JeremyEbert said:I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/prime-%20square-klein%20four%20group.png
JeremyEbert said:I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/prime-%20square-klein%20four%20group.png
JeremyEbert said: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/3Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,2}
Z/4Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,3}
Z/6Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,5}
Z/8Z
(s+k)^2 = {1,4,1,0}
s^2 = {0,1,4,1}
n = {1,3,5,7}
Z/12Z
(s+k)^2 = {1,9,4,0}
s^2 = {0,4,9,1}
n = {1,5,7,11}
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}
(s+k)^2 + s^2 = 1 (mod 12) ** only modulo 12. ? additive inversion of the identity element to its multiplication group Z/12Z*?
((s+k)^2 - s^2)^2 = 1 (mod 12)
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).
DonAntonio said:It seems to be that only, or perhaps mainly, you know what you're talking about: "Klein four group geometry"? "A group geometry"?
If time passes by and you get no answer, perhaps it's time to give more background, books, papers for people to know what you mean. It may be that
what you call "some group geometry" is known by another name to someone else...
DonAntonio
JeremyEbert said:Thank you for your reply DonAntonio.
Let me preface my responses with the fact that I am self taught so I may be describing things incorrectly or may be over looking something obvious, but that is why I am here, to learn. Also, I quote wiki a lot. Sorry for that.
wiki:
Geometric group theory
"the study of finitely generated groups via exploring the connections between algebraic properties of such groups and topological and geometric properties of spaces on which these groups act (that is, when the groups in question are realized as geometric symmetries or continuous transformations of some spaces)."
The geometry I speak of relates to the points in Euclidean space defined by (x,y,z) where
x=s=(n-k^2)/(2k)
y=sqrt(n)
z=(s+k)
I had hoped that this would provide more detail:
http://dl.dropbox.com/u/13155084/Given%20a%20divisor%20k.pdf
The reason I say "Klein four group geometry" is because that group seems to fit best or "mainly" as you stated.
DonAntonio said:Well, it still looks a little odd: geometric group theory is a well-known, pretty advanced subject within group theory. I studied
a little of it no less than with Prof. Iliyah Ripps while in graduate school. Undoubtedly this could be a rather tough subject for
a non-mathematician.
What you're talking about, though, seems to be something else, related, as above, to group theory, geometry, number theory, etc., but in
a different way, apparently...
The link you wrote looks interesting but if you can I'd like to see books, papers, etc. about that in order to decide whether it is
something I can mess with (interesting, level, etc.) or not.
DonAntonio
Anti-Crackpot said:For whatever it's worth Don Antonio, I used Jeremy Ebert as an example of potentially undiscovered genius in a paper submitted last fall to two professors at Columbia University (Sociology Department).
He has already rediscovered the Dirichlet Divisor Sum geometrically, and there is no telling how far he could go with proper support from experts. It's a tough trick to pull to go backwards from simply "getting it" to understanding how it is that you "got it."
DonAntonio said:Uneducated intelligence can easily be wasted in vain, in particular in realms such as mathematics where so often a hefty base is
needed to build upon it.
If he wants support then he'd rather go to some Maths Depts. in some university and approach some people there.
DonAntonio
JeremyEbert said:As far as number theory goes, one relation is the Divisor Summatory Function equivalence.
[itex]D(n)=\sum_{k=1}^{\lfloor\sqrt{n}\rfloor}\left(2 \cdot \left\lfloor \frac{n-k^2}{k}\right\rfloor+1\right) [/itex]
http://en.wikipedia.org/wiki/Divisor_summatory_function
also the Coupon collector's problem
http://en.wikipedia.org/wiki/Coupon_collector's_problem
related:
http://dl.dropbox.com/u/13155084/divisor%20semmetry.png
http://dl.dropbox.com/u/13155084/DSUMv2.htm
http://dl.dropbox.com/u/13155084/prime.png
http://math.stackexchange.com/quest...ntation-for-the-divisor-summatory-function-dx
JeremyEbert said:I've shown this to one of my employees who is also a Math teacher at a local College. I'm really interested in what your take on this is Don Antonio.
JeremyEbert said:I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/prime-%20square-klein%20four%20group.png
Anti-Crackpot said:A natural question to ask:
"What was the response of the math teacher?"
Hopefully he or she even knew what it was, but, based on anecdotal experience, I would not bank on it, tho', in principle, the basic formulation for D(n) is junior high school, if not elementary school, level. It's just a quotient table when you get down to it, albeit a quotient table that currently confounds all modern techniques available with regards to our ability to fully understand it...
- AC
chiro said:Hey JeremyEbert.
I'm a little lost as to what you are trying to ask. Do you want specific kinds of comments on your diagrams and your PDF or do you have some specific question you had in mind?
You have put in a lot of effort and I don't want to see it go to waste by asking undirected questions. Are you trying to solve something in particular? Are you trying to investigate something for a particular reason? Are you looking for comments on a particular issue?
If the above is the case then this would help us give a more directed answer and know exactly what to focus our attention on to initiate the conversation.
JeremyEbert said:Chiro,
Thanks for your interest. I see you've had some post that deal with quantum. I've noticed that the geometry of this equation produces a type of parabolic coordinate system.
http://dl.dropbox.com/u/13155084/CircleRecusion.png
http://mathworld.wolfram.com/ParabolicCoordinates.html
This coordinate system seems to be of importance in the quantum world when it comes to angular momentum.
http://www.ejournal.unam.mx/rmf/no546/RMF005400609.pdf
With the similarities between the Riemann zeros and the quantum energy levels of classically chaotic systems and the parabolic coordinates created by this geometry, it seems like this might be another connection between primes and the quantum word. What do you think?
chiro said: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.
Anti-Crackpot said: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_0 + r_1)
Q = (r_0 * r_1)
D = (r_0 - r_1)^2 = P^2 - 4Q
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
DonAntonio said:Well, it still looks a little odd: geometric group theory is a well-known, pretty advanced subject within group theory. I studied
a little of it no less than with Prof. Iliyah Ripps while in graduate school. Undoubtedly this could be a rather tough subject for
a non-mathematician.
What you're talking about, though, seems to be something else, related, as above, to group theory, geometry, number theory, etc., but in
a different way, apparently...
The link you wrote looks interesting but if you can I'd like to see books, papers, etc. about that in order to decide whether it is
something I can mess with (interesting, level, etc.) or not.
DonAntonio
Anti-Crackpot said: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
JeremyEbert said: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 = nZ/2Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,-1}
Z/3Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,2}
Z/4Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,3} Z/6Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,5} Z/8Z
(s+k)^2 = {1,4,1,0}
s^2 = {0,1,4,1}
n = {1,3,5,7} Z/12Z
(s+k)^2 = {1,9,4,0}
s^2 = {0,4,9,1}
n = {1,5,7,11} 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} (s+k)^2 + s^2 = 1 (mod 12) ** only modulo 12. ? additive inversion of the identity element to its multiplication group Z/12Z*?
((s+k)^2 - s^2)^2 = 1 (mod 12)
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).
JeremyEbert said: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/wikipedia/commons/a/a5/ComplexSinInATimeAxe.gif
chiro said: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.
Anti-Crackpot said: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 said: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 said:Nice! The connection to the Fano plane opens up all kinds of wonderful things.
JeremyEbert said: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/PL3D2/P_Lattice_3D_2.html
http://en.wikipedia.org/wiki/Minkowski_space
JeremyEbert said: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/Barycentric%20coordinates%20on%20triangles.txt