- PROJECTIVE AND DESCRIPTIVE GEOMETRY 369
Mobius is best remembered for two concepts, the Mobius transformation, and
the Mobius band. A Mobius transformation, by which we now understand a map-
ping of the complex plane into itself, æ t-> w, of the formw = a" "*~ â ad - be ^ 0,
cz + d
can be found in his 1829 paper on metric relations in line geometry. He gave such
transformations with real coefficients in terms of the two coordinates (x,y), the
real and imaginary parts of what we now write as the complex number z, and
showed that they were the most general one-to-one transformations that preserve
collinearity. The Mobius band is discussed in Section 4 below.
2.10. Julius Pliicker. A number of excellent German, Swiss, and Italian geome-
ters arose in the nineteenth century. Their work cannot be classified as purely
projective geometry, since it also relates to algebraic geometry. As an example, we
take Julius Pliicker (1801-1868), who was a professor at the University of Bonn
for the last 30 years of his life. Pliicker himself remembered (Coolidge, 1940, p.
144) that when young he had discovered a theorem in Euclidean geometry: The
three lines containing the common chords of pairs of three intersecting circles are
all concurrent. Plucker's proof of this theorem is simplicity itself. Suppose that
the equations of the three circles are A = 0, Â = 0, C = 0, where each equation
contains x^2 + y^2 plus linear terms. By subtracting these equations in pairs, we get
the quadratic terms to drop out, leaving the equations of the three lines containing
the three common chords: Á — Â=¼, A — C = 0, B-C = 0. But it is manifest
that any two of these equations imply the third, so that the point of intersection of
any two also lies on the third line.
Plucker's student Felix Klein (1926, p. 122) described a more sophisticated
specimen of this same kind of reasoning by Pliicker to prove Brianchon's theorem^17
that the opposite sides of a hexagon inscribed in a conic, when extended, intersect
in three collinear points. The proof goes as follows: The problem involves two sets,
each containing three lines, six of whose nine pairwise intersections lie on a conic
section. The conic section has an equation of the form q(x, y) = 0, where q(x, y) is
quadratic in both ÷ and y. Represent each line by a linear polynomial of the form
ajX + bjy + Cj, the jth line being the set of (x, y) where this polynomial equals zero,
and assume that the lines are numbered in clockwise order around the hexagon.
Form the polynomial
s(x,y) = (aix + biy + cl)(a 3 x + b 3 y + c 3 )(a 5 x + b 5 y + c 5 )
- ì(á 2 £ + b 2 y + C2)(a4X + b^y + c 4 )(aex + fay + ce)
with the parameter ì to be chosen later. This polynomial vanishes at all nine
intersections of the lines. Line 1, for example, meets lines 2 and 6 inside the conic
and line 4 outside it.^18
Now, when y is eliminated from the equations q(x,y) = 0 and s(x,y) = 0,
the result is an equation t(x) = 0, where t(x) is a polynomial of degree at most 6
in x. This polynomial must vanish at all of the simultaneous zeros of q(x, y) and
s(x, y). We know that there are six such zeros for every ì. However, it is very easy
(^17) Klein called it Pascal's theorem.
(^18) This polynomial is the difference of two completely factored cubics, by coincidence exactly the
kind of polynomial that arises in the six-line locus problem, even though we are not dealing with
the distances to any lines here.