Klein four-group geometry

chiro

Well oddly enough, there has been shown to be a link between the Riemann Zeta function and Quantum Field Theory.

I remember reading about this ages ago from work associated with Alain Connes, but I don't have any deep knowledge, only superficial awareness of the fact. I would continue your work if this is what you are working on, because if you want ways to solve the Zeta problems and connect it with general quantum phenomena (discrete structures, diophantine systems and anything involving some kind of discrete system or finite-field) then that would be extroadinarily useful.

Here are two results from a google search involving the Zeta function and quantum field theory:

http://www.ricam.oeaw.ac.at/specsem/...nload/Moch.pdf

http://arxiv.org/pdf/hep--th/0701284.pdf

Anti-Crackpot

Chiro,

Give me any 6 Lucas Sequences with P's and Q's constructed from 4 integer variables (r_0, r_1, r_2, r_3) in toto and if the ratio between successive terms of those six sequences at the limit as n approaches infinity is an integer, then those same P's and Q's can be used in a very simple, regular and consistent manner to construct a quartic and its resolvent cubic that will have integer roots.

P = (r_x + r_y)
Q = (r_x * r_y)
D = P^2 - 4Q (Discriminant)

Minimal Start Terms:
U-Type Lucas Sequence = 0, 1
V-Type Lucas Sequence = 2, P

Recursive Rule:
-Q*a(n) + P*a(n+1) = a(n+2)

In other words, seems to me not only that there is a very clear linkage between the mathematics of the quartic (and thereby V4, the Klein Four-Group...) and Diophantine systems, but that this linkage may help explain how Jeremy is uncovering what he is uncovering.

If you are unclear as to what I am referencing, I can show you how to map those P's and Q's to the quartic. It took me all of about a day to work out the "translation" once Jeremy got me thinking about it.

- AC

chiro

I'm not sure what a Lucas sequence is. I only know the basics of number theory, but I think if you show me something like an external web-page or something that clarifies your ideas I can take a look at it.

JeremyEbert

Don Antonio,
Here is a link with some useful information regarding this although this site seems to be political in nature.
http://wlym.com/~animations/ceres/In...ciprocity.html

JeremyEbert

Nice AC! I'll have to play with this!

JeremyEbert

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

given the divisors of 12 have a different property than the divisors of 24, is there a link to equal temperament?

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

http://upload.wikimedia.org/wikipedi...InATimeAxe.gif

JeremyEbert

mind typo, s+k=radius:

s=(n-k^2)/(2k)
k=versine
sqrt(n)=sine
s=cosine

Anti-Crackpot

Consider the form:

C(a(n+2), j) (Binomial Coefficient)
a(n + 2) = x*a(n) + y*a(n+1) + z

The Lucas Sequences are just a special case of the above where:
j = 1
z = 0
x = r_0 * r_1 = Q
y = r_0 + r_1 = P

Take, for example, r_0 = phi and r_1 = -1/phi
r_0 * r_1 = -1 = Q
r_0 + r_1 = 1 = P

Start terms: 0, 1 (U-Type Lucas Sequence) gives the Fibonacci Series
Start terms: 2, 1 (V-Type Lucas Sequence) gives the Lucas Series

The recursive rule is -(-1)*a(n) + 1*a(n+1) = a(n + 2)

(sqrt (P^2 - 4Q) +/- 1)/2 = (sqrt(1^2 - 4(-1)) +/- 1)/2 = phi, 1/phi

You can read more about Lucas Sequences on Wolfram MathWorld or on Wikipedia. All kinds of identities follow from the maths and it's pretty simple stuff actually. http://mathworld.wolfram.com/LucasSequence.html

Lucas Sequences are also related to Carmichael's Theorem which involves the introduction of new prime factors into integers associated with recursively based sequences. Thus, for example, Mersenne Numbers (2^x - 1) [which are a Lucas Sequence] can only be prime where x is prime.

- AC

Anti-Crackpot

You've got a clock that goes to 12. In step increments you can go 1, 2, 3, 4, 6 or 12 at a time to end up back at 12. These are the subgroups of 12. Now take the totient of 1, 2, 3, 4, 6, 12 equal to 1, 1, 2, 2, 2, 4. These are the number of elements in each subgroup. Add them together and what do you get? 1 + 1 + 2 + 2 + 2 + 4 = 12. This holds for any integer, not just 12.

But what if you want to make sure you hit every number on your clock rather than skipping over, for instance, 1, 3, 5, 7, 9 and 11?

There are 4 ways to do that, which is the essence of the term automorphism aka "self-mapping." You can step forward by 1 or 7 and back by 1 or 7, which is the same as stepping back by 5 or 11 or forward by 5, or 11. Thus, 1, 5, 7, 11.

This fact underlies the maths of the Circle of Fifths of the Western Musical System which is based on the Perfect 5th. 12 steps, 7 notes at a time and there you are, from C through G, D, A, E, B, F# etc. and back to C after 84 notes.

I'll take a look at the link you posted, but, in short, the mathematics of Automorphism Group Z(12) underlies all of Western Music. Think about that in relation to what you posted:

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}

Restate n...
n = 1, 5, 7, 11, 24 - 11, 24 - 7, 24 - 5, 24 - 1

Two times, not 1 time, around the clock for a 720 degree rotation comprised of 168 notes, "coincidentally" the number of symmetries associated with the Fano plane. Whoever thought to divide the day and week the way it's divided, I might add, seems to have been a natural group theorist.

- AC

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]

JeremyEbert

so s+k is the scalar from the unit circle. s+k = (n+k^2)/(2k)

[itex]\frac{n+k^2}{2k} \cdot \frac{-k + i \sqrt{n}}{k + i \sqrt{n} }[/itex]

alternative form:

[itex]\frac{(\sqrt{n} + i k)^2}{2k }[/itex]

an interesting result of this equation is the additive inverse of the divisor k is -k produces the quotient and vice versa.

12/3=4
where n=12, k=3 the result is 0.5 + i sqrt(12)
where n=12, k=-3 the result is -0.5 + i sqrt(12)
where n=12, k=4 the result is -0.5 + i sqrt(12)
where n=12, k=-4 the result is 0.5 + i sqrt(12)

Anti-Crackpot

Glad to have been of some help. If you'd like to see how the Fano plane fits in to Minkowski SpaceTime, John Baez discusses this in week 219. Here's an excerpt...

"So, there's an interesting but complicated relation between hyperbolic geometry over Z/7 and the Fano plane. How does the Klein quartic curve fit in? There's more to this side of the story than I've managed to absorb, so I'll just say a few words - probably more than you want to hear. For more detail, try my Klein quartic curve webpage.

There are 48 nonzero lightlike vectors in 3d Minkowksi spacetime, but if you take one of them and apply elements of PSL(2,Z/7) to it, you get an orbit consisting of only 24. These 24 guys correspond to the 24 heptagons in the heptagonal tiling of the Klein quartic curve! In other words, PSL(2,Z/7) acts in precisely the same way."
http://math.ucr.edu/home/baez/week219.html

Bear in mind that if John Baez hasn't been able to fully absorb it, no one's expecting you to either. But good to know that the relationships are there.

- AC

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

JeremyEbert

AC,
You have made reference to my models and Minkowski space before. I think you were spot on.

[Click ok then key sequence = 4,d,space]

https://dl.dropbox.com/u/13155084/PL...tice_3D_2.html

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

Anti-Crackpot

Funny, Jeremy, but you are quite likely the only person in the world other than myself who recognizes the potential significance of what you just posted, which, unfortunately, cannot be said aloud. Enough said. Many thanks for this.

- AC

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.

Btw, if you're going to play around with roots of the quartic, a natural place to begin would be with either 1, 5, 7, 11 or (24 - 1, 5, 7, 11). Two other sets of roots that are rather interesting are:

12, 19, 29, 59 & 11, 4, 40, 10. Perform any binary operation excepting division upon any two of those terms and you'll have an integer that divides the Monster Group. In fact, the first 4 roots (12, 19, 29, 59) gets you 14 of the 15 supersingular primes as factors if you combine in all possible ways. The only supersingular prime missing is 23, which is where the second set of roots comes in. Add those terms to the first set of roots and you get a multiple of 23.

Should you ever take the next step (to the quintic), then 11, 12, 19, 29, 59 is pretty "nifty," since if you add, subtract and multiply out all possible combinations and then multiply all the terms together, you get a subgroup of the Monster.

Combine 12, 19, 29, 59, 11, 4, 40, 10 in all possible ways and multiply out and here's what you get:

2^53 * 3^24 * 5^17 * 7^10 * 11^8 * 13^3 * 17^2 * 19^3 * 23^4 * 29^2 * 31 * 41 * 47 * 59^2 * 71

That's a pretty big number, about 17780 times the estimated number of atoms in the observable universe, and the Monster just happens to be one of it's subgroups.

- AC

Anti-Crackpot

As a perhaps overly simplistic observation since we're just talking about the basic multiplication table here...
5 + 8 + 9 + 8 + 5 = (9 - (-2))^2 + (9 - (-1))^2 + (9 - (0))^2 + (9 - (1))^2 + (9 - (2))^2

This symmetry property holds for all even rows. Odd row entries have 1st differences from the central terms equal to a Pronic Number.

I would tell you how to relate this table to the divisors of the p^k*q^(n - k) triangle, but I know you already know that.

Interesting though, how the very structure of the basic multiplication table embeds the tetrahedron, one of the most important structural forms in nature (i.e. what do water and diamonds have in common?).

- AC