Talk:Seven-dimensional cross product/Archive 3

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Is the X-product tied to a 2D plane in 3D the way it's tied to a 2D plane in 7D?

Lounesto points out (page 97) "In the 3D space a×b = c×d implies a, b, c, d are in the same plane, but in R⁷ there are other planes than the linear span of a and b giving the same direction as a and b". In addition, the 7D cross product doesn't behave like a vector in 3-D: it is invariant only under G₂, a set of symmetry operations far smaller than SO(7), while in 3D the full SO(3) can be used. These different properties give the finger to the notion that the cross product we are looking at is describing a plain vanilla 2-D plane embedded in a 7-space.

These oddities should appear in the article with their implications for the cross product and the connection of a cross product to a 2D plane. In 3-D the cross-product has a clear connection to rotations, and on an infinitesimal basis can generate rotations about its axis. These very interesting properties from both a mathematical and physical standpoint apparently are not carried over to 7D.

These limitations are worthy of a subsection to make clear the distinctions, and possibly provide some additional insight into the cross product. Brews ohare (talk) 15:29, 27 May 2010 (UTC)

I'm not sure what you mean by "These different properties give the finger to the notion that the cross product we are looking at is describing a plain vanilla 2-D plane embedded in a 7-space". The cross product does not describe a 2D plane, in 3 or 7D.

One way to understand Lounesto's point is in 7D planes (and rotations) have 21 degrees of freedom. As lines and vectors have only 7 then a map from planes to vectors will lose information, and multiple planes with map to the same vector. Only in three dimensions do planes and lines have the same dimensionality, and are related through the Hodge dual so there's as 1-1 correspondence.--JohnBlackburne^words_deeds 15:44, 27 May 2010 (UTC)

Nice description. So what happens with the connection to infinitesimal rotations? Apparently an attempt to use cross products causes separate rotations to become lumped together as the same, even though they aren't? In other words, this cross product of two vectors is pretty useless language for a description of what goes on in 7D, eh? One will have to develop a different way to handle these matters. How does that impact the formulation of (say) conserved quantities. While in 3D these can be related to SO(3) and simply described with cross products, in 7D they will be described by SO(7) and can't be summarized in terms of the 7D cross product.

I don't know: I've come across infinitesimal rotations, in e.g. looking at the connections to Lie Algebra, but it's not something I've had reason to really get to know. That link doesn't work for me BTW.--JohnBlackburne^words_deeds 16:18, 27 May 2010 (UTC)

Maybe a few words about the total futility of description using cross products in 7D could be contrasted with the very useful description constructed with cross-products in 3D. Brews ohare (talk) 15:57, 27 May 2010 (UTC)

I'm not sure what you mean.--JohnBlackburne^words_deeds 16:18, 27 May 2010 (UTC)

Hi John, here it is again, or try googling CLE Moore: Rotations in Hyperspace ; Proceedings of the American Academy of Arts and Sciences, Volume 53 (1918). As you know, symmetries lead to conservation laws. Where a rotational symmetry exists, a conservation law appears. If the cross-product identifies the symmetry it can be associated with it. But if the 7D cross product is subject to an aliasing problem (to rephrase your remarks), that isn't going to work. Brews ohare (talk) 16:37, 27 May 2010 (UTC)

Irrelevant 3D stuff

I just removed long footnote that had nothing to do with the article. If something's not suitable for the article that does not mean it should be added among the references - that section is meant to be a list of sources. Stick to information about the 7D cross product, drawn from sources that are about the 7D cross product. We are not short of good references for this article, there's no need to try and include information from wholly irrelevant sources.--JohnBlackburne^words_deeds 18:45, 27 May 2010 (UTC)

Hi John: I've rewritten this section to be more suitable. Lounesto is still there, but the more general reference for this equation is added. Brews ohare (talk) 19:05, 27 May 2010 (UTC)

Reverted as that source says nothing about the 7D cross product. Please stop introducing incorrect information from irrelevant sources --JohnBlackburne^words_deeds 19:33, 27 May 2010 (UTC)

Hi John. I thought we had straightened that out: there is no 3D "stuff" here. if one defines x = x₁ e₁ + x₂e₂ + ... in a space of any dimension n, the Binet-Cauchy identity always applies as does the Lagrange identity. In particular:

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\sum _{i=1}^{n-1}\sum _{j=1+1}^{n}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\ .$

Of course, a basis set {e_i} is necessary to determine the components that enter into the summation. If one comes up with the appropriate multiplication table:

$\mathbf {\hat {e}} _{i}\mathbf {\times \ {\hat {e}}} _{j}=\sum c_{ij}^{k}\mathbf {\hat {e}} _{k}\ ,$

with the correct structure factors {c^k_ij} (as can be found only in 3D and in 7D), then the right-hand side with the summation can be expressed as the magnitude of a vector perpendicular to x and y that we can call x × y (or $\mathbf {x\otimes y}$, if one doesn't wish to prejudge the properties). That is what Lounesto is using under the label "Pythagorean theorem".

Let me inquire, do you agree with these remarks? They all are documented, I believe. Brews ohare (talk) 20:21, 27 May 2010 (UTC)

Documented where ? The only sources you provided were for 3D.--JohnBlackburne^words_deeds 20:24, 27 May 2010 (UTC)

HI John: I'll take your response as suggesting the above is not correct. I'll provide sources for you again.

First, I've provided sources for the Binet–Cauchy identity, namely Eric W. Weisstein (2003). "Binet-Cauchy identity". CRC concise encyclopedia of mathematics (2nd ed.). CRC Press. p. 228. ISBN 1584883472. You have looked at this before, but not closely enough to see his Equation 1 that clearly is an n dimensional formula. As Weisstein says, this identity becomes the Lagrange identity when two of the four vectors are made the same, and this is the equation displayed above. I hope there is no trouble with this result or its generality, extending to n dimensions? See also Robert E Greene and Steven G Krantz (2006). "Exercise 16". Function theory of one complex variable (3rd ed.). American Mathematical Society. p. 22. ISBN 0821839624.

Second: Weisstein also displays the simplified version for n = 3 where he says this form is often itself called the Lagrange identity. Formally, it appears identical to the n=7 case used in this WP article, but although Weisstein provides the formula for general n and for n = 2 & 3, he did not provide n=4.

Third: Nonetheless, there is no doubt that with the appropriate multiplication table, the summation is indeed ||x × y||². As you know, there is much discussion in the literature pointing out that such a multiplication table is possible only in 3-D and in 7-D. An example is of such discussion is: Silagadze, Z.K. (2002). "Multi-dimensional vector product" (PDF). arXiv:math/0204357. {{cite journal}}: Cite journal requires |journal= (help)

You'll recall we both made spreadsheets to prove our multiplication rules worked and my own also checked Lagrange's formula. I doubt you have a problem with that: You have supplied such a multiplication table in this article yourself, and you quote a simple form for it from Lounesto. The article Octonion has another version sourced to Shestakov

Fourth: Putting this result into the Lagrange identity, we have

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\|\mathbf {x\times y} \|^{2}\ .$

This is identically the equation in use by Lounesto.

If you have further requests for support, please let me know just what points need more detail. Brews ohare (talk) 21:39, 27 May 2010 (UTC)

The problem is none of the sources are about the cross product in 7D. To take e.g. this source on the Binet Cauchy identity no-where does it mention the above formula, or say anything about 7D. It gives a general algebraic formula then relates that to a vector equation in 3D. But that is particular to 3D and does not generalise. It's also a different equation to the one above even in 3D. So such a source adds nothing to our understanding of the 7D cross product.--JohnBlackburne^words_deeds 21:42, 27 May 2010 (UTC)

Hi John: It's a moving target. First you didn't see Binet–Cauchy identity applied in n-D. Now it applies in n-D, 2,-D and 3-D but not in 4-D. Weisstein refers explicitly to the cross product in 3-D but the obvious substitution of Σx_ie_i × Σy_je_j and the Silagadze paper seemingly are beyond your understanding. Hmm. Well such shenanigans may be fun, but they are not serious. Brews ohare (talk) 22:00, 27 May 2010 (UTC)

it's an "obvious substitution" to you maybe, but even if your reasoning were correct that would be OR, which is not allowed. Hence you need sources on the topic, which is the seven-dimensional cross product. We already have more than enough good sources for such a short article, but if you can find more then they can be added and used to improve the article. But we can't use sources on entirely different topics connected by unsourced OR.--JohnBlackburne^words_deeds 22:21, 27 May 2010 (UTC)

So labeling of this equation as "Pythagoras' theorem" will not be changed, though no source on the planet uses this terminology but Lounesto, and even though we both know that it is the 7-D form of the Binet-Cauchy identity. And although Pythagoras' theorem is about sums of squares in n dimensions, and is not about cross-products that have no meaning outside 3- and 7-dimensions. We cannot even put a footnote in to reassure the reader that Pythagoras' theorem doesn't change just for 7-D. Got it, John: we cannot drop Lounesto's inappropriate labeling of the 7-D form of the Binet-Cauchy identity because personally you like it. Brews ohare (talk) 23:41, 27 May 2010 (UTC)

It's not the 7D form of the Binet-Cauchy identity. You keep claiming this but have provided no evidence of this, e.g. relevant sources. Until you do we should not be rewriting the article with misleading footnotes based on your original research or synthesis. Further I don't know what you've got against Lounesto. It's by far the best source on the subject as although it does not contain a proof it covers most of the topic very well, as well as explaining all the theory it uses in earlier chapters. But as I've already written: if you have a source on the 7D cross product you prefer please include it.--JohnBlackburne^words_deeds 08:25, 28 May 2010 (UTC)

John:

It is established that

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\sum _{i=1}^{n-1}\sum _{j=1+1}^{n}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\ ,$

is called the Lagrange identity (a specialization of the Binet-Cauchy identity) in n-dimensions. For example see this.

On the other hand, the equation Lounesto calls the "Pythagorean theorem" is

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\|\mathbf {x\times y} \|^{2}\ ,$

and is being applied here to 7 dimensions.

In 3D, Weisstein, Eq. 3 displays the Lagrange identity in this form using the cross-product. Of course, to express the Lagrange identity in terms of the cross product one needs to define the cross product, and that can be done only in 3 and in 7-dimensions. To define it in 7-D, one uses Lounesto's multiplication table for the $\mathbf {{\hat {e}}_{i}\times {\hat {e}}_{j}} \,$. If the resulting components are plugged into Lagrange's identity, you get the same thing as Lounesto's equation, of course. (You know that, we did it in spreadsheets and David proved it algebraically).

So you might explain whether you personally disbelieve this connection, or are simply insisting upon WP:RS and WP:OR as a rather pedantic application of WP rules to suppress even a footnote . You don't need even to establish this connection between Lagrange and Lounesto's equation to justify a note to the reader that this is an idiosyncratic use (read: use peculiar to Lounesto, and no-one else) of the term "Pythagorean theorem" , which is normally taken as a sum of squares in n-dimensions, including n=7 of course. See Douglas; §3.10. (Such a note was there before all this started, BTW, but you removed it in its entirety). Brews ohare (talk) 20:48, 28 May 2010 (UTC)

Why is is "idiosyncratic"? You understand it, I understand it, and given that it's half way through a post-graduate level maths textbook I suspect all readers of the book will understand it. As for "no-one else" you've yet to provide another source on the seven-dimensional cross product that puts it differently. That's reliable sources - a Google spreadsheet and David's incorrect "proof" are not by any stretch of the definition reliable sources.--JohnBlackburne^words_deeds 21:54, 28 May 2010 (UTC)

John, try a little harder please. Don't put words in my mouth. Address the two points: do you believe Lounesto's equation is not equivalent to the Lagrange identity in 7-D? Do you think the typical reader will intuit that Lounesto's "Pythagorean theorem" is obviously connected to the standard Pythagorean theorem, so no note would help the reader? Brews ohare (talk) 22:10, 28 May 2010 (UTC)

I don't believe it is, no. I've not seen it in any source on the 7D cross product, and I'm not convinced by David's shaky logic. As for Pythagoras's theorem I think anyone with a degree in maths, the level of mathematics this is at, will be able to see the correspondence between

|x × y|² = |x|² |y|² − (x ⋅ y)²

and Pythagoras's theorem, i.e.

a² = c² - b²

without any further help. --JohnBlackburne^words_deeds 22:31, 28 May 2010 (UTC)

So, you do not think:

$\sum _{i=1}^{6}\sum _{j=i+1}^{7}(a_{i}b_{j}-a_{j}b_{i})^{2}=\sum _{i=1}^{7}\left(\mathbf {a\times b} \right)_{i}^{2}\ .$

Perhaps you really think there is an error in the spreadsheet not discovered for any a and b that has been tried?

Of course, what we believe is not terribly relevant to WP, but if you actually think there is equivalence it might make you a bit more receptive, eh?

As Lagrange's equation applies for all a and b, and as one side of Lagrange's identity and one side of Lounesto's are identical, it seems hard to avoid the conclusion that the other sides are equal as well. Brews ohare (talk) 00:16, 29 May 2010 (UTC)

John, It seems to me that we are all agreed that,

|x × y|² = |x|² |y|² − (x ⋅ y)²

holds in 3D, and that it is the special case of Lagrange's identity in 3D, and that it is also Pythagoras's theorem.

We are also all agreed that this equation holds in 7D.

The disagreement seems to be that in 7D, I see this equation as being the Lagrange identity only, and not the Pythagorean identity, whereas you see it the other way around. Hence there are two issues to be ironed out.

(1) The Jacobi identity limits the sine relationship to 3D. Hence the equation above does not convert to the Pythagorean identity in 7D.

(2) Then there is the issue of proving that the equation above is the special case of the Lagrange identity in 7D. You keep stating that my proof is wrong. It's not wrong. I have said quite correctly that in 7D, the left hand side involves seven z^2 terms. These expand into 252 terms when multiplied out distributively. 168 of these terms mutually cancel and that leaves 84 terms. The 84 terms reduce to 21 squared terms in brackets. These 21 squared terms are exactly as is required by the Lagrange identity. That is not original research. It is a simple case of illustration by example of an established mathematical equation. Try it out for yourself. But I warn you that you will get a headache while trying to cancel out the 168 terms (never mind doing the initial expansions). I had to get my applied maths professor to do it for me because I kept making errors. He provided me with an amazing sheet of paper with all the 168 terms, with cancellation dots in pencil above each of them. (actually, the 168 terms are in duplicate so it really becomes a mutual cancellation of two lots of 42 terms within a set of brackets, multiplied by 2. That is 168 terms in total.) David Tombe (talk) 11:58, 29 May 2010 (UTC)

(1) No it doesn't: the relation to a² = c² - b² works in 3 and 7 dimensions. Neither depends on the Jacobi identity.

(2) I've yet to see a plausible proof: the above formula free paragraph is a good example of why proofs are done using mathematical steps and formulae. If as you say you could not do even do it without your maths teacher's help it's unlikely you understand it well enough to produce a proper proof. But as I've already written it doesn't matter as a proof by you or from Brews is original research, so WP is no place for it.--JohnBlackburne^words_deeds 12:29, 29 May 2010 (UTC)

No John, The reason why I contacted the applied maths professor was because I couldn't get the cancellation correct. There were so many terms involved, (252), that I kept making trivial errors while writing the subscripts. There is no need to infer that I didn't understand the issue. The issue is about multiplying out the equation,

|x × y|² = |x|² |y|² − (x ⋅ y)²

in its orthogonal components and proving that the distributive law holds. And that reminds me to tell Brews that we can actually set up a cross product of this kind in any dimensions as long as it is an odd number, but that only 3 and 7 will work in the above Lagrange equation. We can set up a 5D cross product of mutually orthogonal vectors, but it won't satisfy the Lagrange identity, and hence it won't satisfy the distributive law, and so it will be practically useless. In fact, I was very surprised to discover that the 7D cross product satisfies the Lagrange identity. All the sources only show the proof in the 3D case, which is quite basic, and it is such that I would have been highly sceptical about the idea that this proof could be applied in 7D also. You do recall that I initially objected to the idea that the above equation holds in 7D. You convinced me otherwise by using numerical substitution and so I made the extra effort to get to the bottom of the matter. I then realized that it involved the exact same cancellation as I had got the applied maths professor to do for me back in 1993. And I then realized when I dug out the old notes, that it was the same equation. The thing about the 7D cross product which makes it special is the fact of that amazing mutual cancellation of the 168 terms. That doesn't happen in any other dimensions.

On the other point, I agree with you that Lounesto is a very good source for this topic. But he seems to have overlooked the link between the sine relationship and the Jacobi identity. He knows that the Jacobi identity doesn't hold in 7D and he knows that the Jacobi identity is linked to lie algebras, which are in turn linked to rotations. But he hasn't then made the connection back to angle. I have serious doubts therefore that Pythagoras's theorem will hold in 7D.

I think that an additional name on top of 'Pythagoras's theorem' is required to describe the equation above, and the obvious choice is the Lagrange identity. David Tombe (talk) 19:02, 29 May 2010 (UTC)

a=b & b=c means a=c

It is established that

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\sum _{i=1}^{n-1}\sum _{j=1+1}^{n}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\ ,$

is called the Lagrange identity (a specialization of the Binet-Cauchy identity) in n-dimensions. For example see this.

Lounesto's "Pythagorean theorem" is:

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\|\mathbf {x\times y} \|^{2}\ ,$

and is being applied here to 7 dimensions. Restricting ourselves to 7-D vectors x and y, both the Lagrange identity and Lounesto's equation have identical left-hand sides. Consequently their right-hand sides also are identically the same in this 7-D case. In other words:

$\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}=\|\mathbf {x\times y} \|^{2}\ ,$

applies in 7-dimensions for every pair of 7-D vectors x and y.

It follows that in 7-dimensions Lounesto's equation is exactly the same as Lagrange's identity. Brews ohare (talk) 14:02, 29 May 2010 (UTC)

By the same argument the Pythagorean trigonometric identity and the geometric progression (0.5 + 0.25 + 0.125 + ...) are both = 1. So the Pythagorean trigonometric identity and the progression are "exactly the same". But they are not, they are two different things that have the same value. In mathematics generally just because two things are equal to something simpler (in this case the square of xy sin θ) does not mean they are the same formula, or the same identity. To deduce so is synthesis. As you clearly do not have a source for your reasoning perhaps now is the time to drop it.--JohnBlackburne^words_deeds 14:39, 29 May 2010 (UTC)

John, you are just a bit too close to this right now. The relation above is an identity that holds for all vector arguments. I am sure that you'd agree that if the relation f(x, y ) = g(x, y) holds for all permissible x and y, then f≡g. Your comparison is not applicable. The statement stands. Brews ohare (talk) 14:54, 29 May 2010 (UTC)

To elaborate:

$f(x)=x^{2}+2x+1\ ;\ \mathrm {and} \ \ g(x)=(x+1)^{2}\ ,$

differ only in the way they are evaluated, it is proper to say f and g are the same function because they map every x into the same value, whether you call that value f or call it g. Likewise,

$f(\mathbf {x,\ y} )=\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2};\ \mathrm {and} \ \ g(\mathbf {x,\ y} )=\sum _{i=1}^{7}\|(\mathbf {x\times y} )_{i}\|^{2}\ ,$

differ only in the order the terms are added up. Regardless of whether you calculate using the components of the cross product (using g ), or evaluate it using the components of the vectors themselves (using f ), you get

$f(\mathbf {x,\ y} )=g(\mathbf {x,\ y} )=\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}\ .$

That is, precisely the same number results for each pair (x, y), regardless of which function one chooses: f or g. Therefore, f = g.

It follows that in 7-dimensions Lounesto's equation (using g ) is exactly the same as Lagrange's identity (which uses f ). Brews ohare (talk) 17:40, 29 May 2010 (UTC)

A consequence of equivalence is that Lounesto could make f = g by definition of the cross product, and deduce his "Pythagorean equation" from Lagrange's identity. Brews ohare (talk) 18:30, 29 May 2010 (UTC)

Brews, This seems to be a clash between the pure mathematician and the applied mathematician.

John has already satisfied himself that,

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\|\mathbf {x\times y} \|^{2}\ ,$

holds in seven dimensions on the grounds that if we use this equation as a starting point, the likes of Silagadze have come up with some very high powered pure maths type proof that such an equation can only hold in 3 or 7 dimensions. Hence John needs no more convincing. But most people would never comprehend such a pure maths proof. Even applied mathematics university students wouldn't necessarily understand it. They will be looking for something more illustrative. They will easily be convinced in the simple 3D case, but the 7D case does not look very convincing from the outset. Indeed, no other case apart from 3 and 7 works. What John seems to be overlooking is the fact that this equation is the Lagrange identity. With his pure maths proof, he has by-passed all need to mention the Lagrange identity. But applied mathematicians will commence all of this from the Lagrange identity explicitly. My applied maths professor began like Lounesto and Silagadze with the equation above as a desired starting point. But then he took the generalized 'n' dimensional Lagrange identity and demonstrated that only odd number dimensions work. Then he further demonstrated that 5D didn't work. He then did the big 168 term cancellation to show that 7D did work. He then said that he spoke to some pure mathematicians and that they said that there are 'group theoretic' reasons why it only works in 3 and 7 dimensions.

You are adopting the applied maths approach and you have identified that the problem is one of justifying the equation,

$\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}=\|\mathbf {x\times y} \|^{2}\ ,$

That brings you to the big 252 term expansion that was deleted a couple of days ago. It is not original research. It is mere substitution into an established equation and it is absolutely correct. David Tombe (talk) 19:17, 29 May 2010 (UTC)

Yes, rather than an abstract argument, one can simply construct the result (an applied math approach, you say). If one requires of the cross-product that

$\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}=\sum _{i=1}^{7}\|(\mathbf {x\times y} )_{i}\|^{2}\ ,$

you are requesting the double summation on the left be re-expressed on the right as a sum of only seven combinations of the many products on the left. It also is required that the cross product be orthogonal to its arguments. Both can be done in 7-D if the components of the cross product are found using Lounesto's multiplication table for the basis vectors:

$\mathbf {\hat {e}} _{i}\mathbf {\times {\hat {e}}} _{i+1}=\mathbf {\hat {e}} _{i+3}\ ,$

with {1, 2, 3, 4, 5, 6, 7} permuted cyclically and translated modulo 7. That is what both John and I did separately using a google spreadsheet. David, you had the courage to do it algebraically.

However, my argument is neither the pure nor the applied maths approach. It consists of the very basic logic that if :

$h(\mathbf {x,\ y} )=\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}$

and also

$h(\mathbf {x,\ y} )=\sum _{i=1}^{7}\|(\mathbf {x\times y} )_{i}\|^{2}\ ,$

for all pairs x, y, then

$\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}=\sum _{i=1}^{7}\|(\mathbf {x\times y} )_{i}\|^{2}\ .$

(Here it is not significant to the argument, but it happens that: $h(\mathbf {x,\ y} )=\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}\ .$)

That conclusion is a consequence of such simple high-school logic that I don't think it aspires to the name "mathematics". Brews ohare (talk) 22:09, 29 May 2010 (UTC)

The literal use of a syllogism like a=b & b=c → a=c is not WP:OR or WP:SYNTH either. Brews ohare (talk) 15:57, 30 May 2010 (UTC)

A note would aid the reader

The article carries over the practice of Lounesto in using the term "Pythagoras' theorem" to refer to the equation:

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\|\mathbf {x\times y} \|^{2}\ .$

Using the ancillary relations:

$\mathbf {x\times y} =\|\mathbf {x} \|\ \|\mathbf {y} \|\sin \theta \ \ \mathrm {and} \ \ (\mathbf {x\cdot y} )=\|\mathbf {x} \|\ \|\mathbf {y} \|\cos \theta \ ,$

one obtains:

$1-\cos ^{2}\theta =\sin ^{2}\theta \ ,$

which is sometimes called the fundamental Pythagorean trigonometric identity. There is, therefore, a connection between Lounesto's equation and the Pythagorean trigonometric identity. The label suggests some connection.

The common use of the term "Pythagorean theorem" is different, however. In n-dimensions the basic formulation is provided by Douglas, p. 60:

$\|\sum _{i=1}^{n}f_{i}\|^{2}=\sum _{i=1}^{n}\|f_{i}\|^{2}\ ,$

where f is an n-dimensional vector in an n-dimensional space. Clearly, the practical implications of this equation differ from those of Lounesto's equation. One such difference is that the Pythagorean theorem is a quadratic relation involving a single vector, while Lounesto's is a quartic equation involving a pair of vectors.

Whatever steps one might take to connect Lounesto's equation to the Pythagorean theorem, it is clear that some kind of bridge connects the two and must be understood. I'd suggest that the reader would benefit from a footnote suggesting that the label here as applied to Lounesto's equation is heuristic in nature, and there is no intention with this labeling to suggest that this equation somehow is a replacement or equivalent for Pythagoras' theorem in 7-dimensions.

It also would be helpful if the note indicated this equation is in fact Lagrange's identity in 7-dimensions, which notification would provide the correct context to further pursue this equation and to connect it to spaces of other dimensions. Brews ohare (talk) 14:50, 29 May 2010 (UTC)

Again, do you have a source for this "fact" ? --JohnBlackburne^words_deeds 15:38, 29 May 2010 (UTC)

You respond only to the last sentence, which is the topic in this thread. Brews ohare (talk) 15:45, 29 May 2010 (UTC)

In the 3D case, I'm with Lounesto. The equation is actually the Lagrange identity, but the reason why it is chosen as a desired starting point is because it carries the soul of Pythagoras's theorem. This does however need to be clarified in the main article. As regards the 7D case, it is not Pythagoras's theorem. But it does carry the spirit of something similar to Pythagoras's theorem in 7D. So overall, I'm not too bothered about Lounesto's choice of terminology. I do however think that elaboration, clarification, and an additional name would be to the benefit of the readers. David Tombe (talk) 19:27, 29 May 2010 (UTC)

I've asked Brews this but I'll try again: do you have a source on the 7D cross product for this. Your own working, or your maths professor's, does not qualify - it's original research. Drawing on that and odd facts from 3D to make a case is synthesis. We need a source. The information can then be included in the article with a straightforward reference; no need for long expository footnotes making tangential arguments, like the one I removed. --JohnBlackburne^words_deeds 20:27, 29 May 2010 (UTC)

Again, John, your remark belongs in this thread. Here the point is simply that a short footnote to clarify that the labeling as "Pythagoras' theorem" is oblique and not intended to suggest that the above sum-of-squares is being supplanted. Brews ohare (talk) 20:32, 29 May 2010 (UTC)

You've read my point here, I'm not going to repeat it yet again - it's you that's been starting multiple threads to try and prove your point, so don't expect me to guess which is the correct thread to answer in. But I can't see anything anywhere that answers my point. Which source on the 7D cross product is it from? If it's not from a source it's original research and synthesis that has no place here.--JohnBlackburne^words_deeds 23:12, 29 May 2010 (UTC)

John; your comment belongs in the earlier thread here. The present thread refers to the possible confusion engendered by referring to an expression involving a cross product as "Pythagoras' theorem", which expression is very unlike the normal sum-of-squares statement of Pythagoras' theorem, which applies in n-dimensions, while the cross-product applies (of course) only in 3-D and 7D. However, to accommodate you, I have answered your comment directly in a new section immediately below. Brews ohare (talk) 14:25, 30 May 2010 (UTC)

The role of Lagrange's identity should be made clear

John: I moved your comment to this section where discussion of your question can be isolated and you will not have to concern yourself with separating two distinct matters. The discussion of your question first began here. The sources are provided, and the logic “a = b; b = c ; therefore a = c” is stated. If you have trouble with the logic, the sources for the premises, or the conclusion, please indicate them precisely, and I will attempt to clarify. Here is your comment:

You've read my point here, I'm not going to repeat it yet again - it's you that's been starting multiple threads to try and prove your point, so don't expect me to guess which is the correct thread to answer in. But I can't see anything anywhere that answers my point. Which source on the 7D cross product is it from? If it's not from a source it's original research and synthesis that has no place here.--JohnBlackburne^words_deeds 23:12, 29 May 2010 (UTC)

The object of the discussion is to establish the role of Lagrange's identity in limiting the cross product.

The discussion begins with Lounesto's equation (labeled "Pythagorean theorem"), and with Lagrange's identity, one source is Weisstein, another is Boichenko et al.

I regret that you have not had time to look at these sources and gain some familiarity with Lagrange's identity. You claim it is irrelevant in that it doesn't refer directly to the cross product, but of course it is framed for n-dimensions, and the cross product is not defined except in 3 and 7-dimensions.

Lounesto's "Pythagorean theorem" is used as part of the definition of the cross product as:

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\|\mathbf {x\times y} \|^{2}\ ,$

and is being applied here to 7 dimensions.

Lagrange's identity in 7-dimensions is:

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\ ,$

(a specialization of the Binet-Cauchy identity) . For example see this.

John, please take time to notice that both Lounesto's equation and Lagrange's identity hold for every pair of vectors x, y in 7-space. Please take time also to notice that Lagrange's identity has the same left-hand side as Lounesto's equation.

Therefore, as pointed out in great detail earlier, the right-had sides also are equal. That makes:

$\|\mathbf {x\times y} \|^{2}\ =\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\ .$

(Notice how straightforward this formulation is in 3-D.) In other words, Lounesto could equally begin instead with the above logical equivalent as his definition. It has no relation to Pythagoras' theorem.

Doubtless, the thought will arise that although the above could be done, it is not what Lounesto chose to do, and therefore the Lagrange relation is not pertinent here. On the other hand, as the Lagrange relation holds true, whether you ignore it or not, it is apparent that a successful cross product will satisfy

$\|\mathbf {x\times y} \|^{2}\ =\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\ ,$

no matter how it originates. Differently said, Lounesto's procedure is logically equivalent to the above, but that fact can be established only if one is aware of Lagrange's identity.

My aim in this protracted discussion has been simply to point out that it is not "Pythagoras' theorem" that is the key in establishing the 7-D cross product, and that is not the real content of Lounesto's equation:

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\|\mathbf {x\times y} \|^{2}\ ,$

but the Lagrange theorem, and Lounesto's approach based upon defining the cross product as

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\|\mathbf {x\times y} \|^{2}\ ,$

has really very little to do with enforcing a condition upon the cross product imposed by Pythagoras' theorem and much more to do with imposing upon the cross product the necessary constraint imposed by Lagrange's identity.

Consequently, Lounesto's label for his equation as "Pythagoras' theorem" alludes to a near irrelevancy in defining the cross product, and not the real source of restrictions upon it, which stem instead from Lagrange's identity. Brews ohare (talk) 13:38, 30 May 2010 (UTC)

I'd like to see somewhere in the article a mention of Lagrange's identity and the equivalence of Lounesto's definition of the cross product to one based upon:

$\|\mathbf {x\times y} \|^{2}\ =\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\ ,$

This last equation is a clear illustration of the restrictions upon the cross product, as one can see directly that there are many terms on the right that must be summarized in only the seven components of the cross product on the left. These restrictions imposed by Lagrange's identity clearly play an important role in deciding whether a multiplication table can be found. It is a service to the reader to alert them to this identity, important in many arenas actually, that plays such an important role here in determining the cross product. Brews ohare (talk) 13:32, 30 May 2010 (UTC)

Brews, Just as a point of interest, a multiplication table can be found in 5D or in any odd number of dimensions,

z = x × y	x × y
i	j×m, l×k
j	i×l, k×m
k	i×j, l×m
l	i×m, j×k
m	i×k, j×l

But it's only in 3D and 7D that the Lagrange identity will be satisfied. Hence, a cross product can exist in any odd number of dimensions and so can the Lagrange identity. But the two only intersect at 3 and 7 dimensions. That intersection is Pythagoras's theorem in the 3D case. In the 7D case it does not appear to be Pythagoras's identity because the Jacobi identity restricts the sine relationship to 3D. David Tombe (talk) 22:17, 30 May 2010 (UTC)

Proposed amendment

The following is a proposal for a brief insertion into the existing article. The surrounding text already present is included to show the flow is not interrupted.

Characteristic properties

A cross product in an n-dimensional Euclidean space V is defined as a bilinear map

V × V → V

such that

x ⋅ (x × y) = y ⋅ (x × y) = 0,     (orthogonality)

|x × y|² = |x|² |y|² − (x ⋅ y)²     (Pythagorean theorem)

for all x and y in V, where x ⋅ y denotes the Euclidean dot product. The first property states that the cross product is perpendicular to each of its arguments. The second property determines the magnitude of the cross product. The second property determines the magnitude of the cross product. An alternative expression for the magnitude is in terms of the angle θ between the vectors, which in higher dimensions is given by:

x ⋅ y = |x||y| cos θ.

From the Pythagorean trigonometric identity and the second property the norm |x × y| is therefore:

|x × y| = |x||y| sin θ.

This is the area of the parallelogram with edges x and y, as measured in the 2-dimensional subspace spanned by the vectors.

( Insert→ ) By substitution for the right-hand side using Lagrange's identity Weisstein, Boichenko et al.:

$\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}=\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\ ,$

the second equation defining the cross product also can be written as:

$\|\mathbf {x\times y} \|^{2}\ =\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\ .$

This formulation shows clearly the challenge for the cross-product to succeed in capturing the many terms on the right in only the seven components of a single cross-product vector. ( ←End of insert ) It has been shown that nondegenerate nontrivial cross products with 2 factors exist only for n = 3 and n = 7. ...

Of course, the added references would be properly formatted using the cite book template. Brews ohare (talk) 16:27, 30 May 2010 (UTC)

(edit conflict)Again, do you have a source for that, i.e. a source relevant to this article on the 7D cross product? That Lagrange's identity exists and is defined in n-dimensions is not at issue. It's the leap you make that it's important for the 7D cross product that needs a source.--JohnBlackburne^words_deeds 16:46, 30 May 2010 (UTC)

John: OK, we're converging on the issue here. I take it that you agree that Lagrange's identity applies to n=7, as it does to all n. I take it you agree that it enables the second form of definition for the cross product, namely:

$\|\mathbf {x\times y} \|^{2}\ =\sum _{i=1}^{6}\sum _{j=1+1}^{7}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}\ .$

The issue outstanding is whether it matters that Lagrange's identity or its consequence above applies. Have I got it, John? Brews ohare (talk) 17:33, 30 May 2010 (UTC)

It clearly doesn't matter that if you do a long and tedious calculation (which according to David is so difficult he "kept making errors") you can show that the thing on the left equals the thing on the right. If it did matter, i.e. if it were interesting or useful, it would be in a source. If it were called "Lagrange's identity" in 7D it would be in a source.

On the other hand Pythagoras's theorem is clearly relevant. Not only is the formula clearly of the form a² = c² - b², but when you know that the dot product is xy cos θ you can immediately deduce that |x × y| is xy sin θ from the related Pythagorean trigonometric identity. This is all that's needed: the second property establishes the magnitude of x × y, and it doesn't get much simpler than xy sin θ. Most importantly it's sourced.--JohnBlackburne^words_deeds 20:16, 30 May 2010 (UTC)

John, your remarks make no sense. When equation “a=b” is sourced and equation “b=c” is sourced, there is neither need to source “a=c” nor to derive it. As spelled out at length above, it is grade-school logic, as you know of course. Your arguments are parody. Brews ohare (talk) 05:49, 31 May 2010 (UTC)

WP:SYN says:

"A and B, therefore C" is acceptable only if a reliable source has published the same argument in relation to the topic of the article.

so, again, where is your source? And your source for this:

This formulation shows clearly the challenge for the cross-product to succeed in capturing the many terms on the right in only the seven components of a single cross-product vector.

?--JohnBlackburne^words_deeds 07:36, 31 May 2010 (UTC)

Brews, earlier you wrote: "The literal use of a syllogism like a=b & b=c ? a=c is not WP:OR or WP:SYNTH either."

Look at the latter where it says:

• Do not combine material from multiple sources to reach or imply a conclusion not explicitly stated by any of the sources. If one reliable source says A, and another reliable source says B, do not join A and B together to imply a conclusion C that is not mentioned by either of the sources. This would be a synthesis of published material to advance a new position, which is original research.

Now replace A with "a=b", B with "b=c" and C with "a=c". This is a school book example of WP:SYNTH. DVdm (talk) 07:51, 31 May 2010 (UTC)

DVdm: If we were dealing with something like the example given in WP:OR where a conclusion actually is drawn, you'd have a point. But we're not. There is no judgment on my part involved. It is straightforward substitution, done routinely in all mathematical reasoning. Pardon me if I take you at face value as expressing your real (and final) understanding of the matter. Brews ohare (talk) 15:06, 31 May 2010 (UTC)

Yes John, it is indeed a challenge for the cross-product to succeed in capturing the many terms on the right in only the seven components of a single cross-product vector. That was the challenge which I put to you in April. But unfortunately it turned out that we were working at cross purposes, because you accepted it in 7D but only in a particular uniform. It's a challenge alright. The seven terms which you mention are all squared terms each with six components. That multiplies out to 252 terms, and 168 of those are mutually cancelling. It's time for you to rise to that challenge, and then you will see for yourself how easy it is to make trivial mistakes with all the subscripts etc. In 1993, I couldn't do it, and I sent it to my applied maths professor because I wasn't sure whether or not the problem was that it couldn't actually be done, or that I was making silly mistakes. I only realized last month that what I was doing in 1993, which was trying to prove the distributive law for the 7D cross product, was actually one and the same problem that we were discussing here.

In fact, I'll give you a start. Take,

z1 = x2y4-x4y2+x5y6-x6y5+x3y7+x7y3

z2 = x3y5-x5y3+x6y7-x7y6+x4y1-x1y4

z3 = x4y6-x6y4+x7y1-x1y7+x5y2-x2y5

z4 = x5y7-x7y5+x1y2-x2y1+x6y3-x3y6

z5 = x6y1-x1y6+x2y3-x3y2+x7y4-x4y7

z6 = x7y2-x2y7+x3y4-x4y3+x1y5-x5y1

z7 = x1y3-x3y1+x4y5-x5y4+x2y6-x6y2

Square each of the z terms and you will get 252 terms in total. 168 of these will mutually cancel. In fact, the 168 terms will be 2x84 terms, with each group of 84 containing two groups of 42 mutually cancelling terms. The remaining uncancelled 84 terms can then be reduced to the 21 squared terms. David Tombe (talk) 18:41, 31 May 2010 (UTC)

Reconsideration of 7-D cross product?

This dialogue is transferred from John Blackburne's Talk page. Brews ohare (talk) 19:38, 3 June 2010 (UTC)

Hi John:

In the article Cross product is the section Cross product#Alternative formulation based upon the paper by WS Massey. This is the approach he uses for the 7-D cross product.

In the cross-product article also is the subsection Cross product#Lagrange's identity, which follows (for the 3D cross product) exactly the paradigm I have suggested for the 7-D cross product. In particular, it combines the documented relations:

$\|\mathbf {a} \times \mathbf {b} \|^{2}=\|\mathbf {a} \|^{2}\|\mathbf {b} \|^{2}-(\mathbf {a} \cdot \mathbf {b} )^{2},$

and

$\sum _{1\leq i<j\leq n}\left(a_{i}b_{j}-a_{j}b_{i}\right)^{2}=\|\mathbf {a} \|^{2}\ \|\mathbf {b} \|^{2}-(\mathbf {a\cdot b} )^{2}\ ,$

to obtain the cross product in terms of components of a and b (changed to 7D below):

$\|\mathbf {a} \times \mathbf {b} \|^{2}=\sum _{1\leq i<j\leq 7}\left(a_{i}b_{j}-a_{j}b_{i}\right)^{2}\ .$

I wonder if this 3-D example might have persuaded you to support including this result in the article Seven-dimensional cross product? Brews ohare (talk) 16:03, 3 June 2010 (UTC)

Nothing's changed: we need a source for this, i.e. one on the 7D cross product. If you can point to a source that "this result" is from it will be clear how it relates to what's there already. Otherwise it's original research, so should not be included.--JohnBlackburne^words_deeds 16:09, 3 June 2010 (UTC)

(edit conflict) and in future please take time to review your comments before posting, not after. It is most annoying that I have to edit my reply twice because you've edited yours in the interim.--JohnBlackburne^words_deeds 16:09, 3 June 2010 (UTC)

OK, John. It is still my opinion that no source is necessary. Both the equations:

$\|\mathbf {a} \times \mathbf {b} \|^{2}=\|\mathbf {a} \|^{2}\|\mathbf {b} \|^{2}-(\mathbf {a} \cdot \mathbf {b} )^{2},$

which I might shorthand as:

$x(a,b)=y(a,b)\ \mathrm {for\ every\ choice\ of} \ a,b,$

and

$\sum _{1\leq i<j\leq 7}\left(a_{i}b_{j}-a_{j}b_{i}\right)^{2}=\|\mathbf {a} \|^{2}\ \|\mathbf {b} \|^{2}-(\mathbf {a\cdot b} )^{2}\ ,$

which I might shorthand as:

$z(a,b)=y(a,b)\ \mathrm {for\ every\ choice\ of} \ a,b,$

are documented in the cross product article and in previous discussion between us. The result I'd like to see in the 7-D cross product article is then:

$x(a,b)=z(a,b)\ \mathrm {for\ every\ choice\ of} \ a,b\ ,$

which I cannot understand as a "new" result unsupported by sources, because a simple mathematical substitution of equivalent results is simply Routine calculation, and not at all Synthesis to advance a position: it doesn't involve my judgment on an issue, but merely replaces one expression of a mathematical quantity with another one in different form, sourced as being identical. Brews ohare (talk) 19:38, 3 June 2010 (UTC)

Doing algebra using seven dimensional vectors is not "routine calculations" by any stretch of the definition - just look at the examples given. So it is original research and synthesis.--JohnBlackburne^words_deeds 20:09, 3 June 2010 (UTC)

John: The point here is content-independent, which is why it is not WP:OR. The functions x(a,b), y(a,b) and z(a,b) can be any functions whatsoever. Regardless of their form, the mathematical statement that x(a,b) = y(a,b) = z(a,b) for all a, b requires x(a,b) = z(a,b) for all a,b. That is a matter of mathematical usage, and to quarrel with it is to quarrel over grammar, not content. Brews ohare (talk) 20:27, 3 June 2010 (UTC)

For example, T. Koetsier says “In order to prove nowadays the equality of two functions f and g of one variable, one must prove that ∀(x) f(x) = g(x).” That is, f(x) = g(x) is true for all x for which f and g are defined. Clearly, if f(x) = g(x) and g(x) = h(x), then f(x) = h(x). Adding more arguments doesn't change anything. Brews ohare (talk) 22:07, 3 June 2010 (UTC)

Resolution

It turns out to be trivial to show the Gram determinant form of cross product is exactly the same as the Lagrangian form:

$\|\mathbf {a} \times \mathbf {b} \|^{2}=\sum _{1\leq i<j\leq 7}\left(a_{i}b_{j}-a_{j}b_{i}\right)^{2},$ "Lagrangian" or component form

$={\frac {1}{2}}\sum _{i,j=1}^{n}\left(a_{i}b_{j}-a_{j}b_{i}\right)^{2},$

$=\sum _{i}a_{i}^{2}\sum _{j}b_{j}^{2}-\sum _{i,j}a_{i}a_{j}b_{i}b_{j}=\|\mathbf {a} \|^{2}\|\mathbf {b} \|^{2}-(\mathbf {a\cdot b} )^{2}\$ "Gram determinant" or "Pythaogorean" form

In other words, the expression of the cross product in component form using the Lagrange identity contributes nothing new to the situation, and is trivially identical with the "Pythagorean theorem" or "Gram determinant" formulation of the cross product. Consequently, either can be used at will. Brews ohare (talk) 05:55, 6 June 2010 (UTC)

It's algebraic substitution, not OR

Brews, It's OK. As far as I am concerned, you don't need a source for this information. It is not original research and it is not synthesis. The equation,

$\sum _{1\leq i<j\leq n}\left(x_{i}y_{j}-x_{j}y_{i}\right)^{2}=\|\mathbf {x} \|^{2}\ \|\mathbf {y} \|^{2}-(\mathbf {x\cdot y} )^{2}\ ,$

is public domain. In 3D the left hand side becomes,

(x₂y₃-x₃y₂)² + (x₃y₁-x₁y₃)² + (x₁y₂-x₂y)² which happens to equate to z₁² + z₂² + z₃² in the cross product.

In 7D the left hand side becomes,

(x₂y₄-x₄y₂)² + (x₃y₇-x₇y₃)² + (x₆y₅-x₅y₆)² + (x₁y₄-x₄y₁)² + (x₃y₅-x₅y₃)² + (x₆y₇-x₇y₆)² + (x₁y₇-x₇y₁)² + (x₂y₅-x₅y₂)² + (x₄y₆-x₆y₄)² + (x₁y₂-x₂y₁)² + (x₃y₆-x₆y₃)² + (x₅y₇-x₇y₅)² + (x₁y₆-x₆y₁)² + (x₂y₃-x₃y₂)² + (x₄y₇-x₇y₄)² + (x₁y₅-x₅y₁)² + (x₃y₄-x₄y₃)² + (x₂y₇-x₇y₂)² + (x₁y₃-x₃y₁)² + (x₂y₆-x₆y₂)² + (x₄y₅-x₅y₄)²

which happens to be equal to z₁² + z₂² + z₃² + z₄² + z₅² + z₆² + z₇² in the cross product. It's just a question of taking,

z1 = x2y4-x4y2+x5y6-x6y5+x3y7+x7y3

z2 = x3y5-x5y3+x6y7-x7y6+x4y1-x1y4

z3 = x4y6-x6y4+x7y1-x1y7+x5y2-x2y5

z4 = x5y7-x7y5+x1y2-x2y1+x6y3-x3y6

z5 = x6y1-x1y6+x2y3-x3y2+x7y4-x4y7

z6 = x7y2-x2y7+x3y4-x4y3+x1y5-x5y1

z7 = x1y3-x3y1+x4y5-x5y4+x2y6-x6y2

and squaring each of the z terms. You will get 252 terms in total. 168 of these will mutually cancel. In fact, the 168 terms will be 2x84 terms, with each group of 84 containing two groups of 42 mutually cancelling terms. The remaining uncancelled 84 terms can then be reduced to the 21 squared terms. That is not original research. It is pure algebraic substitution. We have plenty of sources for the 3D case. We don't need any sources in order to apply the same logic in 7D when it is manifestly obvious to everybody that the same principle applies there too. David Tombe (talk) 21:07, 3 June 2010 (UTC)

Recent revisions

In the current version several useful points were dropped that appear in the earlier version:

• The squared magnitude of the cross-product is the Gram determinant of the vectors. That connection to another article is useful, as is the connection of the Gram determinant to the squared area of the parallelogram defined by the two vectors.
• The introduction of the angle θ which was carefully referred to Hildebrand previously and connected to the Cauchy-Schwarz inequality has simply been sloughed over in the new version.
• The useful expression of the cross product in terms of three-dimensional cross products has been dropped.

$\mathbf {x} \times \mathbf {y} =\sum _{i=1}^{7}\pi _{i}(\mathbf {x} )\times \pi _{i}(\mathbf {y} ).$

There is advantage in keeping these items: they provide additional background for the reader, they connect the article to other topics on WP, and they take very little space or explanation. Brews ohare (talk) 03:43, 1 July 2010 (UTC)

The article had a number of problems, mostly formatting but also editorial so I've tried fixing it: making the formatting and notation consistent, separating out the arguments for existence/references to the proof, removing duplication and fixing some errors.

The Gramian matrix is an unnecessary complication in two dimensions as it simplifies to (xy)² - (x⋅y)², i.e. the expression given which depends only on the dot product and norm/length. It's useful for generalising to more than two vectors but not for two - that's how Lounesto uses it for example. Other than that it duplicated the information given a few lines down about the area of a parallelogram (and again unnecessarily stated the general case of that).

I missed out saying what the angle is so have added that in, with a link in case anyone does not know what an angle is. There wasn't anything about the Cauchy-Schwarz inequality and I don't see how it's related.

That expression was unclear: it said "the sum of seven 3-dimensional cross products" but did not specify them, i.e. did not say what 3-dimensional space they were over, what handedness (the "right handed rule") they use, etc.. To fully specify all that would be quite difficult and as a calculation method it's far more complicated than any of the others, because of the repeated projections. The article still gives four or five ways to calculate the product, all simpler and clearer.

--JohnBlackburne^words_deeds 09:43, 1 July 2010 (UTC)

The point of mentioning the Gram matrix is to make that connection; that is, to indicate that this particular dot-product expression has deeper roots, and is not an isolated invention. A major value of WP is its capacity to link across related subjects, enabling readers to see connections between a topic they are looking at and other topics.

I'm a bit surprised that you are unaware of the connection of angle to the Cauchy-Schwarz inequality; at any rate I don't see the reference to Hildebrand restored.

If you find the expression of the cross product in terms of three-dimensional cross products would benefit, just provide a source rather than deleting the connection. Brews ohare (talk) 11:19, 1 July 2010 (UTC)

But it doesn't have "deeper roots", i.e. it doesn't depend on the Gram matrix. The gram matrix simplifies to the right hand side, but there are many things that simplify to the RHS, all of them by definition more complex. If they help illustrate the 7D cross product it makes sense to include them - e.g. if they provide some geometric insight - but otherwise it's just an irrelevant algebraic coincidence.

I really don't see in the version before I changed it any mention of the Cauchy-Scwarz inequality. But it's of little use anyway as it only shows the RHS is non-negative, correct but not enough to show anything, and trivially true from the LHS being |x × y|².

The expression in terms of three-dimensional cross products had no source, and was not mentioned in the sources I consulted (if it were it might be straightforward to improve it so it makes sense).

--JohnBlackburne^words_deeds 12:21, 1 July 2010 (UTC)

Apparently we disagree, largely because we are looking at the matter from different perspectives. (i) Gram determinant: the Gram determinant obviously is an object in the mathematical universe, it has a well-known interpretation as the volume of a solid object, and it has a WP article. IMO it is helpful to the reader to see this peculiar equation is not an invention, but an example of a class of such formulas. That linkage, again IMO, is a major function of WP: let readers know of connections they may be unaware of. (ii) the definition of angle is well-understood by yourself, but it is certainly reasonable to provide the definition and a source to that definition for any reader that needs a bit of of a jog in this department. (iii) Obviously the expression in terms of 3D cross products did not arise from the blue. I'd undertake to track a source down if that really satisfied your objections, and did not simply raise some new ones in an unending list. Brews ohare (talk) 20:53, 3 July 2010 (UTC)

The unit vector notation

I was wondering would it be better for the 21 multiplications to use the i, j, k, l, m, n, and o notations for the unit vectors, rather than e1, e2, e3, e4, e5, e6, and e7? A reader who has never heard of the seven dimensional cross product before usually likes to see it explicitly stated in the language that they are familiar with when using the 3D cross product. It makes it easier to grasp the concept. What do you say? David Tombe (talk) 17:07, 1 July 2010 (UTC)

As it was it said "the familiar i, j, and k, unit vector notation" then added four more letters without explaining them, and was inconsistent with the rest of the article. The e₁, ... notation are much preferred for higher dimensions: they generalise to any dimension, while the alphabet starting at i only goes up to 18. If you used more letters you would not be able to use them for other things – (x, y, z) for example – without confusion, while still others like l and o are easily confused with numbers. Also formulae like

$\mathbf {e} _{i}\times \mathbf {e} _{i+1}=\mathbf {e} _{i+3}$

are only possible with e_i notation. For these reasons higher dimensions (and even lower dimensions once you start using them) are usually described using e_i or things like them. It's also consistent: it means the different ways of calculating the product use the same notation, which is much clearer as it's easier to see that they're all the same product (not automatic in seven dimensions).--JohnBlackburne^words_deeds 18:03, 1 July 2010 (UTC)

John, That's fair enough. It's actually alot clearer now than it was before, but I'm going to re-arrange the relative positions of the 21 operations so as to have all the e1's along the top row etc. The idea will be to emphasize the fact that e1 can be the product of three different pairs from the other six. I'll do that sometime over the weekend because it's going to be a very fiddly job. David Tombe (talk) 17:17, 2 July 2010 (UTC)

There's no need to do that. It already presents that information clearly in the vector expansion, i.e. this

${\begin{aligned}\mathbf {x} \times \mathbf {y} =(x_{2}y_{4}-x_{4}y_{2}+x_{3}y_{7}-x_{7}y_{3}+x_{5}y_{6}-x_{6}y_{5})\,&\mathbf {e} _{1}\\{}+(x_{3}y_{5}-x_{5}y_{3}+x_{4}y_{1}-x_{1}y_{4}+x_{6}y_{7}-x_{7}y_{6})\,&\mathbf {e} _{2}\\{}+(x_{4}y_{6}-x_{6}y_{4}+x_{5}y_{2}-x_{2}y_{5}+x_{7}y_{1}-x_{1}y_{7})\,&\mathbf {e} _{3}\\{}+(x_{5}y_{7}-x_{7}y_{5}+x_{6}y_{3}-x_{3}y_{6}+x_{1}y_{2}-x_{2}y_{1})\,&\mathbf {e} _{4}\\{}+(x_{6}y_{1}-x_{1}y_{6}+x_{7}y_{4}-x_{4}y_{7}+x_{2}y_{3}-x_{3}y_{2})\,&\mathbf {e} _{5}\\{}+(x_{7}y_{2}-x_{2}y_{7}+x_{1}y_{5}-x_{5}y_{1}+x_{3}y_{4}-x_{4}y_{3})\,&\mathbf {e} _{6}\\{}+(x_{1}y_{3}-x_{3}y_{1}+x_{2}y_{6}-x_{6}y_{2}+x_{4}y_{5}-x_{5}y_{4})\,&\mathbf {e} _{7}.\\\end{aligned}}$

shows how the e₁ term is produced from the other six, and so on. The point of the 21 operations is to show the symmetry of the product modulo 7 and by groups of 3, which also makes it easier to see that the product is not unique. It's the same as Lounesto, page 96, except as we're not constrained by page count I've written it out in full.--JohnBlackburne^words_deeds 21:58, 2 July 2010 (UTC)

John, what I have just done is fully consistent with your intentions here. By the way, I do believe that the 21 operations should appear higher up in the article, even before the issue of properties and caracteristics. General readers may like to see the end product first because it is likely that they will want to quickly compare the concept to how they understand the 3D cross product. You and I can both follow the line of reasoning as it is presented in this article, but it did take me quite a while to do so. I suggest we make it easier for future readers to do so. David Tombe (talk) 14:47, 4 July 2010 (UTC)

Using label ‘Pythagorean theorem’ to describe the Gram determinant

At present, the equation:

$|\mathbf {x} \times \mathbf {y} |^{2}=|\mathbf {x} |^{2}|\mathbf {y} |^{2}-(\mathbf {x} \cdot \mathbf {y} )^{2}$

is labeled with the notation (Pythagorean theorem). The basis for using this label is that the author Pertti Lounesto labels this equation this way in his book ‘Clifford Algebras and Spinors’.

It may be noted that:

• (i) Lounesto provides absolutely no reason for using this label, and never discusses this choice in any manner whatsoever.
• (ii) Nowhere is this description of this equation used in any other book by any other author anywhere.
• (iii) The Pythagorean theorem describes the magnitude of a vector as the square root of the sum of the squares of its pairwise orthogonal components, which has no apparent connection to this equation.
• (iv) This equation is well known as the Gram determinant of the vectors x and y.
• (v) It also is well known that the Gram determinant is related to the square of the area of the parallelogram defined by the vectors x and y, which is exactly the role of this equation in this context in determining the magnitude of the cross product.
• (vi) Berger ‘Geometry I, Volume 2’ lists the magnitude of the cross product as one of its defining properties and states:

$\|x_{1}\times \ \cdots \times \ \ x_{n-1}\|=\left(\mathrm {Gram} (x_{1},\ \cdots \ ,\ x_{n-1})\right)^{1/2}\ .$

Other authors do likewise: e.g. Exercise 1.76 in Nasar, Gallier, Problem 7.10, Part 2 (p vectors in a space with n ≥ p) Gallier refers to this equation as the "Lagrange identity".

• (vii) Lounesto, p. 98 also uses the Gram determinant label to describe this equation in his §7.5 : ‘Cross products of k vectors in ℝⁿ’.

In view of these facts that show this labeling is

• (i) used only by one author and without justification,
• (ii) has no apparent connection to the Pythagorean theorem, and is therefore useless and misleading,
• (iii) can readily be replaced by a reference to the Gram determinant, for which there is clear application in this context,

it would seem reasonable to either (i) provide an explanation for the use of this label so the reader knows why it is used, or (ii) remove this label and substitute a reference to the Gram determinant as done by Berger, or (iii) simply remove the label altogether, or change it to something else, e.g. the ‘Lagrange identity in 3-D and 7-D’. What is unreasonable is to leave this label in place without explanation for its use.

John, if you insist upon making no changes in this regard, I suggest that this issue be prepared for an RfC. Brews ohare (talk) 14:36, 4 July 2010 (UTC)

Brews, I believe that Lagrange identity would be the best all round name for this equation. I'm fully sympathetic with Lounesto's use of the name Pythagoras in relation to the 3D case. In fact, in the 3D case, I would prefer Pythagoras to Lagrange. But in relation to the 7D case, the use of the term Pythagoras is alot more grey, and as you rightly point out, Lounetso seems to be unique in using this term in the circumstances. David Tombe (talk) 14:51, 4 July 2010 (UTC)

(edit conflict)this is about an entirely different cross product, the n-1 in n dimensions version. this is not even in the body of a text, it's an exercise, again nothing to do with this topic. The only one on the 7D cross product is Louneso, except he doesn't use the Gram determinant for it either, it's used in a later section. It's not needed for the 7D cross product and is just a further complication: You can't use it directly, all you can do is simplify it to the form that is already given, which as it's a familiar form in itself, as Pythagoras's theorem and the same as the identity in 3D, there's no need to supply a more complex thing that simplifies to it.

The topic, which is the 7D cross product, is covered very well in a number of sources already referenced in the article. There is no need to use sources on entirely unrelated topic. Of course they will use a different approach, different notation and different reasoning appropriate to their subject and readership, but to include them will just lead to a confusing and difficult to follow article. No-one is going to consult "Solved Problems in Electromagnetics" to learn about the 7D cross product, and I don't know how you came across it when looking for references on this subject.

David, no. your two examples illustrate my point perfectly. There are multiple more complex things that simplify algebraically to the expression given. Do we need to give them all, with multiple sources, even though they are useless until simplified to what's already there? What's there is straightforward and clear. I'm sorry if you don't understand it and think it's "grey" but introducing more complex expressions, unsupported by the sources on the 7D cross product, will make it even more difficult to understand.--JohnBlackburne^words_deeds 15:21, 4 July 2010 (UTC)

John: I find you to ramble here. If you look at Lounesto §7.5 : ‘Cross products of k vectors in ℝⁿ’ and put n=7 and k=2 that is what I'd recommend here. Likewise, Gallier, Problem 7.10, Part 2 (p vectors in a space with n ≥ p) with n=7 and p=2. I just don't understand your attachment to the label ‘Pythagorean theorem’, whose applicability is explained neither by Lounesto or by yourself in this article. Your discarding of all other sources, all of which follow a different path, is not convincing, and this labeling of the equation is misleading and confusing, most especially with absolutely no explanation of why the label is used. Brews ohare (talk) 15:42, 4 July 2010 (UTC)

(edit conflict)Again both those sources are on the more general case, of n not two vectors, and only Lounesto is on the 7D cross product. As for "Pythagorean theorem", to me it says the quantities |x||y|, |x×y| and x⋅y are related by the theorem, so form a right angled triangle. One angle of the triangle is the angle between the vectors so it can be formed with two sides parallel to the vectors, i.e. "Pythagorean theorem" has both geometric and algebraic meaning, and it relates also to the Pythagorean trigonometric identity later. It's not required for much of the theory but is useful as a way to think about what is essentially a geometric construct.--JohnBlackburne^words_deeds 15:51, 4 July 2010 (UTC)

The Pythagorean theorem says c² = a² + b²; if I try to recast the equation in this form, c = ||x||²||y||² and a = ||x×y||² while b= (x·y)². Question 1: How is that seen to be an application of Pythagoras' theorem? Question 2: What explanation of it are you going to put into the article so it makes a modicum of sense to the general reader (or to me)? Brews ohare (talk) 16:00, 4 July 2010 (UTC)

The fact is, it cannot be done unless one posits ||x × y|| = ||x|| ||y|| sin θ instead of the posit ||x × y||² = ||x||²||y||² -(x·y)². And, if you do that, the connection is to the Pythagorean trigonometric identity, not to Pythagoras' theorem. It also would require a reorganization of the article to derive the relation ||x × y||² = ||x||²||y||² -(x·y)², which still would not be the "Pythagorean theorem". Brews ohare (talk) 16:12, 4 July 2010 (UTC)

This "reverse" approach is not favored by Lounesto, Massey or, in fact, any author. All prefer to start with the Gram determinant form. That probably is because norms and dot-products are more fundamental than the notion of "angle" in n-dimensions. In fact, angle is defined in terms of dot-product. Brews ohare (talk) 16:27, 4 July 2010 (UTC)

No it's not. The angle in n dimensions is defined as in two or three dimensions. It's related to the dot product and so can be calculated using it, but it does not depend on it for definition. An alternative approach is to determine the simple rotation that rotates from one vector to the other and take its log. Or as in 3D if you had a protractor in e.g. 7D you could simply measure the angle. And there's no need to derive ||x × y||² = ||x||²||y||² -(x·y)². It's a condition require for the cross product, so it's simply asserted, then other things are derived from it. It's not the only condition that could be used, it's just the most useful one. --JohnBlackburne^words_deeds 16:36, 4 July 2010 (UTC)

John: You are creating conflict where none exists. As noted in my remarks, one can assume ||x × y||² = ||x||²||y||² -(x·y)² (that is what ‘posit’ means) and in fact that is what Lounesto, Massey and (it seems) most (if not all) others do. As for defining angle by a rotation, that is not the standard approach in n-dimensions, which is to define it in terms of the dot product, as you know. It would seem that determining an axis of rotation in 7D for the 2D cross product is not straightforward. However, all of this is arguing around the point, and not addressing it. The point is, again, that Pythagoras' theorem has nothing to do with it, no matter which way you go. And no-one except Lounesto (and John Blackburne) ever mentions Pythagoras' theorem in connection with the magnitude of the cross-product. Brews ohare (talk) 19:01, 4 July 2010 (UTC)

John, If this was only about the 3D case, I'd be supporting you. But it's about the 7D case. It seems that the clash between us in this regard originates with your belief that Pythagoras's theorem is a theorem of 2D space, whereas I believe it is a theorem of 3D space. But let's put that aside for a minute. Let's recall our discussion over at the 'Curl' article. We were basically in agreement that curl could not exist in 7D, at least in a way that would connect to geometry. I of course could actually set up a 7D binary curl based on the idea of making up a differential operator as d/si + d/tj + d/duk +d/dvl + d/wm + d/dxn + d/dyo and distributing it out over a 7D vector as per the rules of the 21 operation table. But as you have agreed already, this operation would have lost all connection with geometry.

Likewise, I believe that this 7D cross product has lost all connection with geometry, sine, Pythagoras etc. David Tombe (talk) 19:11, 4 July 2010 (UTC)

RfC: Does ‘Pythagorean theorem’ appropriately designate the definition of the magnitude of a vector cross product?

Does the label ‘Pythagorean theorem’ appropriately designate the definition of the magnitude of a vector cross product?

At present, the equation:

$|\mathbf {x} \times \mathbf {y} |^{2}=|\mathbf {x} |^{2}|\mathbf {y} |^{2}-(\mathbf {x} \cdot \mathbf {y} )^{2}$

is labeled with the notation (Pythagorean theorem). The basis for using this label is that the author Pertti Lounesto labels this equation this way in his book ‘Clifford Algebras and Spinors’. However, it may be noted that:

• (i) Lounesto provides absolutely no reason for using this label, and never discusses this choice in any manner whatsoever.
• (ii) Nowhere is this description of this equation used in any other book by any other author anywhere.
• (iii) The Pythagorean theorem describes the magnitude of a vector as the square root of the sum of the squares of its pairwise orthogonal components, which has no apparent connection to this equation.

Because this labeling is:

• (i) used only by one author and without justification or explanation,
• (ii) has no apparent connection to the Pythagorean theorem, and is therefore useless and misleading,

is it reasonable to either (i) provide an explanation for the use of this label so the reader knows why it is used, or (ii) remove this label and/or substitute something else? Brews ohare (talk) 20:10, 4 July 2010 (UTC)

Remarks

In a subsequent section of the same book, Lounesto, p. 98 labels the equation as ‘Gram determinant’, a designation also used by several other authors, and a designation in keeping with the standard interpretation of Gram determinant of two vectors as the squared area of their contained parallelogram. Brews ohare (talk) 19:26, 4 July 2010 (UTC)

As noted by me above, but reproduced here to save others having to find it, I can't see the problem with using "Pythagorean theorem" as

• It's sourced, from a good source on the 7D cross product (of which there are few - it's not like there are dozens of sources that don't use this)
• It's not as if there's another better name for it
• It matches the algebra of the Pythagorean theorem, i.e. a² + b² = c²
• It matches the geometry: a right triangle can be constructed with the sides in the formula with two sides parallel to the two vectors, so the angle between them is an angle of the triangle
• This leads directly to the Pythagorean trigonometric identity using this triangle, and so to the xy sin θ form of the magnitude

I would say it is not discussed in the source as the algebraic and geometric properties are obvious, or at least straightforward and so helps the reader visualise the cross product, in the source as here. I don't see what the problem with it is.--JohnBlackburne^words_deeds 20:28, 4 July 2010 (UTC)

• The label is applied by Lounesto, but not discussed, justified, or motivated in any way. Is this really what ‘a source’ means? All sources (including Lounesto) derive the same results from the same equations, while absolutely none of them (except Lounesto) uses the Pythagorean label for the dot-product equation.
• No construction is described in the article (or in Lounesto) indicating the dot-product formulation is the same assumption as the Pythagorean equation a² + b² = c².
• John has things exactly backwards: it is the posit that ||x × y|| = ||x ||y|| sin θ that introduces the triangle, and that assumption taken as an axiom can be run backward to derive the form in terms of the dot product by use of the Pythagorean trigonometric identity. If one wishes to run things backward, that's what the article should do. However, no author does it this way.
• If ‘the properties’ are obvious to John, maybe he can help the reader by describing them in the article? Brews ohare (talk) 21:56, 4 July 2010 (UTC)

The bottom line: whatever the connection of the dot-product equation to Pythagoras' theorem, the two things are not the same thing, and the label confuses at least some readers by suggesting that somehow these two different things are the same thing. Brews ohare (talk) 22:04, 4 July 2010 (UTC)

Comment. It does seem a little confusing: it may follow from the Pythagorean Theorem, but that's not the same as saying it is the Pythagorean Theorem. Nor is the connection immediate, since it relates a square (i.e. an area) to something more complex. If it's possible to produce a good diagram that actually shows the triangle, then it might be justified. -- Radagast3 (talk) 08:01, 5 July 2010 (UTC)

I'm not sure I could draw a good diagram of it, but the triangle has sides xy, xy sin θ and xy cos θ. If it is drawn in the plane of x and y, so one of them lies along the hypotenuse, then scaling that side by the length of the other puts it along the hypotenuse and the other two sides are then projections of this parallel and perpendicular to the to the second vector (which is therefore parallel to the other side with length xy cos θ). The squares are part of the Pythagorean formula, i.e. it's not about areas it's about the ratio of the lengths or magnitudes.--JohnBlackburne^words_deeds 08:43, 5 July 2010 (UTC)

Well, xy, xy sin θ and xy cos θ gives a Pythagorean triangle all right, but doesn't immediately seem relevant to $|\mathbf {x} \times \mathbf {y} |^{2}=|\mathbf {x} |^{2}|\mathbf {y} |^{2}-(\mathbf {x} \cdot \mathbf {y} )^{2}$. It all makes sense after the next few lines, but as it stands, the "Pythagorean theorem" comment just adds confusion. -- Radagast3 (talk) 11:43, 5 July 2010 (UTC)

Comment the formula is a few logical steps from what most people regard as the Pythagorean Theorem, so it seems fine to say that Lounesto has called it Pythagorean Theorem, but not to say that it is PT. I see no compelling need to give the equation a name especially where it introduces confusion.--Salix (talk): 11:22, 5 July 2010 (UTC)

Comment We have a three way dispute here. Brews opposes the term 'Pythagoras' in both the 3D case and the 7D case. John supports the use of the name 'Pythagoras' in both the 3D case and the 7D case. I support the use of the name 'Pythagoras' only in the 3D case. Hence if this article was about the 3D cross product, I'd be quite happy to leave it at 'Pythagoras's theorem'. It may not technically be Pythagoras's theorem, but I can fully understand Lounesto's poetic licence in the circumstances. But this article is about the 7D cross product, and in the 7D case, the fact that the equation holds at all seems to be somehwhat of an anomaly. It is based on a miraculous cancellation of 168 terms on distribution. I'm not as yet sure what this is trying to tell us. There is clearly something of the spirit of Pythagoras's theorem there, yet it falls short of being the full Pythagoras's theorem. Lounesto seems to have overlooked the fact that the Jacobi identity restricts the sine relationship to the 3D case. I seem to recall that the Encylopaedia Britannica source where I first read about the 7D cross product did not actually give a name for the equation in question. It merely stated the equation as a desired condition, without giving it a name. It may be that in order to settle this issue that we will have to simply remove the name altogether and present the equation without a name. David Tombe (talk) 11:40, 5 July 2010 (UTC)

Comment: As it stands the label is confusing. It seems to falsely indicate that the identity is an immediate consequence of the Pythagorean theorem. The result is that the reader is left trying to puzzle out a mental proof of the identity with no success. The article on the 3-dimensional cross product starts with the length as xy sin θ, from which the identity follows easily, but this article presents the material in a different order. In any case, the terminology used in articles should reflect the most common usage in the literature and that doesn't appear to be the case here. There is only one source that uses the terminology and inconsistently at that. It's also questionable how much benefit there is in having a label, so including it despite the potential confusion it may cause does not seem like a good idea.--RDBury (talk) 11:39, 5 July 2010 (UTC)

Rather than remove "Pythagorean theorem" it has been replaced with Gram determinant, with Lounesto as a source. This makes no sense as Lounesto is the source that has "Pythagorean theorem" as a label for that expression. The Gram determinant isn't used at all for the 7D cross product - it is used in a later section when discussing generalisations to products of more than two vectors. No other source uses it as a label for this expression either.--JohnBlackburne^words_deeds 09:05, 6 July 2010 (UTC)

It is interesting that yourself, a mathematician, refuses to see an argument for general integers k, namely Lounesto, p. 98 cannot understand that it applies to the case k = 2. Lounesto, in seeking a general formulation, says: ‘a natural thing to do is to consider a vector valued product a₁ × … × a_k satisfying:

$\|\mathbf {a_{1}\times \ \dots \ \times a_{k}} \|^{2}=\det(\mathbf {a_{i}\cdot \ a_{j}} )\ \ \mathrm {Gram\ determinant}$. ’

He goes on to say ‘The solution to this problem is that there are vector valued cross products in ’ a variety of cases including the case of k=2 n=7. That is extremely clear. Contrary to your assertions, this point of view is also adopted by other authors, for example Gallier and several others you pooh-poohed earlier because they mentioned the approach in exercises , rather than the main text (for example, Nasser, Exercise 1.76, p. 14).

These matters could be discussed at length in the article under a section "Generalization to k vectors", where Lounesto could be quoted verbatim in the article if that is your preference. That might be a good idea anyway.

In any event, John, you haven't a leg to stand on here, and are presuming to distort the position in the literature. Brews ohare (talk) 11:49, 6 July 2010 (UTC)

I have settled this matter by introducing a section called "Generalizations". Brews ohare (talk) 12:27, 6 July 2010 (UTC)

Bringing the 21 operations to the introduction

The first thing that any curious reader will want to see is what the 7 dimensional cross product looks like when expressed in the same language as the familiar 3 dimensional cross product. Previously the article somewhat resembled Rolf Harris painting one of his masterpieces. We were all kept in suspense as regards what it's all leading up to. David Tombe (talk) 22:09, 4 July 2010 (UTC)

I've moved it back: as per MOS:LEAD the introduction should be an accessible overview of the subject, so should avoid detailed technical content. More generally it makes sense to present the properties first before giving the detailed expression of the product, which anyway is over limited use: most of the results can be derived from the defining properties, as they clearly do not depend on the particular version of the product used.

There were problems too with the changes made while it was moved. First it was changed to a different product: there is not just one 7D cross product but many, and changing it made other parts of the article wrong and the whole article inconsistent. Also the table was unclear - the cross product is not associative so the order matters, but in a table like that the order is unclear. Better to simply write out the products as there's no limit to space. The information about the "completely antisymmetric tensor" also had to be removed as it too was about a different product.--JohnBlackburne^words_deeds 11:34, 6 July 2010 (UTC)

A Wikitable is a much clearer and more elegant presentation of the multiplication rule. The rule shown is that from Octonion which has the value that it agrees with the standard cross product from 3 dimension in making i x j = k etc. That alone makes the order of factors clear, but a simple note can add further clarity (it was already given in terms of the ε_ijk rule from the source summarizing the table) without abandoning the table for a messy term by term presentation. And, of course, it agrees with the Octonion article. Your arguments favor rewriting the article to suit this table, not the reverse. Brews ohare (talk) 13:13, 6 July 2010 (UTC)

(edit conflict)It's not clearer: as noted above the table was unclear and renders the rest of the article incorrect. Whether it's more elegant and the term by term expansion is "messy" is entirely subjective, and I would remind you again that changing the formatting just because you like it a different way is according to ArbCom unacceptable.--JohnBlackburne^words_deeds 13:21, 6 July 2010 (UTC)

As noted above, the table is not unclear in any way, and you have not presented any argument to support that view. As pointed out, the rule behind the table is presented algebraically, so no confusion is possible, and an example of the first component is provided as further aid to interpreting the table. This table agrees with Octonion, is sourced, and is more readily understood because it incorporates the standard unit vector i × j =k relations for the first 3 unit vectors.

The Lounesto form for the multiplication rule is left intact, so the remainder of the article still makes perfect sense. What would improve matters and be very helpful, would be an explanation of how the two rules both can be shown to be valid, a point that you could make that would help the reader considerably, and a very useful addition. How about that, John? Brews ohare (talk) 13:35, 6 July 2010 (UTC)

Written out fully it incorporates the product i × j = k: just take any such triplet of vectors. It's unclear because the product is anticommutative so depending which way you do the multiplication the result is different. But again, even without that objection changing the format just because you like it a different way is unacceptable.

As for some information on how the different versions of the 7D cross product are related I don't think it's necessary but if you think it's useful it's in the sources so feel free to add something.--JohnBlackburne^words_deeds 14:20, 6 July 2010 (UTC)

John: Perhaps you are unfamiliar with the standard notation ijk to represent unit vectors along the x- y- and z-axes? This corner of the table is the same as in 3-D. I don't think Lounesto's rule has that similarity.

Yes, I'd like to add the information relating different rules. Perhaps you can help me with that? How would you go about it? Brews ohare (talk) 14:31, 6 July 2010 (UTC)

It's all in the sources so I'd start with them - I assume you are familiar with them.--JohnBlackburne^words_deeds 14:45, 6 July 2010 (UTC)

John, The table should go in the introduction because a reader will want to quickly see the end product. This is after all an encyclopaedia and not a pure maths textbook. A pure maths textbook has a different purpose from that of an encyclopaedia. A pure maths textbook will be aimed at students who are being trained in logical processes and who are accustomed to building up to an end result. David Tombe (talk) 16:45, 6 July 2010 (UTC)

According to MOS:LEAD the lead should provide an accessible overview, so avoid too much technical content such as that. It's also unclear (what is i' × i or j x i for example), is yet another version of the 7D cross product (different from the other two) and uses inconsistent formatting with the rest of the article. As already noted above the i, j, k notation is generally reserved for 3D, as it generalises poorly to higher dimensions: for the same reason we use (x, y, z) in 3D but use e.g. (x₁, x₂, x₃, ,,, x_n) for n dimensions.--JohnBlackburne^words_deeds 17:36, 6 July 2010 (UTC)

John, I agree that we shouldn't have too much technical content in an introduction, but not to the extent of eliminating the main item itself. The seven dimensional cross product is the very set of 21 operations itself, and as such it should appear in the introduction in some form or other. The 3D cross product is familiar in i, j, and k notation. The 7D cross product is not familiar at all, and the fact that it exists at all comes somewhat as a surprise, even to many graduates. The first thing that anybody will want to know is 'what does it look like in relation to the 3D cross product?'. The encyclopaedia article needs to address that question right in the introduction. David Tombe (talk) 17:55, 6 July 2010 (UTC)

It isn't "the very set of 21 operations" that you give. That is just one of many possible sets that satisfy the conditions, and without that being specified what's there is incorrect and misleading. Most calculations done with the 7D cross product do not involve the term by term expansion - they can't as if you e.g. proved something from one such rule you could not be sure if it were valid for all others. It's also inconsistent with both versions given later on (the one that was there, the one that Brews added), does not give all the products you can form, and uses inconsistent formatting. It also seems to be unsourced.--JohnBlackburne^words_deeds 18:27, 6 July 2010 (UTC)

I've Wikitabled an ijklmno form of the multiplication table, not the same one David suggested, which is yet another form. I don't like this approach because it leads to considerable duplication, or an inelegant splitting of the topic and revisiting it later. Also, the numbered-subscript form allows a simple summary expression in terms of algebra, while the ijklmno form does not. I'd favor the multiplication be internally cross-linked to the later section, and not be explicitly listed in the introductions. That approach avoids these difficulties and still directs the reader to an explicit listing. Brews ohare (talk) 20:40, 6 July 2010 (UTC)

John, The version which I put into the introduction was identical to your own version. I simply replaced e1 with i, e2 with j etc. because I wanted to use the same notation which is used in the more familiar 3D cross product. The 21 operations were put in as an example of what the 7D cross product looks like. It was never intended to be the unique example, and if I didn't make that clear, it would have been easy to have re-worded it to make it clear.

Brews, since you wish to avoid duplication, I suggest that you only have one table. Since both you and John prefer the e1, e2, notation, then go for that notation, but I do believe that a table is necessary in the introduction. A reader will want to see at a glance an example of what the 7D cross product might look like. The article as it stands after the introduction is good as a pure maths textbook. It builds up to the final product in a series of arguments. But the '21 operation table' is the natural starting point for the encyclopaedia reader. The starting point for the 3D cross product is the i, j, k inscription that Sir William Rowan Hamilton made at Brougham Bridge in 1843. The first thing that any reader wants to see is how that set of relationships between i, j, and k, expands into 7 dimensions. David Tombe (talk) 23:40, 6 July 2010 (UTC)

(edit conflict)I doubt very much that any reader will want to start with one of the many different ways of calculating the product. As I've already said it's of no use for proving anything as, unlike in 3D, it's not unique. So if you use one particular expansion to prove a result you can't be sure it will work for all others. That's why if you want to prove anything or understand the product fully you need to start from the defining properties and proofs. More generally this is postgraduate, not high school, algebra so would normally be done abstractly, independent of the basis vectors, even independent of the dimension. It's useful to write down one expansion to show one exists, after the general existence of the product has been proved, but I don't see the need for any more. And it's best to do so with consistent notation - you yourself wrote "it's a lot clearer" after I made the formatting and products consistent.--JohnBlackburne^words_deeds 00:24, 7 July 2010 (UTC)

I'd guess that David is representative of some readers that want to see a concrete version of the cross product up front. Therefore, I moved the table to the Introduction and put in some caveats about non-uniqueness of the table. Then by way of a parallel development, I made a table for Lounesto's version for the later section. The two versions are contrasted in the section on Fano planes. Brews ohare (talk) 04:00, 7 July 2010 (UTC)

You might find this Fano plane link interesting. Brews ohare (talk) 04:09, 7 July 2010 (UTC)

John, The version of the 21 operations which you did was very good. Brews's table was also very good, and it happens to be in the exact format that I used to work out what a 7D cross product would look like when I first read about it in Encyclopaedia Britannica. If the two of you prefer to use e1, e2, e3, then that's fine with me. I will not object, although I would have preferred i, j, and k. But the most important thing is that the readers can see at a glance what the 7D cross product looks like. The 21 operation table, whether in your preferred format, or mine, or Brews's, tells a reader at a glance practically everything that they may ever ask about 7D cross product. Yes, this is a postgraduate topic, but the job of an encyclopaedia is to make the best attempt possible at making such advanced topics accessible to a more general readership. It's more about letting the readers have a basic idea of what the subject entails, rather than having a full understanding of it. But having said that, there is no reason why we can't do the latter also. In this case, the latter has already been reasonably well done.

In many respects, there is an analogy between the idea of putting the 21 operation table in the introduction and putting a picture of the Four-handed chess board in the introduction of that article. One can see at a glance what it's all about as compared to the more familiar version. But with words alone, it takes some explaining. You might say that the 21 operation table is the diagram. David Tombe (talk) 09:53, 7 July 2010 (UTC)

More on multiplication tables

John, I wonder if you could beef up the discussion of Fano planes? For instance, is it true that I could simply label the nodes in the Fano diagram any way I like and get a valid multiplication table? How many valid multiplications tables (Fano diagrams) are out there? Brews ohare (talk) 17:36, 7 July 2010 (UTC)

If you want it added I suggest you do it yourself: I have not come across Fano planes in relation to the 7D cross product. And there are far more important things to do in this article such as fix the errors, unclear mathematics and unencyclopaedic writing and presentation added by you and David in the last few days.--JohnBlackburne^words_deeds 23:33, 7 July 2010 (UTC)

John: Your comments about the Fano plane strike me as a bit off. You are doubtless aware that the Fano plane is used often in terms of the Octonions, and its role here is very similar indeed, as must be very clear to you as mathematician used to seeing patterns. It also is clear that there are many multiplication tables, and that they're all connected through the Fano diagram. It would be nice to tidy that up a bit.

I don't agree with you that the most important things left to do in this article are to "correct" contributions and "clarify" mathematics by David and I, which, I am afraid, simply means forgetting the general reader and writing for the specialist only. It would be nice if you could try to collaborate and pack up the put-downs and snide remarks. Brews ohare (talk) 05:07, 8 July 2010 (UTC)

I don't see how it is "snide" to state the fact that I've not come across Fano planes used for the 7D cross product. Perhaps you could post the source that relates them for my and other editors benefit – it will be easier then for another editor to look at this. --JohnBlackburne^words_deeds 06:59, 8 July 2010 (UTC)

John: The word "snide" means "slyly disparaging" and refers to your attitude, although your phrase “fix the errors, unclear mathematics and unencyclopaedic writing and presentation added by you and David” is more accurately described as insulting. Brews ohare (talk) 14:03, 8 July 2010 (UTC)

More generally I've already pointed out the problems with the tables and having multiple product rules I've tried fixing this but you simply [1] reverted to a broken and confusing version again without reason. I've just restored a correct version version, please don't change it again to a broken version or without good reason (as already noted changing the format just because you prefer it another way is simply unacceptable).--JohnBlackburne^words_deeds 06:59, 8 July 2010 (UTC)

A multiplication table is hardly a “broken and confusing version” when compared to a listing of some products. A table is easier to use and shows the pattern of the multiplication rule at a glance, something that cannot be said of your listing. Brews ohare (talk) 14:03, 8 July 2010 (UTC)

No it doesn't. The product is non-commutative so the order matters. The order is clear when you write e₁ × e₂ = e₄, it's not at all clear from a table like that. Writing out the product also lets you organise them so the symmetry and periodicity is obvious: the rule e_i × e_{i + 1} = e_{i + 3} stands out immediately. The same is not true of either table. The shading only shows the product is anticommutative, but otherwise looks random suggesting that anticommutativity is the only symmetry, which is clearly not the case. --JohnBlackburne^words_deeds 15:20, 8 July 2010 (UTC)

John: the rule e_i × e_{i + 1} = e_{i + 3} is a particular symmetry well-expressed by this formula, and that's about it. This rule suffices to determine the table, but not directly because it applies only for adjacent unit vectors. For example, e₁ × e₃ = e₇ can be found from this rule only by applying some identities. The table is helpful because such products can be found by inspection, without the need for manipulations. Brews ohare (talk) 17:21, 8 July 2010 (UTC)

John, I don't agree with your objections. However, your multiplication table is good, and so I will support its retention in the main body of the text. Equally I support the retention of Brews's table in the introduction. I see Brews's table as playing the role of 'the diagram' for the article. In fact, Brews's table ought to be beefed up somewhat into diagram format. When a general encyclopaedia browser is familiar with a topic such as chess or 3D cross product, they will have a picture of it in their minds. In the case of 3D cross product, that picture is the plaque on the wall at Brougham Bridge. When a more complex variant of such basic themes is ever mentioned, the first thing that a reader will want to have is a mental picture of how the extended concept looks in comparison to the standard well known theme. David Tombe (talk) 10:24, 8 July 2010 (UTC)

"In the case of 3D cross product, that picture is the plaque on the wall at Brougham Bridge." I don't agree with this statement at all. When I think of the 3D cross product I think of a bilinear map that takes any pair of vectors to a third vector perpendicular to both of them. In my opinion this is what the concept of "cross product" is about, and that should be in the introduction- not one possible multiplication table. Actually, I wonder if it might be a good compromise to put the multiplication table in a side bar like the chess board in the chess article that David brought up. Holmansf (talk) 13:23, 8 July 2010 (UTC)

Holmansf: Of course you have stated one of the two general requirements for a cross product, the other being the relation for its magnitude. However, you might agree that these statements of properties that appear in the section "Definition" are not immediately and intuitively translatable to a multiplication table of any kind. That is why a concrete example in the intro may be useful in the Intro, especially to the naive reader. It is clearly pointed out along with the table that it is not unique and that the properties in the "Definition" section are primary. Brews ohare (talk) 14:03, 8 July 2010 (UTC)

I'd like to repeat my suggestion that the multiplication table in the intro be moved to a side bar near the introduction (maybe sidebar isn't the right term- I'm not a wikipedia expert). What do you think about that? Holmansf (talk) 15:15, 8 July 2010 (UTC)

I implemented this suggestion by formatting the Wikitables to float at the right of the text. Brews ohare (talk) 17:26, 8 July 2010 (UTC)

Holmansf, Regarding the plaque at Brougham Bridge, you are right. I'm actually disappointed that Hamilton didn't inscribe i×j=k, j×k=i, k×i=j, because that was the most interesting aspect of his inspiration. David Tombe (talk) 18:14, 8 July 2010 (UTC)

The distributive law

The sentence “These unit vectors can be multiplied out distributively in both three and seven dimensions” appears in the section Expansion in unit vectors. Clearly, it is true, but what exactly is the point in bringing it up in this context? Maybe a little elaboration would illuminate its relevance? Brews ohare (talk) 17:56, 7 July 2010 (UTC)

Brews, It certainly can be elaborated on. In fact I was coming to that very point next. If you look further up the talk page, you will see that the issue of the distributive law has been raised in the past. The distributive law holds in the case of the 3D cross product and also the 7D cross product, but in a typical 3D cross product course, this fact would be proved using geometry. As the article stands right now, there is nothing that overtly deals with the proof that the distributive law holds for the 7D cross product. But we need to know that the distributive law is valid before we can multiply two vectors out distributively in terms of their unit vector components.

So something important is missing from the article. I have argued further up that the prooof that the distributive law holds is in fact the same thing as proving that the Lagrange identity holds. In fact, that is the full significance of desiring that the Lagrange identity is one of the defining properties. If the Lagrange identity doesn't hold, as it wouldn't in the case of a 5D cross product, then the distributive law will not hold, and so we could not multiply a 5D vector out distributively using the 5D cross product. David Tombe (talk) 00:00, 8 July 2010 (UTC)

The distributive property is part of the definition of a cross product- a cross product is bilinear. You then use this as part of the definition of a particular cross product when you write down a multiplication table. That's what the comment refers to I believe (ie. the multplication table is extended to arbitrary vectors by using bilinearity). Probably the wording should be changed, IMO. Holmansf (talk) 13:23, 8 July 2010 (UTC)

Holmansf: Perhaps your suggestion is that “These unit vectors can be multiplied out distributively in both three and seven dimensions” is meant to tell the reader how general vector multiplication is effected using the multiplication rule for unit vectors? Brews ohare (talk) 14:07, 8 July 2010 (UTC)

I changed the sentence to say that. Brews ohare (talk) 14:36, 8 July 2010 (UTC)

Yes that is what I meant. I am changing the paragraph again because I think it is still confusing. In particular the stuff about vectors from different planes giving the same product is in the wrong place. Holmansf (talk) 15:15, 8 July 2010 (UTC)

Brews: In regard to the different planes thing- to me it just seems like a random comment that serves to distract and possibly confuse. Why do you think it's important? Why do you think it should be placed right there? That section is about how to define and calculate cross products in 7D using an orthonormal basis, not about contrasts between the 7D and 3D cross products. Holmansf (talk) 16:33, 8 July 2010 (UTC)

Hi Holmansf: This comment was put there originally by Blackburne. If you can find a better home for it, go ahead. It is important because it marks a very different behavior from the 3D case, and suggests that physical interpretation of the 7D cross product may be a problem, because it is a many-to-one mapping. Brews ohare (talk) 16:59, 8 July 2010 (UTC) I moved it; see what you think. Brews ohare (talk) 17:03, 8 July 2010 (UTC)

Actually it looks like David Tombe added it here

http://en.wikipedia.org/w/index.php?title=Seven-dimensional_cross_product&action=historysubmit&diff=371254844&oldid=371253997

Before this edit it made sense IMO. I'm going to change to better reflect the old version. Holmansf (talk) 17:29, 8 July 2010 (UTC)

Holmansf, Before that edit was made, the lack of uniqueness referred to the fact that for every plane spanned by two vectors, there is more than one other vector perpendicular to that plane. I changed it to refer to the lack of uniqueness of the cross product itself, in that any unit vector can be the product of different pairs from amongst the remaining six. For example look along the top row of the table below that sentence and you can see that the unit vector e1 is the cross product of three distinct pairs from the other six.

Both aspects of lack of uniqueness are correct, but only the latter specifically relates to the cross product, as in the cross product being the vector that is the product of two other vectors. David Tombe (talk) 17:49, 8 July 2010 (UTC)

Holmansf: There are two aspects involved here. Your changes are fine, including revision of the header, but I've reinserted the paragraph further down that points out different vectors have the same cross product, which seems to me to be not quite the same thing as saying there are multiple normals for each plane, and fits in as a discussion of the multiplication table itself. Brews ohare (talk)

BTW, if a plane is described by two vectors that lie in it, the statement that there are multiple vectors orthogonal to it is a bit obvious in 7 dimensions, because there are 5 vectors orthogonal to the two defining the plane. Brews ohare (talk) 18:02, 8 July 2010 (UTC)

That's it Brews, there are two aspects. But only the latter is important. The former can only become important if we assume a geometrical connection. And just as there is no geometrical connection with the 7D curl, neither is there a geometrical connection with the 7D cros product. The aspect which I dealt with is the one that is important in relation to the distributive law, and that is the one that is relevant for the multiplication table. And we cannot assume the distributive law just because of bilinearality. The distributive law follows from the Lagrange identity, and that is why we choose the Lagrange identity as one of the defining properties. David Tombe (talk) 18:05, 8 July 2010 (UTC)

Holmansf: I apologize, but I removed the two leading sentences. One is a digression (non-uniqueness) and is better described later (as now it is) and the other is a trivial unrelated observation (multiple normals for a plane: a 2D space in a 7D world). Brews ohare (talk) 18:14, 8 July 2010 (UTC)

Hmm, well maybe the stuff about non-uniqueness should not go there, but I think it should go somewhere. Let me elaborate. The non-uniqueness that is important is the non-uniqueness of the cross product itself. This means that there are actually different maps from V x V to V that satisfy the requirements to be called a cross product, and these maps are not just the same as each other up to a factor of -1 (ie. up to signs). It doesn't mean that a particular cross product is not injective in some sense. The non-uniqueness of the cross product itself stems from the fact that given a plane spanned by say the first two basis vectors e_1 and e_2 you could choose any unit vector perpendicular to this plane (infinitely many in 7D!) to be the product e_1 x e_2 and then continue to build a cross product based on that initial choice (that's probably not the only choice required ...). Thus there are actually infinitely many possible 7D cross products. This is in contrast to the 3D case where there are only 2 unit vectors perpendicular to the plane spanned by e_1 and e_2, and so only two choices of cross product.

With these comments in mind I am removing the other material we have been discussing.

Also, David, what do you mean by "distributive law?"Holmansf (talk) 18:39, 8 July 2010 (UTC)

Holmansf, On non-uniqueness, I was focusing on the fact that vector e1 can be the cross product of three different pairs from amongst the other six. That is the only aspect of non-uniqueness that is important as regards the multiplication table. On the distributive law, a vector a is given by a1i + a2j + a3k + a4l + a5m + a6n + a7o and a vector b is given by b1i + b2j + b3k + b4l + b5m + b6n + b7o. In order to work out a×b as per the chosen multiplication table, we need to know that the operation a×b is distributive. If the operation satisfies the Lagrange identity,

$|\mathbf {a} |^{2}|\mathbf {b} |^{2}-(\mathbf {a\cdot b} )^{2}=|\mathbf {a\times b} |^{2}$

then it will be distributive. This can be checked by distributing it out in the terms of the Lagrange identity. It only works in 3 and 7 dimensions. But I see that you have assumed the distributive law on the basis of bilinearality. I can't see how you can do that, because if you could, you could then define a 5D multiplication table and multiply it out distributively only to discover that it would clash with the Lagrange identity. David Tombe (talk) 19:20, 8 July 2010 (UTC)

David, the assumption that a cross product is bilinear (part of the definition), implies that it is distributive. Are you saying that you can omit the bilinearity from the definition and still have the same result?

Also, if you really feel strongly about it you can put the stuff about "vector e1 can be the cross product of three different pairs from amongst the other six" back in and I won't remove it again. However, I don't see why it's important. Holmansf (talk) 20:07, 8 July 2010 (UTC)

Holmansf: Your language “However, unlike in three dimensions, there is no unique (up to sign) cross product defined in seven dimensions. This is because in seven dimensions there are more than two unit vectors perpendicular to any given plane.” is inadequate. Obviously, having more than two vectors perpendicular to a plane does not deny uniqueness of the cross product, unless one cannot pick which one. As you can see in concrete terms from the multiplication tables, any unit vector is the cross product of three choices for its constituent pairs. That is a consequence of both requirements: magnitude and perpendicularity. There are therefore three planes with the same cross product, and this cross product is unique. What isn't unique is there is not a single pair of vectors (a single plane) leading to this cross product. Brews ohare (talk) 20:10, 8 July 2010 (UTC)

You have it the wrong way round. Don't try and derive or understand anything from any particular multiplication rule, as they are not unique, so anything you work out from one rule will be different for another.

The uniqueness of the 3D cross product arises as any two non-parallel vectors define a plane, and there's only one vector, its normal (and its negative), perpendicular to that plane. This does not happen in 7D as there are five independent dimensions perpendicular to each plane, and an infinity of directions, so an infinity of possible 7D cross products - though it's usual to only consider those that can be written in terms of simple products of the basis elements. In addition it's easy to write down different products that satisfy the definition, so to demonstrate the product isn't unique.User:JohnBlackburne (talk) 20:26, 8 July 2010 (UTC)

John: Yes, I see that there are an infinity of directions perpendicular to a plane made up of any combination of the five unit vectors perpendicular to that plane. However, I see that every multiplication table has the property that any unit vector can act as the cross product of three and only three other pairs. Any independent vectors x and y in the plane of e_i e_m will have a cross product involving e_i × e_m and therefore parallel to this cross product, which cross product is therefore certainly unique to this plane. What am I missing? Brews ohare (talk) 20:41, 8 July 2010 (UTC)

It would appear that a vector's being perpendicular to a plane is not sufficient to make it a cross-product of vectors in that plane. Brews ohare (talk) 20:47, 8 July 2010 (UTC)

Well perhaps it should be explained differently. However, I think you are missing the point. Given any unit vector v perpendicular to the plane spanned by e_1 and e_2, you could define a cross product such that e_1 x e_2 = v. Therefore there are at least as many distinct seven dimensional cross products as there are unit vectors perpendicular to the plane containing e_1 and e_2, which is to say infinitely many.

I think some of the confusion here is coming from the imprecise use of the term "cross product." I am trying to use "cross product" to consistently refer to the map from V x V to V (this is the definition given in the article ...). You and David use "cross product" to refer to the image of a certain pair of vectors under a particular version of the cross product (the map). Holmansf (talk) 21:04, 8 July 2010 (UTC)

Holmansf: I do not follow your reasoning. Given any multiplication table whatsoever, and selecting the e₁ e₂ of that table, there is one and only one vector e₁ × e₂. Consequently there is one and only one normal to that plane that is in the direction of a cross-product to that plane (apart from sign). The cross product is not simply a map; it has two defining properties. Don't we agree on this? Brews ohare (talk) 21:16, 8 July 2010 (UTC)

Perhaps you are suggesting something different: for any vector V I can find a plane perpendicular to it and construct a basis such that e₁ × e₂ is parallel to V? Then what you are saying amounts to this: one can construct an infinity of coordinate systems. That is not a statement about cross products. Brews ohare (talk) 21:21, 8 July 2010 (UTC)

Let me try to put it into your language. For any V perpendicular to the plane containing e₁ and e₂ there is a multiplication table that defines a cross product such that e₁ × e₂ = V. Since there are infinitely many such V, there are infinitely many possible 7D cross products. Holmansf (talk) 23:26, 8 July 2010 (UTC)

In short, the paragraph:

Notice that any one of the seven unit vectors can result from the cross product of three distinct pairs (representing three distinct planes) from among the other six. For example, from the first row of the listing above one finds e₁ is given by e₂ × e₄, e₃ × e₇, and e₅ × e₆. Thus, unlike the 3-dimensional cross product, the same 7-dimensional cross product can result from multiplying pairs of vectors residing in different planes.

is beyond controversy, a simple observation based upon the multiplication table, and should be reinserted in the discussion of the table. Brews ohare (talk) 20:25, 8 July 2010 (UTC)

No, you can't look at one product rule (or multiplication table) and deduce a general property about the product. What you're seeing is something quite different, that multiple pairs of vectors multiply to give the same vector. This happens in 3D too, they just have to be in the same plane, not three.--JohnBlackburne^words_deeds 20:57, 8 July 2010 (UTC)

John, you are mistaken about what is claimed here. All that is said is that for this multiplication table, three pairs lead to the same cross product, a fact nobody can possibly dispute. Then it is said that in general the possibility exists for the same cross product to result from several planes. Again, this is incontrovertible as a concrete example is presented where exactly this happens. There is no false extrapolation beyond the example involved here. Brews ohare (talk) 21:09, 8 July 2010 (UTC)

The cross product in 7D can be considered as a map from planes to lines, i.e. from bivectors to vectors, as every pair of vectors has a plane/bivector associated with it, and the map from these vectors to bivectors is bilinear.

But while in 3D to get from the bivectors to vectors you take the dual you can't do this in 7D as bivectors and vectors aren't dual. Instead bivectors form a 21-dimensional space. So a map from the bivectors to vectors will naturally associate three directional bivectors with each direction vector. So vectors in three planes multiply to give each vector. But this has nothing to do with uniqueness: even if there were just one product all of the above would be true.--JohnBlackburne^words_deeds 21:18, 8 July 2010 (UTC)

John: Your response is oblique and unnecessarily abstract. From the given table (and any other table, in fact) a unit vector is the cross product of three other pairs of unit vectors. There is nothing complicated about this: it means that the cross-product of vectors in any of three planes all are parallel to the same direction. Period. Hence, given we have here an example where this occurs, the statement can be made that in general the possibility exists for the same cross product to result from several planes. There is simply no argument, however abstract, that can contradict this point. If you wish, you can try to put this observation is a general context, but that will not controvert the example, just illuminate it. Brews ohare (talk) 21:31, 8 July 2010 (UTC)

I am not even sure what you mean. The 7D cross product is the whole product, so it makes no sense to talk of the "the same cross product" referring to different parts of this whole. And again, arguing from observations made of a particular rule will get you nowhere as nothing about the general product can be proved from it. More importantly arguing from your own observations is OR and strictly forbidden. So unless you have a source it has no place here. --JohnBlackburne^words_deeds 21:39, 8 July 2010 (UTC)

John: There is no OR in looking at the first line of your listed multiplication rule and observing that three different cross products lead to e₁. Calling this "parts of the whole" seems to suggest that one can talk only of general x × y and not about particular vectors like the unit vectors, which of course means a multiplication table and the distributive law are not useful ideas. And finally, suggesting that a concrete instance cannot be taken as proof of the possibility of existence of such an instance is simply a logical error. Brews ohare (talk) 21:54, 8 July 2010 (UTC)

Holmansf, The reason why I made the change which you highlighted in the link above, is because the kind of non-uniqueness that was mentioned in the older version was not relevant to the issue which followed on. Furthermore, I don't like getting involved in the subject of 'planes' in relation to the 7D cross product. The 7D cross product is pure algebra with no connection to geometry. But if you do insist on considering 'planes' enclosed by two unit vectors, then of course it is a trivial fact that the remaining 5 unit vectors will all be orthogonal to that plane, even the ones which aren't actually the cross product of the two vectors which define that plane. But this piece of trivial information has got nothing to do with the multiplication table, and the section was about the multiplication table and its operation under the distributive law as between two vectors x and y. That's why I changed it to the more relevant point about uniqueness, which is that that each unit vector is uniquely given by three different pairs from the remaining six, within the context of a particular multiplication table. I am not overly worried whether that material is restored or not because it is manifestly obvious from looking at the multiplication table. But the other aspect of non-uniqueness, if it needs to be mentioned at all, should be mentioned elsewhere in the article.

the 7D cross product is very geometric: it's definition is in terms of orthogonality and area, and most of its properties have geometric interpretations. 7D space may not have much practical use but Euclidian geometry works as well in 7D as in 3 or 4D.--JohnBlackburne^words_deeds 21:39, 8 July 2010 (UTC)

As regards the distributive law, it's best to look at my reply to Brews in the section below. David Tombe (talk) 21:22, 8 July 2010 (UTC)

John, It's better not to intersperse your replies. Anyway, how come that over on the 'curl' page, you were quite comfortable with the idea that a 7D curl would have no geometrical significance? David Tombe (talk) 21:45, 8 July 2010 (UTC)

I was just pointing out a flaw in your reasoning, which if you take it on board might help you understand the broader picture. And curl doesn't exist in 7D, as you know, so there's nothing to have geometric significance and so to relate to here.--JohnBlackburne^words_deeds 21:58, 8 July 2010 (UTC)

John, I can define a 7D curl based on the multiplication tables for the 7D cross product. I can set up a 7D differential operator and multiply it out distributively on a 7D vector and call it a 7D curl. But it will have no more connection to geometry than the 7D cross product of this article. What's the difference? David Tombe (talk) 22:07, 8 July 2010 (UTC)