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I'm not a physicist and don't understand their perspective on these matters. I say it is mathematically unsound not because it isn't rigorous. I'm not expecti
by yequalsx 12y ago
I'm not a physicist and don't understand their perspective on these matters. I say it is mathematically unsound not because it isn't rigorous. I'm not expecting rigor.
However, I do expect well founded definitions and operations. To define an operation on two vectors and not point out that it isn't a vector while stating that the geometric algebra is different "from all other associative algebras" is misleading. Associative algebras are closed under the operations they have. So something ought to be said about the peculiarity of talking about associative algebras while not having an operation that is closed.
The only paper I've read from Hestenes is the paper in question and the perspective I have is that of a mathematician. My comment only dealt with this paper and from my perspective things were muddled a bit. He talks about confusion of physics students in graduate school on the topic of vectors. I'm assuming such students will eventually have to deal with tensor products and the exterior algebra when they study general relativity. Hence my belief that one should talk about the larger algebraic structure from the get go rather than glossing over the fact that the geometric product of a two vectors is not a vector and the mystery of squaring a wedge product.
Scalars + bivectors are isomorphic to the complex numbers provided the base field is R and the isomorphism is as vector spaces not as algebras or rings.
- jacobolus 12y agoScalars + bivectors (obviously talking about those generated by geometric products of vectors in R2) are isomorphic to the complex numbers, as a ring.
- yequalsx 12y agoI didn't read enough of the paper to see how bivector multiplication is defined. I'm assuming everything is in the exterior algebra and thus made my comment with that assumption.
- dullcrisp 12y agoI had the same reaction as you, but then I skipped to the end where he said that mathematicians call this a Clifford algebra[0], and with that it all made sense. I agree with jacobolus that mathematicians are clearly not the target audience of this paper, so the mathematical soundless is left as an exercise to the reader if they are so inclined. [0] http://en.wikipedia.org/wiki/Clifford_algebra#Introduction_and_basic_properties http://en.wikipedia.org/wiki/Clifford_algebra#Introduction_a...