A triple product identity
John Baez
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?
Noam Elkies
Depends what you mean by "elegant". It can certainly be done
quickly via the identity A x (B x C) = (A.C) B - (A.B) C:
B' x C' = B' x (A x B) = (B'.B) A - (B'.A) B = [ABC] A
(using the standard shorthand [ABC]=det(A,B,C)=A.(BxC), and noting
that B'.A=0 because B' = A x something), from which it follows that
A' . (B' x C') = [ABC] A.A' = [ABC]^2.
But there's surely an even more conceptual proof if you admit
concepts such as duality and exterior powers.
--Noam D. Elkies (elk...@math.harvard:edu)
Dept. of Mathematics, Harvard University
Kevin Brown
On 11 Jul 1997 17:24:18 -0400, ba...@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?
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. Anyway, this leads directly to the handy formula for the
cross product of two cross products
(AxB)x(CxD) = B[(CxD).A] - A[(CxD).B]
Using this, your quantity can be expressed as
A'.(B' x C') = (BxC).((CxA)x(AxB))
/ \
= (BxC).( A[(AxB).C] - C[(AxB).A] )
\ /
Remember that in general the quantity (AxB).C equals the volume of the
parallelepiped with the vectors A,B,C forming adjacent edges. From
this it's clear that (AxB).A = 0, because two of the three edges are
identical. Thus the right-most term in the preceeding expression
drops out, and your quantity is simply (BxC).A[(AxB).C], which, in
view of associativity of scalar multiplication over the dot product,
is the same as [(BxC).A][(AxB).C]. Both factors equal the volume of
a parallelepiped with edges A,B,C, so the product equals [A.(BxC)]^2.
Christian Ohn
John Baez <ba...@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).
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
William F. Hammond
> 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
ilias kastanas 08-14-90
@
@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?
Using U x (V x W) = (U.W) V - (U.V) W , B' x C' (which might
be called A'') is = ((C x A).B) A = (A.(B x C)) A; that does it. Of
course B'' = (A.(B x C)) B and C'' = (A.(B x C)) C too.
Ilias
David Chatterjee
>
>> B' = C x A
>> C' = A x B
>
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 ($) equal one. More
generally, both sides are quadratic in each term a, b, c and we are
surely done. (But I imagine the original "brute force" proof looked
much the same. And you could argue that there's nothing here that
isn't in the Hammond nullstellensatz proof.)
David Chatterjee.
John Baez
In article <5q9vsi$d...@rc1.vub.ac.be>, Christian Ohn <ch...@ulb.ac.be> wrote:
>John Baez <ba...@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).
>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.
Timothy Murphy
ba...@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: t...@maths.tcd.ie
tel: +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
Timothy Murphy
ba...@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.
Kevin Brown
> 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
Chris Hecker
ksb...@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
Harry Gaines
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
William C Waterhouse
In article <5q68a2$5...@schauder.mit.edu>, ba...@math.mit.edu
> 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
David Petry
>In article <5q68a2$5...@schauder.mit.edu>, John Baez <ba...@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.
>>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.
ws...@aicom.com
On 14 Jul 1997 17:56:36 +0100, Timothy Murphy <t...@maths.tcd.ie>
wrote:
>ba...@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.
>
In my opinion, the concept of tensor seems complicate the original
question. The "brute force" proof, in that case, is not less elegant!
wai-shun,
University of British Columbia
John Rickard
John Baez (ba...@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
Chuck Wegrzyn
If you are looking for a particularly elegant and easy to understand
proof, pick up a book on differential forms. The prove is very easy.
Chuck Wegrzyn
weg...@garbagedump.com
Chris Hecker wrote in article ...
>ksb...@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.
>
[snip]
>