# Angular velocity is $\dot{g}$ carried to the identity element of the group

I was reading the example below from Arnolds book I can't really understand why the angular velocity is $$\dot{g}$$ carried to the identity element of the group. I would appreciate if someone who understand it could explain.

## 2 Answers

\begin{aligned}G^{T}\, G=I_{3}\\ \Rightarrow \\ \dfrac {dG^{T}}{dt}\,G+G^{T}\, \dfrac {dG}{dt}=0\\ \text{with}\\ \dfrac {dG}{dt}=\sum \dfrac {\partial G}{\partial g_{i}}\ \dfrac {dg_{i}}{dt}\\ \Rightarrow \\ \begin{pmatrix} \tilde \omega \end{pmatrix}^{T}+\begin{pmatrix} \tilde \omega \end{pmatrix}=0\\ \begin{pmatrix} \tilde \omega \end{pmatrix}=G^{T}\,\dfrac {dG}{dt}=G^T\,\sum \dfrac {\partial G}{\partial g_{i}}\ \dfrac {dg_{i}}{dt}\end{aligned}

Where

$$G$$ orthonormal matrix

$$\tilde \omega$$ skew matrix. thus $$\tilde{\omega}^T=-\tilde{\omega}$$

Given a group element $$g(x)$$ the quantity $$\omega= g^{-1} \frac{\partial g}{\partial x^\mu}dx^\mu$$ is the associated Maurer-Cartan form. Whenever you have a parameter-dependent group element any calculus associated with it is much simpler if you use this combination. It's particulaly useful when yo want compute the angular velocity of a rigid body in terms of the time derivatives of the Euler angles. The reason for "carrying back to the origin" is that the resultant matrix is an infinitesimal element of the group -- i.e an element of the Lie algebra.