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.