By Hubert Shutrick


In this first section a summary of some of the basic definitions and properties that are needed are given. More details can be found here for example.

In the projective plane, a point is determined by a non-zero vector of homogeneous coordinates \mathbf{u}=(x, y, z) so, for each real number r \neq 0 , r \mathbf{u} determines the same point as \mathbf{u} . Given two points \mathbf{u}_{s} and \mathbf{u}_{t} , the line that joins them can be described by

\mathbf{u} = s \mathbf{u}_{s} + t \mathbf{u}_{t}

where s and t are homogeneous parameters for the line. It can be arranged that any third point on the line has parameters (1, 1) by scaling the vectors that represent the other two points. Three new points on the line give another parametric representation with parameters s' and t' and the relation between them is of the linear relation

\pmatrix{s' \\t'} = \pmatrix{a & b \\ c & d}\pmatrix{s \\ t}

where ad bc \neq 0 . This set of permitted coordinate transformations is essentially the projective structure of the line. An invariant of the structure is cross-ratio: it is a property of the four points that is independent of parameter choice.

A bijective mapping from one line to another is a projectivity if it is an isomorphism of the projective structure. Hence, a bijection p: l \rightarrow l' will be an isomorphism if it takes points with parameters (s,t) on l to (s',t') on l' linearly like a change of coordinates.

The images of three points under a projectivity define it completely, for instance, the columns of the matrix are the parameters of the images of (1,0) and (0,1) and they can be scaled so that their sum is the image of (1,1). Cross-ratio will be preserved by a projectivity and, conversely, since three points and a cross-ratio determine the fourth, a bijection will be a projectivity if it preserves cross-ratio.

The immediate example of a projectivity is a projection from one line to another from a point that is not on either. A projectivity between two different lines that is not a projection, one that does not leave their common point fixed, can always be realized as the composite of two projections using an intermediate line.


A projectivity from a line to itself is a projective automorphism.


An involution is a projective automorphism that is its own inverse so the matrix \mathbf{A} of the mapping satisfies

\mathbf{A}^{2} =\pmatrix{a^{2} + bc & b(a+d) \\ c(a+d) & d^{2} + bc} = r\mathbf{I}

where \mathbf{I} is the unit matrix and r\neq0. The condition for an involution is then d = -a. By scaling \mathbf{A} so that its determinant is 1 and we can assume that r = \pm 1 . Since the eigenvalues of A must satisfy \lambda^{2} = \pm 1, there are two cases, \lambda = \pm 1 or \lambda = \pm i. In both case, the eigenvectors parametrize two fixed points of the automorphism but in the second case they are parametrized by complex numbers. Pairs of points X and X' that correspond under the involution divide the two fixed points F_{1} and F_{2} harmonically since

[F_{1},F_{2};X,X'] = [F_{1},F_{2};X',X]

implies that the cross-ratio is -1. Conversely, for two points on a line, the pairs that divide them harmonically are in involution since three points and a cross-ratio determine a fourth point linearly.

Pencils of Conics

A special case where involutions occur, is when a pencil of conics intersect the line. A conic is determined by equating a homogeneous second degree polynomial f in x, y and z to zero. Two conics with polynomials f_{p} and f_{q} give a pencil of conics

pf_{p} + qf_{q} = 0

where p and q are homogeneous parameters for the pencil and it consists of all the conics that pass through their four points of intersection. Any other point that is not one of their four common points has just one of the conics passing through it determined by solving for p and q . Each point on a line that does not pass through one of the four common points determines then a conic that intersects the line in a new point. We shall prove that this correspondence is an involution.

If the parametric equations for the line are substituted into the equation for the pencil, we get a homogeneous polynomial f(p,q,s,t) that is linear in p,q and quadratic in s,t. The condition that the quadratic equation

as^{2} + 2bst + ct^{2} = 0

has double roots is the usual b^{2} = ac , a quadratic equation in p,q. Let us assume that its roots are real and take the conics of the pencil that correspond to them as a new base for the pencil so that they are given by (p,q) = (0,1) and (p,q) =(1,0). These two conics are tangent to the line at the points of intersection and are fixed points of the correspondence.

Next we take new parameters for the line such that (s,t) = (0,1) and (s,t) =(1,0) are the fixed points. The polynomial f then becomes simply pt^{2} + qs^{2} and the point (s,t) =(1,1) gives (p,q) =(1,-1) so the corresponding points are (1,1), (1,-1) dividing the fixed points harmonically. However, any point on the line except the fixed ones can be given parameters (1,1) so all corresponding pairs divide the fixed points harmonically implying that it is an involution.

If the two fixed points are imaginary, the calculation can be made in just the same way to show that it is an involution.

Involuton given by two of its pairs

It will be seen in the next sections that pairs of all involution are intersections of conics in a pencil and therefore are determined by the unique pencil of quadratic equations that are given by inserting the parametric equation of the line into the equations of the conics. This suggests a simple way of calculating the involution given two of its pairs. Working with non-homogenous parameters, if two given pairs are (s,s') and (t,t') and we seek u' for given u, then the quadratic equations that have the three pairs as roots are linearly dependent since they are members of the pencil. Therefore,

\left|\matrix{1 & -(s + s') & s s'\\1 & -(t + t') & t t'\\ 1 & -(u + u') & u u' }\right| = 0

which is an equation for u'.

In particular, the parameters u of the double points will be the solutions of the quadratic equation for u' = u in the determinant.

Apollonian circles

This section shows how involutions relate to the Coaxial Circles Theorem. The applet in that article illustrates the pencils of circles described here.

Given an involution on a line, consider the set of circles whose diameters are the segments joining the pairs in the involution. Such a set is a pencil of conics. All circles pass through the imaginary circular points at infinity (1,i,0),(1,-i,0) and the other two common points of the set can be real or imaginary. The line containing the common points is the common radicle axis of pairs of circles in the pencil.

Real case

Consider the pencil of circles that pass through (1,0,1) and (-1,0,1) in the euclidean plane and their intersection with the y-axis. The involution takes the point (0,r,1) to (0,-1/r,1) because of the intersecting chords theorem. The two fixed points are given by r = -1/r so they are (0,i,1) and (0,-i,1).

Imaginary case

Consider the involution on the x-axis given by the pencil of circles through (0,i,1) and (0,-i,1). Their equations are

(x-a)^{2} + y ^{2} = a^{2} 1

with centre (a,0,1) and the double points are ( 1,0,1) given by the point circles when the radius is zero.

All circles in one of the pencils are orthogonal to all in the other.

Special case

Consider the pencil of circles with centre the origin intersecting, once more, the x-axis. The involution is obvious: (a,0) goes to (-a,0). The fixed points are the origin and the point at infinity on the x-axis which give the harmonic division. The origin is a point circle in the pencil and is the intersection of the degenerate conic consisting of the lines that connect it to the circular points at infinity. In fact the circular points at infinity are double points of each conic in the pencil the lines connecting the origin to them are tangent to each conic in the pencil. There are no finite radical axes.

Butterfly Proof

An example of theorems that can be proved using these results is the butterfly theorem. The pencil of conics that pass through A,B,C,D in the diagram include the circle and the pair of lines AB,CD, whose intersection H is a double point of the involution on PQ and the other double point must be the point at infinity since H is the midpoint of PQ. Hence, it is also the midpoint of XY because AD,BC is another degenerate conic in the pencil.


Maxwell, E. A. The Methods of Plane Projective Geometry Based on the Use of General Homogeneous Coordinates, Cambridge University Press, 1946