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In geometric algebra, for deriving the rotations formula in $ \mathbb R^3 $ I see the following steps ($ \mathbf i $ is the plane of rotation, $ \mathbf u $ is the vector to be rotated, with components $ \mathbf u_{\parallel} $ in the plane of rotation and $ \mathbf u_{\bot} $ perpendicular to it, $ \mathbf v $ is the resultant vector):

$$ \mathbf v = \mathbf u_{\parallel} e^{\mathbf i\theta} + \mathbf u_{\bot} $$ $$ = \mathbf u_{\parallel} e^{{\mathbf i\theta}/{2}}e^{{\mathbf i\theta}/{2}} + \mathbf u_{\bot}e^{{-\mathbf i\theta}/{2}}e^{{\mathbf i\theta}/{2}} $$ $$ = e^{{-\mathbf i\theta}/{2}}\mathbf u_{\parallel} e^{{\mathbf i\theta}/{2}} + e^{{-\mathbf i\theta}/{2}}\mathbf u_{\bot}e^{{\mathbf i\theta}/{2}} $$ $$ = e^{{-\mathbf i\theta}/{2}}\mathbf u e^{{\mathbf i\theta}/{2}} $$

I am unable to understand step (3).The hint given says that the geometric product of two orthogonal vectors is anti-commutative, which is clear enough, but how does that extend to this case where we have a product of a vector and a multivector ($e^{{\mathbf i\theta}/{2}}$)?

edit: I was able to verify the equality by taking $ \mathbf i = e_1e_2 $ and $ \mathbf u_{\parallel} = ae_1 + be_2 $, but stll not able to understand it geometrically.

user997712
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This can be seen by breaking the exponential down into trig functions.

$$u_\parallel e^{i\theta/2} = u_\parallel (\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})$$

Since $u_\parallel$ lies entirely in the plane $i$, it should be clear that $i u_\parallel = -u_\parallel i$. Either way, this is a vector orthogonal to $u_\parallel$ lying in the plane $i$, but the order of multiplication determines whether you get a particular vector or its negative (based on what wedge product of $u_\parallel$ and $i u_\parallel$ yields a bivector with the same orientation as $i$).

When you do that, you get the following:

$$ (\cos \frac{\theta}{2} - i \sin \frac{\theta}{2})u_\parallel = e^{-i \theta/2}u_\parallel$$

Welcome to the wide world of geometric algebra; I hope you continue to find answers to your questions here.

Muphrid
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  • Thank you for the answer. I now realize that i have not covered multiplication of vectors with bivectors (or pseudoscalars), that is why it doesn't make sense to me yet. The book only expects verification of the step, not the proof. – user997712 Jun 16 '13 at 19:57
  • I see. Well, you can imagine that the bivector $i$ is the product of two orthogonal vectors, $i = u_\parallel w$ for some $w$, and then you can use associativity of the geometric product. What book are you using, by the way? – Muphrid Jun 16 '13 at 20:11
  • Yes, that's what i ended up doing for the verification. I am using Linear and Geometric Algebra by Alan Macdonald and finding it really good. – user997712 Jun 16 '13 at 20:14
  • Indeed, Macdonald's books are a very accessible first introduction to the subject. Good luck! – Muphrid Jun 16 '13 at 20:37