Rotor of a rotor in Geometric Algebra I was reading Geometric Algebra for Physicists, by Doran and Lasenby, and, in section 5.5.2, they calculate the Thomas Precession.
However, at a certain point, they have the exponential of an exponential and lower it, can someone explain how is that done in detail?
I show the calculations here:
$$
n=\mathrm{e}^{-\omega t I \sigma_{3}} \boldsymbol{\sigma}_{2}=R_{\omega} \boldsymbol{\sigma}_{2} \tilde{R}_{\omega} \tag{5.150}
$$
where $R_{\omega}=\exp \left(-\omega t I \sigma_{3} / 2\right) .$ We now have
$$
\mathrm{e}^{\alpha n / 2}=\exp \left(\alpha R_{\omega} \sigma_{2} \tilde{R}_{\omega} / 2\right) \overset{??}{=}  R_{\omega} R_{\alpha} \tilde{R}_{\omega} \tag{5.151}
$$
where
$$
R_{\alpha}=\exp \left(\alpha \sigma_{2} / 2\right)
$$
How does one perform the last step in (5.151)?
 A: The third term in Eq (5.151) stems from two definitions:

*

*The definition of the exponential operator on the multivector $A$ is,
$$
\exp\left(A\right)\triangleq\sum_{j=0}^\infty\frac{A^j}{j!}=1+A+\frac12A^2+\frac16A^3+\cdots\tag{1}.
$$

*Rotors obey $R_\omega \tilde R_\omega=1$.

Thus, if you use Equation (1) with $A=R_\omega\boldsymbol\sigma_2\tilde R_\omega$ and point #2, you will find,
$$
\exp\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)=R_\omega \tilde R_\omega+R_\omega\boldsymbol\sigma_2\tilde R_\omega+\frac{1}{2}\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)^2+\frac{1}{6}\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)^3+\cdots\tag{2}
$$
For the higher order terms, you can expand out the products to find that each of them will obey,
$$
\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)^j=\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)\cdots\left(R_\omega\boldsymbol\sigma_2\tilde R_\omega\right)=R_\omega\boldsymbol\sigma_2^j\tilde R_\omega\tag{3}
$$
due to repeated use of point #2.
Hence, you can combine Equations (2) and (3) here along with the associative property of the geometric product to write Eq (5.151) as,
\begin{align}
\exp\left(\alpha R_\omega\boldsymbol\sigma_2\tilde R_\omega/2\right)&=R_\omega\left(1+\alpha\boldsymbol\sigma_2/2+\frac{1}{2}\left(\alpha\boldsymbol\sigma_2/2\right)^2+\frac{1}{6}\left(\alpha\boldsymbol\sigma_2/2\right)^3+\cdots\right)\tilde R_\omega \\
&=R_\omega\mathrm{e}^{\alpha\boldsymbol\sigma_2/2}\tilde R_\omega \\
&= R_\omega R_\alpha \tilde R_\omega 
\end{align}
