in front of him. In times to come, the moon will have receeded still further from the earth and will be too small to blot out the sun, and there will no total eclipses (just annular ones).

Thirdly, there is the close link between Poincaré Dodecahedral Space and a beautiful mathematical object which exists in four dimensional space: the 120-cell or dodecaplex. In four dimensions there exist highly symmetrical solids (or rather, hypersolids) which are the analogues in 4 dimensions of the regular polyhedra in 3 dimensions. Their king, the 120-cell, will be the subject of future posts.

The 120-cell is linked in turn to the remarkable sphere-packing lattice in 8 dimensions known as E8, which is linked in turn to the biggest of the exceptional Lie groups, also called E8. Some people think that this group may hold the key to particle physics (see, for example, Garrett Lisi's Exceptionally Simple Theory of Everything). Clearly, it would be 'kinda neat' if the particles and forces of the universe turned out to be modelled on the 'ultimate' Lie group - and even better if this was somehow tied in with the topology of the universe.

The first link in this chain (120-cell to E8) will be the subject of my next post.

The cosmos is big, the speed of light is small

Back to astronomy - the reason why we got interested in Poincaré Dodecahedral Space (PDS, for short) in the first place. Is it possible that our cosmos, this actual universe that we're in, is topologically a PDS? Well, we don't see images of our back-view (twisted by 36°) hanging out there in space - not anywhere at all close by, at least. Certainly not close enough for me to appear as a human being, seen from behind. If anything 'I' would appear as a galaxy, or a cluster, or a supercluster of galaxies...

It seems clear that there is no 'repetition' as we look out into the cosmos, on a scale that is less than say a hundred million light years (i.e. the diameter of our local supercluster of galaxies). But on scales larger than that it is difficult to be so sure. One has to think rather carefully about what our image would actually look like. Not only would it be 'us' as seen from an unfamiliar point-of-view, but it would also be 'us' as we were several hundred million years ago. There would be no shortcut for the light coming to us from the distant image.

Is there any possibility of seeing two images of the same object which have the same 'date-stamp', i.e. such that the light which reaches us left both images at the same epoch? Yes, if the two images are equidistant from us. Imagine an object that is just half-way between me and my image. Then the object will have an image that is halfway between me and the second image of myself that I see by looking behind me - and the two images of my object will be equidistant from me.

Now, the cosmological (constant-curvature) PDS is constructed by identifying opposite faces of a non-Euclidean dodecahedron. Where am I in the dodecahedron? The at first sight surprising answer is that I might be anywhere - and it really doesn't matter. The space of the PDS is absolutely homogeneous; the curvature is the same everywhere, there are no special points. Admittedly space is not quite so isotropic - or 'the same in all directions' - as it was when space was topologically trivial. At any point there are now special directions: the six pairs of directions in which I see images of myself (or could do so, if I could see far enough).

The point is, that the edges and vertices of the dodecahedron are artifices of the construction by which we conceived of the PDS. Now we have got there, we can discard the original dodecahedron: it is no more than a map of the actual space. Other maps are possible, and there is nothing to stop me using a map which is centred on me.

To get from the space in which I find myself to the map, I need to measure out a regular dodecahedron, centred on myself, of such a size (and orientation) that any two opposite pentagonal faces are images of one another, in other words, identical.

Then, any object which happens to be located on one of the faces of this mapping-dodecahedron, will have a second image on the opposite face, and both images will be equidistant from me. I will get two different views of the same object, at the same moment of its existence.

Given that the scale of this dodecahedron is likely to be of the order of hundreds of millions if not billions of light-years, there is a danger that it will be too big for me to see any repeated images at all. This will be the case if a sphere drawn with me as centre, and radius the distance travelled by light since the origin of the universe (approximately 14 billion light years), is small enough to fit inside the dodecahedron without touching any of its faces.

There is also the possibility that this sphere is just about the same size as the dodecahedron, so that it intersects each of its twelve faces in a circle (see the picture below). This is the possibility that Roukema and his colleagues investigated. It is only objects lying within these 12 circles in the sky that can have double images (in opposite directions), and they must be far enough away that both the objects and their images lie within the visibility sphere (also called the cosmological horizon) - the sphere centred on me whose radius is the age of the universe times the speed of light. (The point is that if the object is too close to me, its image will be too far away to be seen by me.)

The visiblity sphere just beginning to burst through
the cosmic dodecahedron

In this scenario, only objects that are nearly as old as the universe are going to have double images. But there's not much we can see out there - with the notable exception of the primaeval plasma which emits the microwave background radiation, and at which (with the help of the WMAP sattellite) we have been gazing so intently during the last few years.

In other words, if the visibility sphere is only just big enough to intersect the dodecahedron, the only thing of which we are likely to be able to see a double image is the primaeval plasma. The glowing plasma is very nearly as old as the universe itself, so to find the double images we need to look in the direction of the 12 circles, above-mentioned, in which the visibility sphere intersects the dodecahedron. If Roukema's hypothesis is correct, when we look at one of these circles and its opposite number (which is also its topological image), we will be looking at the same bits of the primaeval plasma, and they should look the same so far as their tiny temperature variations are concerned.

Testing the theory is a question of looking at all possible opposing circles and testing them for correlations (not forgetting the 36° twist!). The unkowns are the orientation of the dodecahedron, and the size of the circles of intersection (which depends on the exact relationship between the size of the dodecahedron and the age of the universe).

In the latest (January 2008) paper by Roukema et al. (link) the configuration of the dodecahedron which yields the best correlations turns out to be such that the face-centres are at the following positions (specified by galactic longitude l and latitude b; I have also converted to Right Ascension and Declination - using Nasa'sCoordinate Converter - so that you know where to look in the sky; and have added the nearest constellation in each case):

Face 1 l=117° b=020° RA=22h17  Decl=81°  Cepheus
Face 2 l=184° b=062° RA=10h47  Decl=37°  Leo Minor
Face 3 l=046° b=049° RA=15h57  Decl=28°  Corona B.
Face 4 l=060° b=-013° RA=20h30  Decl=17°  Dolphin
Face 5 l=125° b=-042° RA=01h00  Decl=21°  Andromeda
Face 6 l=176° b=-004° RA=05h20  Decl=30°  Taurus

Note that the five faces 2..6 are arranged symmetrically around face 1. The remaining six face-positions are antipodal to the six given ones. (You may find it easier to think in terms of vertices of an icosahedron instead of face-centres of a dodecahedron. The angular distance between adjacent vertices is 63.43°, or tan^-1(2).)

The circles that give the best correlation have an angular radius of 11°. Even if the data were entirely random, there would no doubt be a dodecahedron, and a circle radius that yielded a maximum for the correlation. The question is, whether the quality of the optimum correlation is such as to make one believe that there really is a 'signal' - indicating the PDS topology - in all those masses of data from WMAP, rather than just noise. Much of the latest paper from Roukema et al. is devoted to trying to convince us that there really is a signal. I haven't yet understood the argument. But to quote from the paper's conclusion:

Do we really live in a Poincaré Dodecahedral Space? Further constraints either for or against the model are certainly still needed, but the evidence in favour of a PDS-like signal in the WMAP data does seem to be cumulating.

'Bye for now!

What is needed is Non-Euclidean geometry

Suppose for a moment that regular dodecahedra did fit together to 'tessellate' 3-D space. Each vertex would be the common property of 4 dodecahedra, and four edges would converge at each vertex. Their configuration would be just like what is called a tetrahedral bond in chemistry - think for example of the carbon atom surrounded by 4 hydrogens in methane. Looking down one of the bond directions, the other three are equally spaced with apparent angles of 120 deg between them. If the carbon is at the origin, we can take the hydrogens to be at (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1). The cosine of the angle between any two of the bonds will be -1/3, giving us the angle as 109.471° = α. Here we have another measure of the failure, in actual fact, of dodecahedra to fit snuggly together. For the angle at the vertex of a regular pentagon is only 108°. Another case of 'angle deficit'.

But hold on - so far our unspoken assumption has been that we are working in Euclidean geometry. In non-Euclidean geometry, the angles of a triangle add up to less than 180° (in hyperbolic, negative-curvature geometry), or more than 180° (elliptic, or positive-curvature geometry) - and the more so, in both cases, the larger the area of the triangle. The same will go for for the internal angles of a pentagon - so what we need to get a pentagon with angles of 109.471° is elliptic geometry. Altogether, the angles will exceed 540° by 7.355° or 0.128 of a radian, implying that the area of the pentagon must be 0.128 times the square of the radius of curvature - or just 0.128 if we take the radius of curvature as one unit.

We can obtain the length of the side of the pentagon, by doing a bit of spherical trigonometry (which applies in elliptic geometry). The second fundamental formula of spherical trigonometry goes as follows:

cos(A) = - cos(B) cos(C) + sin(B) sin(C) cos(a).

Here A, B, C are the three angles of a triangle, and a is the length of the side opposite A (again we assume the radius of curvature is one unit). The formula is usually thought of as determining the angle A when a, B and C are known. However, it can also be used to determine a when A, B and C are known. In the present case, A is 72°, B and C are each α/2. We obtain

cos(a) = ( cos(72) + cos²(α/2))/sin²(α/2) = 0.9635

hence a = 0.2709.

In flat space, the volume of a dodecahedron whose side is a, is (2 + 3.5 φ) times a³ (here φ stands for the golden number), giving us in the present case 0.1524. This is of course only an approximation (an underestimate, in fact), for the dodecahedron about which we are now talking exists in positively curved space.

It is interesting to compare this with the volume comprised in a 3-sphere of unit radius, which is just 2π²: enough to fit in approximately 2π²/0.1524 ≈ 129 of our dodecahedra.

Poincaré Dodecahedral Space - topology

Suppose that 3 adjacent windows of my dodecahedral cell (see previous posts), having a vertex in common, are coloured blue, red and yellow. Then the images I see through these windows will also have three coloured faces. Here is a picture of what I will see (click on the image for better resolution).

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

Poincare Dodecahedral Space (contd.)

View through the front window of my dodecahedral cell. Note that the back window of my image's cell matches the front window of mine. And that my image's front window is twisted by 36 degrees relative to mine.

View through two adjacent windows of my dodecahedral cell (click on the image to get better resolution). The edges of one pentagonal face of my cell are marked. Convinve yourself that the image cells are marked correctly.

View from the right-hand image, looking out of the blue window towards the left-hand image. On the left, the blue front window of the right-hand cell, with the marked pentagon abutting on it. On the right, how the blue window of the left-hand image appears. Note the rotation of 36 degrees.

Are we in Poincare Dodecahedral Space after all?

For a few years, people have been floating the idea that our world exists inside a space that has 'exotic topology' - that is, a space that joins up with itself in peculiar ways.

I hadn't paid much attention, until my eye was caught by an article in New Scientist magazine (12th January 2008) entitled 'Our finite, wrap-around universe'. It reported on a recent paper by Boudewijn Roukema, of Nicholas Copernicus University (how appropriate!) in Poland. (Feeling strong? You can see the paper at www.arxiv.org/abs/0801.0006 . Click on 'PDF' under where it says 'Download' in the top right hand corner.)

Roukema identifies 12 spots in the starry sky, arranged symmetrically, so that the angle between any two of these spots matches the angle between two faces of a regular dodecahedron (another way of putting it - the points could be the vertices of a regular icosahedron).

These special points have the property, that, if you could look far enough in the direction of one of these spots, you would see the universe beginning to repeat itself. It would be a bit like looking at yourself between two parallel mirrors - except that, in that scenario, you see as many images of the back of your head as of the front (i.e. - your face). In Roukema's scenario, you see only the back of your head. The image of yourself that you see is facing the same way as you. You are not looking in a mirror; you are located in a convoluted space, such that light leaving the back of your head goes on a long journey round the universe and enters in at your eyes in the usual way.

Now for the catch: your image is facing the same way as you (forwards!) but doesn't seem to be standing upright. She or he is leaning over as if in some sort of strange kaleidoscope - at an angle of 36°!

Beyond the first image you can see a second - twisted through another 36°. And so on, unto the tenth image, which is once more as upright as you are.

Now turn your head and look in the direction of another of Roukema's spots. Again you see an image of youself, with turned head, from the back - and twisted through 36°.

Why 36°?

(What follows does not answer the question Why?, at least not directly. Think of it as fleshing out the question...) Suppose you were standing in a large dodecahedral cell, with glass walls. Your image would then stand in a similar dodecahedral cell. What the rotation of 36° allows, is that the front wall of your dodecahedron matches the back wall of your image's cell. If you imagine your cell expanding, eventually it would be so big that it could actually fit onto its image!

The same applies to the image that you saw by turning your head: there is another cell out in that direction, also capable of fitting together with your cell, here.

Now for a question that may occur to you (if you have any trace of the mathematical disease in your blood): we've seen that the two image cells, seen in different directions, fit onto adjacent faces of the original cell, here, that the real you (?) is standing inside. One wonders: do they fit onto each other?

Well, that opens a can of worms... I think we'd better leave it 'til tomorrow.