Problem: A triple product identity Suppose A, B, and C are vectors in R^3 and we define A' = B x C B' = C x A C' = A x B Then we have the triple product identity A' . (B' x C') = ( A . (B x C) )^2. or in symmetric notation [A',B',C'] = [A,B,C]^2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - From - Mon Jul 14 14:48:40 1997 From: Christian Ohn Newsgroups: sci.math.research Subject: Re: A triple product identity Date: 13 Jul 1997 07:25:06 GMT Organization: Brussels Free Universities VUB/ULB Lines: 12 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: <5q9vsi$dhi@rc1.vub.ac.be> References: <5q68a2$535@schauder.mit.edu> Originator: dan@orion.math.uiuc.edu John Baez wrote: : Suppose A, B, and C are vectors in R^3 and we define : A' = B x C B' = C x A C' = A x B : Then if I'm not mistaken, the triple product A' . (B' x C') is the : square of the triple product A . (B x C). Using the identity X x (Y x Z) = (X.Z) Y - (X.Y) Z, it takes one line: (BxC).{(CxA)x(AxB)} = (BxC).{[(CxA).B]A-[(CxA).A]B} = [(CxA).B][(BxC).A] Christian From - Mon Jul 14 14:49:54 1997 From: hammond@csc.albany.edu (William F. Hammond) Newsgroups: sci.math.research Subject: Re: A triple product identity Date: 13 Jul 1997 18:52:16 +0000 Organization: Dept of Math & Stat, SUNYA, Albany, NY Lines: 40 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: References: <5q68a2$535@schauder.mit.edu> Originator: dan@orion.math.uiuc.edu In article <5q68a2$535@schauder.mit.edu> baez@math.mit.edu (John Baez) writes: > Suppose A, B, and C are vectors in R^3 and we define > A' = B x C > B' = C x A > C' = A x B > Then if I'm not mistaken, the triple product A' . (B' x C') is the > square of the triple product A . (B x C). I checked this by brute > force, but if it's true there should be some elegant way to show it. Let f(A, B, C) = {A, B, C} = A . (B x C) g(A, B, C) = {BxC, CxA, AxB}. It is, I trust, familiar to all that f is characterized, over any field of scalars up to a multiplicative constant as a non-degenerate alternating tri-linear form. And we might as well ask whether the identity g = f^2 holds for all complex vectors A, B, C, the point being that we can invoke Hilbert's NullStellenSatz in that case. It is easy to see that g is homogeneous of degree 2 in each variable and that g is symmetric. Moreover, it is easy to check that g vanishes whenever f does, i.e., whenever the subspace of C^3 generated by A, B, C has dimension less than 3 since then the subspace of C^3 generated by A', B', C' has dimension no greater than 1. So by Hilbert's NullStellenSatz (since f is an *irreducible* homogeneous polynomial of degree 3 in 9 variables), g is divisible by f. And the quotient h = g/f is a polynomial of degree 3 in 9 variables that is homogeneous of degree 1 in each variable, i.e., h is tri-linear. Moreover, g symmetric, f skew-symmetric --> h skew-symmetric --> (in characteristic zero) h alternating. Hence, h = c f, for some constant c. One checks that c = 1 by looking at the standard basis I, J, K. -- Bill Hammond From - Mon Jul 14 22:44:21 1997 From: davidc@fly.dpmms.cam.ac.uk (David Chatterjee) Newsgroups: sci.math.research Subject: Re: A triple product identity Date: 14 Jul 1997 12:04:10 GMT Organization: Cambridge University, Pure Mathematics and Mathematical Lines: 57 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: <5qd4jq$5d8@lyra.csx.cam.ac.uk> References: <5q68a2$535@schauder.mit.edu> Originator: dan@orion.math.uiuc.edu >In article <5q68a2$535@schauder.mit.edu> baez@math.mit.edu (John Baez) writes: > >> Suppose A, B, and C are vectors in R^3 and we define > >> A' = B x C >> B' = C x A >> C' = A x B > >> Then if I'm not mistaken, the triple product A' . (B' x C') is the >> square of the triple product A . (B x C). I checked this by brute >> force, but if it's true there should be some elegant way to show it. So far I've seen several replies using the same vector identity (and one using more sophisticated language from Bill Hammond). You might like this simple duality-based proof that doesn't require remembering the formula: perhaps this counts as elegant, perhaps not. Instead of vectors we use forms. The question of course implies we have a metric, and thus a volume 3-form, vol. By *a I mean the Hodge dual of the p-form a, ie. such that a ^ *a = vol . A scalar triple product of three vectors amounts to the number * ( a ^ b ^ c ) coming from three 1-forms (ie. the volume spanned by the three). The problem restated is to show that for three 1-forms a, b, c, it holds that the square of this volume ( * ( a ^ b ^ c ) )^2 ($) equals the number * ( *(b^c) ^ *(c^a) ^ *(a^b) ) ($) ie. the volume spanned by a', b', c' dual to the vectors of the question. (Sorry for using ^ as both exterior multiplication and the power two.) But this is obvious: first suppose that a, b, c are orthonormal (and ordered so that vol = a ^ b ^ c ). Then for instance *a = b^c *(b^c) = a and we find in short order that both sides of ($) equal one. More generally, both sides are quadratic in each term a, b, c and we are surely done. (But I imagine the original "brute force" proof looked much the same. And you could argue that there's nothing here that isn't in the Hammond nullstellensatz proof.) David Chatterjee. From - Tue Jul 15 10:41:29 1997 From: baez@math.mit.edu (John Baez) Newsgroups: sci.math.research Subject: Re: A triple product identity Date: 14 Jul 1997 10:34:31 -0400 Organization: MIT Department of Mathematics Lines: 36 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: <5qdddn$b66@schauder.mit.edu> References: <5q68a2$535@schauder.mit.edu> <5q9vsi$dhi@rc1.vub.ac.be> Originator: dan@orion.math.uiuc.edu In article <5q9vsi$dhi@rc1.vub.ac.be>, Christian Ohn wrote: >John Baez wrote: >: Suppose A, B, and C are vectors in R^3 and we define >: A' = B x C B' = C x A C' = A x B >: Then if I'm not mistaken, the triple product A' . (B' x C') is the >: square of the triple product A . (B x C). >Using the identity X x (Y x Z) = (X.Z) Y - (X.Y) Z, it takes one line: > >(BxC).{(CxA)x(AxB)} = (BxC).{[(CxA).B]A-[(CxA).A]B} = [(CxA).B][(BxC).A] This is basically the "brute force" method that I used, although I write pretty big so it took me more than one line. I had been hoping for either a purely geometrical proof or a proof using the fact that the triple product is a determinant --- or both, since the determinant is just the volume of a parallipiped. Yue Hu showed via email: we may assume A,B,C linearly independent, and let V = A . (B x C) and V' = A' . (B' x C'). Then the bases (A,B,C) and (A',B',C') are dual up to a factor of V: A' . A = V, A' . B = 0, A' . C = 0 and so on. We can write all these equations as one matrix equation: ( A'1 A'2 A'3 ) ( A1 B1 C1 ) ( B'1 B'2 B'3 ) ( A2 B2 C2 ) = V Id ( C'1 C'2 C'3 ) ( A3 B3 C3 ) Taking the determinant, the desired result follows. One can also reinterpret this as a geometrical proof. From - Tue Jul 15 20:48:40 1997 From: Timothy Murphy Newsgroups: sci.math.research Subject: Re: A triple product identity Date: 14 Jul 1997 17:56:36 +0100 Organization: Dept. of Maths, Trinity College, Dublin, Ireland. Lines: 42 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: <5qdlo4$8d4@salmon.maths.tcd.ie> References: <5q68a2$535@schauder.mit.edu> Originator: dan@orion.math.uiuc.edu baez@math.mit.edu (John Baez) writes: >Suppose A, B, and C are vectors in R^3 and we define >A' = B x C >B' = C x A >C' = A x B >Then if I'm not mistaken, the triple product A' . (B' x C') is the >square of the triple product A . (B x C). I checked this by brute >force, but if it's true there should be some elegant way to show it. >Does anyone know one? Doesn't it follow fairly easily from the fact that all invariant tensors in the non-cartesian case (ie invariant under GL(n,k)) can be constructed from the epsilon tensors and the delta tensor (as shown eg in Weyl "The Classical Groups") ? It follows that an invariant tensor of type (0,r) must be a linear combination of tensor products of epsilon tensors. Now if more generally you take B' = C_1 x A, C' = A_1 x B_1, the triple product must be a sum of products of triple products (like [A,A_1,B][B_1,C,C_1]). When A_1 = A, etc, the only product which does not vanish is [A,B,C][A,B,C]. Hence the given triple product must be a scalar multiple of [A,B,C]^2. Taking the special case A=i, B=j, C=k it follows that the scalar multiple is 1. -- Timothy Murphy e-mail: tim@maths.tcd.ie tel: +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland From - Tue Jul 15 20:49:01 1997 From: Timothy Murphy Newsgroups: sci.math.research Subject: Re: A triple product identity Date: 14 Jul 1997 18:32:18 +0100 Organization: Dept. of Maths, Trinity College, Dublin, Ireland. Lines: 36 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: <5qdnr2$9qv@salmon.maths.tcd.ie> References: <5q68a2$535@schauder.mit.edu> Originator: dan@orion.math.uiuc.edu baez@math.mit.edu (John Baez) writes: >Suppose A, B, and C are vectors in R^3 and we define >A' = B x C >B' = C x A >C' = A x B >Then if I'm not mistaken, the triple product A' . (B' x C') is the >square of the triple product A . (B x C). I checked this by brute >force, but if it's true there should be some elegant way to show it. >Does anyone know one? I gave a slightly inaccurate answer to this just now. I should have said that it follows from the fact that the only tensors invariant under SL(n,R) [not GL(n,R)] are those constructible from the epsilon tensors and the delta tensor. Hence the only such tensors of type (0,r) (ie multilinear maps from r vectors to the scalars) are linear combinations of products of epsilon tensors. It follows that the triple product in question must be a scalar multiple of [A,B,C]^2. -- Timothy Murphy e-mail: tim@maths.tcd.ie tel: +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland From - Tue Jul 15 20:49:15 1997 From: ksbrown@seanet.com (Kevin Brown) Newsgroups: sci.math.research Subject: Re: A triple product identity Date: Tue, 15 Jul 1997 00:58:45 GMT Organization: Seanet Online Services, Seattle WA Lines: 24 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: <33cabe2a.52075@news.seanet.com> References: <5q68a2$535@schauder.mit.edu> Originator: dan@orion.math.uiuc.edu On 11 Jul 1997 17:24:18 -0400, baez@math.mit.edu (John Baez) wrote: > Suppose A, B, and C are vectors in R^3 and we define > A' = B x C B' = C x A C' = A x B > Then if I'm not mistaken, the triple product A' . (B' x C') is > the square of the triple product A . (B x C). I checked this > by brute force, but if it's true there should be some elegant > way to show it. Does anyone know one? The triple product A.(BxC) corresponds to the volume of a parallelepiped with adjacent edges A,B,C. The volume of this solid is V = |A||B||C| S^(2/3) S'^(1/3) where S is the product of the sines of the pairwise angles between the vectors A,B,C, and S' is the product of sines of the dual vectors A',B',C'. Likewise the volume of the dual parallelepiped is V' = |A'||B'||C'| S'^(2/3) S^(1/3) The magnitudes of the primed vectors are |B||C|sin(B_C), |A||C|sin(A_C), and |A||B|sin(A_B), so we have V' = (|A||B||C|)^2 S'^(2/3) S^(4/3) = V^2 From - Tue Jul 15 20:49:33 1997 From: checker@netcom.com (Chris Hecker) Newsgroups: sci.math.research Subject: Re: A triple product identity Date: Tue, 15 Jul 1997 07:15:52 GMT Organization: Netcom On-Line Services Lines: 24 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: References: <5q68a2$535@schauder.mit.edu> <33c8101f.32386107@news.seanet.com> Originator: dan@orion.math.uiuc.edu ksbrown@seanet.com (Kevin Brown) writes: >It's certainly true, as it follows from the ubiquitous triple product >identity > Ax(BxC) = B(A.C)-C(A.B) >I don't know of a particularly elegant proof of this identity - it's >usually just derived in texts by crunching it out, component-wise, in >a few lines. The book Dynamic Analysis of Robot Manipulators, a Cartesian Tensor Approach (I think), by Balafoutis and Patel has a proof of this that's not a grind, but I'm not sure I'd call it elegant. I can't remember the specifics and the book is at work, but it's based on the identity: a'b' = b tensor a - 1 a.b where ' is the skew-symmetric operator (is this the Hodge star in exterior calc? I haven't read that book yet ;). This identity is based on a few other identities that I can't remember. Anyway, I don't think they drop down to indices, but they might have had to do some icky proofs with the Levi-Civita alternating tensor and the Kronecker delta. Like I said, I don't think it was elegant. Chris From - Wed Jul 16 11:28:57 1997 From: Harry Gaines Newsgroups: sci.math.research Subject: Re: A triple product identity Date: Tue, 15 Jul 1997 10:16:37 -0400 Organization: Honeywell Inc. Lines: 25 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: <33CB8645.6F76@stpete.honeywell.com> References: <5q68a2$535@schauder.mit.edu> Originator: dan@orion.math.uiuc.edu John Baez wrote: > > Suppose A, B, and C are vectors in R^3 and we define > > A' = B x C > B' = C x A > C' = A x B > > Then if I'm not mistaken, the triple product A' . (B' x C') is the > square of the triple product A . (B x C). I checked this by brute > force, but if it's true there should be some elegant way to show it. > Does anyone know one? The representation of dot and cross products that lends itself best to such algebraic manipulations employs index notation of tensor analysis including the Kronecker(sp?) delta and the permutation symbol. In 3-D there are only two identities to learn: contracting two indices of a product of two permutation symbols gives twice a Kronecker delta; and, contracting on one gives the difference of two products of two deltas. I'm not familiar with current books on tensor analysis but the classic _Applications of . . . by McConnell is probably still available from Dover. See Chapter I. Harry From - Wed Jul 16 11:29:23 1997 From: wcw@math.psu.edu (William C Waterhouse) Newsgroups: sci.math.research Subject: Re: A triple product identity Date: 15 Jul 1997 18:26:29 GMT Organization: Penn State, Department of Mathematics Lines: 28 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: <5qgfcl$ith$1@dodgson.math.psu.edu> References: <5q68a2$535@schauder.mit.edu> Reply-To: wcw@math.psu.edu Originator: dan@orion.math.uiuc.edu In article <5q68a2$535@schauder.mit.edu>, baez@math.mit.edu (John Baez) writes: > Suppose A, B, and C are vectors in R^3 and we define > > A' = B x C > B' = C x A > C' = A x B > > Then if I'm not mistaken, the triple product A' . (B' x C') is the > square of the triple product A . (B x C). I checked this by brute > force, but if it's true there should be some elegant way to show it. > Does anyone know one? Let f be the linear transformation whose matrix (in basis i,j,k) has columns A,B,C. The "triple product" then is the determinant. The cross products involve various 2 by 2 minors, and it's easy to check that the matrix with columns A', B', C' is in fact the matrix of the second exterior power of f (in basis j ^ k, k ^ i, i ^ j). Thus we're asking about the determinant of the second exterior power. Now the exterior power takes composites to composites, so the function sending f to the det of its second exterior power is multiplicative. It thus has to be a power of the original determinant. You can tell which power it is by testing on f = cI. William C. Waterhouse Penn State From - Thu Jul 17 13:43:23 1997 From: petry@accessone.com (David Petry) Newsgroups: sci.math.research Subject: Re: A triple product identity Date: Wed, 16 Jul 1997 15:29:51 GMT Organization: AccessOne Lines: 33 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: <5qisvc$knb$1@kanga.accessone.com> References: <5q68a2$535@schauder.mit.edu> <5q936g$oar$1@news.fas.harvard.edu> Originator: dan@orion.math.uiuc.edu >In article <5q68a2$535@schauder.mit.edu>, John Baez wrote: >>Suppose A, B, and C are vectors in R^3 and we define >>A' = B x C >>B' = C x A >>C' = A x B >>Then if I'm not mistaken, the triple product A' . (B' x C') is the >>square of the triple product A . (B x C). I checked this by brute >>force, but if it's true there should be some elegant way to show it. >>Does anyone know one? Here's a rather elegant way to show it. The triple product of three vectors A,B,C is just the determinant of the matrix M(A,B,C) which has columns equal to A, B and C. It's easy to show that M(A', B', C') = transpose (1/M(A,B,C)) det(M(A,B,C)), from which the desired result follows immediately. This generalizes to higher dimensions. Let the dimension be N. Instead of the vector product of two vectors, we use the vector product of N-1 vectors, and instead of the triple product, we use the N-product of N vectors, which is just the determinant of the matrix with each column equal to one of the vectors. Then given N vectors, we can form N vector products, each vector product being the product of all the vectors except one. Then in English, we have the following identity: The N-product of the N vector products is equal to the N-product of the N vectors raised to the N-1 power. From - Fri Jul 18 10:48:32 1997 From: jrr@atml.co.uk (John Rickard) Newsgroups: sci.math.research Subject: Re: A triple product identity Date: 17 Jul 1997 13:15:38 +0100 (BST) Organization: ATM Ltd Lines: 31 Approved: Daniel Grayson , moderator for sci.math.research Message-ID: References: <5q68a2$535@schauder.mit.edu> X-Server-Date: 17 Jul 1997 12:15:41 GMT Originator: dan@orion.math.uiuc.edu John Baez (baez@math.mit.edu) wrote: : Suppose A, B, and C are vectors in R^3 and we define : : A' = B x C : B' = C x A : C' = A x B : : Then if I'm not mistaken, the triple product A' . (B' x C') is the : square of the triple product A . (B x C). I checked this by brute : force, but if it's true there should be some elegant way to show it. I don't know if this is equivalent to any of the other solutions posted. Note that, provided one distinguishes between covariant and contravariant vectors -- so that the cross product of two vectors of the same variance has the opposite variance -- everything here depends only on the volume element and not on the metric (written in tensor notation, there are epsilon_ijk and epsilon^ijk but no g_ij or g^ij). Thus A' . (B' x C') is invariant under any volume-preserving linear transformation. If A, B, and C are linearly independent, then there is a volume-preserving linear transformation taking A, B, C to A1, B1, C1 provided A . (B x C) = A1 . (B1 x C1); thus, in the linearly independent case, A' . (B' x C') is a function of A . (B x C): dimensionality shows that the former is proportional to the square of the latter, and checking a single case determines the constant of proportionality. The case where A, B, C are linearly dependent follows by continuity, of course. -- John Rickard Subject: Cubic Acres and Last Month's Most Interesting Footnote Date: 17 Sep 1998 22:42:16 GMT From: weemba@sagi.wistar.upenn.edu (Matthew P Wiener) K R Brownstein "Cubic Acres" AM J PHYS 66#8, Aug '98, p739 contains a proof that given a parallelepiped, the scalar triple product of the three _area_ vectors is, up to sign, the volume squared. (As is well-known, the scalar triple product of the edge vectors is, up to sign, the volume.) The author concludes by pointing out that Scrooge McDuck often described the wealth in his vaults in terms of "cubic acres". For the incredulous reader, he includes a footnoted reference: [^2] Walt Disney, Uncle Scrooge Four Color #456 (Uncle Scrooge #2) (Western Publishing, Racine WI, 1953), p2. Brownstein conjectures Scrooge's wealth is thus actually the square of what every one has thought it was all along. -- -Matthew P Wiener (weemba@sagi.wistar.upenn.edu)