$50US prize for first clean isometric isomorphism between polysign P4 and RxC
Tim BandTech.com
I attempt to promote the polysign numbers
http://bandtechnology.com/PolySigned
which are nearly field algebras in any specified dimension and support
emergent spacetime geometry from arithmetic. For P4 the RxC similar
space seems to have become a sticking point of the old school whereby
the polysign numbers can be avoided by steerage over to associative
algebra studies.
The standard refutation seems to be that the polysign construction is
contained within existing associative algebra work and the higher sign
systems will always mimic copies of R and C so that the four-signed
numbers P4 will mimic RxC. This statement does seem to hold up to
isomorphism, but this bending of the P4 arithmetic product does not
necessarily lend itself to understanding the system any better so long
as the bending is not specified. So the next stage of that study is to
map an isometric isomorphism.
A $50 prize deliverable as a money order to anywhere possible (or if
you prefer a personal check) for solving this puzzle. $50 is not much
money. Out at the bar here in the northeast one night of drinking will
waste away your prize money so the excitement should be more focused
on the problem at hand. I will post your work at my website and credit
you or if you prefer I can link to your own content.
This offer stands to anyone who can instantiate an isometric
isomorphism between P4 and RxC which is consistent under product.
The layout of this construction should include a transformation T such
that
y( P4 ) <--> x( R ), z( C )
e.g. T( y ) = x, z
such that
T( y1 y2 ) = T( y1 ) T( y2 ).
The right hand side arithmetic product should be clearly defined on
RxC which is the part of the puzzle that remains troubling.
The product y1y2 (in P4) is clearly defined and extends from the usual
properties of the real numbers but in a more general form called
polysign. P4 is the four-signed numbers, which are three-dimensional
in ordinary spatial terms. The polysign side of this puzzle is fully
disclosed on my website.
In addition I have started an analysis of this problem at
http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html
which to me suggests an infinite progression though I have not proven
this.
Even if a convergent series is discovered I will grant the prize.
The work should be demonstrated either computationally or as a paper
proof.
It should be expressed clearly and redundantly enough for a simpleton
to understand.
Also if a paper proof is given its results should be computable for
values y1 and y2 in P4.
- Tim
victor_me...@yahoo.co.uk
> This has been an ongoing debate by a few here.
Hardly a debate.
> I attempt to promote the polysign numbers
> http://bandtechnology.com/PolySigned
Yawns all round!
> the polysign numbers can be avoided by steerage over to associative
> algebra studies.
Your "polysign numbers" are associative (and commutative)
algebras over R.
> numbers P4 will mimic RxC. This statement does seem to hold up to
> isomorphism,
Exactly!
> but this bending of the P4 arithmetic product does not
bending??
> A $50 prize
That's not quite in Mr Clay's league :-(
> This offer stands to anyone who can instantiate an isometric
> isomorphism between P4 and RxC which is consistent under product.
I already indicated which real quadratic form on R x C
corresponded to your distance on P_4.
> In addition I have started an analysis of this problem at
> http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html
Lots of silly pictures, but no actual maths :-(
> It should be expressed clearly and redundantly enough for a simpleton
> to understand.
Or even you!
Tim BandTech.com
<victor_meldrew_...@yahoo.co.uk> wrote:
> On 19 Feb, 15:56, "Tim Goldenballs" <tttppp...@yahoo.com> wrote:
> I attempt to promote the polysign numbers
> > http://bandtechnology.com/PolySigned
> > algebra studies.
> Your "polysign numbers" are associative (and commutative)
> algebras
> > numbers P4 will mimic RxC. This statement does seem to hold up to
> > isomorphism,
> > but this bending of the P4 arithmetic product does not
> > A $50 prize
> > This offer stands to anyone who can instantiate an isometric
> > isomorphism between P4 and RxC which is consistent under product.
>
> I already indicated which real quadratic form on R x C
> corresponded to your distance on P_4.
>
> > In addition I have started an analysis of this problem at
> > http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html
> > It should be expressed clearly and redundantly enough for a simpleton
> > to understand.
This prize remains open since Victor's rendition is lacking a
transform. I've attempted to ascertain whether it is the transform on
my website, but his communications have been obscurant.
If Victor has indeed specified the information then that is his gain.
Unfortunately as you can see in his post here he merely harkens back
to a post of a 'distance function' which carries no transform with it,
though I have been nice enough to state the current problem reusing
his own (x,z) notation. I do encourage you Victor to specify your
solution more clearly since there are several others who have been
confounded by what could seem a simple puzzle. I myself do see the
resemblance of RxC with P4 and with the help of Hero van Jindhelt
exposed the P2xP3 image (RxC) as an 'identity axis' and its
accompanying rotational plane orthogonal to it. If we were to widen
this topic out we could formulate this as a first instance of
orthogonal geometry in the terms of polysign. Hence polysign number
theory would naturally build the orthogonal coordinate system though
it would do so under a stipulation not unlike a brane-theoretic
construction, accept that the polysign build is already in terms of
general dimension, with spacetime already emergent at the prebrane
level, or at the least built into it via the general dimensional
progression. This form of emergent spacetime places demands on the
physics that will follow. The most challenging of these that I have
come accross is a challenge to isotropic space. If I were to go on
here in a standard and normal form of communication this post would
turn into a chapter and unfortunately I remain repeating my words over
and over here, to the point of boredom, as Victor himself can see
clearly. I thank Victor for giving me some attention though he places
challenge after challenge on me. One of his obscure references I never
even got to looking up because the first one was all in German and it
took me too long just to get through a few details. Victors
communications are cryptically such and I am too primitive to
understand his level of banter. This I admit of many areas of
mathematics and have my fair share of criticism for all that is built
on top of the real number as fundamental. It may well be that all of
it needs to be rechecked from a magnitudinal standpoint. What is a
magnitudinal standpoint? It is something very primitive so far back in
fundament as to feed the great philosophers on their leads to the real
number, to the complex number, and so on into quaternions and the
flood of number systems since. While the polysign could be swamped in
this flood by people such as Victor or Mariano who hold greater
credibility than I do, I know that polysign are correct and so I am
comfortable working on and on at this problem. To date this is my
strongest attempt on usenet. There are very few usenet prizes I
believe yet I may be forming the future state of an open mathematics
community. The state of communication amongst respected mathematicians
may be so low and of such a military structure that the censors being
granted an upperhand have stifled the entire genre. By splitting and
splintering off every little bit into subjugation the large picture
has been avoided and the 'high' quality as refinement may be cast off
as merely tangential variations. This is certainly my gut feeling on
much of academic mathematics. To those who see the beauty in polysign
I ask you to address this higher concern of mine which puts me at odds
with so many. Who am I to speak so big? Who are you to put me down?
Let's go both ways and see if a quality discussion can be had.
Certainly fear of physics and philosophy will not be a healthy sign
and I may ignore such people in the future who pride themselves on
'pure' mathematics.
- Tim
victor_me...@yahoo.co.uk
>
> This prize remains open since Victor's rendition is lacking a
> transform.
All the information is in my previous postings...
> unfortunately I remain repeating my words over
> and over here, to the point of boredom,
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz....
> I am too primitive to
> understand
OK!
<rest of 58-line paragraph snipped>
amy666
P4 = R x C
the confusion is the following :
x is not a " product " or " extension "
as Golden might have guessed.
further , indeed P4 =/= R times C as demonstrated on his site.
further confusion : cartesian product = / = cartesian coordinates.
Golden has to realise :
his notation might be confusing to some ,
and similarly
he might be confused by more or less standard notation.
it works both ways.
basicly when Timothy talks about R x C , he actually means R x R x R.
when he talks about P2 x P3 , then he is talking about R x R x R.
and then P4 = R x C.
unfortunately , i dont see how we can merge both notations into a consistant easy whole new notation.
although this might get close :
C^n = P(2n + 1)
i claim the prize for this :
C^n = P(2n + 1)
:)
tommy1729
Tim BandTech.com
No Tommy. the little 'x' in RxC is a cartesian product, not an
arithmetic product. Certainly I know enough of mathematics to make
this distinction and I'm surprised that you would think so little of
me.
>
> further , indeed P4 =/= R times C as demonstrated on his site.
>
> further confusion : cartesian product = / = cartesian coordinates.
Errr.... I don't see this one Tommy. Upon instantiating a Cartesian
product the components have inherently been granted independence in an
informational way as much as the matter of geometric orthogonality.
So if I take two undefined domains A and B and I let some little a be
in A and b in B then when I construct
A x B
I do inherently allow a notation such as
(a,b)
or
(a1,b1)
to mean one element in the constructed space, their order being
completely relevant. If I have no spec for either an element's value
in A or its value in B then I would argue that the construction AxB
was ill-defined. When these A and B are in the reals then we clearly
see instances such as
( - 12.23, + 2.34 )
as valid elements in AxB
For RxC though we would instantiate elements like
( - 1.23, + 2.34 - 1.21 i )
I do see the direct similarity to RxRxR
( - 1.23, + 2.34, - 1.21 )
and so the only relevant difference is in the definition of an
arithmetic product. These arithmetic products on geometrical spaces
are not generally practiced as natural behaviors. For instance we do
not generally multiply a spoon by a bowl in standard geometrical
space. We are off in a domain that is more esoteric than ordinary
spacetime. This is a deep puzzle for the extraction of physics from
the polysign, yet we do see product relationships in classical force
equations, so that if we were to consider charge then we do get closer
to this geometrical multiply, but still not close enough to call it
good enough.
>
> Golden has to realise :
>
> his notation might be confusing to some ,
>
> and similarly
>
> he might be confused by more or less standard notation.
>
> it works both ways.
>
> basicly when Timothy talks about R x C , he actually means R x R x R.
>
> when he talks about P2 x P3 , then he is talking about R x R x R.
>
> and then P4 = R x C.
>
> unfortunately , i dont see how we can merge both notations into a consistant easy whole new notation.
>
> although this might get close :
>
> C^n = P(2n + 1)
>
> i claim the prize for this :
>
> C^n = P(2n + 1)
>
> :)
>
> tommy1729
These mathematicians sure are a cryptic lot. So let's look at Tommy's
latest for n=1:
C = P(3) (or just P3 in my standard notation)
Ok. That one is good. Next
CxC = P5
Well, the dimension is good, so maybe... Next though Tommy is the part
of the puzzle that I've been trying to get Victor to address. To check
your claim if I say
"Oh, Ok Tommy, so what about a P5 product ( @ 2 - 3 # 2 )( @ 1 *
5 ) ?"
I know from doing enough polysign that this is child's work in the P5
domain:
@ 2 * 10 - 3 # 15 # 2 + 10
= * 8 - 1 # 15 + 8
in reduced unordered form. Now for Tommy's version in CxC...
This is why we need a transform! Tommy, to prove your CxC claim for P5
what is the Transform T that will bring
T(P5) -> CxC ?
I've given you the P5 arithmetic here. Now it needs to be done in CxC,
otherwise there is no significance to this claim. My mathematical
reason tells me that the arithmetic product for a space CxC is no
better defined than it is for a space RxRxRxR and this then leaves you
to define the product in CxC, yet since you've only come to define it
to match the product in P5 (which does naturally exist) then after all
this work the significance will be when that product is formed as for
two elements z1 and z2 whose first complex parts are z11 and z21 and
second complex parts are z12 and z22 such that their arithmetic
product is defined as
( z11, z12 )( z21, z22 ) = ( z11 z21, z12 z22 )
or some such fairly simple combination. Yet still we will need that
damn transform. So it's two things needed: the transform and the
definition of the product in the proposed equivalent space. Then we
simply run the result in P5 through the transform and we should get
the same output as the transformed inputs to the newly defined
product.
- Tim
Tim BandTech.com
<victor_meldrew_...@yahoo.co.uk> wrote:
> On 19 Feb, 17:39, "Tim Goldenballs" <tttppp...@yahoo.com> wrote:
> > This prize remains open since Victor's rendition is lacking a
> > transform.
> All the information is in my previous postings...
This is all that I have from previous posting of yours:
"We do know; at least we who are competent to have worked it out
do. It's easy to write down the distance function on R x C
corresponding to yours on P_4, viz. the square of your
distance corresponds to the map (x,z) |--> (1/3)(x^2 + 2|z|^2)."
Unfortunately I do not understand how this relates to the arithmetic
product problem.
If it were a correction factor on the simple RxC product then we would
see an instance like
( + 1 # 1 )( + 1 # 1 ) = + 2 # 2
working with your correction factor on the product resultant right?
This particular value should be easy if you've got x mapped to U(+1#1)
where U() is the unit vector.
For this instance z is zero.
where x is the distance of +1#1 which is
sqrt( 1 + 1 - 2/3 ) = sqrt( 4/3 ) = R(y), C(y) = 0.
In other words the value in P4 +1#1 maps to (sqrt(4/3), 0 ) in RxC.
Next we square this since this is the simple instance that I've chosen
to work with:
( sqrt(4/3), 0 )( sqrt(4/3), 0 ) = ( 4/3, 0 )
which assumes you are using the simplest product definition in RxC.
Now clearly your distance of the resultant is 4/3.
The distance of the P4 resultant +2#2 is
sqrt( 2 + 2 - (2/3)( 4 )) = sqrt( 4/3 )
So now I want your function to match up these values
(1/3)(x^2) = (1/3)(4/3) = 4/9.
You say "the square of your distance corresponds to the map"
but when I square the P4 native distance I see
4/3 .
I am sorry to have used a value with so much redundant numerical
content, but it is a simple enough instance to decode your formula
with. Again I would go back to a transform T() which maps P4 to RxC
and a product defined in RxC. Here I've been defining both and have no
agreement from you as to whether these are even what you are using.
Perhaps by using a third value for the product resultant we could
avoid more confusion so that in P4 we have
y1 y2 = y3
and the corresponding RxC system
(x1,z1)(x2,z2) = (x3,z3)
The focus is on the values y3 and the claimed similar (x3,z3).
It seems to me that your factor of 1/3 for this instance is wrong, so
I've disproven this attempt at matching up the P4 product with RxC as
an isometry. I'm sorry Victor but the prize money cannot be yours
unless there is some mistake going on, but already I'm left trying to
decode your own cryptic communications and I don't feel good about
what I've written here because what you wrote and are defending is not
clearly stated.
This is a standard catch in human interaction and even game theory. A
good player can only play to the quality of his opponent. We both
think that we are good but clearly a mess has been made. Things should
be crystal clear and any good mathematician would take a care as to
how he puts a piece of information that he will stand by as you are
doing.
- Tim
hagman
> This has been an ongoing debate by a few here.
> I attempt to promote the polysign numbers
> http://bandtechnology.com/PolySigned
> which are nearly field algebras in any specified dimension and support
> emergent spacetime geometry from arithmetic. For P4 the RxC similar
> space seems to have become a sticking point of the old school whereby
> the polysign numbers can be avoided by steerage over to associative
> algebra studies.
>
> The standard refutation seems to be that the polysign construction is
> contained within existing associative algebra work and the higher sign
> systems will always mimic copies of R and C so that the four-signed
> numbers P4 will mimic RxC. This statement does seem to hold up to
> isomorphism, but this bending of the P4 arithmetic product does not
> necessarily lend itself to understanding the system any better so long
> as the bending is not specified. So the next stage of that study is to
> map an isometric isomorphism.
Let's see:
As a set, P4 consist of all expressions of the form
- a + b * c # d
where a,b,c,d are non-negative real numbers.
To avoid confusion with standard meanings of +, -, *, I write
(a,b,c,d)
instead.
Addition in P4 is defined by
(a,b,c,d) + (e,f,g,h) = (a+e, b+f, c+g, d+h)
where the pluses on th RHS are the usual addition for non-negative
reals.
Multiplication in P4 is defined by
(a,b,c,d) * (e,f,g,h) = (ah+bg+cf+de, ae+bh+cg+df, af+be+ch+dg, ag
+bf+ce+dh)
where again addtion and multiplication on the RHS is standard.
By repeatedly multiplying with itself, we observe that
(1,0,0,0)^2 = (0,1,0,0)
(1,0,0,0)^3 = (0,0,1,0)
(1,0,0,0)^4 = (0,0,0,1)
OTOH, addition in RxC is defined by
(a,z) + (a',z') = (a+a',z+z') and multiplication by
(a,z) * (a',z') = (a*a',z*z') (all operators have their usual meaning)
This encourages us to map (1,0,0,0) to (-1,i) and the other
units to powers thereof. Mopr precisely, define the following map
phi: P4 -> R x C
(a,b,c,d) |-> (-a+b-c+d, a*i -b -c*i +d)
First we need to show that this is well-defined - we have not yet
mentioned the cancellation identity (x,x,x,x)=0.
But indeed phi(x,x,x,x) = (-x+x+x+x, x*i-x-x*i+x) = (0,0)
and in general (because phi is R-linear when viewed as a map
R^4 -> R x C) phi(a+x,b+x,c+x,d+x) = phi(a,b,c,d).
Thus phi is indeed *well-defined*.
We now have a map of sets phi: P4 -> RxC.
In fact phi(a,b,c,d) = (0,0) implies -a+b-c+d = 0 and a-c=0 and b-d=0.
From this it follows that a=b=c=d, i.e. (a,b,c,d)=0.
Hence phi is an *injective* map.
We observe that phi is onto, for if (a,z) in R x C is arbitrary,
then let u = max{Re(z),0}, v = max{-Re(z),0}, w = max{Im(z),0}, x =
max{-Im(z),0}.
Then u,v,w,x are non-negative numbers and z = (u-v)+i*(w-x).
Hence
phi(w,v,x,u) = (-w+v-x+u, z).
Let b = a-(-w+v-x+u).
If b>=0, then
phi(w,v+b/2,x,u+b/2) = (a,z)
and if b<0, then
phi(w-b/2,v,x-b/2,u) = (a,z)
hence phi is *surjective*.
Next note that
phi((a,b,c,d) + (e,f,g,h)) = phi(a+e, b+f, c+g, d+h)
= (-(a+e)+(b+f)-(c+g)+(d+h), (a+e)*i - (b+f) - (c+g)*i + (d+h))
= (-a+b-c+d, a*i -b -c*i + d) + (-e+f-g+h, e*i-f-g*i+h)
= phi(a,b,c,d) + phi(e,f,g,h).
Hence phi is *compatible* *with* *addition*.
Next
phi((a,b,c,d) * (e,f,g,h))
= phi(ah+bg+cf+de, ae+bh+cg+df, af+be+ch+dg, ag+bf+ce+dh)
= (-(ah+bg+cf+de)+(ae+bh+cg+df)-(af+be+ch+dg)+(ag+bf+ce+dh),
(ah+bg+cf+de)*i-(ae+bh+cg+df)-(af+be+ch+dg)*i+(ag+bf+ce+dh))
= ( (-a+b-c+d)*(-e+f-g+h), (a*i-b-c*i+d)*(e*i-f-g*i+h) )
= (-a+b-c+d, a*i-b-c*i+d) * (-e+f-g+h,e*i-f-g*i+h)
= phi(a,b,c,d) * phi(e,f,g,h)
Hence phi is *compatible* *with* *multiplication*.
Until now we have thus shown that phi is a ring isomorphism.
The norm of an element (a,b,c,d) of P4 is defined as
sqrt( aa + bb + cc + dd - (2/3)(ab + bc + cd + de + ac + bd)).
(This definition is based on a "belief" uttered in a usenet
discussion).
To avoid square roots, consider the expression
||(a,b,c,d)||^2 = aa + bb + cc + dd - (2/3)(ab + bc + cd + de + ac
+ bd).
Likewise the first definition of a norm on R x C that comes to mind
is defined by
||(x,z)||^2 = r^2 + z*\bar{z} = r^2 + Re(z)^2 + Im(z)^2.
However, this norm is not so natural as it seems.
Any choice of positive factors alpha, beta allows us to
define
||(x,z)||^2 = alpha * r^2 + beta *(Re(z)^2 + Im(z)^2)
and (since RxC has zero divisors and ||.|| cannot be multplicative
anyway)
there's no specific reason to prefer one choice over the others --
after all
they all define the same topology on RxC.
Let's calculate this for phi(a,b,c,d):
||phi(a,b,c,d)||^2 = ||(-a+b-c+d, a*i-b-c*i+d)||^2
= alpha*(-a+b-c+d)^2 + beta*((a-c)^2+(d-b)^2)
= alpha*(a^2+b^2+c^2+d^2+2(ac+bd-ab-ad-bc-cd)) + beta*(a^2+c^2-2ac
+d^2+b^2-2bd)
= (alpha+beta)*(a^2+b^2+c^2+d^2) -2(alpha*(ab+ad+bc+cd) + (beta-
alpha)*(ac+bd))
It thus turns out that with the specific choice alpha=1/3, beta=2/3,
we have
||phi(a,b,c,d)|| = ||(a,b,c,d)||
We have thus proved the
THEOREM.
P4 is isomorphic as a metric R-algebra to the RxC with norm
||(r,z)|| = sqrt( 1/3 * r^2 + 2/3 * (Re(z)^2+Im(z)^2) ).
hagman
achille
as that of u, u^2, u^3, 1 in any formal algebra where u^4 = 1. The
identity -1+1*1#1 = 0 indicates P4 is nothing but R[u]/(1+u+u^2+u^3),
a quotient ring of the polynomial ring R[u].
Since 1+u+u^2+u^3 factors into p(u)q(u) where p(u) = 1+u, q(u) = 1+u^2
and p(u), q(u) relative prime to each other.
P4 ~ R' = R[u]/(p(u)q(u)) ~ R[u]/(p(u)) x R[u]/(q(u))
To construct the isomorphism between R[u]/(p(u)q(u)) and R[u]/(p(u)) x
R[u]/(q(u))
explicitly, we use the fact p(u) q(u) relative prime implies for some
polynomial
m(u), n(u),
1 = m(u) p(u) + n(u) q(u),
If we use [ g ]_f to denote the equivalent class of g(u) in R[u]/(f
(u)),
identifies like
( [ m p ]_pq )^2 = [ ( m p )^2 ]_pq = [ m p ( 1 - n q ) ]_pq = [ m
p ]_pq
( [ n q ]_pq )^2 = [ ( n q )^2 ]_pq = [ n q ( 1 - m p ) ]_pq = [ n
q ]_pq
[ 1 ]_pq = [ m p ]_pq + [ n q ]_pq
[ m p ]_pq [ n q ]_pq = [ m n p q ]_pq = [ 0 ]_pq
allows us to show the map
( [ a(u) ]_p, [ b(u) ]_q ) -> [ a(u)n(u)q(u) + b(u)m(u)p(u) ]_pq
is the desired isomorphism. In our case,
1 = (1/2(1-u))(1+u) + (1/2)(1+u^2)
and it is obvious
R[u]/(p(u)) = R[u]/(1+u) ~ R,
R[u]/(q[u]) = R[u]/(1+u^2)~ C, ( with u <-> i in C ).
the desired isomorphism between R x C and R' is given by
(a,b+ic) \in R x C -> [ a (1/2(1+u^2)) + ( b + c u )(1/2(1-u^2)) ]
_pq
= [ (a+b)/2 + c/2 u + (a-b)/2 u^2 - c/2 u^3 ]
_pq \in R'
or (LHS of <-> in regular notation, RHS of <-> in P4 notation)
(1, 0) ~ [ 1/2(1+u^2) ]_pq <-> -(1/2)*(1/2)
(0, i) ~ [ 1/2(u-u^3) ]_pq = [ (1/2) + u + (1/2)u^2 ]_pq <-> -
(1/2)+1*(1/2)
amy666
well if A x B is intended as a couple (a,b)
then indeed (R,C) is not iso to P4.
i already agreed upon that within the past , and i wont take it back.
but a huge "if".
depends on notation and interpretation.
on both sides.
this reminds me of set theory issues ...
i do not think little of you timothy.
and actually you know that.
and in some cases ive actually used your polysigned , including notation.
the issue with notations is ;
its tied to tradition and previous beliefs.
( just like set theory sigh )
regards
tommy1729
hagman
Yes, in fact this has been around here for a few years and the
isomorphism is valid even though TimGolden starts with non-negative
magnitudes only.
This time I merely tried to obtain the isomorphism (which is known
since 2003 at least) P4 = RxC from basic principles.
The avid reader will observe that it is almost impossible
not to find the isomorphism, at least with R[X]/(1+X+X^2+X^3).
In order to get rid of the theme of polysign numbers once and forever,
I took the freedom to compile a few results in a LaTeX document
here:
http://www.von-eitzen.de/math/PolysignNumbers.pdf
Contents:
*) Isomorphsims Pn = R[X]/(1+X+...+X^{n-1})
*) Isomorphisms Pn = C x C ... x C ( x R?)
*) Natural metric induced on Pn producing a regular simplex.
But since even today the isomorphism P4 = R x C is doubted
by others, I have little hope ...
hagman
hagman
Well maybe not (R,C), but R x C is iso P4. :)
Tim BandTech.com
May I take a copy of your paper for my own website?
Or if you do wish to make a fairly permanent link that would suffice.
Certainly I do have a response for you, particularly the points which
you wipe over yet which you do care to state clearly.
I deeply appreciate your serious response.
You give me pause for concern.
As I consider carefully what you have written
I know that you will understand that the amount of time it takes me to
respond and the level of confidence with which I can do so are coupled
as they are for you in your asymmetrically diaposed position. Yet the
symmetry of the system remains for both of us, and so as no mistakes
are appointed by you to polysign I am deeply grateful of your own
exposition in the terms of existent mathematics. As to the audience
whom you are capable of addressing who is it exactly? Shall we
consider that the gorilla conjecture can be brought up a level? Which
audience may I address in my simplicity of language? I do struggle to
follow your document and so my own acceptance of my own willful
ignorance should be noted, though again the weakness of my position
may merely bring your own actions down to size. How is it that you
have quashed polysign numbers? In a broth so thick as to mention the
chinese remainder theorem, which sends me off to do some research; I
am baffled. Yet I do credit you with great understanding and
mathematics far beyond my own simple capacity.
I have gotten your fullname and I guess a link to your own site which
I can only get a sneezly gist of in my bitter subterfuge of self
denial,
not to mention my lack of ability to understand any language other
than English, which deserves a fair back-sneezle.
I do love that you have yielded such a strong confrontation. I have
been seeking this out for a very long time.
The first principle that I would wish to address does go straight into
geometry.
I do see you and Tommy with a cryptic conversation on ordered pairs
versus a Cartesian product by pair.
I have little doubt of my own understanding. I admit that I am blind
to the subtle differentiation that you two have.
My own understanding does seem to leave me in an information theoretic
approach.
This boils down to questioning the fundaments of geometry and
graphical representation.
An argument can be made that the polysign behavior
sum over s ( s x ) = 0
which for any n-signed number system yields
- 1 + 1 ... @ 1 = 0
is inherently a statement of simplex geometry.
Indeed this tight coupling is all that I have to work from, for
without it the simplex geometry is gone.
Thus this simple first principle is at a behavioral level beyond the
algebra.
It is a render operation.
It is the graphics.
It is the representation.
It is not the algebra.
Never is this first law necessary to perform any computation.
Until the render is needed this operation need not exist.
Thus as you may not see the informational context is of an underlying
principle of accumulation.
While this principle may be argued by some as timelike I am happy to
argue that it is as much spacelike.
The render...
Please do consider it and how often and when you apply it.
Sum over s ( s x ) = 0,
- Tim
achille
>
> Yes, in fact this has been around here for a few years and the
> isomorphism is valid even though TimGolden starts with non-negative
> magnitudes only.
> This time I merely tried to obtain the isomorphism (which is known
> since 2003 at least) P4 = RxC from basic principles.
> The avid reader will observe that it is almost impossible
> not to find the isomorphism, at least with R[X]/(1+X+X^2+X^3).
> In order to get rid of the theme of polysign numbers once and forever,
> I took the freedom to compile a few results in a LaTeX document
> here:
> http://www.von-eitzen.de/math/PolysignNumbers.pdf
> Contents:
> *) Isomorphsims Pn = R[X]/(1+X+...+X^{n-1})
> *) Isomorphisms Pn = C x C ... x C ( x R?)
> *) Natural metric induced on Pn producing a regular simplex.
> But since even today the isomorphism P4 = R x C is doubted
> by others, I have little hope ...
>
> hagman
>
Nice paper, too bad there isn't an IgNobel Prize for mathematics ;-p
hagman
I went for the IgFields medal ;)
victor_me...@yahoo.co.uk
>
> THEOREM.
> P4 is isomorphic as a metric R-algebra to the RxC with norm
> ||(r,z)|| = sqrt( 1/3 * r^2 + 2/3 * (Re(z)^2+Im(z)^2) ).
This is a very nice summary of the situation. But perhaps
a more natural way of seeing this isometry is to realize
that we should be considering R x C as the *-fixed points
of the R-algebra C x C x C with involution defined by
(z_1, z_2, z_3)* = (z_3*, z_2*, z_1*)
where z_j* is the complex conjugate of z_j. The images
of Goldenballs's units are (i^r,i^{2r},i^{3r})
for r in {0,1,2,3}.
Under the quadratic form |z_1|^2 + |z_2|^2 + |z_3^2|
it's plain these 4 points form the vertices of a regular
tetrahedron. Scaling by the factor 1/3 makes the distance to
the origin 1.
Exactly the same game works for all of
Mr Goldenballs's P_n.
Tim BandTech.com
> On 23 Feb., 07:27, achille <achille_...@yahoo.com.hk> wrote:
>
>
>
> > On Feb 23, 4:39 am, hagman <goo...@von-eitzen.de> wrote:
>
> > > Yes, in fact this has been around here for a few years and the
> > > isomorphism is valid even though TimGolden starts with non-negative
> > > magnitudes only.
The very need to state the term
"non-negative magnitude"
is a state of mind which I do care to criticize.
As such I reenjoy the concept of the swiss army knife and the chisel.
But for you Hagen I would have to raise the knife to a loaded tool
chest I think. It's easy to get carried away with the tools. When such
a simple construction as polysign comes along what is the value of
remaining in the toolchain?
Come now, let's be fair. If I were to reconstruct your number system
using your number system with a high degree of complexity wouldn't you
laugh at me?
All of my work would have been done just to chase my tail around.
Who is wagging the dog here? Is it I or is it Hagen? Clearly my
construction is compact and to the point. In that the P2 construction
builds the real numbers then the information contained within the
definition of real numbers that is common to the laws of polysign can
be removed from detail x in sx. This leaves magnitude a very raw
concept with no need of declaring such silliness as
"non-negative magnitude"
for under the polysign construction there never was any negative
magnitude. Magnitude is inherently signless. Clearly then it is the
simplest of the concepts of continuum that we are discussing as
magnitude. Sign then is relegated to the discrete portion s of sx. It
has been tradition to claim that there are two signs; not three, not
four, not even one...
It has been tradition to take the magnitude of a real number or a
complex number
| A |
rather than build up these higher number systems from magnitude. The
way of seeing the system under polysign is an altered state and it
does cause the old system to sway, especially when emergent spacetime
has been sitting there for eons waiting to be discovered with no
signal detected from your shop.
I must reread your paper again but for now I am settling into The
Curly Haired Hen. La Poule a Poils reminds me of my own front porch
where I have a hen that is roosting for the winter.
You know, you've been so quiet and now out I come with prize money and
you are barking up a storm, winning and then refusing the prize money.
If I get your mailing address I may just send you the check anyway.
- Tim
> > > This time I merely tried to obtain the isomorphism (which is known
> > > since 2003 at least) P4 = RxC from basic principles.
> > > The avid reader will observe that it is almost impossible
> > > not to find the isomorphism, at least with R[X]/(1+X+X^2+X^3).
> > > In order to get rid of the theme of polysign numbers once and forever,
> > > I took the freedom to compile a few results in a LaTeX document
> > > here:
> > > http://www.von-eitzen.de/math/PolysignNumbers.pdf
> > > Contents:
> > > *) Isomorphsims Pn = R[X]/(1+X+...+X^{n-1})
> > > *) Isomorphisms Pn = C x C ... x C ( x R?)
> > > *) Natural metric induced on Pn producing a regular simplex.
> > > But since even today the isomorphism P4 = R x C is doubted
> > > by others, I have little hope ...
- T, BandTech.com
Tim BandTech.com
> On Feb 22, 3:39 pm, hagman <goo...@von-eitzen.de> wrote:
>
> > On 21 Feb., 13:40, achille <achille_...@yahoo.com.hk> wrote:
>
> > > On Feb 21, 2:24 am, hagman <goo...@von-eitzen.de> wrote:
>
> > > > On 19 Feb., 16:56, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>
> > > > > This has been an ongoing debate by a few here.
> > > > > I attempt to promote the polysign numbers
> > > > > http://bandtechnology.com/PolySigned
was here but google is truncating so I am too.
> > > > > A $50 prize deliverable as a money order to anywhere possible (or if
> > > > > you prefer a personal check) for solving this puzzle. $50 is not >
> > The multiplication table of -1,+1,*1,#1 (in P4 notation) is the same
> > > as that of u, u^2, u^3, 1 in any formal algebra where u^4 = 1. The
> > > identity -1+1*1#1 = 0 indicates P4 is nothing but R[u]/(1+u+u^2+u^3),
> > > a quotient ring of the polynomial ring R[u].
>
> > Yes, in fact this has been around here for a few years and the
> > isomorphism is valid even though TimGolden starts with non-negative
> > magnitudes only.
> > This time I merely tried to obtain the isomorphism (which is known
> > since 2003 at least) P4 = RxC from basic principles.
lately is the isometric form and attempting to discuss what exactly
the correction factor that you label 'norm' and Victor called
'distance function' should be interpreted as. I do accept the form and
will when I get to coding enter it into my code as a HagenDazs or some
such named T3HeitzenCorrectionFactor.
A graphic of the correction factor will expose an ellipsoid won't it?
> > The avid reader will observe that it is almost impossible
> > not to find the isomorphism, at least with R[X]/(1+X+X^2+X^3).
> > In order to get rid of the theme of polysign numbers once and forever,
are no claims of any deep mistake on my part in constructing them.
Their simplicity is outstanding wouldn't you say? That three lines
sum over s ( s x ) = 0
s1 x1 + s1 x2 = s1 ( x1 + x2 )
(where '+' is superposition) can build out all of these systems,
including the real numbers and the complex numbers themselves. Hagen
you seem such a philosphical character yet your communications are
always edge on, sharp, and short. I conclude that your own cryptic
mind has not fully decoded yet, or that you are working with me from
the other side.
My own lack of ability is fair proof of the simplicity of the polysign
in that it builds to such complexities as you are assimilating.
Whereas Hamilton would have claimed that there are no 3D complex
numbers, I will maintain that P4 are they, and that spacetime support
is near at hand to one who can accept this P4 freak show. As you have
validated the polysign family as compatible with some high level form
of mathematics then I would point this out clearly as a confrontation
to your claim to have put polysign numbers to bed. On the one hand
they are childishly simple, yet on the other you've rebuilt them out
of the sacred real and complex numbers, and with a fair amount of
detail required. I guess that you do understand them and see that I am
looking for a fight rather than simply wanting to slide them in to
some journal that I will struggle to gain access to and then let them
slide out of sight into the stacks of history. Yet they should stand
freely on their own
> > I took the freedom to compile a few results in a LaTeX document
> > here:
http://www.von-eitzen.de/math/PolysignNumbers.pdf
> > Contents:
> > *) Isomorphsims Pn = R[X]/(1+X+...+X^{n-1})
> > *) Isomorphisms Pn = C x C ... x C ( x R?)
> > *) Natural metric induced on Pn producing a regular simplex.
> > But since even today the isomorphism P4 = R x C is doubted
> > by others, I have little hope ...
I do accept the isomorphism and now the isometry too, though I still
wonder whether the RxC product could be declared such that the norm
disappears.
I have been lazy lately and haven't done much work, but I do see some
nice projects brewing.
- Tim
>
> > hagman
>
> Hello Hagen.
>
> May I take a copy of your paper for my own website?
> Or if you do wish to make a fairly permanent link that would suffice.
> Certainly I do have a response for you, particularly the points which
> you wipe over yet which you do care to state clearly.
> I deeply appreciate your serious response.
> You give me pause for concern.
> As I consider carefully what you have written
> I know that you will understand that the amount of time it takes me to
> respond and the level of confidence with which I can do so are coupled
> as they are for you in your asymmetrically diaposed position. Yet the
> symmetry of the system remains for both of us, and so as no mistakes
> are appointed by you to polysign I am deeply grateful of your own
> exposition in the terms of existent mathematics. As to the audience
> whom you are capable of addressing who is it exactly? Shall we
> consider that the gorilla conjecture can be brought up...
>
> read more »
Tim BandTech.com
I failed to catch this detail earlier. There is a lack of distance
conservation in the inputs under your mapping phi() here.
I've finally gotten back into my code and have implemented your
distance correction that you conclude with, however I've not yet
implemented this transformation of the inputs.
As I try to make sense of this I go back to the definitions of
isometry and so forth and wonder if my requirement:
"This offer stands to anyone who can instantiate an isometric
isomorphism between P4 and RxC which is consistent under product."
is sensible since the concept of isometric isomorphism seems to be
merely a matter of two spaces without any operators acting.
You have taken the freedom to morph both the input space and the
output (product) space by two different functions, even though they
are the same underlying spaces, i.e. for values in P4 their product
will remain in P4, and likewise for RxC. I do see that there is some
attention to differing distance functions within isometry, but then
too each space already comes with prebuilt distance functions.
You know we went over this a long time ago and I am sorry it's taken
me so long long to catch this detail again here. I will carry on and
enter your input transform phi() at which point the work I have
already coded should hopefully check out OK. In your final theorem
statement beneath here there is no statement of "isometric
isomorphism" so I hope to get your own interpretation of these
details. As I recall the last thread that I raised a alarm bell on
your phi() mapping went dead. This one's been dead for awhile. We'll
see if I can raise it.
- Tim
Tim BandTech.com
> On Feb 20, 2:24 pm, hagman <goo...@von-eitzen.de> wrote:
This has been an ongoing debate by a few here.
I have decided to reopen the $50 US prize whose worth in just a few
short months may be meaningless if the capitalists decide to hold
firm. There are many pretty domains to inhabit and like them we have
our choice in the math world of which domain we wish to inhabit, at
least for a while as we ponder the domain that is existence.
The requirements that I have given have not been strictly met by Hagen
Von Eitzen and so I refresh my initial post here. I am sorry if I have
misworded the summary position as "isometric isomorphism" for I am no
longer certain that such an expression has any meaning as a means of
equivalence between P4 and RxC. The equivalence mechanism which I have
proposed is generic and requires no extraordinary distance function
manipulations, for the product ought to be capable of holding its own.
If Hagen wants to claim the prize he may successfully convince me, but
meanwhile I do refresh the $50 US prize money for the solution that I
seek and which was clearly stated in my initial post which follows:
I attempt to promote the polysign numbers
http://bandtechnology.com/PolySigned
which are nearly field algebras in any specified dimension and support
emergent spacetime geometry from arithmetic. For P4 the RxC similar
space seems to have become a sticking point of the old school whereby
the polysign numbers can be avoided by steerage over to associative
algebra studies.
The standard refutation seems to be that the polysign construction is
contained within existing associative algebra work and the higher sign
systems will always mimic copies of R and C so that the four-signed
numbers P4 will mimic RxC. This statement does seem to hold up to
isomorphism, but this bending of the P4 arithmetic product does not
necessarily lend itself to understanding the system any better so long
as the bending is not specified. So the next stage of that study is to
map an isometric isomorphism.
A $50 prize deliverable as a money order to anywhere possible (or if
Androcles
"Tim BandTech.com" <tttp...@yahoo.com> wrote in message
news:942a53fd-7392-4a74...@z15g2000yqm.googlegroups.com...
> On Apr 2, 8:37 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:
>> On Feb 20, 2:24 pm, hagman <goo...@von-eitzen.de> wrote:
> This has been an ongoing debate by a few here.
>
> I have decided to reopen the $50 US prize whose worth in just a few
> short months may be meaningless if the capitalists
Off topic for a sci newsgroup. Take it where it belongs.
amy666
>
> "Tim BandTech.com" <tttp...@yahoo.com> wrote in
> message
> news:942a53fd-7392-4a74-9278-9ac16fb9790b@z15g2000yqm.
> googlegroups.com...
> > On Apr 2, 8:37 am, "Tim BandTech.com"
> <tttppp...@yahoo.com> wrote:
> >> On Feb 20, 2:24 pm, hagman <goo...@von-eitzen.de>
> wrote:
> > This has been an ongoing debate by a few here.
> >
> > I have decided to reopen the $50 US prize whose
> worth in just a few
> > short months may be meaningless if the capitalists
>
> Off topic for a sci newsgroup. Take it where it
> belongs.
>
>
algebra off topic in a math forum ??
thats new !!
victor_me...@yahoo.co.uk
> On Apr 2, 8:37 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:> On Feb 20, 2:24 pm, hagman <goo...@von-eitzen.de> wrote:
>
> This has been an ongoing debate by a few here.
>
> I have decided to reopen the $50 US prize
> The requirements that I have given have not been strictly met by Hagen
> Von Eitzen and so I refresh my initial post here.
You mean you aren't going to give "hagman" fifty of
your worthless currency units after all?
You mean sod! :-(
leland....@gmail.com
> On Apr 2, 8:37 am, "Tim BandTech.com" <tttppp...@yahoo.com> wrote:> On Feb 20, 2:24 pm, hagman <goo...@von-eitzen.de> wrote:
>
> This has been an ongoing debate by a few here.
>
<snip>
> This offer stands to anyone who can instantiate an isometric
> isomorphism between P4 and RxC which is consistent under product.
> The layout of this construction should include a transformation T such
> that
> y( P4 ) <--> x( R ), z( C )
> e.g. T( y ) = x, z
> such that
> T( y1 y2 ) = T( y1 ) T( y2 ).
> The right hand side arithmetic product should be clearly defined on
> RxC which is the part of the puzzle that remains troubling.
Hagen already did that for you, you apparently couldn't quite see
that. His work was quite elegant and clear, but since you seem to
desire things spelled out extremely explicitly:
Suppose y1 = -a+b*c#d where a,b,c,d are in RR. The Hagen shows the
isomorphism follows via the map T(y1) |-> ( b+d-a-c , d-b + (a-c)i ).
Now, if y2 = -r+s*t#u where r,s,t,u are in RR, then if I follow your
polysign ideas correctly we have:
y1y2 = +ar*as#at-au
*br#bs-bt+bu
#cr-cs+ct*cu
-dr+ds*dt#du
which if we collect terms (or signs -- whatever the right term is) we
get:
y1y2 = -(au + bt + cs + dr)+(ar + bu + ct + ds)*(as + br + cu + dt)#
(at + bs + cr + du)
where the '+' signs inside braces is the usual addition of positive
real numbers. By definition of the map we have (using square braces
for clarity):
T(y1y2) = ((ar + bu + ct + ds) + (at + bs + cr + du) - (au + bt + cs +
dr) - (as + br + cu + dt) , (at + bs + cr + du) - (ar + bu + ct + ds)
+ [(au + bt + cs + dr) - (as + br + cu + dt)]i )
Alternatively we have:
T(y1)T(y2) = ( b+d-a-c , d-b + (a-c)i )( s+u-r-t , u-s + (r-t)i )
= ( (b+d-a-c)(s+u-r-t) , (d-b + (a-c)i)(u-s + (r-t)i) )
Which expands out to (using square braces for clarity):
T(y1)T(y2) = ( (ar + bu + ct + ds) + (at + bs + cr + du) - (au + bt +
cs + dr) - (as + br + cu + dt) , (at + bs + cr + du) - (ar + bu + ct +
ds) + [(au + bt + cs + dr) - (as + br + cu + dt)]i )
Thus we have T(y1y2) = T(y1)T(y2) as expected. Hagen showed more than
this, of course, including the isometry using an appropriate metric on
RR x CC, and much more elegantly. In short, you owe Hagen $50.
Tim BandTech.com
Yes, I agree to what you say here. Hagen has refused the money, but I
would still like to pay him. And yes he has gone far above and beyond
my request in his pdf
http://www.von-eitzen.de/math/PolysignNumbers.pdf
much of which I am confused by. I don't understand how a polynomial
constructs a multidimensional system. That is what seems to be going
on starting at section three.
I guess my thinking was partly that the transform phi() which Hagen
presented is offensive to the geometry which I have assumed to be
Euclidean in my own attempts. In other words since RxC is a 3D space
the distance of the transform phi() could maintain the distance
presented it in P4. As I process this attachment that I have I see
that clearly Hagen's way must be the right way to maintain the simple
isolated product in RxC, but that another way might be stated which
does maintain the distance of the source space and actually looks very
likely to exist. Yet the arithmetic product in RxC will then appear
more contorted. This solution would expose a series I believe, as
exposed by this analysis:
http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html
which does conserve distance in the transform from P4 to RxC, which I
call T3 since RxC matches the tatrix format of the polysign
dimensional progression P1,P2,P3,...
Taking the freedom to use two distance transforms to solve the problem
feels suspicious to me, but it works and I have implemented the P4
version of Hagen's work into my code.
Still I am curious what a distance conserving version would do to the
arithmetic product definition in RxC so I offer up even another prize
to expose this. You seem so bothered Leland, but I am happy that you
follow along and am happy to be corrected by you. Thanks for putting
your own two cents in here. At some level all of this is just a
diversion to me, but it seems to be drawing out some sort of result as
evidenced by Hagen's excellent work. The polysign numbers are
extremely primitive. They build the real numbers and the complex
numbers, on which Hagen's work relies. They yield spacetime through
their curious distance behaviors in P4+. I don't feel that these
features are well born out by Hagen's paper, but that is his agenda
and I respect that. Especially to overlook the curious behaviors of P1
and the time correspondence is a missed opportunity to lead into
physical phenomena. This is where the polysign paradigm leads and
signals of coherence are present, but they are hidden under the shroud
of the math that Hagen purveys, which treats the real number as
fundamental. Under the polysign paradigm the sign symbols in front of
every 'real' number are not fundamental but are evidence of
substructure and this theory is born out by the polysign construction.
- Tim
leland....@gmail.com
I would suggest that this is a point that you need to work on. It's
great that you're interested in exploring the properties of polysign
numbers. Having ideas and exploring their mathematical consequences is
a great thing to do. An important point, however, is that polysign
numbers can be described very neatly within a standard commutative
algebra framework (or via group algebras). This point is important
because it allows for generalisation and more efficient explication --
the existing framework exists for a reason: it is both flexible, and
powerful. What seems difficult to discern from polysign numbers (say
the isomorphism between P4 and RR x CC) is often straightforward to
see and prove in the usual framework. Now, not knowing the usual
framework of commutative algebra is not a bad thing: lots of people
don't know it, and don't have a need to know it. However, it has been
pointed out to you several times that a good understanding of such
things (rings, polynomial rings, group rings, etc.), and associated
theorems, will go a long way toward allowing you to see and prove more
regarding your polysign numbers. You really should be investing the
effort to learn exactly what it is people like Hagen are talking
about.
The best place to start is a decent introductory book on modern
algebra (or abstract algebra). I believe Dover publications have a
number of offerings at very reasonable prices (ranging from $13 to
$35). Alternatively your local library may have some suitable books.
There is even a fair range of material on-line for free; you could
try:
http://www.math.miami.edu/~ec/book/
or for something a little more advanced try:
http://www.math.uiuc.edu/~r-ash/Algebra.html followed by
http://www.math.uiuc.edu/~r-ash/ComAlg.html
or just do some searching for on-line course notes. It will be fun and
interesting and it will give you a whole range of far more powerful
tools and techniques for you to use in your studies.
Tim BandTech.com
> or just do some searching for on-line course notes. It will be fun and
> interesting and it will give you a whole range of far more powerful
> tools and techniques for you to use in your studies.
Well I started into your first link and do like it. The 'relation' as
it is presented there is new to me. There are so many areas of math.
Maybe abstract algebra will feel more valuable than some other
branches I've tried.
Do you think that the polynomial path to dimensionality is so
difficult to describe? Does it lead from the square root of a negative
real as a different entity from the square root of a positive real?
Whether the multidimensional space ought to be granted first, then
build out the number systems as subsystems seems a possibility. Is it
as if a blindfold is lifted when we go from the real numbers to the
complex numbers? Then lifting this blindfold again we should land in
RxC, etc.? The polysign progression suggests a different pattern.
Unfolding the blindfold we see a reality correspondence in the
progression
P1 P2 P3 | P4 P5 ...
Thanks to the distance behavior of the arithmetic product there is a
breakpoint beyond P3. How this can be translated into physics is
mostly what I attempt. Spacetime with unidirectional time is embodied
by the polysign progression. This simply has not been exposed by the
mathematics that you purvey.
I appreciate your guidance. I'll work some more on abstract
algebra.
- Tim
amy666
R[x]/( 1 + x + x^2 + x^3 )
its quite trivial to see this is equivalent to P4.
R[x] is the magnitude.
and "/( 1 + x + x^2 + x^3 ) " is just the cancellation law , where 1 , x , x^2 and x^3 are the different signs.
regards
tommy1729
( btw expressions like "/( 1 + x + x^2 + x^3 ) " are the reason multisections and polysigned are so strongly related , as has always been said by me and galathaea )
Lalo Torres
The collection D of all split complex numbers z = x + y j for x, y ∈ R forms an algebra over the field of real numbers. Two split-complex numbers w and z have a product wz that satisfies N(wz) = N(w)N(z). This composition of N over the algebra product makes (D, +, ×, *) a composition algebra.
A similar algebra based on R2 and component-wise operations of addition and multiplication, (R², +, ×, xy), where xy is the quadratic form on R², also forms a quadratic space. The ring isomorphism
D -> R²
x + yj |--> ( x + y , x - y )
relates proportional quadratic forms, but the mapping is not an isometry since the multiplicative identity (1, 1) of R² is at a distance √2 from 0, which is normalized in D.
source : https://en.wikipedia.org/wiki/Split-complex_number
It has certain resemblance with:
with https://en.wikipedia.org/wiki/Quasi-isometry#Examples
The character · is mnemonic to indicate component of the pluses and minuses.
In the right side, component-wise (A,Z) units
In the left, the P4 units
Alternatively, i character can be treated as sign.
(1/√2)(+1 @ #1 ) = (·+ 1) ≊ (-1,0)
(1/√2)(-1 @ *1 ) = (·- 1) ≊ (+1,0)
(1/√2)(-1 @ (*1)' ) = (+· i) ≊ (0,+i)
(1/√2)(+1 @ (#1)' ) = (-· 1) ≊ (0,-1)
(1/√2)(*1 @ (-1)' ) = (-· i) ≊ (0,-i)
(1/√2)(#1 @ (+1)' ) = (+· 1) ≊ (0,+1)
In his web, Tim give four of the six units, only the units of imaginary axis of the complex component is missing.
Note that the modulus of A units and the Z units is different, but, that is corrected with a correction factor.
Also, note that multiply any A unit with any Z unit yield 0, that is, they are zero-div couples.
In (A,Z) units can not be arithmetically mixed before the spatial display...
... but this is not the case of P4...
Note that if we add an A unit with a Z unit, a √2 factor appears.
For example (-1,0) with (0,+i) :
(·- 1) @ (+· i) = (1/√2)(-1 @ *1 ) @ (1/√2)(-1 @ (*1)' ) = (1/√2)( -1 @ *1 @ -1 @ (*1)' )
(·- 1) @ (+· i) = (1/√2)(-2) = (1/√2)( -( (√2)(√2) ) ) = -√2 = (@√2)(-1)
(·- 1) @ (+· i) = (@√2)(-1) <--- here is the -√2 factor
(1/√2)((·- 1) @ (+· i)) = -1 <---
The justification of mixing components through addition in P4 is exactly, the zero-divs.
The Zero divisors are other way to say "compartmentalize in components".
Note that the logics of the corrected factors of H. Von Eitzen to regularize the tetrahedron is impeccable.
Now, certainly numbers can be obtained through (A,Z),
since they both are dependent over a,b,c,d (magnitudes or non-negative reals)
The addition and product developed by H. Von Eitzen is also impecable.
P4_display(a,b,c,d) = - a + b * c # d
A_Z_display(a,b,c,d) = ( -a+b-c+d , a*i -b -c*i +d )
but, as an analogy, not different that getting numbers out of R² relative to the split-complex numbers.
Isomorphism, absolutly, isometric, mmmm...
https://en.wikipedia.org/wiki/Taxicab_geometry#/media/File:Manhattan_distance.svg