Klein four-group geometry

Anti-Crackpot

Jeremy, just as a thought, here is an equation form that might be worth looking in to just to see what comes of it. It's an "abuse" of the Ramanujan theta function, substituting primes p and q for a and b (where |ab| < 1) and setting the start of the range at 0, not negative infinity:

[itex]\sum^{∞}_{0}p^{T_n}q^{T_{-n}}[/itex]

For example, set p to 2 and q to 3 and you get the below sequence as first differences:
2^(n(n+1)/2)*3^(n(n-1)/2)...
http://oeis.org/A081955
2, 24, 1728, 746496, 1934917632, 30091839012864, 2807929681968365568, 1572081206902992767287296...
# Divisors of first differences 2, 8, 28, 77... = (T_n + 1) * (T_-n + 1)

SUM = 2, 26, 1754, 748250, 1935665882, 30093774678746 ...
# divisors= 2, 4, 4, 32, 8, 32, 32, 32, 4, 128...

Curious if any interesting patterns would emerge. For instance, is that a coincidence that the # of divisors up above are all powers of 2? No idea actually, but it does seem a bit odd. Will have to see what happens with the 11th term at the very least :-)

If nothing else, it will get you thinking about the Ramanujan Theta Function in relation to your model. (Which I suspect is, in fact, related)

- AC

Note: Turns out the 11 th term has 16 divisors. Also a power of 2. The numbers are getting pretty big pretty fast, so not sure how many more terms I can check. Last term checked: 1.2872079903823517043082155685251539206949300954 × 10^46

JeremyEbert

Quote by Anti-Crackpot

This is a work of art Jeremy. Quite literally. And to think it's got the Dirichlet Divisor Sum embedded within. Very cool.

Its related to Julian Voss-Andreae's "Father and Mother"

"Father and Mother of the series Spin Family (2009) by physicist-turned-sculptor Julian Voss-Andreae"
http://en.wikipedia.org/wiki/File:%2..._Family%22.jpg

Anti-Crackpot

Quote by JeremyEbert

Its related to Julian Voss-Andreae's "Father and Mother"

"Father and Mother of the series Spin Family (2009) by physicist-turned-sculptor Julian Voss-Andreae"
http://en.wikipedia.org/wiki/File:%2..._Family%22.jpg

You read my mind. That's exactly the piece I was thinking of.

JeremyEbert

Quote by JeremyEbert

AC,

I'm working on some notes about this.

http://oeis.org/A003991

"Consider a particle with spin S (a half-integer) and 2S+1 quantum states |m>, m = -S,-S+1,...,S-1,S. Then the matrix element <m+1|S_+|m> = sqrt((S+m+1)(S-m)) of the spin-raising operator is the square-root of the triangular (tabl) element T(r,o) of this sequence in row r = 2S, and at offset o=2(S+m). T(r,o) is also the intensity |<m+1|S_+|m><m|S_-|m+1>| of the transition between the states |m> and |m+1>. For example, the five transitions between the 6 states of a spin S=5/2 particle have relative intensities 5,8,9,8,5. The total intensity of all spin 5/2 transitions (relative to spin 1/2) is 35, which is the tetrahedral number A000292(5). [Stanislav Sykora, May 26 2012]"

https://dl.dropbox.com/u/13155084/Ba...0triangles.txt

So one of the 4 Quantum "Spatial and angular momentum numbers" Ms where spin S = n-0.5

relative intensities = k(n+1-k)

http://en.wikipedia.org/wiki/Quantum...mentum_numbers

"the spin-raising operator is the square-root"

the square root produces the Klein four group from (n-k^2)/(2k)

AC has some findings for hbar (reduced Planck's constant) that fix right in as well I beleive.

interesting...

JeremyEbert

a binomial distribution link as well

http://epr.cm.utexas.edu/WhatIsEPR.html

"where N is the number of equivalent nuclei and I is the spin. It is important to note that this formula only determines the number of lines in the spectrum, not their

relative intensities. Coupling to a single nucleus with spin n/2 gives (n + 1) lines each of equal intensity.

2NI + 1 = 2(1)(n/2) + 1 = n + 1 lines

For example, coupling to a single vanadium nucleus (I = 7/2) will result in a spectrum of eight lines all of equal intensity.

Simulated EPR spectrum showing coupling to one nucleus (/I/ = 7/2)
Coupling to n equivalent nuclei, each with spin ½ again gives (n + 1) lines,

2NI + 1 = 2(n)(1/2) + 1 = n + 1 lines

but, since there are multiple nuclei interacting, the relative intensities of the lines follow the binomial distribution shown below.

# of Equivalent Nuclei Relative Intensities
1 1:1
2 1:2:1
3 1:3:3:1
4 1:4:6:4:1
5 1:5:10:10:5:1
6 1:6:15:20:15:6:1

An example of this is shown by the EPR spectrum of the radical anion of benzene, [C6H6•]-, in which the electron is delocalized over all six carbon atoms and therefore

exhibits coupling to six equivalent hydrogen atoms. As a result, the EPR spectrum shows seven lines with relative intensities of 1:6:15:20:15:6:1. Similar distributions

can be derived for n equivalent nuclei with spins greater than ½."

JeremyEbert

Quote by JeremyEbert

AC,

I'm working on some notes about this.

http://oeis.org/A003991

"Consider a particle with spin S (a half-integer) and 2S+1 quantum states |m>, m = -S,-S+1,...,S-1,S. Then the matrix element <m+1|S_+|m> = sqrt((S+m+1)(S-m)) of the spin-raising operator is the square-root of the triangular (tabl) element T(r,o) of this sequence in row r = 2S, and at offset o=2(S+m). T(r,o) is also the intensity |<m+1|S_+|m><m|S_-|m+1>| of the transition between the states |m> and |m+1>. For example, the five transitions between the 6 states of a spin S=5/2 particle have relative intensities 5,8,9,8,5. The total intensity of all spin 5/2 transitions (relative to spin 1/2) is 35, which is the tetrahedral number A000292(5). [Stanislav Sykora, May 26 2012]"

https://dl.dropbox.com/u/13155084/Ba...0triangles.txt

Some great info on Angular Momentum States.

*"According to the laws of quantum mechanics spin angular momentum is
not only quantized in magnitude, but in orientation relative to the z axis in
physical space.Therefore, the vector representing spin can only assume certain
orientations relative to the z axis in spin vector space. The quantum number
Ms specifies the possible orientation of a given angular momentum in space."
....

*"Multiplicity, M, of state with quantum number S = 2S + 1"

*"Pythagorean Relationships and Spin Angular Momentum"

http://www.columbia.edu/itc/chemistr...em/spin/03.pdf

JeremyEbert

for a set of complex numbers Z

z=(((n+1)/2)-k) + i(k(n+1-k)) whose (x/y) coordinates fall on a circle(x/y) or sphere(x/y/z) of radius (n+1)/2

so essentially those are points of a circle or sphere under quantization, integer and half-integer points.

a sphere would fall in line with the "inverse square law" perfectly for Spin, Energy and Force relative intensity coordinates.

make sense?

Anti-Crackpot

Quote by JeremyEbert

So one of the 4 Quantum "Spatial and angular momentum numbers" Ms where spin S = n-0.5

relative intensities = k(n+1-k)

http://en.wikipedia.org/wiki/Quantum...mentum_numbers

Reverse the sign so that you're adding a square rather than subtracting it...

k(n+1+k)

Set a(0) = 0 and a(n + 1) = 2a(n) + n(mod 2)sgn(n-1)
a(n) = 0, 0, 0, 0, 1, 2, 5, 10, 21...
Without the 3 leading zeroes, that's this progression here: http://oeis.org/A000975

Then, for range 0 ≤ k ≤ 8 and n = a(n), you'll get the following set of values:

0, 2, 6, 12, 24, 40, 72, 126, 240

I haven't worked out the algebra, but you can be pretty sure the recursively defined progression up above is this one here (derived accidentally):
A029929 n(n+ceiling(2^n/12))
REFERENCES C. Muses, The dimensional family approach in (hyper)sphere packing..
http://oeis.org/A029929
Next values: 468, 960, 2002...

Via Wikipedia
http://en.wikipedia.org/wiki/ADE_classification
ADE Classification
Algebro-geometrically, McKay also associates E6, E7, E8 respectively with: the 27 lines on a cubic surface, the 28 bitangents of a plane quartic curve, and the 120 tritangent planes of a canonic sextic curve of genus 4.[14][15] The first of these is well-known, while the second is connected as follows: projecting the cubic from any point not on a line yields a double cover of the plane, branched along a quartic curve, with the 27 lines mapping to 27 of the 28 bitangents, and the 28th line is the image of the exceptional curve of the blowup. Note that the fundamental representations of E6, E7, E8 have dimensions 27, 56 (28·2), and 248 (120+128), while the number of roots is 27+45 = 72, 56+70 = 126, and 112+128 = 240.

Related progression:

A002336 Maximal kissing number of n-dimensional laminated lattice.
http://oeis.org/A002336

Imagine a crazy universe where lattices actually had something to do with physics and the optimal division of space had something to do with its converse, the optimal (recursively defined) packing of space. :-)

- AC

UPDATE:
k(n+1+k) = kn + k + k^2
Set n to 1/12*(2^k + 2(-1)^k - 6)
Thus, ceiling [k(1/12*(2^k + 2(-1)^k - 6) +1+ k)] = 0, 2, 6, 12, 24, 40, 72, 126, 240... etc.

NOTE that ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1+ n)] does not "equal" n(n+ceiling(2^n/12)), but you'd never know if you just looked at the integer solutions since the progressions with respect to that seem to be one and the same at least for n>-1.

Now reverse the sign back...
xxxxxx ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1- n] = 0, 0, -2, - 6, -8, -10, 0, 28, 112, 306...
... and take note of the 112 that appears in position 8. 112 + 128 = 240 (which follows since 128 = 2*8^2)

Anti-Crackpot

Quote by JeremyEbert

for a set of complex numbers Z

z=(((n+1)/2)-k) + i(k(n+1-k)) whose (x/y) coordinates fall on a circle(x/y) or sphere(x/y/z) of radius (n+1)/2

so essentially those are points of a circle or sphere under quantization, integer and half-integer points.

a sphere would fall in line with the "inverse square law" perfectly for Spin, Energy and Force relative intensity coordinates.

make sense?

Can you walk me through this, Jeremy? Also, any graphics or specific numbers as examples would be great. I'm a bit of a dummy and need to see things visually at times.

Anti-Crackpot

Quote by Anti-Crackpot

UPDATE:
k(n+1+k) = kn + k + k^2
Set n to 1/12*(2^k + 2(-1)^k - 6)
Thus, ceiling [k(1/12*(2^k + 2(-1)^k - 6) +1+ k)] = 0, 2, 6, 12, 24, 40, 72, 126, 240... etc.

NOTE that ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1+ n)] does not "equal" n(n+ceiling(2^n/12)), but you'd never know if you just looked at the integer solutions since the progressions with respect to that seem to be one and the same at least for n>-1.

Now reverse the sign back...
xxxxxx ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1- n] = 0, 0, -2, - 6, -8, -10, 0, 28, 112, 306...
... and take note of the 112 that appears in position 8. 112 + 128 = 240 (which follows since 128 = 2*8^2)

Hopefully, it is apparent why I went in to all that, Jeremy.
0, 2, 6, 12, 24, 40, 72, 126, 240
0, 0, -2, - 6, -8, -10, 0, 28, 112

What are the first differences between these two progressions?
0, 2, 8, 18, 32 for the first 5 values. aka 2*n^2 which, of course, gives the maximal number of electrons per shell in the context of relating that both to your formulas and sphere packings.

From the observational/heuristic standpoint, that's a heck of a coincidence don't you think? And the relationship followed organically from what you posted. You really might be on to something...

- AC

JeremyEbert

Quote by Anti-Crackpot

Can you walk me through this, Jeremy? Also, any graphics or specific numbers as examples would be great. I'm a bit of a dummy and need to see things visually at times.

Meee toooo....
here is the visual:
http://dl.dropbox.com/u/13155084/par...spin%205-2.png

JeremyEbert

Quote by JeremyEbert

So one of the 4 Quantum "Spatial and angular momentum numbers" Ms where spin S = n-0.5

relative intensities = k(n+1-k)

http://en.wikipedia.org/wiki/Quantum...mentum_numbers

"the spin-raising operator is the square-root"

the square root produces the Klein four group from (n-k^2)/(2k)

AC has some findings for hbar (reduced Planck's constant) that fix right in as well I beleive.

interesting...

EDIT:Ms where spin S = ((n+1)/2)-0.5

relative intensities = k(n+1-k)

http://dl.dropbox.com/u/13155084/par...spin%205-2.png

JeremyEbert

Quote by Anti-Crackpot

Hopefully, it is apparent why I went in to all that, Jeremy.
0, 2, 6, 12, 24, 40, 72, 126, 240
0, 0, -2, - 6, -8, -10, 0, 28, 112

What are the first differences between these two progressions?
0, 2, 8, 18, 32 for the first 5 values. aka 2*n^2 which, of course, gives the maximal number of electrons per shell in the context of relating that both to your formulas and sphere packings.

From the observational/heuristic standpoint, that's a heck of a coincidence don't you think? And the relationship followed organically from what you posted. You really might be on to something...

- AC

Very nice AC. This fits right in as well.

I noticed an interesting vector a while ago at 45 degrees or pi/4 radians.

parabolic coodinate values at pi/4 = q:

5.8284271247461900976033774484194
23.313708498984760390413509793678
52.455844122715710878430397035775
93.25483399593904156165403917471

q=((1+sqrt(2))*n)^2

u = ((n^2)*6) – q

n = (q*u)^(1/4)

m = ((n^2)*12) – 2q

(2q*m)^(1/2) = 2(n^2) = maximal number of electrons per shell.

visual:
http://dl.dropbox.com/u/13155084/45.png

Anti-Crackpot

I made a a typo in my recent post, Jeremy, but it's an interesting one...

WOLFRAM INPUT #1:
ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1- n)] - n(n+ceiling(2^n/12))

Where n is an integer, you get an integer (2(n*i)^2), else no. Descriptively, you get a series of arcs (think: "quantized" arcs) that trace the general form of a parabola. It's pretty cool. To get your "zeroes," then reverse sign:

WOLFRAM INPUT #2:
ceiling [n(1/12*(2^n + 2(-1)^n - 6) +1+ n)] - n(n+ceiling(2^n/12))

Just thinking out loud here, but I can't help but wonder if there isn't possibly some significance to the 1st differences between these two progressions.

- AC

Anti-Crackpot

FOR REFERENCE
Feel free to add to this list Jeremy, ideally in your next post, so we can start to build a little resource to help you better communicate your ideas to professionals. No need for a thank you (You're welcome in advance.). - AC

Dirichlet Divisor Sum Algorithm
http://dl.dropbox.com/u/13155084/DSUMv2.htm

Divisor Sum Formula + Dirichlet Divisor Sum as Quotient Table
http://math.stackexchange.com/questi...ry-function-dx
(see bottom of page)
See: http://oeis.org/A006218

Prime UNSafe Periods (Related to Cicada Life Cycles)
http://dl.dropbox.com/u/13155084/prime.png
NOTE: This sequence + Dirichlet Divisor Sum totals a Triangular # (T_n)
See: http://oeis.org/A161664
This sequence + "safe" periods also equals a Triangular (T_(n + 1))
See: http://oeis.org/A002541
It follows that 2*unsafe + safe + D(n) is square.

The "Pythagorean Lattice" - Interactive Visual (2-D and 3-D simulation)
https://dl.dropbox.com/u/13155084/PL...tice_3D_2.html
e.g. Click ok then key sequence = 4,d,space (3-D Pattern)
e.g. Click ok then key sequence = 3,v,space (2-D Recursion)

[m] = Control Menu
[ Space Bar ]=Start/Stop

[ 1 ]=NORMAL
[ 2 ]=REVERSE
[ 3 ]=RECURSION
[ 4 ]=PATTERN

[ d ]=3D ON/OFF
[ v ]=2D VIEWS

[ UP/DOWN ]=ZOOM
[ LT/RT ]=RECURSION +-

[ f ]=Faster
[ s ]=Slower
[ r ]=Reset

Circle Recusion
http://dl.dropbox.com/u/13155084/CircleRecusion.png

First visual of hypothesized "Klein Four Group Geometry"
http://dl.dropbox.com/u/13155084/pri...ur%20group.png

Divisor Symmetry n = 12
http://dl.dropbox.com/u/13155084/divisor%20semmetry.png

Mapping to Particle Spin ([itex]m_{s}[/itex])
http://dl.dropbox.com/u/13155084/par...spin%205-2.png

RELATED LINKS

Divisor Summatory Function
http://en.wikipedia.org/wiki/Divisor_summatory_function
Coupon Collector's Problem
http://en.wikipedia.org/wiki/Coupon_collector's_problem

Planck's Constant
http://en.wikipedia.org/wiki/Planck_constant
Quantum Number
http://en.wikipedia.org/wiki/Quantum_number

EXCERPTS

Quote by JeremyEbert

I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/pri...ur%20group.png

Quote by JeremyEbert

Maybe its more than the Klein four group.

Wiki:
"The divisors of 24 — namely, {1, 2, 3, 4, 6, 8, 12, 24} — are exactly those n for which every invertible element x of the commutative ring Z/nZ satisfies x^2 = 1.
Thus the multiplicative group (Z/24Z)× = {±1, ±5, ±7, ±11} is isomorphic to the additive group (Z/2Z)3. This fact plays a role in monstrous moonshine."

I checked. Only the divisors of 24 produce the specific symmetry such that:

s=(n-k^2)/(2k)
s+k =(n+k^2)/(2k)
(s+k)^2 - s^2 = n

Z/2Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,-1}

...

Z/24Z
(s+k)^2 = {01,09,16,12,01,09,04,00}
s^2 = {00,04,09,01,12,16,09,01}
n = {01,05,07,11,13,17,19,23}

...

In the Klein Four-Group the quadratic residues of (s+k)^2 and s^2 are permutations of the symmetric group of order 4 (S4).

http://www.physicsforums.com/showpos...60&postcount=4

Quote by JeremyEbert

Nice! The connection to the Fano plane opens up all kinds of wonderful things.

as far as the equivalence relation
s=(n-k^2)/(2k)
k=versine
sqrt(n)=sine
s=cosine

this can be simplified to:
SIN(theta) = sqrt(n)/(s+k)
COS(theta) = s/(s+k)

which reduces to a very simple:

[itex]\frac{-k + i \sqrt{n}}{k + i \sqrt{n} }[/itex]

Quote by JeremyEbert

for a set of complex numbers Z

z=(((n+1)/2)-k) + i(k(n+1-k)) whose (x/y) coordinates fall on a circle(x/y) or sphere(x/y/z) of radius (n+1)/2

so essentially those are points of a circle or sphere under quantization, integer and half-integer points.

a sphere would fall in line with the "inverse square law" perfectly for Spin, Energy and Force relative intensity coordinates.

make sense?

Quote by JeremyEbert

here is the visual:
http://dl.dropbox.com/u/13155084/par...spin%205-2.png

ALSO...

Quote by JeremyEbert

My model shows:

Dirichlet Divisor function
sum ((2*floor[(n - k^2)/k]) + 1) where k=1 to floor[sqrt(n)]
= A006218 http://oeis.org/A006218

Cicada function
sum ( ((n-k^2)/(2k)) - (((n-k^2)/(2k)) mod .5) ) * -2 where k=1 to n
= A161664 http://oeis.org/A161664

I would think this would be of great interest to professional mathematicians, maybe not...

Anti-Crackpot

Quote by Anti-Crackpot

FOR REFERENCE
Feel free to add to this list Jeremy, ideally in your next post, so we can start to build a little resource to help you better communicate your ideas to professionals. No need for a thank you. (You're welcome in advance.) - AC

Dirichlet Divisor Sum Algorithm
http://dl.dropbox.com/u/13155084/DSUMv2.htm

Divisor Sum Formula + Dirichlet Divisor Sum as Quotient Table
http://math.stackexchange.com/questi...ry-function-dx
(see bottom of page)
See: http://oeis.org/A006218

Prime UNSafe Periods (Related to Cicada Life Cycles)
http://dl.dropbox.com/u/13155084/prime.png
NOTE: This sequence + Dirichlet Divisor Sum totals a Triangular # (T_n)
See: http://oeis.org/A161664
This sequence + "safe" periods also equals a Triangular (T_(n + 1))
See: http://oeis.org/A002541
It follows that 2*unsafe + safe + D(n) is square.

FYI, Jeremy, you may know this already, but others almost certainly don't...

Let S(n) = Safe Period
Let D(n) = Dirichlet Divisor Sum
Let U(n) = Unsafe Period

2S(n) + 1D(n) + 3U(n) is First Pentagonal
1S(n) + 2D(n) + 3U(n) is Second Pentagonal

Add and subtract at will in combination with n and 2n + 1 to generate equivalencies. It becomes clear, with a little experimentation, the recursive nature of the Dirichlet Divisor Sum as well as the likelihood of a relationship to Euler's summing technique for identifying primes based on Generalized Pentagonal Numbers (i.e. his hypothesis on that which I believe has never been proven is very likely to be true). What comes next depends on what came before and the "metronome" marking the "beat" is (at least partly) five-sided in nature...

- AC

These relationships follow since:
T_(n-1) = S(n) + U(n)
T_(n) = D(n) + U(n)
n^2 = S(n) + D(n) + 2U(n)

And, for good measure...
T_(n-1) + T_(n) + n^2 = 2n^2

Anti-Crackpot

Quote by JeremyEbert

I checked. Only the divisors of 24 produce the specific symmetry such that:

s=(n-k^2)/(2k)
s+k =(n+k^2)/(2k)
(s+k)^2 - s^2 = n

Jeremy, would you mind providing definitions for n, s and k?
n is clearly the automorphisms of the divisors of 24, but beyond that I am trying to decode in a precise way how you are getting your s's and k's.

Btw, in Lucas Sequence Notation where P = (s + k) and Q = (s * k) and D = P^2 - 4Q, the Discriminant, those 3 statements above can be reformulated, in order, via simple algebraic manipulation...

Q = (s * k) = ((n - k)^2)/(2)
P = (s + k) = ((n + k)^2)/(2k)
Q = (s * k) = ((n - k)^2)/(2)

sqrt (P^2 - 4Q), by the maths of Lucas Sequences, it follows, will always be an integer.

Also, could you clarify the below "claim" (as mathematicians call it) that "the square root produces the Klein four group from (n-k^2)/(2k)" (the first quoted mathematical statement above). What's apparent to you is not apparent to me. Don't forget that I'm self-taught too :-)

Quote by JeremyEbert

So one of the 4 Quantum "Spatial and angular momentum numbers" Ms where spin S = n-0.5
relative intensities = k(n+1-k)
http://en.wikipedia.org/wiki/Quantum...mentum_numbers

"the spin-raising operator is the square-root"

the square root produces the Klein four group from (n-k^2)/(2k)

- AC

P.S. As a little piece of mathematical trivia: D(P) = Q = 35 where P = (5 + 7) and Q = (5 * 7). What's the the next integer for which that relationship holds? Is it a common occurrence? Or an uncommon one?

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JeremyEbert	for a set of complex numbers Z z=(((n+1)/2)-k) + i(k(n+1-k)) whose (x/y) coordinates fall on a circle(x/y) or sphere(x/y/z) of radius (n+1)/2 so essentially those are points of a circle or sphere under quantization, integer and half-integer points. a sphere would fall in line with the "inverse square law" perfectly for Spin, Energy and Force relative intensity coordinates. make sense?