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Cayley transform
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==Quaternion homography== In the [[four-dimensional space]] of [[quaternion]]s <math>a+b\vec{i}+c\vec{j}+d\vec{k}</math>, the [[versor]]s :<math>u(\theta, r) = \cos \theta + r \sin \theta </math> form the unit [[3-sphere]]. Since quaternions are non-commutative, elements of its [[projective line over a ring|projective line]] have homogeneous coordinates written <math>U[a,b]</math> to indicate that the homogeneous factor multiplies on the left. The quaternion transform is :<math>f(u,q) = U[q,1]\begin{pmatrix}1 & 1 \\ -u & u \end{pmatrix} = U[q - u,\ q + u] \sim U[(q + u)^{-1}(q - u),\ 1].</math> The real and complex homographies described above are instances of the quaternion homography where <math>\theta</math> is zero or <math>\pi/2</math>, respectively. Evidently the transform takes <math>u\to 0\to -1</math> and takes <math>-u \to \infty \to 1</math>. Evaluating this homography at <math>q=1</math> maps the versor <math>u</math> into its axis: :<math>f(u,1) =(1+u)^{-1}(1-u) = (1+u)^*(1-u)/ |1+u|^2.</math> But <math>|1+u|^2 = (1+u)(1+u^*) = 2 + 2 \cos \theta ,\quad \text{and}\quad (1+u^*)(1-u) = -2 r \sin \theta .</math> Thus <math>f(u,1) = -r \frac {\sin \theta}{1 + \cos \theta} = -r \tan \frac{\theta}{2} .</math> In this form the Cayley transform has been described as a rational parametrization of rotation: Let <math>t=\tan\phi/2</math> in the complex number identity<ref>See [[Tangent half-angle formula]]</ref> :<math>e^{-i \varphi} = \frac{1 - ti}{1 + ti} </math> where the right hand side is the transform of <math>ti</math> and the left hand side represents the rotation of the plane by negative <math>\phi</math> radians. ===Inverse=== Let <math>u^* = \cos \theta - r \sin \theta = u^{-1} .</math> Since :<math>\begin{pmatrix} 1 & 1 \\ -u & u \end{pmatrix}\ \begin{pmatrix} 1 & -u^* \\ 1 & u^* \end{pmatrix} \ = \ \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \ \sim \ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \ ,</math> where the equivalence is in the [[projective linear group]] over quaternions, the [[inverse function|inverse]] of <math>f(u,1)</math> is :<math>U[p,1] \begin{pmatrix} 1 & -u^* \\ 1 & u^* \end{pmatrix} \ = \ U[p+1,\ (1-p)u^*] \sim U[u(1-p)^{-1} (p+1), \ 1] .</math> Since homographies are [[bijection]]s, <math>f^{-1} (u,1)</math> maps the vector quaternions to the 3-sphere of versors. As versors represent rotations in 3-space, the homography <math>f^{-1}</math> produces rotations from the ball in <math>\R^3</math>.
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