Re: Treating Magnitude as Fundamental

Hi hagman. I'm not comfortable with all of your notation but it looks
like you are close. The geometry is not imposed. It is inherent and I
will make a note to you below where this thought occurred to me.

Furthermore I will quibble with you about your usage of the term
'reals' within the construction. This is because P2 are the reals and
upon deleting away the sign mechanics of the product definition and
the superposition rules of the polysign definition all that is left is
magnitude. We cannot build the reals from the reals. I did say quibble
and beyond that I am happy to flex to you definition for this thread
and will try to keep up with your notation.

Are you afraid to declare that this system generalizes sign?
That's alright if you are, so is just about everyone.
I would not get overly concerned about the roots of unity argument.
There is something there but dimension is a far more pervasive part of
the system since these nth roots of unity are in n-1 dimensional
space. Of equal importance an algorithm that takes (-1)^m where m is a
progression will cover the dimensions of the space. I think this is
the simpler way to look at it.

On Sep 14, 2:07 am, hagman <goo...@xxxxxxxxxxxxx> wrote:
On 13 Sep., 13:51, "Timothy Golden BandTechnology.com"



<tttppp...@xxxxxxxxx> wrote:
On Sep 13, 5:12 am, Hero <Hero.van.Jind...@xxxxxx> wrote:

Hero wrote:
lwal...wrote:

What about P4? When Tim first posted
about the polysigned numbers here
at this newsgroup years ago, it was
Robin Chapman who pointed out that
P4 is isomorphic to R x C. Mr.
Chapman proved this by using the
following isomophism:

-1 becomes (-1,i)
+1 becomes (1,-1)
*1 becomes (-1,-i)
#1 becomes (1,1)

and then use componentwise multiplication
in the ring R x C.

I did know this one, it's nice to know.
The four points ( -1, +1, *1 and #1) form a tetraeder in 3D.

With Tim's coordinates it's a tetra with edges of equal length.
Robin has a stretch factor in the second component of sqrt( 3/2), so
it's not an equal edged tetra.

With friendly greetings
Hero

I'm pretty sure that the anonymous author has a memory mismatch. I
think he is actually talking about information from Gene Ward Smith.
The only thread that I find Robin Chapman on is
http://groups.google.com/group/sci.math/msg/6e202537299206c9
which was very early on. And I really don't think that I've spoken
with Chapman about four-signed whereas Gene has made the author's
specific claim and I have refuted it near
http://groups.google.com/group/sci.math/msg/65e4bea31dd8a0f1
I'm now reasonably certain that this corrects the author
lwal...@xxxxxxxxx's statement.

I still do refute the claim that P4 can be stated cleanly in RXC.
I have studied this problem carefully yet have not resolved the
discrepancy.
In this regard it is an open problem.
If anyone has a P4 equivalent product I am happy to test it through
the same code base that generates the graphics on my website
http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html
Caution! the gifs on this page overload a small computer. I must have
some bugs in them which I have not corrected yet. I apologize but
better yet would be to fix them. I am a lazy one man band and the cord
to these cymbals is broken. If you know gifs and palette issues
perhaps you could help me out. I guess they are not compressed
properly. They are made in libgd from a 24 bit color map series of
frames.

Nice to hear from you Hero.

-Tim

If I get it right, you essentilly start with P, the set of positive
reals (or possibly
the set of positive rationals or of some other subfield F of R),
for me I call this magnitude(x).
add symbols s_0, ..., s_{n-1} such that
1) s_0 = 1
2) s_i * s_j = s_k if i+j = k mod n
3) s_0 + ... + s_{n-1} = 0
Because of 3), s_0 has an additive inverse and we may therefore forget
that we started
only with P instead of R (or Q or F).
Thus we can use standard notation to see that you construct the ring
TIM_n(F) := F[s_0,s_1,...,s_{n-1}]/(s_0-1, s_0+...+s_{n-1}, s_i s_j
- s_k | i+j=k mod n)
for some natural number n>0.
Since s_i = (s_1)^i, this can be simplified to
TIM_n(F) = F[s]/(1+s+...+s^{n-1}, s^n-1) = F[s]/(1+s+...+s^{n-1})
Especially,
TIM_1(F) = F[s]/(1) = 0
TIM_2(F) = F[s]/(1+s) = F
TIM_3(F) = F[s]/(1+s+s^2)

I'm sorry to say that I do not understand the expressions above. I
think it is the usage of the slash mark '/' that I don't understand.

For F=R, we find TIM_3(F) = R[s]/(1+s+s^2) = C by mapping s |-> (third
root of unity).

Here is the point that I noted above. This mapping is only in
hindsight. The polysign construction is freestanding. It is directly a
consequence of the generalization of sign wrt the superposition and
product that P3 and C correspond; nothing more. A transform must be
established but this is merely a frame of reference.

In general, we have a surjective ring homomorphism TIM_n(R) -> C by
mapping
s|->(n^th root of unity), but this is not an isomorphism.
If we take F=Q instead, the standard map TIM_n(Q) -> C, s |-> n^th
root of unity
is injective (i.e. an isomorphism with its image field Q[n^th root of
unity])
because we have (n-1)-dimensional Q-spaces on both sides -- provided
that n is prime
(we need that phi(n)=n-1).

Did I get it right so far?
hagman

I hope you got that right. The construction is very direct under my
own rendition and the indirect nature of your language is difficult
for me. The truth is that sign can be generalized so no different than
you teach a grade schooler about two signs out of thin air you can
also teach them three signs. Beyond this there are a series of
consequences and open problems that deserve study and I do welcome
your own rendition though I may criticize it as elitist jargon. I
don't mean offense to you. It seems that you truly think this way and
there are others like you so I am deeply grateful of your attention
and ability to put this so that others can take it in. Thanks and
carry on. There is a puzzle in expressing the P4 product in terms of
RxC that seems to be coming up again here and if you like a challenge
I welcome this pursuit.

-Tim

.

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