Talk:Seven-dimensional cross product/Archive 2

Rotation
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)
 * 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 wordsdeeds 21:49, 17 April 2010 (UTC)

Please add some discussion to the article. Brews ohare (talk) 22:07, 17 April 2010 (UTC)
 * 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 wordsdeeds 12:57, 18 April 2010 (UTC)

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)

Use of sinθ
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 &times; y|2 = |x|2 |y|2 &minus; (x &middot; y)2

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


 * 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)


 * 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)
 * 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)


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


 * 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)


 * 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)

Pythagorean identity
Here is a source (page4) which explains how to expand the right hand side of,


 * x &times; y|2 = |x|2 |y|2 &minus; (x &middot; 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)
 * 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 wordsdeeds 12:51, 18 April 2010 (UTC)


 * 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)

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)


 * 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)
 * 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)
 * As regards the (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)


 * 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)

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


 * 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)


 * 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)

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)

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)

Or: &emsp; &emsp; z12 = (x2y3-x3y2)2;&emsp; 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)

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)


 * 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)
 * 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 wordsdeeds 11:36, 19 April 2010 (UTC)

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&middot;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)

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)
 * 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 wordsdeeds 15:27, 19 April 2010 (UTC)

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;&emsp; 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)
 * 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 wordsdeeds 16:54, 19 April 2010 (UTC)

Hi David: I was a bit hurried. What I've shown is that ΣΣi<j(xiyj -xjyi)2 -x&middot; 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)
 * I've made a spreadsheet that does the entire calculation:
 * 
 * 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 wordsdeeds 23:28, 19 April 2010 (UTC)

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 &minus; (X &middot; 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, 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 &minus; (X &middot; 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 &times; Y|2 = |X|2 |Y|2 &minus; (X &middot; 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)


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

John Blackburne's conflicting positions over two articles
At the cross product article, John Blackburne is adamant that,


 * $$ |\mathbf{a} \times \mathbf{b}|^2 + |\mathbf{a} \cdot \mathbf{b}|^2 = |\mathbf{a}|^2 |\mathbf{b}|^2$$

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)

Spreadsheet to test Pythagorean theorem for 7D X-product z = x × y for arbitrary x & y
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)
 * 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 wordsdeeds 19:25, 20 April 2010 (UTC)

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,


 * $$\biggl( \sum_{k=1}^n a_k^2\biggr) \biggl(\sum_{k=1}^n b_k^2\biggr) - \biggl(\sum_{k=1}^n a_k b_k\biggr)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2,$$

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


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

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

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,. David Tombe (talk) 06:43, 21 April 2010 (UTC)

Commutator product
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 wordsdeeds 17:26, 18 April 2010 (UTC)

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)

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)

Question for John Blackburne
John, The equation,


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

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,


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

holds in seven dimensions? David Tombe (talk) 11:10, 22 April 2010 (UTC)
 * 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 wordsdeeds 11:22, 22 April 2010 (UTC)

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,


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

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)
 * David: My take is that Blackburne says the equation,
 * $$|\mathbf{x}|^2 |\mathbf{y}|^2 -|\mathbf{x} \cdot \mathbf{y}|^2 = |\mathbf{x} \times \mathbf{y}|^2$$
 * 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)

Lagrange's identity etc.
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 wordsdeeds 13:53, 26 May 2010 (UTC)


 * 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)
 * 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 wordsdeeds 14:08, 26 May 2010 (UTC)

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)
 * 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 wordsdeeds 16:14, 26 May 2010 (UTC)

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)


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

Pythagorean theorem
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:


 * $$ \|\mathbf V \|^2 = \sum \|\mathbf v_k \|^2  \ ,$$

where
 * $$ \mathbf V = \sum \mathbf v_k \, $$

and the vk 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:
 * $$ \mathbf x \otimes \mathbf y = \sum _{i, \ j} x_i y_j \mathbf {\hat e_i} \otimes \mathbf {\hat ej} = \sum_{i, j, k} c_{ij}^k x_i y_j\mathbf{\hat e_k} \, $$

and substitute this result into the general form of Lagrange's identity, then the special form of Lagrange's identity:
 * $$\| \mathbf x \otimes  \mathbf y \|^2    = \|\mathbf x \|^2  \|\mathbf y \|^2  -( \mathbf{ x \cdot y } )^2 \, $$

is found only in 3-dimensions and in 7-dimensions and even there only for some very particular multiplication tables ck ij. The connection to Pythagoras' theorem is not immediate. If angle is introduced using the dot product:
 * $$(\mathbf{x \cdot y}) = xy \ \cos \theta \, $$

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


 * $$\| \mathbf x \otimes \mathbf y \|^2 = (xy)^2 \ \left(1 - \cos^2 \theta \right) \ . $$

Only at this point is any relation to Pythagoras found, namely via the Pythagorean trigonometric identity:
 * $$ \sin^2 \theta + \cos^2 \theta = 1 \ . $$

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:


 * $$\| \mathbf x \otimes \mathbf y \| = xy \ \sin \theta \ . $$

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


 * 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)
 * 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)
 * 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)
 * 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)


 * 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 wordsdeeds 17:59, 26 May 2010 (UTC)
 * 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)

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)
 * 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 wordsdeeds 19:41, 26 May 2010 (UTC)

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)
 * 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 wordsdeeds 20:21, 26 May 2010 (UTC)


 * 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)
 * 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<sub style="margin-left:-2.0ex;">deeds 20:46, 26 May 2010 (UTC)

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)
 * 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<sub style="margin-left:-2.0ex;">deeds 23:30, 26 May 2010 (UTC)
 * 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)


 * (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<sub style="margin-left:-2.0ex;">deeds 10:24, 27 May 2010 (UTC)


 * (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)

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)


 * 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<sub style="margin-left:-2.0ex;">deeds 14:41, 27 May 2010 (UTC)


 * 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)
 * 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<sub style="margin-left:-2.0ex;">deeds 15:07, 27 May 2010 (UTC)

Rotations
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)
 * 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<sub style="margin-left:-2.0ex;">deeds 18:08, 26 May 2010 (UTC)
 * 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)
 * 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<sub style="margin-left:-2.0ex;">deeds 18:28, 26 May 2010 (UTC)

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

Exercise: Show that the Jacobi identity,


 * $$\mathbf {a \times ( b \times c)} + \mathbf {b \times ( c \times a)} + \mathbf {c \times ( a \times b)} = 0 \, $$

together with:


 * $$\mathbf { a \times b} = ab\  f(\theta)\mathbf {\hat n} \ ,$$

where θ is the angle between the vectors and $$\mathbf{ \hat n}$$ is perpendicular to the vectors a and b, requires that
 * $$f(2 \theta) = 2\cos \theta \ f(\theta) \, $$

so that


 * $$f(\theta)=\sin\theta\ ,$$

taking
 * $$ f(\pi/2)=1 \ .$$

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,
 * $$\mathbf{a \times ( b \times c)} = abc\ f(\theta)\  \mathbf{\hat a \times \hat n }\, $$
 * $$\mathbf {c \times ( a \times b)} = abc\ f(\theta)\ \mathbf{ \hat c \times \hat n} \, $$

and so,
 * $$ \mathbf{( \hat a + \hat c )\times \hat n } \ f(\theta) = \mathbf {\hat b \times \hat n }\  f(2\theta) \ . $$

Now,
 * $$\mathbf { \hat a + \hat c} = 2\cos\theta \ \mathbf {\hat b} \, $$

and so:
 * $$ 2\cos\theta\ f(\theta) = f(2\theta) \ . $$

Since
 * $$ \sin(2\theta) = 2\cos \theta \sin \theta \, $$

it follows that:
 * $$\frac{\sin (2 \theta)}{\sin \theta} \ f(\theta) = f(2 \theta) \, $$

or, for arbitrary values of θ:
 * $$ g(\theta) = g(2\theta) \, $$

wheθre:
 * $$ g(\theta) = \frac{f(\theta)}{\sin\theta} \ . $$

Hence:
 * $$ g(\theta) =\mathrm{ constant } \, $$

and so:
 * $$ f(\theta) = \sin \theta,\ \mathrm{taking}\ f(\pi/2) = 1\ . $$ David Tombe (talk) 19:40, 26 May 2010 (UTC)
 * That would be OR but it's not even correct maths and contains numerous elementary mistakes. Please take your maths homework elsewhere. -- JohnBlackburne words<sub style="margin-left:-2.0ex;">deeds 19:45, 26 May 2010 (UTC)

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)


 * 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<sub style="margin-left:-2.0ex;">deeds 20:15, 26 May 2010 (UTC)

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)


 * 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)


 * 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)


 * 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)
 * 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)
 * 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<sub style="margin-left:-2.0ex;">deeds 14:08, 27 May 2010 (UTC)

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)
 * 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<sub style="margin-left:-2.0ex;">deeds 15:11, 27 May 2010 (UTC)

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 :
 * "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)
 * 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)


 * 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)

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)


 * 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)


 * 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)