Talk:Lagrange's identity

The special case of seven dimensions
It stated in the article that the general form,


 * $$\biggl( \sum_{k=1}^n x_k^2\biggr) \biggl(\sum_{k=1}^n y_k^2\biggr) - \biggl(\sum_{k=1}^n x_k y_k\biggr)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (x_i y_j - x_j y_i)^2,$$

reduces to,


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

in the special case of 3 dimensions. This is also true for 7 dimensions. In the seven dimensional case, we get 21 terms on the right hand side. These terms take the form,


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

for z12 - z72, and based on the relationships from the seven dimensional cross product,

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

The article has been amended to reflect this fact. It is wrong to say that the above reduction only applies in 3D. David Tombe (talk) 07:42, 21 April 2010 (UTC)
 * Why does it? You have provided no source for this, just some very flawed reasoning. What you've written above does not even come close to a mathematical proof, and even in it's incomplete state contains some basic mathematical errors that you'd expect a high school student to make. So, until you come up with a proper source for it, it does not belong in the article.-- JohnBlackburne wordsdeeds 13:09, 21 April 2010 (UTC)


 * That's very strange John. You just spent about five days at the talk page of seven dimensional cross product arguing that it holds in 7D. In fact you even provided sources in the form of Silagadze and Lounesto which were adamant that the Pythagorean identity holds in both 3D and 7D. David Tombe (talk) 14:57, 21 April 2010 (UTC)
 * Yes, the Pythagorean identity holds in 3D and 7D. Or more precisely it is a condition of the cross product in 7D. But that is a different article: that and Lagrange's identity are not the same thing.-- JohnBlackburne wordsdeeds 17:38, 21 April 2010 (UTC)

John, You are now playing on words. The equation,


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

whatever name you like to call it, holds in both 3 and 7 dimensions and that is exactly what you were arguing over at the seven dimensional cross product page. You cannot wriggle out of this by supplying an alternative definition for the Pythagorean identity which in fact is more accurate. The point is that you were using the name Pythagorean identity for the equation,


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

I didn't like that usage of the word, but I didn't quibble about it. Now you are trying to wriggle out of this by jumping to the more accurate usage of the term Pythagorean identity. David Tombe (talk) 00:08, 22 April 2010 (UTC)
 * Here is z1, from the 7D cross product article:
 * $$z_1 = (x_2y_4 - x_4y_2 + x_3y_7 - x_7y_3 + x_5y_6 - x_6y_5)$$
 * Please show how you get from that to what you claim is z12 (copied from above):
 * z12 = (x2y4-x4y2)2 + (x3y7-x7y3)2 + (x6y5-x5y3)2
 * then perhaps you will see for yourself how they are not the same in 7D.-- JohnBlackburne wordsdeeds 07:45, 22 April 2010 (UTC)

John, I'll gladly discuss your question, but only after you have first answered the question below on both this page and on the talk page of seven dimensional cross product. The question is very simple, and it only requires a 'yes' or 'no' answer.

Does the equation,


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

hold in seven dimensions? David Tombe (talk) 11:04, 22 April 2010 (UTC)


 * Yes - although as I note above it's not so much that it holds, it's more that it's proposed as a condition the cross product must satisfy. From it and the other properties it's shown that a product only exists in 3 and 7 dimensions. For other reasons there are multiple products in 7D. Meanwhile please look at the two lines above. One is the expression for z1 from the 7D cross product page, the other is what you claim is z12. It is clear that if the first equation is squared you do not get the second equation, so it is completely wrong, or at least it is something completely different. So they are not the same, as you claim.-- JohnBlackburne wordsdeeds 11:18, 22 April 2010 (UTC)

I've contributed a subsection on 7D that is rigorously sourced to Lounesto. Yes, the cross product must satisfy this relation, and it is a property proposed for any cross product. That means the Pythagorean theorem is holds in 3D and in 7D, as Lounesto says point blank. Hopefully, the wording of the new subsection makes all this clear. Brews ohare (talk) 14:50, 22 April 2010 (UTC)
 * But that's the Pythagorean theorem, not Lagrange's theorem. Nowhere in Lounesto (I have the book in front of me) does it refer to Lagrange's identity. The two are only the same in 3D, so it has no place here. David tried to prove that they are the same but his maths is simply wrong: again here is David's calculation
 * z12 = (x2y4-x4y2)2 + (x3y7-x7y3)2 + (x6y5-x5y3)2
 * And here is z1, from the 7D cross product article:
 * $$z_1 = (x_2y_4 - x_4y_2 + x_3y_7 - x_7y_3 + x_5y_6 - x_6y_5)$$
 * It is clear the above squared does not equal what David has, so the two things are not the same.-- JohnBlackburne wordsdeeds 15:26, 22 April 2010 (UTC)


 * John: There are a multitude of possible multiplication tables for the 7D cross-product. You have proposed one at Seven-dimensional cross product. David another. My spreadsheet here is based upon an equally valid, but different one found at Octonion. So this little algebraic exercise doesn't mean much unless the same multiplication table is used.


 * As you very well know from the intro to this article, Lagrange's identity is:


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


 * Nonetheless, you are proposing (using an algebraic instance based on incompatible multiplication tables) that Lagrange's identity does not reduce in 7D to:


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


 * Right? My spreadsheet calculates the Lagrange terms and shows the Pythagorean identity and the Lagrange identity are identical for all x and y in 7D. Brews ohare (talk) 15:57, 22 April 2010 (UTC)
 * Or at least, identical for all x and y that I have tried - a rather large and varied set. Brews ohare (talk) 16:01, 22 April 2010 (UTC)
 * Your spreadsheet calculates the cross product in 7D. This is Lagrange's identity. The reason it works in 3D is the dual relation between vectors and bivectors, as Lagrange's identity is the same as
 * $$(a \cdot a)(b \cdot b) - (a \cdot b)^2 = (a \wedge b) \cdot (a \wedge b).$$
 * In 3D a ∧ b and a × b are dual, so the above can be written in terms of the cross product, as the article says. But this only works in 3D. In 7D there is no such dual relation so the above formula cannot be written in terms of the cross product in 7D.-- JohnBlackburne wordsdeeds 16:26, 22 April 2010 (UTC)

John, Now that you have admitted that it holds in seven dimensions, why did you make this revert ?

On your question to me, I made a mistake in segregating the equation into seven parts. I'm not going to cover up the mistake and it's important that these issues are corrected, but it didn't have any bearing on the wider issue as regards the fact that the equation holds in seven dimensions. The seven zundefined2 terms do indeed equate to the twenty-one xy terms on the right hand side. The correct equation should look like this,

z12 + z22 + z32 + z4:2 + z52 + z62 + z72 = (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

The question which we were tackling was how to relate the 21 terms to the magnitude of a cross product. I had my doubts, having been mislead by two wikipedia articles into thinking that this could only be done in 3D. You finally convinced me with your numerical examples. It then became a case of trying to rationalize how the 21 xy terms related to the 7 zundefined2 terms. But we agreed in the end that it works in 7D. And yes, I shouldn't have assumed that the individual zundefined2 terms in the equation can be paired off with their family members. Your point above, in that respect was correct. The equation only holds over the entire set of relationships. But we still don't know why you are resisting the correction at this article and at the cross product article. These two articles should now read that the equation,


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

holds in both 3 and 7 dimensions, and you have been deliberately obstructing this correction. David Tombe (talk) 16:15, 22 April 2010 (UTC)

Summary of 7D case
JohnBlackburne:

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

or, equivalently, Jones, page 4:


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

=\sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2\ ,$$

The generalized Pythagorean theorem states:


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

= |\mathbf { a\ \times\ b}|^2\ ,$$

John, you believe that the Lagrange's identity holds true for any dimensionality of space, and the Pythagorean theorem only when dimensionality of the space is n = 3 or n = 7. I believe that is consensus, and supported by Lounesto.

Please let me know if you disagree.

Given these two beliefs, the issue, then, as to whether the Lagrange identity is the same as the Pythagorean identity, is whether:


 * $$|\mathbf { a\ \times\ b}|^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2\ ?$$

If the answer is yes, the two are the same. If the answer is no, the two are different.

If as you believe, John, the two are different when n = 7, then there are two distinct identities satisfied by the components of a and b when n = 7.

Is that your position? Brews ohare (talk) 17:56, 22 April 2010 (UTC)

BTW, my spreadsheet here calculates both sides of this equation for arbitrary a and b in 7D, and finds them both to be the same for every case that I have tried. Brews ohare (talk) 18:36, 22 April 2010 (UTC)

Formal proof
Jones, page 4 has shown Lagrange's identity is equivalent to:
 * $$|\mathbf a|^2|\mathbf b|^2 - (\mathbf { a \cdot b})^2

=\sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2\ ,$$

One possible multiplication table in 7-space is:
 * $$\mathbf {e_i \times e_j} = \varepsilon _{ijk} \mathbf{e_k} \, $$

where $$\varepsilon _{ijk}$$ is a completely antisymmetric tensor with a positive value +1 when ijk = 123, 145, 176, 246, 257, 347, 365. Consequently,


 * $$ \mathbf{a \times b} = \Sigma_i \Sigma_j a_i b_j \varepsilon _{ijk}\mathbf{e_k} \ . $$

Using this cross product, the generalized Pythagorean theorem is established as follows:


 * $$|\mathbf{a \times b}|^2 = \Sigma_k \left( \Sigma_i \Sigma_j a_i b_j \varepsilon _{ijk}\Sigma_p \Sigma_q a_p b_q \varepsilon _{pqk} \right) \ $$


 * $$=\Sigma_i a_i \left[\left(\Sigma_j\Sigma_k\varepsilon _{jki}b_j\Sigma_p \Sigma_q \varepsilon _{pqk}a_p b_q\right)\right] $$
 * $$=\mathbf {a \cdot} \left[ \mathbf {b \ \times} \left( \mathbf {a\ \times \  b} \right)\right] \ . $$

Using identities from the 7D cross product article for the triple cross product with only two different vectors:


 * $$=\mathbf {a \cdot} \left[

|\mathbf {b}|^2 \ \mathbf  a

- \left(\mathbf {b \cdot a }\right) \mathbf b

\right] \, $$
 * $$=|\mathbf a|^2|\mathbf b|^2 -\left(\mathbf{a \cdot b}\right)^2 \ . $$

Combining this with Jones' result:
 * $$ |\mathbf{a \times b}|^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2\ .$$

I think that does it. In 7D the Lagrange's identity and the generalized Pythagorean theorem are the same condition. Brews ohare (talk) 21:33, 22 April 2010 (UTC)

An alternative approach is to contract the εijkεpqk as sums of δipδjq etc. and avoid the use of the triple product identity. Brews ohare (talk) 21:52, 22 April 2010 (UTC)
 * I can see a couple of problems with your maths, mostly that εijk is not the usual Levi-Civita symbol: it is a shorthand for the operation and restricted to a limited range of indices (which depend on your choice of 7D cross product), so you can't do algebra with it as in 3D.
 * But you've only asserted that:
 * $$|\mathbf{a \times b}|^2 = |\mathbf a|^2|\mathbf b|^2 -\left(\mathbf{a \cdot b}\right)^2 \ . $$
 * it would be surprising if you got anything different - you could have copied it from the 7D cross product article. That these two things are equal does not make the identity the same. E.g. consider
 * $$ \tan \theta = \frac{\sin \theta}{\cos \theta}$$
 * and
 * $$ \tan \theta = \frac{2\tan \frac{\theta}{2}}{1 - tan^2\frac{\theta}{2}}$$
 * These are obviously equal, but that does not make them the same identity: the second is a half angle formula, but the first one is not.
 * In the same way just because this, one way of writing Lagrange's identity:
 * $$|\mathbf a|^2|\mathbf b|^2 - (\mathbf { a \cdot b})^2 =\sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2\ ,$$
 * and this, the Pythagorean condition on the cross product in 7D:
 * $$|\mathbf{a \times b}|^2 = |\mathbf a|^2|\mathbf b|^2 -\left(\mathbf{a \cdot b}\right)^2 \ . $$
 * are equal does not mean the second is Lagrange's identity in 7D. If there were a source that they are the same then we could use it to support the changes, otherwise it is synthesis.-- JohnBlackburne wordsdeeds 22:09, 22 April 2010 (UTC)

John: First, no confusion has been made between the 3D epsilon tensor and the 7D one. Second, if the contraction of the epsilon tensors is used to form products of Kronecker deltas, exactly the same summation formula appears with the components just as they appear in the Lagrange form. Third, the spreadsheet shows the two are identical component wise.

Your final comment is that if A = B and B = C, then A = C is synthesis. In other words, why bother? Brews ohare (talk) 22:39, 22 April 2010 (UTC)
 * The maths is all in 7D surely, so there should be no 3D Levi-Civita symbols, but it's not important as it's a result we already know. My point was, with the trig example, that two things being equal does not make them the same identity. Lounesto does not connect them in seven dimensions (he does in 3). No source says they are the same identity in 7D, as they are not, and to deduce the connection yourself is synthesis. -- JohnBlackburne wordsdeeds 23:01, 22 April 2010 (UTC)


 * Again, the use of the symbol εijk is carefully defined for 7D in the exposition, and does not refer to the 3D Levi-Cevita symbol.


 * Whatever form a × b takes it is a polynomial in {(aibk)2}. Likewise, the sum :$$ \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2\ .$$ is a polynomial in {(aibk)2}. Can you really suggest that they are not the same polynomial when equality applies for arbitrary choices of {ai} and {bj}? I believe such a claim is disprovable in principle.


 * Do you care if you are right or not? Or is this really about strict enforcement of the WP challenge of unsourced material, regardless of common sense? Brews ohare (talk) 23:59, 22 April 2010 (UTC)

Complicated algebra; is it needed?
Brews, I agree that it has been established to everybody's satisfaction that,


 * $$\biggl( \sum_{k=1}^n x_k^2\biggr) \biggl(\sum_{k=1}^n y_k^2\biggr) - \biggl(\sum_{k=1}^n x_k y_k\biggr)^2 = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (x_i y_j - x_j y_i)^2,$$

reduces to,


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

in the special cases of both 3 and 7 dimensions. Establishing this relationship in 3D is a relatively straightforward business, and that is perhaps why this article and the cross product article gave the misleading information that it only holds in 3D. But it does hold in 7D even though proofs of that fact are anything but straightforward. It's far from immediately obvious that the 21 terms on the right hand side in the 7D case expand into 84 terms, and that these 84 terms are in fact the 252 terms that we get when we square,


 * $$z_1 = (x_2y_4 - x_4y_2 + x_3y_7 - x_7y_3 + x_5y_6 - x_6y_5)$$

for all $$z_1$$,- - - -$$z_7$$, and that within these 252 terms we have three groups of 84 terms, two of which are mutually cancelling. That is not straightforward mathematics. If we look at the case of 5D, the 252 terms reduce to 80 terms. 60 of those terms reduce to the 15 terms of the Lagrange identity, but we are then left with an extra 20 which means that the 5D cross product does not match with the Pythagorean identity. Strangely, it only works for 3 and 7.

And so it's correct that this article now mentions the 7D case, and I support the changes that introduce this fact into the article. As regards terminologies, the equation,


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

could rightfully be called either the Lagrange identity or the Pythagorean identity. David Tombe (talk) 03:48, 23 April 2010 (UTC)

Hi David: Well, you are right that the algebra could be a mess. The form:


 * $$ \mathbf{a \times b} = \Sigma_i \Sigma_j a_i b_j \varepsilon _{ijk}\mathbf{e_k} \, $$

or, for the norm squared:


 * $$|\mathbf{a \times b}|^2 = \Sigma_k \left( \Sigma_i \Sigma_j a_i b_j \varepsilon _{ijk}\Sigma_p \Sigma_q a_p b_q \varepsilon _{pqk} \right) \, $$

can be manipulated using the contraction of the epsilon tensor as sums of Kronecker deltas to produce the result. This epsilon tensor is not the Levi-Civita symbol in 3-D, but that defined by Lev Vasilʹevitch Sabinin, Larissa Sbitneva & I. P. Shestakov. An alternative is to use the representation of the 7D cross product as a sum of 3-dimensional cross products using the standard Levi-Civita symbol.

However, algebra to one side, Blackburne's suggestion is that the two sourced results that no-one disputes, namely:


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

and


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

=\sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2\ ,$$

do not imply:


 * $$|\mathbf{a \times b}|^2 =\sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2\ .$$

Of course, if the equality held only for a particular pair of vectors a & b, that concern would be warranted. However, in fact, the equality holds for every possible pair of 7 vectors a & b. Inasmuch as we are dealing with only the space of 7-vectors, I'd say that if function f and function g mapping the 7-space to the scalars both produce the same scalar for every permissible argument, they are the same function for all intents and purposes. Moreover, in this case, both f and g are not general functions, but both are linear combinations of exactly the same variables, namely {aiajblbm}.

So, although the algebra is complicated, I don't think it even is necessary. Blackburne's concern is unfounded. Brews ohare (talk) 15:05, 23 April 2010 (UTC)

Proof
The second form of the right-hand looks much nicer if you drop the superfluous condition i<>j (the terms corresponding to i=j are zero).

It is then also much easier to prove the identity by expanding the squares (in the rhs) and separating the sums.

Or am I missing something? 11:08, 29 July 2016 (UTC) — Preceding unsigned comment added by 110.23.118.21 (talk)

j not equal to i
Whats the use of j not equal to i. Staticgloat (talk) 08:31, 25 January 2021 (UTC)