Talk:Seven-dimensional cross product/Archive 1

Some questions 2008
Is this an Exterior product or something else? --Salix (talk): 19:19, 27 September 2008 (UTC)

Why can't cross products be defined in fifteen dimensions? The article should provide at least some explanation, even if it's obvious, because nothing's really obvious at this level. Eebster the Great (talk) 02:36, 8 December 2008 (UTC)

Multi-dimensional vector product by Silagadze
I've read the attached reference to Multi-dimensional vector product by Z.K.Silagadze. The quote of interest would be: "From a view point of composition algebra,the vector product is just the commutator divided by two. According to Hurwitz theorem the only composition algebras are real numbers,complex numbers,quaternions and Octonions. Quaternions produce the usual three-dimensional vector products. The seven-dimensional vector product is generated by Octonions." I'll try to put this in with a tag for improvement. 86.80.122.213 (talk) 00:01, 3 April 2009 (UTC) LCV

Nevermind, looks like it's already there. 86.80.122.213 (talk) 00:03, 3 April 2009 (UTC) LCV

Why only three and seven?
Eebster asks, 'why can we not have a fifteen dimensional vector cross product?'. Well at least in part, the answer lies with the distributive law. The only reason why the vector product is so useful is because it can be used to describe certain problems in three dimensions, and that we can multiply the individual components out. If the operation did not obey the distributive law, it would have no practical use, because we could not numerically multiply the individual components out.

As it so happens, the seven dimensional cross product also obeys the distributive law.

As regards the deeper question of 'why only three and seven?', I don't know. I read in a 1970's Encyclopaedia Britannica that they were in the process of deriving a theorem to prove that it could only exist in three and seven dimensions, but that the theorem was very lengthy and complicated, and that it was not yet completed.

All I can say is that I once tried to multiply it out in five dimensions and it failed the consistency test. If you want to set up a fifteen dimensional set of operators and try it out, you are welcome. But you will need lots of paper, and lots of patience, and your eyes will be sore at the end of it. The chances are that you will make at least one silly mistake which will ruin the whole purpose of the exercise. David Tombe (talk) 03:17, 29 December 2009 (UTC)


 * The reason is given in the last section: the space above it must be a normed division algebra, and these only exist in 1, 2, 4, 8 dimensions. So a cross product can only exist in 0, 1, 3, 7 dimensions. Except the first two are too trivial, so only in 3 and 7 dimensions. --JohnBlackburne (talk) 10:18, 29 December 2009 (UTC)

John, I think it needs alot more explaining than that. Why only three and seven? And you reply "the space above it must be a normed division algebra, and these only exist in 1, 2, 4, 8 dimensions". Where's your proof that these exist only in 1, 2, 4, and 8 dimensions? You've stated the facts themselves as the proof. David Tombe (talk) 14:48, 29 December 2009 (UTC)


 * WP:NOT, in particular WP:NOTTEXTBOOK, suggests you are looking in the wrong place for proofs. The article has some good references so you could consult those, or try and prove it yourself (you will learn a lot more that way). Or ask at the reference desk --JohnBlackburne (talk) 14:58, 29 December 2009 (UTC)

John, I'm not going to try and prove it myself. The Encyclopaedia Britannica in the 1970's said that the proof was extremely complex and as yet not completed. So it would be a tall order to suggest that I could just simply prove it myself. Your first reply made out that you knew the proof and that the proof was a one liner. I think that we both know that that is not the case. David Tombe (talk) 15:08, 29 December 2009 (UTC)


 * I would disregard the Encyclopaedia Britannica if it's so old. Most of the theory on it is based on today has been developed significantly in the last two decades. And where did I say "one liner" ? --JohnBlackburne (talk) 15:21, 29 December 2009 (UTC)

OK John, There are a number of points to be tackled here. A 1970's Encyclopaedia Britannica is not old. Things have not moved on that much since the middle of the nineteenth century when Hamilton's colleagues were getting into difficulties with the 16 dimensional equivalent to octonians. Encyclopaedia Britannica is written by professionals. Maths articles in the Britannica will be written by recognized academics.

An encyclopaedia is not a textbook. It's purpose is to give basic knowledge to interested readers. So we must not lose track of the fact that wikipedia is also an encyclopaedia and not a textbook. Encyclopaedia readers don't want to read 'pure maths speak'. And the so-called explanation in the main article for why we can only have the cross product in 3 and 7 dimensions, not only fails to explain it, but it is written in the language of 'pure maths speak'.

Your reply above,

''the space above it must be a normed division algebra, and these only exist in 1, 2, 4, 8 dimensions. So a cross product can only exist in 0, 1, 3, 7 dimensions.''

is 'pure maths speak' and it fails to address the very question that it is supposed to be addressing. Few readers, even with a university mathematics education could make any sense out of that statement.

The truth of the matter is that we can in fact have a cross product in 5 dimensions. But it will not obey the distributive law, and as such it will be effectively useless. The 7 dimensional cross product does however obey the distributive law.

We know that it is generally accepted that we can only have the cross product in 0, 1, 3, and 7 dimensions. But unless that proof that they were talking about in the 1970's Britannica is now finally completed, then we don't have a formal proof of this fact as such. We are no further on than we were in the days of Hamilton in the 1840's.

So I put in a sentence regarding the importance of the distributive law as regards the reason why we can only have it in 0,1,3, and 7 dimensions. I used a language that was going to be comprehensible to the average high school student. But you deleted it, and referred me to the incomprehensible 'pure maths speak' in the last section, which didn't even answer the issue in question.

I've always had a problem with pure mathematicians. Applied mathematicians can also be problematic. Applied mathematicians often go off the rails as regards linking their maths to the real world. But pure mathematicians tend to off the rails as regards linking their maths to maths. I've always felt that pure mathematicians think that they are not really in this world, and that they relish 'pure maths speak' because of its inaccessibility to the rest of humanity.

When writing these maths articles, please try to have some consideration for the curious readers who will not be educated in 'pure maths speak'. David Tombe (talk) 04:19, 1 January 2010 (UTC)


 * I've added two new refs with two different proofs, both of which you can read online - the proofs are only one and a half and two pages, but they are from 2004 and 2009 so it's understandable an encyclopaedia from 1970 does not know about them. Each contains other references you could check for other proofs. WP is not the place for the proofs themselves but they are a good reference for anyone like yourself wanting more information. -- John Blackburne (words ‡ deeds) 09:16, 1 January 2010 (UTC)

John, OK. Now we are beginning to make progress. You have provided two links for the main article which purport to give proofs for the question that has been raised above. You now acknowledge, contrary to your initial comments above, that no proof already existed in the article, but merely a statement of the facts.

One of these proofs that you have supplied may well be the proof that the 1970's Encyclopaedia Britannica was referring to. I think it is now important that you make a direct note in the main article, regarding the issue of 'proof', and then draw specific attention to these links which you have kindly provided.

But we are still lacking something. The proofs in question are very very complex, and will only be understood by the top echelons in the pure mathematics community. We need something else for the poor old applied mathematicians like myself to be able to relate to. I like to explain things to people in a way that they can at least partially understand. As I said above, it is actually possible to set up a 5 dimensional cross product. But because it won't obey the distributive law it will be useless. It's important to give this layman's explanation for the benefit of the average reader. We need some statement near the top of the article to the extent that it's only in 1, 3, and 7 dimensions that a vector product will satisfy the distributive law, and hence be of any practical use.

And one final point. You do realize that the second reference that you supplied confirms exactly what I said on the edit that you deleted. It says that we can have the vector cross product in 0,1,3, and 7 dimensions. You deleted my edit which said that, and said that the 0, and 1, cases were wrong.

While I have sympathy with your own private views regarding the 0 dimensional case, and to a lesser extent regarding the 1 dimensional case, I was merely writing what I had read in the Encyclopaedia Britannica. You will maybe now be slowly beginning to realize that your wholesale reversion of all my edits was somewhat premature. David Tombe (talk) 10:52, 1 January 2010 (UTC)


 * The proofs are as they are. They are short and straightforward but are dense and assume a familiarity with techniques beyond the level of this article. But in general WP is not the place for proofs, so so beyond providing the links there's not much we can do. As they are from this decade I am sure the Encyclopaedia Britannica writers in 1970 were not aware of them.


 * As for 0 and 1 dimensions you can derive them but they are not cross products. In one dimensional space all "vectors" are parallel, so sin θ == 0. In 0 dimensions you don't have numbers. The products in each case are just identically 0: trivially satisfying the conditions but not cross products, or meaningful products in any sense. You can define trivial products like this in any dimension but non-zero so non-trivial ones that satisfy the conditions only exist in 3 and 7 dimensions. -- John Blackburne (words ‡ deeds) 11:20, 1 January 2010 (UTC)


 * OK, I've clarified all that in the article, summarising the above and making the refs inline so the proofs are associated with where they're mentioned. -- John Blackburne (words ‡ deeds) 11:32, 1 January 2010 (UTC)

(We just got an edit conflict, so the below may now be out of date)

John, I would agree with you about the zero dimensional case. As for the one dimensional case, I also have sympthay for your point of view. But I was just stating what it said in the sources. The sources say 0, 1, 3, and 7.

I rationalize with it all as follows. The one dimensional case is plain simple real number algebra. It is not technically a cross product, but at least it is something. I can't imagine any kind of algebra in zero dimensions. The three dimensional case is the familiar cross product which works excellently with standard three dimensional Euclidean geometry. Using it in conjunction with the partial differential spatial operator leads to the extremely useful 'curl' operator. I do believe it was Maxwell who coined the terms curl, div, and grad. Finally, we have the mysterious seven dimensional vector product which works, but which doesn't tie in with any geometry that the human mind can comprehend. I await the day when we see a practical application of the seven dimensional case.

Finally add the extra imaginary factor of the square root of minus one to any of these three, and we get (1) complex conjugate pairs (2) quaternions, and (3) octonians. The 1,3,7 then becomes a 2,4,8. This points to a series based on powers of 2. So why not a 15 dimensional vector product as the man rightly asks above?

As for the proof of why only 1, 3, and 7, it may well be too complicated for wikipedia but it is important to write in the main article that proofs have been attempted, and then to give the links to these purported proofs. I can't vouch for the authenticity of the proofs, and I doubt if many editors will be able to do that. Whether or not these proofs are correct is something that only the pure mathematics community will deep down know amongst themselves. But nevertheless, the official state of existing knowledge is that we trust the pure mathematics community and that they claim that proofs exist, and that these proofs are in the links which you have kindly provided. David Tombe (talk) 11:45, 1 January 2010 (UTC)


 * Yep. That's the idea. You've now drawn attention to the proofs, if they are proofs. But we have to trust them. Now what about something for the layman and the poor old applied mathematicians? What about an explanation as to why a 3D case is so useful, but why a 5D case is useless, and why the 7D case might be useful? David Tombe (talk) 11:50, 1 January 2010 (UTC)

But it does not exist in 1 dimension. From either
 * $$\mathbf x \times \mathbf y = \mathrm{Im}(\mathbf{xy}) = \frac{1}{2}(\mathbf{xy}-\mathbf{yx}).$$

or
 * |x &times; y| = |x||y|sin &theta;

it is identically zero. The first equation as x and y would be complex numbers which commute so the product is zero. The second equation more obviously as in 1D there is only one direction so &theta; is zero or π and sin &theta; is zero. This is not particular to 1D - you can define a trivial cross product that results in the zero vector in any dimension, including 5D. The sentence "Nontrivial binary cross products exist only in 3 and 7 dimensions" covers this pretty well.

As for why not 15, the 16 dimensional Sedenions are not a Normed division algebra, which is required for the construction given to work. But the proofs also show that it's only possible in 3 and 7 dimensions more directly. As for use, it does not have one that I'm aware of, apart from to answer the question "does the cross product exist in any other dimensions?", though a lot of interesting theory is associated with it. I recommend Lounesto's book, which has a chapter on it. -- John Blackburne (words ‡ deeds) 12:26, 1 January 2010 (UTC)


 * John, The one dimensional case is i×i = 0. And the equation |x &times; y| = |x||y|sinθ is only a result which follows in the special case of three dimensions. That needs to be made clear in the article. The article currently states the equation |x &times; y| = |x||y|sinθ in a way that gives the wrong impression that it is general for all dimensions that the cross product holds in.


 * The situation has been confused by the fact that Hamilton was working on the premises that i, j, and k, were extensions of the imaginary concept. So Hamilton's quaternions differed slightly from the later Gibbs's cross product, in that for Hamilton i×i = -1, whereas for Gibbs i×i = 0. Cross product is real. But Hamilton was thinking of the 3D bit as being imaginary. Hamilton also adds an extra real number algebra to the situation, making the 3D vector cross product into a 4D quaternion.


 * Hence we have a 1, 3, and 7 when we use the Gibbs style relationship i×i = 0, and we get a 2, 4, and 8 when we use the Hamilton style with the extra real number, and where i×i = -1.


 * But whatever, the cross product does exist in 1 dimension, and the sources say so. You need to go by what the sources say. This article still needs some touching up.


 * And we need explanations for the layman, the high school students, and the applied mathematicians. We need to state in plain English, that the reason why only 1, 3, and 7 work, is because they are the only ones that obey the distributive law. David Tombe (talk) 10:47, 2 January 2010 (UTC)


 * It exists but is trivial, as in zero dimensions. There's nothing interesting to say about it beyond what's already said in the lede and last section.


 * As for explanations for the layman the reasons why the cross product only exists in 3 and 7 dimensions are given in the proofs, 2 of which you can read following links - it's nothing to do with the distributive law. I don't think these can be summarised in a way that would make them more accessible, nor do I think WP is the place to do so. -- John Blackburne (words ‡ deeds) 10:59, 2 January 2010 (UTC)

John, You are now backtracking. When I first added mention of the zero and one dimensional cross products to the main article, as per the sources, you made a mass revert boldly stating that there is no cross product in zero or one dimensions. While I am sympathetic to your own private views as regards the zero dimensional case, I was merely writing what was in the sources. As regards the one dimensional case, I have shown you above that it does exist, and you have now agreed that it exists. So we are now finally making some kind of progress.

Wikipedia is an encyclopaedia with the aim of bringing general knowledge to a wide readership. Those proofs that you have supplied are fine for the pure maths community, and I'm glad that you have provided them. I intend to study them soon. But we need something for other levels as well. Where do you get the idea that it has got nothing to do with the distributive law? It has got everything to do with the distributive law. I can easily set up a cross product in 5 dimensions. But it will not obey the distributive law and so we cannot therefore multiply out the components, and so it is useless. Are we not allowed to state this fact in the article? Does everything in the article have to be shrouded in 'pure maths speak' so that only a pure maths elite can read it? David Tombe (talk) 11:15, 2 January 2010 (UTC)


 * Do you have a reference for that ? You cannot use a result you've proved otherwise, as per WP:NOR. And the proofs I've found, with references, suggest it's far more complex than "will not obey the distributive law". -- John Blackburne (words ‡ deeds) 12:08, 2 January 2010 (UTC)

John, The distributive law issue is not a proof as such. It's a qualitative explanation as to why a 5D cross product would not work. The proofs that you have provided are new stuff. But it has been generally accepted since the 1840's that outer product only works in 1, 3, and 7 dimensions? What was the basis of that general acceptance? Well didn't Hamilton's friend WT Graves try it out practically in 15 (16 actually because he was doing octonians)? And didn't he get stuck? Why did he get stuck?

You are a mathematician. You know yourself that an algebra is only an algebra if it satisfies basic laws like the commutative law, the associative law, and the distributive law. We also know that cross product is anti-commutative and non-associative. So if it is non-distributive, then it is useless. But 1,3, and 7 dimensions are not non-distributive. That is the whole point behind them. The 3D cross product can be multiplied out in its i, j, and k component form, hence making it into a useful tool. And you can easily set up a 5D arrangement yourself with i, j, k, l, and m and then multiply it out as if the distributive law worked. You will find that it crashes in 5D.

I don't think that OR comes into it. When we do the cross product in applied maths, one of the first things that we do is to prove that it obeys the distributive law. It's a fundamental fact of why the 3D cross product is a useful tool in applied mathematics.

Let's go back to the 1960's before those proofs came out. What would your explanation have been as to why the 5D cross product didn't work? You would have been telling people that it only works in 1, 3, and 7. They would have asked 'why not 5 or 15?'. What would you have replied? David Tombe (talk) 15:27, 2 January 2010 (UTC)


 * What? I've added two proofs for the existence of cross products only existing in 3 and 7 dimensions. I've no interest in finding any more. Please feel free to spend five minutes with Google looking for them yourself. -- John Blackburne (words ‡ deeds) 15:36, 2 January 2010 (UTC)

John, You have missed the point. The readers want a layman's explanation as to why 5D or 15D doesn't work. David Tombe (talk) 15:57, 2 January 2010 (UTC)


 * There isn't one: the proofs though short are hardly layman's proofs. But as I wrote you are welcome to add one, properly referenced, if you can findd one.-- John Blackburne (words ‡ deeds) 16:01, 2 January 2010 (UTC)

John, There is one. It's the one that you deleted. It's the fact that only 1, 3, and 7 obey the distributive law. No course on vector product is complete without showing that it obeys the distributive law. David Tombe (talk) 17:02, 2 January 2010 (UTC)


 * You mean this ? I'm happy for you to add it back in with a proper reference. So far you've only indicated it's your own work, which we can't include as per WP:NOR. -- John Blackburne (words ‡ deeds) 17:12, 2 January 2010 (UTC)

Well John, that really is rich. You blatantly disregard the sources when it comes to the issue of the fact that cross product works in 0,1,3, and 7 dimensions. The sources are quite clear on that point, and you have just provided such a source yourself. But nevertheless, you have decided that we are not allowed to mention the 0 and 1 dimensional cases in the article because you personally do not believe in them. You deleted my amendment where I added mention of the two trivial cases of 0 and 1 dimension, while stating that they are wrong.

But when it comes to something much more basic, such as the distributive law, then you are demanding sources. You can see yourself, just by looking at the cross product article that the distributive law is a central feature of the cross product. But it appears that while you accept that cross product doesn't work in 5D, the penny has never yet dropped with you that the reason for this is that the 5D case does not obey the distributive law!

You see John, I'm an applied mathematician. I use these operations, and I inquire into why they are useful. I don't get involved in terminologies such as 'normed division algebra' or 'Clifford algebra'. To say that 5D doesn't work because it is not a 'normed division algebra' does not give a reason that is of any use to the applied mathematician, or the high school student, or the layman.

We need to be fair to the wider readership. We need to end all these silly games of shrouding the topic inside a defensive ring of 'pure maths speak'. Cross product is a subject which is of interest to a wider range of scientists than merely the pure maths elite. It is simply not fair that the article should be owned by the pure mathematicians, and disguised in 'pure maths speak'.

Anyway, I'll leave you to it and you can write the article whatever way you like. David Tombe (talk) 04:57, 3 January 2010 (UTC)


 * We need to make the article as accessible as possible, but not at the expense of accuracy, or more precisely verifiability. So if there's a proof it does not work in 5D because of the distributive law, and there's a reference for it, we can include it. Otherwise we have to make do with the maybe less accessible but well sourced proofs that are there now. -- John Blackburne (words ‡ deeds) 10:23, 3 January 2010 (UTC)

John, You are getting confused between the issue of (1) The reason why it doesn't work in 5D, and (2) The proof of the fact that it only works in 1D, 3D, and 7D. These are two different issues and you are confusing them with each other.

The reason why it doesn't work in 5D is because the distributive law doesn't work in the 5D case. It has got nothing to do with the issue of proof. No proof is necessary to show that the distributive law doesn't work in 5D, other than to show that the proof of the distributive law in 3D, which is in the article, doesn't work when you try it out with a 5D case. You are getting your train of logic all wrong. The distributive law is an essential feature of the cross product. And we can have a cross product in 1, 3 and 7 dimensions. The reason we can't have it in 5 dimensions is because the distributive law will not hold in 5 dimensions. This is such a fundamental fact that it seems to have been totally missed by the pure mathematicians. If you prove that it only works in 1, 3, and 7 dimensions, you still haven't explained why it works.

Once again, I should remind you that it is you who is ignoring the sources. You have decided to impose your own opinion on the article and avoid listing the 0, and 1 dimensional cases, as per the sources. David Tombe (talk) 11:55, 3 January 2010 (UTC)


 * I'm not ignoring any sources, you have still to provide any -- John Blackburne (words ‡ deeds) 11:57, 3 January 2010 (UTC)

John, You are ignoring the sources. The sources say that it works in 0,1,3, and 7 dimensions. The article only mentions 3, and 7. I added in the 0, and 1. You removed that edit stating that 0, and 1 are wrong. David Tombe (talk) 12:00, 3 January 2010 (UTC)


 * The article does mention the trivial 0 and 1 dimensional products, and I think is as clear about them as it can be. But I was asking in particular for a source for your assertion that "The reason why it doesn't work in 5D is because the distributive law doesn't work in the 5D case."-- John Blackburne (words ‡ deeds) 12:42, 3 January 2010 (UTC)


 * I have added Pertti Lounesto's hard source and explicitised the number of factors. Do you think there is an effective way to stop David from disrupting this talk page? This is getting rather tiresome. DVdm (talk) 13:05, 3 January 2010 (UTC)

John, We seem to be working on 'cross talk pages'. On the cross product page the statement reads The cross product is only defined in three or seven dimensions. I added mention of the zero and one dimensional cases, and you removed that edit contrary to what the sources say. So while you are continuing to maintain that your edits are in keeping with the sources, then it is impossible to have a rational discussion with you on this topic. It seems to me that you never before realized the actual reason why 5D doesn't work. It is wrong to say that 5D doesn't work because it is not a 'normed division algebra'. That is a statement of the facts, but it is not a reason. The actual reason is in fact so basic that the pure mathematicians don't seem to have noticed it. 5D doesn't work because it doesn't obey the distributive law, and hence we can't multiply out the components. I don't intend to discuss the issue of whether or not such a trivial fact needs a source, with somebody who so blatantly disregards sources in other respects. So go ahead and write the article how you wish. But do have some consideration for the readers who are not skilled in 'pure maths speak' and who are reading with an inquiring mind.David Tombe (talk) 14:57, 3 January 2010 (UTC)


 * You say "5D doesn't work because it doesn't obey the distributive law" but the proofs are far more complicated than this. Which is why I suspect you are wrong and so why you need something other than your own assertion that it's a "trivial fact" to support it. The article is currently well sourced and is not going to be improved by any editor inserting unsourced assertions like this.-- John Blackburne (words ‡ deeds) 15:40, 3 January 2010 (UTC)

John, The proofs, if they are actually proofs, are about proving that we can only have cross products in 0, 1, 3, and 7 dimensions. That is a completely separate issue from why we can have a cross product in 0, 1, 3 or 7 dimensions, but not in 5 or 15 dimensions. You are confusing these two issues. The second issue is about practicality. Can you not see the distinction? David Tombe (talk) 16:07, 3 January 2010 (UTC)


 * It's not about practicality, which can be determined once we have the sourced information to evaluate. But without a source for your assertion it's impossible to evaluate at all. Please provide one. -- John Blackburne (words ‡ deeds) 16:10, 3 January 2010 (UTC)

Forget it John, I'm not going to bring the debate down to sources with somebody who blatantly disregards sources. You can't have it both ways. Clearly you had never before considered the issue of why we cannot have a 5D cross product. David Tombe (talk) 16:14, 3 January 2010 (UTC)


 * You might not have noticed but my last change to the page was to provide two sources for the proof that there's a non-trivial cross product in 3 and 7 dimensions only. It is you that has yet to produce sources for the assertions you want to insert. -- John Blackburne (words ‡ deeds) 16:28, 3 January 2010 (UTC)


 * This is useless. I propose we give David the last word if he wants it. The disruption is then ended by simple silence. DVdm (talk) 16:38, 3 January 2010 (UTC)


 * (reply to David after edit conflict) But there is where the debate should be, are the statements in an article backed up by reliable sources. If so is fine to include it, if not then the statement should be removed. So the question is which source can be use to back up that 5 and 15 dim cross products do not obey the distributive law? --Salix (talk): 16:41, 3 January 2010 (UTC)

We're working on 'cross talk pages'. Ideally we should be discussing this on the more general Talk:cross_product page. JohnBlackburne removed this edit of mine contrary to what it says in the sources. Hence it is impossible to have a rational discussion about this topic with people who blatantly disregard sources, yet at the same time demand sources to back up basic facts which are not in dispute, but which have been overlooked by pure mathematicians who are so blinkered by 'pure maths speak' that they have lost track of what the operator is all about in the real world. So you can all take it or leave it. If you don't want to have that line about the distributive law in the article, then so be it. But I want to leave you all with this thought.

If, as you all claim, it has got nothing to do with the distributive law, then why can we not have a cross product in 5 dimensions? Why does it not work? I can easily set one up. But why does it not work? A proof that cross product only works in 1, 3, and 7 does not actually explain why it doesn't work in 5. Talking the language of 'normed division algebra' or 'Clifford algebra' merely states the facts, but it doesn't tell the lay reader why why can't actually have a 5D cross product.

Any university course on the 3D cross product begins by proving that it obeys the distributive law and the Jacobi identity. Apply that same proof to a 5D case and it will not work. Apply it to a 7D case and it will work.

And one more important point. The 7D case is a misnomer. It is not a truly seven dimensional concept relating to 7D space in the sense that the 3D cross product relates to 3D space. The 7D cross product only involves 3 components in any given operation.It has no connection with either 3D geometry or 7D geometry if such a thing even exists.

And there I will rest my case and leave you all to it. David Tombe (talk) 04:27, 4 January 2010 (UTC)


 * "The 7D case is a misnomer" => An article talk page is not a soapbox, nor is it a moot court, but thank you for having decided to finally rest your case. DVdm (talk) 08:31, 4 January 2010 (UTC)

The Lagrange Identity
John, The equation,


 * x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2

is only a special 3D case of the Lagrange identity. The more general form of the Lagrange identity is more complicated. You have decided unilaterally that the special 3D version is now the more general n-dimensional version. You are are quite wrong.

Do you never discuss changes on the talk page before you do your reverts? David Tombe (talk) 08:40, 5 January 2010 (UTC)


 * If a change is obviously incorrect I will revert it. If I think it's adequately described in the edit summary I'll initially leave it at that.


 * But as you've raised it here: that formula is a condition of the cross product in 3D and 7D. With orthogonality it's one of the two conditions that the cross product has to satisfy. So to say it is does not hold in 7D is wrong, and contradicted by the article at the top of the same section.-- John Blackburne (words ‡ deeds) 08:47, 5 January 2010 (UTC)

John, How can you possibly use the argument that it is 'contradicted by the article at the top of the same section'? Since when has material in a wikipedia article been considered as a reliable source such as to back up its own errors?

The Lagrange identity in three dimensions cannot possibly be a condition for the cross product in seven dimensions. How could you say such a thing? You have got alot to learn about this subject. David Tombe (talk) 08:57, 5 January 2010 (UTC)


 * Just read the article, especially the 8th line. If you need sources for it check e.g. the references. -- John Blackburne (words ‡ deeds) 09:04, 5 January 2010 (UTC)

No John, it doesn't work like that. The 8th line is wrong as regards the 7D case. Whoever wrote that was copying from a source about the 3D case. So let's not play silly games of quoting a wikipedia sentence as proof of its own veracity.

This article needs overhauled, because whoever wrote it didn't think it through properly. It's no good telling me to go and read the references. You show me the page and line in those references that says that the special 3D case of the Lagrange identity can be extrapolated to any dimension. David Tombe (talk) 09:17, 5 January 2010 (UTC)


 * Lounesto, page 42, outlines the argument. -- John Blackburne (words ‡ deeds) 09:27, 5 January 2010 (UTC)

John, You only have to look at page 4 and the top of page 5 in this weblink. It is quite clear that the lagrange identity in the form above, only applies in three dimensions. David Tombe (talk) 14:02, 5 January 2010 (UTC)


 * That page says nothing about higher dimensions. I recommend you get a copy of Lounesto's book, "Clifford Algebras and Spinors". It's an excellent book and has a chapter entirely on "The Cross Product" and a later one on Octonions that also discusses the 7D cross product. -- John Blackburne (words ‡ deeds) 14:52, 5 January 2010 (UTC)

John, The last three lines on page 4, reading on into page 5 makes it quite clear that the Lagrange Identity,


 * x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2

only holds in the special case of 3 dimensions. The information in the main article is therefore wrong because it applies the Lagrange Identity to the seven dimensional vector cross product. David Tombe (talk) 09:23, 15 April 2010 (UTC)


 * The last three lines on page 4 demonstrate the 3d case. It says nothing about other dimensions. DVdm (talk) 09:44, 15 April 2010 (UTC)

DVdm, Page 4 is about the Schwarz inequality which holds in all dimensions. The Lagrange Identity is a special 3D case of the Scwharz inequality.David Tombe (talk) 10:04, 15 April 2010 (UTC)
 * David, that chapter is about the product in three dimensions. It says nothing about the seven dimensional case so is no use as a source for this article.-- JohnBlackburne wordsdeeds 10:13, 15 April 2010 (UTC)

John, Yes, the chapter is about the 3D cross product. It shows how the Lagrange identity is derived as a special 3D case of the n dimensional Shwarz inequality. The 3D Lagrange identity is then utilized in the 3D cross product in order to estabish the relationship a×b = absinθ. Are you seriously trying to say that the 3D Lagrange identity can be used in the 7D cross product? David Tombe (talk) 10:29, 15 April 2010 (UTC)
 * And where in this article does it mention the Lagrange identity?-- JohnBlackburne wordsdeeds 10:39, 15 April 2010 (UTC)

John, It's clearly mentioned in the section 'Characteristic properties'. The Lagrange identity is this, David Tombe (talk) 10:48, 15 April 2010 (UTC)
 * x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2


 * You'll find the n-dim identity formulated on page 25 of this one. DVdm (talk) 10:57, 15 April 2010 (UTC)

DVdm, Yes, but we specifically need the 3D version to establish that a×b = absinθ. I never doubted that there was an 'n'dimensional version. But the 'n'dimensional version can't be used to prove that a×b = a b sinθ. My original point in all of this was that, should be listed along with those properties of the 3D cross product that do not hold in the 7D case. At the moment, the main article lists it as a characteristic property that does hold in the 7D case. I believe that this is an error in the main article. David Tombe (talk) 11:04, 15 April 2010 (UTC)
 * x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2


 * You believe wrongly. The "angle" θ can be defined as the thing that satisfies |a×b| = |a| |b| sinθ. Any first year math student knows this. This has been explained to you more than once.
 * By the way, you will find another ref to the identity in n-dim in this one. DVdm (talk) 11:17, 15 April 2010 (UTC)

Dvdm, Let's assume as you say that it's a simple matter of defining "angle" θ through the equation |a×b| = |a| |b| sinθ. We're then agreed that the 3D Lagrange identity must follow. The question then is, how can a 3D identity be compatible with a 7D operator? I'm not denying the existence of a 7D Lagrange identity, but it is specifically the 3D Lagrange identity that is used to establish the relationship |a×b| = |a| |b| sinθ. It's the compatibility between a 3D identity and a 7D operator that you need to address. David Tombe (talk) 14:14, 15 April 2010 (UTC)


 * David, concerning your statement "I'm not denying the existence of a 7D Lagrange identity", above you have said:
 * "The equation ... is only a special 3D case of the Lagrange identity. The more general form of the Lagrange identity is more complicated." => it is not.
 * "It is quite clear that the lagrange identity in the form above, only applies in three dimensions." => It does not.
 * "...makes it quite clear that the Lagrange Identity ... only holds in the special case of 3 dimensions." => It does not.
 * "The Lagrange Identity is a special 3D case of the Scwharz inequality." => It is not.
 * "... we specifically need the 3D version to establish that a×b = absin?." => We don't.
 * "I never doubted that there was an 'n'dimensional version." => You did. Four times in this section alone."
 * If we are supposed to continue assuming good faith from your part, I hope you understand that assuming no clue is our only remaining option here. Please stop it. Thank you. DVdm (talk) 14:21, 15 April 2010 (UTC)

DVdm, I think that we're going to have to leave it to future readers to decide who doesn't have a clue. Meanwhile I still maintain that the main article contains an error by virtue of assuming that a certain 3D identity applies to a 7D operator. David Tombe (talk) 17:07, 15 April 2010 (UTC)


 * DVdm: Reading this thread over, I find that Tombe's specific argument, stated succinctly immediately above, has not been addressed. It is an unfortunate tactic to exercise a violation of WP:AGF yourself, rather than engage in the thought needed to address the issue: to repeat Tombe's argument: a certain identity verified in 3D has been applied to a 7D operator without reference to any consideration that might establish it's extension to 7D, and despite specific stated reservations indicating that the 3D proof has elements that cannot apply in 7D. Either a suitable source can be found, or it cannot. Brews ohare (talk) 09:05, 16 April 2010 (UTC)


 * Nothing "has been applied to" anything in the article. No "3D proof" is present in the article. DVdm (talk) 09:21, 16 April 2010 (UTC)


 * The sources are in the article, all of them online if you want to review them yourself. As always I recommend Lounesto's book, mostly as it explains both the three and seven dimensional cases very well in a few short pages, though it does not have a proof. (he references this proof, which is not in the article but behind a paywall so I can't look at it).
 * As for David's argument his main mistake is identifying this
 * |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2
 * as the Lagrange identity. It is, but only in three dimensions. In seven it does not simplify like this. Lounesto calls the above the "Pythagoras theorem", which makes some sense if you think of the various things geometrically but I don't think would add anything to the article. Here the above is simply a condition. We say we want a product that satisfies it as well as orthogonality and antisymmetry, then deduce (or point to the proofs that show it) that non-trivial products only exist in 3 and 7 dimensions.-- JohnBlackburne wordsdeeds 21:38, 16 April 2010 (UTC)

John, You have just pointed out what I have been trying to point out all along, which is that
 * |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2

is a special 3D case. And in doing so, you have tried to infer that I have been saying the opposite and that that is 'my mistake'. My very first sentence in this section was,

The equation,

*|x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2

is only a special 3D case of the Lagrange identity.

Now that we are finally both agreed that this identity is restricted to 3D, can we get back to the original argument which was that we cannot apply a 3D identity to a 7D operator? The 3D Lagrange identity should be listed alongside the Jacobi identity as an identity which holds only in the 3D cross product and not in the 7D cross product. David Tombe (talk) 02:24, 17 April 2010 (UTC)

The source Z K Silagadze Multi-dimensional vector product, states that the identity:


 * |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2

applies in ℝn over the real numbers with the standard Euclidean scalar product. This statement contradicts Tombe.

This result is derived from the assumed property set forth as a desideratum of the vector cross product that:


 * |x &times; y| = |x| |y| ;&emsp; &emsp; provided that &emsp; (x &middot; y) = 0

using


 * |x &times; y| =  |( x- ( x &middot; y ) / | y|2 y ) &times; y |

He proceeds to show that


 * ''x &times; ( y &times; z) =y(x &middot; z) - z(x &middot; y)

then holds only for dimension n = 3, the case n = 1 being uninteresting. This statement is probably what Tombe is referring to as restricted to 3-D. If this condition is abandoned, n must satisfy:


 * n (n - 1) (n - 3) (n - 7) = 0,

in order that a necessary ternary identity be satisfied.

There is no need to introduce the notion of the Lagrange identity (and the article doesn't do this). There also is no need to introduce the notion of the angle θ, through sinθ, which the article does do. I'd suggest the angle θ be dropped from the article.

Silagadze goes on to mention the vector product as connected to the commutator divided by two in a composition algebra, and the connection to the Hurwitz theorem as limiting such algebras to real numbers, complex numbers, quaternions and octonions. Quaternions lead to the usual 3D vector products, and the 7D vector product is generated by octonions. The other two aren't interesting. Presently the article does not make these connections.

Apparently there are physical applications of octonions: Silagadze suggests the sources: Okubo & Dixon & Feza Gürsey, Chia-Hsiung Tze.

My take from reading this source is that everybody here is a bit right and a bit wrong, and a more collaborative atmosphere would lead to a better article. Brews ohare (talk) 05:00, 17 April 2010 (UTC)


 * Brews, Thanks for that information. It seems then that while Silagadze considers the expression,


 * |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2


 * to be applicable in all dimensions, the source which I supplied above makes it clear that it only applies as a special 3D case of the Schwarz inequality. This is exactly what I had feared. I feared that this would end up as an issue of contradiction between sources. Certainly if it applies in 7D then the angle formula a×b = absin will follow automatically, but if it doesn't apply in 7D, then the angle formula will not follow.


 * It is the applicability of the angle formula which I was challenging in relation to the 7D cross product. What we really need to do now is to establish the definitive answer as to whether the equation,


 * |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2


 * is restrictied to 3D. I have one source which says that it is restricted to 3D, and also, the wikipedia article which John Blackburne has quoted above confirms this. Ironically however, John also quotes Lounesto who seems to think that this identity applies to the 7D case, and you have just produced another source which seems to say likewise.


 * On a more general note, I think that we are all agreed that the cross product only holds in 0, 1, 3, and 7 dimensions. And I think that we are all agreed that the 0D case is less than trivial and in reality meaningless. The 1D case is simply i×i = 0 and it is trivial and of no interest. The 3D case is a full cross product which is highly useful. The 7D case is not a full cross product because some of the identities that hold in the 3D case don't hold in the 7D case. We are all agreed about what some of these identities are. It seems that the only thing that remains to be resolved is whether or not the particular identity,


 * |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2


 * falls into that category of identities that hold in 3D but not in 7D. I am of the opinion that it does fall into that category, and as such the angle relationship a×b = absin should be removed from the article, and the identity above should be moved to join alongside with the Jacobi identity as one of the identities that doesn't hold in 7D. David Tombe (talk) 09:42, 17 April 2010 (UTC)


 * David you are misunderstanding the use of the formula:
 * |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2
 * In 3D, after the cross and dot product have been defined, it is an identity relating them. It is also Lagrange's identity, but only in 3D - see Lagrange's identity. This identity gives the magnitude of the cross product as |a||b| sin θ. The other defining properties of the cross product are orthogonality (a × b is orthogonal to both a and b) and antisymmetry, as well as it being a bilinear with vector result.
 * In 7D we do not prove the above formula: we instead state it as a property of the product we want to find, the "cross product" in 7 dimensions. We then find a product that satisfies this and the other properties. Finally we show (though this is not in the article but in the references) that this is the only non-trivial cross product of two vectors other than the familiar 3D version.-- JohnBlackburne wordsdeeds 10:12, 17 April 2010 (UTC)

John, I agree with everything that you have said in your first paragraph above. I also understand the principle which you are advocating in your second paragraph. You are stating properties which you wish to hold, one of them being the formula in question, and then claiming that the 7D cross product is one of the outcomes (along with the 3D cross product of course). But I am having difficulty in reconciling the two. How can we start with a property which we already know to be a 3D property, and expect it to be a property of a 7D operator? I am not a pure mathematician and it's possible that I may be overlooking something here. But what you seem to be saying is that the formula,
 * |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2

is a special 3D case of the Lagrange Identity (which I agree with), but that it can also exist in this exact form in 7D outside of the context of the Lagrange identity. That is where I am having the difficulty. You may be right, but I thought that I would raise the matter because it is crucial as regards the issue of the validity of the angle formula a×b = absin in the 7D case. David Tombe (talk) 10:48, 17 April 2010 (UTC)


 * We don't expect it to be true, we assert it. We say "this works in 3D, is there any other dimension it works in?" and as
 * x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2
 * is a property of the product in 3D was say that it must be a property of the product in other dimensions. This article gives one such product, in seven dimensions. The proofs in the references prove this is the only non-trivial dimension other than three.
 * As for Lagrange's identity this
 * $$(a \cdot a)(b \cdot b) - (a \cdot b)^2 = (a \wedge b) \cdot (a \wedge b).$$
 * Is a general form of it, i.e. the form in all dimensions expressed in terms of simple products. In 3D is is the same as
 * x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2
 * Because the exterior/wedge product can be associated with the cross product as it's dual. This is not true in seven dimensions as vectors and bivectors are not dual. There is a relationship between them, given in e.g. Lounesto, but it's rather more complex.-- JohnBlackburne wordsdeeds 11:58, 17 April 2010 (UTC)

Silagadze does propose Blackburne's approach of postulating what one wants in a cross product and then finding in what dimensions it can be found. In 7 D, Silagadze's postulates force abandonment of:


 * ''x &times; ( y &times; z) =y(x &middot; z) - z(x &middot; y)

which holds in 3D only, and which may agree with Tombe in these terms: Postulating one of two cross-product properties means that in 3D both of the two properties will apply, while in 7D only one of the two will be true.

A point of Silagadze, which is not being looked at here, is that he takes this postulate approach as standing independent of another approach that defines the cross product in the context of composition algebra. He regards composition algebra as a realization of the postulate methodology, but appears to believe other realizations may be possible. From the composition algebra approach, the cross product is defined, not in terms of properties as above, but as:


 * ''x &times; y =$1/2$ ( x y − y x )

Then the Hurwitz theorem provides the dimensional restrictions. As I understand Silagadze, he is suggesting that there are various possible starting points, but the resulting dimensional restrictions end up the same.

This debate here on WP may really be one of what axioms are selected as the starting point, and which approaches are the more restrictive? Brews ohare (talk) 13:28, 17 April 2010 (UTC)


 * John and Brews, OK let's then assert that,
 * |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2
 * holds in any dimensions, and then set out to find which operators satisfy this assertion. But in doing so, let's bear in mind that this assertion, when in 3D, is the 3D case of more complicated 'n'D relationships such as the Lagrange identity and the Schwarz inequality. On that basis, I would expect that any operators satisfying this assertion would have to be in 3D only. But that is only my own assumption. Silagadze apparently finds this assertion to be compatible with a 7D cross product. I suspect that Silagadze has made an oversight, in that his more primitive assumptions are already restricted to 3D.
 * Now I don't doubt that proofs exist which restrict cross products to 0,1,3 and 7 dimensions. I fully accept that a 7D cross product 'of sorts' exists. But do any of the proofs listed in the references actually utilize the relationship,
 * |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2 ?
 * I have scanned through a few of these proofs and I can't see where that relationship comes into it, but do please correct me if I am wrong on that point because I only did a scan read.
 * Ultimately I think that the problem here lies with trying to define 'angle' in 7D. It's no good defining angle through dot product and cosine. An angle only has a cosine if it exists in the first place.
 * I'll need to take a closer look now at exactly what Silagadze has done, and I need to see exact quotes from the proofs in the references which directly link this relationship to the 7D cross product. I just can't see how a relationship that is so heavily connected with 3D can in any way be extended to 7D. David Tombe (talk) 14:17, 17 April 2010 (UTC)

I'll paraphrase to see if I got your point. Please correct me if I have slipped up. (i) The Lagrange identity is valid in any number of dimensions. (ii) |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2 is the same as the Lagrange identity in 3D. Now where do these facts take us? If we propose to find whether this 3D "coincidence" can be made to occur in other dimensions, it appears that it can, in 7D, but the resulting cross-product doesn't satisfy all the 3D identities, only some of them. That seems to be possible: of course, only the details will decide whether it really is true. As I understand it, Silagadze went through the details to his satisfaction and found all was well. As I understand it, your instincts are to disbelieve his analysis. Of course, your instincts may be quite valid, and personally I hold some regard for your instincts. However, without actually support by detailed analysis, instinct remains only a motivation for more work, I'd guess. What do you think?

My guess is that there are several logical bases for introducing the cross product. Silagadze's is one. Forcing the wedge product definition of cross product in 3D to other dimensions is another. There may be more, and one of them may fit your instincts. If so, you have a viable alternative that suits your intuition. Brews ohare (talk) 14:46, 17 April 2010 (UTC)


 * Brews, Ultimately if we have a source which directly links |x &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2 to the number 7, then we'll have to let it go at that. It sounds like your Silagadze might be that source. Is it possible that you can give a brief outline as to how Silagadze makes the linkage to the number 7 without using any 3D assumptions? David Tombe (talk) 15:00, 17 April 2010 (UTC)
 * Silagadze uses a different starting point: |a × b| = |a| |b| if a.b = 0. Geometrically this is "the cross product's magnitude is the same as the area of the rectangle with sides a and b, if they are perpendicular vectors". This looks like a special case, so a weaker condition than the Pythagorean one, as geometrically that one says the magnitude of the cross product is the same as the parallelogram with sides a and b, and a rectangle is just a parallelogram with perpendicular sides. But he then derives the more general rule from the specific, so showing they are equivalent. All this is without reference to the dimension, i.e .the properties are true in any dimension, if the cross product exists. He then shows that such a product only exists in 3 and 7 dimensions.-- JohnBlackburne wordsdeeds 16:21, 17 April 2010 (UTC)

John, Thanks for that information. I know that we cannot use wikipedia articles as sources, but I'd like to hear your comments on this passage which I have copied from the wikipedia article cross product.


 * The following identity also relates the cross product and the dot product:
 * :$$ |\mathbf{a} \times \mathbf{b}|^2 + |\mathbf{a} \cdot \mathbf{b}|^2 = |\mathbf{a}|^2 |\mathbf{b}|^2.$$
 * This is a special case of the multiplicativity $$\scriptstyle |\mathbf{vw}| = |\mathbf{v}| |\mathbf{w}|$$ of the norm in the quaternion algebra, and a restriction to $$\scriptstyle\mathbb{R}^3$$ of Lagrange's identity.

As regards Silagadze, all his fundamental postulates are 3D based. Even though he then shows that these axioms satisfy n(n-1)(n-3)(n-7) = 0, it has already been established from the outset that n=3. The equation n(n-1)(n-3)(n-7) = 0 came from other references and there was no establishment of the fact that n could equal 7, because it had already been established as being particularly equal to 3. All that Silagadze established was that if n doesn't equal 3 then it could equal 7. But we already know from the fundamental axioms that n does equal 3. David Tombe (talk) 17:07, 17 April 2010 (UTC)
 * No, they are not "3D based". Only later in his proof does he show that the dimension must be 3 or 7 (the other vales 0 and 1 not representing products). The earlier properties are stated and manipulated without assuming the dimension.-- JohnBlackburne wordsdeeds 17:18, 17 April 2010 (UTC)

John, I am only working from a secondary transcription of the Silagadze source which states clearly that the fundamental axioms are all 3D. But I haven't actually seen the primary source. But let's say for the sake of argument that the primary source is silent on the issue. Silagadze does openly state that the vector triple product only holds in 3D. That deals with the volume of a parallelepiped. What would make the Pythagorean relationship any different in that respect? David Tombe (talk) 17:27, 17 April 2010 (UTC)
 * The Silagadze paper is online, and linked in the article. No-where does is say the fundamental axioms are all 3D. I suggest you read it as it's perfectly clear.-- JohnBlackburne wordsdeeds 17:35, 17 April 2010 (UTC)

John, I don't think that that will be necessary. As I have already said, why would the vector triple product be restricted to 3D and not also the Pythagorean relationship? Is it just because the volume of the parallelepiped is too obviously 3D? I've already shown you sufficient evidence that the identity in question is heavily linked up to 3D. It wouldn't be within the natural order of things for it to be applicable to a 7D operator. It's now become obvious that this is yet another of these modern controversies that isn't going to be resolved on wikipedia and so I think that we had better close the discussion down before somebody ends up 'in Seine'. But thanks anyway for your participation. You have been very helpful in drawing my attention to some of the underlying issues. David Tombe (talk) 17:47, 17 April 2010 (UTC)

Is the Silagadze formal approach separate from the wedge product approach?
My understanding of Silagadze's paper is that he begins with some postulated properties of the cross product, inspired by the 3D example, and then shows that these properties can be obtained in 7D, although some additional properties of the 3D cross product are not carried over.

My understanding also is that the cross product and wedge product as related in 3D can be carried over to other dimensions only if n=7, and this approach has similar limitations, but not arrived at the same way as Silagadze, but through the limitations of composition algebra and the Hurwitz theorem.

I take it that Silagadze believes this last to be a particular embodiment of his approach.

So here are two questions:
 * Are we to believe that the "cross product" can be arrived at in several ways that boil down to the normal concept of cross product in 3D, but which do not obey all the 3D behaviors in 7D?
 * The cross product is defined like this, and generally accepted. Some products of the 3D product are not true in 7D, but they are not needed for the definition. You could define other products which do not satisfy all the conditions but they would not be cross products and so not as interesting.
 * Is this general 7D cross product unique, regardless of how we go about it, or may there be several cross products, obtained by retaining a different subset of 3D behaviors as defining properties of the cross product? Brews ohare (talk) 19:10, 17 April 2010 (UTC)
 * It's not unique and the article says so, at the start of this section, before giving one of the ways to do it. Another way, as seen in Lounesto, is define a trivector, again not unique, which when multiplied by a ∧ b gives the cross product. In three dimensions the trivector, i.e. the pseudoscalar, is unique up to a scale factor.-- JohnBlackburne wordsdeeds 19:44, 17 April 2010 (UTC)

Characteristic properties
This section states as a property that the norm of the cross product should be the area of the corresponding parallelogram and links to the 2D article parallelogram. Unfortunately, that leaves the reader wondering what an n-dimensional parallelogram is, and how its area is defined. Is this a circular relation as we can't define such an area in n-dimensions without referring to a cross product?

In any event, some guidance and probably a reference is necessary here. Brews ohare (talk) 21:40, 17 April 2010 (UTC)

And, it seems likely that parallelograms can be defined of a variety of dimensions (also this), and therefore are more flexible than cross products. What about the area of parallelograms in dimensions where cross products can't be defined? Brews ohare (talk) 21:52, 17 April 2010 (UTC)


 * It's defined in 7D the same way as in 3D: as the shape on the plane spanned by the vectors. Such a plane always exists if the vectors are not parallel, and it's a standard 2D Euclidian plane, so angles, lengths, shapes etc. all have their familiar meanings - and you don't use the cross product to define a parallelogram in 2D.-- JohnBlackburne wordsdeeds 21:53, 17 April 2010 (UTC)

Hi John: As the definition is related in a fundamental manner to the parallelogram in the article, something is needed to explain what that means in n (or n-1) dimensions. Brews ohare (talk) 22:06, 17 April 2010 (UTC)
 * I see you added a reference on parallelograms but the reference is not about parallelograms: the author is, as he states on the first page, using the word "parallelogram" instead of "parallelepiped", and that is what his chapter is about, the volume of parallelepipeds. I have therefore removed it.-- JohnBlackburne wordsdeeds 17:05, 18 April 2010 (UTC)
 * Again, the reference is misleading. It links to a chapter titled "Volumes of parallelograms", but the author means something totally different by this, as he explains on the first page: the discussion is all about parallelepipeds. On the same page it says

"a 3-dimensional parallelogram could lie in a plane and itself be a 2-dimensional parallelogram, for instance. PROBLEM 8–1. Consider the “parallelogram” in R3 with “edges” equal to the three points ... Draw a sketch of it and conclude that it is actually a six-sided figure in the x − y plane."


 * So a parallelogram can be a six-sided plane figure ! And this is all on the first page. It seems a poor source to use for parallelograms as the author has a very different understanding of them from you or me.-- JohnBlackburne wordsdeeds 18:00, 18 April 2010 (UTC)