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
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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
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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
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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
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>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
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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
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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
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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
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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
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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
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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
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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
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>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)
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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)