Re: Polysigned A^A

On Jan 28, 12:21 pm, "Tim BandTech.com" <tttppp...@xxxxxxxxx> wrote:
On Jan 26, 11:43 am, Mariano Suárez-Alvarez



<mariano.suarezalva...@xxxxxxxxx> wrote:
On Jan 26, 2:20 pm, "Tim BandTech.com" <tttppp...@xxxxxxxxx> wrote:

On Jan 20, 1:21 pm, Mariano Suárez-Alvarez

<mariano.suarezalva...@xxxxxxxxx> wrote:
On Jan 20, 4:18 am, lwal...@xxxxxxxxx wrote:

On Jan 19, 3:44 pm, Mariano Suárez-Alvarez

<mariano.suarezalva...@xxxxxxxxx> wrote:
On Jan 19, 7:48 pm, amy666 <tommy1...@xxxxxxxxxxx> wrote:
n = 2 ?
there are no non-zero zero-divisors in P_2.
( nor in P_3 at least not after applying the reduction rule )
since you below mention 4 variables , i assume you meant P_4 and either made a typo or are confused.
That's just a typo. As you observed, I'm using e0, e1, e2, e3....

I'm torn between whether I should use Golden's notation to
discuss polysigned numbers, or Mariano's. On one hand,
Golden is the inventor of polysigned numbers, so I ought
to defer to him. On the other, with Mariano's, it's easier
to see how to multiply, say, (e1)(e2) = e3, rather than
remember that (-1)(+1) = *1. Indeed, with Mariano's we
have that (ei)(ej) is either e(i+j) or e(i+j-4).

So I will use both. Now in that case, it's easy to find a
pair of zero divisors in P4. Using Golden's notation:

(#1 +1)(#1 -1)
= #1 -1 +1 *1 = 0

Using Mariano's notation, we have:

(e0+e2)(e0+e1)
= e0+e1+e2+e3 = 0

I assume that what Mariano wrote above is equivalent. So
we have a = #1 +1 = e0+e2 and b = #1 -1 = e0+e1

No it is not equivalent. You have given *another*
factorization of zero. The thesis in tommy's "theorem"
is valid for some zero divisors and invalid for others.

Then
a b = 0
so that both a and b are zero-divisors. But you should
be able to check very easily that a is not a polynomial
in b nor b is a polynomial in a.

Au contraire. Consider, using Mariano's notation, the
polynomial b^2-2b+2:

b = e0+e1
b^2-2b+2 = (e0+e1)^2-2(e0+e1)+2e0

(Notice how I identified 1, the multiplicative identity
of R, with e0, the multiplicative identity of P4. Golden
does this all the time by writing 1 = #1 in P4.)

b^2-2b+2 = (e0+e1)^2 -2(e0+e1) +2e0
= e0^2+2e0e1+e1^2 -2e0-2e1 +2e0
= e0 +2e1 +e2 -2e1
= e0 +e2
= a.

So a = b^2-2b+2. (This is a bit more cumbersome if we
attempt to use Golden's notation.)

But then again, note that _every_ polysigned number
is a polynomial in e1.

There is very little one can say of interest in the
arithmetic of Pn for n >= 2, for it is isomorphic
as a real algebra as a direct product of copies of R
and of C.

Is it? Golden, of course, regularly admits that P3 is
isomorphic to C. Robin Chapman, many years ago, found
an isomorphism between P4 and RxC, defined as:

#1 = e0 = (1,1)
-1 = e1 = (-1,i)
+1 = e2 = (1,-1)
*1 = e3 = (-1,-i)

But Golden claims that this isn't really an isomorphism
between P4 and RxC. This is because he envisions the
elements #1, -1, +1, and *1 (the ei) to be 3-dimensional
vectors which are vertices of a regular tetrahedron. So
in particular, the difference between any two of them
should be equal. But in RxC, we have:

#1 = e0 = (1,1) = (1,1,0) in R^3
-1 = e1 = (-1,i)= (-1,0,1) in R^3
+1 = e2 = (1,-1)= (1,-1,0) in R^3

norm(e0-e1) = sqrt((1+1)^2+(1-0)^2+(0-1)^2) = sqrt(6)
norm(e0-e2) = sqrt((1-1)^2+(1+1)^2+(0-0)^2) = 2.

So Golden won't accept any isomorphism unless it's an
isometry as well, so that:

norm(ei-ej) = k delta(i,j) for some constant k
(delta means Kronecker delta).

What about P5? I've already found some zero divisors
for P5. Letting phi be the golden ratio, % be the
fifth sign, and e0 thru e4 be Mariano's notation, I
have found:

a = %phi +1 *1 = (phi)e0+e2+e3
b = %phi -1 #1 = (phi)e0+e1+e4

Then ab = 0.

Meanwhile, I attempted to find an isomorphism, in
the spirit of Chapman's for P4, between P5 and CxC. In
fact, since Chapman's used the fourth roots of unity,
I tried using the fifth roots of unity. If we let xi
be a nontrivial fifth root of unity, then how about:

%1 = e0 = (1,1)
-1 = e1 = (xi,xi^2)
+1 = e2 = (xi^2,xi^4)
*1 = e3 = (xi^3,xi)
#1 = e4 = (xi^4,xi^3)

But both Golden and an impartial mathematician (not
Chapman, for he had long left sci.math by this time)
assured me that this is _not_ an isomorphism.

Mariano, if you would please tell us what the real
isomorphism between P5 and CxC (or is it RxRxC) is? And
also, does it meet Golden's additional requirement that
it be an isometry, with e0 thru e4 representing vectors
that form, in R^4, the vertices of the regular
pentachoron (i.e., the 4-simplex)?

Let C(n) = { e0, e1, ..., e(n-1) } be the cyclic
group of order n. Let R be the field of real
numbers, and let RC(n) be the real group algebra
of C(n), which is commutative.

Let w = e0 + e1 + ... + e(n-1) be the element
of RC(n) which is sum of the elements in C(n).
One can check trivially that the ideal I generated
by w in RC(n) is of dimension 1. Consider the
algebra A(n) = RC(n)/I.

According to Maschke's theorem, the algebra RC(n)
is semisimple, so that its quotient A(n) is also
semisimple.

Within the wiki on Maschke's theorem I see this stipulation:
"If (V, ρ) is a finite-dimensional representation of a finite group
G over a field of characteristic zero"
so I see a reliance upon field contingencies that we know are not true
of polysign.

I have no idea what "field contingencies" are.

In any case, my post explains that you can obtain the
polysign numbers as a quotient of the group algebra of
the cyclic group with coefficients in the field of real
numbers.

The theorem of Maschke, as stated in Wikipedia, is:

Let G be a finite group and K a field whose characteristic
does not divide the order of G. Then KG, the group algebra
of G, is a semisimple algebra

Take for G a finite cyclic group, and for K the field R
of real numbers. The characteristic of R is zero, which
does not divide the order of G, so we can conclude that
RG, the real group algebra of G, is semisimple.

Yet my own ability to grasp the group theoretic terminology is poor.
I wonder if you can comment on this field clause Mario?
I see this wording as referring to (V,p) as the field rather than G.

Not at all. In fact, it does not refer to G as the field either.

My own puzzlement begins back at 'finite field'- can these be
continuous?

I have no idea what "continuous" means in this context.

In any case, that sentence in Wikipedia is absolutely irrelevant
in your context: it simply observes that in Maschke's theorem
there is no need for the field K to be infinite.

As I read it they cannot be- this is more like Galois theory where a
finite number of elements are in play whereas with polysign we deal in
continuous magnitude within the fundament. So I wonder if this can
apply. The marriage of discrete sign with continuous magnitude may put
polysign outside of this area's reach.

I do not know what this paragraph means.

-- m

OK I'm pretty far off in understanding your path. I am picking up bits
and pieces here and there, but am far from putting them together
nicely. Is it as if to say that we may as well deal in elemental
values
s x
for the solution to the division problem? This is acceptable I think
except when we have a denominator with a sum. So for instance to
suppose that in P4
- a + b * c # d
-----------------
s x
is the same as
- a + b * c # d
--- @ --- @ --- @ ---
s x s x s x s x
where we've simply split out the fraction('@' meaning superposition),
thusly arriving back at elemental values since whatever s is we can
certainly find
s1 / s2
in P4 via the following table:
s1 s2 s1/s2
- - #
- + *
- * +
- # -
+ - -
... (through sixteen possibilities of P4)
which is merely the multiplication table transposed, the modulo-4 sum
of the second and third column being represented in the first column,
# interpreted as @ (the zero sign).
The magnitude of the resultants are just
a/x, b/x, c/x.
The inverse modulo sum then would need a symbol to state that cleanly
within the
s x
elemental notation. Some guy just recently suggested that rather than
( s1 x1 )( s2 x2 ) = ( s1 @ s2 )( x1 x2 )
just use
( s1 x1 )( s2 x2 ) = ( s1 s2 )( x1 x2 )
which is a nice simplification and consistent with this new need since
then where product is implied by the juxtaposition we can use the
traditional '/' operator symbol to reverse it. Without wrecking the
notation I do think the form
s1 / s2
is consistent in either notation so long as it is understood that this
is tied back to a modulo sum when the '/' operator is used with signs.

This is merely a microstep back a few paces from where you all are
getting to but it is stated simply and is a solid footing. In studying
the sums of such divisors we will see that the resultant is a sum of
discrete rotations, each weighted by the magnitude of the divisor. One
generalization would allow the form
z z z z
--- @ --- @ --- @...

[snip]

I have no idea what is the relationship between this
post of yours and the one I wrote and to which
you are replying.

-- m
.

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