I've been (slowly) working my way through a book on geometric algebra and have found one part particularly confusing.

I can understand the equation $e_1e_2=\exp(e_1e_2 \pi/2)$

Where the substiution $e_1e_2=i$ has been made ($e_1$ and $e_2$ are orthongonal unit vectors and $i^2=-1$)

But I cannot understand understand the following ($R_\phi$,$R_\theta$ are rotors):

If $R_\phi=\exp(-e_1e_2\phi/2$) then:

$R_\theta=\exp(-R_\phi e_2e_3R_\phi^\dagger\theta/2)=-R_\phi\exp(e_2e_3\theta/2)R^\dagger_\phi$

How are the rotors removed from the exponent?


1 Answer 1


Expand the expression in terms of sines and cosines.

$$R_\theta = \cos \frac{\theta}{2} - R_\phi e_{23} R_\phi^\dagger \sin \frac{\theta}{2}$$

Since $R_\phi R_\phi^\dagger = 1$, you can rewrite this as

$$R_\theta =R_\phi \cos \frac{\theta}{2} R_\phi^\dagger - R_\phi \left(e_{23} \sin \frac{\theta}{2} \right)R_\phi^\dagger $$

It should be clear, at this point, that you can gather the sine and cosine back into an exponential, but now the rotors $R_\phi, R_\phi^\dagger$ are on the outside.

Edit: this is a common result when dealing with rotations. For instance, this result tells you that a series of rotations about a rigid body's intrinsic axes can be converted into rotations about an extrinsic, fixed set of axes, in reverse order.

  • $\begingroup$ Thanks Murphid I hadn't realized this expansion was valid. Given that the replacement $e_1e_2=\exp(e_1e_2\pi/2)$ had already been made. $\endgroup$
    – Aaron
    Dec 25, 2014 at 0:19

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