Associative Composition Algebra/Homographies

The concept of homography has already been introduced as Möbius transformations on the projective binarion line. In fact, the concept has been extended to screw displacements using quaternions, where non-commutativity of the algebra has been accommodated.

Since the associative property is a requisite of a mathematical group, this text has required AC algebras to have it so that the algebras have a multiplicative group. Furthermore, associativity, and the fact that multiplication distributes over addition, are used in the following application of matrix multiplication:

Proposition: On an associative composition algebra, the homography $$\begin{pmatrix} p & q \\ r & s \end{pmatrix}$$ is well-defined on the projective line.

With u taken from the group of units of A, (ua, ub) are the homogeneous coordinates of a point in the projective line P(A). One writes :$$(a,b) \sim (ua, ub)$$ and ~ is an equivalence relation on A x A; for instance, it is a transitive relation because of associativity.
 * $$(ua,\ ub) \begin{pmatrix}p & q \\ r & s \end{pmatrix} = ((ua)p + (ub)r, (ua)q + (ub)s) $$
 * $$= (u(ap) + u(br), u(aq )+ u(bs)) = (u (ap + br), u(aq + bs)) .$$

These equalities, involving the matrix product on the right, show that the result of the matrix transformation does not depend on the representative (a,b) from an equivalence class of the relation.

The condition $$aA + bA = A$$ requires the pair (a,b) to be sufficient to generate A: they must not both lie in a proper subalgebra. The projective line is
 * $$P(A) = \{U[a:b]: aA + bA = A \},$$ where U[a: b] represents the equivalence class of (a, b).

Embedding, far and near translations
A canonical embedding of A into P(A) is given by
 * $$E: \ A \rarr P(A) \ \text{by} \ a \mapsto U[a:\ 1] .$$

If ab = 1, then $$U[a:\ 1] \sim U[1:\ b]$$ since a &isin; U. For such a,
 * $$E(a)\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \ = \ E(a^{-1}), $$

showing that $$\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} $$ moves the elements of U &sub;A to the equivalence class of U[a−1: 1], thus extending the multiplicative inverse map to P(A).
 * $$U[0: \ 1] \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} = U[1: \ 0]$$

is referred to as a point at infinity, but unless A is a division algebra, it is not the only element of $$P(A) \backslash E(A) .$$

For a positive real number p, the action of $$\begin{pmatrix}p & 0 \\ 0 & 1 \end{pmatrix}$$ on E(A) agrees with the dilation a → pa acting in A. Furthermore, inner automorphisms are extended by homographies:
 * $$U[a:\ 1] \begin{pmatrix}u & 0 \\ 0 & u \end{pmatrix}\ = \ U[au:\ u] \ \sim \ U[u^{-1} au:\ 1].$$

With the canonical embedding z → [z : 1 ] of A into P(A), the transformation $$\begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix} $$ is a near translation. The other embedding z → [ 1 : z ] takes the origin to [ 1 : 0 ], sometimes written $$\infty$$ since it is a point at infinity with respect to the canonical embedding. The transformation
 * $$[ z : 1 ] \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix} = [ z : z t + 1 ] $$ is a far translation because of its effect on the second embedding:
 * $$[ 1 : z ] \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix} = [ 1  : z + t ]  .$$

Proposition : Suppose $$\begin{pmatrix}a & c \\ b & 1 \end{pmatrix} \in SL(2, A) .$$ Then the matrix is a product of a far translation with a near translation.
 * proof: 1 = a &minus; bc implies a = 1 + bc. Then
 * $$\begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ b & 1 \end{pmatrix} =\begin{pmatrix} 1+bc & c \\ b & 1 \end{pmatrix} = \begin{pmatrix} a & c \\ b & 1 \end{pmatrix} .$$

Conformal spacetime transformations
In 1910 reference was made to "conformal transformations of spacetime" by Harry Bateman and Ebenezer Cunningham, though the method of description was by differential geometry of transformations respecting Maxwell's equations of electromagnetism. Using M &sub; B to represent spacetime, and the AC algebra B for homographies on P(B) to represent transformations, a general conformal transformation can be written:
 * $$g = \begin{pmatrix}pu & b \\ a & v \end{pmatrix} .$$ The more commonly noted subgroups are an affine group (b = 0), the Poincaré group (p = 1 and b = 0), and the Lorentz group (p = 1 and a = b = 0).

There are 15 degrees of freedom in g: p is one, a and b contribute four each, while u and v contribute six.

In particular, $$\begin{pmatrix}u & 0 \\ 0 & u \end{pmatrix}$$ with u = exp(a r) generates the orthogonal group O(3) in the Lorentz group, and
 * $$\begin{pmatrix} v & 0 \\ 0 & 1/v \end{pmatrix}$$ with v = exp(b hr) generates the boosts, per the exercises in the last chapter.

Exercises

1. Find the coordinates of elements of the projective line over the field with two elements.

2. For g extending translation, rotation, and inversion, find {x : xg = x}, the fixed point set of g.

Cross ratio
On the real projective line the homography
 * $$\begin{pmatrix}-1 & 1 \\ a & b \end{pmatrix}$$ takes [a: 1] to [0: 1] and [b: 1] to [1: 0].

The numbers between them have positive real values. The midpoint of the interval (a,b) goes to [1: 1].

For commutative rings (binarions here), there is a cross ratio homography which maps a sufficiently distinct triple from the ring to the projective line over the Galois field Z/2Z, which is contained in the projective line over any ring. But in the non-commutative case (quaternions here), the homography which separates points p and q is only conditionally normalizable. Indeed, compose
 * $$\begin{pmatrix}t-q & 0 \\ 0 & t-p \end{pmatrix}$$ with the separation of p and q:


 * $$[t:1] \rightarrow [t-p:\ t-q] \rightarrow [(t-p) (t-q) : \ (t-q) (t-p)] \ne [1:1] .$$

Exercise: Show that $$t(q-p) + (p-q)t + qp - pq = 0$$

is sufficient to provide a normalized homography mapping of {p, q, t} to {[0 :1], [1 :0], [1, 1] }.