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.