# Commutator SO(3) - what am I doing wrong?

1. Jul 13, 2012

### BWV

reading that the commutator of rotations on two orthogonal axes is i * the rotation matrix for the third axis

but if I commute this

\begin{pmatrix}\mathrm{cos}\left( \theta\right) & -\mathrm{sin}\left( \theta\right) & 0\cr \mathrm{sin}\left( \theta\right) & \mathrm{cos}\left( \theta\right) & 0\cr 0 & 0 & 1\end{pmatrix}

with this

\begin{pmatrix}1 & 0 & 0\cr 0 & \mathrm{cos}\left( \theta\right) & -\mathrm{sin}\left( \theta\right) \cr 0 & \mathrm{sin}\left( \theta\right) & \mathrm{cos}\left( \theta\right) \end{pmatrix}

I get (using innerproduct function in Maxima)
\begin{pmatrix}0 & \mathrm{cos}\left( \theta\right) \,\mathrm{sin}\left( \theta\right) -\mathrm{sin}\left( \theta\right) & -{\mathrm{sin}\left( \theta\right) }^{2}\cr \mathrm{cos}\left( \theta\right) \,\mathrm{sin}\left( \theta\right) -\mathrm{sin}\left( \theta\right) & 0 & \mathrm{cos}\left( \theta\right) \,\mathrm{sin}\left( \theta\right) -\mathrm{sin}\left( \theta\right) \cr {\mathrm{sin}\left( \theta\right) }^{2} & \mathrm{cos}\left( \theta\right) \,\mathrm{sin}\left( \theta\right) -\mathrm{sin}\left( \theta\right) & 0\end{pmatrix}

which has a determinant of zero and therefore not part of SO(3)

obviously I am not getting something, but don't see it. Any help is much appreciated

2. Jul 13, 2012

### qbert

you are confusing the Lie group and the Lie algebra. The commutation
relation applies to the Lie algebra (ie the generators of the Lie group).

3. Jul 15, 2012

ah, thanks.