Hopf fibration to rotation matrices

Question

Is there a more direct mapping from Hopf fibration to 3D rotation matrices without going through quaternions as an intermediate representation?

See: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2896220/pdf/nihms-152034.pdf

Answer 1

A number of years ago, I worked out the explicit formula for the double covering homomorphism $SU(2)\to SO(3)$. (I am sure this must be in various textbooks and papers, but I do not know a reference.) Here is the formula: For complex numbers $a,b$ with $|a|^2+|b|^2=1$, we have

$$\psi\left(\begin{bmatrix} a&b \\ -\bar b&\bar a\end{bmatrix}\right) = \begin{bmatrix} \tfrac12(a^2+\bar a{}^2-b^2-\bar b{}^2) & \tfrac i2(a^2-\bar a{}^2+b^2-\bar b{}^2) & -(ab+\bar a\bar b) \\ \tfrac i2(\bar a{}^2-a^2+b^2-\bar b{}^2) & \frac12(a^2+\bar a{}^2+b^2+\bar b{}^2) & i(ab-\bar a\bar b) \\ a\bar b + \bar ab & i(a\bar b-\bar ab) & |a|^2-|b|^2\end{bmatrix}.$$

Of course, we have $(a,b)\in S^3\subset\Bbb C^2=\Bbb R^4$. If you set $a=x_1+ix_2$ and $b=x_3+ix_4$, you can easily work out the matrix in $SO(3)$ in terms of $x_1,x_2,x_3,x_4$ and convert, using the formulas in the paper, the Hopf fibration coordinates directly into a rotation matrix, as you desire.

I worked this out using the Hopf fibration, but it should work out to be the adjoint representation of the unit quaternions, $SU(2)$, on its ($3$-dimensional) Lie algebra.