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John, The equation,

• |x × y|² = |x|² |y|² − (x · y)²

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

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

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

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

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

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

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

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

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

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

John, You only have to look at page 4 and the top of page 5 in this weblink [1]. It is quite clear that the lagrange identity in the form above, only applies in three dimensions. David Tombe (talk) 14:02, 5 January 2010 (UTC)[reply]

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

John, The last three lines on page 4, reading on into page 5 [2] makes it quite clear that the Lagrange Identity,

• |x × y|² = |x|² |y|² − (x · y)²

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

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

DVdm, Page 4 is about the Schwarz inequality which holds in all dimensions. The Lagrange Identity is a special 3D case of the Scwharz inequality.David Tombe (talk) 10:04, 15 April 2010 (UTC)[reply]

David, that chapter is about the product in three dimensions. It says nothing about the seven dimensional case so is no use as a source for this article.--JohnBlackburne^words_deeds 10:13, 15 April 2010 (UTC)[reply]

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

And where in this article does it mention the Lagrange identity?--JohnBlackburne^words_deeds 10:39, 15 April 2010 (UTC)[reply]

John, It's clearly mentioned in the section 'Characteristic properties'. The Lagrange identity is this,

• |x × y|² = |x|² |y|² − (x · y)²

David Tombe (talk) 10:48, 15 April 2010 (UTC)[reply]

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

DVdm, Yes, but we specifically need the 3D version to establish that a×b = absinθ. I never doubted that there was an 'n'dimensional version. But the 'n'dimensional version can't be used to prove that a×b = a b sinθ. My original point in all of this was that,

• |x × y|² = |x|² |y|² − (x · y)²

should be listed along with those properties of the 3D cross product that do not hold in the 7D case. At the moment, the main article lists it as a characteristic property that does hold in the 7D case. I believe that this is an error in the main article. David Tombe (talk) 11:04, 15 April 2010 (UTC)[reply]

You believe wrongly. The "angle" θ can be defined as the thing that satisfies |a×b| = |a| |b| sinθ. Any first year math student knows this. This has been explained to you more than once.

By the way, you will find another ref to the identity in n-dim in this one. DVdm (talk) 11:17, 15 April 2010 (UTC)[reply]

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

David, concerning your statement "I'm not denying the existence of a 7D Lagrange identity", above you have said:

• "The equation ... is only a special 3D case of the Lagrange identity. The more general form of the Lagrange identity is more complicated." => it is not.
• "It is quite clear that the lagrange identity in the form above, only applies in three dimensions." => It does not.
• "...makes it quite clear that the Lagrange Identity ... only holds in the special case of 3 dimensions." => It does not.
• "The Lagrange Identity is a special 3D case of the Scwharz inequality." => It is not.
• "... we specifically need the 3D version to establish that a×b = absin?." => We don't.
• "I never doubted that there was an 'n'dimensional version." => You did. Four times in this section alone."

If we are supposed to continue assuming good faith from your part, I hope you understand that assuming no clue is our only remaining option here. Please stop it. Thank you. DVdm (talk) 14:21, 15 April 2010 (UTC)[reply]

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

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

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

The sources are in the article, all of them online if you want to review them yourself. As always I recommend Lounesto's book, mostly as it explains both the three and seven dimensional cases very well in a few short pages, though it does not have a proof. (he references this proof, which is not in the article but behind a paywall so I can't look at it).

As for David's argument his main mistake is identifying this

|x × y|² = |x|² |y|² − (x · y)²

as the Lagrange identity. It is, but only in three dimensions. In seven it does not simplify like this. Lounesto calls the above the "Pythagoras theorem", which makes some sense if you think of the various things geometrically but I don't think would add anything to the article. Here the above is simply a condition. We say we want a product that satisfies it as well as orthogonality and antisymmetry, then deduce (or point to the proofs that show it) that non-trivial products only exist in 3 and 7 dimensions.--JohnBlackburne^words_deeds 21:38, 16 April 2010 (UTC)[reply]

John, You have just pointed out what I have been trying to point out all along, which is that

|x × y|² = |x|² |y|² − (x · y)²

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

The equation,

*|x × y|² = |x|² |y|² − (x · y)²

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

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

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

• |x × y|² = |x|² |y|² − (x · y)²

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

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

• |x × y| = |x| |y| ;    provided that   (x · y) = 0

using

• |x × y| = |( x- (x · y) / | y|² y ) × y|

He proceeds to show that

• x × ( y × z) =y(x · z) - z(x · y)

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

• n (n - 1) (n - 3) (n - 7) = 0,

in order that a necessary ternary identity be satisfied.

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

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

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

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

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

• |x × y|² = |x|² |y|² − (x · y)²

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

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

• |x × y|² = |x|² |y|² − (x · y)²

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

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

• |x × y|² = |x|² |y|² − (x · y)²

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

David you are misunderstanding the use of the formula:

|x × y|² = |x|² |y|² − (x · y)²

In 3D, after the cross and dot product have been defined, it is an identity relating them. It is also Lagrange's identity, but only in 3D - see Lagrange's identity#Lagrange's identity and vector calculus. This identity gives the magnitude of the cross product as |a||b| sin θ. The other defining properties of the cross product are orthogonality (a × b is orthogonal to both a and b) and antisymmetry, as well as it being a bilinear with vector result.

In 7D we do not prove the above formula: we instead state it as a property of the product we want to find, the "cross product" in 7 dimensions. We then find a product that satisfies this and the other properties. Finally we show (though this is not in the article but in the references) that this is the only non-trivial cross product of two vectors other than the familiar 3D version.--JohnBlackburne^words_deeds 10:12, 17 April 2010 (UTC)[reply]

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

|x × y|² = |x|² |y|² − (x · y)²

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

We don't expect it to be true, we assert it. We say "this works in 3D, is there any other dimension it works in?" and as

|x × y|² = |x|² |y|² − (x · y)²

is a property of the product in 3D was say that it must be a property of the product in other dimensions. This article gives one such product, in seven dimensions. The proofs in the references prove this is the only non-trivial dimension other than three.

As for Lagrange's identity this

Is a general form of it, i.e. the form in all dimensions expressed in terms of simple products. In 3D is is the same as

|x × y|² = |x|² |y|² − (x · y)²

Because the exterior/wedge product can be associated with the cross product as it's dual. This is not true in seven dimensions as vectors and bivectors are not dual. There is a relationship between them, given in e.g. Lounesto, but it's rather more complex.--JohnBlackburne^words_deeds 11:58, 17 April 2010 (UTC)[reply]

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

• x × ( y × z) =y(x · z) - z(x · y)

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

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

• x × y =1⁄2 (x y − y x)

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

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

John and Brews, OK let's then assert that,

• |x × y|² = |x|² |y|² − (x · y)²

holds in any dimensions, and then set out to find which operators satisfy this assertion. But in doing so, let's bear in mind that this assertion, when in 3D, is the 3D case of more complicated 'n'D relationships such as the Lagrange identity and the Schwarz inequality. On that basis, I would expect that any operators satisfying this assertion would have to be in 3D only. But that is only my own assumption. Silagadze apparently finds this assertion to be compatible with a 7D cross product. I suspect that Silagadze has made an oversight, in that his more primitive assumptions are already restricted to 3D.

Now I don't doubt that proofs exist which restrict cross products to 0,1,3 and 7 dimensions. I fully accept that a 7D cross product 'of sorts' exists. But do any of the proofs listed in the references actually utilize the relationship,

• |x × y|² = |x|² |y|² − (x · y)² ?

I have scanned through a few of these proofs and I can't see where that relationship comes into it, but do please correct me if I am wrong on that point because I only did a scan read.

Ultimately I think that the problem here lies with trying to define 'angle' in 7D. It's no good defining angle through dot product and cosine. An angle only has a cosine if it exists in the first place.

I'll need to take a closer look now at exactly what Silagadze has done, and I need to see exact quotes from the proofs in the references which directly link this relationship to the 7D cross product. I just can't see how a relationship that is so heavily connected with 3D can in any way be extended to 7D. David Tombe (talk) 14:17, 17 April 2010 (UTC)[reply]

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

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

Brews, Ultimately if we have a source which directly links |x × y|² = |x|² |y|² − (x · y)² to the number 7, then we'll have to let it go at that. It sounds like your Silagadze might be that source. Is it possible that you can give a brief outline as to how Silagadze makes the linkage to the number 7 without using any 3D assumptions? David Tombe (talk) 15:00, 17 April 2010 (UTC)[reply]

Silagadze uses a different starting point: |a × b| = |a| |b| if a.b = 0. Geometrically this is "the cross product's magnitude is the same as the area of the rectangle with sides a and b, if they are perpendicular vectors". This looks like a special case, so a weaker condition than the Pythagorean one, as geometrically that one says the magnitude of the cross product is the same as the parallelogram with sides a and b, and a rectangle is just a parallelogram with perpendicular sides. But he then derives the more general rule from the specific, so showing they are equivalent. All this is without reference to the dimension, i.e .the properties are true in any dimension, if the cross product exists. He then shows that such a product only exists in 3 and 7 dimensions.--JohnBlackburne^words_deeds 16:21, 17 April 2010 (UTC)[reply]

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

The following identity also relates the cross product and the dot product:

:

This is a special case of the multiplicativity of the norm in the quaternion algebra, and a restriction to of Lagrange's identity.

As regards Silagadze, all his fundamental postulates are 3D based. Even though he then shows that these axioms satisfy n(n-1)(n-3)(n-7) = 0, it has already been established from the outset that n=3. The equation n(n-1)(n-3)(n-7) = 0 came from other references and there was no establishment of the fact that n could equal 7, because it had already been established as being particularly equal to 3. All that Silagadze established was that if n doesn't equal 3 then it could equal 7. But we already know from the fundamental axioms that n does equal 3. David Tombe (talk) 17:07, 17 April 2010 (UTC)[reply]

No, they are not "3D based". Only later in his proof does he show that the dimension must be 3 or 7 (the other vales 0 and 1 not representing products). The earlier properties are stated and manipulated without assuming the dimension.--JohnBlackburne^words_deeds 17:18, 17 April 2010 (UTC)[reply]

John, I am only working from a secondary transcription of the Silagadze source which states clearly that the fundamental axioms are all 3D. But I haven't actually seen the primary source. But let's say for the sake of argument that the primary source is silent on the issue. Silagadze does openly state that the vector triple product only holds in 3D. That deals with the volume of a parallelepiped. What would make the Pythagorean relationship any different in that respect? David Tombe (talk) 17:27, 17 April 2010 (UTC)[reply]

The Silagadze paper is online, and linked in the article. No-where does is say the fundamental axioms are all 3D. I suggest you read it as it's perfectly clear.--JohnBlackburne^words_deeds 17:35, 17 April 2010 (UTC)[reply]

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

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

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

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

So here are two questions:

Are we to believe that the "cross product" can be arrived at in several ways that boil down to the normal concept of cross product in 3D, but which do not obey all the 3D behaviors in 7D?

The cross product is defined like this, and generally accepted. Some products of the 3D product are not true in 7D, but they are not needed for the definition. You could define other products which do not satisfy all the conditions but they would not be cross products and so not as interesting.

Is this general 7D cross product unique, regardless of how we go about it, or may there be several cross products, obtained by retaining a different subset of 3D behaviors as defining properties of the cross product? Brews ohare (talk) 19:10, 17 April 2010 (UTC)[reply]

It's not unique and the article says so, at the start of this section, before giving one of the ways to do it. Another way, as seen in Lounesto, is define a trivector, again not unique, which when multiplied by a ∧ b gives the cross product. In three dimensions the trivector, i.e. the pseudoscalar, is unique up to a scale factor.--JohnBlackburne^words_deeds 19:44, 17 April 2010 (UTC)[reply]

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

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

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

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

Hi John: As the definition is related in a fundamental manner to the parallelogram in the article, something is needed to explain what that means in n (or n-1) dimensions. Brews ohare (talk) 22:06, 17 April 2010 (UTC)[reply]

I see you added a reference on parallelograms but the reference is not about parallelograms: the author is, as he states on the first page, using the word "parallelogram" instead of "parallelepiped", and that is what his chapter is about, the volume of parallelepipeds. I have therefore removed it.--JohnBlackburne^words_deeds 17:05, 18 April 2010 (UTC)[reply]

Again, the reference is misleading. It links to a chapter titled "Volumes of parallelograms", but the author means something totally different by this, as he explains on the first page: the discussion is all about parallelepipeds. On the same page it says

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

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

Often the cross product in 3D is related to rotation; shouldn't rotation be a subsection here too? Brews ohare (talk) 21:44, 17 April 2010 (UTC)[reply]

In 3D the cross product is related to relation as it is the dual of the bivector, and the bivector is related to rotation: usually the cross product can be replaced with an exterior product to simplify the maths (or at least simplify it if you understand bivectors).

In 7D bivectors still are related to rotations, but they form a 21-dimensional linear space and are not dual to, or simply related to, vectors in 7D.--JohnBlackburne^words_deeds 21:49, 17 April 2010 (UTC)[reply]

Please add some discussion to the article. Brews ohare (talk) 22:07, 17 April 2010 (UTC)[reply]

I don't see how it would help: the cross product in 7D is not related to rotations, as it's not related to many things. I would say there's no point mentioning it unless its notable in some way, such as it was once thought to be the case but now is not. The 7D cross product is far less useful and interesting than the quantity of our discussions here imply, as it's really a fairly obscure mathematical oddity, of interest mostly as it relates to other little used bits of mathematics like higher dimensional geometric algebra and the octonions.--JohnBlackburne^words_deeds 12:57, 18 April 2010 (UTC)[reply]

John, I would agree with you that the 7D cross product is a mathematical oddity which is not very useful in practice. But it is neverthless of interest to readers for the very purpose of enabling them to ascertain how its usefulness compares to that of the 3D cross product. I studied this subject in depth and I came to this page in January to fix it up. You seem to be intent on making it out to be more useful than it really is, by wanting to insist contrary to simple natural reasoning, that it is commensurate with the Pythagorean identity. I suggest that you get out a pen and a piece of paper and apply the Pythagorean relation to 7D and see how you get on. David Tombe (talk) 14:41, 18 April 2010 (UTC)[reply]

It adds no information to introduce sinθ in this section, and it has raised questions on this Talk page. Just what the "inner angle" is between x & y in n-dimensions has not been presented in the article, and seems to me to be a confusing side issue, less clear than the preceding statement in the article:

|x × y|² = |x|² |y|² − (x · y)²

which has a ready interpretation in n-dimensions. Brews ohare (talk) 00:44, 18 April 2010 (UTC)[reply]

Brews, I maintain that if you were to multiply the right hand side of this equation out in 7D that you would get 21 xy-xy terms. This would reduce to a triplication of 7×z^2 terms within the context of a 7D cross product. Your equation would then become unbalanced by a factor of 3. However, if you do the same thing in 3D, it will check out perfectly. David Tombe (talk) 07:57, 18 April 2010 (UTC)[reply]

What an "inner angle" is between x & y in n-dimensions is now presented in the article. DVdm (talk) 08:50, 18 April 2010 (UTC)[reply]

DVdm: The role of sinθ in a 2D parallelogram is not a question, although it is what you have addressed. What needs explanation is (i) the definition of a parallelogram in 6D, and (ii) the meaning of sinθ in 6D. Personally, answering these questions is irrelevant and all reference to both should be removed from the article as nothing is added. The explanation of parallelogram in 6D and sinθ in 6D is a non-intuitive waste of time. Brews ohare (talk) 14:00, 18 April 2010 (UTC)[reply]

Please stop disrupting this tak page. DVdm (talk) 14:21, 18 April 2010 (UTC)[reply]

Why is my suggestion about content of this page characterized as a disruption instead of being responded to with your ideas why this irrelevant material has some value to this article? Surely the notion of the area of a n-dimensional parallelogram and its connection to an angle is non-intuitive, therefore requires explanation, and as it is not needed, all of that extra description would be simply digression without value. Brews ohare (talk) 16:54, 18 April 2010 (UTC)[reply]

We don't need n-dimensional parallelograms. There is a simple 2-dim paralellogram in the 2-dim subspace spanned by the vectors. I have removed the irrelevant remark again. Please stop re-inserting it. DVdm (talk) 18:42, 18 April 2010 (UTC)[reply]

Here is a source[4] (page4) which explains how to expand the right hand side of,

|x × y|² = |x|² |y|² − (x · y)²

in any dimensions. Put in seven dimensions and we get 21 terms. It is quite clear that the Pythagorean identity only holds in 3D. DVdm avoids commenting on the issue and John Blackburne, rather than commenting, immediately goes to the introduction of the main article and writes that the Pythagorean identity holds with the 7D cross product, when it is quite obvious to everybody that this is not true. The article aleady contains this piece of information and so one would have thought that the last thing to do on discovering that the information may not be correct would be to repeat the information in the introduction in the knowledge that it was being objected to by one editor. So much for collaborative editing. David Tombe (talk) 12:28, 18 April 2010 (UTC)[reply]

That source is about the cross product in 3D: it deduces the Schwarz inequality in general, then shows that in 3D the result gives the expression above. Nowhere does it work out a 7D result, so it is irrelevant to this article which is on the seven-dimensional cross product.--JohnBlackburne^words_deeds 12:51, 18 April 2010 (UTC)[reply]

Commenting on the issue:

• In junior high school we learned that (a-b)^2 = a^2 - 2ab + b^2, which is an equation with 1 term on the lhs and 3 terms on the rhs -- or, in David Tombe's lingo, an equation "unbalanced by a factor of 3".
• In 3 dimensions the lhs has 3 quadratic z-terms, which when expanded produce 9 terms -- The equation is "unbalanced by a factor of 3".
• In 7 dimensions the lhs has 7 quadratic z-terms which when expanded produce 21 terms -- The equation is "unbalanced by a factor of 3".

DVdm (talk) 13:01, 18 April 2010 (UTC)[reply]

DVdm, The equation only applies in 3D. If we expand the right hand side in 7D as per the formula at the end of page 4 in the source which I provided, there will be 21 terms and the equation will no longer be an equation. Try it out for yourself. Expand the right hand side in 7D. You will end up with 7 × (z^2) terms in triplicate. This of course reflects the fact that in the 7D cross product, we have unit vectors i,j,k,l,m,n, and o, and that each of these is the product of three different pairs of the rest. For example, i might be equal to j×k, l×m, and n×o. In total there are 21 operations.

In 0D, there are 0 operations. In 1D, there is 1 operation which is i×i = 0. In 3D there are 3 operations and the subject is well understood and widely used. In 4D there are 6 operations. In 5D there are 10 operations. In 6D there are 15 operations, and in 7D there are 21 operations. Hence only in 3D does the right hand side of the equation above balance with the left hand side. And for your interest, in the even dimensions, we can't even relate them to any cross product type relationship, ie. we can't get a z^2 from the x and y. For the odd dimensions we can always get a cross product type relationship, but it is only in the case of 0, 1, 3, and 7, that this cross product will obey the distributive law. Hence in 5D, the operation is totally useless. In 7D none of the geometrical relationships hold. That means neither the Jacobi identity, the vector triple product, nor the Pythagorean identity hold in 7D. David Tombe (talk) 14:27, 18 April 2010 (UTC)[reply]

David Tombe, please stop disrupting this tak page. I am sorry, but your comments really show that you have no clue. Please try to accept that. DVdm (talk) 15:36, 18 April 2010 (UTC)[reply]

David: As I understand Silagadze, he postulates the Pythagorean identity, and then deduces that besides n=3, n=7 will work. That is, the restriction to 7D is one of the consequences of requesting the property given by the Pythagorean identity. Are you suggesting that it is absurd even to postulate the hypothetical possibility of the Pythagorean identity in 7D? In other words, it is internally contradictory on the face of it? That seems not to be true in 3D, so why can't we entertain the hope (provisionally) that it might be true in more dimensions? We then, according to Silagadze, will discover that indeed this postulate is not viable except for the cases n=3 or 7. Brews ohare (talk) 17:07, 18 April 2010 (UTC)[reply]

As regards the [5] (page4), this author proves that the left-hand side of the Pythagorean identity must be positive or zero. We know already that for n=3 it is zero. It seems likely, though I haven't looked at it carefully, that for n=7 Silagaze's approach would show that n=7 is the only other case for which the left-hand side is zero. What do you think about that? Brews ohare (talk) 17:21, 18 April 2010 (UTC)[reply]

Brews, Take a look at the last line of the derivation on page four and look at the right hand side. It is a summation over (XiYj-XjYi)^2. He then shows that in the special case of 3D we get (X1Y2-X2Y1)^2 + (X1Y3-X3Y1)^2 + (X2Y3-X3Y2). Under the terms of the 3D cross product this becomes Z3^2 + Z2^2 + Z1^2. Hence we get the Pythagorean relationship. However, if we substitute in seven dimensions, instead of having the three (XiYj-XjYi)^2 terms on the right hand side, we will have 21 terms. Under the rules of the 7D cross product, these terms will converts to Z^2 terms, but we will have 3[Z1^2 + Z2^2 + Z3^2 + Z4^2 + Z5^2 +Z6^2 +Z7^2]. The factor of three means that we have lost the Pythagorean relationship. If we were to do it for 5D, we would have a factor of 2. But in addition to that, the 5D cross product would not satisfy the distributive law. For 4D and 6D it would end up a dog's dinner. So the fact is that the Pythagorean identity is unequivocally a 3D affair. 7D is only special by virtue of the fact that we can actually have a cross product of sorts that obeys the distributive law. So you are correct in your believe that I think it is an absurdity to assert a 3D identity to exist in any dimensions and then set out to find which dimensions are compatible. There are however proofs listed in the sources of the main article which purport to prove that cross products only exist in 0,1,3,and 7 dimensions. These proofs are all very modern even though the result itself has been known since the 1840's. One should not confuse a proof of this issue with the other issue of whether or not a particular cross product is commensurate with the Pythagorean identity. David Tombe (talk) 17:49, 18 April 2010 (UTC)[reply]

Brews, I have removed an irrelevant remark which you added. Rationale in the edit summary. DVdm (talk) 15:36, 18 April 2010 (UTC)[reply]

DVdm, I'll take your phony allegation of disruption as meaning that you have done the substitution in 7D and realized that you have been wrong all along. And we'll leave it for future readers to decide who the disruptive elements have been in this debate, which in other circumstances might have proved to have been a very interesting and instructive debate. The correct answer is very easy to establish for anybody who wishes to perform the substitution in 7D into the contentious equation. David Tombe (talk) 17:14, 18 April 2010 (UTC)[reply]

I have no other option that repeating my request to stop disrupting this talk page. Also please stop accusing me of making a phony allegation. The history of this talk page speaks for itself -- you have no idea what you are talking about or dealing with here. This is the second time you propose to "leave it for future readers", so please do leave it for future readers now. Thank you. DVdm (talk) 18:06, 18 April 2010 (UTC)[reply]

Response to D Tombe: Hi David; I gather that your analysis is that simple algebra suggests that in 7D, in the Pythagorean theorem a different numerical factor must be introduced. Silagadze's derivation, his Eq 4, depends only upon the cross product being orthogonal to its constituents, and that for normal constituents |A × B| = |A| |B|. I'll check into it, but it looks pretty basic. Where do you think a fallacy creeps in? Brews ohare (talk) 18:34, 18 April 2010 (UTC)[reply]

I agree that there are 21 squared terms in the cross product, but I'm unsure how to relate them to the z's. For example, do we take:

z₁² = (x₂y₃-x₃y₂)² + (x₃y₄-x₄y₃)² +(x₄y₅-x₅y₄)² +(x₅y₆-x₆y₅)² +(x₆y₇-x₇y₆)²

Then each z has five terms. That won't work: there are too many terms. Besides, the z's aren't orthogonal. We need each z to have 3 terms. Then the 21 components of the cross product will have 7 z-terms and the Pythagorean theorem will hold with no extra numerical factor. Brews ohare (talk) 19:20, 18 April 2010 (UTC)[reply]

Or:     z₁² = (x₂y₃-x₃y₂)²;  z₂² = (x₃y₄-x₄y₃)²;

That won't work either: each z has 1 term, and there are then 21 z-components instead of 7. Brews ohare (talk) 19:37, 18 April 2010 (UTC)[reply]

I suspect that when the issue of how to express orthogonal z′s is straightened out, the proof that n=7 will be at hand. Brews ohare (talk) 19:41, 18 April 2010 (UTC)[reply]

Brews, Your second suggestion is correct, and yes, there will be 21 z-components. We will have 3[z₁² + z₂² +z₃² +z₄² +z₅² +z₆² +z₇² ]. Each z-component will be in triplicate due to the fact that each unit vector in the 7D cross product can be the product of three different pairs of the other 6. For example, unit vector i could be equal to the products j×k, l×m, and n×o. So the 21 components actually only contain 7 different z directional components. This of course puts a factor of 3 unto the right hand side of the equation which scuppers the Pythagorean identity. And yes, we can indeed have a cross product in 7D but this particular equation has got nothing to do with the proof of that fact. This equation only proves that the 7D cross product does not fit with the Pythagorean identity. David Tombe (talk) 04:41, 19 April 2010 (UTC)[reply]

David: if you want to prove the Pythagorean identity wrong it's easy. Find a counter-example. I.e. pick two vectors, work out their 7D cross product using one of the methods in the article, and if it's magnitude is not equal to ab sin θ then the cross product does not satisfy the Pythagorean identity. Because the product is bilinear you can choose two unit vectors - then the magnitude needs only be compared with sin θ, making it easier still. If you are sure it is wrong it should be easy to find a counter example and post it here. If you cannot, because every cross product you try has magnitude ab sin θ, then logically the cross product satisfies the identity.--JohnBlackburne^words_deeds 11:36, 19 April 2010 (UTC)[reply]

Hi David and John: Following up on this suggestion I made a spreadsheet to calculate the sum of squared differences in David's source, p. 4 for two 7-D vectors x & y. Letting x=(1, 0, 0, 0, 0, 0, 0) and y=(0, 1, 0, 0, 0, 0, 0) or any other pair with entries of 1 or 0 such that x·y = 0 the spreadsheet produces |x × y|² = |x|²|y|². As all choices of x &y of this type span the 7D space, I'd say that establishes for me that for orthogonal vectors the Pythagorean theorem works. Assuming I've set up the sum of squared differences correctly, would you agree that this experiment proves the point? Brews ohare (talk) 14:53, 19 April 2010 (UTC)[reply]

BTW, for nonorthogonal vectors that I have tried, |x × y| < |x||y|, so a θ can be found such that |x × y| = |x||y| sinθ, not that θ means anything. That's only a spot check, so it's not definitive. However, Silagadze's derivation indicates that proof for orthogonal vectors is sufficient to establish the general case. Brews ohare (talk) 15:20, 19 April 2010 (UTC)[reply]

Yes, Silagadze proves the general case from the special case of x ⋅ y = 0, so if you are satisfied |x × y| = |x||y| is true for all orthogonal x and y then the Pythagorean identity holds for all x and y. Of course it's not me that needs convincing.--JohnBlackburne^words_deeds 15:27, 19 April 2010 (UTC)[reply]

John, That's exactly what I just did above. I substituted seven dimensions into the right hand side. Watch carefully. Brews, you were on the right tracks when you wrote,

z₁² = (x₂y₃-x₃y₂)²;  z₂² = (x₃y₄-x₄y₃)²

All we have to do is to write out all 21 terms, but you weren't sure how to do this. It's based on this set of correspondences,

If x and y are 2 and 4, 3 and 7, or 6 and 5, then z is 1

If x and y are 1 and 4, 3 and 5, or 6 and 7, then z is 2

If x and y are 1 and 7, 2 and 5, or 4 and 6, then z is 3

If x and y are 1 and 2, 3 and 6, or 5 and 7, then z is 4

If x and y are 1 and 6, 2 and 3, or 4 and 7, then z is 5

If x and y are 1 and 5, 3 and 4, or 2 and 7, then z is 6

If x and y are 1 and 3, 2 and 6, or 4 and 5, then z is 7

This will enable you to convert your 21 terms into 7 × z-terms, and the right hand side will become,

3[z₁² + z₂² +z₃² +z₄² +z₅² +z₆² +z₇² ]

The factor of 3 will scupper the Pythagorean relationship. In fact for 4D we will have a factor of 1.5, for 5D we will have a factor of 2, for 6D we will have a factor of 2.5, and for 7D we will have a factor of 3. However, only the odd dimensions allow us to set up a set of relationships as I have just illustrated above. And even in the case of 5D, the final result will not obey the distributive law. It seems that it is only 3 and 7 which can both produce a table of correspondences as above, and also satisfy the distributive law. But we must also conclude that it is only in the case of 3D that the Pythagorean relationship holds, because it's the only case where the coefficient on the right hand side of the equation is unity. David Tombe (talk) 16:17, 19 April 2010 (UTC)[reply]

You misunderstood me David. You don't need to do any algebra, just supply the two vectors for which the Pythagorean identity does not hold: if it is always wrong by a factor of 3 then you should be able to use any non-trivial data to disprove it.--JohnBlackburne^words_deeds 16:54, 19 April 2010 (UTC)[reply]

Hi David: I was a bit hurried. What I've shown is that ΣΣ_i<j(x_iy_j -x_jy_i)² -x· y ≥ 0, with equality for normal vectors. That applies regardless of what x, y are chosen. However, I haven't shown that this expression is the norm of x × y. Silagadze suggests that if this expression is not the norm of some x ⊗ y then ⊗ ≠ ×. So I'm only half way. I'll look harder. Brews ohare (talk) 16:56, 19 April 2010 (UTC)[reply]

I've made a spreadsheet that does the entire calculation:

[6]

Type the two vectors in the coloured areas at the left, and it works out the cross product, inner product, magnitude, then compares two sides of the Pythagorean identity. Whatever vectors I type in the two sides are identical (the diff is zero) so the identity holds. All the maths is there in the spreadsheet, it's not very pretty but it's a long time since I used a spreadsheet and the first time I've used Google docs. --JohnBlackburne^words_deeds 23:28, 19 April 2010 (UTC)[reply]

John, I followed your argument perfectly except for one bit. You didn't show how you actually esblished the 7D cross product in order to determine its magnitude. Let's now recap on the entire subject from the beginning. In 1843, Hamilton establishes the basis of the 3D algebra‡,

i = j×k

j = k×i

k = i×j

Lets already assume the 3D scalar dot product and then expand the expression,

|X|² |Y|² − (X · Y)²

for vectors X = x1i + x2j +x3k, and Y = y1i + y2j +y3k

Using the formula on page 4 of the source which I supplied [7], this comes out to be equal to,

z₁² + z₂² +z₃²

Since this is equal to the magnitude of vector X×Y, we have then proved that,

|X×Y| = XYsinθ

Shortly after Hamilton's discovery in 1843, his friend John T. Graves decides to try out the 7D algebra‡,

i = j×l, k×o, and n×m

j = i×l, k×m, and n×o

k = i×o, j×m, and l×n

l = i×j, k×n, and m×o

m = i×n, j×k, and l×o

n = i×m, k×l, and j×o

o = i×k, j×n, and l×m

However, since each vector can be obtained in three different ways, the expression,

|X|² |Y|² − (X · Y)²

expands to,

3[z₁² + z₂² +z₃² +z₄² +z₅² +z₆² +z₇²]

The factor of 3 scuppers the Pythagorean relationship. At the top of page 5 in my source, it is made quite clear that the equation,

|X × Y|² = |X|² |Y|² − (X · Y)²

only holds in 3D, and also in the wikipedia article cross product, and also in the wikipedia article Lagrange identity. On your spreadsheet, you seemed to pull the numerical magnitude of |X×Y| out of a hat.

The conclusion is that the 7D cross product does not link to geometry, and that neither the Jacobi identity, the vector triple product, nor the Pythagorean identity apply, whereas they all do in the case of the 3D cross product. And as such, it is an absurdity to commence the 7D cross product from the standpoint of the Pythagorean identity.

‡ Hamilton and Graves were actually working in 4D and 8D. But the relationships of interest within quaternions and octonians are 3D and 7D, and in later years Gibbs and Grassman concentrated on these aspects. David Tombe (talk) 03:44, 20 April 2010 (UTC)[reply]

John, I've checked it out and your numbers are correct. But we still have to resolve the dilemma about the factor of 3. It would appear that rather than having z1 in triplicate, that we have three separate values that add together, as like,

z₁² = (x₂y₄-x₄y₂)² + (x₃y₇-x₇y₃)² + (x₆y₅-x₅y₃)²

That would get rid of the factor of three and it means that you are right, and that the equation,

|x × y|² = |x|² |y|² − (x · y)²

holds in both 3 and 7 dimensions. I'm still not convinced however about the issue of angle applying in seven dimensions, but that is another matter. David Tombe (talk) 05:58, 20 April 2010 (UTC)[reply]

At the cross product article, John Blackburne is adamant that,

only holds in 3D. So was I until earlier today. I based my adamancy partly on that article and partly on other sources. And I argued the point on this page for the last four days. Eventually after working it out from first principles, I concluded that I was wrong, and that in fact, the above equation can also hold in 7D. As you can read above, I conceded the point to John Blackburne. And as such, I went to the cross product article to make the appropriate correction.

Unfortunately however, it now seems that John Blackburne holds a different view on different pages. On this page, he believes that the above equation holds in both 3D and 7D. But over on the cross product page, he believes that it only holds in 3D.

I will now stick by my latest findings that it can hold in 7D, and the argument from first principles is shown above. This is a classic case where conflicting sources in maths need to be resolved by establishing the correct result from first principle arguments. And while my priority is to make the articles correct for the benefit of the readers, I will now withdraw from both the cross product articles until such times as other editors establish exactly what John Blackburne's position is. David Tombe (talk) 12:41, 20 April 2010 (UTC)[reply]

An on-line Google spreadsheet using Google docs can be found here, following the lead by Blackburne. It finds z = x × y for arbitrary vectors, checks that z is normal to both x & y, and compares the norm |z| with the Pythagorean formula. The formula works for all x and y I have tried. Brews ohare (talk) 19:10, 20 April 2010 (UTC)[reply]

Yes, seems to work. I did not think to add an orthogonality check to mine so yours is more complete. A shame we can't add them to the article as that would be original research, but the maths is all there and it's more instructive for the reader to understand it and do it themselves.--JohnBlackburne^words_deeds 19:25, 20 April 2010 (UTC)[reply]

Yes, we know that the numbers work, and we are now all agreed on that issue. But let's not lose track of what the original argument was about. The argument from the outset was over the issue of the conversion of Lagrange's identity in the general form,

into the particular form which is being referred to as the Pythagorean identity,

I had been wrongly saying that the particular form only works in 3D. I was mislead by two other wikipedia articles (cross product and Lagrange's identity), and by the bottom of page 4 of this source [8]

After working it out from first principles above, I finally agreed with John Blackburne that this conversion also applies for n=7. So the dispute on this article is now over. We are all agreed that the Pythagorean identity holds in both 3 and 7 dimensions. It is wrong however to say that it holds in all dimensions because we can only have a cross product in 3 and 7 dimensions.

The new problem is now the fact that when I went to correct the other two wikipedia articles at cross product and Lagrange's identity, John Blackburne reversed his position and undid the corrections, stating in the edit summary that it only works for n=3. So on this page, John Blackburne is agreeing that it works for both 3 and 7, whereas on those other two pages, he is saying that it only works in 3. See here [9], [10]. David Tombe (talk) 06:43, 21 April 2010 (UTC)[reply]

I've removed this as it's not relevant to the 3D or 7D cross product: it's an entirely different product generally most useful when used with bivectors, not vectors. It uses a similar symbol, ×, but as the source notes "there is little danger of confusing them in formulas, since we will use the commutator product only when one of the arguments is a bivector".--JohnBlackburne^words_deeds 17:26, 18 April 2010 (UTC)[reply]

John: You are right about the source, but not about the suggestion. I am aiming at description of Silagadze's remarks, where he says:

“So far we only have shown that seven dimensional vector product can exist in principle. What about its detailed representation?”

He continues:

“Namely, for any composition algebra with unit element e we can define the vector product in the subspace orthogonal to e by x × y = 1⁄2(xy-yx). Therefore, from the viewpoint of composition algebra, the vector product is just the commutator of divided by two.”

He appears to leave open the possibility of different realizations, though none are given.

Instead of deleting this discussion, you might try to re-express it in a form more suitable to yourself. Brews ohare (talk) 18:13, 18 April 2010 (UTC)[reply]

I'd add that IMO Silagadze's approach is a general one that reduces to Hurwitz' theorem only for this particular realization as a composition algebra. That is, the notion of 7D cross product is not co-extensive with composition algebras. Brews ohare (talk) 18:20, 18 April 2010 (UTC)[reply]

John, The equation,

is listed over on the main article as a property of the seven dimensional cross product. You argued all last week that you believed this equation to be valid in seven dimensions. I believe now that you are correct in that respect. But in order for me to be able to compare your position here to your position on the talk page at Lagrange's identity, I need to ask you for a straight 'yes' or 'no' answer to the question,

Do you believe that the equation,

holds in seven dimensions? David Tombe (talk) 11:10, 22 April 2010 (UTC)[reply]

David stop posting the same question in multiple places, and stop mischaracterising my posts. I have already given clear reasons at Talk:Lagrange's identity why you are mistaken.--JohnBlackburne^words_deeds 11:22, 22 April 2010 (UTC)[reply]

John, as regards your latest edits, by all means call it the Pythagorean identity if you want to. But don't then come to talk page at Lagrange's identity claiming that the Pythagorean identity holds in 3 and 7 dimensions and link to an equation that wasn't the equation that we were talking about. On the talk page at Lagrange's identity it appeared that you were trying to say that the equation,

has got nothing to do with the Pythagorean identity. This is yet another example of you putting on a different face for a different page. David Tombe (talk) 16:22, 22 April 2010 (UTC)[reply]

David: My take is that Blackburne says the equation,

has nothing to do with Lagrange's identity, not Pythagoras'. Lounesto has clearly named this equation the Pythagorean theorem, although that nomenclature contradicts much mathematical usage for this term, which uses this name for the squared norm of a vector as the sum of squares of the orthogonal components (also called Parseval's identity). IMO, Blackburne's concern that Pythagoras' equation is not the same as Lagrange's identity in 7D has been disposed of at this link. Brews ohare (talk) 16:22, 23 April 2010 (UTC)[reply]

I just removed this as it made no sense. The first part, the equation, is already given as the defining property of the cross product, as is the distributive property from bilinearity. The rest seems to be original research, but it's difficult to tell what it's trying to show as it made no sense. It's unsourced as the only source was to a chapter on the cross product in 3D, so there's little chance anyone could improve it.--JohnBlackburne^words_deeds 13:53, 26 May 2010 (UTC)[reply]

IIRC, he has been pushing this on a number of talk pages, including yours. Since you weren't really interested in pursuing this, perhaps he had decided that his research really belongs in this article. DVdm (talk) 14:01, 26 May 2010 (UTC)[reply]

Yes, I saw it on mine. Could not make sense of it then and still can't now. But even if made presentable the article is not the place for unsourced OR. If it's correct and at all notable it will be in a source somewhere.--JohnBlackburne^words_deeds 14:08, 26 May 2010 (UTC)[reply]

John, It wasn't original research, and it didn't have any bearing on the controversy about angle that is being discussed on another page.

The problem here seems to be that you are approaching this whole issue as a pure mathematician, whereas I am approaching as an applied mathematician. I am fully aware of the fact that the equation in question is a defining equation for the cross product. That is all covered in an earlier section. I am fully aware that the likes of Silagadze and Lounesto can prove that this defining equation restricts the cross product to 3 and 7 dimensions. And we are all agreed that the distributive law holds for both cross products.

But some readers have commented on the absence of any formal proof of the fact that the distributive law holds. In my applied maths notes, there is such a proof for the 3D case. If you apply that same proof to the seven dimensional case, it expands into 252 terms. We need to cancel out two pairs of 84 terms, and then reduce the remaining 84 terms into 21 squared terms.

As regards stating that the equation in question holds in 3 and 7 dimensions, I accept that the pure mathematicians have found a way to prove this fact. But an applied maths reader is likely to want to see an illustration. I could have done an illustration for the 3D case because it is rather trivial. Instead, I supplied a source. I then went on to apply the exact same logic to the 7D case and pointed out that it involves 252 terms, and that the proof is simultaneous with the proof that the distributive law holds.

Rather than deleting the section, it would have been more beneficial if somebody could have converted the cumbersome equation into a more preferable format. Maths typing is not my speciality.

It's all very well that you fully understand this subject. But try to put yourself into the frame of mind of a casual high school student who knows about the 3D cross product and who wants to see more clearly how this could be extrapolated to 7D. Think about the questions that might be going through his mind. I was trying to address the kind of questions that might be asked. David Tombe (talk) 15:59, 26 May 2010 (UTC)[reply]

If it's not OR then please provide the non-WP source it is from. Then maybe someone can work out what it means, as what you added made no sense.--JohnBlackburne^words_deeds 16:14, 26 May 2010 (UTC)[reply]

John, It made perfect sense. The distributive law is proved in the process because we have to distribute the i, j, and k components, or the i, j, k, l, m, n, and o components while proving the equation. David Tombe (talk) 09:59, 27 May 2010 (UTC)[reply]

Please take your OR elsewhere and stop disrupting this talk page. DVdm (talk) 10:05, 27 May 2010 (UTC)[reply]

I'd suggest that the label "Pythagorean theorem" attached to the relation:

|x × y|² = |x|² |y|² − (x ⋅ y)²

(as is done by Lounesto) be replaced. This label is ambiguous. It could be taken to imply that this statement is equivalent to Pythagoras' theorem, or that it somehow replaces Pythagoras' theorem in 7-dimensions. Without exception, in every dimensionality of space, Pythagoras theorem is universally:

where

and the v_k are orthogonal components. The stated equation:

|x × y|² = |x|² |y|² − (x ⋅ y)²

is more correctly designated as a special form of Lagrange's identity.

To bore you a bit, if we define:

and substitute this result into the general form of Lagrange's identity, then the special form of Lagrange's identity:

is found only in 3-dimensions and in 7-dimensions and even there only for some very particular multiplication tables c^k _ij. The connection to Pythagoras' theorem is not immediate. If angle is introduced using the dot product:

this relation can be substituted into the special form of Lagrange's identity as:

Only at this point is any relation to Pythagoras found, namely via the Pythagorean trigonometric identity:

Doubtless this was what Lounesto had in mind with his labeling, and not to indicate a replacement for the Pythagorean sum of squares.

The Pythagorean trigonometric identity then leads to:

If you are in agreement, I'll change this labeling accordingly in the article. Brews ohare (talk) 17:35, 26 May 2010 (UTC)[reply]

What you think "Lounesto had in mind", is entirely your WP:SYNTH and doesn't belong here. DVdm (talk) 17:40, 26 May 2010 (UTC)[reply]

DvDM: Can you kindly try to be helpful instead of searching for non-sequitors to start arguments over? Nobody cares what Lounesto thought, and it is not an issue. The issue is that the labeling he has used as a shorthand designation of an equation is ambiguous and misleading, and there is no need that this label be be used here. Brews ohare (talk) 17:53, 26 May 2010 (UTC)[reply]

If indeed "Nobody cares what Lounesto thought, and it is not an issue", then a statement like "Doubtless this was what Lounesto had in mind with his labeling, and not to indicate a replacement for the Pythagorean sum of squares" is entirely out of line. Please open a blog somewhere to speculate about what Lounesto had in mind. This is not the place for such. DVdm (talk) 18:01, 26 May 2010 (UTC)[reply]

You can reject my categorical statement that this observation I made about Lounesto's thoughts is peripheral, and that it is not the subject here. But why? Why make catty remarks about blogs?. Brews ohare (talk) 18:08, 26 May 2010 (UTC)[reply]

(edit conflict) Why is it ambiguous ? All that line does is impose a condition on the magnitude of x × y. The words "Pythagorean theorem" guides the reader to one way of thinking of it: the different parts are in the ratios of the ratio of the sides of a right angled triangle, one angle the same as the angle between the vectors. But this is a bit more work to establish formally - the vectors are not sides of such a triangle for example - so too much should not be read into it, and it doesn't say anything about the Pythagorean theorem. I would prefer to use "Pythagorean identity", which was there at one point, as that more directly relates to the cos and sin below, but David objected to it and removed it.--JohnBlackburne^words_deeds 17:59, 26 May 2010 (UTC)[reply]

Hi John: Ambiguity doesn't suggest that a correct interpretation can't be found, but that also incorrect interpretations may result. I can speak for myself that I was led completely astray by this labeling initially. You may say I am a dope or that I represent only small group of readers. But a change of labeling would avoid this problem for all readers, eh? Could you help? Brews ohare (talk) 18:08, 26 May 2010 (UTC)[reply]

I've been in mixed minds about Lounesto's use of the name 'Pythagoras' in relation to that equation. It is more commonly referred to as the Lagrange identity, but even that is not an ideal name in the circumstances, because it pre-empts the point which is being made. 'Pythagoras identity' captures the spirit and purpose of the equation much better for the circumstances, and indeed it is an absolutely accurate name, but for the 3D case only. Lounesto is nevetheless in a minority in using this name. In some sources, no name is used.

What about labelling it with the dual name Pythagorean identity/Lagrange identity? David Tombe (talk) 19:27, 26 May 2010 (UTC)[reply]

The ratios of the magnitudes relate the same way in 3 and 7 dimensions, and the angle is defined the same way, so the name is as accurate there as here. As for Lagrange's identity I don't see how that's relevant - do you have any sources that say it's the more common name for this expression?--JohnBlackburne^words_deeds 19:41, 26 May 2010 (UTC)[reply]

John, You're entitled to your opinion as regards the fact that angle works just as well in 7D as in 3D. But I do suggest that you look at the link between 'sine of angle' and Jacobi identity that I supplied in the section below. On your other point, I can certainly supply sources that call the equation in question the Lagrange identity, but I can't supply any sources that will specifically state that this is a more common name than the Pythagorean identity. Can you supply any sources which specifically state that the Pythagorean identity is the more common name? At any rate, you should have seen that I was suggesting a compromise by using both names. David Tombe (talk) 19:54, 26 May 2010 (UTC)[reply]

It's not my opinion it's fact: angles are defined identically in three and seven dimensions, it just you can do more with them in seven. And it's not the name of the equation, no more than "orthogonality" is the name of the previous one. It's a property, of in this case the magnitudes relative to each other: they are related as the sides of a right angled triangle, so satisfy Pythagoras's theorem.--JohnBlackburne^words_deeds 20:21, 26 May 2010 (UTC)[reply]

This source refers to the same equation Lounesto calls the Pythagorean theorem by the name Lagrange's identity. This source calls it Lagrange's vector identity. A more general version is called Lagrange's identity in this source, third from bottom of list of equations. I can find nobody aside from Lounesto that calls this equation "Pythagoras' theorem". This google book search] shows roughly 900 citations that refer to Pythagoras as sum of squares. The name Pythagoras' theorem in n-dimensions can be found here and here, here & so forth. Brews ohare (talk) 20:28, 26 May 2010 (UTC)[reply]

All of those are from three dimensional vector algebra where as the first explicitly mentions "the n = 3 case is equivalent to (it)". They say nothing about it more generally or in seven dimensions.--JohnBlackburne^words_deeds 20:46, 26 May 2010 (UTC)[reply]

Look again John: (i) The name Pythagoras' theorem in n-dimensions can be found here and here, here & so forth. All are in n-dimensions and refer to Pythagoras' theorem as sum of squares. No reference refers to Lagrange's equation as "Pythagoras' theorem" except Lounesto. This source states Lagrange's identity in Eq. (1) in n dimensions, and later says that the restricted form also is often referred to as Lagrange's identity itself. Lagrange's identity provides the n-dimensional version too. Krantz states the n-dimensional form (p. 22) and says the Cauchy-Schwarz inequality is a consequence of it. If Lounesto is indulged in "inventing" a label for his unorthodox 7D "Pythagoras' theorem" based upon it's 3D connotations, the other sources identifying the Lagrange's identity in 3 D are an even better starting point for a 7D label having much usage and properly distinguishing this result from Pythagoras' theorem. Brews ohare (talk) 22:04, 26 May 2010 (UTC)[reply]

How is he inventing anything? The values of |x × y|, |x||y| and x⋅y are such that if a triangle were formed with side lengths equal to those values it would be a right angled triangle, so the values satisfy Pythagoras's theorem. Furthermore the angle between the vectors is one of the angles of the triangle, immediately giving the magnitude of the cross product as xy sin θ, which with the similar expression for the dot product give the Pythagorean identity. Straightforward and to me clear.--JohnBlackburne^words_deeds 23:30, 26 May 2010 (UTC)[reply]

John: Here are the facts: (i) Lounesto's label misled me, so yes, it can be misleading. (ii) No-one other than Lounesto refers to the equation this way (ii) The arguments to support his usage are all 3D arguments; no n-D arguments exist. (iii) Many authors call this the Lagrange identity, not only in 3D but in n-D. Predominant usage favors calling this one the Lagrange identity. (iv) There is a well established usage for the term Pythagorean theroem, namely the sum of squares, and Lounesto's label is not referring to this traditional usage. Don't you think you are a bit unrelenting on this one? Brews ohare (talk) 00:18, 27 May 2010 (UTC)[reply]

• (i) If you misunderstood it that is your problem, it is perfectly clear to me, as I described above; but do introduce other sources on the 7D cross product if you find ones that are more accessible to you
• (ii) It's nothing to do with 3D, Pythagoras's theorem is defined in 2D and the triangle is a 2D triangle
• (iii) That makes no sense as the cross product is not defined in n-dimensions, only 3 and 7; I suspect you are misunderstanding the sources
• (iv) (|x × y|)² + (x⋅y)² = (|x||y|)², i.e. a² + b² = c², Pythagoras's theorem--JohnBlackburne^words_deeds 10:24, 27 May 2010 (UTC)[reply]

(i) If I misunderstood it is my problem - yes it is, but not only my problem. Others will find this challenge to their understanding of Pythagoras as sum of squares to be perplexing and will wonder if in 7D something odd happens to Pythagoras' theorem. If it is labeled as Lagrange's identity, that won't happen.

(ii) You have not responded to Lounesto being the only author that uses this designation.

(iii) John, it is the Lagrange identity that applies in any dimension n. However, it takes on a special form in 3-D and in 7-D, the one displayed by Lounesto. The fact that the general Lagrange identity simplifies in these two cases is not a pretext for changing its name to Pythagoras' theorem in these cases. In fact, the simplified form in 3D is called Lagrange's identity by many many authors, and is the same form as in 7D. There is no misunderstanding on my part.

(iv) Your little algebra exercise has been presented before in my initial exposition in this subsection in a manner that shows clearly the various steps you have swept together blurring the initial starting point of Lagrange's identity. Please do me the courtesy to read what I initially wrote here.

(v) You have raised the notion that all that is involved here is a plane embedded in a 7-space. That is a bit elliptic inasmuch as 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. Brews ohare (talk) 14:33, 27 May 2010 (UTC)[reply]

John, Regarding your point number (iv), that equation, which you have called Pythagoras's theorem, only holds in 3 and 7 dimensions, yet in your point number (ii), you have said that Pythagoras's theorem is defined in 2 dimensions. As regards your point number (iii), The Lagrange identity is defined in all dimensions. The equation at your point number (iv) is simply the Lagrange identity in the special case of 3 and 7 dimensions. There is good reason to refer to the equation (|x × y|)² + (x⋅y)² = (|x||y|)², as either the Lagrange identity or the Pythagorean identity. And there is also good reason based on this information to assume that Pythagoras's theorem is narrowed down to 3 and 7 dimensions. But the Jacobi identity narrows it down even further to just 3D. David Tombe (talk) 14:14, 27 May 2010 (UTC)[reply]

Pythagoras's theorem is defined in 2D, but that means it works in any 2D subspace of 3D or 7D - such as the plane spanned by the two initial vectors. The angle between the vectors is one of the angles of the triangle, so the Pythagorean trig identity can be deduced from it. This would work in any dimension, but the cross product only works in 3D and 7D so the term (|x × y|)² is only defined in those dimensions. No contradiction: simple and elegant mathematics.--JohnBlackburne^words_deeds 14:41, 27 May 2010 (UTC)[reply]

Blackburne: You have responded to David, but not to my comment before his. Pythagoras' theorem on sum of squares is applicable in spaces of any dimension. It is not restricted to 2-D or to 2-D subspaces. It is Lagrange's identity in its special form using the cross-product that is restricted to 3-D and 7-D. You can supplement it with the Pythagorean trigonometric identity, but that doesn't make Lagrange's identity into Pythagoras theorem. The two together can produce the sin θ form for the magnitude of the cross product.

You have yet to respond to Lounesto being the only author to call this special form of Lagrange's identity by the name "Pythagoras' theorem". Brews ohare (talk) 14:58, 27 May 2010 (UTC)[reply]

I did - I wrote that if you are unhappy with Lounesto and have some other source on the 7D cross product that you prefer then add it to the article, and we can use it to improve the article. So far though your arguments seem based on 3D and so irrelevant sources.--JohnBlackburne^words_deeds 15:07, 27 May 2010 (UTC)[reply]

John Blackburne: I wonder if you might consider a bit further elaboration of the subsection "Rotations". The idea of rotations in 7-D appears to have some non-intuitive aspecs, mentioned briefly by Lounesto. For example, the Jacobi identity does not hold, and so the cross product cannot form a Lie algebra. Yet I seem to recall that Lie algebras and rotations are closely allied. The thought occurs to me that the angle θ introduced with the dot product may not be the angle that x must be rotated through to align with y. Further, Lounesto says that x × y has the same direction as some c × d vectors where c and d are not in the same plane as x & y. How is θ to be connected to rotation of x into y and about what axis?? Thank you, John. Brews ohare (talk) 17:59, 26 May 2010 (UTC)[reply]

Seems fine to me. Rotations work in all dimensions, not just three, but are much more complicated than you suggest. E.g. a general rotation in three dimension has three planes and three angles of rotation. That the cross product in seven dimensions is not rotationally invariant is interesting but not that surprising, and it's even more interesting that the group under which it is invariant is G₂ as it's an application for that exceptional group, but I don't know you can say much more than that.--JohnBlackburne^words_deeds 18:08, 26 May 2010 (UTC)[reply]

John; of course it seems fine for you, but what about casting some light for those of us that aren't familiar with all this and wonder what the absence of a Jacobi identity means in this context? Brews ohare (talk) 18:14, 26 May 2010 (UTC)[reply]

I don't see how it's related, except that they're both properties of the 7D cross product, so are related by being in the same article.--JohnBlackburne^words_deeds 18:28, 26 May 2010 (UTC)[reply]

John, The Jacobi identity and the sine of an angle are related. See this exercise,

Exercise: Show that the Jacobi identity,

together with:

where θ is the angle between the vectors and is perpendicular to the vectors a and b, requires that

so that

taking

Solution: Take a, b, and c to be coplanar and choose the angle between a and b, and between b and c to be θ, and the angle between a and c to be 2θ. Then,

and so,

Now,

and so:

Since

it follows that:

or, for arbitrary values of θ:

wheθre:

Hence:

and so:

David Tombe (talk) 19:40, 26 May 2010 (UTC)[reply]

That would be OR but it's not even correct maths and contains numerous elementary mistakes. Please take your maths homework elsewhere. --JohnBlackburne^words_deeds 19:45, 26 May 2010 (UTC)[reply]

John, You have been making alot of allegations of errors today. Can you point out the errors in the above proof? It's not OR and I didn't derive this proof. It was in my applied maths notes from 1979. David Tombe (talk) 19:57, 26 May 2010 (UTC)[reply]

My apologies, that is correct, but not very relevant. It uses a more complex formula to derive a simpler one which can be derived more easily in other ways. That the Jacobi identity does not hold here means you can't use a similar proof here, but the magnitude is the same, by definition.--JohnBlackburne^words_deeds 20:15, 26 May 2010 (UTC)[reply]

There is no other theorem I know of that provides a general proof of this sort without invoking dimensionality of the space. The only other approach I know of is to use Lagrange's vector identity, and that requires 3D or 7D. Because this proof of David's requires the Jacobi identity, however, it (non-obviously) is restricted to 3-D. Brews ohare (talk) 20:36, 26 May 2010 (UTC) My suspicion is that there is no simple and direct connection between the cross product in 7D and rotations in 7D just for the same reason: no Jacobi identity. Brews ohare (talk) 20:39, 26 May 2010 (UTC)[reply]

The standard way to prove that the cross product contains a sine relationship is to use the cosine relationship in the Pythagorean/Lagrange identity. But that of course assumes that the cosine relation holds in the dot product. The importance of the Jacobi identity is to yield the sine relationship independently from the cosine relationship.

If we ignore the Jacobi identity and then accept the sine relationship for any dimensions in the Lagrange identity, the concept of angle goes skew from the rotation axis. Basically, outside of 3D, 'angle' becomes merely an algebraic construct which loses its connection with rotation, and Pythagoras's theorem loses its connection with geometry. David Tombe (talk) 10:07, 27 May 2010 (UTC)[reply]

Tombe, your comments show that you have WP:NOCLUE. Angles and rotation make perfect sense in 2D. Please take your painful ignorance elsewhere. You are severely disrupring this talk page. DVdm (talk) 10:12, 27 May 2010 (UTC)[reply]

DVdm: It is WP:Uncivil to denigrate editors as "having no clue" and exhibiting "painful ignorance" and saying their good faith efforts on this talk page are "disrupting". Quite evidently you don't agree with David, but you have two better choices: (i) simply ignore D Tombe's contributions to discussion or (ii) actually try to make constructive comments. The choice you have made is actually capable only of disruption of the atmosphere on this page and can only assist a downward spiral. Brews ohare (talk) 13:46, 27 May 2010 (UTC)[reply]

It's been my experience (with some editors) that rewriting an article to address issues on the Talk page, even where one disagrees or judges there to be misunderstanding, results in a clearer presentation that anticipates reader confusion. Pooh-poohing objections rather than discussing them squelches opportunity for a more accessible article useful to a wider audience. The "sound bite" approach to rejection and reversion of contributions is widespread on WP, but not a good method. Brews ohare (talk) 14:01, 27 May 2010 (UTC)[reply]

I'm afraid DVdm is right: the angle is most easily understood in 2D but it works in all higher dimensions. What's more it works identically, e.g. the formula for calculating it, arccos (a⋅b / ab), is independent of dimension, simple rotations through an angle work the same way, etc.. This has been repeatedly pointed out to David, with references to places where it is clearly explained, but he has simply ignored this and kept on repeating clearly incorrect mathematics. That he is wrong justifies the WP:NOCLUE; that he comes back again and again with the same nonsense to the same talk page is simply disruptive.--JohnBlackburne^words_deeds 14:08, 27 May 2010 (UTC)[reply]

Hi John. Personally I do not believe use of Noclue has ever accomplished anything more than a momentary venting for the editor that uses it. The notion that repeated attempts by David to get his point across have not been successful doesn't mean that the audience has listened carefully and responded thoughtfully. The glib reiteration that 2D planes in 7D are not different from 2-D planes in 3-D does not address at all the peculiarities mentioned by Lounesto and does not explain what the loss of the Jacobi identity means. It is just being glib to suggest that fundamental differences in 7-D magically are of no importance when one looks at the cross product. A civil discussion could well improve this article, whether or not David ultimately accepts the result. Brews ohare (talk) 14:46, 27 May 2010 (UTC)[reply]

We've had the discussion, or at least I and others have said how it works repeatedly. But that David either ignores or does not understand us, and repeatedly states the same erroneous basic facts long after he's been shown to be wrong, makes it impossible to argue with him. As for the Jacobi identity it means nothing, other than what's stated in the article.--JohnBlackburne^words_deeds 15:11, 27 May 2010 (UTC)[reply]

I don't think there is anything uncivil about pointing to WP:NOCLUE as long as we are supposed to assume good faith. Let's look at some evidence from [11]:

• "If we try to contemplate a 2D space in isolation, we are effectively playing the game of 'let's pretend'."
• "I would say that we can't properly conceive of the idea of a 2D space any more than we can conceive of the idea of a 5D space."
• "I just don't think that we can assume that Pythagoras's theorem in its classical form can automatically be generalized to 'n' dimensions"
• "In an 'n'D inner product space, I see Pythagoras's theorem as being merely a definition."
• "Anything that we assume about a 2D space is based on our observations of 2D geometry in a 3D space."
• "It's impossible to know anything at all about the realities of a purely 2D space, because the idea is purely imaginary."
• "We can certainly discuss it (the concept of an angle in 2-dim space). But it will all be pure speculation that will no doubt be heavily prejudiced by our knowledge of a 2D plane in a 3D space."
• "But we should not assume that these defined 'n' dimensional Pythagoras's theorems, which are purely mathematical constructs, should be equated with the very real Pythagoras's theorem, which is actually a proveable theorem in 3D space."
• "A purely mathematical 2D space has go no connection whatsoever with areas or geometry."
• "We cannot assume that a purely mathematical 2D space has got any connection with a 2D plane in a 3D space."
• "Only 3D space can be linked to Euclidean geometry,"
• "The problem is because I can't imagine any such concept as a plane geometrical 2D plane in the absence of a third dimension. If we want to simply assume that such a 2D plane can exist, and then import all the rules and visualizations from a 2D plane in a 3D space, then of course I would have to concede that we can have angle. But we will run into trouble when we discover that we can't use the cross product to describe rotational phenomena"
• "You seem to be assuming that a mathematical 2D space can be represented by a 2D plane as we understand such in a 3D space. Are you confident that you can make that assumption?"
• "You are still making the assumption that a purely 2D space can be represented by plane geometry, whereas in fact it is merely an algebraic contruct."

Yes, this is merely on user talk page, but after all the efforts that have been made to explain things to David Tombe, we are not allowed to categorize this under WP:NOCLUE, then I think the only other option is to assume bad faith and interpret this as trolling. DVdm (talk) 15:42, 27 May 2010 (UTC)[reply]

DVdm: Most of the statements above are quite defensible. And it is clear that what David is after is clarification of how 3D space is differentiated from other dimensional spaces. That is clearly a question many, many readers will ask. A clear comparison is yet to be found in this article, or on this Talk page, or on WP. It's just contentious to say your alternative to assigning a label WP:NOCLUE is to assume bad faith and make accusations of trolling. Personally, I would never accuse Tombe of bad faith. He has his own intuitive view of many matters, and actually such an intuitive examination, even if outside the box, provides some insight into a better formulation of the article and to examples that you might never otherwise think of. Brews ohare (talk) 17:19, 27 May 2010 (UTC)[reply]

He was topic banned from editing physics articles for exactly the same reasons, just like you were, and i.m.o. the two of you should be banned from editing mathematics related articles - broadly construed. Wherever either one of you show up, there is trouble. DVdm (talk) 17:30, 27 May 2010 (UTC)[reply]

DVdm: Your notion that "where we show up, there is trouble" seems to indicate dislike for open back-and-forth. There is no "trouble" here; just comparing notes, and all of it polite except when you began with the WP:NOCLUE, "trolling", now escalated to "bans from editing". You don't like discussion beyond the sound-bite format. I am sorry to bother you, but so far the two of us haven't actually discussed substance: only attutude. Maybe working upon the article would prove more tolerable, eh? Brews ohare (talk) 19:48, 27 May 2010 (UTC)[reply]

DVdm, Regarding the quotes by myself which you have listed above, I don't think that you will find anything there which represents an unconventional point of view. I have a letter dated 29th April 1993 from Lars Mahinske at the editorial division of the Encyclopaedia Britannica. The letter provides a reply from one of their advisers in relation to a query which I had made about their maths article from where I first learned about the seven dimensional cross product. I will give you an exact quote from this reply. The reply indicates that he has picked me up wrongly and assumed that I was suggesting that the geometry in 7D was the same as the geometry in 3D. This is a common problem in correspondence. Although I can't remember what I wrote on the original query, my guess is that I asked them how the 7D case relates to geometry. Anyway, here is the relevant quote from the reply,

But again he/she has made an unwarranted assumption: that the vector product in 7 dimensions should have the same geometric properties as that in 3. The article doesn't say that, merely that a list of certain algebraic properties must hold.

This tells us that it is a standard belief amongst the mathematical establishment that the 7D cross product loses its connection with geometry. Hence if Pythagoras's theorem holds in 7D, the interpretation will shift. David Tombe (talk) 20:14, 27 May 2010 (UTC)[reply]

Assuming good faith, it's not a surprise to me that you don't think that I "will find anything there which represents an unconventional point of view." And again, a statement like "This tells us that it is a standard belief amongst the mathematical establishment that {whatever} ..." following the phrase you quoted, is WP:SYNTH of the WP:CLUELESS kind. Sorry, but we can't help you with this. O.t.o.h. assuming bad faith, I'm supposed "not to feed the troll". So either way there's nothing for me to comment. My apologies to the other contributors for having brought this up here. DVdm (talk) 20:48, 27 May 2010 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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:

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:

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 , 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)[reply]

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

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:0204357 {{arxiv}}: Check arxiv value (help). {{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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

John:

It is established that

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

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 . 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)[reply]

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)[reply]

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)[reply]

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)[reply]

So, you do not think:

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)[reply]

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)[reply]

(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)[reply]

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)[reply]

It is established that

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

Lounesto's "Pythagorean theorem" is:

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:

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)[reply]

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)[reply]

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)[reply]

To elaborate:

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,

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

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)[reply]

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)[reply]

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

John has already satisfied himself that,

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,

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)[reply]

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

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:

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 :

and also

for all pairs x, y, then

(Here it is not significant to the argument, but it happens that: )

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)[reply]

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)[reply]

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

Using the ancillary relations:

one obtains:

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:

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)[reply]

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

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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:

and is being applied here to 7 dimensions.

Lagrange's identity in 7-dimensions is:

(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:

(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

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:

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

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)[reply]

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:

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)[reply]

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)[reply]

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)^[1]

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:^[2]

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.^[3]

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

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

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)[reply]

(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)[reply]

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:

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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:

and

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

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)[reply]

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)[reply]

(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)[reply]

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

which I might shorthand as:

and

which I might shorthand as:

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:

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Resolution

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

"Lagrangian" or component form

"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)[reply]

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,

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)[reply]

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.

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

At present, the equation:

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:

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)[reply]

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)[reply]

(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)[reply]

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)[reply]

(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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

At present, the equation:

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)[reply]

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)[reply]

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)[reply]

• 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)[reply]

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)[reply]

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)[reply]

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)[reply]

Well, xy, xy sin θ and xy cos θ gives a Pythagorean triangle all right, but doesn't immediately seem relevant to . 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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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:

. ’

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

(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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

(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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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.

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 [12] 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)[reply]

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 comnplex 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)[reply]

"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)[reply]

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)[reply]

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)[reply]

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)[reply]