# Associative Composition Algebra/Homographies

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:

$(ua,\ ub){\begin{pmatrix}p&q\\r&s\end{pmatrix}}=((ua)p+(ub)r,(ua)q+(ub)s)$ $=(u(ap)+u(br),u(aq)+u(bs))=(u(ap+br),u(aq+bs)).$ 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. The above 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. One says that the transformation is well-defined.

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).

In the non-commutative cases (quaternions), this construction using homogeneous factors multiplying on the left, and matrix transformations on the right, brings this projective structure within reach.

A canonical embedding of A into P(A) is given by

$E:\ A\rightarrow P(A)\ {\text{by}}\ a\mapsto U[a,\ 1].$ If ab = 1, then $U[a,\ 1]\sim U[1,\ b]$ since a ∈ 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 ⊂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 the point at infinity, but unless A is a division algebra, it is not the only element of $P(A)\backslash E(A).$ The action of${\begin{pmatrix}1&0\\t&1\end{pmatrix}}$ on E(A) can be verified to agree with the translation aa + t acting in A. Similarly, for a positive real number p, the action of ${\begin{pmatrix}p&0\\0&1\end{pmatrix}}$ on E(A) agrees with the dilation apa acting in A. Furthermore, inner automorphisms are extended by homographies:

$U[a,\ 1]{\begin{pmatrix}u&0\\0&u\end{pmatrix}}\ =\ U[au,\ u]\ \sim \ U[u^{-1}au,\ 1].$ August Moebius first popularized homographies in the context of A = C, the division binarions. 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 ⊂ 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).

Some insight into the role of component b in g is gained from the product

${\begin{pmatrix}0&1\\1&0\end{pmatrix}}{\begin{pmatrix}1&0\\a&1\end{pmatrix}}{\begin{pmatrix}0&1\\1&0\end{pmatrix}}\ ={\begin{pmatrix}0&1\\1&0\end{pmatrix}}\ {\begin{pmatrix}0&1\\1&a\end{pmatrix}}={\begin{pmatrix}1&a\\0&1\end{pmatrix}}.$ Due to the conjugation with the inverter operator, the transformation is a translation at infinity.

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.

3. Suppose $(b-a)^{-1}$ exists in a AC algebra. Find a homography that sends U[a,1] to U[1,0] and U[b,1] to U[0,0].

4. Find the image of U[c,1] under the homography above. When is there a homography to move it to U[1,1]?

5. Show that when A is a division AC algebra, the homography group is triply transitive (any three points can be mapped to any other three).