Re: Multiplication and Division in Triplets; division by zero-sized vectors

From: Robin Chapman (rjc_at_ivorynospamtower.freeserve.co.uk)
Date: 10/14/04 

Date: Thu, 14 Oct 2004 10:35:35 +0100

Roger Beresford wrote:

> "I can only add and subtract [triplets]." (W.R.Hamilton, speaking to
> his son in the 1840's.)
>
> Triplet Multiplication.
> Define A={a1,a2,a3}, B={b1,b2,b3} and their triplet product AB={a1 b1
> +a3 b2 +a2 b3, a2 b1 +a1 b2 +a3 b3, a3 b1 +a2 b2 +a1 b3}, where a1
> etc. are real or complex numbers (this should also work with
> quaternion and octonion numbers, but I have not tested it). Triplet
> Multiplication uses the C3 group as a multiplication table.

So this is the group algebra of C_3 over R or C.

> Conserved properties.
> For any triplet X={x1,x2,x3}, the functions "Xs" and "Xp",
> Xs={x1+x2+x3}, Xp={((x1-x2)^2 +(x2-x3)^2 +(x3-x1)^2)/2}, are conserved
> properties (I call them "sizes") for triplet algebra, i.e. ABs =As Bs,
> ABp =Ap Bp.

If K is a field of characteristic zero, then the group algebra
K C_3 is isomorphic to K x K(omega) or K x K x K according
to whether K contains a root of x^2 + x + 1 = 0 or not. Here omega is
a primitive cube root of unity. Your maps X |--> X_s and X |--> X_p
are related to the projections of K C_3 onto the factors
of this decomposition. In all cases X_s is the projection of
X onto the first factor.

In the case where K = R, then X_p = pi(X) sigma(pi(X))
where pi is the projection from R C_3 to R(omega) = C
and sigma is complex conjugation: more explicitly
pi(x_1,x_2,x_3) = x_1 + x_1 omega + x_2 omega^2.

In the case where K = C, then X_p = pi_2(X) pi_3(X)
where pi_2 and p_3 are the projections from R C_3 to R(omega) = C
defined by
pi_2(x_1,x_2,x_3) = x_1 + x_1 omega + x_2 omega^2 and
pi_3(x_1,x_2,x_3) = x_1 + x_1 omega^2 + x_2 omega.

> Xs & Xp are the factors of the determinant of the C3 multiplication
> table with x1 etc. mapped onto the indices.

Frobenius created the theory of group characters in his study
of the factorization of the determinants of multiplication tables of
arbitrary finite groups.

>
> (Triplet division. First version, B/A=B.Ainv)
> Define Xinv={1/(3Xs) +(2x1-x2-x3)/(3Xp), 1/(3Xs) +(2x3-x1-x2)/(3Xp),
> 1/(3Xs) +(2x2-x1-x3)/(3Xp)}

As this shows, an element of K C_3 is invertible iff its
projections to K and or K(omega) are all nonzero.

>
> If one of As and Ap is zero, A is a "zero-sized" triplet, and Ainv
> appears to involve division by zero.

"appears"? Such elements of the group ring are not invertible.

> (As=0, Ap=0 is the trivial
> vector, {0,0,0}.) Recall that division by natural numbers, or by
> integers, creates a dividend and a remainder. This idea is used to
> avoid division by zero in a second version of triplet division. It
> creates a dividend BAd and a remainder BAr, B/A =BAd +BAr, by
> "projecting" the dividend onto a constrained sub-space where either
> BAds=0 (if As=0) or Badp=0 (if Ap=0), and "ejecting" a remainder with
> the other constraint. If neither As nor Ap is zero, the remainder is
> zero; if both=0, the dividend is zero and the remainder is all of B:-
>
> (* Triplet division, second version, B/A=BAd + BAr *)
> Which[
> As==0 And Not(Ap==0),
> BAd=B.{(2a1-a2-a3)/(3Ap), (2a3-a1-a2)/(3Ap), (2a2-a1-a3)/(3Ap)},
> Ap==0 And Not(Ap==0),
> BAd=B.{1/(3As) , 1/(3As), 1/(3As)},
> As==0 And Ap==0,
> Print["Division by {0,0,0}];BAd={0,0,0},
> Not(As==0)And Not(Ap==0)(*i.e. Else*),
> BAd=B.{1/(3As) +(2a1-a2-a3)/(3Ap), 1/(3As) +(2a3-a1-a2)/(3Ap),
> 1/(3As) +(2a2-a1-a3)/(3Ap)}
> ];
> BAr=B-A.BAd

This seems to be analogous to the Moore-Penrose inverse of matrices.
There the M-P inverse A^+ of A satisfies A A^+ A = A.
In your division the inverse 1/A_d has the property that
A(1/A_d) is an idempotent in K C_3. But there
is still something a bit aribtrary and unsaisfying about this:
A(1/A_d) is not necessarily the generator of the ideal in K C_3
generated by A. For example let A = (1, omega, omega^2) in C C_3.
Then A_s = A_p = 0 so your algorithm gives an "inverse" of zero.
But this can be obviated. We can define A^+ as follows for K = C
(I'll omit the case K = R). Let A = (a_1, a_2, a_3),
b_0 = a_1 + a_2 + a_3, b_1 = a_1 + a_2 omega + a_3 omega^2
and b_2 = a_1 + a_2 omega^2 + a_3 omega.
Let c_j = 1/b_j if c_j =/= 0 and c_j = 0 if b_j = 0.
Let A^+ = (1/3) (c_0 + c_1 + c_2, c_0 + c_1 omega^2 + c_2 omega,
c_0 + c_1 omega + c_2 omega^2).
Then A A^+ is the idempotent generator of the ideal generated
by A, and BA^+ is a solution of AX = B provided such a solution exists.
(Your inverse BA_d lacks this property in general).

More generally one can consider a general group algebra C G
over a finite group, and define a Moore-Penrose type "inverse"
by decomposing CG as a product of matrix algebras and applying
Moore-Penrose inversion in each. I was going to suggest this with
C replaced by an arbitrary characteristic zero field K, but that
would require Moore-Penrose inverses in matrix algebras over skew
fields. I don't know if there are such things .... anyone?
Of course there are no problems if, as here G is abelian, 

-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9"
Francis Wheen, _How Mumbo-Jumbo Conquered the World_