Talk:Clifford algebra

Early thread

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)

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)

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Char 2

It would improve the readability of this article if all mention of characteristic two would be banned to it's own section. I think.

Section on Clifford algebras and rigid body kinematics

I would like to add a section after the Clifford algebra#Examples: real and complex Clifford algebras that constructs planar quaternions, quaternions and dual quaternions used to represent planar displacements, spatial rotations and spatial displacements of rigid bodies as Clifford algebras. I will use vectors in R³ with the usual Euclidean metric and the negative sign for the Clifford multiplication, and select the even sub algebra to obtain quaternions. A similar construction, but using a degenerate metric yields quaternions that represent planar displacements, and finally the same construction on R⁴ with a degenerate metric yields dual quaternions. I hope this is viewed as a useful contribution to this article.Prof McCarthy (talk) 03:29, 16 August 2011 (UTC)[reply]

I just added the section on quaternions and propose to added similar sections for dual quaternions and planar quaternions under the same heading.Prof McCarthy (talk) 05:57, 16 August 2011 (UTC)[reply]

The material on rigid body kinematics might better appear at Geometric algebra, or a subarticle of that, which deals with some of the geometric interpretations and uses to which Clifford algebras can be put. The convention there is to use a positive sign for a Euclidean metric, a negative one for additional hyperbolic directions (hence my edit yesterday, for which I apologise, not having seen you were proposing to use the opposite convention). Jheald (talk) 11:15, 17 August 2011 (UTC)[reply]

Not a problem at all, I appreciate the help. The goal here was to show how Euclidean geometry and the construction of a Clifford algebra yield quaternions and dual quaternions. The information on rigid body kinematics appears in the articles on screw theory and dual quaternions. I understand the issue that the current version of this article does not allow for a degenerate quadratic form and in fact implies that they are not used. I am not enthusiastic about moving this material to the article on Geometric algebra, but if that is preferred I will do it. Prof McCarthy (talk) 17:12, 17 August 2011 (UTC)[reply]

One more thing, part of the reason for adding this explicit Clifford algebra based construction was a controversy that can be found on the discussion page of the article on dual quaternions, in which someone's web-page proposed that the dual unit may not commute with the quaternion units. This made it into the Wikipedia article and confused several editors, and I am sure many readers as well. Prof McCarthy (talk) 17:18, 17 August 2011 (UTC)[reply]

Fair enough. I'd be interested in your assessment of the Conformal Geometric Algebra proposed by Hestenes and various others. This essentially adds a second non-spatial direction to the dual quaternions, in a way that appears to generally neaten up the system, allowing translations to be represented using the same "sandwich" form as rotations (and rotations about a point), also happening to integrate in the whole of inversive geometry and the "screws by reflection" approach currently mentioned in passing in the screw theory article; and as a side bonus giving very tidy product-of-blades algebraic forms for spheres and circles and their intersections. It's currently on my slate of articles to aim to work up (eventually), though I'm a bit heavily committed in real life just at the moment. But if you're working on the screw theory article, it probably be worth examining there, particularly in the context of comparing and contrasting with the dual quaternion approach. Jheald (talk) 10:18, 18 August 2011 (UTC)[reply]

Thank you for your patience with this contribution on quaternions and dual quaternions. I have met David Hestenes and several of his colleagues over the years and understand their enthusiasm for Geometric Algebra. They have taken quite a broad view that extends far beyond my range. Regarding Conformal Geometric Algebra, I can say that in the particular case of embedding spatial displacements in four dimensional space such that translations in a hyperplane are approximated by rotations, effectively about axes at infinity, actually works quite well. Furthermore, the Clifford algebra constructed for this system yields Clifford's original biquaternions. However, I am not able to judge the extensions of this idea that Geometric Algebra allows. As for screws by reflection, this is an important tool in the geometry of relative spatial displacements, but I have been slow to update that particular section of the article. I will pay more attention to it. Prof McCarthy (talk) 13:37, 18 August 2011 (UTC)[reply]

Section Properties: Incorrect inclusion of GA assumption?

All the sub-sections of section Properties appear to make the assumption that a Clifford Algebra Cℓ(V,Q) has a unique subspace that may be identified with V. This may be the distinction between a Clifford Algebra and a Geometric algebra (GA) - in the latter a chosen subspace is identified as V. I believe such an identification is an additional assumption of GA and not part of CA, and all the sub-sections only make sense with the explicit identification of a sub-space to be treated as the "vector" subspace. I think that concepts/assumptions from GA have been allowed to confuse the matter in this article on CA. Consequences of dropping this GA assumption include:

Sub-section Relation to the exterior algebra: there is no unique mapping Cℓ(V,Q) → Λ(V).
Sub-section Grading: Without identification of the subspace V, the grade of an element in Cℓ(V,Q) cannot in general be determined. Also, reflection through the origin is defined on V, and this operation changes its effect according to the choice of V. The decomposition into even and odd eigenspaces is accordingly not unique.
The remaining sub-sections Antiautomorphisms and The Clifford scalar product also rely heavily on this assumption, with similar problems.

I am sure this is going to be contested, so I'll illustrate this using a concrete example. If we use Cℓ_0,2(R), we know that if treated as graded (i.e. we identify two elements of the algebra to be a basis for V, these two elements e₁ and e₂ have grade 1, and the element e₁e₂ has grade 2. We know, however, that Cℓ_0,2(R) is isomorphic to quaternions, and the elements i = e₁, j = e₂, k = e₁e₂ are isomorphic under cyclic rotations. Thus the grade of any of these elements is undefined. Another way of thinking of it is that multiplication can be fully defined by a Cayley table of orthogonal basis elements of Cℓ(V,Q), we can always choose n elements from this table that all anticommute and can thus be identified as a basis for V, and that this choice is never unique for dimension higher than for complex or split-complex numbers.

My suggestion is to remove the entire section, as it does not belong in the CA article. An alternative is to replace section Properties with a simple statement to the effect that GA is a CA with the addition of an identified subspace to be V. This distinction between CA and GA should be highlighted in the GA article. Quondum (talk) 09:14, 25 September 2011 (UTC)[reply]

I think that that distinction is WP:OR, unless you can find a source that makes it. As previously discussed elsewhere, I'm pretty sure that the signature of a Clifford algebra is considered an essential property of the algebra, so that Cℓ_2,0(R) and Cℓ_1,1(R) are considered different algebras, even though an isomorphism can be constructed between them. I thought I had looked up and posted some references from the CA literature, identifying the grade structure and signature as part of the defining properties of each algebra; but it seems I never did. Jheald (talk) 10:33, 25 September 2011 (UTC)[reply]

While your point about WP:OR is certainly valid, it seems to me that it applies equally well in the opposite direction relating to the distinction between Clifford algebras. Your interpretation includes a mathematically pointless distinction/limitation at the level of the algebra, and excludes an algebra from the category of a Clifford algebra until you have defined the mapping i and quadratic form Q for that algebra (thus excluding quaternions etc., which are adequately defined and useful, but have no identifiable subspace V, and contradicting the description "They can be thought of as one of the possible generalizations of the complex numbers and quaternions"). If i is significant in the distinction, then not only are Cℓ_2,0(R) and Cℓ_1,1(R) different, but so are an infinite number of varieties of Cℓ_2,0(R) from each other. This does not fit with the description as the "freest" algebra in the introduction. Thus: do you (or anyone else) know of any references that may clarify this, so that the article as it stands is not WP:NOR? Quondum (talk) 12:34, 25 September 2011 (UTC)[reply]

I'll try to remember to look out some references. From what I recall, from what I looked at before, the standard books that survey Clifford algebra in purely algebraic terms, without any geometric discussion, did present the signature as part of the defining properties of each algebra.

I suppose what you choose to put in or not into the definition of an object or structure depends in part upon what you are interested in -- i.e. what sort of propositions you want to draw out about it. One of the sources of interest for the algebraists is the notion of Bott periodicity, or more generally relating some of the characteristic properties of different algebras to each other in turns of their different signatures. For this therefore, you're very much talking about the properties of the algebra of the particular signature as the item of interest; with any isomorphism with a CA of a different signature a secondary property of only incidental interest.

Book definitions of Clifford Algebra, and also of particular Clifford Algebras, tend to start with the definition of a scalar product for the base-space elements, which can then be used to construct the whole alegbra by closure. From that perspective, the signature is where you start from, and therefore part of the basic identification of the particular algebra. Jheald (talk) 15:16, 25 September 2011 (UTC)[reply]

I think that, for the time being (until someone digs up references), we'll have to agree to disagree. I understand how the construction starts from a specific signature, base space etc., but many of the stated resultant properties only make sense if one develops amnesia about the route (signature etc.) used to construct a particular algebra, implying that (probably less intuitive) definitions that do not start with any specific signature or generating vector space will be possible. Generally only mathematicians will care about this nicety, and I'm unfortunately a mathematician at heart but not by training - very frustrating.

You say "what you choose to put in", but this does not make sense as one must have consensus on what one means by mathematical terms. My gut says there must be a mathematical consensus on this, but we do not seem to have anyone really knowledgeable contributing to this discussion.

I suggest any further discussion on the subject on one of our user talk pages (probably mine: it is me championing this). Quondum (talk) 19:32, 25 September 2011 (UTC)[reply]

Just to clarify: by "what you choose to put in", by "you" I was considering a hypothetical mathematician developing a subject, rather than you personally or me or a wikipedian writing an article -- so my point was that "how the mathematicians who have developed the subject have framed their definitions" tends to reflect the propositions they are then going to put those objects to; with CA a particular focus of interest has been the patterns in the relationships between different CAs that are implied by the differences in their signatures, so generally I think the signature has been included as part of the defining properties of the particular CA. Jheald (talk) 22:18, 25 September 2011 (UTC)[reply]

This is an open invitation to anyone interested in contributing to the resolution of what exactly should be presented in this respect in the article to debate the matter on my talk page. Quondum (talk) 17:51, 27 September 2011 (UTC)[reply]

Here's a definition from Lounesto 2001 (one of the refs), p 190

Definition: An associative algebra over F with unity 1 is the Clifford algebra Cl(Q) of a non degenerate Q on V if it contains V and F = F ⋅ 1 as distinct subspaces so that

(1) x² = Q(x) for and x ∈ V

(2) V generates Cl(Q) and an algebra over F

(3) Cl(Q) is not generated by any proper subspace of V'

He notes below the definition that "The above definition gives a unique algebra only for non-degenerate quadratic forms Q "

So the algebra does contain the field, and can be defined in terms of the signature. Furthermore each non-degenerate signature gives a unique algebra, which implies that if you're given an algebra defined another way you can deduce the signature and V provided the signature is non-degenerate.

The difference between geometric algebra and Clifford algebra is that in geometric algebras the field is R or C and degenerate signatures are not allowed, as given in the article. The latter is clearly an important factor: as someone with more of a practical interest in such algebras than a theoretical one I tend to only deal with geometric algebras as the only ones I'm interested in are real geometric algebras.--JohnBlackburne^words_deeds 20:42, 28 September 2011 (UTC)[reply]

All my comments apply in non-degenerate, real Clifford algebras. I think that the references are being misinterpreted, because a whole pile of contradictions result if one assumes that distinct choices of V define distinct Clifford algebras, or equivalently, that a specific subspace V of Cℓ(Q) has a privileged status. The definition you have quoted does not support the position that the subspace V can be isolated as having a privileged role. Ask yourself the following: is Hamilton's quaterionic algebra a Clifford algrbra? If so, what is a basis for its generating vector space V – {i,j}, {j,k} or {k,i}? If not, why are Clifford algebras presented as a "possible generalization" of, amongst others, quarternions? Saying that distinct generating vector spaces yield distinct algebras by definition is akin to saying the choice of a different generator element for ℤ_p^* yields distinct rings. Quondum (talk) 08:18, 29 September 2011 (UTC)[reply]

Reviewing the above, it is evident that the interpretation of the word "unique" is a problem. You (JohnBlackburne) seem to be interpreting it to mean "distinct from others", i.e. that every choice of V and Q produces a different algebra. I interpret it to mean "the only", i.e. that there is only one algebra that can be generated using that choice, and that this is not necessarily distinct from an algebra generated from a different starting point. This distinction in interpretation is crucial. Quondum (talk) 05:48, 1 October 2011 (UTC)[reply]

Most common, at least for those with a primary interest in their geometrical applications, is to identify the quaternions with Cℓ⁺_3,0(R), the even sub-algebra of the 3D real geometric algebra Cℓ_3,0(R). This identification places the quaternions i, j and k each on an exactly equal standing. Of course, as with any even Clifford sub-algebra, the sub-algebra, being closed, can be identified as isomorphic to a Clifford algebra in its own right, in this case Cℓ_0,2(R). It may be interesting to ask what it is that the (one-sided) transformation represents that maps Cℓ_0,2(R) -> Cℓ_0,2(R) (-j) in the context of Cℓ_0,2(R), which has the effect of replacing ij with i as the pseudo-scalar for the system. But that I will have to leave as an "exercise for the reader". Jheald (talk) 09:15, 1 October 2011 (UTC)[reply]

You have described the use of Clifford algebras in a geometric setting, and I agree with your statement as far as it goes. That is however not what the article is about - it should serve as a reference of the mathematical objects called Clifford algebras. I think perhaps it will make things easier if we speak of "equivalent algebras" (being isomorphic as abstract algebras in every respect), a term in ([Algebra/Clifford Algebras]). Then my statements becomes: there is in general no way of uniquely identifying a 1-vector subspace that respects the isomorphism that keeps being mentioned, and consequently no choice of psuedo-scalar, way of grading the algebras, Grassmann product etc. that respects the isomorphism. I like to think of Clifford algebras as the core algebraic structure that is preserved by the isomorphism, rather than as something about which a defining characteristic must be ignored to allow such an isomorphism. And I have a feeling I'd not be the only one. Quondum (talk) 17:17, 1 October 2011 (UTC)[reply]

A mathematical object can be isomorphic to another mathematical object without the two being the same mathematical object.

As our article on isomorphism puts it, "If there exists an isomorphism between two structures, the two structures are said to be isomorphic." The article continues: "In a certain sense, isomorphic structures are structurally identical, if more minute definitional differences are ignored."

Here the "more minute definitional differences" would be the difference in signatures, i.e. the difference in the identified base space. The difference does not prevent the two algebras being isomorphic, but it does cause them to be identified as different mathematical objects, different algebras.

Every definition I have seen of particular Clifford algebras, even in works with no interest in geometrical applications, is similar to JohnBlackburne's above. It simply makes sense for what mathematicians want to do with them, because mathematicians want to speak about the relationships between Clifford algebras of different signatures. Jheald (talk) 23:46, 1 October 2011 (UTC)[reply]

I could concede your point about the consensus of what is meant when we speak of a Clifford algebra on pragmatic grounds, but it is clear that there is no clarity generally on the distinction between what I'll call the "core structure" (that which is reserved by the isomorphisms) and what you are referring to - e.g. see Split-complex numbers#Algebraic properties which apparently equates the isomorphic algebras. I'll retract and replace my contention with the observation that unfortunately ambiguity reigns; me trying to clarify the picture is not going to work.

I would thus like to reframe what is worrying me: There is a core structure with some features from which we can make some interesting deductions, and I'm sure there is value in explicitly identifying/distinguishing this core algebra normally lost in the ambiguity, perhaps in a dedicated section. Some interesting questions only become apparent within its framework. One such feature is the difference between the (3,1) and (1,3) signatures used for Minkowski space, both of which seem to work perfectly well, but which are non-isomorphic even in the core algebra. The question arises whether one signature is incorrect, or whether restrictions placed on the physical quantities restores isomorphism between them, which would itself be mathematically interesting. Example: Can most or all physics equations be phrased only in terms of the core algebra? Apparent counterexamples to consider include the ubiquitous derivative operator ∇ and the fact that most quantities seem to be confined to certain subspaces (vectors, bivectors (EM fields), spinors (fermions) etc.). Quondum (talk) 09:32, 2 October 2011 (UTC)[reply]

Is this line correct?

Not being familiar with the field, I wonder if someone who knows it could verify that the following line from the article is correct:

C\ell _{n}(\mathbb {C} )\cong C\ell _{p,q}(\mathbb {R} )\otimes \mathbb {C} \cong C\ell (\mathbb {C} ^{p+q},Q\otimes \mathbb {C} ).

I ask because the notation seems to vary within the line. That may just be because of using the definition of Cl sub n which appears just above this line, but I'm confused by the change of notation. It might be clarifying to split it into two separate statements. The present form is not bad exposition if you are familiar with the material, but for me, at least, as an outsider the combination of notations is unclear. Dratman (talk) 23:31, 16 December 2011 (UTC)[reply]

That line certainly had a misplaced ")" because "quadraticform⊗ℂ" does not have meaning, and also Q is described as a real quadratic form, with positives and negatives in signature, but then the Q was next to copies of ℂ, then tensored with ℂ. I surmised they must have meant to use reals inside instead. I'm not very familiar with complexification but this seems plausible. Rschwieb (talk) 01:54, 17 December 2011 (UTC)[reply]

My interpretation is that the ")" was not misplaced, but that the "Q⊗ℂ" was intended to mean the complexification of Q, i.e. the analytic extension of Q to ℂ. To me this interpretation makes sense, but the use of the notation for a tensor product to denote this complexification seems inappropriate.

The development of the argument seems to have been lost, even though it is a simple one. Perhaps

C\ell _{n}(\mathbb {C} )=C\ell _{n,0}(\mathbb {C} )\cong C\ell _{p,q}(\mathbb {C} )=C\ell _{p,q}(\mathbb {R} )\otimes \mathbb {C} \,

where the isomorphism step is demonstrated by the substitution z_j ↦ iz_j, j ≤ p. Would this be more obvious? — Quondum^t_c 05:58, 17 December 2011 (UTC)[reply]

You could very well be right, I have no idea what the author was thinking. Until we have an idea of a reference clarifying this line, we might want to tag it and leave it alone. I don't think we should mix "p,q" into the complex Clifford algebra notation, because I think the intention is that "p,q" should indicate the signature of the form, but the complex Clifford algebras do not have signatures with both 1's and -1's. Rschwieb (talk) 13:31, 17 December 2011 (UTC)[reply]

I'm starting to think the point about complexification may be too far afield from the main topic. It all seemed to make sense up to that point. I am not sure the tensor product of two vector spaces is really necessary to demonstrate here. Also, the book I've been reading about Geometric Algebra (Dorst, Fontijne & Mann, Geometric Algebra for Computer Science, Elsevier/Morgan-Kaufmann 2007) keeps emphasizing that they always use reals and count that as a virtue. Their attitude would not necessarily be relevant, except that the section we are discussing is introduced by "Main Article: Geometric Algebra" Dratman (talk) 14:37, 17 December 2011 (UTC)[reply]

Well spotted. On the other hand, the section heading is a mismatch with the "main article". In the context of Clifford algebras, dealing with the Clifford algebras over ℂ does make sense, and the point about such algebras being trivially isomorphic when non-degenerate is probably worth making. I do agree that the point about the complexification of a real algebra can be removed. This suggests a restructuring of the section (splitting into real and complex examples?).

On notation, the notation Cℓ_p,q(ℂ) does get used, and it is useful when you are dealing with a specific algebra. See, for example, Dirac algebra, which is defined as Cℓ_1,3(ℂ). — Quondum^t_c 15:06, 17 December 2011 (UTC)[reply]

We seem to be in agreement, but I am not qualified to make any substantive changes here, so hope someone else will delicately remove the bit about complexification. Dratman (talk)

Done. — Quondum^t_c 16:26, 17 December 2011 (UTC)[reply]

I can see the convenience of using the p,q notation to indicate that it is the complexification of the Clifford algebra signature p,q. Is there a reference that does that? Rschwieb (talk) 19:21, 17 December 2011 (UTC)[reply]

Drat. I could have sworn I'd seen the notation in some paper by a reputable author, but now I cannot find its use anywhere (except in Wikipedia articles). It seems most authors point out the isomorphism of all nondegenerate quadratic forms on Cⁿ and then they simply use the standard Cℓ_n(C) or some equivalent notation. They thus leave the burden on the reader to keep track of which basis vectors to multiply by i. In my browsing I did find a notation that I liked used by many authors: a subscripted tensor product of algebras, e.g. Cℓ_p,q(R) ⊗_R C to denote complexification. This is more specific, and seems to denote replacement of the R field by the following algebra. When something like Cℓ_p,q(R) ⊗ C occurs, we should consider putting the R subscript in. — Quondum^t_c 11:52, 18 December 2011 (UTC)[reply]

Elaboration

The passage: " It turns out that every one of the algebras Cℓ_p,q(R) and Cℓ_n(C) is isomorphic to a matrix algebra over R, C, or H or to a direct sum of two such algebras." Has a major problem and a minor problem. Firstly, it leaves open the possibility that a Clifford algebra may be the direct sum of a matrix ring over C and a matrix ring over H. Using A⊕A is meant to clarify this. Secondly, the use of "over F" is overloaded here: a standard interpretation of "a matrix algebra over F" means that you are talking about the ring of matrices with entries from F as an F algebra. In our case here, we never talk about an H algebra, and in a few cases the matrix ring over C is an R algebra rather than a C algebra. Saying "with entries from" is both descriptive of the object and neutral as to whether the field is R or C.

My revision doesn't fully succeed in bringing the sentence in line with the standard interpretation, but I felt any more qualifications would be overcomplicated. (Eg. ... or the matrix algebra M(C) over R or...) In any case I think it's less likely to mislead the reader about the base field. Rschwieb (talk) 17:00, 17 December 2011 (UTC)[reply]

Sorry, I missed the first point - that one's worth having clear. The second point does not seem to be making an intelligible distinction between overloads to me: you seem to be describing the same thing, so what you're saying is going over my head. When we are dealing with a matrix ring with H entries, is it not also an H-algebra? When is a matrix ring over C not a C-algebra? A separate point: we may want to say "full matrix ring". — Quondum^t_c 18:41, 17 December 2011 (UTC)[reply]

I just meant "over F" has two meanings: 1)"Matrix algebra over F": as in "entries from F"; and 2) "algebra over the field F" as in the definition of an F-algebra, that the algebra is an F vector space.

To address your questions, you could treat the matrix ring over H as an H algebra, but you should be careful because official definitions of algebras over noncommutative rings vary. More to the point, H-algebras never arise in the classification of real and complex Clifford algebras, as the algebra groun field is always real or complex.

A matrix ring over C is always a C algebra, but several of the Clifford algebras are isomorphic to the matrices over C as R algebras (but not as C algebras). I think putting "full matrix ring" is a good idea: go ahead. Rschwieb (talk) 19:11, 17 December 2011 (UTC)[reply]

That the definitions are different is obvious, but that they lead to anything different is less so. That H is not a field does complicate the use of the wording: one does have to assume a definition expanded to an algebras over rings. I guess my intuition was that any matrix algebra with elements of field K must be a K-algebra, and I suppose the generalization to non-commutative rings like H seems so natural (there are different definitions?) that I felt this would be implied, which of course I shouldn't do: it hasn't been defined. As to classification, an algebra can be over many things, e.g. every C-algebra is an R-algebra, and we can call it by either name. It makes more sense in classification to use a standard field. In my mind a shortcoming of Classification of Clifford algebras is: that for (and only for) even n, Cℓ_n(C) is a real Clifford algebra is not even mentioned but left for the curious reader to discover by inspection, is not outlining the classification properly. In conclusion, I fully concede your point. I'm sure initially I interpreted exactly "a matrix with entries" in the phrase "a matrix algebra over". You might want to cast your eye over Matrix ring for exactly this (ab)use of the phrase. — Quondum^t_c 04:33, 18 December 2011 (UTC)[reply]

Q

"Freest algebra generated by V satisfying vv = Q(v)1"

Hmm, does this define multiplication in Cl(V)? If that is what is happening, then it could be pointed out. Besides, the "1" on the far right must be the identity in Cl(V)? In order to make sense of expressions like uv, do I have to go through the bilenear form? YohanN7 (talk) 18:42, 8 September 2012 (UTC)[reply]

It's an associative algebra, so it inherits the product, sum and properties of them from that. And a unital algebra, so yes, it has an identity and that's it. Apart from that the algebra is generated by considering the products and sums of the algebra and seeing how many things can be generated. Using rules like the following from the article

uv+vu=2\langle u,v\rangle

(a scalar)

expressions such as abab can be reduced by e.g. rearranging and replacing anything that looks like vv with the scalar Q(v). Eventually you find that for an n-dimensional Vector space the Clifford algebra has 2ⁿ dimensions.--JohnBlackburne^words_deeds 19:45, 8 September 2012 (UTC)[reply]

Thanks for clarifying. I still think that the introduction can be made clearer. At that point (the intro) we don't have an associative unital algebra. We have only a vector space V and a quadratic form Q on it.

My suggestion is that it is pointed out that in vv = Q(v)1, the product on the left is that of the would-be algebra and the 1 on the right it's multiplicative unit. But it's no big deal. It's a good article by the way. YohanN7 (talk) 20:27, 8 September 2012 (UTC)[reply]

I'm not too sure what you mean – the algebra is stated as being constrained to being a unital associative algebra, so that is axiomatic. I have updated the article for clarity as per the remaining suggestion. — Quondum 22:00, 8 September 2012 (UTC)[reply]

Reading this again, I'm beginning to wonder why we have "freest" instead of "biggest". "Free" here has a pretty specific meaning that is not as accessible as "biggest". Rschwieb (talk) 22:48, 8 September 2012 (UTC)[reply]

I like the edit. Perhaps redundant information for a mathematician, but it helps a first time reader. YohanN7 (talk) 09:02, 9 September 2012 (UTC)[reply]

Not redundant, merely pedantic. Mathematicians seem to be happy with abuse of notation (as in uv + vu = 2⟨u, v⟩) and, as in this case, leaving unstated inferences that can readily be made by a mathematician. The encyclopedic context demands more complete and precise statements than a mathematician would be happy with (and may often prefer).

On "freest" vs. "biggest", I see that there is no obvious way to work out what is meant. Should this not be linked somewhere, at least? — Quondum 09:59, 9 September 2012 (UTC)[reply]

'Freest' to me means the system that results from freely using the sum and product of the algebra to form all possible products. I.e. i.e. just a common English word that describes one way to generate the Clifford algebra. As such it there's no need to link it – the most sensible target would be algebra which is already linked via associative algebra.--JohnBlackburne^words_deeds 11:18, 9 September 2012 (UTC)[reply]

I didn't like "biggest" at all (far too vague, most such algebras are infinite anyway) and "freest" not much better. How about "most general"? Deltahedron (talk) 12:09, 9 September 2012 (UTC)[reply]

"Most general" did come to mind, and it is a step down from "freest", but I think it will still leave a lot of people scratching their heads. Since it's already as vague as "biggest" is, I thought we may as well go with "biggest" for broadest appeal. Rschwieb (talk) 13:02, 9 September 2012 (UTC)[reply]

Surely what is meant is in effect with products being distinct ("newly generated") whereever possible within the constraints imposed by the axioms. Every other possible ("smaller") structure would presumably have to be a quotient of this structure. Is it not possible to find some English term for the algebra-of-which-all-others-must-be-quotients? (unless I've got my wires crossed, of course: I'm flying by the seat of my pants, here) — Quondum 13:52, 9 September 2012 (UTC)[reply]

Your wires seem to be OK! "Free" could be paraphrased as "free of restrictions", actually meaning "as free of restrictions as is possible". The free algebra has no restrictions (unless you are counting the things that make it an associative algebra), and the Clifford algebra has the one restriction v⋅v=Q(v)1, but no other restrictions. Rschwieb (talk) 19:15, 9 September 2012 (UTC)[reply]

This is a nice remark. Although publishing papers in J. of Algebra, Ill. J. of Math., ... on Clifford- and Weyl algebras and their unifications, it never came to my mind, that free means exactly this and is not just a word like big. So its natural to drop every other wording and write only free in this article and then (free means free of restrictions). Everybody will understand this, no more scratching! Who has invented this notation? One should meantion him as well. In french it is libre (Bourbaki). — Preceding unsigned comment added by 130.133.155.68 (talk) 17:27, 24 September 2012 (UTC)[reply]

I can appreciate the problems you guys have formulating this. While I think that I have managed to convince myself that I understand the definition mathematically, it's not easy to come up with a spot on wording. One reason is the "characteristic property". Whenever there is a "characteristic property" of whatever one can on the one hand speak of "A whatever" and on the other hand, if one can construct an example, then one can appeal to the characteristic property and speak of "THE whatever". This is probably a practice that obscures the layman from otherwise perfectly understandable material.

Have you ever had the feeling that you in one blow understood almost everything you didn't understand for years? Well, I've had that feeling today. Clifford Algebras pretty much put Einsteins enegy-momentum equation in the same home as the Dirac Matrices. The Lie Algebra so(3;1)sits pretty inside it too - and a couple of scarier looking things that usually are decorated with plenty of indices.

This has to be one of the most important math articles. It's good and it can get better. YohanN7 (talk) 21:21, 9 September 2012 (UTC)[reply]

I have a long list of epiphanies I wish I could have :) Rschwieb (talk) 23:49, 9 September 2012 (UTC)[reply]

Use in Physics

From a mathematical point of view this is a nice article, especially that description of universality. From the physical point of view there is missing something. Before mentioning the Dirac matrices, one should mention the Pauli-matrices as well, since they are a representation of the Clifford-algebra in an Euclidean R³ (the Duffin-Kemmer-Petiau matrices of physics are not). And creation and annihilation operators of quantum Fermi-Dirac statistics are another example. — Preceding unsigned comment added by 130.133.155.68 (talk) 16:57, 24 September 2012 (UTC)[reply]

Clifford group

Hi!

I think it would help if the maps in the exact sequences were given names and/or descriptions. My guess is that the first two are basically trivial injections, and that the last one is trivial too (everything->identity operator on V)? The second to last one follows from the canonical map from V to Cl(V)? The field K is the same thing as k times the identity in Cl(V) for all k in K? Blue links to kernel and exact sequence would probably help too. Given that the subject is presented in such generality (the "characteristic 2 disclaimer is everywhere"), and with some degree of claim of completeness, unnecessarily compact notation is uncalled for. As is, parts of the text cares a lot more about "not saying anything provably wrong" and "not forgetting special cases" than "explicitly saying whats right". I am not asking for a readable exposition for an eighth-grader. But, I guess I am asking for painfully spelled out definitions and plenty of blue links. There are things that are wrong in a strict sense (i.e. there is no norm defined through a (even non-degenerate) non-positive definite quadratic form), but these errors pass through the otherwise meticulous "correctness" filter because the abuse of notation and terminology is so obvious to the experienced. Such errors are generally caught too by those knowing something already about the subject. If these implicit definitions and understandings are slightly beyond the present level of understanding (or need reminding of), then they are unlikely to be picked up correctly by the reader.

I know that this is asking for a lot, but I feel the article and the subject is worth it. (I did have concrete suggestions too;)) YohanN7 (talk) 13:32, 4 January 2013 (UTC)[reply]

While we're on this topic, I want to throw out there that it looks like "Clifford group" could probably be taken and transplanted into its own page. It is not a vital organ to explaining Clifford algebras, so I think we should consider splitting it out. Rschwieb (talk) 14:53, 4 January 2013 (UTC)[reply]

I agree that it makes sense to move this to its own article. As a group, it stands on its own; it does not require the context of a Clifford algebra. —Quondum 22:10, 1 March 2014 (UTC)[reply]

Signature in Physics section

In the complex case, there is no Sylvester's law of inertia. But there are still forms with different signatures. They are, of course, basis-dependent. Since the starting point is spacetime (real vector space) with metric signature $(3, 1)$ or $(1, 3)$ , and one always ends up with something isomorphic to $C ℓ 4 (C) ≅ M n (C)$ , which is what physicists use, and this needs to be explained. Either you complexify spacetime (which doesn't affect the signature), or you complexify the Clifford algebra. The alternative of changing basis in complexified spacetime (the $(x, y, z, ict)$ -convention) to get signature $(4, 0) = 4$ is out of fashion.

Also note that $C ℓ 3, 1 (C)$ and $C ℓ 4 (C)$ are isomorphic, not equal.

As of now, the article is plain wrong in that section (Physics). YohanN7 (talk) 15:07, 3 September 2014 (UTC)[reply]

As of this morning, it was $C ℓ 1,3 (C)$ as it has been for a long time. I'll change back. YohanN7 (talk) 15:12, 3 September 2014 (UTC)[reply]

And oh, the distinction between $C ℓ 1,3 (C)$ and $C ℓ 4 (C)$ isn't meaningless at all. Saying that is totally the same as saying that there is no distinction whatsoever between space and time. All physicists maintain the distinction between space and time in the Clifford algebra. YohanN7 (talk) 15:32, 3 September 2014 (UTC)[reply]

For those who see mention of $C ℓ 1,3 (C)$ as impossible, one has to go with the alternative that I suggested in my previous edit. First $C ℓ 1,3 (R)$ or $C ℓ 3,1 (R)$ , then complexify. But note:(!) Even so, you will get different gamma matrices, with different properties from those of $C ℓ 4 (C)$ . Remember, a Clifford algebra has a distinguished subspace whose basis depends on the original basis. This basis (in a few variants) is imposed by nature, and is not on equal footing with just any basis resulting in only plus-signs in the quadratic form. YohanN7 (talk) 15:55, 3 September 2014 (UTC)[reply]

I am not quarrelling with what is said here about complex algebras, the real case is significant and seems to have been neglected. And to say that "one always ends up with something isomorphic to

C ℓ 4 (C) ≅ M n (C)

, which is what physicists use" seems to be an overly sweeping statement, or otherwise defines "physicist" very narrowly. —Quondum 16:10, 3 September 2014 (UTC)[reply]

You have a point. "... what physicists use" should be "... what most physicists dealing with particle physics and relativity use". In practice in those subjects, which are the subjects where Lorentz transformations arise, you don't get very far without leaving

C ℓ 1,3 (R)

(or

C ℓ 3,1 (R)

). For instance, as soon as you need some sort of inversion (which is not to be found in the identity component of the Lorentz group

O(3, 1)

), you actually cannot (AFAIK) do without the complexification without becoming artificial. More abstractly, the thingies that the Clifford algebra operate on are almost always complex vector spaces. If the space of operators is to be complete, you simply need all of

M n (C)

, but with the spacetime metric intact, i.e. the gamma matrices square to both +1 and -1 to maintain sanity.

Then there are surely other areas of physics where

C ℓ p,q (R)

is the only appropriate alternative. There are probably also those who insist on real algebras in particle physics too, but I guess you know their ways better than I do. Could you write something? YohanN7 (talk) 16:56, 3 September 2014 (UTC)[reply]

(ec) YohanN7 I'll have to pass on the subjective points ("what all physicists maintain" and "saying there is no distinction between space and time" and the entire final paragraph) for now. I am not aware of any extraneous layers added by physicists, I'm only aware of the mathematical differences. I'd love to see these ideas spelled out, but at this point you are just asserting them by fiat and then excoriating me for not agreeing with you.

Let me explain why I say the distinction is meaningless mathematically. It's bad form to suggest complex spaces have a signature like the real ones do. There's only one form, and only one (isoclass of) Clifford algebra! I guess physicists have a use for a "latent image" of the real metric on a complexified algebra(?) That would be fine, but in that case we need to explain that is what is going on. As written, nobody is going to catch on to this interpretation that the algebra is suddenly "different" if you scale one of its basis elements. Why not start with the real case and then say "and physicists also find the complexification of this algebra useful in the following ways..."?

My main concern is the notation

C ℓ n,m (R)

, which suggests this idea that there is some mathematical distinction for complex spaces. If there is a distinction, it's purely in interpretation. I do not recall seeing it before, and I haven't found any usage of it at the moment. It certainly is in conflict with the Clifford Algebra Structure page. Where is this notation "a thing"? Rschwieb (talk) 17:26, 3 September 2014 (UTC)[reply]

I am not excoriating you. Yes, I am asserting some things flat out (it's a talk page), but I do my best to back up the assertions that I think need to be backed up. You say distinction between different, but isomorphic algebras is meaningless. Yes here we disagree. But since you refuse to acknowledge arguments from physics (you just choose to pass, simpler that way?), I don't see how we can discuss at all. I don't shy away from math in this discussion.

What you will find in the physics literature is occasional mention of Clifford algebras. One time out of N where N is big (bigger than the number pf physics books I have read), they will mention real and complex Clifford algebras. But they use them, promise. They do not at all hesitate to plug in a factor of

i

both here and there. And, they rarely change basis in spacetime in any other way than to keep the (1, 3)-signature when the form is restricted to real spacetime (light-cone coordinates do occur). I don't agree that people won't see that retaining the "signature" after complexification in one end or the other is keeping the usual spacetime basis. After all, someone wrote it that way, and it has been around for some time.

I know that there is only one isoclass. I did spell that out in my edit that you reverted, at least that the involved algebras were isomorphic. I know about Sylvester's law of inertia, but I didn't know that signature of a form was undefined for complex spaces, just that it (can I refer to it even if it doesn't exist?) is basis dependent and can be brought to a standard form by change of basis, but this is precisely what we don't want to do.

Insisting on

C ℓ 3,1 (R)

will not work as explained above (inversions of various sorts). Insisting on

C ℓ 4 (C)

will not work either because of the "subjective point" that space and time, tough connected, are of different nature. Physicists really have a use for a "latent image" of the real metric on a complexified algebra(!) Do I really have to dig up sources?

I'm neither a mathematician nor a physicist, but I tend to side more and more with the mathematician. But for real pretending that a doughnut and a coffee cup are the same things is not helpful, especially in a section about applications in physics where a particular Clifford algebra is used, not an equivalence class.

I want to remind you that I didn't start this, and that

C ℓ 1,3 (C)

has been in the article since 2004. I also don't understand your last sentence (your main concern). Surely, you mean

C ℓ n,m (C)

. Then I'm with you. We have to invent a new notation for this, because

C ℓ 4 (C)

will convey the wrong idea (and the wrong group).

Lastly, it is okay if you intersect the above post (don't forget to indent and sign

)YohanN7 (talk) 19:15, 3 September 2014 (UTC)[reply]

And the solution, of course, is to write

C ℓ 1,3 (R) C

.

Two things. First, that signature notation for some Clifford algebras over the complex numbers is also in the Spinors section. Is this notation standard? If so, it would be nice if a short discussion about this notation was added. If not, I don't think we should be creating non-standard notation for wiki articles. Second, there is a completely real (Majorana basis) 4 by 4 matrix representation of the dirac algebra with signature (3,1). Because there are 16 basis elements in this algebra, this covers the full space of 4x4 real matrices, and so Cl_3,1(R) is isomorphic to the algebra of the 4 by 4 real matrices. I think my edit avoided all this discussion of arbitrary signature choice in a complex algebra, although I understand what YohanN7 is trying to say because in practice physicists use these primarily with complex valued wavefunctions, and so on. Saladforked (talk) 23:59, 3 September 2014 (UTC)[reply]

The two things, the first first. The signature is a problem, and I agree of course, we shouldn't invent non-standard notation. We still should not forget that the quadratic form comes from spacetime in the complexified version. Put it this way, the vector space over which the Clifford algebra is defined isn't a neutral

C 4

, it is a

C 4

together with a very privileged set of bases, those related by Lorentz transformations. A mere complexification should not hide this fact. I like writing

C ℓ 1,3 (R) C

(or

C ℓ 3,1 (R) C

). It retains the signature and implies the correct Clifford algebra.

Secondly, The Majorana basis doesn't work physically. It has physical implications, and (not surprisingly), can treat only Majorana fermions whose wave functions can be treated as real in any Lorentz frame. You have to admit that pretending that the physicists work with another structure than they really do just because there is insufficient notation to cover it would be rather silly.

An even more compelling reason (than the inversions I have mentioned) for a complex algebra is this: The physicist convention for Lie algebras is that everything is multiplied by

i

. The sigma matrices are, in the

(3, 1)

-convention defined as

\sigma ^{\mu \nu }=-{\frac {i}{4}}[\gamma ^{\mu },\gamma ^{\nu }],

where

{γ μ \in Cℓ 3,1 (R): μ = 0,1,2,3}

are the corresponding gamma matrices. Note the factors of

i

. The sigmas,

{σ μν \in Cℓ 3,1 (R) C

} don't belong to the real algebra. They have the commutation relations of the Lorentz Lie algebra,

[\sigma ^{\mu \nu },\sigma ^{\rho \tau }]=i(\eta ^{\tau \mu }\sigma ^{\rho \nu }+\eta ^{\nu \tau }\sigma ^{\mu \rho }-\eta ^{\rho \mu }\sigma ^{\tau \nu }-\eta ^{\nu \rho }\sigma ^{\mu \tau }).

The spin Lie algebra sitting inside the Clifford algebra is the main motivation for using Clifford algebras. Insisting on a real Clifford algebra would make the article incompatible with the majority of the physics literature, and seems to be motivated solely by that there is insufficient notation available. YohanN7 (talk) 07:30, 4 September 2014 (UTC)[reply]

YohanN7 There's quite a bit above here that I only scanned, but the upshot seems to be we understand each other now. (And yes, it looks like I cut and pasted something with an R instead of a C. You guessed right.) I think using the complexification notation is a great idea, and I had a look at the latest edit and things look great.

By the way, I'm interested in seeing (if you have an example in mind) an illustrative example of leveraging basis elements of a complex space that square to -1. That sounds like an interesting benefit that might normally be passed over in pure math texts. If you've got one, please toss it on your talkpage or my talkpage or wherever and if not, thanks for reading :) Rschwieb (talk) 13:30, 4 September 2014 (UTC)[reply]

And sorry to induce you to make rationalizations about your barbs. They not beneath me either, and I'm as inclined as anyone to respond in kind to such things. That's also part of my reason for not reading the block above too closely. Sorry for that and for any previous friction above. Rschwieb (talk) 13:48, 4 September 2014 (UTC)[reply]

How boring wouldn't life be without a little argument once in a while? I have, by the way, straightened up a section (that I wrote) in Gamma matrix too. The present problem wasn't there, but I apparently had believed that any Clifford algebra is complex at the time I wrote it. YohanN7 (talk) 16:08, 4 September 2014 (UTC)[reply]

Saladforked : I took a look at Spinor and didn't see any occurences of

C ℓ 1,3 (C)

anywhere. I only see

C ℓ n (C)

which is fine by me. Hopefully I didn't cause too much confusion with my typo about

C ℓ 1,3 (R)

being a problem. It was a copy and paste error I didn't proofread carefully enough. The single index indicating dimension is standard, and the split-index used for real spaces is just an added feature made possible by the properties of metric signatures. Rschwieb (talk) 13:48, 4 September 2014 (UTC)[reply]

Rschwieb : The mentions are in the spinor section in this article. YohanN7 (talk) 14:35, 4 September 2014 (UTC)[reply]

I like the edits. And as you say, it would be unfair to limit a discussion of the physics just to avoid notational issues. You not only addressed the physics much better, but also found a clarifying notation. I had to laugh at "How boring wouldn't life be without a little argument once in a while?", because this is the most reasoned and productive discussion I have seen on wiki. Nice to know there are smart reasoned editors on here. Saladforked (talk) 02:23, 5 September 2014 (UTC)[reply]

What I wrote about the Majorana basis isn't really true. It works as well as any basis, but the "reality condition" becomes empty unless the spinors describe Majorana fermions. They may exist actually exist by the way. I don't know anything myself, but I think I have seen somewhere that it has been speculated that neutrinos are Majorana fermions. (Don't hold me too responsible for this statement. I may recall wrong.)

Also,

C ℓ 1,3 (C)

occurs in gamma matrix as well. YohanN7 (talk) 11:17, 5 September 2014 (UTC)[reply]

YohanN7 : Not related to the article, but in case you are curious, you recall correctly about neutrinos and a Majorana mass term. Some motivation can be boiled down to simply the notation "that which is not forbidden is mandatory" in quantum mechanics. The experiments looking for Neutrinoless double beta decay are searching for this. A Majorana mass term can also be motivated by the Seesaw_mechanism which can explain why the neutrinos we see have such incredibly small mass. Saladforked (talk) 00:16, 7 September 2014 (UTC)[reply]

On the comical side: I wrote that the

(x, y, z, ict)

-convention is "out of fashion" – and I can back it up

. There is actually a whole section in Gravitation by MTW that proclaims the death of $ict$ , the reason being that it hides the indefiniteness of the metric. YohanN7 (talk) 11:55, 5 September 2014 (UTC)[reply]

The rephrasing in terms of the complexification of

C ℓ 1,3 (R)

, and denoting it

C ℓ 1,3 (R) C

is, IMO, a substantial improvement. The "reality condition" (which I'm going to take as that the coefficients of the basis must be real for real vectors) is a crucial additional constraint in the arguments that go before, and is sort of implied by the explicit complexification. I'd like to see this notation used elsewhere where the signature is considered significant (i.e. where there is a reality condition, suggested by the use of the confusing notation

C ℓ 1,3 (C)

). —Quondum 15:26, 5 September 2014 (UTC)[reply]

I see this is an old discussion, and to be honest, I did not read through the whole of it. But I think what is currently mentioned in the Physics section is kind of misleading. There should not be an appearance of $C ℓ 4 (C)$ in the first place -- it has nothing to do with the algebra of the gamma matrices. The $γ i$ 's are, abstractly speaking, generators of $C ℓ 1,3 (R)$ (which has already been mentioned). It's just that the standard representation of the $γ i$ 's, the Gamma matrices, are as a sub-algebra of $M 4 (C)$ (the matrix algebra of 4x4 complex matrices). But even then, as I understand it, the representation has nothing to do with $C ℓ 4 (C)$ in particular. The algebra of the gamma matrices is still a real algebra (Lorentz matrices, represented as 4x4 matrices of reals, get multiplied to the gamma matrices under Lorentz transformation -- thus the coefficients of the gamma matrices are all reals), representing $C ℓ 1,3 (R)$ . I think this should be made clear in that section. - Subh83 (talk | contribs) 17:25, 16 July 2015 (UTC)[reply]

That is from the point of view of mathematics(?). Most physicists don't give a damned about

C ℓ 1,3 (R)

being real and some prefer the other convention for the metric, so that

C ℓ 3,1 (R)

would be the real algebra, and perhaps much preferable, since otherwise you'd have to work with $8 \times 8$ -matrices in order to have what the hardcore mathematician would call a Clifford algebra – as I have been made painfully aware. Calling the space spanned by the signature (1, 3) gammas a Clifford algebra is fine in theoretical physics (that's what they do), but not here in this article, which is a math article. Oops, memory fading. This is a problem with

C ℓ 4,1 (R)

contra

C ℓ 1,4 (R)

. Moreover, by standard physicist convention, the spin Lie algebra representation of the Lorentz group (in either signature) is not embedded in the real algebra, and the spin Lie algebra representation is the main reason that physicists use the Clifford algebra to begin with, barring a few GA enthusiasts. Complexification solves all of these problems; the resulting algebra is isomorphic to

C ℓ 4 (C)

, and that is what the article correctly says. YohanN7 (talk) 19:08, 16 July 2015 (UTC)[reply]

And this is why there's ongoing confusion about Majorana vs Dirac spinors in some of the mustier corners of physics. These two correspond to two distinct ways of creating products of Weyl spinors; there actually seem to be even more ways, some of which are weird and unfamiliar and would have remained so but for it getting dredged up and proposed as dark matter. Getting the algebra correct (in this article), and getting the CPT symmetries correct, allowing a standard reference, would be worthy. 67.198.37.16 (talk) 00:33, 22 May 2024 (UTC)[reply]

Use of "reversal"

In Élie Cartan (1966), The Theory of Spinors, I found the term reversal being defined as an improper rotation, whereas in this article it is defined as a particular antiautomorphism, something that is entirely different. The latter sense presumably alludes to the effect of reversing the ordering a product of 1-vectors. The former (as an improper rotation) might be more interesting in the context of geometric algebra. Should we clarify the multiple use of the term (or even remove it)? Also, I find it strange that Élie Cartan is not cited in this article. —Quondum 19:54, 20 November 2014 (UTC)[reply]

First Line

In my opinion the sentence "Clifford algebras are a type of associative algebra" is not the most appropriate for signifying the intended meaning. I propose

          "Clifford algebras constitute/comprise a subset of associative algebras"

or the simple

          "a Clifford algebra is an associative algebra"

SoSivr (talk) 14:19, 23 December 2014 (UTC)[reply]

The "simple" version above has the disadvantage that it seems to assert that every associative algebra is a Clifford algebra. I've updated the statement to say something more meaningful/definitive. —Quondum 20:35, 20 December 2016 (UTC)[reply]

char(K) = 2

In the article we read

In particular, if char(K) = 2 it is not true that a quadratic form determines a symmetric bilinear form.

But that's not true. Each quadratic form $q$ gives rise to a symmetric bilinear form $b$ via $b(x,y):=q(x,y)-q(x)-q(y)$ . That's so by the definition of a quadratic form. --Jobu0101 (talk) 12:15, 20 December 2016 (UTC)[reply]

That statement does need revision to be useful.

A correct statement of what I presume was meant would be:

In particular, if $char(K) = 2$ it is not true that a quadratic form determines a symmetric bilinear form such that $q(x) = b(x, x)$ .

However, a more informative and correct statement would be to the effect of:

For char K = 2, there is no natural one-to-one correspondence between quadratic forms and symmetric bilinear forms

For the time being, I've modified the statement via "determines" → "uniquely determines" to make it true (assuming naturalness); it should still be fixed because even this is technically untrue without the naturalness qualification, or a statement in terms of a specific derivation. Does someone have a reference that addresses this in a clean way? —Quondum 20:09, 20 December 2016 (UTC)[reply]

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Hopf algebra?

The tensor algebra has two coalgebras on it, one of which is a Hopf algebra. Is the universal construction of the Clifford algebra from the the tensor algebra compatible with the Hopf algebra? If so, can this be articulated? If not, why not, and can further deformations be made to make it compatible? 67.198.37.16 (talk) 20:59, 4 May 2019 (UTC)[reply]

Traditional definition

Lars Ahlfors and W.K. Clifford assume basis elements are imaginary units with negative one for their squares. See Lars Ahlfors (1985) "Mobius transformations and Clifford numbers", (pages 65 to 73 in Differential Geometry and Complex Analysis, H.E. Rauch Memorial Volume, I. Chavel & H.M. Farkas editors, Springer books). Expansion of the category of Clifford algebras to include basis elements that square to plus one came with those who would include split-complex numbers and split-quaternions. These latter algebras are composition algebras and are an ill-fit for traditional Clifford algebra. — Rgdboer (talk) 02:47, 2 December 2019 (UTC)[reply]

James Byrnie Shaw (1907) Synopsis of Linear Associative Algebra, page 84 at Internet Archive. — Rgdboer (talk) 01:11, 6 December 2019 (UTC)[reply]

"BRD algebra" listed at Redirects for discussion

An editor has asked for a discussion to address the redirect BRD algebra. Please participate in the redirect discussion if you wish to do so. _signed,Rosguill ^talk 19:58, 30 January 2020 (UTC)[reply]

Notation "Cℓ"

I am inclined to replace the notation "Cℓ" with "Cl" due to font issues, notwithstanding its use in some texts. In the context, there is no ambiguity. Any objections? —Quondum 17:25, 29 March 2020 (UTC)[reply]

Done —Quondum 03:21, 10 April 2020 (UTC)[reply]

Quaternion Quadratic Form

In the section on constructing quaternions, the article says the quadratic form is the negative of the normal Euclidean metric, but looking at the page on geometric algebra, and the bilinear form, that doesn't seem correct. Can someone help me understand this, or correct it if it's incorrect? Porg656 (talk) 19:09, 25 January 2023 (UTC)[reply]

I guess, you are right, and it should be changed. Quadratic form is positive here.

May be one should write another section about implementation of quaternions as factor of the algebra Cl_{0,3}(R) (factor by equations e_i * e_j = e_k, i,j,k = 1,2,3 or 2, 3, 1 or 3,1,2).

But here, in this section, quaternions are constructed as even subalgebra of Cl_{3,0}(R) . Grecksun (talk) 08:38, 12 February 2023 (UTC)[reply]

Some unmentioned examples

In the section "Examples: real and complex Clifford algebras", it should also be mentioned that:

$Cl 1,0 (R)$ is isomorphic to the algebra of split-complex numbers;
$Cl 2,0 (R) ≅ Cl 1,1 (R)$ is isomorphic to the algebra of split-quaternions;
$Cl 3,0 (R) ≅ Cl 1,2 (R)$ is isomorphic to the algebra of biquaternions.

129.104.241.214 (talk) 01:39, 11 November 2023 (UTC)[reply]

Yes, this seems correct. 67.198.37.16 (talk) 00:15, 22 May 2024 (UTC)[reply]

Properties of the Lipschitz group

It should be mentioned that every element of the Lipschitz group is homogeneous, and in fact, equal to the product of at most $\mathrm {dim} (V)$ vectors with non-zero quadratic form. I'm just not adding it because I'm not sure on the specifics (does this work for nondegenerate forms? what about char 2?) – viiii (talk) 03:28, 14 December 2023 (UTC)[reply]