Matrix expression for elements of $SO(3)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:25:12Z http://mathoverflow.net/feeds/question/70154 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70154/matrix-expression-for-elements-of-so3 Matrix expression for elements of $SO(3)$ mathstudent 2011-07-12T17:33:39Z 2011-07-15T08:21:09Z <p>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.</p> http://mathoverflow.net/questions/70154/matrix-expression-for-elements-of-so3/70160#70160 Answer by David Speyer for Matrix expression for elements of $SO(3)$ David Speyer 2011-07-12T18:10:31Z 2011-07-15T08:21:09Z <p>Here is the standard <a href="http://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation" rel="nofollow">quaternion</a> answer: Given $(a,b,c,d)$ such that $a^2+b^2+c^2+d^2=1$, the matrix <code>$$\begin{pmatrix} a^2+b^2-c^2-d^2&amp;2bc-2ad &amp;2bd+2ac \\ 2bc+2ad &amp;a^2-b^2+c^2-d^2&amp;2cd-2ab \\ 2bd-2ac &amp;2cd+2ab &amp;a^2-b^2-c^2+d^2\\ \end{pmatrix}$$</code> 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.</p> http://mathoverflow.net/questions/70154/matrix-expression-for-elements-of-so3/70161#70161 Answer by Will Jagy for Matrix expression for elements of $SO(3)$ Will Jagy 2011-07-12T18:24:43Z 2011-07-12T18:55:41Z <p>To give something explicit in sine and cosine,</p> <p>$$<br> \left( \begin{array}{ccc} \cos\theta \cos\psi &amp; -\cos\phi \sin\psi + \sin\phi \sin\theta \cos\psi &amp; \sin\phi \sin\psi + \cos\phi \sin\theta \cos\psi \\ \cos\theta \sin\psi &amp; \cos\phi \cos\psi + \sin\phi \sin\theta \sin\psi &amp; -\sin\phi \cos\psi + \cos\phi \sin\theta \sin\psi \\ -\sin\theta &amp; \sin\phi \cos\theta &amp; \cos\phi \cos\theta \end{array} \right)<br>$$</p> <p>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. </p> http://mathoverflow.net/questions/70154/matrix-expression-for-elements-of-so3/70190#70190 Answer by MTS for Matrix expression for elements of $SO(3)$ MTS 2011-07-12T23:51:55Z 2011-07-12T23:51:55Z <p>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 $S^3 \subseteq \mathbb{C}^2$ <code>$$SU(2) = \left\{ \begin{pmatrix} a &amp; -\overline{b} \\ b &amp; \overline{a} \end{pmatrix} : |a|^2 + |b|^2 = 1 \right\}$$</code> 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 $SO(3)$.</p> <p>Now consider the orthonormal basis for $\mathfrak{su}_2$ given by <code>$$e_1 = \begin{pmatrix} i &amp; 0 \\ 0 &amp; -i \end{pmatrix}, e_2 = \begin{pmatrix} 0 &amp; 1 \\ -1 &amp; 0 \end{pmatrix}, e_3= \begin{pmatrix} 0 &amp; i \\ i &amp; 0 \end{pmatrix},$$</code> let <code>$$x = \begin{pmatrix} a &amp; -\overline{b} \\ b &amp; \overline{a}\end{pmatrix},$$</code> and write down the adjoint action of $x$ on $e_1,e_2,e_3$. For instance you get <code>\begin{align} xe_1x^{-1} &amp; = \begin{pmatrix} i(|a|^2 - |b|^2) &amp; 2ia\overline{b} \\ 2i\overline{a} b &amp; -i(|a|^2 - |b|^2) \end{pmatrix} \\ &amp; = (|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}</code> 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.</p> <p>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 $$a_{mine} = a + i b, \quad b_{mine} = c + id.$$</p>