Let U and V be two lines in the projective plane, defined by their
dual coordinate vectors. Consider a set of lines defined by:
Theorem: The cross ratio of any four lines of a pencil is invariant under collineations.
Cross ratios of collinear points and coincident lines are linked as follows:
Theorem: The cross ratio of four lines of a pencil equals the cross ratio of their points of intersection with an arbitrary fifth line transversal to the pencil (i.e. not through the pencil's centre) -- see fig. 3.1.
In fact, we already know that the cross ratios of the intersection points must be the same for any two transversal lines, since the lines correspond bijectively to one another under a central projection, which is a collineation.
The simplest way to establish the result is to recall the line intersection formulae of section 2.2.1. If U and Vare the basis 3-vectors for the line pencil, and L is the transversal line, the intersection points of L with U, V and are respectively , and . In other words, the coordinates of a line in the basis are the same as the coordinates of its intersection with L in the basis . Hence, the two cross ratios are the same.
To establish the result, we will consider a particular transversal line:
the line at infinity. Let
ax0 + bx1 + cx2 = 0 be the
equation defining U and
dx0 + ex1 + fx2 = 0 the one
corresponding to V. Therefore the one associated to
If we consider two reference points M and N being the intersection with
line U and V, i.e. the points with coordinates
Another way to prove the result is to show that cross ratios of lengths
along the transversal line can be replaced by cross ratios of angle
sines, and hence are independent of the transversal line chosen (c.f fig. 3.1):
Theorem (Möbius): Let be any four lines intersecting in O, and be any four points respectively on these lines, then
This 19th century result extends gracefully to higher dimensions. To
prove it, let
(a,b, 1), (x,y, 1) and (u,v, 1) be the normalized
affine coordinate vectors of O, Ai and Aj. Then