# Associative Composition Algebra/Quaternions

William Rowan Hamilton's real quaternions H and biquaternions B are constructed from pairs of division binarions or bibinarions, respectively. These operations are defined:
    ${\text{Multiplication:}}\ (w,x)(y,z)=(wy-xz^{*},\ wz+xy^{*}),$
    ${\text{Conjugation:}}\ (w,x)^{*}=(w^{*},-x),$
    ${\text{so that}}\ N(w,x)=(w,x)(w^{*},-x^{*})=ww^{*}+xx^{*}=N(w)+N(x).$
A third quaternion algebra Q = split-quaternions is a variant of H and a subalgebra of B. The following chapter explores split-quaternions through exercises.
H and B were both described by W. R. Hamilton in his Lectures on Quaternions (1853). AC algebra Q was described by James Cockle and called coquaternions. For a time H, B, and Q had special profiles in their use as AC algebra, but matrix rings were exploited in the twentieth century to provide linear representations for them, and thus absorb them into the larger study of linear algebra. Indeed, Q is ring isomorphic to M(2,R), the 2 × 2 real matrices, and B is ring isomorphic to M(2,C), the 2 x 2 complex matrices. Representation of H uses the context in B. In the Linear Algebra the idea of composition is visible with the determinant of a matrix, which has a similar property.

## Division quaternions

In the notation of Hamilton, with $w=a+bi,\ z=c+di,$ (w,z) is written a + bi + cj + dk, where the products $ij=k=-ji,\ jk=i=-kj,\ ki=j=-ik$ can be confirmed, and noted for anticommutativity. The set {i, j, k} has been taken as the basis of space in presentations of kinematics, mechanics, and physical science.
Furthermore, $i^{2}=j^{2}=k^{2}=-1.$ In fact,
    $(q\in H\land q^{2}=-1)\equiv (q=xi+yj+zk\land x^{2}+y^{2}+z^{2}=1),$
so there is a sphere S$^2$ of imaginary units in H.
Say that u is one of them, then the complex arithmetic of Euler's formula gives $e^{au}=\cos a+u\sin a.$ In the quaternion context, e$^{au}$ is a versor, and versors are the points of elliptic space, a geometry entirely devoted to rotations. W. K. Clifford was an exponent of elliptic geometry, and much more, until his flame was extinguished at age 34.
For vectors in V ⊂ H, anticommutivity means perpendicularity:
    $\forall a,b\in H\ \ ab+ba=0\equiv (a,b\in V\land a\perp b).$
Lemma: if a and b are square roots of minus one and a ⊥ b, then aba = b.
    proof: $0=a(ab+ba)=a^{2}b+aba=-b+aba.$
Lemma: Under the same hypothesis, a ⊥ ab and b ⊥ ab.
    proof:$(a\perp ab):\ \ a(ab)+(ab)a=-b+aba=0.$
Let u = exp(θ r) be a versor. There is a group action on H determined by u:

### Conjugation of a vector by a versor

Suppose a pair (a,b) in HxH, not both zero, and a pair (c,d) are related by a non-zero quaternion q through qa=c and qb=d. The relation is denoted (a,b) ~ (c,d). It is an equivalence relation and HxH/~ is a quaternion projective line. The homographies of this projective line are given by matrices from M(2,H). For example,
    $(q,1){\begin{pmatrix}u&0\\0&u\end{pmatrix}}=(qu,u)\thicksim (u^{-1}qu,1).$
The equivalence class for (a,b) is written [a,b]. The mapping $q\mapsto u^{-1}qu$ is called the conjugation of q by u, conventionally taken as a versor. The real part of q is invariant under the conjugation, but it applies to the vector part. The following quaternion arithmetic computation shows that the vector is rotated about the axis of the versor, and by twice its angle:
Note that $u=e^{r\theta }$ commutes with all elements in the plane $\{x+yr:x,y\in R\}.$ Select s from the great circle on S$^2$ that is perpendicular to r. Then rsr = s by the first lemma. Now compute the conjugate of s by u:
    $u^{-1}su\ =\ (\cos \theta -r\sin \theta )s(\cos \theta +r\sin \theta )$
    $=\ (s\cos \theta \ -rs\sin \theta )(\cos \theta +r\sin \theta )$
    $=\ (\cos ^{2}\theta -\sin ^{2}\theta )s+(2\sin \theta \cos \theta )sr$
    $=\ s\cos 2\theta +sr\sin 2\theta ,$ which is rotation by 2 theta in the (s, sr) plane.

### Screw displacement

Linear fractional transformations with quaternions can be demonstrated by considering a kinematic exercise: Given a rotation about the i axis by 2 θ (inner automorphism with versor exp(θ i)) and a desired translation in the j-k plane, find the position of the axis parallel to the i axis where the rotation effects the translation.
The problem can formulated in terms of t = xj + yk, and the transformation first drawing t back to the origin, then preforming the rotation before restoring the position of t:
    $[q,1]{\begin{pmatrix}1&0\\-t&1\end{pmatrix}}{\begin{pmatrix}u&0\\0&u\end{pmatrix}}{\begin{pmatrix}1&0\\t&1\end{pmatrix}}={\begin{pmatrix}u&0\\z&u\end{pmatrix}},$
where $z=tu-ut=2\sin \theta (xk-yj).$
Presume the desired translation is in the j direction at distance a, so the desired image of 0 is aj:
    $[q,1]{\begin{pmatrix}u&0\\uaj&u\end{pmatrix}}=[qu+uaj,u]=[u^{-1}qu+aj,1].$
Set z = uaj and compare j and k coordinates. The k component equation leads to x = a/2 and shows that t must lie on the perpendicular bisector of the segment from 0 to aj (so the radii to 0 and aj are the same). The j component equation leads to $y=-{\tfrac {1}{2}}\cot \theta$ which corresponds to the right triangle with y on the bisector and a/2 as the opposite side, giving $\tan \theta =a/2y.$
The idea of a rotation providing translation by moving the axis of rotation appropriately was described by Mozzi in 1763 and Chasles in 1830, and is considered a feature of Euclidean motions and kinematics. The proposition is stated as the sufficiency of screw displacements to effect the Euclidean group of proper isometries. Screw displacements are rotations in 3-space, possessing an axis of rotation, and the screw motion includes a translation along the axis of rotation. Notice of the sufficiency is attributed variously to Mozzi and Chasles.

## Biquaternions

The AC algebra (B, +, x, * ) has conjugation
    $(w+xi+yj+zk)^{*}=w-xi-yj-yk,\quad w,x,y,z\in C.$
In biquaternions a new imaginary unit h commutes with all the other imaginary units i, j, k, including all r satisfying r$^2$ = − 1. For example, the division binarion w = a + b h, a,b in R.
Suppose now the complex conjugation is invoked:${\bar {w}}=a-bh,\ \ h^{2}=-1.$ as second involution, denoted by an overbar:
    ${\bar {q}}={\bar {w}}+{\bar {x}}i+{\bar {y}}j+{\bar {z}}k\ \ {\text{and}}\ \ M=\{q\in B:q^{*}={\bar {q}}\}.$
The two involutions agree on $M=\{q=a+bhi+chj+dhk:\ a,b,c,d\in \mathbb {R} \}.$
This four-dimensional subspace M was exploited by Ludwik Silberstein (1914) and Cornelius Lanczos (1949) to exhibit a mathematical model of spacetime with speed of light set to one, and admitting Lorentz transformations as conjugation of an event by a versor or hyperbolic versor.
In B, for each square root of −1, r ∈ S$^2$, (hr)$^2$ = +1. Then the plane
    $J=\{x+y(hr)\in B:x,y\in R\}$ is a split binarion algebra with (x + y(hr))* = x - y(hr). In particular
    $u=\exp(ahr)\ =\ \cosh a+hr\sinh a,$ with hyperbolic angle a, is a hyperbola in the plane of R and hr.
Hyperbolic rotation, or squeeze, can be obtained by conjugation with u. Using r and s ∈ S$^2$ ⊂ H as above, then $f(s)=u^{-1}su=$
    $=\ (\cosh a-hr\sinh a)s(\cosh a+hr\sinh a)$
    $=\ (s\cosh a\ -hrs\sinh a)(\cosh a+hr\sinh a)$
    $=\ (\cosh ^{2}a+\sinh ^{2}a)s+(2\sinh a\cosh a)\ hsr$
    $=\ s\cosh 2a\ +hsr\sinh 2a\ ,$
which is hyperbolic rotation of s by a hyperbolic angle 2a in the (s, hsr) plane. The real vector s, outside of M, is found to have a component (sinh a) hsr ∈ M after f.

### Exercises

1. Let f be a mapping on B given by f(s) = v s v, where v = exp(a hr). Show that r ⊥ s implies f(s) = s.
2. Show f(e$^{bhr}$) = exp((2a + b) hr).
3. Interpret f as a mapping on M. Hint: Use terminology of special relativity.
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