Re: Is This System Solvable?

From: M.J.T. Guy (mjtg_at_cus.cam.ac.uk)
Date: 02/19/05 

Date: 19 Feb 2005 15:25:45 GMT

Eugene Shubert <GalileoProject002@everythingimportant.org> wrote:
>
>I'm looking for all sufficiently differentiable real-valued functions
>of three real variables T(R,S,w) defined everywhere except the point
>w=0, that have these properties:
>
>For all X, Y, a, b, such that a is not equal to zero, b is not equal to
>zero, and a+b is not equal to zero, there exists a unique Z=Z(X,Y,a,b)
>such that the following identities are always true:
>
>T(X, Y, a) = T(X, Z, a+b)
>
>T(Y, X, -a) = T(Y, Z, b)
>
>T(Z, Y, -b) = T(Z, X, -a-b)
>
>Note that the uniqueness of Z = Z(X,Y,a,b) is quite remarkable in that
>Z is defined by three different equations!

Actually, it turns out to be less remarkable than appears at first.
When you allow for symmetries, the three equations are really the *same*
equation, seen from different angles.

First, in the interest of symmetry, we'll write c for -(a+b) and reverse
the second and third equations. Then we can see that the equations
are symmetric under any permutation of X, Y, Z. E.g. cyclically
permute X, Y, Z and cyclically permute a, b, c; or swap X, Y and
substitute a, b, c -> -a, -c, -b.

Second, note that the equations aren't really saying very much. They
make no use of the properties of X, Y, Z as reals, and only that a, b, c
are in an abelian group. So we can generalise to assume that X, Y, Z
are from some set S and that a, b, c are from an abelian group A.

Further, no use is made of the actual value of the function T() except
to test for equality. So we can get rid of the function T by expressing
the constraints in terms of the function f: S x S x A x A -> S such
that Z = f(X, Y, a, b). Then f is 1:1 in each of its first two
arguments and the three equations are equivalent to

        f(X, Y, a, b) = f(Y, X, -a, -c) (1)

        X = f(f(X, Y, a, b), -b, -a). (2)

(We seem to have lost the symmetry, but it'll reappear shortly.)
Given such an f() we can construct many functions T() as follows:
Choose a particular c, and choose T(X, Y, -c) arbitrarily subject to
being 1:1 in its second argument. For arbitrary a, T is then defined
by T(X, Y, a) = T(X, f(X, Y, a, -a-c), -c).

f(*, *, a, b) defines a function S x S -> S which is 1:1 on each of its
arguments. (Is there a standard term for such beasties? They turn
up in many places, for example as the multiplication table of a group.
Since when S is finite, they are just Latin Squares, I'll use that
term in the general case.) Such a function can be viewed as a
relation on S x S x S, and such a relation can be viewed as a Latin Square
in six ways. So Latin Squares are grouped in related sets of six.

The relation (1) defines f(*, *, -a, -c) in terms of f(*, *, a, b).
The relation (2) defines f(*, *, -b, -a) in terms of f(*, *, a, b).
In each case, the new function is a 'relative' of the original. Thus
we find six functions forming a related set:

    f(*, *, -a, -c) <--(1)--> f(*, *, a, b) <--(2)--> f(*, *, -b, -a)
           ^ ^
           | |
          (2) (1)
           | |
           v v
    f(*, *, c, a) <--(1)--> f(*, *, -c, -b) <--(2)--> f(*, *, b, c)

Conversely, this is *all* that relations (1), (2) tell us. So the
general function f() is obtained as follows:

Group the elements of ( A \ {0} ) x ( A \ {0} ) in sets of six
{ (a,b), (b,c), (c,a), (-c,-b), (-b,-a), (-a,-c) }. Associate
an arbitrary set of six Latin Squares with each such set.

And that's all.

Mike Guy

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