Talk:Bivector/Archive 2

Archive 1

Archive 2

Roundness in Illustration?

After explaining that the vectors form the sides of a parallelogram, there is an illustration showing several examples and showing one where an ellipse is drawn around both vectors instead. What's with that? —Długosz (talk) 18:55, 3 November 2010 (UTC)

The reason is when the bivector is interpreted as an area the shape is not specified. It's useful to represent it as a parallelogram, as that relates geometrically to the product and to the vectors. But any shape with the same area and orientation (so in the same plane and rotating the same way) represents the same bivector.--JohnBlackburne^words_deeds 19:04, 3 November 2010 (UTC)

Maxwell's equations

The article says that the "traditional 3-vectors" are represented by symbols with lines above them, but the traditional B "vector" is actually an axial vector, not a polar vector, and this deserves special mention since it is rarely taught in physics courses.

Rating

I have updated the Maths rating to B+ (the highest rating available without a Good Article award). I'd say this is the best maths article I have seen - free of unnecessary jargon and Bourbaki virus, clear, comprehensible and informative even to longtime students of the subject and well referenced. I have also updated the Physics rating to B - overall it is excellent, but could use some expansion and the torque animation illustrates the axial vector rather than the bivector formulation. Enon (talk) 22:18, 15 February 2011 (UTC)

I agree that the article is comprehensive and informative, but it contains imprecise statements. To me as a mathematician this makes the article less useful, since it includes statements without due specifications. I see two ways to sort this out: either such statements are removed, or the due technicalities are added. Ideally, both are done, as I have seen in many math articles in Wikipedia: there could be a general description at the beginning of each paragraph, and then precise formulations which the uninterested reader can just skip. An example of what I mean is the statement "bivectors are isomorphic to skew-symmetric matrices". This statement makes no sense unless you specify what structures are isomorphic: is it an isomorphism of vector spaces, of Lie algebras, or of some other structure (also it is poorly formulated). This is not a major issue, but it is disturbing. Similar vague statements appear throughout, and it would be good if the pedagogical specialist could go through the article to make it precise yet comprehensive. 130.238.58.63 (talk) 11:07, 17 February 2012 (UTC)

Interior product

The interior product section is confusing. It says "From associativity a(ab) = a²b, a scalar times b. So ab cannot be a scalar." How is this conclusion derived? Isn't it possible for ab to be a scalar and thus a(ab) to be a vector times a scalar?

Also, it says, "From the law of cosines on the triangle formed by the vectors its value is |a||b|cosθ". I don't think this makes sense at all. What is θ in this context? How does the law of cosines apply? None of these issues are made clear. — Preceding unsigned comment added by 130.216.209.138 (talk • contribs) 06:57:53, 07 May 2012 (UTC)

I have clarified the first point by a minor addition. The second question seems a little like an oversight, since it does say where θ is the angle between the vectors, and it is unclear to me that the law of cosines does not adequately cover this. — Quondum^☏ 08:43, 7 May 2012 (UTC)

The application of the law of cosines is straight forward and does not need to be explained as the section on the law of cosines is well written. The real problem is that there is a mathematical mistake here. The first equation, 1/2 ( ab + ba) = 1/2((a+b)² - a² - b²), is false. If you were to invoke the law of cosines here using a+b as the side opposite to the angle between a and b, there would be a minus sign difference between the LHS and the RHS. The correct way to start this derivation is, 1/2 (ab+ba) = 1/2(a² + b² - (a-b)²). It should then be very clear from the manipulation of the law of cosines that the RHS is simply the dot product between the two vectors a and b. One would just set up a triangle between the 3 vectors a,b and a-b and see the direct application of the law of cosines. There may be questions to this for non flat geometries, but I think that is outside the scope of this derivation. The rest of the derivation would then be consistent. 129.6.136.124 (talk) 20:11, 5 November 2012 (UTC)

Non-definition of geometric product

In the section Formal Definition, something called a _geometric product_ is mentioned. The article says that " For vectors a, b and c, the geometric product on vectors is defined as follows".

But what follows is not a definition.

It is instead a list of properties that one would presumably want a geometric product to satisfy.

It would surely be more desirable to say "a geometric product is a map from blah to blah satisfying these axioms, blah blah blah", and then to construct a example of one, and then perhaps to prove that it is unique (if this is indeed the case). And if it is unique, then we'd actually have a definition of "the geometric product".

Otherwise, it is very painful to try to wade through the article in hopes of getting a crisp, clear definition of just what a bivector is (presumaly it has something to do with a geometric product).

I mean, of course it can be done: one can try to read through and form one's own definition.

But i would respectfully ask for a definition of not more than 100 words for what a bivector is (but i do appreciate the fact that there is an article here, with lots of intriguing suggestions).

Son of eugene (talk) 08:05, 26 August 2012 (UTC)

That is the definition of the product, or a definition. There are many other ways to express it but they all boil down to that. If you follow the link to geometric algebra it defines it as a real Clifford algebra with a non-degenerate quadratic form. At that article it defines a Clifford algebra as a unital associative algebra with a quadratic form, so basically the same definition.

It's also a good definition as it can be used to construct the algebra. It's a straightforward exercise to write down all the elements of a given algebra in two, three or four dimensions, say, from these definitions. All it involves is writing down all possible products then using the rules and others derived from them such as that orthogonal vectors anti-commute to simplify.

Another definition of a bivector is an intuitive one: assert there's a product, say it's not a vector (as it e.g. doesn't transform like one) so a pseudovector, call it a bivector, describe it as the 2D area-element instead of the 1D vector-element, etc.. But that is even less formal and still means it needs to be justified algebraically. It also only works in 3D (at least the reference to pseudovector only makes sense in 3D). It is mentioned further down as a property but isn't used to define it.--JohnBlackburne^words_deeds 11:46, 26 August 2012 (UTC)

Thanks John for kindly replying. I must say that i still do not think it is a definition (formal or otherwise), since surely any definition would at some point involve a sentence of the form "a bivector is ...".

I do appreciate your valid point that one may certainly craft one's own definition by following the references in the article, of which you provide two. My definition would be, perhaps, to let C be a Clifford algebra over a vector space V, and let B be the linear span of ab in C where a and b range over V, and call B a bivector space; then say that a bivector is any element of a bivector space.

That still is not very desirable as it depends on so much machinery. I think it should be possible to define a bivector, or more simply, a bivector space, knowing only what a vector space and a field are. Maybe that's not possible; i certainly don't know. But i do think the article should present a true definition so as at least to not mislead students on what might be accepted in their papers.  :)

No matter what, the authors of the article have performed a work for the public benefit and i salute them for that. Son of eugene (talk) 20:22, 1 September 2012 (UTC)

I have to agree with S-o-e that there is something unsatisfactory in the presentation of this article – the lead and formal definition could both be written to present the concept in a more direct/informative form. For example, the concept that generalises vectors is multivector, not bivector. A list of vague properties is given in the first paragraph of the lead; if this to make it accessible, it should include concrete everyday examples such as the bivector interpretations of torque and/or angular momentum. The formal definition section should not refer to examples used elsewhere in the article. I like the prototype definition given here above, though I would substitite exterior algebra for Clifford algebra (a Clifford algebra always contains an exterior algebra and will acquire the definition of a bivector from it, whereas bivectors are well-defined in any exterior algebra even in the absence of the quadratic/bilinear form necessary in a Clifford algebra). The exterior product is central. This argument suggests that the "formal definition" in terms of geometric algebra is actually inappropriate. I would encourage a rewrite from this perspective; I may make a start, but will not be able to make an extensive revision (my skills tend to be in the detail, not the holistic picture). — Quondum^☏ 05:11, 2 September 2012 (UTC)

Not bivector

Since William Rowan Hamilton introduced biquaternions in his Lectures on Quaternions the term bivector has meant a vector with complex coordinates, that is, an element of C x C x C. The current article here is about 2-vectors in exterior algebra, not bivectors. One can see the term used in Hamilton's sense in a 1907 paper by Ludwik Silberstein. The contents of this article could support another on multilinear or exterior algebra, but its use in this namespace is inappropriate.Rgdboer (talk) 02:51, 17 January 2013 (UTC)

The criterion should be modern use, not historical use. The distinction is also in some sense moot: the two are interchangeable (isomorphic) under some isomorphisms (e.g. the electromagnetic field in Riemann–Silberstein vector). I expect that the use of the term to mean the same as a 2-vector in exterior or geometric algebra has become the dominant use. A mention of the historical usage of the term that you refer to may be reasonable. — Quondum 08:10, 17 January 2013 (UTC)

Thank you for a quick reply and the link to Riemann-Silberstein vector. Unfortunately that article calls E+iB a multivector when it is a complex vector. It also uses the term pseudo-scalar, an article with no references. I appreciate that you view the geometric algebra is "the dominant use", but sloppy mathematics will not do. The content of this article "bivector" is better placed with p-vector where p = 2.Rgdboer (talk) 20:42, 17 January 2013 (UTC)

I agree with Quondum: we follow modern sources and the modern usage is overwhelmingly what's described in this article. You do come across examples of things that are bivectors that are called by other names, but that's normal. It perhaps happens more often with bivectors and multivectors as it's a relatively new branch of mathematics that has been applied to areas of physics that have already been discovered and described, which may have other names. But that does not stop them being bivectors.--JohnBlackburne^words_deeds 21:03, 17 January 2013 (UTC)

What Quondam said.

The thing that matters is how bivector is used now. If we look at books published in the last 15 years, there is a clear and very prevalent meaning, which is the meaning discussed in the article. Jheald (talk) 21:07, 17 January 2013 (UTC)

It is not modern to give a special name to a 2-vector. What's next, a special name for a 2-form. Texts like Flanders Differential forms uses the concept of "space of p-vectors" with p < or = n. Giving a name to the case p=2 is not done. Furthermore, there is no "interchange" or isomorphism. Squares of basis vectors in exterior algebra are always zero but not in complex space. Rgdboer (talk) 22:31, 29 January 2013 (UTC)

But that is what has been done: mathematics settled on the name 'bivector' for 2-vectors many years ago. Perhaps the most compelling argument for doing so is bivectors are real things: many quantities encountered in physics are better described using bivectors than vectors. Example include many rotational quantities such as angular velocity, angular momentum, torque; electricity and magnetism when unified in the electromagnetic tensor (also called the Maxwell bivector); rotors (quaternions in 3D) are generated by exponetiating bivectors, and quaternions themselves are better described as having scalar and bivector parts than scalar and vector parts.--JohnBlackburne^words_deeds 03:47, 30 January 2013 (UTC)

Note that the use of the term bivector for an antisymmetric tensor of order two goes back at least as far as Cartan, The Theory of Spinors (English translation 1966), p. 19, where a section "Bivectors and infinitessimal rotations" follows directly after a section introducing "Multivectors". Google books confirms that the same word, "bivecteur", appears in the original 1938 French edition, [1]. Jheald (talk) 19:23, 9 March 2013 (UTC)

Going still further back, Google books finds "bivecteur" in an 1897 book on differential geometry, being defined as the exterior product of two vectors: Cesare Burali-Forti, Introduction à la géométrie différentielle, p. 14. [2] Jheald (talk) 19:45, 9 March 2013 (UTC)

And further back: Giuseppe Peano uses the term "bivettore" in Italian in this sense in an 1888 book on geometric calculus, Calcolo Geometrico secondo l'Ausdehnungslehre di H. Grassmann, preceduto dalle operazioni della logica deduttiva, Fratelli Bocca Editori, Torino, 1888, pp. XI, 171. (webpage with download), eg sec 24 on p. 55 "Def. Il produtto di due vettori dicesi bivettore"; English translation published in 2000 [3]. Peano again presents the concept in an 1893 book Lezioni di analisi infinitesimale [4]. This Italian 1991 page gives a summary of Peano's approach and terminology [5].

Annals of Mathematics published a review [6] of Peano's book in 1891, along with another book on Grassmann's system by Edward Wyllys Hyde (1843-1930), including on the last page "In the configuration of the second species, the product of two vectors ... plays a part analogous to that of the vector in the first species. It is called a bivector by Prof Peano ("plane vector" by Prof. Hyde)". So the appearance of the term in English can be dated to at least 1891, and in Italian to at least 1888. Jheald (talk) 11:20, 10 March 2013 (UTC)

No. The exterior square v ∧ v of any vector is always 0. It can be non-zero only if scalars do not commute, which is not the case for fields (including complex numbers). Incnis Mrsi (talk) 19:44, 28 April 2013 (UTC)

Jheald has shown good evidence. The 1938 book by Elie Cartan is key. After 1966 it became available in English. Geometrie Sponorielle (1973) by Max Morand supported Cartan's initiative. Considerable parts of these works take place in the arena of 2x2 complex matricies, precisely the algebra of biquaternions. Expropriation of the term bivector seems not to have been protested. Some efforts of Ronald Shaw have confounded the two uses. The state of the History section of this article might be improved by Jheald's references.Rgdboer (talk) 20:29, 22 April 2013 (UTC)

Actually the key book -- in Europe anyway -- is probably the one by Peano. This book, according to our article, is what made Grassmann's exterior algebra come to be seen as mathematically respectable. So it's no surprise that we find Peano's terminology being taken up in Italy, Germany and France, which were the countries where most interest in Grassmann's work developed, while the Anglo-Saxon world became increasingly focussed on the vector algebra of Gibbs and Heaviside.

There was one 1920s French book I found (that unfortunately I can't seem to find again) that even explicitly characterised the two schools as the "vector" system and the "bivector" system, according to whether you had a vector product or a bivector product of two vectors. Jheald (talk) 22:00, 22 April 2013 (UTC)

Exterior algebra vs. Clifford algebra

The article is even worse than you thought, guys. While you conduct an argument about primogeniture and related stuff, the "definition" confuses two related, but different things: the Clifford product u v = u ⊗ v − u⋅v and the exterior product 2(u ∧ v) = u ⊗ v − v ⊗ u = u v − v u. This misunderstanding leads to the present gibberish in the Bivector #The space Λ2ℝn subsection which contradicts itself about whether bivectors form a linear subspace, and in the next subsection which shows its dimension (a triangular number) without explanation where you got it from. The Clifford product of two vectors is a bivector plus a scalar. Only for pairs of orthogonal vectors are the Clifford product and the exterior product the same thing. Thanks to Special:WhatLinksHere/Euclidean subspace which caused me not only to browse, but to examine this article. Incnis Mrsi (talk) 19:44, 28 April 2013 (UTC)

I've fixed the 'The space Λ2ℝn' section. The dimensionality of the even subalgebra's a pretty standard and straightforward result, though I don't have a source for it: the only things I can find point to special cases, e.g. in 2, 3, 4 dimensions. This is just how the sources I'm familiar with work: they from basics prove results in 2, 3, 4 dimensions but don't do much for a general number of dimensions.--JohnBlackburne^words_deeds 20:35, 28 April 2013 (UTC)

For a Clifford algebra one can orthogonalise to produce an orthogonal set of basis vectors e_i. The number of basis bivectors e_ij is then straightforwardly n(n-1)/2, since there are n ways to choose i, (n-1) to choose j ≠ i, divided by two because e_ij = - e_ji. Jheald (talk) 22:18, 28 April 2013 (UTC)

Given two Clifford objects of homogeneous grades m and n, their exterior product is simply the m+n grade part of the Clifford product. Jheald (talk) 22:24, 28 April 2013 (UTC)

LoL… do you think that Clifford algebra is a Z-graded algebra, do you really? It would be fairly naïve. Incnis Mrsi (talk) 05:45, 29 April 2013 (UTC)

Oops, I understand the thought, it was just poorly expressed. There is no “Clifford objects of homogeneous (Z-)grades”. But there is a (linear only) isomorphism between the exterior algebra (which is Z-graded) and Clifford algebra (which is not), where respective multiplications are related as Jheald wrote above. Incnis Mrsi (talk) 05:52, 29 April 2013 (UTC)

Clifford algebras may not be Z-graded algebras (only Z₂ graded algebras), so perhaps technically one should only talk about the even-odd grading of a Clifford algebra. (Lounesto pp. 42-44). Nevertheless, because of the trivial relationship between the basis n-vectors of the Clifford algebra and those of the external algebra, it is common to talk about the grade (or dimensional grade, if one wants to be absolutely explicit) of each part of an element of a Clifford algebra, and to denote the grade-m part by <A>_m (eg Hestenes & Sobczak pp. 1 & 4; Doran et al, p. 23; Dorst et al, p. 44 (exterior algebra), pp 150-151 (extension to Clifford algebra); Perwass, p. 57; etc). Jheald (talk) 09:18, 29 April 2013 (UTC)

Wikipedia policy on tech articles

I'm pretty sure this article is a violation of Wikipedia:Make technical articles understandable. Decora (talk) 18:04, 29 July 2013 (UTC)

Conceivably, though your comment is not particularly helpful. The topic is a subtopic of a larger topic (e.g. exterior and geometric algebras), and as such it is difficult to give an overview in isolation, withoiut the associated concepts. It can help for someone relatively new to the topic to edit it for readababilty: someone very familiar with the topic cannot see the hurdles faced by others. — Quondum 19:43, 29 July 2013 (UTC)

Decora, look, people have been striving to make all mathematics articles accessible which should go without saying... but as Quondum says, Clifford algebra has quite a steep learning curve. I'm not really an editor of this article, but recently I have been trying to try and clarify the term "bivector" with its links to the cross product and exterior algebra in the contexts of tensors and angular momentum in relativity. It's not just this article, but entire collections of articles that must transfer the ideas, and it is happening slowly but surely... M∧Ŝc²ħεИ_τlk 20:13, 29 July 2013 (UTC)

Maxwell equations

The vacuum means no sources, do the compact form is dF = 0, not dF = J. — Preceding unsigned comment added by 132.3.61.81 (talk) 03:28, 21 August 2013 (UTC)

No, this F (note boldface) is a bivector, not a differential form. Its ∂F corresponds to d ⋆F in the 2-form formalism. But I feel that the current version obfuscates what happened with another (d F = 0) equation. I can infer that the left side of ∂F = J is inherently a Clifford field and the equality states, inter alia, that coefficients of all four triple products (e₁ e₂ e₃, e₁ e₂ e₄, e₁ e₃ e₄, e₂ e₃ e₄) must be zero since the right-hand side is defined as a vector, but this does not follow obviously from the text. Incnis Mrsi (talk) 06:25, 21 August 2013 (UTC)