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Klein four-group geometry

 
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Apr30-12, 08:38 PM   #1
 

Klein four-group geometry


I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/pri...ur%20group.png
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May1-12, 06:26 AM   #2
 
Quote by JeremyEbert View Post
I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/pri...ur%20group.png
I'm working on the explination.
http://dl.dropbox.com/u/13155084/Giv...ivisor%20k.pdf
May1-12, 02:54 PM   #3
 
Quote by JeremyEbert View Post
I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/pri...ur%20group.png
A better visual.
http://dl.dropbox.com/u/13155084/pri...ur%20group.png
May4-12, 04:38 PM   #4
 

Klein four-group geometry


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).
May4-12, 06:33 PM   #5
 
Quote by JeremyEbert View Post
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).


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
May9-12, 08:15 AM   #6
 
Quote by DonAntonio View Post
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
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/Giv...ivisor%20k.pdf

The reason I say "Klein four group geometry" is because that group seems to fit best or "mainly" as you stated.
May9-12, 09:03 AM   #7
 
Quote by JeremyEbert View Post
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/Giv...ivisor%20k.pdf

The reason I say "Klein four group geometry" is because that group seems to fit best or "mainly" as you stated.


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
May9-12, 09:33 AM   #8
 
Quote by DonAntonio View Post
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
I'll see if I can find any real reference material for you.

It seems to me to be related to Pythagorean Triples. Here is something I wrote quite a while ago when I started down this path. It's very armature I know but it was my first attempt to make since of the relations I was noticing. It might help give a little background I guess.

http://dl.dropbox.com/u/13155084/Pyt...%20lattice.pdf
Another visual:
http://dl.dropbox.com/u/13155084/CircleRecusion.png
May9-12, 11:21 AM   #9
 
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/questi...ry-function-dx
May9-12, 05:07 PM   #10
 
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."
May9-12, 05:22 PM   #11
 
Quote by Anti-Crackpot View Post
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."


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
May10-12, 06:36 AM   #12
 
Quote by DonAntonio View Post
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
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.
May10-12, 01:00 PM   #13
 
Quote by JeremyEbert View Post
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/questi...ry-function-dx


A related number theory paper:

Note the geometry of Gauss circle problem.

http://www.homepages.ucl.ac.uk/~ucahipe/Lfunctions.pdf
May11-12, 03:55 AM   #14
 
Quote by JeremyEbert View Post
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.
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
May11-12, 04:11 AM   #15
 
Quote by JeremyEbert View Post
I think this a map of the Klein four-group geometry. Comments?
http://dl.dropbox.com/u/13155084/pri...ur%20group.png
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.
May11-12, 01:17 PM   #16
 
Quote by Anti-Crackpot View Post
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
Radio silence thus far.
May11-12, 01:41 PM   #17
 
Quote by chiro View Post
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.
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?
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