Talk:Multiple cross products

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Is this text original to wikipedia? Could the mass of HTML be explained by copyvio, or just over-ambitious layout? — Sverdrup 19:49, 17 Mar 2004 (UTC)

Looks like format problems to me. Probably this page can be turned into something comprehensible by noting the way to write multiple cross products by means of the exterior algebra and Hodge star operator.

Charles Matthews 19:54, 17 Mar 2004 (UTC)

I'm not an expert, but I thought of Einstein notation — Sverdrup 19:55, 17 Mar 2004 (UTC)

Basically the same thing: the star raises or lowers the index.

Charles Matthews 19:57, 17 Mar 2004 (UTC)

Is this useful at all? It describes how to express a cross product of n terms as a sum of 2n-1 terms. Just multiplying one cross product at a time would be much faster. Andris 00:42, 18 Mar 2004 (UTC)

I'd like to finish thinking about what is said, before coming to a conclusion. Structure matters, not just computation. Charles Matthews 08:12, 18 Mar 2004 (UTC)

Probably this should all be just an application of a standard identity for the Levi-Civita symbol.

Charles Matthews 21:59, 6 Nov 2004 (UTC)

Warehousing long discussion from the page. Charles Matthews 17:16, 26 Nov 2004 (UTC)

Something like

A×((B×C)×D)

might be calculated by utilising the determinant pattern three times. Doing the determinants over and over is the complicated way: they can be worked out directly in the following way.

A = (a1, a2, a3) = a<1>.

The last term means each component has an a, the first component is numbered 1 and the numbers go up 1 for each component.

So

A X B = (a2.b3 - a3.b2 , a3.b1 - a1.b3 , a1.b2 - a2.b1) = ab<23-32>

That is, every term in every component has an a and b in it so they go out the front in order, then you use the pointed bracket to show how the index numbers are distributed, and an up arrow to show how to get from the first component to the second to the third. Addition cycles around the index numbers:

1+1 = 2 , 2+1 = 3, 3+1 = 1.

That is, 1-2-3 is considered a cyclic order.

Repeated crossing on the right produces a sequence of vectors

A , A X B, (A X B) X C , ((A X B)X C)X D, ...

where

A = a<1>¤
A X B = ab<23 - 32>¤
(A X B) X C = abc< 313 - 133 - 122

+212>¤

( A X B) X C) X D = abcd< 1213 -

2113 - 2333 + 3233 - 2322 + 3222 + 3112 - 1312>¤

Aside from being a way of avoiding the need to write the other components, the pointy bracket's main purpose is to facilitate a straightforward operation of converting a vector into a vector crossed on the right, in other words we are going to use it to find an operation R such that:

R(A) = A X B, R(A X B) = (A X B) X C , etc.,

and the R( ) that turns out to work is the one that adds an additional letter on the right just before the pointy bracket, lowers each index number of each index term by 1 and adds an additional term of -2 to each term on the right, then raises each index number of each index term by 1 and puts a 3 at the end of each term. Denote it by

R( ) = 3/-2]( )

so the lower side of the slash is the number you put at the end of the lowered index terms and the upper is the one for the raised terms accordingly.

An example to check this all.

R(A X B )
= 3/-2](A X B )
=3/-2](ab<23 - 32>¤)
=abc 3/-2](<23 - 32>¤)
=abc<31 3- 13 3+ -12 2- - 21 2>¤
=313 - 133 - 122 + 212 >¤
= (A X B) X C