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Higher dimensional cross products

  1. Mar 13, 2014 #1
    I've always heard that the cross product only exists in a well defined way for 3 and 7 dimensions. From my own reading I've found that a cross product in 3 dimensions is nothing more than the product of two quaternions with only pure imaginary components (that is, the real part is zero). Likewise, a seven dimensional cross product is simply the product of two octonions where again the real part is zero. But why can't this process simply continue? Why not construct the 16 component sedenians and define the 15 dimensional cross product to be the product of two sedenians with zero real parts? Is there some property of sedenians that disallows this?
     
    Last edited: Mar 13, 2014
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  3. Mar 13, 2014 #2

    micromass

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    Let's begin by defining what a cross product is supposed to be. A cross product is a function

    [tex]\times:\mathbb{R}^n\times\mathbb{R}^n\rightarrow\mathbb{R}^n[/tex]

    such that

    1) ##\times## is bilinear
    2) ##\mathbf{x}\cdot (\mathbf{x}\times\mathbf{y}) = (\mathbf{x}\times \mathbf{y})\cdot\mathbf{y} = 0##
    3) ##|\mathbf{x}\times\mathbf{y}|^2 = |\mathbf{x}|^2|\mathbf{y}|^2 - (\mathbf{x}\cdot \mathbf{y})^2##

    We don't demand anything about a Jacobi identity, which won't be satisfied anyway except in the case ##n=3##.
    As you know, in the cases ##n=3## and ##n=7##, there is such a cross product (but it might not be unique in the case ##n=7##). In particular, we can define the cross product by identifiying ##\mathbb{R}^7## with the imaginary octonions and then set

    [tex]\mathbf{x}\times \mathbf{y} = \textrm{Im}(\mathbf{x}\mathbf{y}) = \frac{1}{2}(\mathbf{x}\mathbf{y} - \mathbf{y}\mathbf{x}).[/tex]

    Something similar works with ##n=3## and the imaginary quaternions.
    Now, why doesn't it work with ##n=15## and the imaginary sedenions? Well, sedenions do not form a division ring. Even worse, there are nonzero sedenions whose product is zero. For example,
    [tex](e_3 + e_{10})(e_6 - e_{15}) = 0[/tex]
    With this, it is easy to see that the third property is not satisfied. Indeed, we set ##\mathbf{x} = e_3 + e_{10}## and ##\mathbf{y} = e_6 + e_{15}##, which are imaginary sedenions.
    [tex]\mathbf{x}\times \mathbf{y}= \mathbf{0},~\mathbf{x}\cdot \mathbf{y} = 0[/tex]

    Now, this of course only shows that our naive choice of cross product will not work, but perhaps there is some other choice that does work. This can be proven not to be the case. Indeed, if ##\times## is a cross product on ##\mathbb{R}^n##, then it can be proven that ##\mathbb{R}^{n+1}## is a normed division algebra by setting
    [tex](a,\mathbf{x})(b,\mathbf{y}) = (ab-\mathbf{x}\cdot \mathbf{y}, a\mathbf{y} + b\mathbf{x} + \mathbf{x}\times \mathbf{y})[/tex]
    But a famous theorem by Hurwitz shows that the only normed division algebras are ##\mathbb{R}##, ##\mathbb{C}##, the quaternions and the octonions. See http://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(normed_division_algebras [Broken])
     
    Last edited by a moderator: May 6, 2017
  4. Mar 13, 2014 #3
    Thanks so much. That makes it abundantly clear.
     
  5. Mar 14, 2014 #4

    mathwonk

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    In another sense, there is in fact a cross product in every dimension, satisfying a generalization of all 3 stated properties, provided one allows more than 2 factors in the product. I.e. for vectors in n space, it is a product of n-1 vectors. This is discussed on pages 84-85 of Spivak's Calculus on Manifolds. The product is essentially a choice of a vector normal to the hyperplane spanned by the given n-1 arguments, or zero if they are not independent. It has direction chosen to give the usual orientation of n space, when combined with the argument sequence, and length chosen to satisfy the analog of property 3.
     
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