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I'm reading the Oersted Medal Lecture by David Hestenes to improve my understanding of Geometric Algebra and its applications in Physics. I understand he does not start from a mathematical "clean slate", but I don't care for that. I want to understand what he's saying and what I can do with this geometric algebra.

On page 10 he introduces the unit bivector i. I understand (I think) what unit vectors are: multiply by a scalar and get a scaled directional line. But a bivector is a(n oriented) parallellogram (plane). So if I multiply the unit bivector i with a scalar, I get a scaled parallellogram?

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    If someone would be so kind to tag this with a new tag `geometric-algebra`, I'd appreciate it.2012-07-30
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    Using **i** in (quantum) physics for anything else than $\sqrt{-1}$, should be forbidden by law!2012-07-31
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    Usage of $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ for quaternions (with $\mathbf{i}^2=\mathbf{j}^2=\mathbf{k}^2=\mathbf{ijk}=-1$) is quite common, and given that the subset of quaternions of the form $a+b\mathbf{i}$ is isomorphic to the set of complex numbers, with the quaternion $\mathbf{i}$ mapping to the imaginary unit $\mathrm{i}$, the quaternions can be considered an extension of the complexes, thus using the same symbol makes sense. Now IIRC in 3D geometric algebra the bivectors plus scalars are isomorphic to the quaternions, thus i,j,k for bivectors seems somewhat justified.2012-07-31
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    @celtschk OK. But Hestenes maintains quaternions are just an aspect/transformation/... of the more general geometric algebra (at least in sofar they are used in Physics). I can't deduce any geometric meaning from quaternions either, so although insightful, it doesn't help me much (probably why you made it a comment anyways `;-)`).2012-07-31
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    Actually my comment was an answer to the comment by @draks – basically that in some sense Hestenes *did* use that symbol for the imaginary unit (of course it is definitely *a* square root of $-1$ in the geometric algebra, because its square is $-1$). BTW, I think a bivector is better imagined as a small circle than as a parallelogram because it is invariant under rotations in its own plane.2012-07-31
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    @celtschk OK I agree, in this case. But when I scroll down and look at the indices of $\gamma$...2012-07-31
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    @draks that is not geometric algebra. That's Dirac algebra which only makes sense in Dirac theory. Geometric algebra promises (I'm still learning) to be applicable/useful everywhere, not just in the case of $\gamma$s. And an index $i$ is nothing evil.2012-07-31
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    Ok Ok, I'm guilty: I don't like **i** as index in any case.2012-07-31

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