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