From: Ramakrishna Kakarala Newsgroups: sci.math Subject: Cross-product in n-dimensions (n>3)? Date: 19 Nov 1995 23:11:30 GMT Is there a N-dimensional version of the cross-product that is usually defined for 3-D vectors? In particular, if x and y are any two N-d unit vectors, I am looking for a vector operation [x,y] such that = = 0 || [x,y] || = 1 - ^2 with <> denoting the inner product. Does such an operation exist? Any help on this is appreciated. -- Ramakrishna Kakarala, Ph.D. Lecturer Department of Electrical and Electronic Engineering University of Auckland Private Bag 92019 Auckland, New Zealand Ph. +64-9-373-7599 x 8198 Fax: +64-9-373-7461 E-mail: r.kakarala@auckland.ac.nz ============================================================================== From: David Ullrich Newsgroups: sci.math Subject: Re: Cross-product in n-dimensions (n>3)? Date: 21 Nov 1995 18:30:32 GMT >Why? Because the only finite division algebras which have a dimension >sufficiently large to support a cross product at all have order 4 and 8 >and there are no others. The cross-product must have dimension 1 less >than that of the division algebra. If I can scrape the time together, >I will put together a fuller explanation of this. Uh, we're talking about different objects here. I'm guessing that your "cross product" is the cross product of two vectore - if you want to say that there's one of those in R3 and in R7 and nowhere else fine. Seems to me that one of the most useful aspects of the cross product in R3 is the fact that u X v is orthogonal to u and to v. There certainly is an (n-1)-linear operator X in Rn such that X(u_1,...u_(n-1)) gives a vector orthogonal to each of u_1 through u_(n-1). And you can write down a formula for it "just like" the "determinant" formula for the cross product in R3 - if e1, ... en is the standard basis for Rn then X(u_1,...,u_(n-1)) is the (formal) determinant of the nxn "matrix" that has the "vector" (e1,...en) for its first row and then has the u_j's for the other rows. -- David Ullrich Don't you guys find it tedious typing the same thing after your signature each time you post something? I know I do, but when in Rome... ============================================================================== From: David Shield Newsgroups: sci.math Subject: Re: Cross-product in n-dimensions (n>3)? Date: 23 Nov 1995 06:06:48 GMT bobs@mathworks.com (Bob Silverman) wrote: >In article <48q9nl$c82@news.cis.okstate.edu>, >David Ullrich wrote: >: It's not called the cross product, but the gizmo in n dimensions that's >:analogous to the cross product takes n-1 vectors and gives a vector orthogonal >:to all of them (you can set n=2 here as well). >: If you look at the formula for the cross product in 3 dimensions in terms >:of that determinant the obvious generalization does this. >: >Actually, the cross product as defined in 3 dimensions also exists in >7 dimensions and in NO OTHER. > >Why? Because the only finite division algebras which have a dimension >sufficiently large to support a cross product at all have order 4 and 8 >and there are no others. The cross-product must have dimension 1 less >than that of the division algebra. If I can scrape the time together, >I will put together a fuller explanation of this. > >-- >Bob Silverman >The MathWorks Inc. >24 Prime Park Way >Natick, MA I have thought of the generalisation of cross product as the operation which from two vectors (A1, A2, ... , An) and (B1, B2, ... , Bn) produces the n-by-n skew-symmetric matrix C with C(i,j) = AiBj - AjBi. This matrix has n(n-1)/2 _independent_ entries; and it is _only_ for n = 3 that this number is equal to n, so that the matrix can reasonably be interpreted as itself a vector in the original space. But this preserves different properties (e.g., A x B = - B x A) from the ones that originally interested Ramakrishna Kakerala. I look forward to finding out about the 7-D generalisation in Bob Silverman's fuller explanation. David Shield d.shield@bendigo.latrobe.edu.au ============================================================================== 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