The Transwedge Product
5 comments
·May 23, 2025senderista
elengyel
I don't think you're missing anything. One of the goals of this article is to demonstrate how geometric algebra is really just another part of exterior algebra anyway. However, performing transformations in spacetime with the geometric product might make some things easier to understand, but I'm not making any strong claims about that. This article contains a small glimpse of what I'm talking about: https://terathon.com/blog/relativistic-quaternions.html
hamish_todd
Since Geometric Algebra == Clifford Algebra, Michael Atiyah, Roger Penrose, and Paul Dirac disagree with you.
senderista
If you're referring to spinors, those can be formalized without Clifford algebras (e.g. as fiber bundles). Can you show me any papers from these scientists that explicitly reference Clifford algebras?
null
I haven't seen any reason to believe that geometric algebra offers any real advantages for physics over ordinary exterior algebra and differential forms, and almost all theoretical physicists seem to agree with me. Am I missing something?