Matrix form for polysign product

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Timothy Golden BandTechnology.com

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May 28, 2006, 11:04:46 AM5/28/06
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This thread is motivated by a discussion at:

http://groups.google.com/group/sci.physics.relativity/msg/c551631f5c8b61f6
.
We are getting into pure math and this is a clean start on the problem
that will hopefully draw in some feedback from you.

The polysign construction is covered at:
http://bandtechnology.com/PolySigned/PolySigned.html

We would like a matrix expression that will do the polysign product. I
am slightly turned off by this approach because of the simplicity of
the native math that we are about to garble but since the lingua franca
is block language it must be done.

We can take two Cartesian values x1, x2 of similar dimension and
transform them to polysigned values. We can then take their product in
the polysigned domain. We can then transform this back to a cartesian
value x3.

So we know that this product
x1 x2 = x3
is possible and exists in any whole dimension. The challenge is
expressing it purely in the matrix system. Rather than use P4 I think
we should use P3 as an example since the number of terms will be
simplified. Since the process is general it will extend naturally.

The following notation will be used:

An explicit Cartesian value will be represented in the form:
[ Cn X0, X1, X2, ..., X(n-1) ]
where X are real values and are the usual vector form in n dimensions.
An explicit polysigned value will be represented in the form:
[ Pm S0, S1, S2, ..., S(m-1) ]
where S are magnitudes that inherently hold their sign as their
position.
S0 is the identity sign, S1 is first sign, etc.
If we instantiate two 2D Cartesian x's x1 and x2 we'll show their
components as:
x1 = [ C2 x10, x11 ]
and
x2 = [ C2 x20, x21 ]
so the last index is the component and the first index is an
identifier.
This applies similarly for the polysigned domain.
The transform from Cartesian to polysigned is resolved:
http://bandtechnology.com/PolySigned/CartesianTransform.html
so we can simply take
s1 = x1 t23
s2 = x2 t23
where x1 and x2 are 2D Cartesian, t23 is the 2 x 3 transfrom matrix,
and s1, s2 are in P3. There is a slight ambiguity in that the s values
could come out negative in the math of the transform. If this happens
they can be normalized. The details of this are not problematic and can
be discussed if necessary.

So we have two three-signed values s1 and s2 that came from their
Cartesian counterparts x1 and x2. Their product is defined as:
s1 s2 = s3
where
s30 = s10 s20 + s11 s22 + s12 s21
s31 = s11 s20 + s10 s21 + s12 s22
s32 = s12 s20 + s10 s22 + s11 s21 .
This is covered in slightly different form at:
http://bandtechnology.com/PolySigned/PolySigned.html .
Here the + operator means summation and should not be confused with the
polysign symbolic representation. All of the above values are
magnitudes being multiplied and summed.
The matrix challenge is embodied in the expression of s3 above.
Upon declaring that in matrix notation the desired x3 is
simply:
x3 = s3 t32
where t32 is the inverse transform matrix.

The expression is rotationally variant but allows the product and sum
to take on arithmetic qualities so that the usual distributive,
associative, and commutative laws of the real numbers apply in any
whole dimension.
Does the rotational variance necessarily deny the usage of tensor
notation? In that case the algorthmic matrix operation that is needed
may be new. It has to spin the source vectors around to produce s3.
When that operation is generalized to any dimension the matrix
definition will be completed.
I hope that a matrix guru can get this notation easily. It's not
necessarily embodied by an existing matrix operation.
It really boils down to a matter of notation. This operation is
embodied in the three lines that are the polysigned product. You can
break it out in two ways:

s3 = [ P3 s10s20, s11s20, s12s20 ]
+ [ P3 s12s21, s10s21, s11s21 ]
+ [ P3 s11s22, s12s22, s10122 ]

or alternatively

s3 = [ P3 s10s20, s10s21, s10s22 ]
+ [ P3 s11s22, s11s20, s11s21 ]
+ [ P3 s12s21, s12s22, s12s20 ] .

They are symmetrical extensible representations. I am unaware of the
symbollic representation of such an operation in matrix form. If it
exists already please comment. If not I guess we'll have to name such
an operation and attempt to give it a symbol and definition. I would
call it a spin or flip form. But I am no matrix master so please
comment if you are.

-Tim

Spoonfed

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May 28, 2006, 7:46:37 PM5/28/06
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Timothy Golden BandTechnology.com wrote:
> This thread is motivated by a discussion at:
>
> http://groups.google.com/group/sci.physics.relativity/msg/c551631f5c8b61f6
> .

I'll fill in some of what I thought were major details from that
discussion.

Tim wrote:
> Spoonfed wrote:
> > Tim wrote:
> <Snipping away old stuff.>
> > > Here is the general n-signed product in C++ code:
> > > for( i = 0; i < n; i++ )
> > > {
> > > for( j = 0; j < n; j++ )
> > > {
> > > k = (i+j)%n;
> > > x[ k ] += s1.x[i] * s2.x[j];
> > > }
> > > }


Note, (i+j)%n = (i+j)modulo n

This particular piece of code makes the n-signed numbers
indistinguishable from particular sets of complex numbers. However,
there is another piece to the definition

By definition, P4 represents four vectors lying at angles of
Pi-arccos(1/3) radians from each other. These four vectors cannot lie
in the same plane, thus are not described by any set of four complex
numbers.


In my understanding, this is generalizable, so in one dimension, P2,
you have two vectors lying 180 degrees from one-another,


In 2 dimensions, P3 is three vectors lying 120 degrees from
one-another.


In 3 dimensions, P4 is four vectors lying 109.47 degrees from each
other.

In n dimensions, you have four vectors lying 180-arccos(1/(n-1))
degrees from one another.

Now the final key was that Tim said that multiplication was possible in

any level of dimension. Not cross-product, not dot product, but simple

multiplication. I have never heard of such a thing in 3D, but he's
given a general, reasonable set of rules for simple multiplication in
his C++ code, which reproduces multiplication in 1D and 2D, and seems
to flow into a pattern generalizable to any dimension.


> i={1,0,0}
> j={-1/3, sqrt(8/9),0}
> k={-1/3,sqrt(2/9),sqrt(2/3)}
> l={-1/3,sqrt(2/9),-sqrt(2/3)}


> i+j+k+l=0


> Op0=Transpose[{i,(3j+i)/sqrt(8),(2k+j+i)/sqrt(8/3)}]


> ={{1,0,0},{0,1,0},{0,0,1}}


> Op1=Transpose[{j,(3k+j)/sqrt(8),(2l+k+j)/sqrt(8/3)}]
> ={{-1/3,-sqrt(2/9),-sqrt(2/3)},{sqrt(8/9),-1/6,-sqrt(1/12)},{0,sqrt(3/4),-1­/2}}


> Op2=Transpose[{k,(3l+k)/sqrt(8),(2i+l+k)/sqrt(8/3)}]
> {{-1/3,-sqrt(2/9),sqrt(2/3)},{-sqrt(2/9),-2/3,-sqrt(1/3)},{sqrt(2/3),-sqrt(­1/3),0}}


> Op3=Transpose[{l,(3i+l)/sqrt(8),(2j+i+l)/sqrt(8/3)}]
> ={{-1/3,sqrt(8/9),0},{-sqrt(4/9),-1/6,sqrt(3/4)},{-sqrt(2/3),-sqrt(1/12),-1­/2}}


> I have verified this in Mathematica, but I can't copy and paste into a
> text editor, so errors may have cropped up in the re-typing.


> Anyway, I verified with these operators that
> Op0.i=i
> Op0.j=j
> Op0.k=k
> Op0.l=l


> Op1.i=j
> Op1.j=k
> Op1.k=l
> Op1.l=i


> Op2.i=k
> Op2.j=l
> Op2.k=i
> Op2.l=j


> Op3.i=l
> Op3.j=i
> Op3.k=j
> Op3.l=k


> Det[Op0]=Det[Op2]=1
> Det[Op1]=Det[Op3]=-1


Op0 is identity corresponding to multiplication by #1 in Tim's P4
notation
Op1 (multiplying by -1) and Op3 (multiplying by *1) involving
reflection, rotation and scaling, and Op2 (multiplying by +1) is a
proper transformation involving rotations and scaling, but no
reflection.

Timothy Golden BandTechnology.com

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Jun 1, 2006, 6:55:10 AM6/1/06
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I'm unburying this thread in the hopes that you will respond.
The form:

s30 = s10 s20 + s11 s22 + s12 s21
s31 = s11 s20 + s10 s21 + s12 s22
s32 = s12 s20 + s10 s22 + s11 s21 .
is extensible to any dimension and provides a commutative product
definition.
Is there an existing definition of this matrix form?
The sources s1 and s2 are native to the polysigned domain so that their
unit vectors form a nonorthogonal n-1 dimensional simplex coordinate
system subject to the following extensible rule:
[ 1, 1, 1 ] = 0
In the three component (P3) form above this system generates the
complex numbers.
In P2 it generates the real numbers. P4 and above are poorly
understood.
In trying to talk matrix language the system suffers a breakage in P4
where rotational variance affects the product so that the math is not
tensor compatible. Yet this simple patterned and extensible expression
looks very much like a matrix form.
What is it?

Spoonfed

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Jun 4, 2006, 11:04:22 AM6/4/06
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I only understand a little of your terminology, but I think you are
asking for these.

/s1-s3\ = /1 1/(2 Sin[th2])\/r1\
\s2-s3/ \0 2/(2 Sin[th2])/\r2/

and

/r1\ = /1 Cos[th2]\/s1-s3\
\r2/ \0 Sin[th2]/\s2-s3/

Where th2 = Pi - ArcCos[1/2]

And once we get in the {s1-s3,s2-s3} domain:

i=/1 0\
\0 1/

j=/0 -1\
\1 -1/

k=/-1 0\ (k won't be used in the solution, but it's here for
reference.)
\-1 1/
/ \
{i,j} = | / 1 0\ |
| \ 0 1/ |
| |
| / 0 -1\ |
| \ 1 -1/ |
\ /

So multiplying {s00,s01,s02} by {s10,s11,s12} is the same as
multiplying
{s00-s02,s01-s02,0} by {s10-s12,s11-s12,0}.

({i,j}.{s00-s02,s01-s02}).{s10-s12,s11-s12}

(Note you can switch the order of the 2X1 matrices, but the 2X2X2
matrix must be multiplied first)

{i.j}.{s00-s02,s01-s02}=
(s00-s02) */1 0\
\0 1/
+(s01-s02)*/0 -1\
\1 -1/

=/s00-s02 s02-s01\
\s01-s02 s00-s01/

Then we dot this matrix with {s10-s12,s11-s12} to get...

(s00-s02)(s10-s12)+(s02-s01)(s11-s12)+(s01-s02)(s10-s12)+(s00-s01)(s11-s12)

simpifying to

s00 (s10+s11-2 s12)+s01 (s10-2 s11+s12)+s02 (-2 s10+s11+s12)

...at which point I need to take a break. A check is still needed--How
does this compare to answers achieved with Polysigned method or by
complex notation? I see I've failed to keep the indexes the same, for
one thing.

Timothy Golden BandTechnology.com

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Jun 5, 2006, 5:28:11 PM6/5/06
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Spoonfed wrote:
>
> I only understand a little of your terminology, but I think you are
> asking for these.
>
> /s1-s3\ = /1 1/(2 Sin[th2])\/r1\
> \s2-s3/ \0 2/(2 Sin[th2])/\r2/

I don't understand the notational meaning of '/' and '\' in your
equations.
We're working in P3 or 2D so I'm guessing that you're getting your own
proof for complex equivalence here. That should come out of your new
angled Cartesian system here as well.
So I'm guessing and seeing that's what you're doing. Good idea. I
suppose I should be able to decipher the notation but I'm going to let
you tell me.
I'll try to do this your way (or should I say my version of your way)
and see if there is some pattern matching.
So in P3 if we have two points [ P3 a', b', c' ] and [ P3 d', e', f' ]
We force the last component to zero and enter the SP3 domain of real
valued components
These become
[ SP3 a, b, 0 ] and [ SP3 d, e, 0 ]
where a=a'-c', b=b'-c', d=d'-f', e=e'-f' .
Alternatively this will become
[SC2 a, b ] and [SC2 d, e]
where SC2 is a special nonorthogonal Cartesian(is that an oxymoron?) 2D
space with a specific angle of sixty degrees between the axes.
So the product of these two points which have now been expressed in
three subtly different coordinate systems is:
[ SP3 ad, ae + bd, be ]
which is
[ SC2 ad - be, ae + bd - be ] .
So if you slant the coordinate system to sixty degrees instead of
ninety the coordinate system will yield the usual complex number
product using this last formula.The positive real axes is the first
coordinate axis and the positive real is thirty degrees past the
second.
To resolve the components you need to use a paralellogram resolution,
NOT the perpendicular resolution, which is inconsistent.
It is possible to check this doing a few computations like ( i )( i )
but I haven't yet so no verification except algebra. I started checking
(0+1i)(0+1i) but got tired at the reverse transform out of SC2. It
would be nice to demonstrate this. Can you do all this in Mathematica?
It seems very round about compared to my proof:
http://bandtechnology.com/PolySigned/ThreeSignedComplexProof.html
but if you will be convinced probably others will be too so it would be
a nice step and would verify your approach.

>
> and
>
> /r1\ = /1 Cos[th2]\/s1-s3\
> \r2/ \0 Sin[th2]/\s2-s3/
>
> Where th2 = Pi - ArcCos[1/2]
>
> And once we get in the {s1-s3,s2-s3} domain:
>
> i=/1 0\
> \0 1/
>
> j=/0 -1\
> \1 -1/
>
> k=/-1 0\ (k won't be used in the solution, but it's here for
> reference.)
> \-1 1/
> / \
> {i,j} = | / 1 0\ |
> | \ 0 1/ |
> | |
> | / 0 -1\ |
> | \ 1 -1/ |
> \ /
>
> So multiplying {s00,s01,s02} by {s10,s11,s12} is the same as
> multiplying
> {s00-s02,s01-s02,0} by {s10-s12,s11-s12,0}.
>
> ({i,j}.{s00-s02,s01-s02}).{s10-s12,s11-s12}
>

I'm still confused by the notation with / and \.

> (Note you can switch the order of the 2X1 matrices, but the 2X2X2
> matrix must be multiplied first)
>
> {i.j}.{s00-s02,s01-s02}=
> (s00-s02) */1 0\
> \0 1/
> +(s01-s02)*/0 -1\
> \1 -1/
>
> =/s00-s02 s02-s01\
> \s01-s02 s00-s01/
>
> Then we dot this matrix with {s10-s12,s11-s12} to get...
>
> (s00-s02)(s10-s12)+(s02-s01)(s11-s12)+(s01-s02)(s10-s12)+(s00-s01)(s11-s12)
>
> simpifying to
>
> s00 (s10+s11-2 s12)+s01 (s10-2 s11+s12)+s02 (-2 s10+s11+s12)
>
> ...at which point I need to take a break. A check is still needed--How
> does this compare to answers achieved with Polysigned method or by
> complex notation? I see I've failed to keep the indexes the same, for
> one thing.

The equation above looks about right except for some typos. You've
shifted the ID's down to 0 and 1 from 1 and 2 and there is a '-2' where
there should be a '+' in each parenthetical. Also the indices are not
stated so if these are all magnitudes you'd have just a single value at
the evalution of your one line equation. But the form is there.

The product is really very simple and straightforward in its native
format. We are complicating it quite a bit but if you are able to do
these problems in Mathematica under the shifted Cartesian pretext then
it is worthy. We'd have some verification ability since your software
and mine are independent. Does Mathematica handle the nonorthogonal
coordinate system well? What about the product definition? I haven't
tried to see if there is a general product definition coming out of the
forms that we've made in SC but if it exists and can be coded in
Mathematica then you'd have a reasonable system.

-Tim

Timothy Golden BandTechnology.com

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Jun 6, 2006, 10:09:40 AM6/6/06
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Does anyone know of an existing matrix definition for the following
matrix form?

s30 = s10 s20 + s11 s22 + s12 s21
s31 = s11 s20 + s10 s21 + s12 s22
s32 = s12 s20 + s10 s22 + s11 s21 .

It is commutative so that
s1 s2 = s2 s1 .
It is extensible to any dimension.
When the identity
( x, x, ... ) = 0
is invoked the system drops one dimension and derives the real numbers
and complex numbers. It also generates weird high dimensional spaces.
The well behaved members of the family are congruent with spacetime.
The system defines a zero-dimensional space that has time
correspondence.

-Tim

Spoonfed

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Jun 7, 2006, 7:42:52 PM6/7/06
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Timothy Golden BandTechnology.com wrote:
> Spoonfed wrote:
> >
> > I only understand a little of your terminology, but I think you are
> > asking for these.
> >
> > /s1-s3\ = /1 1/(2 Sin[th2])\/r1\
> > \s2-s3/ \0 2/(2 Sin[th2])/\r2/
>
> I don't understand the notational meaning of '/' and '\' in your
> equations.

This is supposed to be a linear algebra expression, with / and \
denoting parentheses on more than one line.

Let me try to give you some pseudocode you can translate into C++

sa = r1 + r2/(2*Sin[th2])
sb = r2 + r2/Sin[th2]
s3=Max(-sb,-sa)
s1=sa+s3
s2=sb+s3

sa=s1-s3
sb=s2-s3
r1=sa + sb*Cos[th2]
r2=sb*Sin[th2]

Wow, I really screwed that one up. 2X2 matrix times 2X1 vector should
give a 2X1 vector. In the end it looks like garbage no matter how you
write it, so have a look at the pseudocode written below. Hopefully a
little less bewildering.


> >
> > ...at which point I need to take a break. A check is still needed--How
> > does this compare to answers achieved with Polysigned method or by
> > complex notation? I see I've failed to keep the indexes the same, for
> > one thing.


'Define two numbers in P3 as ordered triplets {s00,s01,s02} and
{s10,s11,s12}
'Then define ordered pairs {a0,b0} and {a1,b1} based on them as
follows:

a0=s00-s02
b0=s01-s02

a1=s10-s12
b1=s11-s12

'Then we want to form a 2X2 matrix {{m00,m01},{m10,m11}} from a0*i
+b0*j as follows:

'(This is the section of code where we are multiplying a 1X2 by a
2X(2X2) matrix to get a 1X(2X2) 'matrix, or more commonly known as a
2X2 matrix.

m00=a0
m01=-b0
m10=b0
m11=a0-b0

'Now take this new matrix M and multiply it by {a1,b1} to get {a2,b2}

a2=m00*a1+m01*b1
b2=m10*a1+m11*b1

'And to get {s20,s21,s22}

s22=Max[-a2,-b2,0]
s20=a2+s22
s21=a1+s22

>
> The equation above looks about right except for some typos. You've
> shifted the ID's down to 0 and 1 from 1 and 2


Sorry about that.

Spoonfed

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Jun 7, 2006, 8:41:54 PM6/7/06
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I was introduced the other day to GF[4] which bears some vague
resemblance to P3. The multiplication table is the same as for unit
vectors in P3, but the addition properties are a little off.

http://en.wikipedia.org/wiki/Galois_field

Timothy Golden BandTechnology.com

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Jun 8, 2006, 7:40:39 AM6/8/06
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As I look at the tables at the bottom of that wiki page I don't see
much alignment except that GF(3) addition table looks like P3
multiplication table. The reliance on primes seems to be basic to
Galois Fields. In polysign the prime sign systems are very flat in
terms of symmetry of the signs as operators. But there is no
restriction about primes.

P4 and all higher even sign systems are not fields since division does
not necessarily work. The last requirement of the mathematical field
is:

Existence of multiplicative inverses:

For every a ≠ 0 belonging to F, there exists an element
a^−1 in F,
such that
a * a^−1 = 1.

In P4 if we multiply a value by
+ 1 # 1
we will get another value on that same ray. I call this the identity
axis. It exists in every even signed system. In effect what was a zero
becomes a zero axis to the field definition. The dimensionality does
not allow the strict field definition to apply. This does not trouble
me. The polysigned numbers challenge existing mathematics on many
levels. It's because existing mathematics is built upon the real
numbers. The singular zero of this field requirement becomes an axis in
general. I can't change the definition of a field. To quibble over this
as a stopping point is not relevant. The polysigned system behaves on
its own terms.

You've reminded me of
http://groups.google.com/group/sci.math/msg/11ceb7ea329e2a97
Roger Beresford's terplex product exactly matches so he must have the
same matrix form.
It's readily apparent when you look at his unit vector definition.
But again without the identity which drops the actual dimensionality by
one.
We hashed it out on that thread but I have failed to recall it as a
resource. It does exist in other usage and sci.math just isn't a very
good place to query for the info. He has code too:
http://library.wolfram.com/infocenter/MathSource/4894/
His stuff deals in orthogonal bases.
Augmented with the forward and reverse Cartesian transforms it would be
sufficient to do general polysigned math.
This approach would skip over your latest move of SC slanted space. The
identity is somewhat optional to the system. You won't need it except
that equivalent values don't necessarily look the same until reduced.
The inverse transform will take care of that for you as long as you
look at your values in Cartesian. Keep in mind that above P3 the
solutions become rotationally variant so what was isotropic becomes
anisotropic yet maintains linearity. I'm not really comfortable with
the word isotropic. It seems to carry a haze around with it.

In effect your generic Cartesian product is a function that looks like:

Cartesian Product( Cartesian c1, Cartesian c2 )
{ // this does the polysigned product.
// caution: the transform orientation matters.
// |AB| = |A||B| breaks above 2D.
Cartesian s1( c1.dim + 1 ) = PolySignXform( c1 );
Cartesian s2( c2.dim + 1 ) = PolySignXform( c2 );
Cartesian s3( c1.dim + 1 ) = TerplexProduct( s1, s2 );
Cartesian c3( c1.dim ) = InversePolySignXform( s3 );
return c3;
}

Even without the identity function this will work fine. In effect the
identity has been embodied by the transform. The usual vector sum is
the same in either space so there is no trouble there. This should be a
general solution upon generalizing the transform and if you are willing
to work in a particular sign like P4 that part can be hard coded for
now. This might help turn others on to the system.

-Tim

Timothy Golden BandTechnology.com

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Jun 8, 2006, 7:43:26 AM6/8/06
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Oops. This should be 120 degrees, not sixty.

> ninety the coordinate system will yield the usual complex number
> product using this last formula.The positive real axes is the first
> coordinate axis and the positive real is thirty degrees past the
> second.
Above should say the positive imaginary axis is thirty degrees less
than the second SC axis.

Timothy Golden BandTechnology.com

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Jun 8, 2006, 8:14:31 AM6/8/06
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I did the index translation on paper and have verified the match.
You are a bit hush about where you are going. I'm seeing you plugging
into a Lorentz style transform inherently at the specific angle. Out
come the complex numbers. Then what?
On to higher signs? Perhaps you can respond on the other tendril and
let this one die.

Spoonfed

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Jun 10, 2006, 10:17:04 PM6/10/06
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Look closer. The multiplication table for GF(3) corresponds to P2.
GF(4) corresponds to P3.


> P4 and all higher even sign systems are not fields since division does
> not necessarily work. The last requirement of the mathematical field
> is:
>

If multiplication in P4 can be represented by matrix multiplication,
division should be possible through multiplying by the inverse matrix.

> Existence of multiplicative inverses:
>
> For every a ≠ 0 belonging to F, there exists an element
> a^−1 in F,
> such that
> a * a^−1 = 1.
>
> In P4 if we multiply a value by
> + 1 # 1
> we will get another value on that same ray. I call this the identity
> axis.

I thought that #1 and (-1+1*1) identified the identity axis... But
then again, any real-valued eigenvalue associated with a real valued
eigenvector should have this property. I posted a list of these
somewhere either in this or the other thread.

> It exists in every even signed system. In effect what was a zero
> becomes a zero axis to the field definition. The dimensionality does
> not allow the strict field definition to apply. This does not trouble
> me. The polysigned numbers challenge existing mathematics on many
> levels. It's because existing mathematics is built upon the real
> numbers. The singular zero of this field requirement becomes an axis in
> general. I can't change the definition of a field. To quibble over this
> as a stopping point is not relevant. The polysigned system behaves on
> its own terms.
>

Alright, since you brought it up, let's quibble. Perhaps some insight
might be gained. Have you invented a new math, or have you simply
found some interesting properties of the old math?

If the polysigned system behaves on its own terms, then we should
highlight those differences. Nothing to be gained by avoiding the
stopping points; we should approach them directly and stop at them.

To represent zero as a half dimension doesn't make any sense. You say
there is some fundamental difference in P2 between +7-5 and +2. To me
they are simply different representations of the same number.

> You've reminded me of
> http://groups.google.com/group/sci.math/msg/11ceb7ea329e2a97
> Roger Beresford's terplex product exactly matches so he must have the
> same matrix form.
> It's readily apparent when you look at his unit vector definition.
> But again without the identity which drops the actual dimensionality by
> one.
> We hashed it out on that thread but I have failed to recall it as a
> resource. It does exist in other usage and sci.math just isn't a very
> good place to query for the info. He has code too:
> http://library.wolfram.com/infocenter/MathSource/4894/
> His stuff deals in orthogonal bases.
> Augmented with the forward and reverse Cartesian transforms it would be
> sufficient to do general polysigned math.

Cool.

> This approach would skip over your latest move of SC slanted space. The
> identity is somewhat optional to the system.

What are you referring to here?

> You won't need it except
> that equivalent values don't necessarily look the same until reduced.
> The inverse transform will take care of that for you as long as you
> look at your values in Cartesian. Keep in mind that above P3 the
> solutions become rotationally variant so what was isotropic becomes
> anisotropic yet maintains linearity.

i.e. There ARE special directions. Multiplying by #1 maintains the
shape, but multiplying by *1 inverts everything.

> I'm not really comfortable with
> the word isotropic. It seems to carry a haze around with it.
>
> In effect your generic Cartesian product is a function that looks like:
>
> Cartesian Product( Cartesian c1, Cartesian c2 )
> { // this does the polysigned product.
> // caution: the transform orientation matters.
> // |AB| = |A||B| breaks above 2D.
> Cartesian s1( c1.dim + 1 ) = PolySignXform( c1 );
> Cartesian s2( c2.dim + 1 ) = PolySignXform( c2 );
> Cartesian s3( c1.dim + 1 ) = TerplexProduct( s1, s2 );
> Cartesian c3( c1.dim ) = InversePolySignXform( s3 );
> return c3;
> }
>

Hmmm.

> Even without the identity function this will work fine. In effect the
> identity has been embodied by the transform.


Still not quite sure what you mean by the identity.

> The usual vector sum is
> the same in either space so there is no trouble there. This should be a
> general solution upon generalizing the transform and if you are willing
> to work in a particular sign like P4 that part can be hard coded for
> now. This might help turn others on to the system.
>
> -Tim

Whether I'm willing to work is a key question. ;-) At this stage, I
can see it might be worthwhile to put forth some effort into developing
P6, P10, P12, P16, etc. Not the primes--which you've mentioned are
flat, but those with prime-1 signs corresponding to the prime number
based Galois Fields. Just a thought, and it might be totally wrong.

Also, see if you can convince me or anybody that P1 is a construction
worthy of notice. A clue might lie in GF[2] (or might not.)

Timothy Golden BandTechnology.com

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Jun 11, 2006, 8:39:38 PM6/11/06
to
Spoonfed wrote:
> Timothy Golden BandTechnology.com wrote:
> > Spoonfed wrote:
> > > http://en.wikipedia.org/wiki/Galois_field
> > As I look at the tables at the bottom of that wiki page I don't see
> > much alignment except that GF(3) addition table looks like P3
> > multiplication table. The reliance on primes seems to be basic to
> > Galois Fields. In polysign the prime sign systems are very flat in
> > terms of symmetry of the signs as operators. But there is no
> > restriction about primes.
> >
>
> Look closer. The multiplication table for GF(3) corresponds to P2.
> GF(4) corresponds to P3.

Got it. OK. Nice pattern recognition. But I still see a wide divide
between the constructions. I suppose as a cross-study it is relevant to
compare the two. I can see that treating the signs as discrete
operators sends one toward this domain. But let's not forget that the
polysigned system is a continuum. I suppose one could ponder how to
jump to a continuum construction from a finite construction and I
suppose adjacency would be an appropriate tool. As we begin counting we
then realize that scale can be useful and that there is no need to
limit oneself to unit values, though D-theorists seem to be forgoing
that scalar concept:
http://koti.mbnet.fi/mpelt/tekstit/dtheory.htm
There are some surprising congruencies to the polysigned lattice though
the constructions are motivated differently.
What I have absorbed so far of the Galois construction is that it is
very primitive. More so than the polysigned numbers. It's motivation
and consequences are not very clear to me. Also that zero product that
wipes everything away seems pretty destructive. In polysign math the
notion of a zero sign is just the identity sign, which preserves the
state of the operand.
The Galois sum miffs me. Why would like values sum to zero? If
summation is superpostion shouldn't inverse values sum to zero? The
construction will not yield dimensionality.

The polysigned numbers require cancellation to exist via equal amounts
in each orientation. Symmetry and superposition together form the heart
of the polysigned numbers and generate dimensionality.

> > P4 and all higher even sign systems are not fields since division does
> > not necessarily work. The last requirement of the mathematical field
> > is:
> >
>
> If multiplication in P4 can be represented by matrix multiplication,
> division should be possible through multiplying by the inverse matrix.

I'll try to study this approach. The product is not actually a matrix
multiplication though is it?
I'm not seeing it for the moment. I see you have a matrix in the
nonorthogonal Cartesian system. But will the inverse work there?

>
> > Existence of multiplicative inverses:
> >
> > For every a ≠ 0 belonging to F, there exists an element
> > a^−1 in F,
> > such that
> > a * a^−1 = 1.
> >
> > In P4 if we multiply a value by
> > + 1 # 1
> > we will get another value on that same ray. I call this the identity
> > axis.
>
> I thought that #1 and (-1+1*1) identified the identity axis... But
> then again, any real-valued eigenvalue associated with a real valued
> eigenvector should have this property. I posted a list of these
> somewhere either in this or the other thread.

Let's try an example in P4:
( - 1 + 2 ) ( - 1 + 1 # 1 )
= + 1 * 1 - 1 * 2 # 2 + 2
= - 1 + 3 * 3 # 2
= + 2 * 2 # 1 .
Now using the identity axis:
( - 1 + 2 ) ( + 1 # 1 )
= * 1 - 1 # 2 + 2
= + 1 # 1 .
In general the value won't come to +1 # 1. It might be + 1.5 # 1.5 or
it could be - 3.2 * 3.2 but it will always be on this axis. The name
identity axis may be a bad name choice. The word identical is
appropriate but perhaps it should be called the null axis. Any value
multiplied by a value on this axis will come out on this axis. In P6
the values (1,0,1,0,1,0) and (0,1,0,1,0,1) form the identity axis.
This axis is very much like a real line since one side of it has an
inverting property so that in P4:
( + 1 # 1 ) ( + 1 # 1 ) = + 2 # 2 ,
( - 1 * 1 ) ( - 1 * 1 ) = + 2 # 2 ,
( - 1 * 1 ) ( + 1 # 1 ) = - 2 * 2 ,
just as in P2:
( + 1 )( + 1 ) = + 1 ,
( - 1 )( - 1 ) = + 1 ,
( - 1 )( + 1 ) = - 1 .


> > It exists in every even signed system. In effect what was a zero
> > becomes a zero axis to the field definition. The dimensionality does
> > not allow the strict field definition to apply. This does not trouble
> > me. The polysigned numbers challenge existing mathematics on many
> > levels. It's because existing mathematics is built upon the real
> > numbers. The singular zero of this field requirement becomes an axis in
> > general. I can't change the definition of a field. To quibble over this
> > as a stopping point is not relevant. The polysigned system behaves on
> > its own terms.
> >
>
> Alright, since you brought it up, let's quibble. Perhaps some insight
> might be gained. Have you invented a new math, or have you simply
> found some interesting properties of the old math?
>
> If the polysigned system behaves on its own terms, then we should
> highlight those differences. Nothing to be gained by avoiding the
> stopping points; we should approach them directly and stop at them.
>
> To represent zero as a half dimension doesn't make any sense. You say
> there is some fundamental difference in P2 between +7-5 and +2. To me
> they are simply different representations of the same number.

I would not word it quite that way. It's a matter of the operations
that we perform arithmetically. One of them is optional. It is embodied
by the identity
- x + x = 0 (P2) .

The second to last standard field law is:

Existence of additive inverses
For every a belonging to F, there exists an element −a in F, such
that a + (−a) = 0.

If we forgo this law we can still do any arithmetic to the point of an
expression like +7-5.

When we generalize sign we see this odd thing down at P1:
- x = 0 .
By my argument we can still do arithmetic in P1. It's just that when we
go to perform this evaluation we get zero. P1 is zero dimensional and
these qualities are defining what zero dimensional means. The
correspondence to time is remarkable.

As we quibble over this I am claiming a conflict with the flat
defintion of a field. I am not about to argue with mathematicians that
they are wrong. That would be futile. Mathematics as a free language
says that you can build whatever you want. As long as you follow your
rules the results become a measure of the value of those rules.

If the polysigned numbers could speak for themselves in the court of
fields they would say that the inverse field law should be viewed as an
operation that has a more general form than a strict binary inverse.
They also say that the multiplicative inverse may need to exclude more
than just zero. I'm just happy to look at them and see what they do
without bothering to worry about these problems. Traditional
mathematics is eccentric to the real numbers. How many times have I
seen people try to define the polysigned numbers as built from the real
numbers? A lot. Yet it is the polysigned construction that builds the
real numbers and the complex numbers from the same simple rules. They
develop higher dimension systems with a sum and a product that behave
just as they do for the real numbers. That is all new.

I know that you won't find this convincing. We've already been through
this without the field law references. I guess I'm willing to discuss
it endlessly but it sure would be nice to find a nifty way to convince
you. I've given you arithmetic examples. The graphical representation
says a lot. This math is easy to learn, but is not fully developed and
won't be without a lot of work. Still the sum and the product are
complete and they are the grounds for other operations. I still have
not found a general conjugate. It exists in P3 but I can't find it in
P4 or anywhere higher up. Division becomes a lot of work using linear
equations and I haven't gotten far with it. Maybe your inverse matrix
approach will work.


>
> > You've reminded me of
> > http://groups.google.com/group/sci.math/msg/11ceb7ea329e2a97
> > Roger Beresford's terplex product exactly matches so he must have the
> > same matrix form.
> > It's readily apparent when you look at his unit vector definition.
> > But again without the identity which drops the actual dimensionality by
> > one.
> > We hashed it out on that thread but I have failed to recall it as a
> > resource. It does exist in other usage and sci.math just isn't a very
> > good place to query for the info. He has code too:
> > http://library.wolfram.com/infocenter/MathSource/4894/
> > His stuff deals in orthogonal bases.
> > Augmented with the forward and reverse Cartesian transforms it would be
> > sufficient to do general polysigned math.
>
> Cool.
>
> > This approach would skip over your latest move of SC slanted space. The
> > identity is somewhat optional to the system.
>
> What are you referring to here?

SC being 'special Cartesian' where the last component is forced to
zero.
The approach above would simply do the orthogonal Cartesian transform.
It would use the transform vectors at the bottom of:

http://groups.google.com/group/sci.physics.relativity/msg/e4b45e2de0541d59


>
> > You won't need it except
> > that equivalent values don't necessarily look the same until reduced.
> > The inverse transform will take care of that for you as long as you
> > look at your values in Cartesian. Keep in mind that above P3 the
> > solutions become rotationally variant so what was isotropic becomes
> > anisotropic yet maintains linearity.
>
> i.e. There ARE special directions. Multiplying by #1 maintains the
> shape, but multiplying by *1 inverts everything.

I prefer 'rotates' to 'inverts'. I have yet to prove that the sign unit
vectors are the only preservative factors. But I'm pretty sure that is
true. e.g. in P4 any object multiplied by *1 will be a rotated version
of itself with no morphing but a unit vector in the *1-0.2 direction
will rotate and morph the object according to the P4 product
deformation study.

> > I'm not really comfortable with
> > the word isotropic. It seems to carry a haze around with it.
> >
> > In effect your generic Cartesian product is a function that looks like:
> >
> > Cartesian Product( Cartesian c1, Cartesian c2 )
> > { // this does the polysigned product.
> > // caution: the transform orientation matters.
> > // |AB| = |A||B| breaks above 2D.
> > Cartesian s1( c1.dim + 1 ) = PolySignXform( c1 );
> > Cartesian s2( c2.dim + 1 ) = PolySignXform( c2 );
> > Cartesian s3( c1.dim + 1 ) = TerplexProduct( s1, s2 );
> > Cartesian c3( c1.dim ) = InversePolySignXform( s3 );
> > return c3;
> > }
> >
>
> Hmmm.
>
> > Even without the identity function this will work fine. In effect the
> > identity has been embodied by the transform.
>
>
> Still not quite sure what you mean by the identity.

I just mean the cancellation law:
- x + x * x # x = 0 (P4) .


>
> > The usual vector sum is
> > the same in either space so there is no trouble there. This should be a
> > general solution upon generalizing the transform and if you are willing
> > to work in a particular sign like P4 that part can be hard coded for
> > now. This might help turn others on to the system.
> >
> > -Tim
>
> Whether I'm willing to work is a key question. ;-) At this stage, I
> can see it might be worthwhile to put forth some effort into developing
> P6, P10, P12, P16, etc. Not the primes--which you've mentioned are
> flat, but those with prime-1 signs corresponding to the prime number
> based Galois Fields. Just a thought, and it might be totally wrong.
>
> Also, see if you can convince me or anybody that P1 is a construction
> worthy of notice. A clue might lie in GF[2] (or might not.)

I understand. You've certainly spent plenty of time on this
construction and I've learned some things through that process. If it
weren't for the time correspondence I'd have very little to say about
P1 but you can also use it as a starting point to get the whole deal.
So a naive person says, "Wait a minute. Time only goes one way. The
real numbers go two ways. If I just hack off half of them I'll get
time." Then this naive person goes on to wonder about adding a sign to
the real numbers. The puzzle when solved leads to the polysigned
numbers. They are a trivial solution. But now taking what one has
learned about sign in general and applying it back at P1 takes on a
surprise that is congruent with the paradox of time and allows time to
be placed structurally and symetrically with space. No rod or ruler
will measure time directly as they can measure space. It is not
geometrically possible.

Until a physics model arises that gets even more I accede that the
spacetime congruence may be only coincidental. To take this gamble is a
personal choice. The probability of failure is quite high but the
construction is so suggestive to me that I have no trouble with that.
When I found out that P4 and up were broken that was enough for me. I'd
been looking for it and it came out in such a simple principle of
distance. So now the product is the focus of my study since it is the
product that begets the spacetime claim. We see classical particle
models operating under a product relationship (albeit in a reciprocal
space) and so I try to model the space we are in as a particle product
space rather than a Cartesian product space. That takes me up to where
we started. All along the way there is a lot to be filled in. Other
arguments are coming along also like the relative 2D generic particle
scenario. I am fairly comfortable with the notion of an axis being an
inherent quality of a particle for purely geometrical reasons. In a 2D
(P3) particle product scene one is left with the horrible notion of
these angles adding to each other via the product and everything stops
making sense. Particles from far away are modifying the axis just as
much as those nearby. So something is needed just like the reciprocal
space and I have no idea what it is.
Probably that's too cryptic but that is the top of my stack. Down lower
there are the basic math possibilities that may lead somewhere too. To
argue about field laws will not be productive. The raw math is
straightforward. I am interested in convincing anyone to spend time
studying the construction in the hopes that it will prove useful.
Unfortunately it is difficult for people to understand the simplicity
of it. I believe that this is due to the real numbers being the footing
for their internal model.

-Tim

Hello...@126.com

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Jun 12, 2006, 3:03:48 AM6/12/06
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Gene Ward Smith

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Jun 12, 2006, 4:31:10 AM6/12/06
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Timothy Golden BandTechnology.com wrote:
> I'm unburying this thread in the hopes that you will respond.
> The form:
> s30 = s10 s20 + s11 s22 + s12 s21
> s31 = s11 s20 + s10 s21 + s12 s22
> s32 = s12 s20 + s10 s22 + s11 s21 .
> is extensible to any dimension and provides a commutative product
> definition.

If we define addition and the scalar product in the obvious way, this
defines a commutative, associative, and distributive real algebra. So
you know it can only be either
the direct sum of three real factors, or a real and a complex. It turns
out to be the latter. If you take w to be a root of unity, then the
basis elements are {1,w,w^2} and the elements of the algebra can be
equated with a + bw + cw^2. The "a" part, from the 1, is the real
factor and the {w,w^2} the complex factor in R+C.

Timothy Golden BandTechnology.com

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Jun 12, 2006, 2:19:19 PM6/12/06
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What happens when you raise the dimension?
So instead of a 3x3 you could do a 5x5.
The form is extensible.
It seems like the form in general should have been developed and named
some time in the past. Beresford does use it as something he calls
terplex and applies prolifically. I have used it for polysigned numbers
to define their arithmetic product. In the case of polysigned numbers
the actual dimension is dropped one by the law:
( 1, 1, ... ) = 0 .
This makes the coordinate system follow a simplex geometry. The s
positions are literally signs in this construction and their values are
magnitudes. This allows a product and sum definition in any whole
dimension that is commutative, distributive, and associative.

The general product format codes very simply as a software algorithm.
It is a commutative operator that should have been named generically.

-Tim

Gene Ward Smith

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Jun 12, 2006, 3:09:17 PM6/12/06
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Timothy Golden BandTechnology.com wrote:
> Gene Ward Smith wrote:

> > If we define addition and the scalar product in the obvious way, this
> > defines a commutative, associative, and distributive real algebra. So
> > you know it can only be either
> > the direct sum of three real factors, or a real and a complex. It turns
> > out to be the latter. If you take w to be a root of unity, then the
> > basis elements are {1,w,w^2} and the elements of the algebra can be
> > equated with a + bw + cw^2. The "a" part, from the 1, is the real
> > factor and the {w,w^2} the complex factor in R+C.

> What happens when you raise the dimension?
> So instead of a 3x3 you could do a 5x5.

For any odd prime p, you can take a primitive p-th root of unity and
construct a similar
algebra over the rationals Q which is Q+cyclotomic field. Over the
reals, the cyclotomic field part breaks apart into (p-1)/2 copies of C.

> The form is extensible.
> It seems like the form in general should have been developed and named
> some time in the past.

Why? I don't understand what the use of it is. Mostly, fields are
preferred over commutative algebras which are not fields, because the
latter can be understood in terms of the former. Noncomutative
algebras, such as central simple algebras, are another story with a
very rich theory.

Timothy Golden BandTechnology.com

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Jun 12, 2006, 4:28:51 PM6/12/06
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I'm only partially absorbing what you have said about cyclotomic
fields. I didn't mean to restrict the possibility to primes. For
instance a 4x4 could be considered as well. Anyhow you are saying that
C (the complex numbers) will pop out again almost in a raw form.
However the product will wind up being defined accross these. I have
seen this in P4(the four-signed numbers, the 4x4 product form which
comes down to 3D) yet the product is slightly different than the
orthogonal product of its parts. In effect P4 looks like R x C or P2 x
P3, but I don't quite have it yet. There is a small error between the
two and I think there must be some dimensional mixing. But the
resemblance is certainly there.
There is a parity being expressed in the polysigned family in adjacent
dimensions that may be producing something close to cyclotomic
behavior.
You ask why. One simple reason is that I can generate the complex
numbers and the real numbers from a few simple rules using this
product. I can claim that it is the most compact definition, getting
both in one fell swoop and a step. You say commutative algebras are not
fields but I suppose you mean that they are not necessarily fields.
That they may be fields is still a possibility.
I am aware that quaternions are noncommutative and support your very
rich theory statement. They enforce:
| A B | = | A || B |
whereas P4 breaks this conservation law yet allows distributive,
associative and commutative properties just like real number algebra.
The entire polysigned family is very close to satisfying the field
requirements, but some of the statements don't make sense. Like when
the multiplicative inverse excepts zero the polysigned family would
like to except an axis of solutions in the high even signs. Finding
these multiplicative inverses are challenging and have yet to be
resolved as doable or not doable. So far I only know that they are not
doable on the identity axes.

Do you believe the field laws to be perfect? They seem to be eccentric
to the real numbers.

-Tim

tommy1729

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Jun 12, 2006, 5:06:55 PM6/12/06
to
Timothy might be happy I mentioned his polysigned numbers in another topic ? :-)

further to answer a physics question

where did the antimatter go , when the universe was created ??

simple

the big bang is simply an antimatter-black-hole

and bye absorbing antimatter , the matter got repelled ,and still is ...Hubbles constant

what do you think about my critical torus Anthony ?
post there if ya want

greetz
tommy

Gene Ward Smith

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Jun 12, 2006, 5:29:11 PM6/12/06
to

Timothy Golden BandTechnology.com wrote:

> I'm only partially absorbing what you have said about cyclotomic
> fields. I didn't mean to restrict the possibility to primes. For
> instance a 4x4 could be considered as well. Anyhow you are saying that
> C (the complex numbers) will pop out again almost in a raw form.

There's a structure theorem for real algebras which will always make
this likely. Real algebras can have nilpotent elements, like the "dual
numbers", which are R[e] with
e^2=0. There are also infinite-dimensional algebras, like the algebra
of polynomials in n indeterminates, or "symmetric algebra", R[x1, ...,
xn]. However, if your real algebras are finite-dimensional and have no
nilpotent elements, like the ones you are considering, then they are
isomorphic to n copies of the reals plus m copies of the complex
numbers, though the isomorphism is one you have to find.

> However the product will wind up being defined accross these. I have
> seen this in P4(the four-signed numbers, the 4x4 product form which
> comes down to 3D) yet the product is slightly different than the
> orthogonal product of its parts.

Exactly; you have to find out what corresponds to the square root of
minus one in each of these. For instance, if w is a third root of
unity, it's a root of w^2+w+1=0 and (w-w^2)/sqrt(3) will give a square
root of minus one. That tells you how to make the polysign numbers for
n=3 look like real+complex numbers, with the complex numbers written in
the usual way.

> You ask why. One simple reason is that I can generate the complex
> numbers and the real numbers from a few simple rules using this
> product.

Indeed you can, since the 3-dimensional polysign numbers are
Reals+Complexes.

I can claim that it is the most compact definition, getting
> both in one fell swoop and a step.

That I don't see. You are assuming you already know the reals to define
the polysign numbers. You'd have to do something like define them over
the rationals and then define how to complete them.

You say commutative algebras are not
> fields but I suppose you mean that they are not necessarily fields.
> That they may be fields is still a possibility.

Yes, but the only real algebras which are fields are the real and
complex numbers.

> The entire polysigned family is very close to satisfying the field
> requirements, but some of the statements don't make sense.

Yes, they satisfy the commutative associative and distributative laws
for both addition and multiplication, and have no nilpotent elements.


Like when
> the multiplicative inverse excepts zero the polysigned family would
> like to except an axis of solutions in the high even signs. Finding
> these multiplicative inverses are challenging and have yet to be
> resolved as doable or not doable.

I'm afraid the inverse does not always exist; the trouble is you have
zero divisors, which are two nonzero numbers which when multiplied are
zero. An example with the 3-polysign numbers is [1,1,1] x [-2,1,1] =
[0,0,0].

> Do you believe the field laws to be perfect? They seem to be eccentric
> to the real numbers.

I have no idea what that means.

Gene Ward Smith

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Jun 12, 2006, 6:08:53 PM6/12/06
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Gene Ward Smith wrote:

> I'm afraid the inverse does not always exist; the trouble is you have
> zero divisors, which are two nonzero numbers which when multiplied are
> zero. An example with the 3-polysign numbers is [1,1,1] x [-2,1,1] =
> [0,0,0].

I should add that if we set e1 = [1,1,1]/3 and e2=[2,-1,-1]/3, then not
only do we have
e1 x e2 = 0, we have e1^2 = e1 and e2^2 = e2. Therefore, e1 and e2 are
what are called orthogonal idempotents, and can be used to split the 3D
polysign numbers into their real and complex parts:

real part = e1 x number
complex part = e2 x number

Gene Ward Smith

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Jun 12, 2006, 6:16:41 PM6/12/06
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Gene Ward Smith wrote:

I should also add that if you want to pursue this topic in the
mathematical literature, look under "group algebra". The traditional
name being the "cyclic group algebra" R[Cn], where Cn is the cyclic
group of order n.

Timothy Golden BandTechnology.com

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Jun 13, 2006, 7:21:06 AM6/13/06
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No. Well, maybe this is true but what I meant is that P2 are the real
numbers and P3 are the complex numbers. The only difference between the
two constructions is a natural number n.
This comes from the product rule this thread discusses and the
cancellation law. Those are the two rules. Superposition may also need
to be defined. This construction does assume magnitudes (scalars) to
exist so may be construed by traditional mathematics as relying upon
the real numbers. I disagree with this thinking. Magnitude is
fundamental. In this regard the real numbers are built by allowing two
signs to be combined with the scalar concept. Increasing to three signs
generate the complex numbers. The rules are the same for both systems.
They can be extended upward and downward. The family is called the
polysigned numbers. So sign becomes a general discrete entity attached
to scalars. Dimensionality falls out directly. The geometry is unusual.
It is nonorthogonal and perfectly symmetrical. When one considers a
lattice structure stepping around on the lattice is a unidirectional
process. The option to travel back along a segment does not exist above
P2. Instead a loop must be formed if one wishes to return to the same
location. I mention it to help distinguish the geometry from the usual
2D space. When the system is continuous this phenomenon vanishes.


>
> I can claim that it is the most compact definition, getting
> > both in one fell swoop and a step.
>
> That I don't see. You are assuming you already know the reals to define
> the polysign numbers. You'd have to do something like define them over
> the rationals and then define how to complete them.
>
> You say commutative algebras are not
> > fields but I suppose you mean that they are not necessarily fields.
> > That they may be fields is still a possibility.
>
> Yes, but the only real algebras which are fields are the real and
> complex numbers.
>
> > The entire polysigned family is very close to satisfying the field
> > requirements, but some of the statements don't make sense.
>
> Yes, they satisfy the commutative associative and distributative laws
> for both addition and multiplication, and have no nilpotent elements.
> Like when
> > the multiplicative inverse excepts zero the polysigned family would
> > like to except an axis of solutions in the high even signs. Finding
> > these multiplicative inverses are challenging and have yet to be
> > resolved as doable or not doable.
>
> I'm afraid the inverse does not always exist; the trouble is you have
> zero divisors, which are two nonzero numbers which when multiplied are
> zero. An example with the 3-polysign numbers is [1,1,1] x [-2,1,1] =
> [0,0,0].

You've got some misunderstanding here. The three-signed numbers are the
complex numbers so will work. [1,1,1] is zero by definition. So you are
using zero which is excluded from the inverse realtionship. n-signed
numbers are n-1 dimensional due to this relation. Each component of a
number is a scalar. So up in 3D we have four-signed numbers which will
suffer this problem:
( - 2 + 2 ) ( + 3 # 3 ) = 0 .
Neither of these values is zero. However, one of them does lie on the
identity axis.

The polysigned numbers rely upon equal magnitudes in every sign
yielding zero:
P2: - x + x = 0 (the real numbers)
P3: - x + x * x = 0 (the complex numbers)
P4: - x + x * x # x = 0 (start of anisotropic members)


>
> > Do you believe the field laws to be perfect? They seem to be eccentric
> > to the real numbers.
>
> I have no idea what that means.

Math as a religion. The book has been written. One must preserve the
book.
There are few constructions that can go below the existing foundation.
I know it is asking alot for you to go there. Must a scalar be built
from the real numbers? Which is more fundamental? Certainly the scalar
is.

-Tim

Timothy Golden BandTechnology.com

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Jun 13, 2006, 7:23:04 AM6/13/06
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tommy1729 wrote:
> Timothy might be happy I mentioned his polysigned numbers in another topic ? :-)
Certainly.
I'll post there.
-Tim

Spoonfed

unread,
Jun 14, 2006, 10:59:16 PM6/14/06
to
Timothy Golden BandTechnology.com wrote:
> Spoonfed wrote:
> > Timothy Golden BandTechnology.com wrote:
> > > Spoonfed wrote:
> > > > http://en.wikipedia.org/wiki/Galois_field
> > > As I look at the tables at the bottom of that wiki page I don't see
> > > much alignment except that GF(3) addition table looks like P3
> > > multiplication table. The reliance on primes seems to be basic to
> > > Galois Fields. In polysign the prime sign systems are very flat in
> > > terms of symmetry of the signs as operators. But there is no
> > > restriction about primes.
> > >
> >
> > Look closer. The multiplication table for GF(3) corresponds to P2.
> > GF(4) corresponds to P3.
>
> Got it. OK. Nice pattern recognition.

I was lucky enough to have to do a few dozen multiplication and
addition problems in GF(4) for homework, so it didn't take too long to
see the similarity.

> But I still see a wide divide
> between the constructions.

One being finite fields where only 0, -1, +1, *1, exists and the other
being a field containing -x, +y, *z, and all combinations of the three
where x, y, and z are nonnegative real numbers. It is a wide divide.

> I suppose as a cross-study it is relevant to
> compare the two. I can see that treating the signs as discrete
> operators sends one toward this domain. But let's not forget that the
> polysigned system is a continuum. I suppose one could ponder how to
> jump to a continuum construction from a finite construction and I
> suppose adjacency would be an appropriate tool.
> As we begin counting we
> then realize that scale can be useful and that there is no need to
> limit oneself to unit values, though D-theorists seem to be forgoing
> that scalar concept:
> http://koti.mbnet.fi/mpelt/tekstit/dtheory.htm

Interesting, but most of this goes off in a wild angle from my most
successful attempts to understand what little I know of reality.

> There are some surprising congruencies to the polysigned lattice though
> the constructions are motivated differently.
> What I have absorbed so far of the Galois construction is that it is
> very primitive. More so than the polysigned numbers. It's motivation
> and consequences are not very clear to me. Also that zero product that
> wipes everything away seems pretty destructive. In polysign math the
> notion of a zero sign is just the identity sign, which preserves the
> state of the operand.

The textbook I'm reading is called "Fundamentals of Error Correcting
Codes," and I realized I'd better look a little deeper. Found in
chapter 3.3.2, it says "Theorem 3.3.2 The elements of (a base q field)
are precisely the roots of x^q - x."

> The Galois sum miffs me. Why would like values sum to zero? If
> summation is superpostion shouldn't inverse values sum to zero? The
> construction will not yield dimensionality.

I'd been trying to picture some sort of modular arithmetic with vectors
inside a unit circle, where a point on one side of the circle maps to a
point on the opposite side. I just sketched it, and it seems to work
for every sum. I'll bet it works this way with every one of the Galois
Fields. Which brings us right back to where we started, breaking free
of a two-dimensional circle and out into arbitrary-dimensional space.

>
> The polysigned numbers require cancellation to exist via equal amounts
> in each orientation. Symmetry and superposition together form the heart
> of the polysigned numbers and generate dimensionality.
>
> > > P4 and all higher even sign systems are not fields since division does
> > > not necessarily work. The last requirement of the mathematical field
> > > is:
> > >
> >
> > If multiplication in P4 can be represented by matrix multiplication,
> > division should be possible through multiplying by the inverse matrix.
>
> I'll try to study this approach. The product is not actually a matrix
> multiplication though is it?
> I'm not seeing it for the moment. I see you have a matrix in the
> nonorthogonal Cartesian system. But will the inverse work there?
>

Oh my. Remind me when I have time. I think so. We found a way to
represent multiplication by any given Polysign as a matrix. I'm pretty
sure it was an invertible matrix. And I'm pretty sure that each matrix
can be mapped back to a unique (or as unique as possible--considering
that by your concept, different representations of the same polysigned
number actually represent different numbers...) polysigned number.

All matrices with nonzero determinants have inverses. However, the
identity associated with the multiplicative inverse (the way I
calculated them) in this case will be #1 in P4, or *1 in P3, etc. You
define the identity to be a different axis below.

ASIDE: I think the null axis would be a good name for the points lying
along what I consider to be equal coordinates, such as -3+0*0#0 =
-2+1*1#1 = -1+2*2#2

> Any value
> multiplied by a value on this axis will come out on this axis.

OH!!!! I'm starting to get a feel for what you mean now. That's
called an EIGENVECTOR! Well, okay, an eigenvector is a vector whose
direction is not changed when it is multiplied by a quantity called the
eigenvalue which is sometimes real, and sometimes complex. In this
case, you are looking for something that may not be quite the same as
an eigenvector or eigenvalue, (especially since both vector and value
will have the same form) but it's really similar.

I think in P3, (*1)(*1) = *1,
and in P4 (#1)(#1)=#1,
so these would have a similar property

I recommend putting this in a prominent location on your website. Kind
of reminds me of getting rid of the parallel lines postulate for
nonEuclidian geometry.

> As we quibble over this I am claiming a conflict with the flat
> defintion of a field. I am not about to argue with mathematicians that
> they are wrong. That would be futile. Mathematics as a free language
> says that you can build whatever you want. As long as you follow your
> rules the results become a measure of the value of those rules.
>

Well, I read a little bit of the chapter of my textbook on finite
fields, but it has been taking me about two or three readings before it
begins to make sense to me. From the feel of it, it seems like the
Galois field comes more from a set of precisely defined terms and
solutions to polynomial equations.

> If the polysigned numbers could speak for themselves in the court of
> fields they would say that the inverse field law should be viewed as an
> operation that has a more general form than a strict binary inverse.
> They also say that the multiplicative inverse may need to exclude more
> than just zero. I'm just happy to look at them and see what they do
> without bothering to worry about these problems. Traditional
> mathematics is eccentric to the real numbers. How many times have I
> seen people try to define the polysigned numbers as built from the real
> numbers? A lot. Yet it is the polysigned construction that builds the
> real numbers and the complex numbers from the same simple rules. They
> develop higher dimension systems with a sum and a product that behave
> just as they do for the real numbers. That is all new.
>
> I know that you won't find this convincing. We've already been through
> this without the field law references. I guess I'm willing to discuss
> it endlessly but it sure would be nice to find a nifty way to convince
> you. I've given you arithmetic examples. The graphical representation
> says a lot. This math is easy to learn, but is not fully developed and
> won't be without a lot of work. Still the sum and the product are
> complete and they are the grounds for other operations. I still have
> not found a general conjugate. It exists in P3 but I can't find it in
> P4 or anywhere higher up. Division becomes a lot of work using linear
> equations and I haven't gotten far with it. Maybe your inverse matrix
> approach will work.
>

Well, I'm certainly intrigued by the idea, and it will be running
around in my head if I see something else to link to it.

I think some of them might rotate, while others invert. I noticed that
in your sphere animation the swirls get flat and then expand again. I
assume that is going from positive to 0 to negative, thus inverting.
But it also rotated.

Ah, good. I can respect that. Declare your intention to look for
further links in this direction. Some people say that a scientist
should never do this. But that's silly. I remember when I was in Jr.
High and with every problem I hoped against hope that the answer would
come out to be positive instead of negative. It didn't keep me from
getting the right answer, whichever way it turned out to be.

> When I found out that P4 and up were broken that was enough for me. I'd
> been looking for it and it came out in such a simple principle of
> distance. So now the product is the focus of my study since it is the
> product that begets the spacetime claim. We see classical particle
> models operating under a product relationship (albeit in a reciprocal
> space) and so I try to model the space we are in as a particle product
> space rather than a Cartesian product space.

I know that the word Hermitian comes up a lot in discussion of GF[4]
and in quantum mechanics. Maybe I'll eventually see how this all links
together. If I remember right it's a variation on the theme of complex
conjugate. Of course P4 is analogous to GF[5] and I haven't seen much
of that.

> That takes me up to where
> we started. All along the way there is a lot to be filled in. Other
> arguments are coming along also like the relative 2D generic particle
> scenario. I am fairly comfortable with the notion of an axis being an
> inherent quality of a particle for purely geometrical reasons. In a 2D
> (P3) particle product scene one is left with the horrible notion of
> these angles adding to each other via the product and everything stops
> making sense. Particles from far away are modifying the axis just as
> much as those nearby. So something is needed just like the reciprocal
> space and I have no idea what it is.

Reminds me vaguely of discussions of chromodynamic forces. Hmmm, as a
matter of fact it reminds me a LOT of discussion of chromodynamic
forces. Red, Green and Blue quarks, attracted or repelled by a force
which is constant over distance. Green and Blue are equivalent to
anti-red, Green and Red are equivalent to anti-blue, etc.

Okay, that's my trademark pattern match for today.

Spoonfed

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Jun 14, 2006, 11:38:33 PM6/14/06
to

I know it seems important to Tim to have [1,1,1] not equal to zero, and
not allow negative values into his setup. But I think Timothy has come
up with a field which is neither a set of complex numbers nor a set of
real numbers.

But at the very least, Tim has put together a well defined addition,
additive identity, multiplication and multiplicative identity for for
an arbitrary length set of real numbers, i.e. P2 is a group in {R}, P3
is a group in {R,R}, P4 is a group in {R,R,R}, etc. where R is the
reals.

Up until now, the only vector multiplication I knew of that could be
done in arbitrary dimension was the dot product. I don't know whether
this is new to the world, but it is new to me.

Timothy Golden BandTechnology.com

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Jun 15, 2006, 5:08:14 PM6/15/06
to

Hi Jon. Nice to hear from you. This has been one of the most productive
threads to date for me. I get very little work done so communicating
here is an important part of my process. I appreciate that you have
been willing to explore this system. Most people can't do it. The need
to cast judgement and the consequences of that judgement are too much.
If they judge me right or wrong they themselves might be right or wrong
and so the probability of getting success drops to 25%. Thus far no
refutation has denied any of the basics. Just the same there has been
little verification. You've independently demonstrated that the P4
product does not conserve magnitude so that is a nice step. My computer
tells me that it is true and I can verify the math on a piece of paper
to a point but to bother with all of the forward and inverse
transformation is a lot of work. I have been introduced to the term
'cyclotomic' but it seems again to be about finite fields. And the
notion that R and C will keep popping out of the higher spaces does
seem reasonable. So the notion of translating the P4 product over to
( R x C )
is still a going concept. The catch is that the product definition is a
mystery in that space. If that matches some existing mathematics then
the tie could be a nice bonus, but either way the P4 product is the P4
product on its own terms.
I'll try to look into the quark interaction. I made a little mental
step on this angle problem recently. remember that the product is a
force relationship so the result is an acceleration. The same concept
in angle land is actually what is inferred. So the angular product
result would hopefully just be a differential adjustment to the angle.
There is still plenty of haze there but the horror may be a mental
malfunction. The distance notion still is independent of it when the
product is taken literally, so I am still excited about your quark
color statement.

-Tim

Gene Ward Smith

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Jun 15, 2006, 8:04:26 PM6/15/06
to

Timothy Golden BandTechnology.com wrote:
> Most people can't do it. The need
> to cast judgement and the consequences of that judgement are too much.

Here's a free clue: on average, mathematicians probably understand
mathematics better than you do, so if this is directed at the sci.math
readership I think its a misguided missile.

So the notion of translating the P4 product over to
> ( R x C )
> is still a going concept. The catch is that the product definition is a
> mystery in that space.

I've already explained exactly how do to it. There is no catch. It's a
done deal.

If that matches some existing mathematics then
> the tie could be a nice bonus, but either way the P4 product is the P4
> product on its own terms.

I've already explained exactly how this ties into existing mathematics;
if you are really interested you can find out.

Spoonfed

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Jun 15, 2006, 10:22:43 PM6/15/06
to

...or they may have something else at the top of their stack.

> The need
> to cast judgement and the consequences of that judgement are too much.
> If they judge me right or wrong they themselves might be right or wrong
> and so the probability of getting success drops to 25%. Thus far no
> refutation has denied any of the basics. Just the same there has been
> little verification. You've independently demonstrated that the P4
> product does not conserve magnitude so that is a nice step.
> My computer
> tells me that it is true and I can verify the math on a piece of paper
> to a point but to bother with all of the forward and inverse
> transformation is a lot of work. I have been introduced to the term
> 'cyclotomic' but it seems again to be about finite fields.

>From what I can figure, I would say that
*real numbers is to binary
as
*(reduced) polysigned is to Galois Fields.

As for the P4 product not conserving magnitude, I would encourage you
look at your own animation again, and check how multiplying sets of
points on the unit sphere by vectors on the unit sphere yield
quantities inside the unit sphere (and thereby the product has smaller
magnitude than its factors.) Judging strictly from that animation, it
seems to me that the magnitude of the product will always be less than
or equal to the product of the magnitudes. Also the vectors where the
sphere maintains its original shape and the vectors where the sphere is
flat as a pancake are probably worthy of identifying and naming.

> And the
> notion that R and C will keep popping out of the higher spaces does
> seem reasonable.

Now analyzing my own silly motivations for pursuing this whereas nobody
else does. I hate C. Historically there was a pretty big debate about
whether imaginary and complex numbers should be admitted into formal
mathematics. But imaginary numbers had too many good applications. So
even if we have something *completely* equivalent to the set of complex
numbers, and adding absolutely nothing to any branch of physics or
mathematics, I'd still be happy just to know there was a less ephemeral
explanation tha complex numbers.

> So the notion of translating the P4 product over to
> ( R x C )
> is still a going concept. The catch is that the product definition is a
> mystery in that space. If that matches some existing mathematics then
> the tie could be a nice bonus, but either way the P4 product is the P4
> product on its own terms.

I've got a philosophy that before you can come up with a good answer,
you've got to have a good question. First of all you should consider
whether (RXC) is a useful construction or even a usable construction.
Secondly, if you've already found a one-to-one and onto mapping into R
X R X R, will you benefit from finding a mapping into R X C? And
finally, if you think you will benefit, can you take the mapping from R
X R X R into R X C?

> I'll try to look into the quark interaction. I made a little mental
> step on this angle problem recently. remember that the product is a
> force relationship so the result is an acceleration. The same concept
> in angle land is actually what is inferred. So the angular product
> result would hopefully just be a differential adjustment to the angle.
> There is still plenty of haze there but the horror may be a mental
> malfunction. The distance notion still is independent of it when the
> product is taken literally, so I am still excited about your quark
> color statement.
>
> -Tim

The wind is rushing past my ears. You need to find a good college
library, and find a book aimed at somewhere around your level. I still
think what your saying vaguely resembles quantum chromodynamics in that
they both involve three values and their inverses, and also they
resemble each other in that both are concealed by a black haze of
complete incomprehensibility.

Timothy Golden BandTechnology.com

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Jun 16, 2006, 8:27:31 AM6/16/06
to

Right. I've done some of this and I guess I'll publish some graphics
that explore the puzzle.
The real axis is the identity axis or the null axis as we are possibly
renaming it.
The complex plane is exhibited as normal to this line at the origin and
by using a squaring technique I get:

const Poly & P4SecondaryAxis()
{ // this is the proposed P3* embedded in P4 found in CheckP3P2Relation
static Poly s(4);
s[0] = 1.5;
s[1] = 0.75;
s[3]= 0.75;
s.Unitize();
return s;
}

where s[0] is the identity sign and s[1] is the - sign, etc.
This vector is used to build the transforms that allow comparison of
the product in one space versus the other. The graphics are not enough
since they look so much alike.

I'm going to add a meridian to the shell algorithm and see what that
exposes. The spiral shape may be getting a twist (P4) along the real
axis (T3) that is hidden. I'll publish the comparisons so you can see
for yourself the coherent difference. It's like the T3 and P4 are twins
with a mini-me between them. Geeze wouldn't that be something if an
iterative product cancelled it out.

What you say about getting some hard max and min type of values is
true. I once did some of this for the P4 cone but this is more general.
Also the unitary values are in the same class of analysis so all of
that will fall out.

As to the QCD below here I urge you to consider two generic points in a
2D space. There is no need to use the polysign domain. Invoking
relativity on these two generic points will yield the features. That is
to say that by granting each point an independent referential
coordinate system additional qualities are exhibited by generic points
that are consistent with known point particles. No product operation is
needed to get this. Not much at all is needed to do the construction
but a definition of 2D. Relativity does the rest. Each point particle
is its own origin. Each point particle has its own referential zero
angle or unit ray or whatever you want to call it. The information
exhibited jumps to 3D since each particle exhibits independence in this
realm. So a two body problem that is traditionally a one dimensional
solution due to the unitary distance is much more when the
dimensionality is interpreted this way. The congruence to spin axes is
obvious.
Doing the same in one dimension yields a binary that is suggestive of
charge. There is also at least one more bit up in 2D if one allows left
and right handedness. In effect the 1D bit is also handedness. It could
also be construed as a 2-bit solution. Disambiguating the asymmetry of
charge is a puzzle. One would hope that the system would yield a heavy
proton and a light electron but instead these charges look
indistinguishable so I don't think this is a convincing model yet. P1
can only be one-handed so may allow some trickery with its vanishing
act.
Again for me the ultimate topology of spacetime is:
0D + 1D + 2D ... or P1 + P2 + P3 ...
This is the substrate that we (or our particles) are products of under
this model.
The breakpoint at P3 is natural and what is exhibitied on the other
side of it is bizarre.
Whether the bizarre is eliminated or not does not have to be dealt with
yet.
This is really a classical model taken to a new level.

-Tim

> > And the
> > notion that R and C will keep popping out of the higher spaces does
> > seem reasonable.
>
> Now analyzing my own silly motivations for pursuing this whereas nobody
> else does. I hate C. Historically there was a pretty big debate about
> whether imaginary and complex numbers should be admitted into formal
> mathematics. But imaginary numbers had too many good applications. So
> even if we have something *completely* equivalent to the set of complex
> numbers, and adding absolutely nothing to any branch of physics or
> mathematics, I'd still be happy just to know there was a less ephemeral
> explanation tha complex numbers.
>
> > So the notion of translating the P4 product over to
> > ( R x C )
> > is still a going concept. The catch is that the product definition is a
> > mystery in that space. If that matches some existing mathematics then
> > the tie could be a nice bonus, but either way the P4 product is the P4
> > product on its own terms.
>
> I've got a philosophy that before you can come up with a good answer,
> you've got to have a good question. First of all you should consider
> whether (RXC) is a useful construction or even a usable construction.
> Secondly, if you've already found a one-to-one and onto mapping into R
> X R X R, will you benefit from finding a mapping into R X C? And
> finally, if you think you will benefit, can you take the mapping from R
> X R X R into R X C?

Good point. I'm not even sure what the motivation of all of this is
other than translating the polysign language to an existent language.
Either way the polysign language can speak in its own terms. That
another language might accelerate the development is great. Perhaps
that is what Gene is doing, but I'm not convinced that this other
language is as fundamental as the polysign. It's like he's got a swiss
army knife and I've got a chisel.

-Tim

Timothy Golden BandTechnology.com

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Jun 16, 2006, 8:44:40 AM6/16/06
to

Timothy Golden BandTechnology.com wrote:
> As to the QCD below here I urge you to consider two generic points in a
> 2D space. There is no need to use the polysign domain. Invoking
> relativity on these two generic points will yield the features. That is
> to say that by granting each point an independent referential
> coordinate system additional qualities are exhibited by generic points
> that are consistent with known point particles. No product operation is
> needed to get this. Not much at all is needed to do the construction
> but a definition of 2D. Relativity does the rest. Each point particle
> is its own origin. Each point particle has its own referential zero
> angle or unit ray or whatever you want to call it. The information
> exhibited jumps to 3D since each particle exhibits independence in this
> realm. So a two body problem that is traditionally a one dimensional
> solution due to the unitary distance is much more when the
> dimensionality is interpreted this way. The congruence to spin axes is
> obvious.
> Doing the same in one dimension yields a binary that is suggestive of
> charge. There is also at least one more bit up in 2D if one allows left
> and right handedness. In effect the 1D bit is also handedness. It could
> also be construed as a 2-bit solution. Disambiguating the asymmetry of
> charge is a puzzle. One would hope that the system would yield a heavy
> proton and a light electron but instead these charges look
> indistinguishable so I don't think this is a convincing model yet. P1
Well, under a product relationship they don't look identical do they?
(-)(-) = + and (+)(+) = + .
Can the two resultant plus's be distinguished? No, but they came from
different things.
Well the - - is two and the + + is four in non-mod form.
That sort of suggests stepping up to a four-signed system where the -
sign doesn't exist.
Abstract but perhaps worth pondering.
That a product results are in the same space as the original is an
arithmetic concept. The Force product is an entirely different
resultant, just as meters times meters is square meters, but even more
extreme according to the classical force equations.
-Tim

Timothy Golden BandTechnology.com

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Jun 16, 2006, 1:10:16 PM6/16/06
to
I've put a page on my website demonstrating the discrepancy in the flat
product of RxC compared to P4:
http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html
There is a fair amount of transformation to get the comparison. I have
verified that the forward and reverse yield the identity so there is no
conflict there. Just the same I cannot guarantee there to be no math
error.

How's this for a start of the discussion:
The flat RxC product is orthogonal and since its component parts
conserve magnitude in their products so will the result. Now we have a
conflict between what the graph shows and this statement. All of the
sources for the product are unit length vectors, yet they yield non
unit-length vectors. Just as a starting point it may be useful.
Resolving this conflict should help to understand what is the nature of
the R x C product and its look alike the P4 product and the mini-me
difference that they generate.

For more background on these graphs please see:

http://bandtechnology.com/PolySigned/Deformation/DeformationUnitSphereP4.html
and
http://bandtechnology.com/PolySigned/FourSigned.html
The unit shell representation is a survey of the P4 product in its
entirety.
I'll try to get a meridian into the shell algorithm so that any
weirdness there will be exposed.

Spoonfed

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Jun 18, 2006, 5:10:04 PM6/18/06
to

This isn't at all clear to me. It looks like you are defining a
specific Polysigned number to be the secondary Axis. I'm not familiar
enough with C++ to know what represents inputs and outputs of this
code.

I'm not familiar enough with C++ code to

>
> where s[0] is the identity sign and s[1] is the - sign, etc.
> This vector is used to build the transforms that allow comparison of
> the product in one space versus the other. The graphics are not enough
> since they look so much alike.
>

Are you saying the product in P4 is different in some other system?
You would need to define that system because as far as I know there is
no such product in any other system.

> I'm going to add a meridian to the shell algorithm and see what that
> exposes. The spiral shape may be getting a twist (P4) along the real
> axis (T3) that is hidden. I'll publish the comparisons so you can see
> for yourself the coherent difference. It's like the T3 and P4 are twins
> with a mini-me between them. Geeze wouldn't that be something if an
> iterative product cancelled it out.
>

Don't know what meridian means. Have never seen "T3" before.

> What you say about getting some hard max and min type of values is
> true. I once did some of this for the P4 cone but this is more general.
> Also the unitary values are in the same class of analysis so all of
> that will fall out.
>

To the best of my knowledge, that's one of the reasons why eigenvectors
and eigenvalues are of interest.

> As to the QCD below here I urge you to consider two generic points in a
> 2D space. There is no need to use the polysign domain. Invoking
> relativity on these two generic points will yield the features. That is
> to say that by granting each point an independent referential
> coordinate system additional qualities are exhibited by generic points
> that are consistent with known point particles.

In relativity the word "point" is ambiguous and should be avoided. If
you would like a description of relativity theory, I can try to help
you with that. You can speak of objects and their paths, and you can
talk about events and collisions, but "point" presumes a reference
frame.

> No product operation is needed to get this.
> Not much at all is needed to do the construction
> but a definition of 2D. Relativity does the rest. Each point particle
> is its own origin. Each point particle has its own referential zero
> angle or unit ray or whatever you want to call it. The information
> exhibited jumps to 3D since each particle exhibits independence in this
> realm. So a two body problem that is traditionally a one dimensional
> solution due to the unitary distance is much more when the
> dimensionality is interpreted this way. The congruence to spin axes is
> obvious.

I don't know what problem you're trying to solve.

> Doing the same in one dimension yields a binary that is suggestive of
> charge. There is also at least one more bit up in 2D if one allows left
> and right handedness. In effect the 1D bit is also handedness. It could
> also be construed as a 2-bit solution. Disambiguating the asymmetry of
> charge is a puzzle. One would hope that the system would yield a heavy
> proton and a light electron but instead these charges look
> indistinguishable so I don't think this is a convincing model yet. P1
> can only be one-handed so may allow some trickery with its vanishing
> act.

Slow down.

> Again for me the ultimate topology of spacetime is:
> 0D + 1D + 2D ... or P1 + P2 + P3 ...
> This is the substrate that we (or our particles) are products of under
> this model.
> The breakpoint at P3 is natural and what is exhibitied on the other
> side of it is bizarre.
> Whether the bizarre is eliminated or not does not have to be dealt with
> yet.
> This is really a classical model taken to a new level.
>

Take 20 steps back. Ask simple questions. Give simple answers.

When people with swiss army-knives come along don't point your chisel
at them.

Gene Ward Smith

unread,
Jun 18, 2006, 6:09:24 PM6/18/06
to

Timothy Golden BandTechnology.com wrote:
> I've put a page on my website demonstrating the discrepancy in the flat
> product of RxC compared to P4:
> http://bandtechnology.com/PolySigned/Deformation/P4T3Comparison.html

I'm afraid this web page crashed Firefox. Can you simply give an
algebraic expression for whatever product it is you want to talk about?

Timothy Golden BandTechnology.com

unread,
Jun 18, 2006, 9:14:26 PM6/18/06
to
Spoonfed wrote:
> >
> > const Poly & P4SecondaryAxis()
> > { // this is the proposed P3* embedded in P4 found in CheckP3P2Relation
> > static Poly s(4);
> > s[0] = 1.5;
> > s[1] = 0.75;
> > s[3]= 0.75;
> > s.Unitize();
> > return s;
> > }
>
> This isn't at all clear to me. It looks like you are defining a
> specific Polysigned number to be the secondary Axis. I'm not familiar
> enough with C++ to know what represents inputs and outputs of this
> code.

Sorry. I'll take us through the whole thing again here but focus on
this vector. We want to compare the product in P4 versus T3, which is R
x C in polysign. Well, its realy P1xP2xP3 but the P1 part doesn't do
anything. So just think of T3 as P2 x P3 which is R x C. In order to
perform a comparison of these products we have to find the transform
between P4 and T3. We can make an educated guess that the null axis(or
identity axis +1#1) is P2. We know that P3 is orthogonal to P2, but we
don't know how it is oriented. We can rotate through a full revolution
and also change its handedness. Having made the assumption that there
is a P3 we can take any vector in it (but over in P4) and square it:
s2 = s1 s1 .
We have chosen s1 to be an arbitrary vector perpendicular to the null
axis ( +1#1 )
Since we are assuming a complex plane we know that the angle of s2 has
doubled from the +real or *(P3) axis. We simply reflect s2 back from s1
and this vector will be the * axis of the embedded P3 plane. So now we
have defined the plane and it's positive real axis ( or * axis in P3 )
is:
# 1.5 - 0.75 * 0.75
Converted to a unit vector this value becomes a means of transforming
from P4 to T3 along with
# 1 + 1
which is the embedded P2 orientation.
I could print out the actual vectors of P4ToT3() and T3toP4() but I'm
not sure that will be helpful. Let me know if you want them. Anyhow
these two routines embody what has been described and allow the
transform. So now when we multiply a point by another point in P4 we
can also transform to T3 and perform the multiplication in the T3
domain. At this point we can take the results and look in either
system. In this case we are looking in T3, so we take the P4 result and
transform it over to T3. Now, if these two values are equal then the
systems are in agreement. But they are not as evidenced by the
graphics, which take this differencing concept and apply it to the unit
shell. Strangely the difference of the two is another smaller image of
the original product of either space. If you look at the sizes of the
P4 and the T3 you can see that the P4 is larger. Scaling the result
however will not eliminate the error. I can set a scale that will set
the errro to zero for a specific point product, but the scale is not
universal. So the error is still exhibited even after scaling.


> >
> > where s[0] is the identity sign and s[1] is the - sign, etc.
> > This vector is used to build the transforms that allow comparison of
> > the product in one space versus the other. The graphics are not enough
> > since they look so much alike.
> >
>
> Are you saying the product in P4 is different in some other system?
> You would need to define that system because as far as I know there is
> no such product in any other system.

Right, but it looks an awful lot like the R x C independent product.
And Gene has reinforced this prediction so I guess that is why we are
here. His R x C and repetitive claim for higher dimensions is of
interest. You can see that the results are very similar. So no doubt
there is some connection, though it may not be a straightforward one.
Still, I think you are right. As I look around for R x C I find very
little. Supposedly there are methods to make the two come in line. We
have discussed this product comparison back a while ago. I don't know
how important it is to force these things to mesh. Probably it's an
interesting problem.


>
> > I'm going to add a meridian to the shell algorithm and see what that
> > exposes. The spiral shape may be getting a twist (P4) along the real
> > axis (T3) that is hidden. I'll publish the comparisons so you can see
> > for yourself the coherent difference. It's like the T3 and P4 are twins
> > with a mini-me between them. Geeze wouldn't that be something if an
> > iterative product cancelled it out.
> >
>
> Don't know what meridian means. Have never seen "T3" before.
>
> > What you say about getting some hard max and min type of values is
> > true. I once did some of this for the P4 cone but this is more general.
> > Also the unitary values are in the same class of analysis so all of
> > that will fall out.
> >
>
> To the best of my knowledge, that's one of the reasons why eigenvectors
> and eigenvalues are of interest.
>
> > As to the QCD below here I urge you to consider two generic points in a
> > 2D space. There is no need to use the polysign domain. Invoking
> > relativity on these two generic points will yield the features. That is
> > to say that by granting each point an independent referential
> > coordinate system additional qualities are exhibited by generic points
> > that are consistent with known point particles.
>
> In relativity the word "point" is ambiguous and should be avoided. If
> you would like a description of relativity theory, I can try to help
> you with that. You can speak of objects and their paths, and you can
> talk about events and collisions, but "point" presumes a reference
> frame.

That's exactly the point. Take what you have just said literally. The
point particles inherently have an angular measure to the other
particle. There is a conflict here and the angular momentum of point
particles meshes with this angular characteristic. There is no direct
proof. It is anecdotal evidence. Whether you personally reject this
notion is fine. I like it. To me it is a bridge between relativity and
quantum and classical through dimensionality. I predict that the 2D
(P3) product will yield a model of spin. Anyhow I am open to being
wrong but the path is there to explore.

> > Again for me the ultimate topology of spacetime is:
> > 0D + 1D + 2D ... or P1 + P2 + P3 ...
> > This is the substrate that we (or our particles) are products of under
> > this model.
> > The breakpoint at P3 is natural and what is exhibitied on the other
> > side of it is bizarre.
> > Whether the bizarre is eliminated or not does not have to be dealt with
> > yet.
> > This is really a classical model taken to a new level.
> >
>
> Take 20 steps back. Ask simple questions. Give simple answers.

Sorry. There's a few too many issues going on in one thread.
This is just a summary of where I am at. It's not necessary to go
through it all again. You have no need to be burdened with it. Chance
are good that there are errors.

> When people with swiss army-knives come along don't point your chisel
> at them.

That's a good idea.
I see a process of debate. Attacks are acceptable in this domain, no
different than on a chess board. Here an effective attack should be
waged upon the polysign system by others.
As I defend them I learn. And as someone else attacks them they learn
as well. There is no death in this game. We just walk away and call it
a game. Either side is free to walk at any time. This real number issue
does seem to be important. I can hardly believe that people don't see
magnitude as fundamental, and noone seems willing to address the point
except for me. You just keep claiming that the polysign system is built
from the reals, and I keep refuting this. I suppose we are weary of
that. I think the answers to be obvious yet I must address your claim.
Under the algebraic definition there is no need for a Cartesian
product. That is probably the strongest reason to avoid all of the
(x,y,z) type of stuff.

Anyhow the subject of this thread is about the matrix form of the
polysign product.
Now we're off on all sorts of tangents. The P4 product is of interest
since it is a potentially new space so I suppose that is the one to
form a new thread on.

-Tim

Timothy Golden BandTechnology.com

unread,
Jun 19, 2006, 6:01:21 AM6/19/06
to

Sorry about that. There are three animations on that page. They work on
my Firefox but load it pretty badly. The individual images should
probably work OK for you:

The difference:

http://bandtechnology.com/PolySigned/Deformation/T3P4DifferenceStudy.gif
The P4 product:

http://bandtechnology.com/PolySigned/Deformation/AxisDualDeformStudy.gif
The flat R x C product:

http://bandtechnology.com/PolySigned/Deformation/AxisDualTatrixStudy.gif

Anyhow the algebraic will need two transforms, one from P4 to R x C
which I'll call T3 and one inverse of this. So We could call these
P4OfT3 and T3OfP4.
Since we have to compare in one domain we choose T3 arbitrarily.
Instantiate two points in T3: t1, t2 (actuallly rtu1pos etc. in the
code)
Transform these over to P4: s1 = P4OfT3(t1), s2 = P4OfT3(t2) (actually
rsu1pos etc in code)
Take products:
t3 = t1 t2 ( actually t1 in the code... Sorry to confuse but here
these names good)
s3 = s1 s2 ( s1 in code)
Now take s3 back into T3:
ts3 = T3OfP4( s3 )
And finally note the difference:
t3 - ts3
which is nonzero.
This algebra relies on P4OfT3 and T3OfP4 being chosen carefully since
these systems are anisotropic. Even though the unit shell is
symmetrical its product results are not. The fact that the spiral shell
appears similarly in both domains is a strong indication that the
choice is correct. The positive Real is:
+ 1 # 1 (in P4)
The positive real within the complex plane is:
# 1.5 - 0.75 * 0.75 (in P4)
The transforms can be gotten from these. I look at them as projections
from 3D to 3D. This is done in the Cartesian domain. I have verified
that their product yields the identity matrix.
These products are quantified by their behavior over the unit shell.
All other values are scalings of these unit vector results. So the
graphics are providing a survey of the product.

-Tim

Timothy Golden BandTechnology.com

unread,
Jun 19, 2006, 10:24:16 PM6/19/06
to

Hi Jon. Here is the result for the product:
http://bandtechnology.com/tmp/T3P4DifferenceStudySpoonfed.gif
This link may go stale or change meaning as we iterate so not too much
description.
Here are some test values for the product we discussed:

[T3 [C1 0 ][C2 0, 0 ]] TIMES [T3 [C1 0 ][C2 0, 0 ]] = [T3 [C1 0 ][C2 0,
0 ]]
[T3 [C1 1 ][C2 1, 0 ]] TIMES [T3 [C1 0 ][C2 1, 0 ]] = [T3 [C1 0 ][C2 2,
0 ]]
[T3 [C1 0 ][C2 0, 1 ]] TIMES [T3 [C1 0 ][C2 0, 1 ]] = [T3 [C1 0 ][C2
-1, 0 ]]
[T3 [C1 1 ][C2 0, 1 ]] TIMES [T3 [C1 1 ][C2 0, 1 ]] = [T3 [C1 1 ][C2
-1, 2 ]]

If you want more values let me know.
They look good to me.
The notation is a T3 tatrix whose contents are in their Cartesian form.
This means R x C where [C1 x ] is the R and [C2 y, z]is the C as y +
iz.
The graph now has larger error.
I'll try another def if you like.
The result is interesting, but we want to see nothing in the difference
graph.
Oh, here is the product image (no differencing):
http://bandtechnology.com/tmp/T3SpoonfedProduct.gif

General background:
http://bandtechnology.com/PolySigned/FourSigned.html

-Tim

Spoonfed

unread,
Jun 20, 2006, 11:31:57 PM6/20/06
to
Timothy Golden BandTechnology.com wrote:
> Hi Jon. Here is the result for the product:
> http://bandtechnology.com/tmp/T3P4DifferenceStudySpoonfed.gif
> This link may go stale or change meaning as we iterate so not too much
> description.
> Here are some test values for the product we discussed:
>
> [T3 [C1 0 ][C2 0, 0 ]] TIMES [T3 [C1 0 ][C2 0, 0 ]] = [T3 [C1 0 ][C2 0,
> 0 ]]
> [T3 [C1 1 ][C2 1, 0 ]] TIMES [T3 [C1 0 ][C2 1, 0 ]] = [T3 [C1 0 ][C2 2,
> 0 ]]
> [T3 [C1 0 ][C2 0, 1 ]] TIMES [T3 [C1 0 ][C2 0, 1 ]] = [T3 [C1 0 ][C2
> -1, 0 ]]
> [T3 [C1 1 ][C2 0, 1 ]] TIMES [T3 [C1 1 ][C2 0, 1 ]] = [T3 [C1 1 ][C2
> -1, 2 ]]
>
> If you want more values let me know.
> They look good to me.
> The notation is a T3 tatrix whose contents are in their Cartesian form.
> This means R x C where [C1 x ] is the R and [C2 y, z]is the C as y +
> iz.

As far as I know, there is no such product defined in traditional math
for such ordered pairs. Any choice you make as to how to convolve the
pairs is a random decision, and I would encourage you to stop worrying
about this. I want you to look at the eigenvectors and eigenvalues,
which I have given below in R3; I believe you have a transformation
available to convert between R3 and P4.

> The graph now has larger error.
> I'll try another def if you like.
> The result is interesting, but we want to see nothing in the difference
> graph.
> Oh, here is the product image (no differencing):
> http://bandtechnology.com/tmp/T3SpoonfedProduct.gif
>

These are interesting, but they do far too much to make it clear. You
have the path of the red dot moving around in a helical pattern around
the sphere. Instead of doing this, move the red dot straight from the
origin outward along each of the four directions. If I am not
mistaken, moving the dot along the #1 vector (violet), you will have
the sphere simply expand or contract. But I would like to see what
happens as you move the dot straight outward along the -1, +1, and *1
vectors.

The eigenvectors for multiplying by
#1 are {{0, 0, 1}, Eigenvalue=1
{0, 1, 0}, Eigenvalue=1
{1, 0, 0}}, Eigenvalue=1

#1 should preserve the shape of the sphere in all dimensions.

{{1, 1, 1}, {-1, I, -I}, {-1, -1, 1}, {-1, I, -I}}

-1 {{Sqrt[2/3], -(1/Sqrt[3]), 1}, Eigenvalue -1
{(-1 + I)*Sqrt[2/3], (1 + 2*I)/Sqrt[3], 1}, Eigenvalue I
{(-1 - I)*Sqrt[2/3], (1 - 2*I)/Sqrt[3], 1}}, Eigenvalue -I

-1 should reflect the sphere along the plane perpendicular to the first
axis, and I don't know what to do with the complex valued eigenvectors
and eigenvalues. Perhaps it would become apparent if you slid the red
dot along the -1 axis and watched what happened to the sphere.

+1: {{-Sqrt[3/2], 0, 1}, Eigenvalue -1
{1/Sqrt[2], 1, 0}, Eigenvalue -1
{Sqrt[2/3], -(1/Sqrt[3]), 1}}, Eigenvalue +1

This reflects the sphere along the plane perpendicular to the first two
axes and should preserve the coordinate along the third axis.

*1: {{Sqrt[2/3], -(1/Sqrt[3]), 1}, Eigenvalue -1
{(-1 - I)*Sqrt[2/3], (1 - 2*I)/Sqrt[3], 1}, Eigenvalue I
{(-1 + I)*Sqrt[2/3], (1 + 2*I)/Sqrt[3], 1}} Eigenvalue -I

Again, complex eigenvalues and eigenvectors. I believe it has
something to do with rotation.

So my expectation is to see as the red dot moves along the #1 axis, the
sphere will simply expand. For the +1 axis, the sphere will expand in
a somewhat different manner. But as the red dot moves along the -1 and
*1 axes, we should see some rotation.

My coordinates are set up so that vectors are represented (x,y,z). I
just transformed them into modified P4 {#,-,+}. You can see
{Sqrt[3/2], 0, Sqrt[3/2]} appears as an eigenvector for -1, +1, and *1
but not for #1. You have been referring to the vector +1 #1 as an
identity axis. This is that axis, for what it's worth.

{{
{Sqrt[3/2]/2, Sqrt[3/2]/2, Sqrt[3/2]}, {1/(2*Sqrt[2]), 3/(2*Sqrt[2]),
0}, {1, 0, 0}},
{{Sqrt[3/2], 0, Sqrt[3/2]}, {I*Sqrt[3/2], (1 + I)*Sqrt[3/2],
Sqrt[3/2]}, {(-I)*Sqrt[3/2], (1 - I)*Sqrt[3/2], Sqrt[3/2]}},
{{-Sqrt[3/2]/2, Sqrt[3/2]/2, Sqrt[3/2]}, {3/(2*Sqrt[2]),
3/(2*Sqrt[2]), 0}, {Sqrt[3/2], 0, Sqrt[3/2]}},
{{Sqrt[3/2], 0, Sqrt[3/2]}, {(-I)*Sqrt[3/2], (1 - I)*Sqrt[3/2],
Sqrt[3/2]}, {I*Sqrt[3/2], (1 + I)*Sqrt[3/2], Sqrt[3/2]}}}

I am extremely surprised that squaring each point in a sphere yields a
cone. I wish to point out, however that #1 * #1 = #1 and #1 should be
a point on the circle that maps to itself when it is squared. You do
not have the cone intersecting #1 in your diagram. Perhaps you have
mislabeled your axis or I misunderstood which one is the identity.

Timothy Golden BandTechnology.com

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Jun 21, 2006, 11:19:45 AM6/21/06
to

I have produced this:
http://bandtechnology.com/tmp/SignVectorStudy.gif
Nice basic step. Good idea.
Perhaps I'll add some random vectors to this to demonstrate the scaling
effect.
Any vector travelling from the origin outward will provide the scaling
effect. The morphing is constant along these.

Nice catch! You are quite right. It's the zero sign quagmire. The red
axis is always the zero sign which is the identity sign. So the color
mnemonic will be extensible on the color spectrum as ROYGBIV. In P4 red
is #, green is -, blue is +, and violet is *.
I'm still correcting this on my website. The graphs are correct and the
text is wrong. The graphs follow the color mnemonic.

The cone: If you look at the parametric equations for the sphere and
the cone they are not very far off from each other. The two halves of
the original sphere are occupying the same space of the cone, no
different than the squares of the real numbers fold over to the
positive half of the real line. The even signs posess this feature. I
have yet to see it happen on the odd signs. It is related to the
identity axis.

eigenvectors: I'm still trying to grasp the value of them. Do you want
me to transform some of your values? I am happy to do some work with
them.
There are three distinct values in P4 that will preserve themselves
under the self product. There may be more, but the ones that I know of
are:
#1 ,
# 1 + 1 ,
# 1.5 - 0.75 * 0.75 .
The latter two need to be scaled to unity magnitude for them to hold
their magnitude steady under the self product. I'm not sure how this
relates to your eigenvector study. I can hunt for more of these and get
the angles between these three if you think it is important.
The latter two were used to get the R x C transformations which I am
willing to stop worrying about for the moment though I won't be
surprised if that gets awakened again at some point.
The latter two are orthogonal.
Have you studied Clifford algebras? I think that is the closest I can
get to existing continuum products, but they are only associative
algebras. The quaternions are in there. There is no definition for 3D
in them as far as I can gather from:
http://en.wikipedia.org/wiki/Clifford_algebra
and its linked pages. Certainly the polysign construction is very
different from these in any high dimension (3D+). This has been my
understanding and I don't know what to make of the claims from Gene in
this regard. Just wait and see if he comes back with anything I guess.
Anyhow, whatever product anyone wishes to compare P4 to I am happy to
graph out more.

-Tim

Spoonfed

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Jun 21, 2006, 12:37:20 PM6/21/06
to

I was expecting to see rotation or skewing of the figure as we moved
along -x, +x, or *x, I guess I shouldn't have expected this. However,
I think one more change to the animation would help. On the left hand
side, you show a yellow end and a black end of the coil, but on the
right, the coil is all black. To see which direction the coil is
pointing we need it on both sides, and then we can tell whether the end
product has been mirrored by whether the coil has switched handedness.

Eigenvectors and eigenvalues are a really good shortcut to getting a
quick visual concept of a linear transformation if you know how to use
them. Unfortunately, I'm still learning how to use them. I already
transformed them to see what they were in polysigned. It's given me an
idea that we have not chosen the most convenient translation between R3
and P4. Instead, the #1 and +1 vectors should lie in the xy plane so
that the +1#1 vector lies on the x axis, and the -1 and *1 vectors
should lie in the xz plane so the -1*1 vector lies on the -x axis.
This would probably yield much more symmetrical looking operators.

> There are three distinct values in P4 that will preserve themselves
> under the self product. There may be more, but the ones that I know of
> are:
> #1 ,
> # 1 + 1 ,
> # 1.5 - 0.75 * 0.75 .
> The latter two need to be scaled to unity magnitude for them to hold
> their magnitude steady under the self product.

The unit sphere in P4 is, I believe the set (a,b,c,d) where
a^2 + b^2 + c^2 + d^2- 2*(ab+bc+cd+de+ac+bd)/3 =1

This puts (sqrt(3/4),0,sqrt(3/4),0) on the surface of the sphere, and I
believe it maps to the vertex of the cone (3/4,0,3/4,0) when you square
it.

> I'm not sure how this
> relates to your eigenvector study. I can hunt for more of these and get
> the angles between these three if you think it is important.
> The latter two were used to get the R x C transformations which I am
> willing to stop worrying about for the moment though I won't be
> surprised if that gets awakened again at some point.
> The latter two are orthogonal.
> Have you studied Clifford algebras? I think that is the closest I can
> get to existing continuum products, but they are only associative
> algebras. The quaternions are in there. There is no definition for 3D
> in them as far as I can gather from:
> http://en.wikipedia.org/wiki/Clifford_algebra
> and its linked pages.
> Certainly the polysign construction is very
> different from these in any high dimension (3D+). This has been my
> understanding and I don't know what to make of the claims from Gene in
> this regard. Just wait and see if he comes back with anything I guess.
> Anyhow, whatever product anyone wishes to compare P4 to I am happy to
> graph out more.
>
> -Tim

Well, if you believed you'd find something and didn't, I probably won't
find it, believing I won't.

Timothy Golden BandTechnology.com

unread,
Jun 22, 2006, 10:43:30 AM6/22/06
to

Spoonfed wrote:
> > Any vector travelling from the origin outward will provide the scaling
> > effect. The morphing is constant along these.
> >
>
> I was expecting to see rotation or skewing of the figure as we moved
> along -x, +x, or *x, I guess I shouldn't have expected this. However,
> I think one more change to the animation would help. On the left hand
> side, you show a yellow end and a black end of the coil, but on the
> right, the coil is all black. To see which direction the coil is
> pointing we need it on both sides, and then we can tell whether the end
> product has been mirrored by whether the coil has switched handedness.
>
Hi Jon. Happy belated solstice.
I've written code to graph what I call a 'keyed' sphere.
It will have rings ehre the original sign vector positions were.
I concur on the operand sphere coloring also.
I want to rush an image but really there is some work of verification
to do.
Rather than rush I figure I'll get out a reply here to alleviate one
pressure.
The handedness issue may not become apparent in the graphs.
My graphing is non-occluding so everything is see through. The last
pixel written is on top so any path intersections are lying as to their
depth. When you look through the sphere your eye allows it to have two
representations. With no good 3D projection these qualities are going
to be more difficult to expose graphically.

I think we will find that when the sphere crosses over itself in the
coherent projection along the identity axis that that is when the
handedness flips. So half of the space should posess this handedness
flip and that half is determined by the identity axis. any point
resolving on the negative half (negative in the real number line sense
of this peculiar axis) of the identity axis will flip handedness. That
would mean that both -1 and *1 have flipped handedness. This agrees
with your previous results:
http://groups.google.com/group/sci.physics.relativity/msg/84a59e9d01e1d8ab

so when we see the sphere cross itself we have it's rotation pinned
orthogonal to the identity axis and are seeing it turn inside out while
it spins.
How do we measure handedness? I want to go to a physical pair of
tetrahedrons, start with identical signs and do the sign product
manually. But that will only get us the discrete sign operations. Again
I go back to matching up the vertices so that they contact. When they
are out of hand the last two will be four units apart, When they are in
hand the last vertices will be in contact.
Yet I don't have a simple way to detect that algorithmically. To
literally implement that in code could be a bear. It's a neat
fundamental puzzle.

> > The cone: If you look at the parametric equations for the sphere and
> > the cone they are not very far off from each other. The two halves of
> > the original sphere are occupying the same space of the cone, no
> > different than the squares of the real numbers fold over to the
> > positive half of the real line. The even signs posess this feature. I
> > have yet to see it happen on the odd signs. It is related to the
> > identity axis.
> >
> > eigenvectors: I'm still trying to grasp the value of them. Do you want
> > me to transform some of your values? I am happy to do some work with
> > them.
>
> Eigenvectors and eigenvalues are a really good shortcut to getting a
> quick visual concept of a linear transformation if you know how to use
> them. Unfortunately, I'm still learning how to use them. I already
> transformed them to see what they were in polysigned. It's given me an
> idea that we have not chosen the most convenient translation between R3
> and P4. Instead, the #1 and +1 vectors should lie in the xy plane so
> that the +1#1 vector lies on the x axis, and the -1 and *1 vectors
> should lie in the xz plane so the -1*1 vector lies on the -x axis.
> This would probably yield much more symmetrical looking operators.

Yuk. I suppose I could implement this. I don't think you can make the
#1 and +1 go in the same plane when the + 1 # 1 value has been fixed.
Maybe the angles work out though.
Anyhow composite projections are not a problem so that whatever we
graph can be reoriented via a 3 to 3 orthogonal projection.

>
> > There are three distinct values in P4 that will preserve themselves
> > under the self product. There may be more, but the ones that I know of
> > are:
> > #1 ,
> > # 1 + 1 ,
> > # 1.5 - 0.75 * 0.75 .
> > The latter two need to be scaled to unity magnitude for them to hold
> > their magnitude steady under the self product.
>
> The unit sphere in P4 is, I believe the set (a,b,c,d) where
> a^2 + b^2 + c^2 + d^2- 2*(ab+bc+cd+de+ac+bd)/3 =1
>
> This puts (sqrt(3/4),0,sqrt(3/4),0) on the surface of the sphere, and I
> believe it maps to the vertex of the cone (3/4,0,3/4,0) when you square
> it.

This would mean that you have found the native distance function!
And it looks extensible!
Why isn't ad and in there? Oh I see it. You have de where ad is.
I have not tried to check this yet but that will be very easy to do.
And in general dimension too!
This would be a very important piece since a native distance function
alleviates the need to enter the Cartesian domain.

> > I'm not sure how this
> > relates to your eigenvector study. I can hunt for more of these and get
> > the angles between these three if you think it is important.
> > The latter two were used to get the R x C transformations which I am
> > willing to stop worrying about for the moment though I won't be
> > surprised if that gets awakened again at some point.
> > The latter two are orthogonal.
> > Have you studied Clifford algebras? I think that is the closest I can
> > get to existing continuum products, but they are only associative
> > algebras. The quaternions are in there. There is no definition for 3D
> > in them as far as I can gather from:
> > http://en.wikipedia.org/wiki/Clifford_algebra
> > and its linked pages.
> > Certainly the polysign construction is very
> > different from these in any high dimension (3D+). This has been my
> > understanding and I don't know what to make of the claims from Gene in
> > this regard. Just wait and see if he comes back with anything I guess.
> > Anyhow, whatever product anyone wishes to compare P4 to I am happy to
> > graph out more.
> >
> > -Tim
>
> Well, if you believed you'd find something and didn't, I probably won't
> find it, believing I won't.

The funny thing with P3 is that if you aren't given the facts of the
handedness you can't detect them either. When you square a value you
could be on either side. The only detectable reference is a singular
axis. This has implications on the relative point particle model and
how many bits can be gotten from P3. It argues for bilateral symmetry
at an abstract level.
In P4 there is the same effect for - and *, but + and # are detectable.
Even having detected + and # the - and * still go in perfect symmetry
so there is a three-way classification where one of the elements has
twice the population. In P5 we have only one detectable and it doesn't
do anything. The prime signed spaces are transparent in this regard.
The effect still creeps in on the others, P4 being a good example. Is
this related to handedness? I don't see it. P3 doesn't change
handedness under any sign vector product. It may be useful to
investigate your operator determinant on P5 if we don't find a way to
detect handedness natively. Probably the answer is that even-signed
spaces change handedness under product by their identity axis component
and that odd-signed spaces never change handedness.

-Tim

Spoonfed

unread,
Jun 23, 2006, 10:40:44 PM6/23/06
to

Oh, well, that would be alright I guess. I was thinking in terms of
doing it myself if I ever had the time and inclination and you thought
it would be helpful. This is just for going back and forth between
polysigned and cartesian coordinates anyway. Right now it's not really
broke, but I was just tempted to fix it anyway.

> Anyhow composite projections are not a problem so that whatever we
> graph can be reoriented via a 3 to 3 orthogonal projection.
>
> >
> > > There are three distinct values in P4 that will preserve themselves
> > > under the self product. There may be more, but the ones that I know of
> > > are:
> > > #1 ,
> > > # 1 + 1 ,
> > > # 1.5 - 0.75 * 0.75 .
> > > The latter two need to be scaled to unity magnitude for them to hold
> > > their magnitude steady under the self product.
> >
> > The unit sphere in P4 is, I believe the set (a,b,c,d) where
> > a^2 + b^2 + c^2 + d^2- 2*(ab+bc+cd+de+ac+bd)/3 =1
> >
> > This puts (sqrt(3/4),0,sqrt(3/4),0) on the surface of the sphere, and I
> > believe it maps to the vertex of the cone (3/4,0,3/4,0) when you square
> > it.
>
> This would mean that you have found the native distance function!
> And it looks extensible!
> Why isn't ad and in there? Oh I see it. You have de where ad is.

Heh. Yes, and as is my typical behavior, I've spread the derivation
around the house on three or four sheets of looseleaf paper. It's not
pretty, but it definitely forms a pattern.
If you want the general idea consider that the cosine of A in the x
direction is 1, while the cosine of B, C, and D is -1/3, so the x value
is x=A-1/3(B+C+D)
The cosine of B in the y direction is sqrt(8/9) while the cosine of C
and D is -sqrt(2/9) so y=sqrt(8/9)(B-(C+D)/2)
Finally, z=sqrt(8/9)*sqrt(3/4)*(C-D)

To find the Cartesian length, just take sqrt(x^2+y^2+z^2).

In doing this problem I began to think that a different choice of
coordinate system might make the problem easier to tackle. For
instance, if A and B were both in the xy plane, and C and D were both
in the xz plane, it would be easier to calculate.

> I have not tried to check this yet but that will be very easy to do.
> And in general dimension too!
> This would be a very important piece since a native distance function
> alleviates the need to enter the Cartesian domain.
>

I was even wondering if the set of points in P4 which map to themselves
when squared and whether they formed a surface. For instance, this
spherical set, when squared, seems to map to a cone. Another reason
for wanting to change the coordinate system is that it appears from
your diagram that a full sphere maps to a cone all to one side of an
equator of the sphere. I think that that the plane cutting that sphere
marks the dividing plane between negative-like and positive-like
numbers in P4.

I think I follow you up to here. Given two numbers to multiply in P3,
you can rotate around the axis. The direction of +1 is important
because you have to reference your angle from here to figure out how
far to rotate. If you turn the paper over, you still have the +1
vector labeled the same way, but the rotation is reversed. It doesn't
matter, though; it gives the same answer; there is no back or front of
the page, only a line and an angle.

In three dimensions, though, you can tell which side of the paper
you're looking at, and you can tell whether you're looking at a
positive angle on the front of a page, or a negative angle at the back
of a page, or if you're looking at the paper in the mirror, because the
ink is on the front and the letters are all backwards.

> This has implications on the relative point particle model and
> how many bits can be gotten from P3. It argues for bilateral symmetry
> at an abstract level.

That's three things that I don't know the meaning of, right in a row.

> In P4 there is the same effect for - and *, but + and # are detectable.

Which effect is this?

> Even having detected + and # the - and * still go in perfect symmetry
> so there is a three-way classification where one of the elements has
> twice the population.

I don't think I've yet perceived the effect you are talking about here.
Is this anything to do with the negative-like and positive-like
numbers I made up earlier?

>In P5 we have only one detectable and it doesn't
> do anything. The prime signed spaces are transparent in this regard.
> The effect still creeps in on the others, P4 being a good example. Is
> this related to handedness? I don't see it.

Still not sure, but notice the use of imaginary numbers in physics
involve quantities that have some mathematical calculability, but tend
to disappear when measurements are made. I still haven't made any
connection to P4 with anything definitely physical, but you seem to be
making the analogy of + and # to real numbers and - and * to imaginary
numbers?

> P3 doesn't change
> handedness under any sign vector product. It may be useful to
> investigate your operator determinant on P5 if we don't find a way to
> detect handedness natively. Probably the answer is that even-signed
> spaces change handedness under product by their identity axis component
> and that odd-signed spaces never change handedness.
>
> -Tim

Could be. Perhaps Sunday I will spend an hour or two on this.

Timothy Golden BandTechnology.com

unread,
Jun 25, 2006, 10:07:49 AM6/25/06
to

Spoonfed wrote:
> Timothy Golden BandTechnology.com wrote:
> > Spoonfed wrote:
> > > The unit sphere in P4 is, I believe the set (a,b,c,d) where
> > > a^2 + b^2 + c^2 + d^2- 2*(ab+bc+cd+de+ac+bd)/3 =1
> > >
> > > This puts (sqrt(3/4),0,sqrt(3/4),0) on the surface of the sphere, and I
> > > believe it maps to the vertex of the cone (3/4,0,3/4,0) when you square
> > > it.
> >
> > This would mean that you have found the native distance function!
> > And it looks extensible!
> > Why isn't ad and in there? Oh I see it. You have de where ad is.
>
> Heh. Yes, and as is my typical behavior, I've spread the derivation
> around the house on three or four sheets of looseleaf paper. It's not
> pretty, but it definitely forms a pattern.
> If you want the general idea consider that the cosine of A in the x
> direction is 1, while the cosine of B, C, and D is -1/3, so the x value
> is x=A-1/3(B+C+D)
> The cosine of B in the y direction is sqrt(8/9) while the cosine of C
> and D is -sqrt(2/9) so y=sqrt(8/9)(B-(C+D)/2)
> Finally, z=sqrt(8/9)*sqrt(3/4)*(C-D)

It has the necessary symmetry and as I look upon it it seems to be the
only sensible form.
What will the factor do when we we change dimension? It doesn't just go
1/2, 2/3, 3/4, ...
does it? Well, I haven't even verified it yet for P4. I've been
immersed in trying to get the KeyedSphere cleaned up and applied to the
website. I'll verify your distance function next.
It's the most interesting thing that I'll get to do for some time so I
suppose I'm savoring the moment.

>
> To find the Cartesian length, just take sqrt(x^2+y^2+z^2).
>
> In doing this problem I began to think that a different choice of
> coordinate system might make the problem easier to tackle. For
> instance, if A and B were both in the xy plane, and C and D were both
> in the xz plane, it would be easier to calculate.
>
> > I have not tried to check this yet but that will be very easy to do.
> > And in general dimension too!
> > This would be a very important piece since a native distance function
> > alleviates the need to enter the Cartesian domain.
> >
>
> I was even wondering if the set of points in P4 which map to themselves
> when squared and whether they formed a surface. For instance, this
> spherical set, when squared, seems to map to a cone. Another reason
> for wanting to change the coordinate system is that it appears from
> your diagram that a full sphere maps to a cone all to one side of an
> equator of the sphere. I think that that the plane cutting that sphere
> marks the dividing plane between negative-like and positive-like
> numbers in P4.

Right, and that plane is orthogonal to the identity axis.

>
> >
> > The funny thing with P3 is that if you aren't given the facts of the
> > handedness you can't detect them either. When you square a value you
> > could be on either side. The only detectable reference is a singular
> > axis.
>
> I think I follow you up to here. Given two numbers to multiply in P3,
> you can rotate around the axis. The direction of +1 is important
> because you have to reference your angle from here to figure out how
> far to rotate. If you turn the paper over, you still have the +1
> vector labeled the same way, but the rotation is reversed. It doesn't
> matter, though; it gives the same answer; there is no back or front of
> the page, only a line and an angle.
>
> In three dimensions, though, you can tell which side of the paper
> you're looking at, and you can tell whether you're looking at a
> positive angle on the front of a page, or a negative angle at the back
> of a page, or if you're looking at the paper in the mirror, because the
> ink is on the front and the letters are all backwards.

See if you like
http://bandtechnology.com/tmp/SignVectorStudy.gif
The handedness flip is apparent though it is by use of the eye and
again a mathematical measure should be defined. The Cartesian cross
product comes to mind but that is stuck in 3D.

The handedness is not a matter of paper games. It is an argument about
observation. So we start out claiming not to know the identity or
position of the sign vectors. Instead we have only our operators which
in this case are sum, product, and render. I am arguing that you will
never detect any difference between the - vector and the + vector in
P3. Also that you will never detect a difference between the - and *
vector in P4. The detection of the identity vector is possible in all
cases. The detection of the + vector in P4 is also possible, since it's
square will be the # vector. Under these conditions the symmetries are
strict and true. The labels are arbitrary. We can just flip the - and +
sign vector labels in P3 and never know the difference, whether we look
at
a + b,
a a,
a b,
a( b + c ),
etc.
The graphs will be identical.
It is a parallel concept with handedness. Yet of these two systems one
flips handedness( P4 ) and the other( P3 ) does not. P2 does also flip
handedness, and P1 does not.

-Tim

Timothy Golden BandTechnology.com

unread,
Jun 25, 2006, 7:06:18 PM6/25/06
to

Congratulations Jon !!!

I've verified the distance function and found its general form.
The factors go
2/1, 2/2, 2/3, 2/4
for n = 2, 3, 4, ...
The following code has verified this out to large sign:

double SpoonfedDistanceFunction( Poly s )
{
if( s.n == 1 ) return s[0];
double d = 0;
double fac = - 1.0 / ( s.n - 1 );
for( int i = 0; i < s.n; i++ )
{
for( int j = 0; j < s.n; j++ )
{
if( i == j )
d += s[i] * s[j];
else
d += fac * s[i] * s[j];
}
}
return sqrt(d);
}
void testSpoonfedDistanceFunction()
{

for( int j = 2; j < 100; j++ )
{
Poly s(j);

int i = 0;
double d;
while( i++ < 100 )
{
s.Randomize();
s = s * ((double)(i));
d = SpoonfedDistanceFunction( s );
cout << i << " : " << d << s <<"\n";
if( abs( d - i ) > 0.00000001 ) cerr<< "\nGot a mismatch.\n";
}
}
}

Randomize generates a random unit vector. This is scalar multiplied by
an index so that the vector length is the index. They match from P2 to
P100. With your distance function the polysigned code will be quite
improved. It currently relies on the use of the Cartesian domain and so
a transform each time a distance is required. The algorithm confuses
the equation. Since each combination is hit twice the 1/ (n-1) becomes
2/(n-1) as in your equation.
This is really great. I've been wondering about this for some time. The
symmetry of the equation.

-Tim

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