Talk:Geometric algebra

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The lead currently contains a statement "... pseudovector quantities of vector calculus normally defined using a cross product, such as ... the magnetic field". I am not familiar with the magnetic field "normally" being defined in this way, although I'm aware if the Biot–Savart law, which is not only a cross product. The point is that few readers will find this helpful; examples such as this are meant to be familiar. Maybe this particular example (magnetic field) might be better removed? —Quondum 02:04, 13 March 2024 (UTC)Reply

The magnetic field B is the curl of the magnetic potential A, which is the cross product with the del operator: ${\displaystyle \mathbf {B} =\nabla \times \mathbf {A} }$. BrtSaw (talk) 04:53, 31 August 2024 (UTC)Reply

The cross product is not defined on operators in general, and the "del operator" is ill-defined in general. The 'del-cross' notation for the curl operator is merely a handy mnemonic. (The nabla/del symbol is used with a well-defined meaning in geometric calculus, but its similar combinations with binary operators are again simply mnemonic notations.) —Quondum 14:45, 31 August 2024 (UTC)Reply

My understanding is that there are two different communities who study the same object but with different notation. Is this sufficient for the same object to have two different pages?

I'm willing to believe that there is some principled reason for this, but I would like to see it spelled out explicitly, even if just in the talk page. Cooljeff3000 (talk) 16:34, 5 January 2025 (UTC)Reply

I suppose because it is and isn't the same subject at the same time. Yes GAs are specific examples of CAs but the focus is on low dimensional applications over the reals, how much of what you see in this article is in the CA page? So no fork, right? Selfstudier (talk) 17:20, 5 January 2025 (UTC)Reply

The Pfaffian expansion looks interesting, but as far as I can tell, it's based on two papers (by a fellow named Wilmot) that received almost no citations in the 37 years since publication and that essentially appear to have been a thesis of some sort. Of course, this doesn't mean it's wrong --- and it's precisely the sort of expansion I had been looking for, which makes me grateful for Wilmot's effort. However, it probably shouldn't be presented as if it is some standard textbook result. Also, the notation is homegrown. In the form it is written in this wikipedia article, the formula makes no sense. A Pfaffian takes a skew-symmetric matrix as an argument (it's just sqrt(det(M) when the matrix is real), but the argument in the formula is not a matrix. It's a list of scalars. There's some sort of missing step or notation from the papers that didn't make its way into the wikipedia article. Also, the formula and surrounding sentences seem to be lifted directly from a recent paper "Construction of exceptional Lie algebra G2 and non-associative algebras using Clifford algebra" by that same Wilmot. At the very least, the meaning of the expression should be clarified --- in particular, what matrix the Pfaffian has as an argument. It also probably would be a good idea to caveat this in some way since it seems to be one person's project and, as far as I can tell, nobody else is using it. Again, this doesn't mean it's wrong --- it could be a brilliant result that unjustifiably has languished in obscurity --- but it does raise concerns about whether it has been thoroughly vetted. Unfortunately, I lack access to the original papers and cannot work through the details myself --- and the later paper simply refers to those other papers as canon rather than explaining the notation in a self-contained fashion. 73.186.196.76 (talk) 19:29, 9 June 2025 (UTC)Reply

Since this revision in 2013 the section on the geometric product has contained the following derivation step:

the symmetric product can be written as

${\displaystyle {\frac {1}{2}}(ab+ba)={\frac {1}{2}}\left((a+b)^{2}-a^{2}-b^{2}\right)=a\cdot b.}$

I'm not completely confident that I understand the argument that the editor had in mind here, but what does seem clear to me is that this step assumes that ${\displaystyle ab=ba}$, which is not guaranteed by the axioms given further up the page. Theoh (talk) 11:14, 29 September 2025 (UTC)Reply

I'm seeing that this must be related to the polarization identity. I had overlooked the fact that ${\displaystyle (a+b)^{2}=a^{2}+ab+ba+b^{2}}$ so I'm happy with that part now. Does anyone have an opinion on whether the conclusion ${\displaystyle {\frac {1}{2}}((a+b)^{2}-a^{2}-b^{2})=a\cdot b}$ should be attributed to the polarization identity? Theoh (talk) 21:41, 29 September 2025 (UTC)Reply

This illustration is very hard to decipher. I think it is supposed to look like a three-dimensional drawing, with the blue "square" representing a plane orthogonal to the purple vector m, but the angles are just all over the place. Although I enjoy good illustrations, I am not an illustrator, so I can't offer a better alternative, unfortunately. Lasse Hillerøe Petersen (talk) 21:49, 24 November 2025 (UTC)Reply