From: lounesto@dopey.hut.fi (Pertti Lounesto)
Newsgroups: sci.math
Subject: Re: Cross-product in n-dimensions (n>3)? --corrections
Date: 22 Nov 1995 09:03:16 GMT
Ramakrishna Kakarala writes:
The N-d cross-product [x,y] should have the following properties for any
two vectors N-d vectors x and y:
1) linear in both variables: [ax+by,z]=a[x,z]+b[y,z]
[z,ax+by]=a[z,x]+b[z,y]
2) anti-commutative: [x,y]=-[y,x]
3) orthogonality = = 0
4) length: ||[x,y]||^2 = ||x|| ^2 ||y||^2 { 1 - ^2}
If, in addition, [x,y] satisfied the Jacobi condition
[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0
then [x,y] is a Lie algebra bracket that "respected" the norm
of the underlying vector space, as stated in 3) and 4). Is there
such a thing for dimensions n>3?
Ram Kakarala
If means the cosine between the directions x,y, then the answer to
the question is yes, in n=7. [If means the symmetric scalar product,
then the answer is no.] David Ullrich's answer is incorrect; he answered
a modified question posed by himself [Ram Kakarala's corrected question
shows that he intended to have a product of two vectors]. B. Silverman's
answer was correct, but not explicit. Dave Rusin's answer, supporting
Ullrich, was incorrect. If the question is modified in the sense of
Ullrich&Rusin, then the answer of Ulrich&Rusin is incomplete. In the
following I will give the correct and exlicit answer to the original
question, and the complete answer to the modified question [of David
Ullrich and Dave Rusin]. It seems that this question pops up monthly,
and is incorrectly/incompletely answered each time.
There is a cross product of two vectors, satisfying all the usual
assumptions, only in dimensions 3 and 7. In dimension 7 one can
define for e1, e2, ..., e7
e1 x e2 = e4, e2 x e4 = e1, e4 x e1 = e2
e2 x e3 = e5, e3 x e5 = e2, e5 x e2 = e3
.
.
e7 x e1 = e3, e1 x e3 = e7, e3 x e7 = e1
and ei x ej = -ej x ei. The above multiplication table can be condensed
into the form
e_i x e_{i+1} = e_{i+3}
where the indices i,i+1,i+3 are permuted cyclically among themselves
and computed modulo 7.
This cross product of two vectors in R^7 satisfies the usual rules
(a x b).a = 0, (a x b).b = 0 orthogonality
(a x b)^2 + (a.b)^2 = a^2 b^2 Pythagoras/Lagrange
where the second rule can also be written as |a x b| = |a| |b| sin(a,b).
A cross product of two vectors satisfying the above rules (orthogonality
and Pythagoras/Lagrange) is unique to dimensions 3 and 7; it does not
exist in any other dimensions. It can be also defined by quaternion
or octonion products as a x b = _1 where _1 means taking the
1-vector part (or pure=imaginary part) of the product ab of two vectors
a and b in R^n, n=3 or n=7, with R+R^n being either the quaternions H
or the octonions O.
Contrary to the usual 3-dimensional cross product this 7-dimensional
cross product does not satisfy the Jacobi identity
(a x b) x c + (b x c) x a + (c x a) x b = 0
(but satisfies the so called Malcev identity, a generalization of Jacobi).
The 3-dimensional cross product is invariant under all rotations of SO(3),
while the 7-dimensional cross product is not invariant under all of SO(7),
but only under the exceptional Lie group G_2, a subgroup of SO(7).
In R^3 the direction of a x b is unique, up to orientation having two
possibilities, but in R^7 the direction of a x b depends on vectors
defining the cross product; namely (expressed with the contraction "_|"):
a x b = (a ^ b) _| e123 in R^3 when e123 = e1^e2^e3 but
a x b = (a ^ b) _| (e124+e235+e346+e457+e561+e672+e713) in R^7.
Also in R^7 there are other planes than the linear span of a and b
giving the same direction as a x b.
The image set of the simple bivectors a ^ b, where a,b in R^7, is a
manifold of dimension 2.7-3=11 > 7 in the linear space of bivectors
(of dimension 7(7-1)/2=21) while the image set of a x b is just R^7.
So the "identification"
a x b = (a ^ b) _| (e124+e235+e346+e457+e561+e672+e713)
is not a 1-1 correspondence (just a method of associating a vector to a
simple bivector).
Many people neglect the above product when they give their advices to
people asking about the existence of a cross product of two vectors in
higher dimensions, and modify the question by replying to a question
about the existence of a product of k > 2 vectors in higher dimensions
(and give an incomplete answer to their own question).
If we were looking for a vector valued product of k factors, not
just two factors, then one should first try to modify or formalize the
Pythagoras/Lagrange theorem for k factors. A natural thing to do is
to consider a vector valued product a1 x a2 x ... x ak satisfying
(a1 x a2 x ... x ak).ai = 0 orthogonality
(a1 x a2 x ... x ak)^2 = det (ai.aj) Gram determinant.
In this context an answer to the question "in what dimensions there is a
generalization of the cross product" is that there are cross products in
3 dimensions with 2 factors
7 dimensions with 2 factors
n dimensions with n-1 factors
8 dimensions with 3 factors
and no others (except if one allows trivial answers, then there would
also be in all even dimensions a vector product with only 1 factor and
in 1 dimension an identically vanishing cross product of 2 factors).
--
Pertti Lounesto Triality is quadratic