Talk:Clifford algebra

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Mathematics rating: B Class High Priority  Field: Algebra

The nice thing about this page ... is the way of doing the commutative diagram, which is cool.

Otherwise, this is really not up to 2004 standards of WP exposition.

Charles Matthews 16:40, 20 Sep 2004 (UTC)

Okay, I completely rewrote the article. I was completely unsatisfied with the earlier version. My apologies to those that I've annoyed by switching the sign convention. There is still a lot I want to add but the article is getting kind of long as it is. Maybe I can find some better places for some of it. I've already started a new article at classification of Clifford algebras for those issues. Comments on the new version are welcome. -- Fropuff 15:41, 2004 Oct 26 (UTC)

Contents

[edit] Sign convention

Well okay, someone has gone through and switched the sign convention back to v2 = +Q(v). See my comment at Wikipedia talk:WikiProject Mathematics/Conventions -- Fropuff 15:18, 2005 Apr 26 (UTC)

In physics and algebra the v2 = +Q(v) sign convention is almost the only one used. In differential geometry the + and - conventions are about equally common, with the - convention favored mainly by those working in index theory. By the way, the book by Lawson and Michelsohn should not be used as a reference for spinors: it contains some rather bizarre definitions that are different from everyone elses, and there are several errors in the theorems they state. R.e.b. 16:43, 26 Apr 2005 (UTC)

I'm willing to stick with the + sign convention. It is certainly more common in physics. Admittedly, I learned most of what I know about Clifford algebras from a index theorist, so my point of view may be rather narrow.

I knew the definitions in Lawson and Michelsohn were a little different, but I was not aware of the errors. That's very interesting. I may have to rethink what I thought I knew. Thanks for pointing it out. -- Fropuff 16:56, 2005 Apr 26 (UTC)

[edit] Hypercomplex number rewrite, mention of Clifford algebras

Hello, after proposing a rewrite of the hypercomplex number a couple of weeks ago (see Talk:Hypercomplex number) and a little bit of asking around, I've just replaced the article with a rewrite. There was a brief mention about Clifford algebras there before ("The Clifford algebras are another family of hypercomplex numbers." - nothing else), which I expanded a little bit. However, I am not versed with Clifford algebra, and would ask if anyone could have a look at the new hypercomplex number article version and see whether it can be improved. I appreciate any comment or help. Thanks, Jens Koeplinger 22:22, 31 July 2006 (UTC)

Update: A nice and short high-level overview or introduction into Clifford algebras was just given here: http://en.wikipedia.org/w/index.php?title=Hypercomplex_number&oldid=119138089 - Is there any chance to incorporate a similar less-technical overview into the current article, and bring the more rigorous mathematical treatment later? This would follow an approach where articles begin on a high-level and in general terms, and become increasingly more technical the further one reads on. Thanks, Koeplinger 19:58, 31 March 2007 (UTC)

[edit] small notation inconsistency

The section "Spinors" says "suppose that p + q = 2n is even" (discussing spaces with signature (p,q)). I suggest that this definition of n is uncommon: it is much more common to define n = p + q, this being the dimension of the space. Thus, in this same article, in the section "Basis and dimension" we have "If the dimension of V is n...", and just below, in the section "Examples": "...where n = p + q is the dimension of the vector space...". I suggest changing the section "Spinors" to bring this in line with the above, any possible objections? 131.111.8.96 22:18, 19 September 2006 (UTC)

[edit] Confusing Intro

Total noob here. In the intro it is stipulated that v^2 = Q(v). However, Q is a quadratic form -- i.e. it maps VxV to the base field K -- and the algebra's multiplication operation maps VxV to V. In other words, the left side v^2 of the above expression is an element of the algebra, while the right hand side is an element of the base field. Some clarification would be nice. (The same can be said about the subsequent expression uv + vu = 2B(u,v).)

The algebra operation maps VxV not to V but to Cl(V) which is an algebra containing both V and K. It would be better to write v2 = Q(v)1 where 1 is the identity element of the algebra (1 is not an element of V nor is Q(v)1). Actually the base field K can be embedded in the algebra via k\mapsto k\cdot 1 so this is unambiguous. -- Fropuff 05:23, 4 May 2007 (UTC)

The reply to this comment is certainly correct. The confusion is very common. Cl(V) includes both V and the underlying field of V, or at least there are subspaces of Cl(v) naturally isomorphic to both V and the field. This goes for Grassmann algebras too. This confusion blocked me for a long time. I attempt to clarify this on my web site but I think I have not succeeded. I hesitate to modify the article for I don't know how to describe this in the context of the article. I have a dark suspicion that text to enlighten a mathematician will confuse a physicist and vice-versa. NormHardy 04:57, 29 July 2007 (UTC)