[Note appended typo correction -- djr]
==============================================================================
From: Thomas Mautsch
Newsgroups: sci.physics,sci.math
Subject: Re: Cross Product in 7-d?
Date: Mon, 11 Nov 1996 15:28:22 +0100
Kevin Anthony Scaldeferri wrote:
>
> In article <55v9fu$etp@rzsun02.rrz.uni-hamburg.de>,
> Hauke Reddmann wrote:
> >Kevin Anthony Scaldeferri (coolhand@Glue.umd.edu) wrote:
> >: In article <55scd5$hoc@rzsun02.rrz.uni-hamburg.de>,
> >: Hauke Reddmann wrote:
> >: >"True" cross products only exists in dim 3+7.
> >:
> >: Could you explain why 7-d also works? (Also what exactly do you mean
> >: by "true" cross product?)
> >:
> >Ask on sci.math, I'm only a "trivia" expert ;-)
> >(Somehow this is connected to Stokes theorem. Another
> >trivia I remember...I believe in a recent back-issue
> >of Am.Math.Monthly something appeared)
>
> OK, dutifully cross posted to sci.math.
>
> So, math-types, anyone got an explaination for this little fact.
>
> --
> ======================================================================
> Kevin Scaldeferri University of Maryland
>
> "The trouble is, each of them is plausible without being instictive"
Some people go on repeating that in n-dimensional spaces
a cross product is only properly defined if you are given (n-1) vectors.
This might be true in general, but there ARE WAYS to define
a CROSS PRODUCT of TWO vectors a and b
provided that YOUR VECTOR SPACE has ``ENOUGH STRUCTURE''.
To see this let's first take a look at
why there is a cross product on 3-dimensional space:
IT'S BECAUSE FOUR-DIMENSIONAL SPACE CAN BE IDENTIFIED WITH THE
(much discussed) SPACE OF QUATERNIONS
(and 3-d space with the space of imaginary quaternions.
Suppose we are given an orthonormal basis of four vectors in 4-d space;
let us call them
1, i, j, and k.
Define a multiplication on the space by the multiplication table
. | 1 i j k
----------------
1 | 1 i j k
i | i -1 k -j
j | j -k -1 i
k | k j -i -1
Thus we define a multiplication on our 4-d space;
BUT CAREFUL: Multiplication depends on the order of factors:
i.j = k
but j.i = -k .
We've lost commutativity.
A general vector in 4-space (which we will call a quaternion)
is given by the linear combination
x = a . 1 + b . i + c . j + d . k
or simply
x = a + bi + cj + dk,
where a, b, c, d are real numbers (the coordinates of the vector x in
the
given basis 1,i,j,k).
The coefficient a is called its real part
a = Re(x)
and the remaining rest
bi + cj + dk=Im(x)
its imaginary part.
A conjugation is defined by
Conj(x) := Re(x) - Im(x) .
The space of imaginary vectors is just ordinary 3dimansional space and
we will
see that the cross product is simply derived from the product on the
quaternions as follows:
For quaternions x and y define the cross product x X y as
x X y = -1/2 Im( Conj(y).x ) .
On the imaginary part this restricts to
x X y = Im (x.y)
and is just the ordinary cross product
with e.g. the following properties:
(1) x X y is alternating: y X x = - x X y ;
(2) |x X y| = |x \wedge y| .
Now to the generalization to dimension 7 (or 8 respectively).
Here we need the OCTONIONIC STRUCTURE on 8-space:
Write an ortorgonal basis of 8-space as
1, i, j, k, l, il, jl, kl ;
write an arbitrary element (octonion) as
a + b.l
where a and b are quaternions and define a multiplication by
(a+b.l).(c+d.l) := (ac-Conj(d)b) + (da+b.Conj(c)).l .
This time we even loose associativity of the product,
e.g.:
(i.j).l = k.l = kl
i.(j.l) = i.(jl) = (ji).l = -kl .
Anyway, a cross product can be defined on the octonions by the same
formula as in the quaternionic case (the real part of a + b.l is just
Re(a),
the imaginary part the rest Im(a) + b.l).
And the restriction of this cross product to the imaginary octonions is
the one you are looking for.
Properties (1) and (2) still hold in the octonionic case.
A good introduction to octonionic structures can be found in sections
IV.A
and IV.B of the following article
R.Harvey, H.B.Lawson Jr.
Calibrated Geometries
Acta Mathematica 148, 1982
pp. 47 - 156
Hope this helps a bit.
Regards
Thomas
--
Thomas Mautsch
mautsch@mathematik.hu-berlin.de
==============================================================================
From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.physics,sci.math
Subject: Re: Cross Product in 7-d?
Date: Followup-To: sci.math
In article <574rsp$qtk$3@gruvel.une.edu.au>, ibokor@metz (ibokor) writes:
>Kevin Anthony Scaldeferri (coolhand@Glue.umd.edu) wrote:
>: In article <55v9fu$etp@rzsun02.rrz.uni-hamburg.de>,
>: Hauke Reddmann wrote:
>: >Kevin Anthony Scaldeferri (coolhand@Glue.umd.edu) wrote:
>: >: In article <55scd5$hoc@rzsun02.rrz.uni-hamburg.de>,
>: >: Hauke Reddmann wrote:
>: >: >"True" cross products only exists in dim 3+7.
>: >: Could you explain why 7-d also works? (Also what exactly do you mean
>: >: by "true" cross product?)
It has all the "correct" properties.
The product most simply described as the imaginary part of octonion
multiplication.
>: >Ask on sci.math, I'm only a "trivia" expert ;-)
>: >(Somehow this is connected to Stokes theorem. Another
>: >trivia I remember...I believe in a recent back-issue
>: >of Am.Math.Monthly something appeared)
W S Massey "Cross products of vectors in higher-dimensional Euclidean
spaces", AMER MATH MONTHLY, 90 (1983), #10, pp 697-701.
Massey's first theorem is that bilinear maps R^n x R^n -> R^n such that
the result is perpendicular to the factors and the norm of the result
is equal to the area of the parallelogram spanned by the factors is one
of these two products. His second theorem is if we instead assume the
product is continuous, keep perpendicularity as before, but only require
that the product of linearly independent nonzero vectors is nonzero, then
again, we have one of these two products.
>: OK, dutifully cross posted to sci.math.
>: So, math-types, anyone got an explaination for this little fact.
>The final proof of this is due to J.F.Adams. It is connected to the
>"Hopf Invariant 1" problem and so to the fact that the only real
>division algebras have dimension 1, 2, 4 and 8 (sort of).
Massey's theorems do not invoke much in the way of topology.
--
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
==============================================================================
From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener)
Newsgroups: sci.math,sci.physics
Subject: Re: cross products in 4 dimensions
Date: 13 Nov 1996 15:16:59 GMT
In article <56a9j2$l4h@sulawesi.lerc.nasa.gov>, Geoffrey A. Landis In article <55tclo$jsr@newsstand.cit.cornell.edu> Bryan W. Reed,
>breed@HARLIE.ee.cornell.edu writes:
>>In order to pick out a unique (to within minus signs) direction
>>that's perpendicular to all vectors in some linearly independent set
>>S, you need S to have n-1 elements (where n is the dimensionality).
>>That's why you need three vectors in 4 dimensions to form the "cross
>>product." Or whatever the generalization of the cross product is called.
>Cross products really only work in 3 dimensions.
Cross products of the kind described by Bryan Reed work in any dimension
3 and up.
Some consider that cheating. "Real" cross products, the two vectors at a
time sort, work in dimensions 3 *and* 7. The familiar 3-dimensional one
can be thought of as the purely imaginary part of quaternion multiplication.
Similarly, the purely imaginary part of octonion multiplication can be read
as a 7-dimensional cross product. Most of the familiar identities hold in
both cases.
See the nice little paper:
W S Massey "Cross products of vectors in higher-dimensional Euclidean
spaces", AMER MATH MONTHLY, 90 (1983), #10, pp 697-701.
Massey's first theorem is that bilinear maps R^n x R^n -> R^n such that
the result is perpendicular to the factors and the norm of the result
is equal to the area of the parallelogram spanned by the factors is one
of these two products. His second theorem is if we instead assume the
product is continuous, keep perpendicularity as before, but only require
that the product of linearly independent nonzero vectors is nonzero, then
again, we have one of these two products.
--
-Matthew P Wiener (weemba@sagi.wistar.upenn.edu)
==============================================================================
To: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: octonions
From: Pertti Lounesto
Date: 22 Feb 1998 18:07:04 +0200
[deletia -- djr]
at the article of From: Thomas Mautsch
Date: Mon, 11 Nov 1996 there is an error: x X y = -1/2 Im( Conj(y).x )
should be replaced by x X y = 1/2(Conj(y)x-Conj(x)y) = Im(Conj(y)x).
--
Pertti Lounesto http://www.hit.fi/~lounesto