Talk:Cross product/Archive 2

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Archive 1 Archive 2 Archive 3

The seven dimensional cross product

This article needs to be merged with the seven dimensional cross product. The material from the seven dimensional cross product should be brought over to this article, and the latter article deleted. The material from the latter article should be put in a special section entitled 'Seven Dimensional Cross Product'.

Anything to do with the equation |a×b| = |a|×|b|sinθ only relates to the 3D case, and hence should only appear in the sections about the 3D cross product.

Does anybody object if I go ahead and bring the seven dimensional material over to this article? David Tombe (talk) 09:47, 2 January 2010 (UTC)

|a×b| = |a||b|sinθ applies in both three and seven dimensions. It's equivalent to one of the conditions, |a×b|2 = |a|2|b|2 - (a·b)2 from the Pythagorean trigonometric identity. So it's true in both 3 and 7 dimensions.
but as for merging, no that would not be appropriate. This article is on the common vector operation, as per WP:COMMONNAME. When mathematicians and non-mathematicians use the name "cross product" they are almost always referring to this product, not anything else. For most people it's the only one they know. The seven dimensional one is pretty obscure and most people never come across it (I learned of it after finishing my degree). Including it in here would just confuse people, as it's not a mainstream part of vector algebra, which is largely concerned with 3D, and it is also not as well defined as in 3D. --John Blackburne (wordsdeeds) 10:09, 2 January 2010 (UTC)
I fully agree with JohnB. The seven dimensional obscurity deserves an honorable mention under the generalisations section. It already has the mention. DVdm (talk) 11:03, 2 January 2010 (UTC)

John, Fair enough. But the seven dimensional cross product should at least get a more explicit link. As it stands now, we have to guess that the link to 'seven dimensions' will open out to 'seven dimensional cross product'. Anyway, I note that you say that

|a×b| = |a||b|sinθ applies in both three and seven dimensions.

The way I was taught it was that the equation |a×b| = |a||b|sinθ follows as a consequence of a×b satisfying certain conditions such as the Jacobi identity and the distributive law. The proof involved three dimensional geometry.

If you think that the equation |a×b| = |a||b|sinθ holds in either one or seven dimensions, I'd be fascinated if you could explain to me the meaning of the angle θ in either one or seven dimensions. Could we have a seven dimensional square? I read something in one of these two articles about seven dimensional Euclidean geometry. Now there is a concept I had never heard of before! Can you please explain to me what angle θ means in seven dimensions? David Tombe (talk) 11:40, 2 January 2010 (UTC)

Huh? Don't you know that in n dimensions, the angle θ between two vectors is defined as arccos( (a·b) / |a||b| ), which you can write as arcsin( √( 1 - ( (a·b) / |a||b| )2 ) )? Not all that fascinating if you ask me. DVdm (talk) 12:06, 2 January 2010 (UTC)
The angle is defined in seven dimensions as in two and three. You can use e.g. the dot product to calculate it. This can be used to relate the above formula to |a×b|2 = |a|2|b|2 - (a·b)2 – just replace the dot product and use the Pythagorean trigonometric identity.
In one it's a bit less obvious, until you realise that all lines in one dimension are parallel, so θ is zero or π, and so sin θ is always zero. The product is zero always, i.e. trivial.--John Blackburne (wordsdeeds) 12:03, 2 January 2010 (UTC)

John, As regards your answer about the angle θ in seven dimensions, we'll just have to agree to differ. I cannot remotely imagine what an angle means in seven dimensions. Neither can I imagine Pythagoras's theorem in seven dimensions. David Tombe (talk) 12:07, 2 January 2010 (UTC)

An angle between two vectors in 7 dimensions means the arccos of the dot product of the vectors divided by the norms of the vectors. That is what it means in n dimensions for any n. I'm really amazed you didn't know that. DVdm (talk) 12:18, 2 January 2010 (UTC)
It's just an algebraic theorem which can be applied here, using fairly elementary algebra. I suggest you try it before dismissing it.--John Blackburne (wordsdeeds) 12:10, 2 January 2010 (UTC)

DVdm, With cosine involved, I think we're back to tautologies again. Everything to do with cosine begins on a 3D premises. I don't think that the physical concept of 3D space can be extrapolated to 7D that easily. Do we have the same π? Try to imagine seven mutually orthogonal axes. How do we fit 360 degrees in there? I know a pure maths professor who was in my form at school. He is a specialist in 'space in n dimensions'. He couldn't answer when I asked him to explain the meaning of angle in 7 dimensions. He also just happens to be the one who taught me the importance of the distributive law in making a cross product useful. Everything to do with the concept of angle originates within the context of our perception of 3D space. I have often believed that when pure mathematicians extrapolate things to their favourite 'n dimensions' that it is an extrapolation beyond reality. It is were maths leaves the real world. David Tombe (talk) 12:30, 2 January 2010 (UTC)

You really continue to amaze me. DVdm (talk) 12:45, 2 January 2010 (UTC)

DVdm, You can have the cosine. But what does the angle itself mean? David Tombe (talk) 12:48, 2 January 2010 (UTC)

I didn't say anything about cosines. I said something about the arccos of a real number. But anyway, tell your pure maths professor that it produces for instance the "real world angle" in the 2-dim subspace spanned by the two vectors. DVdm (talk) 13:48, 2 January 2010 (UTC)
Inner product space#Norms on inner product spaces --John Blackburne (wordsdeeds) 12:50, 2 January 2010 (UTC)
And WP:RD/MA --John Blackburne (wordsdeeds) 12:53, 2 January 2010 (UTC)
Yes, I think that educating an interested lay person in elementary aspects of vector spaces is a bit off-topic on this talk page. O.t.o.h referring to the reference desk can only point back to where we have been pointing already. DVdm (talk) 13:48, 2 January 2010 (UTC)

It appears to me that you are both overlooking something very fundamental. The 3D cross product involves 3 components in any operation, and it can be used to describe realities in 3D space. The same analogy does not exist with the 7D cross product. In any given operation, we are still only using 3 components, and the linkage to 3D space has been lost. As regards 7D space, if such a thing actually exists, the analogy with the 3D cross product's relationship to 3D space does not exist.

In the sources, any proof that the 3D cross product obeys the relationship |a×b| = |a||b|sinθ will be explicitly in relation to the 3D cross product's linkage to 3D space. There is no proof in the literature that the 7D cross product has got any corresponding equivalent to the equation |a×b| = |a||b|sinθ in relation to 7D space. It was an over extrapolation on the part of whoever wrote the article, to write it in such a way as to imply that |a×b| = |a||b|sinθ was a general result for cross products in 1,3, or 7 dimensions. This was of course no doubt due to the fact that whoever wrote the article was clearly writing it exclusively with the 3D cross product in mind.

What needs to be done it is to make a clear segregation in the article between the 3D case and the 7D case.

While on the subject, let's imagine a 3D rotation. We'll fix the k axis in space and rotate about the k axis in the ij plane. The concept of angle will be understood in that context. Now let's imagine a 7D situation. Once again, we'll fix k in space. But this time we could define rotations in quite a few planes. And if we take the general cross product a×b where a and b are defined in terms of all seven components, then we will be completely lost. Because once we have established the rotation axis, we will have a headache trying to decide which plane to take our angle from.

A truly 7D cross product which would link to a real 7D space by the same analogy with the 3D case, would have to involve all 7 components in any given operation. We do not have a truly 7D cross product in that sense.

The 7D cross product that we do have, is somewhat of a misnomer. It is 7D only in the sense that it involves 7 components. But these seven components do not actually represent unit vectors in any space. David Tombe (talk) 15:07, 2 January 2010 (UTC)

Not much more I can say. The cross product is defined in 3D and 7D. It can be proven to exist in only those dimensions, subject to the restrictions in the definition. All this is on the 7D cross product page. There's no problem with doing it 7D as vectors, angles, etc. are all defined in 7D, identically to 3D (as an angle is a 2D thing it needs defining in more general dimensions). Apart from the two cross product pages there are the links we've given above and the various wikillinks and references on the pages, where it is explained in much more detail at all sorts of levels of expertise. --John Blackburne (wordsdeeds) 15:32, 2 January 2010 (UTC)
"The 7D cross product that we do have, is somewhat of a misnomer." ==> This reminds me of "The speed of light is somewhat of a tautology...".

David, I advise you to avoid this kind of alley. Again, your lack of expertise is not the root problem here but it isn't really helpful, given your obvious failing to know how angles are defined in n-dimensional vector spaces yet attempting to reorganise articles about cross products in 7-dim spaces, and at the same using the talk page to complain about "extrapolations beyond reality" and "where maths leaves the real world". You seem to be insisting to push some point resulting from what looks like a shallow synthesis of a few sources. This is the kind of talk page disruption that brought you in trouble before, so perhaps you shoud be careful. DVdm (talk) 16:51, 2 January 2010 (UTC)

Oh I see. This is disruption is it? In that case I better leave you to it. David Tombe (talk) 17:07, 2 January 2010 (UTC)

A source to back up the idea that no source is required

I've got a source here [1] which more or less backs up my viewpoint that no source is necessary to back up the view that the reason why cross product doesn't work in 5D is because it doesn't obey the distributive law.

The source states that students may enquire as to whether or not a cross product exists in dimensions other than 3. The source states,

This note points out that by proving the elementary propositions and theorem below, students may answer this question themselves.

In other words, if the basic axioms such as distributive law, and Jacobi identity hold, then we will only have a result in 1, 3,and 7 dimensions. As I have been repeatedly saying, this fact is too basic to require a source. But although it is basic,it seems to have been overlooked by some of the editors here.

In matters to do with maths, first principle arguments are often sufficient in the absence of sources. David Tombe (talk) 05:47, 4 January 2010 (UTC)

The note suggests the proof is "elementary" for "students" but does not say how elementary. It also does not give the proof, but "indicates" them. Lastly it says nothing about the distributive law, saying instead that the proofs will need "knowledge of" various things, mostly related to orthonormality. As even if you pay $12 you don't get a proof, just an indication of one, it's a lot less useful as a source than the two proofs on Seven-dimensional cross product which are complete and free. --John Blackburne (wordsdeeds) 09:53, 4 January 2010 (UTC)


"... which more or less backs up my viewpoint that no source is necessary to back up the view that ..." ==> This is probably the best example of a combined case of wp:NPOV, wp:SYN and wp:SOURCE I have ever seen here. That phrase should be taken as an example in various policies and guidelines. DVdm (talk) 10:19, 4 January 2010 (UTC)


John, You were reading the wrong bit. The bit that I was referring to is the quote that I made in italics above, and the sentence immediately following it. I was not talking about any proofs. You are the one who keeps changing the subject to proofs. I was talking about the fact that it is easy to see how only 1D, 3D, and 7D satisfy the basic axioms. One of those basic axioms is the distributive law, and that happens to be the crucial axiom in this issue. But perhaps then you would prefer to put something more vague and general into the main article along the lines of 'the 5D and the 15D cross products don't work because they fail to satisfy the basic axioms'. David Tombe (talk) 11:27, 4 January 2010 (UTC)

The source says nothing about the distributive law--John Blackburne (wordsdeeds) 11:38, 4 January 2010 (UTC)
David, you might consider stopping point-push-disrupting this talk page. You are behaving here (and on Talk:Seven-dimensional cross product) like you were behaving on Speed of light, which earned you a year long ban on physics related articles. What you are doing could lead to an extension to mathematics related articles, or worse. DVdm (talk) 11:49, 4 January 2010 (UTC)


John, The source talks about axioms. The distributive law is one of the axioms of the 'cross product'. A 5D cross product does not satisfy that axiom. That is why we cannot have a 5D cross product. I am very surprised that you didn't know all this. David Tombe (talk) 11:57, 4 January 2010 (UTC)

Distributivity

The derivation of the formula for the cross product depends on the distributive property, but I'm not aware of a simple proof of distributivity. The article should either include a simple proof, or briefly sketch a hard one, or give a reference. But it should at least acknowledge that there's some work to be done.—Preceding unsigned comment added by 129.89.14.247 (talkcontribs) 20:09, 29 January 2010

It doesn't need to be proved as it's part of the definition: it's required to be a Linear map which implies distributivity. This is stated explicitly in the Seven-dimensional cross product article, which is more concerned with that particular generalisation. That together with the orthogonality and the mangnitude being ab sin θ define the cross product entirely. It's not so important in 3D as for most people it's just the product given by the formula, it could be clearer though in 3D so something about the conditions should probably be included. --JohnBlackburnewordsdeeds 20:25, 29 January 2010 (UTC)

The Lagrange Identity in seven dimensions

John, I think you had better explain your reversion. We have just spent a few days arguing about this on the talk page of Seven dimensional cross product. I had been mislead in part because of this article which restricted this particular form,

of the Lagrange identity to 3D. After working it out from first principles I finally conceded that you were correct and that it holds in 7D also. I came here to fix this article in that respect. You have now reversed your position. You had better explain yourself. David Tombe (talk) 10:54, 20 April 2010 (UTC)

There is nothing wrong with either article that needs 'fixing', it is only that you (still, clearly) do not understand it. Lagrange's identity is only equivalent to the above ("Pythagorean") identity in 3D, as its article says. Further this article is about the cross product in 3D, so should not have higher dimensional results inserted at arbitrary points. It links to the 7D version in the lede, has a section on higher dimensions with links further down.--JohnBlackburnewordsdeeds 12:01, 20 April 2010 (UTC)

John, Are you sure you understand the subject? On this page, you are saying that it is 3D only. But over at Seven dimensional cross product you are saying that it is both 3D and 7D. Your argument about this being the 3D page doesn't wash because you also did the same reversion on the more general Lagrange identity page. David Tombe (talk) 12:47, 20 April 2010 (UTC)

This article is not intended to be exclusively about the 3D cross product

John, the reason which you gave for reverting that latest edit doesn't wash. You went to the WikiProject mathematics and you had it confirmed there that the relationship in question applies in seven dimensions. This article is about the cross product in general, even if it concentrates on the more familiar 3D cross product. If a special relationship holds in both 3D and 7D, we cannot state that this relationship is a special 3D case without also mentioning that it is a 7D case as well, because to do so would be misleading. And in that respect, the article as it now stands after your revert, is misleading, because it makes readers think that this is a relationship which is exclusively restricted to 3D. The chances are that whoever first wrote this article didn't think about the 7D case. But we have now established beyond any doubt over at the talk page of seven dimensional cross product talk page that it holds in 7D and so this article needs to be corrected in the light of that information. David Tombe (talk) 15:20, 23 April 2010 (UTC)

This article is a about the product in 3D, as the introduction says, and everything from sections 1 to 7 is written with the assumption the maths is in 3D. The reason for this is the cross product, to mathematicians, means the product in 3D. Most don't know about the product in 7D, and if they do it's given a separate name, so the 3D version can be called "The Cross Product" without ambiguity.
Towards the end it mentions the different generalisations of the cross product - there are more than just the seven dimensional one, depending on how you relax the conditions on the cross product. But the article is not also about them, or any one of them, it is just about the product in 3D.--JohnBlackburnewordsdeeds 15:42, 23 April 2010 (UTC)
To settle this matter, I have added a reference template to the 7D case at the top of this article, and added the 7D case to the disambiguation page for "cross product". Brews ohare (talk) 16:31, 23 April 2010 (UTC)

John, Even if the article is only about the 3D cross product, we still need to make any general information unambiguous and not have readers thinking that a particular result is exclusive to 3D. Adding the information about 7D does not detract from the purpose of the article, whereas ommitting it can spread misinformation. So I will add the information in in brackets. Having said all that, the article shouldn't be only about the 3D cross product. It should be about the 'vector cross product' in general, with an emphasis on the more familiar 3D case. There is really no need to segregate the two articles. They can be written together coherently, pointing out the commonality and then the differences in the introduction, and then doing separate treatments of each. David Tombe (talk) 02:43, 24 April 2010 (UTC)

No, for the reasons above and earlier on this talk page. I recall that you tried to do this before, with no better arguments than now. Please drop it as there is really no point reopening this debate.--JohnBlackburnewordsdeeds 08:40, 24 April 2010 (UTC)

John, You have just reverted an edit and referred me to the talk page. I'm now looking at your explanation and I can't see anything which remotely addresses the issue of why you reverted the edit. The edit in question did two unrelated things. First of all, it segregated the issue of the "Lagrange identity" into a separate sub-section from the "vector triple product". Secondly, it made an important clarification that this form of the Lagrange identity,

holds in both 3 and 7 dimensions. As the article stands now, it is misleading by its omission of any mention of the 7D case. It doesn't matter whether the article is exclusively about the 3D cross product or not. The information regarding the applicability in 7D is so closely related to the topic that its omission is positively misleading. Anybody reading the article as it stands now will easily believe that the above equation only holds in 3D. The purpose of wikipedia is to supply information and not to hold back little secrets for another day. David Tombe (talk) 14:47, 24 April 2010 (UTC)

Incorrect definition

I just restored the previous version of the definition, which is the generally accepted and most widely understood one. It is also the one given by Lounesto, on page 93. On 94 he gives an alternate definition in terms of purely vector products, so he can do algebra with it, but it is not a generally accepted or readily understood definition.--JohnBlackburnewordsdeeds 20:53, 23 May 2010 (UTC)

Hi John: I'm OK with that. What I was trying for was a definition that is easily extended to n-dimensions. Can you help? Brews ohare (talk) 21:02, 23 May 2010 (UTC)
No need for that as long as the text "This article is about the cross product of two vectors in three-dimensional Euclidean space" is sitting on top of the article. DVdm (talk) 21:06, 23 May 2010 (UTC)
As has come up before the place for extending it to 7D, the only dimension it extends to, is the 7D cross product page. That's already linked in the introduction and has a section in Generalizations, so does not need anything else adding. This is a prominent article on an elementary topic, so including more content from the 7D cross product article would I think be inappropriate.--JohnBlackburnewordsdeeds 21:10, 23 May 2010 (UTC)
The "cross product" of n−1 vectors in n-space is well-defined; and the properties on p.779 of the Mathematica Book, together with the "right-hand" rule
(where ei are the unit vectors in the respective directions) should suffice. But that's a different article. The sense in which there are cross products only in dimensions 3 and 7 (with reference) might have a definition which could be extended to multilinear cross products, but I don't have a copy of any such reference with me. — Arthur Rubin (talk) 21:26, 23 May 2010 (UTC)
That's already in the article, under Multilinear algebra. The restriction to 3 and 7 dimensions is for binary products satisfying the criteria, so if you say the crosss product is a × b = c it is only is sensibly and non-trivially defined in those dimensions.--JohnBlackburnewordsdeeds 21:32, 23 May 2010 (UTC)
This article includes short sections on octonion, quaternion, wedge products etc. So it is not strictly 3-D a × b. Maybe the preamble should be extended to describe its scope and its connection to other articles? The connection to cross products of n-1 vectors in n-dimensions should be brought up, for example. Brews ohare (talk) 21:47, 23 May 2010 (UTC)
It's the 67th most popular maths article, and the vast majority of those readers will I think be after the cross product as used in vector algebra, i.e. the 3D one, it's definition and basic properties. The more advanced material should come later, as it now does. Moving it to the top will just confuse many readers, as e.g. those who do maths at school won't have the math background to deal with it.--JohnBlackburnewordsdeeds 22:18, 23 May 2010 (UTC)

Article topic

The article is about the cross product in 3D, i.e. what mathematicians generally understand by the name "cross product". It does not describe the generalisations: it provides links summary-style to most of them, and only the n − 1 way product in n dimensions is described at all.--JohnBlackburnewordsdeeds 22:26, 23 May 2010 (UTC)

John, There is clearly a problem with this article as regards terminologies and disambiguation. The statement at the top that the article is exclusively about the binary operation in 3D is not true. As you have already said yourself, the multilinear algebra section deals with that other concept of cross product which involves multiple vectors in a single operation and extends to the corresponding multiple dimensions. It's a different subject, except in the 3D case, and it is covered in this article. The scope of this article needs to be clarified and the headings amended accordingly. David Tombe (talk) 00:47, 24 May 2010 (UTC)
John: The template claims that the article is about the cross-product in 3-space. However, it has a section on "Generalizations" which has little if anything to do with the 3-D cross product. It mentions normed division algebras, exterior algebra, and multilinear algebra including (n-1)-ary products. There are a few choices: change the template and introduction to describe the article as it actually is, or rewrite the article to fit the present description. Brews ohare (talk) 02:48, 24 May 2010 (UTC)
The article is not about the cross product in 3D "and its generalisations". It briefly mentions some generalisations. This edit was clearly making a disruptive WP:POINT in response to this remark. Please stop doing that. DVdm (talk) 09:03, 24 May 2010 (UTC)

This seems to be a rather unnecessary argument in my opinion, and it stems from the note at the top of the article which restricts the article to the 3D binary operation. I think that we are all agreed that the 3D binary operation is by far the most important and well known case. And I think that we are all agreed that the 3D binary operation should dominate the article. However, there is a section, and rightly so, which deals with the varoius ways of extending this concept to higher dimensions. The note at the top of the article is therefore inaccurate. The problem is not so much a problem with the article itself as a problem with this inaccurate and restrictive note which has been added at the top of the page. In my opinion, the article is about 'cross product'. It should deal with the 3D case first, and then discuss the extensions to higher dimensions further down. That is more or less exactly as it is at the moment. David Tombe (talk) 12:04, 24 May 2010 (UTC)

The note at the top of the article looks perfectly accurate to me. DVdm (talk) 12:30, 24 May 2010 (UTC)

It does not make sense to me

Hi, you cannot define right or left without defining up and down, and it is not in 3D space anyway, but in 2D. Can you see that? Genezistan (talk) 15:26, 25 October 2010 (UTC)

I don't follow. In n≥1 dimensions, I can define "left" to be any direction and "right" to be the opposite direction. There are issues with the cross product in any other than three dimensions, but the page discusses this. Can you rephrase your question? —Ben FrantzDale (talk) 17:46, 25 October 2010 (UTC)
I am trying to visualize what you claim to be in n≥1 dimensions, where you say you can define "left" to be any direction and "right" to be the opposite direction in 2D and I need to have up or down defined, otherwise I have a problem (rotate by 180 degrees to check).
Is that any better?
Regards, Genezistan (talk) 21:55, 25 October 2010 (UTC)
I'm missing something. In R2, I'm assuming we have a coordinate system defined so a vector, x can be distinguished from −x and where −x is equal to x rotated by 180°. Without loss of generality, assume x = [−1,0]. If we assume the space is right-handed, then if x is the "left" direction, then [0,1] would be "up". Is it the assumption of handedness that you are wondering about? You are right that if we don't assume a right-handed coordinate system that we would need to define which way is up versus down in addition to left versus right. The assumption is completely standard, though, so it is usually omitted. Does that help? —Ben FrantzDale (talk) 16:59, 27 October 2010 (UTC)
Yes. Thank you very much. What do you think of the fact that in space there is symmetry, so it does not matter where you put the origo of a co-ordinate system or where you start time from?
Genezistan (talk) 17:24, 27 October 2010 (UTC)
I think you've stumbled on the distinction between a vector space and an affine space. In a vector space, there is one origin and all vectors have their tail there; in affine space, there are points and so-called bound vectors (tangent vectors in the tangent space of points). Subtracting one point from another gives a vector from one to the other. To rephrase your observation, in Euclidean space all tangent spaces are isomorphic—they look the same, so with the exception of algebraic restrictions (such as adding two points being meaningless), the space of points is isomorphic to the space of vectors, so the distiction is usually glossed over in introductory treatments. I find the affine-space treatment more pedantically-satisfying. In more-general spaces such as on manifolds, the distinction between points and tangent vectors becomes critical because points are not in general in a vector space, and each tangent space can look different (e.g., having a different metric tensor). —Ben FrantzDale (talk) 14:23, 28 October 2010 (UTC)
Also, it is possible to rotate such a 2D representation or to mirror it when you lose left or right directions again. What I am also after is the visual representation of mathematical symbols, including numbers. I do not like that in writing for instance 12345, we do not "see, because it is not there" that an individual digit is a cross product of the value of that position (power) and the number iself. I would like to have a notation that reveals such a hidden cross-product.
Genezistan (talk) 04:30, 28 October 2010 (UTC)
Or more precisely I think we should see the cross product of base x power x count explicitely for visualization purposes.
Genezistan (talk) 06:16, 28 October 2010 (UTC)
I am not sure quite what you mean. I agree that it is very important in applied math to think of quantities (numbers, vectors, tensors, etc.) as physical things, not just symbols. However, "...an individual digit is a cross product..." doesn't make any sense to me. I don't know what you mean by "hidden cross-product". —Ben FrantzDale (talk) 14:23, 28 October 2010 (UTC)
Further to that and knowing that in Phyisics where we can visualize qualities and quantities the representation of a triple cross rpoduct could be a sphere with three radiuses not necessarily connected in right angle. In fact we need four vectors and the fourth could be the unit of measure as no count makes sense without a unit of measure!
And since the three vectors are used to represnet directions in space, the fourth should autimatically be time with the difference that it has just one direction and it is positive. So the visualisation could or should be someting similar to what you get with seven cross products.
Genezistan (talk) 07:20, 28 October 2010 (UTC)
In fact the concept of spacetime is also a cross product with motion (and speed to be also considered), if I am not mistaken. What do you think?
Genezistan (talk) 07:08, 28 October 2010 (UTC)
I think you would find geometric algebra and covariance and contravariance of tensors interesting. (Be warned, it is a huge can of worms, but I think it provides the framework you are craving.) As you have seen, in the classical treatment of 3-vectors and cross products, the algebra doesn't mirror the physics: if a quantity is a cross product of vectors, it is an oriented area, and if it is a triple product it is a volume. Geometric algbra addresses this type system. Tensor variance and related concepts address the issues of how these sorts of quantites transform as you change coordinate systems. (For example, a distance gets numerically bigger if you use kilometers rather than miles, but a gradient (e.g., degrees C per distance) gets numerically smaller if you use kilometers rather than miles.) —Ben FrantzDale (talk) 14:23, 28 October 2010 (UTC)

Duplicate matrix material

I raised this here but it was deleted without a reply, so I'll raise it here. The just added material on the matrix representation duplicates what's already there, except less clearly (it's not clear what the significance of the last two lines of formulae is) and with one error (the link to Lagrange's identity is wrong - the formula is a different one with the same name).--JohnBlackburnewordsdeeds 20:16, 31 May 2010 (UTC)

The added material has a plus: it's sourced. The pre-existing material is not. The Lagrange identity for vectors has a few variations. They aren't exactly the same, but are all interrelated: some involve two vectors, some three and some four. It may be too much trouble to sort them all out. Brews ohare (talk) 21:23, 31 May 2010 (UTC)
It's a different Lagrange's identity. I've seen it before, on WP I think, but you provided a link which made it clear: it's the formula
(A × B)⋅(C × D) = (AC)(BD) - (AD)(BC)
"which is sometimes known as Lagrange's identity" so the link to Lagrange's identity is incorrect.
But the matrix content is just a copy of what's already there: The matrices Ta and [a]× are identical and they are used the same way, so the {{cn}} tag you've added is nonsensical as it's the same content you added - you've just added a perfectly good source for it.--JohnBlackburnewordsdeeds 21:50, 31 May 2010 (UTC)
If the source is to be used, the notation of the source is preferable to an unsourced (invented?) notation. I am a bit surprised that the simple substitution A=C and B=D which makes the cited form of Lagrange's identity exactly that of the link happened to escape your notice. This extra generality is sometimes used in naming this identity, sometimes not.Brews ohare (talk) 00:34, 1 June 2010 (UTC)

I added the link to Binet–Cauchy identity, a more general form. Brews ohare (talk) 00:48, 1 June 2010 (UTC)

I've merged the two definitions, preferring the notation which was there before as it's more informative and there's more of it. At the same time I moved the other Lagrange's identity up, so it's clearer how it fits in, using consistent notation for that too, and saying a bit more how they're related.--JohnBlackburnewordsdeeds 13:32, 2 June 2010 (UTC)

Vector quadruple product

According to [2] and [3] the vector quadruple product is (AxB)x(CxD), not (AxB).(CxD). The latter is called a "quadruple scalar product", whereas the former is called "a quadruple vector product" here. I have removed the remark added by 128.205.180.128 (talk · contribs) and which was corrected for caps by John. DVdm (talk) 14:53, 12 September 2010 (UTC)

Being introduced to it by that change I've looked at the article and have created a deletion discussion for it, hopefully to get a consensus over what if anything is the best title for an article on such an identity or identities.--JohnBlackburnewordsdeeds 15:06, 12 September 2010 (UTC)
Heh. I have added a delete !vote at Wikipedia:Articles_for_deletion/Vector_quadruple_product. DVdm (talk) 15:16, 12 September 2010 (UTC)

Clifford algebra

Currently the History section contains this statement:

William Kingdon Clifford combined the algebras of Hamilton and Grassmann to produce Clifford algebra, where in the case of three-dimensional vectors the bivector produced from two vectors dualizes to a vector, thus reproducing the cross product.

Today I added the reference to Clifford's 1878 text Introduction to Dynamic which is explicit on the vector product. The above statement has no reference. Further, the allusion to Clifford Algebras does not improve the discription of cross product's history. The statement appears superfluous.Rgdboer (talk) 03:46, 9 December 2010 (UTC)

Although it does not provide references, the statement is not superfluous in my opinion. It is useful to see that mathematics, at that time, followed two paths, one of which brought the cross product. The other path produced an exterior product which, dualized in 3D, produces a cross product. There's a section in this article about "Cross product as an exterior product", and I think it is nice to have a reference to that section in the history section.
Your edit is very interesting. It seems that 3 years before Gibbs published his notes, Clifford already defined the cross product (although he did not call it cross product). Does it mean that Clifford is the father of cross product? This is weird, as I have read everywhere that Gibbs and Heaviside were the fathers of cross and dot product. Does Clifford refer to Gibbs or Heaviside, in his 1878 text?
Paolo.dL (talk) 09:28, 9 December 2010 (UTC)

For two vectors, the vector part of their quaternion product is the cross product. So Hamilton's earlier product includes the idea. As for Clifford, he represents area, and its orientation, by a vector. This notion he may have gotten from Grassmann, but generally Clifford doesn't give references. Rgdboer (talk) 02:02, 10 December 2010 (UTC)

Who is the father of the cross product?

This sentence was inserted by Rgdboer in the history section:

  • In 1878 William Kingdon Clifford published his Introduction to Dynamic which brought together many mathematical ideas. He defined the product of two vectors to have magnitude equal to the area of the parallelogram of which they are two sides, and direction perpendicular to their plane.

This sentence is very interesting, but it puzzles me. The problem is: who is the father of the cross product? I always thought that Grassmann, Hamilton and Clifford were not able to define a vector product which yielded another vector perpendicular to their plane (what we now call the cross product). Here's why they, as far as I know, were not the fathers:

  1. The quaternion product was defined by Hamilton before the cross product, but quaternions are 4-element vectors. Although the vector part of the quaternion product, which involves only 3 elements, is what now we would call a cross product, Hamilton failed to see it. So, Hamilton is not the father of the cross product.
  2. The exterior product (or wedge procuct) was defined even earlier by Grassmann. In R3 it is the same as a cross product as far as its "scalar components" are concerned. However, its "vector components" are bivectors, not unit vectors. And a bivector, as far as I know, was described and interpreted as a directed parallelogram (a 2-D surface) with an area, not as a directed line segment with a length. I guess that this also means that, if a vector is measured in meters, the exterior product is in square meters. So, Grassmann was not the father.
  3. As far as I know, the Clifford product is similar to, but even more complex than the exterior product, as it takes into account an additional argument, called Q. So, Clifford was going away from the cross product, rather than toward its discovery.

However, the above mentioned sentence inserted by Rgdboer in the history section seems to say something different.

Question: Did Clifford, in 1878, 3 years before the publication of the notes in which Gibbs described for the first time its cross product, really define a "product of two vectors" which returned another vector (i.e. a directed line segment) in the same vector space? This would not be an exterior product or a Clifford product, but a vector product identical to what we now call a cross product! The answer to this question is crucial.

Paolo.dL (talk) 15:58, 10 December 2010 (UTC)

I would not be surprised if he did (though the mentioned source seems not to exist) – the cross product can be derived from the exterior product of Grassmann within geometric algebra by taking the straigtforward dual of the bivector result of the exterior product, and this is only possible with Clifford's geometric algebra, newly discovered in 1878. Geometric algebra includes both Grassmann's algebra and quaternions so can seem more complex, though it can also used to define both from very simple axioms. Gibbs' cross product though was derived from the quite different quaternion product, by taking just the imaginary part, ignoring the insights available with the full quaternion algebra, seemingly unaware of Clifford's deeper discoveries. But that paragraph does seem out of place, as Clifford is mentioned later.--JohnBlackburnewordsdeeds 16:35, 10 December 2010 (UTC)
Mentioned later, but for a different reason. The inserted sentence does not say that you need a dual to obtain a vector. It seems to say that the "product of two vectors" defined by Clifford in 1878 is already a vector (not a bivector). Paolo.dL (talk) 16:57, 10 December 2010 (UTC)


the direction of the cross product by the right-hand rule

In the section of Definition, it says the direction of the cross product by the right-hand rule. But the wiki page of right-hand rule focus the relation of x,y,z-axis of 3-D coordinates. When we consider the cross product a × b , how to determine the rotation from a to b? It seems the rotation angle from a to b in the plane containing a and b must be smaller (0° ≤ θ ≤ 180°)? -ligand (talk) 21:48 May 13,2011 (UTC)

Right hand rule cross product.svg
Yes, as the article says the angle is always measured so it is between 0 and 180 degrees. The angles 0 and 180 can be ignored as the vectors are then parallel and the cross product is zero, so we need only consider 0° < θ < 180°. We can also ignore cases where a or b is zero. The vectors must form two sides of a non-trivial triangle which defines a plane. And the sense of the rotation from a to b in this plane gives the direction, by the right hand rule, as given e.g. by the diagram at right. The right hand rule is a convention: it's not something that can be proved, we just chose from the two possibilities the result given by the right hand rule, so it matches the convention for the coordinate axes.--JohnBlackburnewordsdeeds 22:06, 13 May 2011 (UTC)