Treating Magnitude as Fundamental
Timothy Golden BandTechnology.com
some reason it is buried here and my posts are not showing through on
google's topic list thought the content is present. So I try again
here because I would like to get your input.
This is a really simple construction. Simply consider a magnitude that
sweeps from zero to large values and impose it as m onto an iterated
function:
z(n+1) = - z(n) z(n) @ m
where z is a polysign value and @ implies superposition. The magnitude
essentially takes the zero sign which is right where it should be. The
equivalent in the complex plane (P3) would be
z(n+1) = (e^(i2pi/3)) z(n) z(n) + m .
Plot z(n) for each magnitude m. Iterations of n to 10000 or so will
either get very large, very small, or find some stable orbit. The
increment in m gets pretty good results around 4e-4 but coarser will
certainly generate graphics. The graphics of this procedure are pretty
interesting. The function itself could be alot of different things but
it is the rotational effect in the product that is generating the
interesting detail. The color scheme needs a lot of work but this is
some of the results:
http://bandtechnology.com/PolySigned/MagnitudeSweep
If you don't understand polysign you could try:
http://bandtechnology.com/PolySigned
-Tim
lwa...@lausd.net
<tttppp...@yahoo.com> wrote:
> I've started another thread titled "Magnitude Sweep Functions" but for
> some reason it is buried here and my posts are not showing through on
> google's topic list thought the content is present. So I try again
> here because I would like to get your input.
OK, I've these threads of yours about "polysigned
numbers" over the last few years, but this is the
first time that I've posted in one of them.
We all know how to constuct R, the set of real
numbers, from N, the set of natural numbers.
1. Make N into an additive group, Z.
2. Make Z into a field, Q.
3. Make Q into a complete field, R.
But this is not how the website metamath does
the construction. For metamath uses a different
order to construct R from N:
1. Make N into a multiplicative group, Q+.
(actually metamath uses a script Q here)
2. Make Q+ complete, P (or R+).
3. Add signed numbers to P, R.
In some ways, this actually matches the
historical development of R. For rational
numbers were known to ancient Egyptians
and Sumerians, then the Pythagoreans
introduced irrational numbers such as
root two. Not until Cardano were negative
numbers, along with complex numbers,
formally introduced.
Now what Timothy is doing is generalizing
this third step. The set P (metamath
calls these "unsigned reals," while Tim
calls them "magnitudes") has already
been formed (whether by Cauchy sequences
or Dedekind cuts, it doesn't matter), and
now he is trying to form a new set.
Tim takes P^n, the Cartesian product of
n copies of P (where n is any positive
number), with componentwise addition and
scalar multiplication, and mods out the
set generated by (1,1,...,1). Then this
resulting group he calls Pn. (When n=2,
the group P2 is isomorphic to R, since
if we identify (a,b) with a-b, we have
that (a,b)=(c,d) exactly when a+d=b+c.)
Finally, he defines multiplication as
cyclic convolution, in order to make
Pn into a ring. (This is also a
generalization of multiplication in P2
or R, where (a,b)(c,d)=(ac+bd,ad+bc).)
The reason Tim calls the elements of Pn
the "polysigned" numbers is that he
considers each coordinate of the ntuple
to have a different sign. So just as
(a,b) in P2 denotes a-b (or to be more
precise, +a-b), (x_0,x_1,x_2,x_3) in
P4 is #x_0-x_1+x_2*x_3. He always uses
- as the first sign -- i.e., -1 is the
ntuple (x_0,x_1,x_2,...) where x_1=1
and x_i=0 otherwise. Then + is always
the second sign, * the third sign, and
# the fourth sign. He points out that
in Pn, the nth sign is the same as the
zeroth sign, so that +1 in P2, *1 in
P3, #1 in P4, etc., denotes the
multiplicative identity of the ring Pn.
Since Pn is simply P^n (which is of
course n-dimensional) modded out by
the one-dimensional (1,1,...,1), we
expect Pn to be (n-1)-dimensional.
Tim points out that P1 is simply the
zero ring and P2 is, as already
mentioned, isomorphic to the field R of
real numbers. Then he shows us that P3
is isomorphic to the field C of complex
numbers, as -1, +1, and *1 correspond
to the three cube roots of unity in C.
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. Thus unlike P2 and P3,
P4 is not a field since (-1+1)(-1*1)
equals -1+1*1#1 which is zero. Indeed,
it's easy to show that Pn is never a
field if n is composite.
No one has thus commented on P5. But I
realized that since P3 is related to the
cube roots of unity, and P4 is related to
the fourth roots of unity, perhaps I
could use the fifth roots of unity in
order to decipher P5. And sure enough, I
discovered that P5 is indeed isomorphic
to C^2, by letting -1 in P5 be:
(e^(i4pi/5),e^(i2pi/5))
and defining the other signs similarly.
So P5 is not a field either. In fact, if
we let % be the fifth sign and phi be the
golden mean, then:
(%phi-1#1)(%phi+1*1)
=%phi^2-(phi@1)+(phi@1)*(phi@1)#(phi@1)
(Tim uses @ to denote addition in P --
that is, the set of magnitudes which we
defined before defining Pn.) And since
phi^2=phi@1, we have:
=%(phi@1)-(phi@1)+(phi@1)*(phi@1)#(phi@1)
=0
And thus it appears that Pn cannot be a
field for any n other than 2 and 3. (Of
course, no vector space over R whose
dimension is more than 2 can be a field,
which is why we have noncommutative
quaternions, nonassociative octonions.)
I'm curious as to what would happen if we
formed the polysigned numbers from the
(unsigned) rationals Q, rather than the
unsigned reals (magnitudes) P, to form Qn
rather than Pn. Once again, Q1 would be
the zero ring, Q2 would be the field
of rationals, and Q3 would be isomorphic
to the field Q(sqrt(-3)). Once again,
Qn cannot be a field if n is composite,
but I wonder whether Qp can be a field if
p is a prime greater than 3.
Timothy Golden BandTechnology.com
> I'm curious as to what would happen if we
> formed the polysigned numbers from the
> (unsigned) rationals Q, rather than the
> unsigned reals (magnitudes) P, to form Qn
> rather than Pn. Once again, Q1 would be
> the zero ring, Q2 would be the field
> of rationals, and Q3 would be isomorphic
> to the field Q(sqrt(-3)). Once again,
> Qn cannot be a field if n is composite,
> but I wonder whether Qp can be a field if
> p is a prime greater than 3.
Wow. Thanks for stepping forward.
There is a high quality post
http://groups.google.com/group/sci.math/msg/77653a0e1e43d2bc
which I have verified out to large n and certainly primes were
included.
The author never fully explained it and the book he mentions is
unavailable every time I try to find it. So it is a bit vacuous but no
doubt you might be able to connect with it.
Thank you for your careful rendition and I am in agreement with all
that you say.
I am not so concerned about the field constraints especially since it
is the division clause which is the conflict. Division is still
possible in the higher signs and certainly the algebraic behaviors of
all of Pn are consistent in terms of associative and commutative
behaviors. Yes, division is challenging to do generally, but upon
instantiating a few products we can clearly state that their reversal
is division. The troubling ones are the ones with dimensional
reduction as evidenced in the P4 product survey
http://bandtechnology.com/PolySigned/Deformation/AxisDualDeformStudy.gif
One will never recover the sphere from a line or a plane and these are
some of the possible product results. How this maps into higher
dimension is a level of complexity that may be beyond my simple minded
approach. Still, I do accept that there is a kaleidocopic effect in
dimension. I take some assurance from experiments that these high
dimensional systems are well behaved and yield results similar to the
lesser and better known spaces.
I do not believe that Chapman did get an exact match for the product.
If that is so then I am badly mistaken on
http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html
I will have to go back and find this post. I am impressed by your
memory.
Certainly if there is anything there then I will credit him and
correct this on my website but for me this is still an open problem as
presented in the link above.
Will a zero ring be acceptable to existing math?
That would be great. I do not use the ring terminology since it is
esoteric to many(including myself) but I appreciate your usage of it
and perhaps your rendition will be more appetizing to the mathematical
community.
Thank you for your attention.
-Tim
Hero
>
> 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.
So You have 1D redundancy. This makes it interesting for calculations,
it's a kind of error-correction ( Think of gyro-platforms). And one
has only positive numbers (and zero) - another safety. (when one
treats them as ordered four-tuples).
> Thus unlike P2 and P3,
> P4 is not a field since (-1+1)(-1*1)
> equals -1+1*1#1 which is zero. Indeed,
> it's easy to show that Pn is never a
> field if n is composite.
There are not so many fields possible.
(NB:Tim never claimed this to be a field)
Still, it is a nice multiplication to investigate.
And he gave some beautiful pictures from iterations too.
So, i like Your encouragement of Tim.
With friendly greetings
Hero
Hero
> 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
Timothy Golden BandTechnology.com
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...@lausd.net'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
tommy1729
> 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.
R x C ... true but so are other number systems...
so P4 belongs to the set R x C and is not equal to it.
rather R x C defines a 3rd dimension rather than a number system...
logical since if algebraicly closed R x R then R x R = C
and therefore R x C = R x R x R = R^3 as expected.
if you find this confusing , unbelievable or incomplete
i can assure you I am pretty sure timothy agrees that R x C is to simple to describe P4.
and that R x C simply implies "integer domain" in 3 dimensions.
i have already discussed P4 alot with timothy true email and even in this forum.
i can also assure that i will start a topic soon about such issues wich will go more into detail ...
if timothy doesnt do it before me of course.
(note : this remainds me of certain bogus proofs/disproofs to various set theorems making the same "dimensional" mistakes... my own set theory included , however this is a bit of topic and any set theory is independent of the polysigned numbers )
Mr.
> > > > Chapman proved this by using the
> > > > following isomoRRRRRRRphism:
added the R
> >
> > > > -1 becomes (-1,i)
> > > > +1 becomes (1,-1)
> > > > *1 becomes (-1,-i)
> > > > #1 becomes (1,1)
true for multiplication ...
but breaks down for stuff like zerodivisors and others
> >
> > > > 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.
what ?
> >
> > 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/6e2025372
> 99206c9
> 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/65e4bea31
> dd8a0f1
> I'm now reasonably certain that this corrects the
> author
> lwal...@lausd.net's statement.
yet the author is still a nice guy :-)
>
> I still do refute the claim that P4 can be stated
> cleanly in RXC.
as i expected ...
> I have studied this problem carefully yet have not
> resolved the
> discrepancy.
thinking in 3d or 4d is hard and counterintuitive at times, not to mention hard to imagine or draw.
in fact , as far as i know only perelmann can visualize the 4th dimension...
> In this regard it is an open problem.
a very intresting one...
unfortunately we live far away , else i would have invited you to discuss it.
> 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
equivalence is only possible in terms of couples, if at all.
that is intresting , however P4 is the ultimate form of itself...
>
>
> http://bandtechnology.com/PolySigned/Deformation/P4T3
> Comparison.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
>
regards
tommy1729
hagman
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),
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)
For F=R, we find TIM_3(F) = R[s]/(1+s+s^2) = C by mapping s |-> (third
root of unity).
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
Timothy Golden BandTechnology.com
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.
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
Timothy Golden BandTechnology.com
> so P4 belongs to the set R x C and is not equal to it.
> rather R x C defines a 3rd dimension rather than a number system...
> logical since if algebraicly closed R x R then R x R = C
> and therefore R x C = R x R x R = R^3 as expected.
> if you find this confusing , unbelievable or incomplete
> i can assure you I am pretty sure timothy agrees that R x C is to simple to describe P4.
Dimensionally yes. The catch is we have a product to consider and
there is no automatically defined product in the RxC space. So an
arbittrary definition of a componentwise product comes up but this
choice is arbitrary and is not equivalent though the great Gene Ward
Smith did try to pronounce it so. I am open to a fixup but I have
tried myself and it seems to be more of an iterative self-similar
solution that is needed. Something like a Taylor series approximation.
I like your statement above Tommy and your set theoretic style of
saying it beneath here. We are very near the domain of information and
the issue of orthogonality in an informational domain versus a purely
geometrical domain may have subtleties to do with independence and
dependence. As much as the Cartesian thinker wants it his way it may
not be so in actuality. Representationally there can be no qualms that
there is informational equivalence. The Cartesian product is forgone
in the polysign domains and so a slimmer family of systems is exposed.
> and that R x C simply implies "integer domain" in 3 dimensions.
> i have already discussed P4 alot with timothy true email and even in this forum.
> i can also assure that i will start a topic soon about such issues wich will go more into detail ...
> if timothy doesnt do it before me of course.
> (note : this remainds me of certain bogus proofs/disproofs to various set theorems making the same "dimensional" mistakes... my own set theory included , however this is a bit of topic and any set theory is independent of the polysigned numbers )
> thinking in 3d or 4d is hard and counterintuitive at times, not to mention hard to imagine or draw.
>
> in fact , as far as i know only perelmann can visualize the 4th dimension...
>
> > In this regard it is an open problem.
> a very intresting one...
> unfortunately we live far away , else i would have invited you to discuss it.
> regards
> tommy1729
I hear you Tommy. But here we have a neutral open forum with a sense
of justice. This way is even better yet since I can scratch my balls
and you needn't know.
-Tim
tommy1729
me too...
>
> -Tim
>
about R x C : read my other reply in this topic (same date as this one)
im pretty sure it will ring a bell with timothy golden and hopefully for others too.
regards
tommy1729
tommy1729
looking back at certain replies by certain individuals, i dont find much justice...
i have even been critized for not explicitly defining an infinite series as not meaning 1 + 0 + 0 + 0 + 0 + ...
not to mention the insults for not accepting cantor.
well if you read sci.math often you will have noticed this behaviour before ...
This way is even better yet since I can
> scratch my balls
> and you needn't know.
spare me the details...
but the point is that we have a long history of discussing this, and certain things are easier to explain if we meet ...
since we are outside of the realm of standard notation ( dont accept cartesian and cantor as "gods math" ) and are introducing new concepts for which no symbols have been invented yet ( most symbols refer to cartesian or cantor for historical reasons )
and are not able to say a lot in little time , or make a drawing to clarify;
i think a personal meeting will benefit us both.
>
> -Tim
>
to the heart of the reply:
P4 belongs to C x R.
but so do other number systems ( as pointed out above already ).
so P4 is not neccesarily equal to C x R.
to give an example of a number system that is not P4 yet is C x R :
beresford numbers ( 3d - complex )
also 3d
beresford is essentially the cartesian version of complex in 3d.
a = 1
a^2 = 1
bc = 1
b^2 = c
c^2 = b
and we have the 6 cartesian axes a , -a , b , -b , c , -c
which make a 3rd dimension.
in in the work of professor beresford he proves
beresford is isomorphic to R x C.
logical since
a = 1
a^2 = 1
bc = 1
b^2 = c
c^2 = b
are actually P3
and R = P2
so beresford is isomorphic to P2 x P3 => R x C.
yet beresford =/= P4 !!!
furthermore beresford has a matrixrepresentation and P4 has not ... or at least a very different one (still open problem , yet probably P4 has no matrix representation )
so since P4 AND BERESFORD are both CONSISTANT
--> we must recognize that there are at least 2 kinds of integer domains in 3 dimensions that belong to R x C.
in fact 3rd dimension is isomorphic to R x C , as pointed out in the very beginning of this post by me.
so we can simplify:
--> we must recognize that there are at least 2 kinds of integer domains in 3 dimensions
we must also recognize that using cartesian , cantor and R x C "logic" we dont get an accurate view of dimensionality or integer domains...
writing R x C does not define how many fundamental 3 dimensional integer domains exist.
worse ; it implies only 1 since all 3rd dimensions are R x C and cannot be for instance C x C or
R x R x R ( different from R x C )
and this gets even worse for higher dimensions...
defining the amount of integer domains in 6D as
( we know we need at least one C to be integer domain )
R x R x R x R x C
R x R x C x C
C x C x C
is not working at all !!!
this lists 3 wich is totally incorrect !
if i am not mistaken there is even another 3D number
( also integer domain )
P4 has 4 "signs" / vectors / axes
( and less than 4 is inconsistant in 3d )
beresford has 6 "signs" / vectors / axes
( and more than 6 is inconsistant in 3d ; overdefined space)
so we have ( tommy notation )
P4 = (3,4)
beresford = (3,6)
but there is a number between 4 and 6 !
assuming it is not equivalent to P4 or beresford ( still open question )
3D tommynumbers = (3,5)
which brings questions similar to prevouis "P4 and dimensional number" questions
the tommy notation for an integer domain in 'a' dimensions is
(a,b)
R+ / P0 -> (0,1)
real / P1 -> (1,2)
complex / P3 -> (2,3)
P4 -> (3,4)
3D tommynumbers -> (3,5)
Beresford 3D complex -> (3,6)
P5 -> (4,5)
4D tommynumbers -> (4,6)
relativity numbers -> (4,7)
bicomplex / Beresford 4D complex -> (4,8)
P6 -> (5,6)
5D tommynumbers -> (5,7)
..
of course for b there is the rule that b exists
if and only if
a < b for all a
b =< 2a for all a >= 3
even much stronger !! ;
every fundamental integer domain ( with fundamental meaning P3 and C are the same and is also equal to integer domains that are P3 in "disguise" by using for instance 3 different vectors which also make up the complex plane ; and the analogue in higher dimensions )
is equal to (a,b)
with a < b for all a
b =< 2a for all a >= 3
using the differential equations from prof beresford (easily reconstructable too) one can perform computations in almost any integer domain ...
yet many questions are still open ...
regards
tommy1729
hagman
I don't know what you refer to with your buzzwords.
I gave a more or less complete proof that TIM_3(R)=C (isomorphic as
rings).
And if my rephrasing of your ideas was ok, P3=TIM_3(R).
>
> > 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.
I just tried to simplify your ideas and cast it into standard terms.
hagman
> > 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.
Here: ring modulo ideal.
If A is a ring and I is an ideal then A/I is the ring with elements of
the
form a+I, a in A and operations inherited from A.
If X is an unknown and f a polynomial in X, then you can view
A[X]/(f) as the ring of polynomials in X with coefficients i nA
subject to the condition
that f is zero.
>
> > 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.
Well, '-' is nto out of "thin air", it comes from the desire to solve x
+a=0.
> 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 did not mean to imply anything about "didactic" aspects of multisign
theory, I just wanted to find out what your construct is in standard
terms.
> 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.
Well, apparently the mapping
f: TIM_4(R) -> RxC
s |-> (-1,i)
presented in another post is a ring isomorphism:
It defines a ring homomoprhism because
(1,1) + (-1,i) + (-1,i)^2 + (-1,i)^3 = (1,1)+(-1,i)+(1,-1)+(-1,-
i)=(0,0)
and from
f((s^2+1)/2) = (1,0)
f((1+s)/(1+i)) = (0,1)
f((1+s)/(1-i)) = (0,i)
we see that f is an isomorphism between 3dimensional R-vector spaces.
>
> -Tim
Timothy Golden BandTechnology.com
The puzzle was a matter of trying to decipher the P4 product behavior
in terms of RxC by defining an equivalent product in RxC. So for
instance if I write:
( - 1 # 1 )( + 2 * 3 # 2)
I could transform these into RxC and get the same result upon
transfoming the resultant back into P4:
* 2 # 3 - 2 + 2 * 3 # 2
There are two subproblems to this: how are the reference frames
connected and then what is the product definition. Could it be that if
this does not exist then we can draw a conclusion? RxC sounds pretty
flat. Probably a clean interpretation includes some dimensional mixing
so the RxC product is somewhat complicated. Anyhow such interpretation
is what is sought but as Tommy has said higher up it doesn't matter.
Dimensionality is there and again as Tommy says whether you call it
RxRxR or RxC also does not matter. P4 is a geometry in the usual sense
though it a nonorthogonal basis.
-Tim
lwa...@lausd.net
<tttppp...@yahoo.com> wrote:
> 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
OK, I finally found the thread. The problem is that for some
reason, the thread can't be found in Google Groups. I had to
go to mathforum.org to find the thread.
http://mathforum.org/kb/message.jspa?messageID=512617&tstart=0
In this post, Chapman first mentions the isomorphism between
P4 and R x C. He makes a mistake regarding P4 (he thought
that +1 was the P4 identity, not #1), but Will Twentyman was
there to correct him. Still, Chapman arrived at the conclusion
that P4 is isomorphic to R x C.
This was Tim's response to Chapman (dated October 2003):
http://mathforum.org/kb/message.jspa?messageID=512624&tstart=0
> What is the meaning of the four-signed numbers with their simple
> arithmetical product? Does it have an equivalent like complex math?
> This is a puzzle I hope you will work on. Since the three-signed
> numbers match complex math exactly then there may be some value to
> these four-signed numbers for three-dimensional space, the space we
> all seem to live in.
So now we ask, is P4 isomorphic to R x C?
hagman:
> Well, apparently the mapping
> f: TIM_4(R) -> RxC
> s |-> (-1,i)
> presented in another post is a ring isomorphism:
(snip)
> f((1+s)/(1+i)) = (0,1)
> f((1+s)/(1-i)) = (0,i)
> we see that f is an isomorphism between 3dimensional R-vector spaces.
But hold on a minute. For hagman presents f as a function
with domain TIM_4(R) (or P4 in Tim's own notation) and
codomain R x C. But then the argument of f is given as
(1+s)/(1+i) and (1+s)/(1-i) -- which are not necessarily
in the domain of f at all! For in each case, 1+s, an
element of TIM_4(R) (or #1-1 in P4) is divided by the
_complex_ numbers 1+i and 1-i. How does one divide a
P4 number by a complex number, anyway?
Indeed, by using only TIM_4(R) notation, we see that the
elements of TIM_4(R) are (equivalence classes of) _real_
polynomials (due to the R in TIM_4(R)). But one cannot
divide a _real_ polynomial 1+s by a _complex_ number
like 1+i or 1-i and expect to have a _real_ polynomial
as a result!
At first I thought that hagman had successfully proved
Chapman's 2003 claim that P4 is isomorphic to R x C,
but now I see the flaw in the proof. If hagman can
patch this flaw, then Chapman will be proved correct
after all, and P4 is isomorphic to R x C, and I would
conjecture that Pn is isomorphic to the Cartesian
product of copies of R and C (i.e., R^r x C^c with
r+2c = n-1) for all n.
But if the flaw cannot be fixed, then Tim and t-1729
would be proved correct, and Chapman would be wrong,
and that P4 is indeed a distinct ring from R x C. And
my own claim earlier in this thread that P5 is
isomorphic to C x C would also be wrong, since it is a
generalization of Chapman's claim.
Timothy Golden BandTechnology.com
Nice work digging up this thread. As you can see I wasn't in tune with
the discussion between Chapman and Twentyman just as I am not in tune
with this current analysis. Yet your analysis of hagman is helping me
see a bit. Anyway upon a full definition we will have a set of rules
that establish a reference frame for transformation and a product
definition in representation
HagP4 = ( a, b, c ) where a is real, and b+ic is complex. This is
just a restatement of RxC
and these a,b,c coordinates are returned by a function
HagP4( z )
where z is a four-signed value. Since the product is not defined for
this domain it must be established. In this way the differentiation
between RxRxR and RxC is negligible. Two instances of HagP4
(a1,b1,c1), (a2,b2,c2)
can be combined by a product
HagProduct( (a1,b1,c1), (a2,b2,c2)) = (a3,b3,c3)
and will be consistent with this unindexed four-signed representation:
( - a + b * c # d )( - e + f * g # h ) =
+ ae * af # ag - ah
* be # bf - bg + bh
# ce - cf + cg * ch
- de + df * dg # dh
which is simply the distribution of terms. If such a 3D math were well
known then it would be presented alongside quaternions wouldn't it? So
this is a new construction and especially of interest because it is so
well behaved algebraically in any dimension(unlike existing math). The
extensibility of such a product would also be highly desirable such
that one is not fumbling for the P6 version. If such a mapping does
not exist cleanly then clearly the native tribe is Polysign. You can
pose with your Cartesian equivalent if it exists however its existence
is already so obscured that its value is questionable. Yes it is clear
that the P4 product is rotational in nature and nearly mimics a
primitive RxC product, but some discrepancy is present. I am open to a
solution but I don't think it is nearly so easy as you are thinking.
Perhaps there is an argument in the terms of orthogonal components
here and their supposed independence versus the nonorthogonal
components and their inherent dependence. In effect a small tweak in
any of the P4 components of one operand throws the product out of
whack in all of its resultants. Mimicing this behavior in orthogonal
coordinates will require dimensional mixing and I believe it will be a
iterated form of solution that is cleanest. This is somewhat already
established in my study of the problem and the graphical result:
http://bandtechnology.com/PolySigned/Deformation/T3P4DifferenceStudy.gif
which is the self similar difference of the simplect version of RxC
product where independence is maintained between R and C within the
product:
(a1,b1,c1) (a2,b2,c2)
= ( a1a2, b1b2-c1c2, b1c2+b2c1).
I do not believe that this is HagProduct(). I have forgone the
determination of reference frame but it is covered at my website. It
is pretty straightforward to establish the reference frame.
Any claim of isomorphism can be proven when instantiated. A claim of
existence has been forwarded by several now: Robin Chapman, Gene Ward
Smith, perhaps lwal, and now perhaps hagman. Yet none have
instantiated such a product. So we wait for Hag_P4() and HagProduct().
-Tim
hagman
The canonical multiplication in RxC is given by
(a,b) * (a',b') = (a*a',b*b')
and that corresponds to your P4 muktiplication via the isomorphism
presented.
> So for
> instance if I write:
> ( - 1 # 1 )( + 2 * 3 # 2)
> I could transform these into RxC and get the same result upon
> transfoming the resultant back into P4:
> * 2 # 3 - 2 + 2 * 3 # 2
Indeed, if I got it right,
'#' corresponds to the usual positive numbers
'-' times '-' is '+'
'-' times '+' is '*'
'-' times '*' is '#'
Thus (note that the meaning of '+','-','*' is the usual one on the
right
hand side:
-1 (=s) |-> (-1,i)
#1 (=1) |-> (1,1)
hence - 1 # 1 |-> (0,i+1)
+2 (=2s^2) |-> (2,-2)
*3 (=3s^3) |-> (-3,-3i)
#2 (=2) |-> (2,2)
hence + 2 * 3 # 2 |-> (1,-3i)
We have (0,i+1)(1,-3i) = (0,3-3i).
Is this the same as * 2 # 3 - 2 + 2 * 3 # 2 (or shorter: * 5 # 5 - 2
+ 2)?
Well,
*5 (=5s^3) |-> (-5,-5i)
#5 (=5) |-> (5,5)
-2 (=2s) |-> (-2,2i)
+2 (=2s^2) |-> (2,-2)
hence * 2 # 3 - 2 + 2 * 3 # 2 = * 5 # 5 - 2 + 2 |-> (0,3-3i) as
expected.
> There are two subproblems to this: how are the reference frames
> connected and then what is the product definition.
The product definition in RxC is the canonical one, i.e. component-
wise.
The product definition in R[s] is the canonical one, i.e. from
definition
of polynomials.
> Could it be that if
> this does not exist then we can draw a conclusion? RxC sounds pretty
> flat. Probably a clean interpretation includes some dimensional mixing
> so the RxC product is somewhat complicated. Anyhow such interpretation
> is what is sought but as Tommy has said higher up it doesn't matter.
> Dimensionality is there and again as Tommy says whether you call it
> RxRxR
... in which case the product is not the canonical (i.e. component-
wise)
product. That's okay, but requires us to describe the product
somewhat lengthy.
hagman
Hm, see below.
I might have been typing too fast and thereby intrduced complex
arithmetic
on the left hand side.
To recap, let TIM_4(R) = R[s]/(1+s+s^2+s^3).
This is a 3dimensional R-vector space and so is RxC,
in fact both are R-algebras by virtue of their multiplications.
We can define an R-algebra homomorphism R[s] -> RxC by sending s to
(-1,i).
Since (1,1) + (-1,i) + (-1,i)^2 + (-1,i)^3 = (0,0) holds in RxC,
this homomorphism factors and defines a R-algebra homomorphism
f: TIM_4(R) -> RxC
To show that f is an isomorphism, it is sufficient that
f is an isomorphism when viewed as a linear map, i.e.
to show that the image is a 3dimensional vector space.
But we have
f(1) = (1,1)
f(s) = (-1,i)
f(s^2) = (1,-1)
and these three vectors are R-linearly independent.
Therefore, f is an isomorphism between TIM_4(R) and RxC.
By simple transfer, we obtain an isomorphism between P4 and RxC
if we send #1 to (1,1), -1 to (-1,i), +1 to (1,-1) and *1 to (-1,-i).
I've not checked but assume that the proof by Chapman was essentially
identical to this one.
hagman
Timothy Golden BandTechnology.com
Let's consider just the above line where you say that
*3 -> (-3, 3i )
and let us try to see if you've maintained distance. Clearly
| *3 | = 3 .
But
| ( -3, -3i ) | = sqrt( 9 + 9 ) = 4.89
so this is not consistent. These magnitudes should match.
I'm still open to your interpretation and it looks like you are pretty
careful so I'll way to hear what you have to say about this.
-Tim
Hero
>
> So now we ask, is P4 isomorphic to R x C?
As i said before, Tim is using a generating set of four vectors,
which are forming a tetrahedron with edges of equal length.
Robin is using four vectors, which are forming a tetra, but not with
edges of equal length.
With friendly greetings
Hero
lwa...@lausd.net
> To show that f is an isomorphism, it is sufficient that
> f is an isomorphism when viewed as a linear map, i.e.
> to show that the image is a 3dimensional vector space.
> But we have
> f(1) = (1,1)
> f(s) = (-1,i)
> f(s^2) = (1,-1)
> and these three vectors are R-linearly independent.
>
> Therefore, f is an isomorphism between TIM_4(R) and RxC.
> By simple transfer, we obtain an isomorphism between P4 and RxC
> if we send #1 to (1,1), -1 to (-1,i), +1 to (1,-1) and *1 to (-1,-i).
Well, it appears that hagman has patched up the proof,
and that P4 really is isomorphic to R x C.
Let's review Tim's objection to hagman's proof:
> Let's consider just the above line where you say that
> *3 -> (-3, 3i )
> and let us try to see if you've maintained distance. Clearly
> | *3 | = 3 .
> But
> | ( -3, -3i ) | = sqrt( 9 + 9 ) = 4.89
> so this is not consistent. These magnitudes should match.
Obviously, what's going on here is that Tim is
trying to introduce a metric on both P4 and
R x C, and show that the metrics on the two
rings don't agree, and so the two rings must
not be isomorphic.
Of course, I assume that in algebra, metrics
are irrelevant. In _topology_, they do matter,
but not in algebra.
Still, let us consider the metrics in R x C
and P4 in order to see what's going on.
For R x C, we'll simply use the R^3 Euclidean
metric D((0,0),(x,z)) = sqrt(x^2 + zconj(z)).
(Note: in the following, let D be the metric
in R x C, d be the metric in P4, and f be
Chapman's isomorphism from P4 to R x C.)
Now we look at P4, or more generally Pn. Tim
intends the Pn numbers -1, +1, *1, #1, etc.,
to lie on unit (n-1)-dimensional hypersphere
and indeed, for them to form the vertices of
a regular (n-1)-simplex. Thus he wants -1,
+1, *1, and #1 to lie on the unit sphere and
form the vertices of a regular tetrahedron.
So clearly d(0,-1) = d(0,+1) = d(0,*1) =
d(0,#1) = 1 in P4. Obviously, this doesn't
occur with their images under Chapman's
isomorphism in R x C since the images of
each of those four points is not 1:
D(f(0),f(-1)) = D((0,0),(-1,i))
= sqrt(1^2 + iconj(i))
= sqrt(2)
and similarly for the other four points.
One may say, that's okay -- we're only off
by a constant factor of sqrt(2), so it's
really the same metric.
But what Tim wants us to see are the
distances between the four points -1, +1,
*1, and #1. These correspond to the edges
of a regular tetrahedron, and so they are
all equal. (I believe that the length of
each edge of the regular tetrahedron
inscribed in the unit sphere is 2sqrt(2/3)
but correct me if I'm wrong.)
Now we check the edge lengths of the
tetrahedron in R x C. Recall that
f(#1) = (1,1)
f(-1) = (-1,i)
so D(f(#1),f(-1)) = sqrt(2^2+1^2+1^2)
= sqrt(6).
f(+1) = (1,-1)
so D(f(#1),f(+1)) = sqrt(0^2+2^2+0^2)
= 2.
So the image tetrahedron has one edge of
length two and another of sqrt(6). So we
see that it's clearly not a regular
tetrahedron at all. And so Tim concludes
that f is not truly an isomorphism.
But once again, I don't think that in
_algebra_, one needs to preserve this
distance function in order for f to be
considered an isomorphism.
When dealing with regular tetrahedra in
R^3, it is often pointed out that:
(1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)
are in fact the four vertices of a
regular tetrahedron. And it appears that
if g is a function that assign these four
values to #1, -1, +1, and *1 respectively,
then g is an isomorphism from P4 to R^3
(the multiplication clearly must be done
in R^3, not in R x C) and the distance
is preserved. The edges of tetrahedron
are all 2sqrt(2) -- one can renormalize
this by multiplying by an appropriate
constant if another value is desired.
So perhaps this solution satisfies both
sides -- we've found an isomorphism
between P4 and a real-valued vector space,
but we've maintained Tim's distance
function as well.
Hero
>
> When dealing with regular tetrahedra in
> R^3, it is often pointed out that:
>
> (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)
>
> are in fact the four vertices of a
> regular tetrahedron. And it appears that
> if g is a function that assign these four
> values to #1, -1, +1, and *1 respectively,
> then g is an isomorphism from P4 to R^3
> (the multiplication clearly must be done
This doesn't work out.
How is Your multiplication in R^3 defined?
>...and the distance
> is preserved. The edges of tetrahedron
> are all 2sqrt(2) -- one can renormalize
> this by multiplying by an appropriate
> constant if another value is desired.
>
> So perhaps this solution satisfies both
> sides -- we've found an isomorphism
> between P4 and a real-valued vector space,
> but we've maintained Tim's distance
> function as well.
Let's remember: Robin defined multiplication of RxC as
(a, u ) times ( b, v) = ( a*b, u * v)
With friendly greetings
Hero
tommy1729
but then (a,b)*(a,b) = (a^2,b^2)
and therefore root(x,y) = (root(x),root(y))
however this does not work for x negative ...
lwalkes representation is better but has a similar problem.
so (a,b) fails ; (x,y,z) fails
---> next sets requires min 4 elements
and that is in its simplest form simply P4.
>
> With friendly greetings
> Hero
>
>
>
thanks for spending time on the topic
regards
tommy1729
Timothy Golden BandTechnology.com
I am so happy to have your attention to this subject that I have
little to complain about.
Clearly lwal and hagman (and of course Hero too) are capable and
critical and as far as interaction goes here this is a good thread.
Those nice words aside, is your conclusion consequential? The polysign
construction is very clean and it seems that this supposed replacement
is only half of a replacement. The polysign algebra is carried out on
a specific well understood geometry and your current mapping does not
match. Both are 3D systems. So when we choose points in the 3D space
and map them to each system the mismatch is directly present. In
effect when I pick the point *3 I will not get back (3, -3i) in your
mapping. This actual coordinate in your mapping will be slightly less.
I am open to your work here as being of value. I just am not sure
where it takes us. The words which come to my mind are mostly negative
yet I do not want to be recalcitrant to you. Anyhow it is a mild
negativity and we still have to hear whether hagman is in agreement
with lwal.
It seems that you have divorced the geometry of the system from the
algebra yet these two coexist in the polysign construction. Beyond
puzzling over deeper details of the definition of isomorphism (which
will not be strong ground for me) I also wonder whether your results
can be generalized to Pn?
For people in general there is a powerful resistance to the
generalization of sign. It is a farcical concept and I agree that this
is the case. But since sign is exposed as dimension we can then
replace our usual notion of dimension with this new construction. It
is almost as if a new word should be presented which melds sign and
dimension but this would get too cryptic too quickly. Because the sign
rules are consistent with the real numbers I do believe that the
proper construction context is of generalized sign; polysign.
Especially the form
s x
can be used as elemental where s is sign and x is magnitude and the
rules can be stated in this representation in only a few lines:
s1 x1 + s1 x2 = s1 ( x1 + x2 ) .
( s1 x1 )( s2 x2 ) = ( s1 + s2 ) x1 x2.
Sum over s ( s x ) = 0 .
When we delete these rules (from the reals or from polysign) we are
left with a meager basis that contains no sign component. This is the
magnitude x. Informationally this system is redefining the real
numbers and gets the complex numbers and these higher sign spaces for
free. This has been overlooked by existing mathematics. So I consider
the polysign construction more a discovery than an invention. However
this will be completely unconvincing until one can accept that sign
can be generalized. I am seriously curious how you respond to this
crux and is my projection of your thoughts accurate? Is the higher
mathematical language that you use necessary? I do not think so.
-Tim
tommy1729
yes its a good thread.
no credit is given to me ?
i generalized the concept of polysigned by introducing
my tommy-notation.
i also said R x C only implies a 3rd dimension.
etc...
i dont know how you feel about my tommy-notation.
it is NOT a compeditive theory of the polysign, i sometimes have the feeling you see certain theorems of me as a threath to your own ...
i assure you that is not case.
in fact the polysign are a subset of my concepts and also a good introduction ...
>
> Those nice words aside, is your conclusion
> consequential? The polysign
> construction is very clean and it seems that this
> supposed replacement
> is only half of a replacement.
yes indeed.
The polysign algebra
> is carried out on
> a specific well understood geometry and your current
> mapping does not
> match. Both are 3D systems. So when we choose points
> in the 3D space
> and map them to each system the mismatch is directly
> present. In
> effect when I pick the point *3 I will not get back
> (3, -3i) in your
> mapping. This actual coordinate in your mapping will
> be slightly less.
yes , although they are probably aware of this and only claim a algebraic equivalent rather than a geometrical ...
i think ...
>
> I am open to your work here as being of value. I just
> am not sure
> where it takes us. The words which come to my mind
> are mostly negative
> yet I do not want to be recalcitrant to you.
indeed , it is of significant value , despite our criticism , we do appreciate it...
Anyhow
> it is a mild
> negativity and we still have to hear whether hagman
> is in agreement
> with lwal.
>
> It seems that you have divorced the geometry of the
> system from the
> algebra yet these two coexist in the polysign
> construction. Beyond
> puzzling over deeper details of the definition of
> isomorphism (which
> will not be strong ground for me) I also wonder
> whether your results
> can be generalized to Pn?
i have asked that question too ...
i assume it can , but havent investigated it yet...
maybe when i have more time.
>
> For people in general there is a powerful resistance
> to the
> generalization of sign.
not to tommy1729
what my theory of tommy-notation does is extending not only sign , but generalizing axis and dimension alongside it ...
since signs , dimensions and axes are the only ways to represent integer domain dimensional numbers ( apart from groups , set , matrices and those kind ; !! however they are isomorphic !! )
the tommy-notation (a,b) must cover all sensible dimensional numbers !!
and it is clear that therefore tommy-notation solves or at least represents all integer domain dimensional models and problems
it's only a little step away from polysigned ...
by the majority ...
as galathaea pointed out , projective geometrie is aware of this , however it is not handled as algebra like you do...
So I consider
> the polysign construction more a discovery than an
> invention. However
> this will be completely unconvincing until one can
> accept that sign
> can be generalized. I am seriously curious how you
> respond to this
> crux and is my projection of your thoughts accurate?
> Is the higher
> mathematical language that you use necessary? I do
> not think so.
no , in fact consider questions like these
how many fundamental integer domain 5 dimensional numbers are there ?
how does one divide a 4 D number by a 3D one ?
how does one show a field representation cannot be done in terms of matrices ?
how does one show (proof!) properties about zerodivisors in any integer domain dimension without having a model for dimensional numbers that does not skip any numbers ?
how does one show that the elements of a set of rings defined by countable subsets of dimensional numbers have unique factorization ?
polysigned ( and tommy-notation ) could contribute to making or simplifying proofs.
etc.
>
> -Tim
>
regards
tommy1729
neilist
<snip>
> i think a personal meeting will benefit us both.
Watch ou, Tim. Tommy's coming onto you!!!
lwa...@lausd.net
<tttppp...@yahoo.com> wrote:
> On Sep 16, 5:19 am, Hero <Hero.van.Jind...@gmx.de> wrote:
> > This doesn't work out.
> > How is Your multiplication in R^3 defined?
> > (a, u ) times ( b, v) = ( a*b, u * v)
Guess what? I found an error in my own work. So what I
wrote is not an isomorphism at all! You are correct
that it doesn't work out.
Here's the error:
g(-1) = (+1,-1,-1).
But (+1,-1,-1)(+1,-1,-1) = (+1,+1,+1) = g(#1)
yet in P4, (-1)(-1) = +1, not #1!
So you can disregard the entire post. It is in error.
> we still have to hear whether hagman is in agreement
> with lwal.
I bet hagman would've caught my error.
Note: Chapman's isomorphism between P4 and R x C is
still valid as an isomorphism. But it doesn't preserve
the regular tetrahedron. I'm also curious as to
whether hagman can come up with a way to preserve the
regular tetrahedron.
> I am seriously curious how you respond to this
> crux and is my projection of your thoughts accurate?
I actually find Pn to be a very interesting construct,
a different way of looking at the same thing.
Recall what I said about the history of mathematics,
with irrational numbers (Pythagoras) being accepted
far before negative numbers (Cardano) -- this was
mainly because ever since the Greeks, numbers were
associated with geometry, where negative lengths are
of course impossible.
Here's an interesting link about complex numbers:
http://math.fullerton.edu/mathews/n2003/ComplexNumberOrigin.html
We learn that Cardano, in the 16th century, stumbled
upon complex numbers when trying to solve cubic (not
quadratic!) equations. Suppose Cardano, instead of
coming up complex numbers, decided to solve the
simplest cubic equation, x^3 = 1, by calling its
three solutions -1, +1, and *1. Then it's possible
that we would've had three signs in modern math,
and we'd all be using P3 instead of C!
tommy1729
possible yes.
but of course he already invented the negative and had to build on that , therefore C.
C has a few benefits over P3.
for an integer polynomial : if f(a+bi) is a zero
then so is f(a-bi).
for analytic continuation : if F is analytic for I > 0 or I < 0 then there exists an analytic continuation for all of C.
but apart from that P3 is good too.
regards
tommy1729
Timothy Golden BandTechnology.com
> Guess what? I found an error in my own work. So what I
> wrote is not an isomorphism at all! You are correct
> that it doesn't work out.
> Here's the error:
> g(-1) = (+1,-1,-1).
> But (+1,-1,-1)(+1,-1,-1) = (+1,+1,+1) = g(#1)
> yet in P4, (-1)(-1) = +1, not #1!
There you go. There is a mapping on my website that is even more
similar but it will not reduce to the P4 product easily. I can scrunch
the error into a plane but it does not go away.
> So you can disregard the entire post. It is in error.
Well it is good that you have considered the construction some. There
is much work to do with it. I've put up a cvs source code image:
http://bandtechnology.com/Polysign.cvs.tar.gz
and just the hot directory (simply source code):
http://bandtechnology.com/polysign.tar.gz
This code is GPL'd with a single conspicuous gpl.txt file in the
source directory.
In the tarball is a file called
T3Study.C
which contains most of the work of my P4 comparison with RxC.
Yes, obviously the tetrahedron can be maintained and that will get us
at least a vector space but the product itself will not behave the
same. Still, it is easy to predict an iterative solution. Simply look
at the graphics of this analysis. The self similar error is getting
smaller so in effect a repetitive effort on this error will further
diminish it. In trying a simple scalar correction I did manage to
cancel out the identity axis component so that all of the error is in
the complex plane. At least graphically that was evident. I am of
course open to an error though you can see from the source code it is
pretty carefully made and verified. Even if you do not understand C++
the functional layout is pretty clear.
OK LWalker. My only criticism is that the conception of roots of unity
does not inherently tie in to dimension.
For instance the fifth roots of unity exist in the complex plane.
It is the generalization of sign which does meet dimension. Another
feature that is unique to polysign is that
(-1)^m
where m is natural iterates throughout the space. We do not have an
operator that works this easy or at all in orthogonal systems. Anyhow
even this one is not so valuable as the dimensional relation that is a
direct consequence of
sum over s ( s x ) = 0
for without this all of this other information is trivial.
So still you have dodged a simple question:
Do you believe that this system generalizes sign?
There are others who have come to understand the construction yet
cannot affirm this. I find this fascinating and so I ask you more
directly.
-Tim
hagman
... should be clear from a=1.
> bc = 1
> b^2 = c
> c^2 = b
(in all these examples we should bear in mind that commutativity is
presupposed)
>
> and we have the 6 cartesian axes a , -a , b , -b , c , -c
>
> which make a 3rd dimension.
>
> in in the work of professor beresford he proves
>
> beresford is isomorphic to R x C.
isomorphic as what? as R-algebra, I suppose
>
> logical since
>
> a = 1
> a^2 = 1
> bc = 1
> b^2 = c
> c^2 = b
>
> are actually P3
>
> and R = P2
>
> so beresford is isomorphic to P2 x P3 => R x C.
>
> yet beresford =/= P4 !!!
If Beresford, P4 and RxC are pairwise isomorphic (as R-algebras), in
what respect
is Beresford =/= P4?
Admittedly, the isomorphisms found are not canonical, but you cannot
prove anything
about one that is not true for the other as well.
>
> furthermore beresford has a matrixrepresentation and P4 has not ... or at least a very different one (still open problem , yet probably P4 has no matrix representation )
As per isomorphism, of course P4 has one.
>
> so since P4 AND BERESFORD are both CONSISTANT
>
> --> we must recognize that there are at least 2 kinds of integer domains in 3 dimensions that belong to R x C.
1) As per isomorphism, the structures are the same.
2) What do you mean with "integer domain"?
That term seems to be very important in the rest of your post, but
I don't
know what it means.
I hope this is not a typo for "integral domain":
<1,0> * <0,1> = <0,0> in RxC
( #1 +1 ) ( #1 *1 ) = 0 in P4
(a+b+c) * (b-c) = 0 in Beresford.
>
> in fact 3rd dimension is isomorphic to R x C , as pointed out in the very beginning of this post by me.
The R-algebra RxRxR is not isomorphic to RxC:
You cannot find non-zero x,y,z in RxC such that xy=xz=yz=0, but you
can in RxRxR.
>
> so we can simplify:
>
> --> we must recognize that there are at least 2 kinds of integer domains in 3 dimensions
I agree (if integer domain is replaced by R-algebra), but not
by your reasoning -- rather because we have RxC and RxRxR.
>
> we must also recognize that using cartesian , cantor and R x C "logic" we dont get an accurate view of dimensionality or integer domains...
>
> writing R x C does not define how many fundamental 3 dimensional integer domains exist.
>
> worse ; it implies only 1 since all 3rd dimensions are R x C and cannot be for instance C x C or
>
> R x R x R ( different from R x C )
Ah, now you've seen RxRxR, too.
But what do you mean with "all 3rd dimensions"?
Please define your notation rigorously instead of giving examples.
Especially, since P4 and Beresford are isomorphic as R-algebras,
what is the intrinsic property that produces (3,4) in one case
and (3,6) in the other?
> P5 -> (4,5)
> 4D tommynumbers -> (4,6)
> relativity numbers -> (4,7)
> bicomplex / Beresford 4D complex -> (4,8)
> P6 -> (5,6)
> 5D tommynumbers -> (5,7)
> ..
>
> of course for b there is the rule that b exists
>
> if and only if
>
> a < b for all a
> b =< 2a for all a >= 3
>
> even much stronger !! ;
>
> every fundamental integer domain ( with fundamental meaning P3 and C are the same and is also equal to integer domains that are P3 in "disguise" by using for instance 3 different vectors which also make up the complex plane ; and the analogue in higher dimensions )
Aha, here you seem to imply that we should only classify up to
isomorphism.
Then what makes P4 and Beresfor different?
>
> is equal to (a,b)
> with a < b for all a
> b =< 2a for all a >= 3
>
> using the differential equations from prof beresford (easily reconstructable too) one can perform computations in almost any integer domain ...
>
> yet many questions are still open ...
>
> regards
> tommy1729
hagman
hagman
The exixtence of zero-divisors is often considered a sign of not so
well
behaviour ;)
> The
> extensibility of such a product would also be highly desirable such
> that one is not fumbling for the P6 version. If such a mapping does
> not exist cleanly then clearly the native tribe is Polysign. You can
> pose with your Cartesian equivalent if it exists however its existence
> is already so obscured that its value is questionable. Yes it is clear
> that the P4 product is rotational in nature and nearly mimics a
> primitive RxC product, but some discrepancy is present.
What discrepancy? After all P4 and RxC are isomorphic as R-algebras.
> I am open to a
> solution but I don't think it is nearly so easy as you are thinking.
> Perhaps there is an argument in the terms of orthogonal components
> here and their supposed independence versus the nonorthogonal
> components and their inherent dependence. In effect a small tweak in
> any of the P4 components of one operand throws the product out of
> whack in all of its resultants. Mimicing this behavior in orthogonal
> coordinates will require dimensional mixing and I believe it will be a
> iterated form of solution that is cleanest. This is somewhat already
> established in my study of the problem and the graphical result:
> http://bandtechnology.com/PolySigned/Deformation/T3P4DifferenceStudy.gif
> which is the self similar difference of the simplect version of RxC
> product where independence is maintained between R and C within the
> product:
> (a1,b1,c1) (a2,b2,c2)
> = ( a1a2, b1b2-c1c2, b1c2+b2c1).
> I do not believe that this is HagProduct().
Using b1+i*b1 etc., this becomes the canonical product on the
R-algebra RxC (i.e. componentwise).
> I have forgone the
> determination of reference frame but it is covered at my website. It
> is pretty straightforward to establish the reference frame.
>
> Any claim of isomorphism can be proven when instantiated. A claim of
> existence has been forwarded by several now: Robin Chapman, Gene Ward
> Smith, perhaps lwal, and now perhaps hagman. Yet none have
> instantiated such a product. So we wait for Hag_P4() and HagProduct().
>
> -Tim
I'm sorry, what are you waiting for exactly?
What kind of product do you expect me to instantiate??
I gave a complete proof that P4 and RxC are isomorphic as algebras.
Do you have any problems with that proof?
Do you think there is a gap?
Can you pin it down?
hagman
lwa...@lausd.net
<tttppp...@yahoo.com> wrote:
> So still you have dodged a simple question:
>
> Do you believe that this system generalizes sign?
I believe that Pn does indeed generalize sign.
To be precise, we consider the construction of R
from N as follows:
1. Add (positive) rationals from N to make Q+.
2. Complete Q+ to make P, the set of positive
reals (magnitudes).
3. Add the signed numbers to P to make R.
And so the polysigned numbers are in fact a
generalization of step 3.
tommy1729
with all respect hagman ; i do consider you as an intelligent person however i think you dont get what I am saying at all ...
cant really blame you
i blame cartesian education...
(and perhaps cantor)
you might find it strange but im trying to be as clear as possible.
but it is extremely hard to explain if you dont get it at first.
and i cant compromise on concepts , since they are key to the idea ...
timothy understands better , but not even he knows what i exactly mean with tommy-notation...
( what still amazes me since his theory is part of it )
im not even sure you understand P4 completely...
you are an intelligent person , so there must be something wrong with 'dimensional education'
sure , i and timothy use our own definitions , which might be confusing ...
but then again , we dont have much choice since you cant build on something you dont agree on...
it seems a lot of work to try and clarify all , since you have different comments here ...
but ill try anyway.
> On 15 Sep., 00:30, tommy1729 <tommy1...@gmail.com>
> wrote:
> > timothy wrote:
> > > On Sep 13, 4:42 pm, tommy1729
> <tommy1...@gmail.com>
> > > wrote:
> > > > so P4 belongs to the set R x C and is not equal
> to
> > > it.
> > > > rather R x C defines a 3rd dimension rather
> than a
> > > number system...
> > > > logical since if algebraicly closed R x R then
> R x
> > > R = C
> > > > and therefore R x C = R x R x R = R^3 as
> expected.
> > > > if you find this confusing , unbelievable or
> > > incomplete
> > > > i can assure you I am pretty sure timothy
> agrees
> > > that R x C is to simple to describe P4.
and he confirmed that...
yes , but i wanted to show algebraic closure so i included the square to complete the multiplication table...
still trivial of course.
>
> > bc = 1
> > b^2 = c
> > c^2 = b
and algebraic closure is evident now.
>
> (in all these examples we should bear in mind that
> commutativity is
> presupposed)
yes that why i often rewrite integral domain, just for the ones who just started reading ; and to be complete...
timothy and i dont like non-commutativ algebra and no post has ever been done about it concerning polysigned or tommy-notation
integral domain should always be in your head when reading about polysigned or tommy-notation.
>
> >
> > and we have the 6 cartesian axes a , -a , b , -b ,
> c , -c
> >
> > which make a 3rd dimension.
> >
> > in in the work of professor beresford he proves
> >
> > beresford is isomorphic to R x C.
>
> isomorphic as what? as R-algebra, I suppose
isomorphic to R x C ; integral domain in 3rd dimension; but the point later on is that beresford is not the only 3rd dimensional integral domain isomorphic to R x C.
besides it is R x C quite directly
(axis to axis line correspondance )
since
beresford is isomorphic to P2 x P3 because all axes are indeed the elements of P3 and its "negative"(P2)
and since P3 = C and P2 = R
also note that isomorphic is a vague concept if you dont accept 'cartesian sign logic'.
>
> >
> > logical since
> >
> > a = 1
> > a^2 = 1
> > bc = 1
> > b^2 = c
> > c^2 = b
> >
> > are actually P3
> >
> > and R = P2
> >
> > so beresford is isomorphic to P2 x P3 => R x C.
> >
> > yet beresford =/= P4 !!!
>
>
> If Beresford, P4 and RxC are pairwise isomorphic (as
> R-algebras), in
> what respect
> is Beresford =/= P4?
both are 3D , but beresford has 6 axes and P4 has 4.
also the zerodivisors are different ; even visually if you
draw P4 and Beresford.
so P4 is not just different because the axis or corners are different ; IT TRUELY IS.
there is no mapping from P4 to beresford !
P4 has 2 cones as the area of zerodivisors , whereas beresford has a line and a flat surface
TALK ABOUT DIFFERENCE !
> Admittedly, the isomorphisms found are not canonical,
> but you cannot
> prove anything
> about one that is not true for the other as well.
what ? i can prove all !?
the zero divisors are different
the multiplication is different
the corners are different
the amount of axes are different
everything is different and it is easy to proof.
making a 3D drawing is already sufficient.
and giving a multiplication of both as example too...
>
> >
> > furthermore beresford has a matrixrepresentation
> and P4 has not
do you doubt this ??
this also another difference, yet a bit harder to prove.
... or at least a very different one
> (still open problem , yet probably P4 has no matrix
> representation )
>
> As per isomorphism, of course P4 has one.
NO that is WRONG
P4 has canonical structure of zero divisors and no "negative" axis.
this prevents matrix representation or at least makes it extremely hard ...
(still unproven , feel free to give such a matrix !! )
>
>
> >
> > so since P4 AND BERESFORD are both CONSISTANT
> >
> > --> we must recognize that there are at least 2
> kinds of integer domains in 3 dimensions that belong
> to R x C.
>
> 1) As per isomorphism, the structures are the same.
no , you yourself talked about conical properties for P4 ; and they are not present in beresford...
> 2) What do you mean with "integer domain"?
> That term seems to be very important in the rest
> est of your post, but
> I don't
> know what it means.
> I hope this is not a typo for "integral domain":
im afraid so, although i mean "fundamental" too
meaning for example
vectors ( 2 ; -3+i ; -1-2i )
are considered equal to
vectors ( 1 ; (-1)^2/3 ; (-1)^1/3 ) ( P3 )
since they also give the complex plane
in general if two number systems give the same dimension , are not overdefined AND describe the same zero divisors ( if there are zero divisors )
and of course also algebraicly closed.
then they are called "fundamental".
this should give a good idea of what i mean with integer domain
i admit it started as a typo actually , sorry.
also perhaps confusing with a ring rather than a field.
so actually => ( fundamental ) integral domain.
perhaps integral field is the correct word ?
forgive my bad knowledge of english and my typo's...
> <1,0> * <0,1> = <0,0> in RxC
> ( #1 +1 ) ( #1 *1 ) = 0 in P4
> (a+b+c) * (b-c) = 0 in Beresford.
>
yes ...
so ?
>
> >
> > in fact 3rd dimension is isomorphic to R x C , as
> pointed out in the very beginning of this post by me.
and it still is.
>
> The R-algebra RxRxR is not isomorphic to RxC:
if R x R x R is algebraicly closed of course ...
we are talking about integral domain
else it is
R x R x R is isomorphic to R x cC
where cC is some kind of countercomplex...
but thats off topic here , since we only consider algebraicly closed number systems.
> You cannot find non-zero x,y,z in RxC such that
> xy=xz=yz=0, but you
> can in RxRxR.
not if R x R x R is algebraicly closed AND different from P4 AND Beresford i assume ?
>
> >
> > so we can simplify:
> >
> > --> we must recognize that there are at least 2
> kinds of integer domains in 3 dimensions
>
> I agree (if integer domain is replaced by R-algebra),
> but not
> by your reasoning -- rather because we have RxC and
> RxRxR.
>
> >
> > we must also recognize that using cartesian ,
> cantor and R x C "logic" we dont get an accurate view
> of dimensionality or integer domains...
> >
> > writing R x C does not define how many fundamental
> 3 dimensional integer domains exist.
and it still does not.
> >
> > worse ; it implies only 1 since all 3rd dimensions
> are R x C and cannot be for instance C x C or
> >
> > R x R x R ( different from R x C )
>
> Ah, now you've seen RxRxR, too.
> But what do you mean with "all 3rd dimensions"?
all "fundamental" integral domains
( see above )
>
> >
> > and this gets even worse for higher dimensions...
> >
> > defining the amount of integer domains in 6D as
> >
> > ( we know we need at least one C to be integer
> domain )
> >
> > R x R x R x R x C
> >
> > R x R x C x C
> >
> > C x C x C
> >
> > is not working at all !!!
> >
> > this lists 3 wich is totally incorrect !
> >
> > if i am not mistaken there is even another 3D
> number
3D with 5 axes ...
P4 has 4
Beresford has 6
Tommy has 5
why not ?
a is the dimension , b is the amount of axes
( axes is the plural of axis right ? )
> what is the intrinsic property that produces (3,4) in
> one case
> and (3,6) in the other?
(3,4) is P4
(3,6) is beresford
they are produced by the a and b ; the dimension and the amount of axes
produced ... represented is a better word.
>
>
> > P5 -> (4,5)
> > 4D tommynumbers -> (4,6)
> > relativity numbers -> (4,7)
> > bicomplex / Beresford 4D complex -> (4,8)
> > P6 -> (5,6)
> > 5D tommynumbers -> (5,7)
> > ..
> >
> > of course for b there is the rule that b exists
> >
> > if and only if
> >
> > a < b for all a
> > b =< 2a for all a >= 3
> >
> > even much stronger !! ;
> >
> > every fundamental integer domain ( with fundamental
> meaning P3 and C are the same and is also equal to
> integer domains that are P3 in "disguise" by using
> for instance 3 different vectors which also make up
> the complex plane ; and the analogue in higher
> dimensions )
>
> Aha, here you seem to imply that we should only
> classify up to
> isomorphism.
> Then what makes P4 and Beresfor different?
no this was my first attempt at explaining " fundamental"
>
> >
> > is equal to (a,b)
> > with a < b for all a
> > b =< 2a for all a >= 3
> >
> > using the differential equations from prof
> beresford (easily reconstructable too) one can
> perform computations in almost any integer domain
> ...
> >
> > yet many questions are still open ...
> >
> > regards
> > tommy1729
>
> hagman
>
>
regards
tommy1729
disclaimer : despite agreeing alot with timothy golden, tommy1729 does not garantee that timothy has exactly the same ideas, i only speak for myself , yet i defend timothy's polysigned.
moreover i have to , since my ideas of dimensionality are based upon it and in agreement with it and of course because i feel he is right and polysigned deserves attention.
lwa...@lausd.net
> I'm sorry, what are you waiting for exactly?
> What kind of product do you expect me to instantiate??
> I gave a complete proof that P4 and RxC are isomorphic as algebras.
> Do you have any problems with that proof?
> Do you think there is a gap?
> Can you pin it down?
I believe what Tim wants is to find an isomorphism
that preserves distance. In particular, he
envisions -1, +1, *1, #1 to lie at the vertices of
a regular tetrahedron in 3D space.
But when we consider the images of those four
points via the isomophism, viz. (-1,i), (1,-1),
(-1,-i), and (1,1), we see that they do not form a
regular tetrahedron (when viewed as points of R^3)
for in particular:
d((1,1),(-1,i)) in R x C
= d((1,1,0),(-1,0,1) in R^3
= sqrt(2^2+1^2+1^2)
= sqrt(6)
but
d((1,1),(1,-1)) in R x C
= d((1,1,0),(1,-1,0))
= sqrt(0^2+2^2+0^2)
= 2.
So they don't form a regular tetrahedron
since one edge has length 2 yet another
has length sqrt(6). Tim wants us to find
an isomorphism that preserves distance
(i.e. an isometry) between P4 and R x C.
Timothy Golden BandTechnology.com
Thank you L. Walker.
Nonlinearity is a subject of interest and in that zero divisors
present dimensional consequences such things are not in need of
rejection.
- Tim
hagman
It is clear that there is none as - up to conjugacy - there cannot be
another R-algebra isomorphism between them.
Note that his -1 must map to a 4th root of unity and
+1, *1, #1 must map to the powers of that 4th root
and the first three are R-linearly independant.
THerefore -1 must map to (a,b) such that a^4=b^4=1
and a != 1 and b^2 != 1, i.e. there is oly a choice between
(-1,i) and (-1,-i).
OK, instead of fiddling around with the multiplication, we might
adjust
the metric such that all roots of unity
in RxC have unit length, i.e. define ||(a,b)|| = u*a^2 + v*b*conj(b)
where u,v are positive numbers such that u+v=1 (instead
of making the default choice u=v=1).
Then the squared distance between (-1,i) and (1,1) becomes
u*2^2 + v*(1+i)(1-i) = 4u+2v,
the squared distance between (1,-1) and (1,1) becomes
u*0+v*4 = 4v.
Thus we could choose v=2u, i.e. u=1/3, v=2/3 to make our tetrahedron
regular.
In this sense we could say that P4 "is"
the R-algebra RxC with metric induced by the scalar product
<(a,b),(c,d)> = (a*c + 2*b*conj(d))/3
This looks like a somewhat arbitrary choice, but this is so because
we cannot make that multiplicative, i.e. such that
||(a,b)(c,d)|| = ||(a.b)|| * ||(c,d)|| in general.
IOW, while the generators in P4 have equal distance, distance in P4
does not behave well under multiplication.
hagman
hagman
Oops, this of course fails to be a scalar product both
in the real/bilinear as in the complex/sesquilinear sense.
We better start with the norm as above ...
> This looks like a somewhat arbitrary choice, but this is so because
> we cannot make that multiplicative, i.e. such that
> ||(a,b)(c,d)|| = ||(a.b)|| * ||(c,d)|| in general.
Especially because RxC is not an integral domain.
hagman
In that case you should have added ab=ba=b and ac=ca=c as well.
But then again a=1 says it all.
>
> still trivial of course.
>
>
>
> > > bc = 1
> > > b^2 = c
> > > c^2 = b
>
> and algebraic closure is evident now.
>
>
>
> > (in all these examples we should bear in mind that
> > commutativity is
> > presupposed)
>
> yes that why i often rewrite integral domain, just for the ones who just started reading ; and to be complete...
>
> timothy and i dont like non-commutativ algebra and no post has ever been done about it concerning polysigned or tommy-notation
>
> integral domain should always be in your head when reading about polysigned or tommy-notation.
>
>
>
> > > and we have the 6 cartesian axes a , -a , b , -b ,
> > c , -c
>
> > > which make a 3rd dimension.
>
> > > in in the work of professor beresford he proves
>
> > > beresford is isomorphic to R x C.
>
> > isomorphic as what? as R-algebra, I suppose
>
> isomorphic to R x C ; integral domain in 3rd dimension; but the point later on is that beresford is not the only 3rd dimensional integral domain isomorphic to R x C.
>
> besides it is R x C quite directly
> (axis to axis line correspondance )
> since
>
> beresford is isomorphic to P2 x P3 because all axes are indeed the elements of P3 and its "negative"(P2)
>
> and since P3 = C and P2 = R
>
> also note that isomorphic is a vague concept if you dont accept 'cartesian sign logic'.
Beresford, P4, RxC are sets with structures.
If among these structures I concentrate on addition and
multiplication, I find that
there are mappings e.g. from the underlying set of P4 to that of RxC
such that the mapping
is bijective and it (as well as its inverse) is compatible with
addition and multiplication.
Therefore any statement about the algebraic structure of one transfers
immediately to the other.
As a trivial example, in RxC 1 has an additive inverse, therefore in
P4 #1 has an additive inverse
(in fact that inverse is -1+1*1). And we can find additive inverses
for arbitrary elements by
multiplying with the additive inverse of 1 (or #1).
>
>
>
> > > logical since
>
> > > a = 1
> > > a^2 = 1
> > > bc = 1
> > > b^2 = c
> > > c^2 = b
>
> > > are actually P3
>
> > > and R = P2
>
> > > so beresford is isomorphic to P2 x P3 => R x C.
>
> > > yet beresford =/= P4 !!!
>
> > If Beresford, P4 and RxC are pairwise isomorphic (as
> > R-algebras), in
> > what respect
> > is Beresford =/= P4?
>
> both are 3D , but beresford has 6 axes and P4 has 4.
>
> also the zerodivisors are different ; even visually if you
> draw P4 and Beresford.
The zerodivisors are not different as evident from the isomorphism as
algebras.
In fact, as you always talk about integral domains, how come there are
zerodivisors at all?
>
> so P4 is not just different because the axis or corners are different ; IT TRUELY IS.
>
> there is no mapping from P4 to beresford !
>
> P4 has 2 cones as the area of zerodivisors , whereas beresford has a line and a flat surface
In P4, -a*a (and their opposites +a#a) are a line of zero divisors.
And -b+b-c#c (again together with their opposites *b#b*c+c and
combinations -b+b*c+c,
*b#b*c+c) are a plane of zero divisors. (Gee, that positiveness
restriction makes
talking rather cumbersome).
Where are your cones?
>
> TALK ABOUT DIFFERENCE !
CANNOT! IS NONE!
>
> > Admittedly, the isomorphisms found are not canonical,
> > but you cannot
> > prove anything
> > about one that is not true for the other as well.
>
> what ? i can prove all !?
>
> the zero divisors are different
nope
> the multiplication is different
nope
> the corners are different
but not intrinsic. That's more like choosing a base in a vector space.
And remember that one should *never* choose a base in a vector space,
as you might end up with clumsy matrcies instead of friendly linear
maps ;)
> the amount of axes are different
>
> everything is different and it is easy to proof.
>
> making a 3D drawing is already sufficient.
>
> and giving a multiplication of both as example too...
>
>
>
> > > furthermore beresford has a matrixrepresentation
> > and P4 has not
>
> do you doubt this ??
>
> this also another difference, yet a bit harder to prove.
>
> ... or at least a very different one
>
> > (still open problem , yet probably P4 has no matrix
> > representation )
>
> > As per isomorphism, of course P4 has one.
>
> NO that is WRONG
>
> P4 has canonical structure of zero divisors and no "negative" axis.
>
> this prevents matrix representation or at least makes it extremely hard ...
>
> (still unproven , feel free to give such a matrix !! )
Expressing that is simply extremely convoluted.
I do not consider obscuring any beautiful structure by imposing weird
constrictions a good idea.
>
>
> > > so since P4 AND BERESFORD are both CONSISTANT
>
> > > --> we must recognize that there are at least 2
> > kinds of integer domains in 3 dimensions that belong
> > to R x C.
>
> > 1) As per isomorphism, the structures are the same.
>
> no , you yourself talked about conical properties for P4 ; and they are not present in beresford...
Did I?
Here on sci.math?
>
> > 2) What do you mean with "integer domain"?
> > That term seems to be very important in the rest
> > est of your post, but
> > I don't
> > know what it means.
> > I hope this is not a typo for "integral domain":
>
> im afraid so, although i mean "fundamental" too
But none of the equivalent structures Beresford / P4 / RxC *is* an
integral domain!
>
> meaning for example
>
> vectors ( 2 ; -3+i ; -1-2i )
>
> are considered equal to
>
> vectors ( 1 ; (-1)^2/3 ; (-1)^1/3 ) ( P3 )
>
> since they also give the complex plane
>
> in general if two number systems give the same dimension , are not overdefined AND describe the same zero divisors ( if there are zero divisors )
In what sense are > vectors ( 2 ; -3+i ; -1-2i ) not overdefined?
For example, they are dependent: 7a+4b+2c=0
>
> and of course also algebraicly closed.
I guess you only mean something like closed under addition and
multiplication.
You do know the meaning of algebraically closed?
>
> then they are called "fundamental".
>
> this should give a good idea of what i mean with integer domain
>
> i admit it started as a typo actually , sorry.
>
> also perhaps confusing with a ring rather than a field.
>
> so actually => ( fundamental ) integral domain.
>
> perhaps integral field is the correct word ?
>
> forgive my bad knowledge of english and my typo's...
>
> > <1,0> * <0,1> = <0,0> in RxC
> > ( #1 +1 ) ( #1 *1 ) = 0 in P4
> > (a+b+c) * (b-c) = 0 in Beresford.
>
> yes ...
>
> so ?
>
So all have zero divisors and thus none is an integral domain.
Note that I actually started with the simple pair of zero divisors in
RxC
and pulled them back to P4 and Beresford via the isomorphisms,
so they te isomorphisms) are useful after all.
>
>
> > > in fact 3rd dimension is isomorphic to R x C , as
> > pointed out in the very beginning of this post by me.
>
> and it still is.
>
>
>
> > The R-algebra RxRxR is not isomorphic to RxC:
>
> if R x R x R is algebraicly closed of course ...
None of thes is algebraically closed.
In RxRxR, there is no solution to X^2 + (1,0,0) = (0,0,0)
In RxC, there is no solutionto X^2 + (1,0) = (0,0).
>
> we are talking about integral domain
No, we are not.
Or rather apparently you don't know what you are talking about.
>
> else it is
>
> R x R x R is isomorphic to R x cC
>
> where cC is some kind of countercomplex...
>
> but thats off topic here , since we only consider algebraicly closed number systems.
Apparently we don't.
Or again you do not know what you are talking about.
>
> > You cannot find non-zero x,y,z in RxC such that
> > xy=xz=yz=0, but you
> > can in RxRxR.
>
> not if R x R x R is algebraicly closed AND different from P4 AND Beresford i assume ?
If A,B are rings (or here ususally R-algebras), I (and several more
people) denote
with AxB the ring (or algebra) which has AxB as underlying set and
addition (a,b) + (a',b') = (a+a', b+b')
multiplication (a,b)*(a',b') = (a*a',b*b').
So there's little room for any speculating if's.
>
>
>
> > > so we can simplify:
>
> > > --> we must recognize that there are at least 2
> > kinds of integer domains in 3 dimensions
>
> > I agree (if integer domain is replaced by R-algebra),
> > but not
> > by your reasoning -- rather because we have RxC and
> > RxRxR.
>
> > > we must also recognize that using cartesian ,
> > cantor and R x C "logic" we dont get an accurate view
> > of dimensionality or integer domains...
>
> > > writing R x C does not define how many fundamental
> > 3 dimensional integer domains exist.
>
> and it still does not.
>
>
>
> > > worse ; it implies only 1 since all 3rd dimensions
> > are R x C and cannot be for instance C x C or
>
> > > R x R x R ( different from R x C )
>
> > Ah, now you've seen RxRxR, too.
> > But what do you mean with "all 3rd dimensions"?
>
> all "fundamental" integral domains
>
> ( see above )
No,please, don't let me look back at that place above.
Maybe, but what is an axis?
It looks like I can pick arbitrarily many base points (e.g. in C more
than 3 or four
are possible as well) provided that the origin is in their convex
hull.
>
> ( axes is the plural of axis right ? )
Seems so :)
I don't deny that polysigned deserves attention.
I simply don't get how the simple isomorphisms (as algebraic
structures)
P2 = R
P3 = C
P4 = RxC
are so little accepted.
(And, no, P5 is not CxC as P5 contains a 5th root of unity w such that
1,w,w^2,w^3,w^4 generate
it as R-vector space while CxC does not -- but we can still consider
R[X]/(1+X+X^2+X^3+X^4) as a
more "main-stream" object)
hagman
tommy1729
dear hagman and timothy
i respect both of you however
i have totally different views on distance
for hagman id say he has to understand and think about 4 things ( related to his other reply today )
1) when i say algebraicly closed that is not just closed under addition and multiplication but also root and more generally every polynomial computation and its inverses.
this is the usual meaning of the word.
2) if a 3D number system is commutative , distributive , associativ and algebraicly closed i call that an integral domain. i do think that is the correct word for it, at least thats what i mean with it.
we are talking about fields , not countable rings.
3) a 3D integral domain ALWAYS HAS ZERO-DIVISORS.
in fact every 3+ dimensional number system has zero-divisors.
despite that timothy golden does not like the concept of zero-divisors , he underestimates its use.
it fact ; zero-divisors are one of his best defenses of his theory.
proving the existance of zero-divisors is greatly simplified using his polysigned or math based upon it
(like tommy-notation)
proving zerodivisors in cartesian requires good knowledge of topology and the proof is much longer, yet this is unneccesary...
also related to 4)
4) like i said R x C only implies a third dimension.
a isomorphic to R x C and b isomorphic to R x C does not imply a = b.
you dont seem to agree that P4 and Beresford are much different.
yet you are extremely wrong here.
P4 and BERESFORD ARE VERY DIFFERENT.
and as implied in 3) the difference between p4 and beresford can be immediately seen by analysing the zerodivisors.
beresford is simply P3 on a 3D scale, it is a cartesian 3d complex , whereas P4 is totally different.
P4 is not even cartesian and does not have the same zerodivisors, even the shape of the zerodivisors is different.
considering beresford similar to P4 is hypercartesian thinking ...
and hypercartesian thinking is bad...
thinking too cartesian prevents you from fully understanding dimensionality....
you are educated like that of course.
it part of your youth , education and perhaps even career.
and it works fine in 2D and electric engineering and even relativity , so it seems super.
but it aint, its limits lie exactly there where it works so well.
if you look at the millenium prize problems you will notice 2 of them dealing with 3D and one with 4D (perelmann) ; just to show you writing R x C or C x C for 3d or 4d number systems is insufficient to understand 3D or 4D...
regards
tommy1729
Timothy Golden BandTechnology.com
it.
I do not feel compelled to address hagman's concerns since he has not
addressed mine.
You are sticking to your guns here and I'll keep an eye on this and
put in my two cents where they count, or at least where I think that
they count.
I differ in opinion on the insistence that root is fundamentally
meaningful.
I see it as more relevant that your new g(g(x)) context iterates
through the space when x is minus unity. This is a more fundamental
and direct property which indirectly begets your statement above.
Since we want to work toward fundamentals I reccomend adoption of this
awareness, though its usage is barely existent.
>
> this is the usual meaning of the word.
>
> 2) if a 3D number system is commutative , distributive , associativ and algebraicly closed i call that an integral domain. i do think that is the correct word for it, at least thats what i mean with it.
>
> we are talking about fields , not countable rings.
>
> 3) a 3D integral domain ALWAYS HAS ZERO-DIVISORS.
> in fact every 3+ dimensional number system has zero-divisors.
>
> despite that timothy golden does not like the concept of zero-divisors , he underestimates its use.
>
> it fact ; zero-divisors are one of his best defenses of his theory.
Yes. I will stand by all of my previous statements on zero divisors
and especially though it is cryptic my statement to L. Walker above in
this thread.
> proving the existance of zero-divisors is greatly simplified using his polysigned or math based upon it
> (like tommy-notation)
>
> proving zerodivisors in cartesian requires good knowledge of topology and the proof is much longer, yet this is unneccesary...
Yes, the Cartesian product is not necessary.
>
> also related to 4)
>
> 4) like i said R x C only implies a third dimension.
>
> a isomorphic to R x C and b isomorphic to R x C does not imply a = b.
>
> you dont seem to agree that P4 and Beresford are much different.
>
> yet you are extremely wrong here.
>
> P4 and BERESFORD ARE VERY DIFFERENT.
>
> and as implied in 3) the difference between p4 and beresford can be immediately seen by analysing the zerodivisors.
>
> beresford is simply P3 on a 3D scale, it is a cartesian 3d complex , whereas P4 is totally different.
>
> P4 is not even cartesian and does not have the same zerodivisors, even the shape of the zerodivisors is different.
>
> considering beresford similar to P4 is hypercartesian thinking ...
>
> and hypercartesian thinking is bad...
>
> thinking too cartesian prevents you from fully understanding dimensionality....
Perfect Tommy. Congratulations from me for what it is worth.
Dimensionality is an open problem now.
-Tim
hagman
Good to hear that you want the usual meaning.
However, this means that hardly any Pn is algebraically closed,
esp. in RxC there is no solution of X^2 + (1,1) = (0,0) -- or
similarly
in P4 there is no solution of X^2 #1=0.
>
> 2) if a 3D number system is commutative , distributive , associativ and algebraicly closed i call that an integral domain. i do think that is the correct word for it, at least thats what i mean with it.
>
> we are talking about fields , not countable rings.
Do you have any 3dimensional field in mind?
Hint: The R-dimension of a division algebra is easily shown to be a
power of 2.
And (apart from some remarks in one pos about the looks of Pn if one
started
with positive rationals instead of positive reals -- many rational
versions
of Pn can be embedded int C) countability was never a subject in this
thread.
> 3) a 3D integral domain ALWAYS HAS ZERO-DIVISORS.
... which prevents it from being a field.
So we're *not* talking about fields after all?
> in fact every 3+ dimensional number system has zero-divisors.
Quaternions would be free of zero-divisors, but you require
commutativity.
In that case I guess you are right.
>
> despite that timothy golden does not like the concept of zero-divisors , he underestimates its use.
>
> it fact ; zero-divisors are one of his best defenses of his theory.
>
> proving the existance of zero-divisors is greatly simplified using his polysigned or math based upon it
> (like tommy-notation)
I'm not sure where the simplification is.
In RxC,for example, you can immediately *see* the zero-divisors.
I personally had to find them in P4 via the isomorphism.
>
> proving zerodivisors in cartesian requires good knowledge of topology and the proof is much longer, yet this is unneccesary...
>
> also related to 4)
>
> 4) like i said R x C only implies a third dimension.
>
> a isomorphic to R x C and b isomorphic to R x C does not imply a = b.
but isomorphic, and who cares for more?
>
> you dont seem to agree that P4 and Beresford are much different.
>
> yet you are extremely wrong here.
>
> P4 and BERESFORD ARE VERY DIFFERENT.
>
> and as implied in 3) the difference between p4 and beresford can be immediately seen by analysing the zerodivisors.
The algebra isomorphisms induce a bijection between zro-divisors,
IOW all three systems have the same structure of zero-divisors.
>
> beresford is simply P3 on a 3D scale, it is a cartesian 3d complex , whereas P4 is totally different.
>
> P4 is not even cartesian and does not have the same zerodivisors, even the shape of the zerodivisors is different.
>
> considering beresford similar to P4 is hypercartesian thinking ...
>
> and hypercartesian thinking is bad...
>
> thinking too cartesian prevents you from fully understanding dimensionality....
BTW. what do you mean with Cartesian thinking?
>
> you are educated like that of course.
>
> it part of your youth , education and perhaps even career.
>
> and it works fine in 2D and electric engineering and even relativity , so it seems super.
>
> but it aint, its limits lie exactly there where it works so well.
>
> if you look at the millenium prize problems you will notice 2 of them dealing with 3D and one with 4D (perelmann) ; just to show you writing R x C or C x C for 3d or 4d number systems is insufficient to understand 3D or 4D...
I agree that special things happen in dimension 3.
But I derived my right of writing RxC because an isomorphism to the
subject matter
could be found.
And please note that P5 is *not* isomorphic to CxC!
It's rather R[X]/(1+X+X^2+X^3+X^4)
BTW, this representation makes the existence of zero divisors even
clearer:
The ideal is generated by a polynomial of degree 4, which therefore
cannot
be irrducible over R.
>
> regards
> tommy1729
hagman
<tttppp...@yahoo.com> wrote:
> Hi Tommy. You seem to be taking on the role of arbiter and I welcome
> it.
> I do not feel compelled to address hagman's concerns since he has not
> addressed mine.
Are there any concerns you would like me to address?
hagman
lwa...@lausd.net
> And please note that P5 is *not* isomorphic to CxC!
> It's rather R[X]/(1+X+X^2+X^3+X^4)
I was the one who first proposed that P5 would
be isomorphic to C x C. I used a trick similar
to Chapman's trick for P4 and R x C, using not
the fourth roots of unity but rather the fifth
roots of unity. Map -1 (Tim's notation), s
(hagman's notation) to (e^(4pi/5),e^(2pi/5)).
But if I am wrong, then Tim would be happy to
hear that. So P5 wouldn't be isomorphic to the
Cartesian product of copies of R and C at all.
> BTW, this representation makes the existence of zero divisors even
> clearer:
> The ideal is generated by a polynomial of degree 4, which therefore
> cannot
> be irrducible over R.
I did successfully find a pair of zero divisor
in P5 (even if the isomorphism to C x C is
wrong) anyway. Let -, +, *, #, % be the signs,
and phi be the golden mean:
(%phi-1#1)(%phi+1*1) = 0
This is, of course, related to the factors of
the polynomial 1+s+s^2+s^3+s^4.
Timothy Golden BandTechnology.com
Yes. at 9/15/07 message time we had something like
*3 -> (-3, 3i )
I dispute this mapping since distance is not conserved. Then as L.
Walker points out there is a break in symmetry also. The geometry
should match. This is a basic part of any transformation. I attempt to
be doing physics with this thing and I would criticize the mathematics
that you are working in as denying a fundamental of space. Would
anyone consider working out a physical theory in your RxC space when
it does not match the native geometry? Such errors are confusing to me
and it would seem that you would need another layer on top of the RxC
construction to bring the thing back into whack. Perhaps this piece
does not exist. Even if it does then the cleanest approach is to step
back and preserve the geometry first and accept that the product
definition in RxC is much more complicated. In effect from a physical
point of view your approach is insanity. That is too strong a word
really but what else can I put there? We're talking about basic
geometry and you refuse to stoop to this fundamental level.
I have just read on the internet that the product and the sum are
isomorphic to each other in the reals. The value of isomorphism may be
a tool to understanding however you must also then accept that you
have sacrificed something in achieving that understanding. Geometry is
sacrificed to get your relation. These were geometrically equivalent
spaces until you screwed them up. Coming back here you will see the
perspective of my own analysis:
http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html
which actually looks quite promising and interesting. Both the
similarities and the error suggest a deeper relationship than you have
exposed. The code which performs this analysis is T3Study.C in
http://bandtechnology.com/polysign.tar.gz
I am taunting you a little bit but I do so sincerely.
This message did not post to google from mathforum.org so I repost and
edit so sorry for any confusion if you already read one almost
identical.
- Tim
hagman
<tttppp...@yahoo.com> wrote:
> On Sep 19, 2:57 pm, hagman <goo...@von-eitzen.de> wrote:
>
> > On 19 Sep., 16:27, "Timothy Golden BandTechnology.com"
>
> > <tttppp...@yahoo.com> wrote:
> > > Hi Tommy. You seem to be taking on the role of arbiter and I welcome
> > > it.
> > > I do not feel compelled to address hagman's concerns since he has not
> > > addressed mine.
>
> > Are there any concerns you would like me to address?
>
> > hagman
>
> Yes. at 9/15/07 message time we had something like
> *3 -> (-3, 3i )
> I dispute this mapping since distance is not conserved.
Acknowledged, I was merely interested in the algebraic structure,
not metric.
Especially because I had only read the short summary of polysign
theory given early in this thread -- a definition of a metric
on the set of polysigned numbers was not mentioned there, only
the algebraic structure.
Therefore my isomorphism only dealt with the algebraic structure.
Anyway, I later added in some other branch that by defining a specific
norm on the R-vector space RxC, we can achieve that
-1, +1, *1, #1 are vertices of a regular tetrahedron (i.e. are
equidistant):
As usual, a mapping f: RxC -> R is a norm iff
f(a v) = |a| f(v) for all a in R, v in RxC
f(v + w) <= f(v) + f(w) for all v,w, in RxC
f(v)=0 => v=(0,0)
Especially, if we fix positive real numbers a,b, then
(r,z) |-> sqrt( a r^2 + b z conj(z))
is a norm.
Also note that multiplication with an element (r,z) is an isometry
with respect to that norm if |r|=|z|=1, especially for
the isomrphic images of -1,+1,*1,#1, which sounds nice to have.
Moreover, if we want the isomorphic images of -1,+1,*1,#1 to have unit
length,
all we have to ensure is a+b=1.
Now we compute the distances between
#1 and -1: (1,1) - (-1,i) = (2,1-i) has norm sqrt( 4a +2b )
#1 and +1: (1,1) - (1,-1) = (0,2) has norm sqrt( 0 + 4b )
#1 and *1: (1,1) - (-1,-i)= (2,1+i) has norm srqt( 4a + 2b )
(all other distances are of the same flavour as multiplication
with -1 etc. is an isometry)
Thus we can ensure equidistance if 4a+2b=4b, i.e. b=2a.
Therefore, if we define
|(r,z)| = sqrt((|r|^2 + 2 |z|^2)/3)
we have a norm on RxC such that the four fundamental units have unit
length,
are equidistant and multiplying by them is an isometry.
This may look like an awful choice for a metric, but
a) I prefer this unusual weighted norm together with the standard
multiplication
to the usual norm (from viewing RxC as R^3) but with complicated
multiplication
formulas
b) It is not so unusual at all to have to weigh complex constituents
of a structure twice as much as real constituents; this is a
recurring
theme.
> Then as L.
> Walker points out there is a break in symmetry also.
Since (under the above norm) multiplication by any of -1,+1,*1,#1
is an isometry, I assume any symmetry problems resolved
unless you find another specific issue.
> The geometry
> should match.
I assume the geometry to match since the isomorphism is linear.
> This is a basic part of any transformation. I attempt to
> be doing physics with this thing and I would criticize the mathematics
> that you are working in as denying a fundamental of space. Would
> anyone consider working out a physical theory in your RxC space when
> it does not match the native geometry?
Which is the native geometry of space?
The geometry of RxC is Euklidean.
However, you do not want to model our everyday 3D space with P4, do
you?
Addition may be viewed as translation, but what sould multiplication
correspond to?
Especially, 3D space should be isotropic and not have a preferred axis
an a preferred plane (the zero divisors).
> Such errors are confusing to me
> and it would seem that you would need another layer on top of the RxC
> construction to bring the thing back into whack.
I think the norm defined above should do it.
> Perhaps this piece
> does not exist. Even if it does then the cleanest approach is to step
> back and preserve the geometry first and accept that the product
> definition in RxC is much more complicated. In effect from a physical
> point of view your approach is insanity. That is too strong a word
> really but what else can I put there? We're talking about basic
> geometry and you refuse to stoop to this fundamental level.
>
> I have just read on the internet that the product and the sum are
> isomorphic to each other in the reals. The value of isomorphism may be
> a tool to understanding however you must also then accept that you
> have sacrificed something in achieving that understanding. Geometry is
> sacrificed to get your relation. These were geometrically equivalent
> spaces until you screwed them up. Coming back here you will see the
> perspective of my own analysis:
> http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html
> which actually looks quite promising and interesting. Both the
> similarities and the error suggest a deeper relationship than you have
> exposed. The code which performs this analysis is T3Study.C in
> http://bandtechnology.com/polysign.tar.gz
> I am taunting you a little bit but I do so sincerely.
> This message did not post to google from mathforum.org so I repost and
> edit so sorry for any confusion if you already read one almost
> identical.
No problem as I'm one of them google posters :)
>
> - Tim
Timothy Golden BandTechnology.com
cite your post here as a reference on my website. Thanks for bringing
this to some level that I can understand. This metrical digression is
still puzzling to me. I have a Cartesian transform already, but if the
symmetry is not obeyed then isn't translation somehow screwed up?
I'll have to spend some time on this. Symmetry is somewhat broken
under your rendition. I'm learning something here I guess. Product as
a physical entity is deeply puzzling. Why does a geometry even need
one? Yet it is the product which generates support for spacetime. The
product is also involved in classical force equations, albeit
obscurely. Differential torque may be the most relevant force to
consider and in this way these resultants when viewed as second
derivative (classical) forces will generate torques. This then helps
explain why thermodynamics is so slow to transfer through a material.
The speed of sound conduction is far faster than the speed of heat
conduction so the ordinary sense of atoms vibrating cannot be in
translation. They(atoms or constituents of solid matter) are locked
pretty good into fixed positions but they can still spin. Perhaps a
higher dimension solution could also get here but I do like rotation
enough to stake a claim on torque as product related.
-Tim
tommy1729
wrong , there is a solution in P4 !!
and this is important for understanding P4 !!
P4 is closed.
your cartesian thinking has mislead you.
>
> >
> > 2) if a 3D number system is commutative ,
> distributive , associativ and algebraicly closed i
> call that an integral domain. i do think that is the
> correct word for it, at least thats what i mean with
> it.
> >
> > we are talking about fields , not countable rings.
>
> Do you have any 3dimensional field in mind?
> Hint: The R-dimension of a division algebra is easily
> shown to be a
> power of 2.
and that is the heart of the cartesian mistake.
if only a power of 2 , then we should only have dimensions of 2^n , or stop thinking in terms of R-dimensions in this cartesian way.
and similar problems occur in topology for the same reason.
its kind a like this:
a + bi
hmm lets extend
cartesian way : extending a to countercomplex-like a_2 = -a
and same for b.
leading to the skipping potential extensions ( jumping from 2d to 4d , 4d to 8d etc )
such a mistake in polysign IS IMPOSSIBLE.
and also in tommy-notation.
>
> And (apart from some remarks in one pos about the
> looks of Pn if one
> started
> with positive rationals instead of positive reals --
> many rational
> versions
> of Pn can be embedded int C) countability was never a
> subject in this
> thread.
>
> > 3) a 3D integral domain ALWAYS HAS ZERO-DIVISORS.
>
> ... which prevents it from being a field.
> So we're *not* talking about fields after all?
than how does a 3D algebraicly closed commutative assosiativ distributiv field look like ?
you have a point though.
but then the only commutative distributiv associativ algebraicly closed field would be the 2D plain complex numbers.
timothy and i dont see zero-divisors as a restriction.
>
> > in fact every 3+ dimensional number system has
> zero-divisors.
>
> Quaternions would be free of zero-divisors, but you
> require
> commutativity.
> In that case I guess you are right.
yes , also quaternions are 4D !!
in fact there are only 3 or 4 types of zero-divisor free numbers.
so we cant even generalize quaternions to any dimension.
doesnt work very well for arbitrary dimensional physics...
besides quaternions have already been used alot in physics , not a new idea at all.
>
> >
> > despite that timothy golden does not like the
> concept of zero-divisors , he underestimates its use.
> >
> > it fact ; zero-divisors are one of his best
> defenses of his theory.
> >
> > proving the existance of zero-divisors is greatly
> simplified using his polysigned or math based upon it
> > (like tommy-notation)
>
> I'm not sure where the simplification is.
once you understand its simple.
> In RxC,for example, you can immediately *see* the
> zero-divisors.
> I personally had to find them in P4 via the
> isomorphism.
a confesion :)
>
>
> >
> > proving zerodivisors in cartesian requires good
> knowledge of topology and the proof is much longer,
> yet this is unneccesary...
> >
> > also related to 4)
> >
> > 4) like i said R x C only implies a third
> dimension.
and still does.
> >
> > a isomorphic to R x C and b isomorphic to R x C
> does not imply a = b.
>
> but isomorphic, and who cares for more?
who do you think ?
tim and tom :)
key to understanding our ideas id say.
>
> >
> > you dont seem to agree that P4 and Beresford are
> much different.
> >
> > yet you are extremely wrong here.
> >
> > P4 and BERESFORD ARE VERY DIFFERENT.
and still are
> >
> > and as implied in 3) the difference between p4 and
> beresford can be immediately seen by analysing the
> zerodivisors.
>
> The algebra isomorphisms induce a bijection between
> zro-divisors,
> IOW all three systems have the same structure of
> zero-divisors.
>
> >
> > beresford is simply P3 on a 3D scale, it is a
> cartesian 3d complex , whereas P4 is totally
> different.
> >
> > P4 is not even cartesian and does not have the same
> zerodivisors, even the shape of the zerodivisors is
> different.
> >
> > considering beresford similar to P4 is
> hypercartesian thinking ...
> >
> > and hypercartesian thinking is bad...
> >
> > thinking too cartesian prevents you from fully
> understanding dimensionality....
>
> BTW. what do you mean with Cartesian thinking?
the above replied ... especially 2^n R-algebra thinking
or oversimplifying to R x C.
considering all in 2 directions ( + and - )
it leads to mistakes or at least misunderstandings or insufficient definitions concercerning dimension.
>
> >
> > you are educated like that of course.
> >
> > it part of your youth , education and perhaps even
> career.
> >
> > and it works fine in 2D and electric engineering
> and even relativity , so it seems super.
> >
> > but it aint, its limits lie exactly there where it
> works so well.
> >
> > if you look at the millenium prize problems you
> will notice 2 of them dealing with 3D and one with 4D
> (perelmann) ; just to show you writing R x C or C x C
> for 3d or 4d number systems is insufficient to
> understand 3D or 4D...
>
> I agree that special things happen in dimension 3.
> But I derived my right of writing RxC because an
> isomorphism to the
> subject matter
> could be found.
and in some way thats exactly our point ...
surprising hey ?
:)
> And please note that P5 is *not* isomorphic to CxC!
> It's rather R[X]/(1+X+X^2+X^3+X^4)
>
> BTW, this representation makes the existence of zero
> divisors even
> clearer:
> The ideal is generated by a polynomial of degree 4,
> which therefore
> cannot
> be irrducible over R.
i agree on that.
>
> >
> > regards
> > tommy1729
>
>
>
regards
tommy1729
Timothy Golden BandTechnology.com
mathematically.
I know that I can go to Wikipedia and look it up.
Still, the nomenclature of existing math is being attacked directly by
the polysign construction.
The simplicity of the polysign construction and its suggestive signals
are promising.
We struggle with discrete and continuous thinking throughout physics
and philosophy.
Here we see a simple data structure embodying both with consequences.
I am very happy to have hagman and L. Walker and Tommy studying this
topic.
The polysign construction is apparently overlooked.
Still, it would not surprise me if someone dug out some old manuscript
laying it out no different than it is here. Now we have computers and
animations to help understand these things.
No journal will provide an animation in its paper form.
Until I started graphing P3 operations I had no idea that they were
complex numbers.
Until I lined up the * axis with the real axis it was still invisible.
Such is the nature of the human mind.
Our limitations are profound.
We are caught constructing the truth and therefore all knowledge is
insecure, including the polysign construction. Stumbling along and
being trained on the works of those before us is only half of what
ought to be taught. The other half is the art of construction and how
this can be taught is a puzzle. Clearly an open space must be
established to construct in. This open space is vacuous. From thin air
we can encode definitions and study their consequences. The value of a
construction is its consequences. Diversity in this process is to be
encouraged yet there is only one right answer for every math problem.
It does not feel good to sense this situation. I don't mean to be
gloomy but the ability of the modern mathematician to face this
opening is questionable. I suppose we'll keep going on here and keep
finding some interesting details and enjoy this new thing. Still, if
it is correct (and I believe that it is) then it paints us as in a
gray age. The human's abilities and development are in a progression.
The accumulation is too heavy right now. I'll just keep practicing my
whistle here and see what unfolds. Another rendition will come along
soon I think.
-Tim
Timothy Golden BandTechnology.com
z = - z z @ m
http://bandtech.com/tmp/MagnimatedSimplicity.gif
-Tim
On Sep 11, 1:23 pm, "Timothy Golden BandTechnology.com"
<tttppp...@yahoo.com> wrote:
> I've started another thread titled "Magnitude Sweep Functions" but for
> some reason it is buried here and my posts are not showing through on
> google's topic list thought the content is present. So I try again
> here because I would like to get your input.
>
> This is a really simple construction. Simply consider a magnitude that
> sweeps from zero to large values and impose it as m onto an iterated
> function:
> z(n+1) = - z(n) z(n) @ m
> where z is a polysign value and @ implies superposition. The magnitude
> essentially takes the zero sign which is right where it should be. The
> equivalent in the complex plane (P3) would be
> z(n+1) = (e^(i2pi/3)) z(n) z(n) + m .
> Plot z(n) for each magnitude m. Iterations of n to 10000 or so will
> either get very large, very small, or find some stable orbit. The
> increment in m gets pretty good results around 4e-4 but coarser will
> certainly generate graphics. The graphics of this procedure are pretty
> interesting. The function itself could be alot of different things but
> it is the rotational effect in the product that is generating the
> interesting detail. The color scheme needs a lot of work but this is
> some of the results:http://bandtechnology.com/PolySigned/MagnitudeSweep
> If you don't understand polysign you could try:http://bandtechnology.com/PolySigned
>
> -Tim
hagman
<tttppp...@yahoo.com> wrote:
> Very nice presentation hagman. I'll verify your metric and if it works
> cite your post here as a reference on my website. Thanks for bringing
> this to some level that I can understand. This metrical digression is
> still puzzling to me. I have a Cartesian transform already, but if the
> symmetry is not obeyed then isn't translation somehow screwed up?
> I'll have to spend some time on this. Symmetry is somewhat broken
> under your rendition. I'm learning something here I guess. Product as
> a physical entity is deeply puzzling. Why does a geometry even need
> one?
Geometry does not need that product, but in your P4, multiplication
is one of the fundaments and without it, P4 (including metric) is
nothing
but the metric vector space R^3 -- thus not leaving much room for you
to gain formerly unknown physical insights.
Timothy Golden BandTechnology.com
> On 20 Sep., 23:24, "Timothy Golden BandTechnology.com"
>
> <tttppp...@yahoo.com> wrote:
> > Very nice presentation hagman. I'll verify your metric and if it works
> > cite your post here as a reference on my website. Thanks for bringing
> > this to some level that I can understand. This metrical digression is
> > still puzzling to me. I have a Cartesian transform already, but if the
> > symmetry is not obeyed then isn't translation somehow screwed up?
> > I'll have to spend some time on this. Symmetry is somewhat broken
> > under your rendition. I'm learning something here I guess. Product as
> > a physical entity is deeply puzzling. Why does a geometry even need
> > one?
>
> Geometry does not need that product, but in your P4, multiplication
> is one of the fundaments and without it, P4 (including metric) is
> nothing
> but the metric vector space R^3 -- thus not leaving much room for you
> to gain formerly unknown physical insights.
I don't know if you are still monitoring this thread Hagman but I see
spacetime as
P1 P2 P3 ...
where the behavior of P4+ is indeed broken in terms of distance
conservation under product.
So you see that we have unidirectional time in P1 and the promise of
Maxwell's equations in P2P3.
I do not yet have a full theory built, but the 'physical insights'
already look promising. In effect if we take polysign systems obeying
| A || B | = | A B |
we see spacetime. The ones which do not obey may also be useful, but
they are nonlinear. So dynamics are in order for P4+.
Perhaps my usage of 'nonlinear' here is inappropriate, but you must
see what I mean by it.
I still have not gotten to a computer proof of your metric. I will get
to it and it won't take very long.
Still though, the hole of distance discrepancy cannot be satisfying.
P4 has geometry. It has distance. It has algebra. It seems that
isomorphism is happy to focus its beam on just a few qualities and so
the term is more like an analogy than an equivalence.
Indeed when I look up the term 'isometry' on Wiki I get
"In mathematics, an isometry, isometric isomorphism or congruence
mapping is a distance-preserving isomorphism between metric spaces.
Geometric figures which can be related by an isometry are called
congruent."
I don't care to get too deep into these mathematical terminologies.
Still if I am going to carry on a discussion with mathematicians I
will have to keep working at these. So I point you to a stronger form
which you cannot achieve, or at least have not achieved yet.
-Tim
-Tim
-Tim
Timothy Golden BandTechnology.com
I'm pretty sure that I have an isometric equivalence. This is going to
be really interesting since the T3 tatrix is involved which is the
progression
P1 P2 P3
which can also be represented as:
a11
a21 a22
a31 a32 a33
and has a rotational mime
a11
a21 a22
a31 a32 a33
which is equivalently
a11
a21 a22
a31 a32 a33
and which also carries a zero column of values in rendered (reduced)
form leaving
s1P2 x2a
s1P3 x3a + s2P3 x3b
or equivalently
s1P2 x2a + s1P3 x3a + s2P3 x3b .
However note that this is an anisotropic equation.
Anyhow there must be a shorter path to get that the scalar multiple
that you are getting and I am looking at as an error is actually a P1
component vector. Is it a constant unity? That seems to hard to
believe.
Anyow this reeks of the tatrix form though it is indirect. I was
wanting to manipulate the tatrix pattern to get to the idea that the
real valued components can be cross elements of the tatrix. There are
only three of them. In effect there may be three interpretations from
one tatrix, but as we just saw they are the same form just with values
rotated. These are in some regard isotropic positions so that the
tatrix could be taken more as:
a1 b1 a2
b2 b3
a3
This is some sort of a rotation group I suppose though my usage of the
word group is poor. Anyhow this symmetry of the tatrix should not be
overlooked. Eventually I formalize these loose thoughts into a P1P2P3
isometry that will top your isomorphism. That is unless you beat me to
it first.
-Tim