Re: Is magnitude more fundamental than the real numbers?

In article <1155123799.250626.317520@xxxxxxxxxxxxxxxxxxxxxxxxxxx> "Timothy Golden BandTechnology.com" <tttpppggg@xxxxxxxxx> writes:
....
I believe that I have resolved our apparent disagreement. I've
scrambled together a low grade reciprocal finder (results with large
error) and sure enough the divisor counterexamples that you provide
have no decent reciprocal. I've read the definition of zero divisor and
it is fine though it does not discuss anything about the underlying
structure of the system it is applied to. I suppose that is up to us
and I am now curious what we'll see above p4 in this context.

Zero-divisors are indeed pat of the structure, but much can be said
about zero-divisors without even investigating the underlying
structure (when that is a ring, of course).

The 'other axes' are actually a plane. It is the plane that is
perpendicular to the identity axis that passes through the origin.

Yup, the set of obvious zero-divisors are, using coordinates
(x1, x2, x3, x4), the plane x1 + x3 = x2 + x4 and the line
x1 = x3 & x2 = x4.

We've discussed it a bit before but now I can see it in the context of
division. Just as a multiplication by a point on the identity axis
removes information so does a multiplication by any point in this
plane. Instead of squashing the information into a line it yields two
dimensions of result, but still provides an irreversible effect on any
object. In effect the object has been squashed into a plane and can
never be 'unsquashed' informationally.

Indeed, the product of a zero-divisor and an arbitrary element is a
zero-divisor. That is easy to prove when the product is commutative.

We cannot
get three dimensions back from two. There must be a theorem on
dimensional reduction and zero divisors. It's clear in the
informational context.

There is not necessarily dimensional reduction. There are rings where
every element is a zero-divisor. But in general the zero-divisors form
subspaces of lower dimensionality (assuming we have some dimensional
algebra over the reals, which you have). But it is all quite tricky,
because if a and b are zero-divisors, a + b is not necessarily a
zero-divisor.

The zero
product is composed of one value from each of these parts, those parts
being the identity axis and the plane orthogonal to it on the origin.
There should be a theorem there as well. Upon excluding this plane as
well as the line that I call the identity axis we can have division.

Indeed, you can have division with the elements that are not zero-divisors,
because they all have multiplicative inverses. It is just like in
linear algebra, where a matrix has an inverse if and only if the
determinant is not equal to 0. But if you look at it, you will see
that the structure of the set of matrices with determinant 0 is quite
tricky.

I suppose you are nearing an R x C style equivalent definition of the
P4 product.

Pretty close, but not there yet. The line of zero-divisors should
correspondent with R, and the plane with C. What remains to be done
is finding the exact mapping. (That that is true follows because
in R x C we have also two sets of zero-divisors, one 1D and one 2D,
corresponding to R and C respectively. I think an exact mapping can
be found.)

As I scramble to follow along now I am wondering what you
will find next. And what of P6? Is the zero divisor exclusion the
identity axis and the points orthogonal to it through the origin?

There are many more zero-divisors in P6. I will explain, and that shows
also the reason that I originally thought that Pn might have no zero-
divisors if n is prime. Consider in P6:
(1, 1, 1, 0, 0, 0) * (1, 0, 0, 1, 0, 0) = 0
and
(1, 1, 0, 0, 0, 0) * (1, 0, 1, 0, 1, 0) = 0
These are obvious zero-divisors (you get the other obvious zero-divisors
by circulating the coordinates). That we have two pairs of clearly
different kinds is because 6 = 2 * 3. I do not know whether there are
non-obvious zero-divisors (I am pretty sure, there are none in P4).

What
is a plane in P4 would be a 4D space in P6. This would be a ( 1, n-1 )
type of relation. That is consistent with the zero products, whose
possibility space rises dimensionally.

What you will find (I think) in P6 is that there are four spaces of
zero-divisors.

If the Hopf, Milnor and Kervaire dimensional theorems apply then we
should be able to get a zero divisor structure in P7 as well. But they
suggest that P5 will work perfectly.

In P5 and P7 there are no obvious zero-divisors. So looking for them
can be problematical. But starting at Pn, we can get an (n-1)-
dimensional algebra over R by subtracting the last coordinate from
all coordinates. The last coordinate now is 0. It is now an algebra
over the reals of dimension n-1. So by Hopf, if it is a division
algebra, n-1 should be a power of 2. We excluded already all non-prime
n, now we can also exclude all primes that are not of the form 2^k + 1.
But there is a stronger theorem by Hopf: every commutative division
algebra over k has dimension 1 or 2. (H. Hopf, Ein topologischer
Beitrag zur reellen Algebra, Comment. Math. Helv. 13 (1940), 219-239.)
So we do not even need the theorems by Milnor and Kervaire. What they
did show was that *any* division algebra over R has at most dimension
8, and that those algebras are precisely R, C, Q and O. (But Q is
non-commitative and O is non-associative. If we get further we get
the sedonians, but now there are zero-divisors.)

What remains is looking at P5 and see whether zero-divisors can be found
(there must be some, otherwise it would be a division algebra, and that
can not be the case by the theorem of Hopf.)

and so by the theorem they can only be a division algebra (that is,
division does exist) if n-1 = 1, 2, 4 or 8.

I'm not very good at linear algebra but I'll try to follow along. So
far I don't see how you got the matrix in b. Is this a conjugation
style of division?

No, it is just plain linear algebra notation to denote a set of linear
equations, which division in your systems is: solving sets of linear
equations.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.