# What's the meaning of the unit bivector i?

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. –  rubenvb Jul 30 '12 at 16:54
Using i in (quantum) physics for anything else than $\sqrt{-1}$, should be forbidden by law! –  draks ... Jul 31 '12 at 8:04
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. –  celtschk Jul 31 '12 at 14:48
@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 ;-)). –  rubenvb Jul 31 '12 at 14:56
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. –  celtschk Jul 31 '12 at 15:14

The bivector "i" is the Hestenes thing which corresponds to what is normally called dx wedge dy. This is an antisymmetric 2-tensor with components 1 and -1 at the x,y and y,x positions, and it is represented by a little area square in the x-y plane. This is a differential form.

What Hestenes does to produce geometric algebra is to multiply every vector index by gamma matrices, and these gamma matrices make an antisymmetric algebra with the following law:

$$\gamma_i \gamma_j + \gamma_j \gamma_i = 2g_{ij}$$

Where g is the metric tensor, usually $g_{ij}=\delta_{ij}$ for 3 dimensional (or Euclidean) geometric algebra. In 3d, the $\gamma$ matrices have a standard representation with the Pauli matrices.

This means that when Hestenes is talking about the unit vector, he isn't thinking of it like a unit vector, but as what other people would call the slash of the unit vector, which is the $gamma$ matrices dotted with this unit vector, or in this case, just $\gamma_x$. The $\gamma$ matrices square to 1, and anticommute. So when he multiplies two unit vectors in geometric algebra, he gets

$$\gamma_x \gamma_y = {1\over 2} (\gamma_x \gamma_y - \gamma_y\gamma_x)=\sigma_{xy}$$

Where the last equality is a definition: $\sigma_{ij}$ is defined as the antisymmetric product of $\gamma$ matrices:

$$2\sigma_{ij} = \gamma_i \gamma_j - \gamma_j \gamma_i$$

(most authors omit the factor of 2 on the left hand side). This definition of $\sigma$ is redundant off the diagonal, by antisymmetry of $\gamma$ multiplication. It's just defined this way to make sure that $\sigma_{ii}$ is zero, not 1.

So for Hestense, the unit two-form $dx\wedge dy$ is contracted with sigma, so it is just $\sigma_{xy}$. This means that it has the property that it squares to -1. You can see this by squaring and anticommuting the $\gamma$ matrices.

Hestene's approach hides the gamma matrices. The upside is that this gives you a quick intuition for $\gamma$ algebra without cumbersome notation for the $\gamma$ matrices or explicit matrix representations. The downside is that you don't learn symmetric tensors, and the notation is wildly different from what everyone else uses, with insufficient payoff (as far as I can see) to make up for it. Keeping the gammas explicit makes GA readable to me, but Hestenes prefers to hide them. It's no big deal to translate.

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It seems you avidly hate Hestenes' formulation. I get the parallel with $\gamma$ algebra the way you see it (I first came into contact with geometric algebra when studying the Dirac equation), but in the pdf referenced, he's not talking about $\gamma$ algebra, he's talking Euclidean geometry, where stuff like $\gamma$-matrices (he's not even talking about matrices) are not even relevant. You're giving a non-geometric interpretation IMHO, which is not quite what I'm after. I'm after the interpretation Hestenes describes. –  rubenvb Jul 31 '12 at 9:07
@rubenvb: I don't hate it (although I was annoyed at having to waste an hour before I understood it was gamma algebra), I just know it under a different name, so I get annoyed when I see totally new notation for familiar objects. It's not a parallel with gamma algebra, it's the same algebra! He just makes you learn it early under a different name. –  Ron Maimon Jul 31 '12 at 23:09

Don't be put off by some comments you will find on the internet, it is difficult for people who have spent years learning many formalisms to think of abandoning them; in your case you can acquire in a relatively short time the same effect of all those years (for not just physics but maths, computing, graphics, robotics, it is universal).

See this for GA baked in to a university course at your level.

(Btw, it is better to think of a bivector as a patch of area with a value equal to...)

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A bivector operating on a vector rotates the vector.

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OK. This is part of the meaning I get. The bivector as an operator is a rotor. But the bivector as basis element of the algebra should have a meaning on its own. Is my plane interpretation correct? –  rubenvb Jul 31 '12 at 11:48
This is also in Dirac algebra--- "acting on" means commutators, and the generators of rotations/Lorentz-boosts are the sigma-matrices, or Hestene's bivectors. There really is nothing more to his formalism than gamma algebra under a different name. –  Ron Maimon Jul 31 '12 at 23:10
@rubenvb: I don't have an account here and I don't see a way to comment on an answer without creating one (which I don't want to do) so I may be breaching protocol. Anyway, it sounds as if you're asking if bivectors can be used as a basis and the answer is yes. And yes, you can think of a bivector as an oriented plane. In my introductory calculus-based physics course I introduce students to bivectors as a more sensible alternative to vector cross products. –  Joe Aug 1 '12 at 3:40

The unit bivector represents the 2D subspaces. In 2D Euclidean GA, there are 4 coordinates:

• 1 scalar coordinate
• 2 vector coordinates
• 1 bivector coordinate

A "vector" is then (s, e1, e2, e1^e2).

The unit bivector is frequently drawn as a parallelogram, but that's just a visualization aid. It conceptually more closely resembles a directed area where the sign indicates direction.

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