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).