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I've read a number of introductions to Clifford algebras, and I'm always left with the question of what the geometric product is supposed to mean. The wedge pro
by aesthesia 7y ago
I've read a number of introductions to Clifford algebras, and I'm always left with the question of what the geometric product is supposed to mean. The wedge product and dot product are easy to understand and have obvious interpretations. But other than being a gadget from which you can extract these other products, I don't see what the geometric product is for, or why it should be the primary object of consideration.
- edflsafoiewq 7y agoDespite the name the motivation for the geometric product is principally algebraic, ie. it's useful for doing algebraic manipulation. It does not, AFAIK, possess any geometric meaning outside of special cases. (It's "geometric" in the sense it doesn't depend on a choice of basis I guess.)
- DreamScatter 7y agoActually, it has a lot to do with geometry, in fact all of geometric algebra can be constructed from the geometric product, which serves as a foundation for not just algebra but also geometry. It encodes various geometric properties.
- edflsafoiewq 7y agoYes, it "encodes" them and they can be "constructed" from it, but it does not have a direct geometric interpretation in the way, say, the cross product does.
- orbots 7y agoI've come to see it more as the inverse of a division operator. Like a quaternion can be defined as the ratio of two vectors q = a/b. then qb = a and qc will rotate c the same amount as needed to take b into a. dual quaternions have a similar motivating derivation.
- ajkjk 7y agoI agree with this. Most of the benefits of GA are from embracing the wedge product, not the geometric product. However, I do think there is something there. There is a concept of vector inversion that it makes possible. I have been trying to crystallize this for a long time, though.