Talk:Triple product

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 Field: Geometry

Pseudovectors?[edit]

The stuff about pseudovectors needs to be reviewed by someone who understands it. I think the notation could be improved to clarify things. It is not clear to me what the significance of this is. Dhollm 12:59, 6 May 2007 (UTC)


Can you please clarify what you feel needs to be improved? Also, what do you miss in terms of significance? Thanks. Edgerck 19:19, 6 May 2007 (UTC)


Note: I partly reversed your cuts, which made the article clash with other wikipedia references. I hope to also have made it clearer. Edgerck 19:27, 6 May 2007 (UTC)

index notation[edit]

i think it might be helpful if you included the index notation for the triple product —Preceding unsigned comment added by 128.95.141.33 (talk) 02:10, 26 October 2007 (UTC)

I agree. I believe it can be useful (although the BAC-CAB rule is clear enough), but I can't do it immediately. Paolo.dL (talk) 20:25, 16 January 2008 (UTC)


The indexes with the Levi Civita notation in the final part of the page are wrong. in fact Ax(BxC)= epsilon_ijk a_j epsilon_klm b_l c_m = epsilon_kij epsilon_klm a_j b_l c_m . This change in indexes is in reality a change in the sign of the solution. [1]

Lorebene (talk) 17:00, 27 September 2011 (UTC)

That agrees with Cross product. —Ben FrantzDale (talk) 15:26, 28 September 2011 (UTC)

Pseudovectors and pseudoscalars[edit]

I rewrote the sentences about pseudovectors and pseudoscalars. In my opinion, they were not clear, and one of them was wrong (the "if and only if" in scalar vector product). Please check and let me know if you agree and if you like them. Paolo.dL (talk) 23:42, 16 January 2008 (UTC)

Lagrange's Formula[edit]

Lagrange's formula is currently internally linked, but this leads to a disambiguation page, which in turn leads back to the section on this page. It's completely circular, but I'm not sure what the intention was. Warrickball (talk) 21:24, 12 May 2008 (UTC)

It is just a way to warn the reader that "Lagrange's formula" is ambiguous: it does not refer only to triple product expansion. This is useful information, based on which people may choose to use "triple product expansion", rather than "Lagrange's formula", to avoid ambiguity. Paolo.dL (talk) 07:51, 14 May 2008 (UTC)

Erroneous Equation[edit]

The following was included as a property of triple-products:

"There is also this property of triple products:


\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})=(\mathbf{a}\times \mathbf{b})\times (\mathbf{a}\times \mathbf{c})

" It cannot be correct as the left-hand side of the equality is a scalar and the right-hand side is a vector. Perhaps someone knows what was intended?


The original equation (the correct one), was:

(\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})) \mathbf{a}=(\mathbf{a}\times \mathbf{b})\times (\mathbf{a}\times \mathbf{c})
In fact, there was a fun succession of edits, the first one removed the first opening parenthesis, the next removed the closing parenthesis next to 'a', and at last, Bjordan555 removed the 'a' from it... I don't know why this happened. I've undone the changes and corrected it again. I've changed the parenthesis for brackets to clarify it.

— Preceding unsigned comment added by Gr3gf (talkcontribs) 21:34, 8 June 2009 (UTC)

Another notation[edit]

Can't square brackets be used for the triple product?

[\mathbf{a}, \mathbf{b}, \mathbf{c}] := \mathbf{a} \cdot \mathbf{b} \times \mathbf{c}

...or is it:

[\mathbf{a}, \mathbf{b}, \mathbf{c}] := \mathbf{a} \times \mathbf{b} \cdot \mathbf{c}

(The last is used in Rutherford, D. E. (1965). Vector Methods. University Mathematical Texts. Oliver and Boyd Ltd., Edinburgh. ) Alksentrs (talk) 14:37, 15 June 2009 (UTC)

As in Russian education, the following is considered true: [\mathbf{a}, \mathbf{b}, \mathbf{c}] := (\mathbf{a} \cdot [\mathbf{b} \times \mathbf{c}]). As one has to make two different multiplications - one "cross product", one "dot product" - this is called mixed product.--Q0k (talk) 22:03, 16 November 2009 (UTC)

scalar, vector - and where is mixed ?[edit]

What is \mathbf a \cdot [\mathbf b \times \mathbf c]? What is \mathbf a \times (\mathbf b \cdot \mathbf c)? --Q0k (talk) 04:48, 15 November 2009 (UTC)

The latter is undefined: types do not match. Incnis Mrsi (talk) 23:01, 16 January 2014 (UTC)

Pseudoscalars???[edit]

Under the section Scalar or pseudoscalar the article reads:

"The scalar triple product typically returns a pseudoscalar, although a pseudoscalar is equivalent to a (true) scalar if the (mathematical) orientation of the coordinate system is selected in advance and fixed."

No. The scalar triple product is a mathematical function, not a computer subroutine, and as such it doesn't "return" anything; it takes values. More important, it doesn't have "typical" behavior: It *always* takes a scalar value, defined as the determinant of the matrix (a b c), where a, b, and c form an ordered triple of vectors in R3.

If someone wants to write a different article about some variant of the scalar triple product used in physics, that's one thing. But it's not appropriate to confuse readers about a very standard mathematical function. This entire section should be removed from this article.Daqu (talk) 22:20, 25 September 2010 (UTC)

You are right about "return". As a programmer it makes sense to me but it's not a mathematical term, so I've rewritten it using more mathematical language there and further down. It's still making a valid point, hopefully more clearly now, so there's no need to remove it.--JohnBlackburnewordsdeeds 22:29, 25 September 2010 (UTC)
I believe that articles about programming should be written by people who are expert in the subject of programming. Articles about math, by people expert in math. No one expert in math has ever called the value of the triple scalar product a pseudoscalar. I don't consider a false statement to be a "valid point". The value of a triple scalar product is a number, not a "pseudoscalar".Daqu (talk) 05:16, 26 September 2010 (UTC)
It makes sense to me, but could be clearer so I've rewritten it. The pseudoscalar is a valid and well defined concept in maths and physics, and this is a good example of it. If a reader is not familiar with it they can follow some of the links which go into it in far more detail than I think's appropriate for this article, but it's good to mention it here to fit this product into its various applications and generalisations. --JohnBlackburnewordsdeeds 10:13, 26 September 2010 (UTC)
Pseudoscalar is a well-defined concept in math/physics. Actually, a computer scientist might be more prone than a mathematician to call it a scalar (it's just a float; why give it a fancy type?). If you are working with matrices of scalars, then yea, it's a scalar (as you said, it's just the determinant of the matrix), but if you are using a richer algebra that accounts for more than one coordinate system, then it can't be just a scalar: instead of a matrix you have a tensor and so instead of the determinant you use the Levi-Civita symbol which itself is not a tensor but a pseudotensor again giving you a pseudoscalar. —Ben FrantzDale (talk) 22:08, 26 September 2010 (UTC)

Invariant under rotation[edit]

The statement The scalar triple product is invariant under rotation of the coordinate system. though true seems a little out place. All of vector algebra is invariant under rotation of of the coordinate system - that's the whole point of it. Unless there are objections in the next week or so, I propose to delete this statement. --catslash (talk) 23:50, 15 January 2014 (UTC)

I do not see any problem with reiterating the statement that the triple scalar product is a pseudoscalar, but I think it should be moved to the appropriate subsection. The same formula with R can be used to explain what happens if R is an improper rotation. Incnis Mrsi (talk) 23:01, 16 January 2014 (UTC)
Good idea. --catslash (talk) 22:03, 17 January 2014 (UTC)

new Geometric algebra section[edit]

I have serious problems with the new section. The main result/interpretation of the triple product in GA is

 a ^ b ^ c

Which makes sense in GA (or exterior algebra) and gives a pseudoscalar result. This is given in Triple product#As an exterior product. I don't understand how the new section relates to this, or what other point it is trying to make, while the diagram is simply confusing.--JohnBlackburnewordsdeeds 17:55, 17 January 2014 (UTC)

You are right: I forgot what does “∧” usually mean and, particularly, what is the exterior algebra. No doubt I learned it once, but today missed the difference between “∧” and “anticommutator/2” (denoted as “×” if the geometric algebra article is correct). I thought about anticommutators, but wrote it as “∧”. Does this part of article now make sense? As for the diagram… it is ugly and you may remove it if it degrades the article. Incnis Mrsi (talk) 20:10, 17 January 2014 (UTC)
I though about the aEEa thing. It is called commutator – another mistake. Incnis Mrsi (talk) 22:05, 17 January 2014 (UTC)
'^' is the exterior product/outer product. Exterior algebra is GA with only exterior products/all inner products zero. Exterior algebra is older and perhaps more widely studied, so is worth mentioning even though all the results are found in GA. '×' can be used but it has other meanings in GA so '^' is better and more common. Or one way of distinguishing between them is '×' gives the vector-valued cross product, '^' the bivector-valued exterior product.--JohnBlackburnewordsdeeds 20:28, 17 January 2014 (UTC)

I've removed it. I still could not understand what point it was making while the maths was simply wrong. E.g. it asserts that if a is orthogonal to b and c then the product is zero, which makes no sense. Mutually orthogonal vectors give a non-zero triple product. With the maths wrong and the text unclear it's impossible to fix; pared down to correct maths it would just duplicate content in the rest of the article.--JohnBlackburnewordsdeeds 20:39, 17 January 2014 (UTC)

You miss that there are two triple products in vector calculus. One is (pseudo)scalar and yes, it is non-zero for an orthogonal basis. The second is vector triple product, I said. It is another product. Do you understand at last? Incnis Mrsi (talk) 21:13, 17 January 2014 (UTC)
But that product is already covered, and proved, in the article. It still doesn't make sense. You write (in a footnote so it's not obvious) that b×c is just b^c. But that makes a×(b×c) just a ^ (b ^ c), unless '×' has different definitions at different points. This is why the cross product is avoided in GA, it doesn't generalise, while the geometric product and exterior product are well defined.--JohnBlackburnewordsdeeds 21:32, 17 January 2014 (UTC)
What does the article prove about the vector product? (moving this part to a separate thread) No, you are wrong: for vectors b, c: b×c = bc, but it isn’t a vector but a grade-2 element, a pseudovector, do you understand? It does not make a×(b×c) just a∧(bc) because the former is, by definition from geometric algebra #Extensions of the inner and outer products, a commutator (up to the 1/2 factor) of a vector (grade-1 element) and a pseudovector (grade-2 element)! Whereas the latter is their exterior product, that isn’t 1/2(a(bc) − (bc)a) but is 1/2(a(bc) + (bc)a) because the second argument is even (it’s not a vector). Do you see that the GA formula for ∧ is different from one for ×? It is the thing where I failed last time. IMHO now it is your turn to learn exterior algebra. Incnis Mrsi (talk) 22:05, 17 January 2014 (UTC)

Update: it appeared that my last version forgot to write 1/2 factor in the definition of “×”, although calculation itself is correct. Yesterday was definitely not a good day for me. Incnis Mrsi (talk) 08:32, 18 January 2014 (UTC)

Vector triple product[edit]

The conflict above was created from my desire to fix this deficient section by providing valid explanations. There is no proof of Lagrange's formula. There was only a paragraph that restated the identity with permuted letters and claimed that it is true due to antisymmetry:

It was not clear, derived from which clauses. I commented it out because one can’t transform ×s into ⋅s on the grounds of antisymmetry of the former only. After my small investigation it appeared that Sheepe2004 altered the sense of the paragraph in such way that its initial point (there is an equivalent reformulation of Lagrange's formula) became lost. In any case, even the original version didn’t explain how is Lagrange's formula derived.

The only pre-existing explanation I found is a line in Triple product #Tensor calculus, and it is not especially clear. It states some contraction identity, but does not explain whether is it merely a computational fact or an application of some general rule.

After all this I wrote a new section that served for two purposes. First, it explained why one should expect namely pseudoscalars and vectors, not other types of objects, from a triple product. Second, it attempted to derive Lagrange's formula from Clifford algebra, and I claim it actually derived it up to sign. But it was immediately attacked by JohnBlackburne (see above). What to do now? Will anybody fix the lame section in some way other than I proposed? Incnis Mrsi (talk) 08:02, 18 January 2014 (UTC)

I've just tried putting in a section which does it properly, in I think the best location. First I used the left contraction for the formula, i.e.
\mathbf{a} \;\big\lrcorner\; (\mathbf{b} \wedge \mathbf{c})
not
\mathbf{a} \cdot (\mathbf{b} \wedge \mathbf{c}) or \mathbf{a} \times (\mathbf{b} \wedge \mathbf{c})
which have the problem that 'dot' is only well defined for two vectors while 'cross' is only really defined for two bivectors. It's also unclear how you get from the vector cross product to the GA cross product. The left contraction I think is much more obvious, though I don't have a source for this – the vector triple product doesn't seem to get much coverage anywhere. But the left contraction is both mathematically right and intuitively correct (if you think of the contraction geometrically), and sourced.
After that the proof follows from the characteristic properties of the left contraction (see the other source), and is identical to the result for the cross product. This is an even simpler proof too. I therefore removed the later 'Geometric algebra' section. The only other thing it included was a brief statement of various geometric algebra facts which are explained more fully at that article if readers need some background.--JohnBlackburnewordsdeeds 07:31, 21 January 2014 (UTC)