Question

Hi all. Is there any explicit matrix expression for a general element of the special orthogonal group $SO(3)$? I have been searching texts and net both, but could not find it. Kindly provide any references.

Answer 1

Here is the standard quaternion answer: Given $(a,b,c,d)$ such that $a^2+b^2+c^2+d^2=1$, the matrix $$\begin{pmatrix} a^2+b^2-c^2-d^2&2bc-2ad &2bd+2ac \\ 2bc+2ad &a^2-b^2+c^2-d^2&2cd-2ab \\ 2bd-2ac &2cd+2ab &a^2-b^2-c^2+d^2\\ \end{pmatrix}$$ is a rotation and every rotation matrix is of this form. Note that $(a,b,c,d)$ and $(-a, -b,-c,-d)$ give the same rotation.

Answer 2

To give something explicit in sine and cosine,

$$
\left( \begin{array}{ccc} \cos\theta \cos\psi & -\cos\phi \sin\psi + \sin\phi \sin\theta \cos\psi & \sin\phi \sin\psi + \cos\phi \sin\theta \cos\psi \\ \cos\theta \sin\psi & \cos\phi \cos\psi + \sin\phi \sin\theta \sin\psi & -\sin\phi \cos\psi + \cos\phi \sin\theta \sin\psi \\ -\sin\theta & \sin\phi \cos\theta & \cos\phi \cos\theta \end{array} \right)
$$

Note that three parameters are required. In odd dimension, there is a real eigenvalue. For $SO_n$ this eigenvalue is $+1.$ So there is a fixed vector in some direction. It takes two parameters to specify this point on the unit sphere. The Lie group element is then a rotation around this point. So it takes a third parameter specifying the amount of rotation about that axis.

Answer 3

There is a good way to derive the sort of thing you're looking for: use the double cover $SU(2) \to SO(3)$. $SU(2)$ is diffeomorphic to the 3-sphere $S^3 \subseteq \mathbb{C}^2$ $$ SU(2) = \left\{ \begin{pmatrix} a & -\overline{b} \\ b & \overline{a} \end{pmatrix} : |a|^2 + |b|^2 = 1 \right\} $$ Now $SU(2)$ acts on its Lie algebra $\mathfrak{su}_2$ (which is 3-dimensional) by conjugation. This action preserves the inner product $$ \langle X, Y \rangle = - \frac12 \mathrm{tr}(XY) = \frac12 \mathrm{tr}(X^*Y)$$ (which is a scalar multiple of the Killing form of $\mathfrak{su}_2$, FYI) and hence this gives a homomorphism $SU(2) \to SO(\mathfrak{su}_2) \simeq SO(3)$. (A priori this gives a map to $O(3)$, but $SU(2)$ is connected so the image lands in $SO(3)$.

Now consider the orthonormal basis for $\mathfrak{su}_2$ given by $$ e_1 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, e_2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, e_3= \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, $$ let $$ x = \begin{pmatrix} a & -\overline{b} \\ b & \overline{a}\end{pmatrix},$$ and write down the adjoint action of $x$ on $e_1,e_2,e_3$. For instance you get $$ \begin{align} xe_1x^{-1} & = \begin{pmatrix} i(|a|^2 - |b|^2) & 2ia\overline{b} \\ 2i\overline{a} b & -i(|a|^2 - |b|^2) \end{pmatrix} \\ & = (|a|^2 - |b|^2)e_1 + i(a\overline{b} - \overline{a}b)e_2 + (a \overline{b} + \overline{a}b)e_3. \end{align}$$ This gives you the first column of the matrix representation conjugation by $x$. I'll leave the others to you. But this way you can see where the formulas come from.

This gives exactly David Speyer's answer (possibly modulo some re-ordering of the basis). His four real numbers $a,b,c,d$ would correspond to my complex numbers $a,b$ via $$a_{mine} = a + i b, \quad b_{mine} = c + id.$$