# Arithmetic Geometry

#### mathbalarka

##### Well-known member
MHB Math Helper
Consider a 2-sphere on the real plane equipped with the linear map from the sphere to it's equatorial 2-plane by fixing the topmost vertex of the sphere. This is now an analogue of the Riemann sphere in 3-dimensional space, hence we have the "point at infinity" in addition to the usual reals because of this "wrapping"-like projection. Call the projection $$\displaystyle \zeta$$.

Now, define an operation $$\displaystyle \boxplus$$ over the points of the 2-sphere $$\displaystyle \Gamma$$ by constructing $$\displaystyle A \boxplus B$$ by joining them and finding the intersection of $$\displaystyle AB$$ with the equatorial plane of $$\displaystyle \Gamma$$; call it $$\displaystyle C$$ and construct $$\displaystyle \zeta^{-1}(C)$$ to map it on $$\displaystyle \Gamma$$ again through usual methods.

Note that this creates a group $$\displaystyle \Gamma$$ where the elements are the points on the surface of the sphere. One can easily check that it is also abelian.

Now, as the point at infinity creates a projective version of this arithmetic, it is a projective variety, as well as an abelian one. So if one applies the Mordell-Weil theorem, we get an evidence of finiteness of the generating set.

I am trying to get a structural information of this peculiarly constructed group. I don't have any specific questions at this moment, just a confirmation of my work up to the last statements. It is also very much appreciated if one can derive a strong result concerning this.

Balarka
.

#### Opalg

##### MHB Oldtimer
Staff member
Consider a 2-sphere on the real plane equipped with the linear map from the sphere to it's equatorial 2-plane by fixing the topmost vertex of the sphere. This is now an analogue of the Riemann sphere in 3-dimensional space, hence we have the "point at infinity" in addition to the usual reals because of this "wrapping"-like projection. Call the projection $$\displaystyle \zeta$$.

Now, define an operation $$\displaystyle \boxplus$$ over the points of the 2-sphere $$\displaystyle \Gamma$$ by constructing $$\displaystyle A \boxplus B$$ by joining them and finding the intersection of $$\displaystyle AB$$ with the equatorial plane of $$\displaystyle \Gamma$$; call it $$\displaystyle C$$ and construct $$\displaystyle \zeta^{-1}(C)$$ to map it on $$\displaystyle \Gamma$$ again through usual methods.

Note that this creates a group $$\displaystyle \Gamma$$ where the elements are the points on the surface of the sphere. One can easily check that it is also abelian.

Now, as the point at infinity creates a projective version of this arithmetic, it is a projective variety, as well as an abelian one. So if one applies the Mordell-Weil theorem, we get an evidence of finiteness of the generating set.

I am trying to get a structural information of this peculiarly constructed group. I don't have any specific questions at this moment, just a confirmation of my work up to the last statements. It is also very much appreciated if one can derive a strong result concerning this.
I have a bit of trouble understanding this setup. I think that what you refer to as a 2-sphere is what I would call a 1-sphere because it is a one-dimensional manifold (embedded in two-dimensional space, and commonly known as a circle). If so, then the construction of $A\boxplus B$ is as shown in the diagram below, in which the point $P = (0,1)$ is the unit element of the group. Then $A\boxplus B$ is the point where the line from $P$ to the intersection of the line $AB$ with the $x$-axis meets the circle again. I have the impression that this group structure is not particularly interesting. For example, the inverse of the point $(a,b)$ is just the point $(-a,b)$. The fact that the group operation is associative is not so obvious. It is illustrated by the dashed lines in the diagram and is justified geometrically by Pascal's theorem.

The rational points on the circle, being of the form $\bigl(\frac pr,\frac qr\bigr)$, correspond to Pythagorean triples $(p,q,r)$, though the correspondence is not bijective because the four points $\bigl(\pm\frac pr,\pm\frac qr\bigr)$ all correspond to the same Pythagorean triple. Having said all that, I have no idea whether or how the Mordell–Weil theorem applies to this group $\Gamma$, or whether that says anything interesting about Pythagorean triples. #### mathbalarka

##### Well-known member
MHB Math Helper
Your interpretation of the operations is correct. A complicated proof of associativity can indeed be arrived by a complex bit of coordinate geometry. I didn't make the connection with Pascal's theorem, so I thank you very much for pointing that one out.

About the rational points, yes they are all particularly related with the radius as sqrt(p^2 + q^2) = r and not very much interesting at all, I gave the circle construction as an example here for simplicity and my lack of constructing generality out of it. My main goal is to extend this operation to almost any closed curve, that's why I am asking for the justification of the Mordell-Weil theorem applied at this point.

I think the connection with Mordell-Weil theorem is made by the fact that the particular algebraic variety is projective, due to the fact that the surface is the set of extended reals. A detailed proof without MWT should be worked out, though, for the sake of argument.

PS : One can, however, find at least one interesting property out of it. This curve can approximate any number of the form sqrt(d) with non-perfect square d by applying large multiplications at any initial point. Indeed, the approximation of sqrt(2) by iteration of Euler's sequence is a special case of this. This is, however, none of my more general and more important concerns.

#### Deveno

##### Well-known member
MHB Math Scholar
I've seen this contruction before, here:

AlgTop2: Homeomorphism and the group structure on a circle - YouTube

If I've linked correctly, he also explains why associativity is the most troubling part.

Interestingly enough, this is just the usual group structure on $T$, the unit circle in the complex plane, under complex multiplication (rotate the circle 90 degrees clockwise to see this).

If I recall correctly, at the end of this lecture (or maybe the next one), he goes on to indicate how one can do the same thing for other algebraic curves, such as a parabola, or a hyperbola, but doesn't go into all the details.

A VERY dim memory comes back to me on how one can carry out such a procedure for many other algebraic curves, such as elliptic curves, as discussed here:

Elliptic curve - Wikipedia, the free encyclopedia

#### mathbalarka

##### Well-known member
MHB Math Helper
I have not seen the video (because of shortage of time), but will see it soon, it seems to have something relevant.

Deveno said:
Interestingly enough, this is just the usual group structure on T, the unit circle in the complex plane, under complex multiplication (rotate the circle 90 degrees clockwise to see this).
Indeed. I realized it more than a year ago and that is exactly what made me think a lot about these things. I believe this group, not nearly general comparing what I am considering, is not quite trivial.

About elliptic curves, note that these operations are NOT the same, not at all. ECs use a decent addition theorem that any line intersects a elliptic curve touches at least two more points on the curve in the projective plane.

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#### mathbalarka

##### Well-known member
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The lecture refers to Lemmermeyer & Shirali, which is, as it seems, not available online.

#### mathbalarka

##### Well-known member
MHB Math Helper
Fortunately, I had time to view the lecture extensively there. If one gives one close look at that, one can see that my operation has nothing to do with mine, neither does the papers by Shirali.

BTW, to those who do not wish to view the video by Wildberger by going through all the basics of topology and algebraic geometry, try this : http://www.ias.ac.in/resonance/Volumes/13/10/0916-0928.pdf

#### Turgul

##### Member
I have not thought about this for very long, but there are a couple of things that struck me.

First, as described it's not quite a group because you have not described how to add points to themselves nor how to add points whose connecting line is parallel to the "$x$-axis" (and how to do this is important, as it's closely tied to finding inverses), and additional tedious work then would still be needed to decide addition involving such points was associative. But these concerns should be easy enough to deal with.

More fundamentally, this group is not projective. You are attempting to give a group structure to a circle, which is an affine variety. Your group law is built on the set $x^2+y^2 = 1$, say, using the line $y=0$ and the point $(0,1)$. I have not checked the details, but it is likely that the group structure works out over an arbitrary field $k$. However, the best you can hope for is an affine group stucture, and there are only two 1D affine algebraic groups! That is to say, if this group is a nice algebro-geometric object at all, it must be either $k$ or $k^*$ for general reasons, and it doesn't take much thought to see that $k^*$ makes more sense (has an element of order 2).

#### mathbalarka

##### Well-known member
MHB Math Helper
Turgul said:
First, as described it's not quite a group because you have not described how to add points to themselves nor how to add points whose connecting line is parallel to the "x-axis" (and how to do this is important, as it's closely tied to finding inverses), and additional tedious work then would still be needed to decide addition involving such points was associative. But these concerns should be easy enough to deal with.
I haven't described them, yes, but surely that doesn't mean they cannot be?

For the sake of clarification I describe them here :

Doubling a point can be done by drawing a tangent on A and then doing the same reverse mapping by the point at infinity to the circle.

For the second, if such two points exists (they surely do), they are inverses of each other.

For the third, I have mentioned that I have derived a proof of assosiativity, nevertheless, it's too large to fit in here.

It's been a year since I though about it and posted it in MMF, so surely I have done every possible check for ambiguities.

Turgul said:
More fundamentally, this group is not projective. You are attempting to give a group structure to a circle, which is an affine variety.
Indeed, I agree. As this is an affine group, it is a subgroup of some general linear group.

BTW, what do you mean by $$\displaystyle k^*$$?

#### Opalg

##### MHB Oldtimer
Staff member
... if this group is a nice algebro-geometric object at all, it must be either $k$ or $k^*$ for general reasons, and it doesn't take much thought to see that $k^*$ makes more sense (has an element of order 2).
I think it's probably the case that the group is not "a nice algebro-geometric object". It is, in the case where the Pascal line is the line at infinity (as in the video link at comment #4) and the group structure is then the usual group operation on the circle. But in mathbalarka's construction the Pascal line is the "equatorial" diameter of the circle, and the group structure is not the same at all. For example, the two points at the ends of the equatorial diameter are idempotent.

#### mathbalarka

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MHB Math Helper
Opalg said:
I think it's probably the case that the group is not "a nice algebro-geometric object".
As I see it from an experimental perspective, I neither claim it nice nor trivial. If you believe the later, you must have a reason to do so, I am sure?

Opalg said:
For example, the two points at the ends of the equatorial diameter are idempotent.
Indeed, that is true, and also that the sum of those two end points is the only undefined object on the group.

#### Turgul

##### Member
If you are considering your construction over a field $k$, I use $k^*$ to mean the multiplicative group of the field under multiplication. For example, $\mathbb{Q}^*$ is the multiplicative group of all nonzero rational numbers.

As I said, adding the special cases is not that big of a deal, in principle. As Opalg points out though, the two points on your line intersecting your circle actually prohibit this from being a group at all! The best you can do is give a group structure to the circle minus two points, which is (algebro-geometrically) isomorphic to the line minus one point. As this is an affine algebraic group with the underlying space isomorphic to the line minus a point, the only possible group you can get is the multiplicative group $k^*$. That is to say, if you look at the real points, the group you get is isomorphic to $\mathbb{R}^*$ and if you look at the rational points, you get the group $\mathbb{Q}^*$.

Geometrically that makes some sense. If you are looking at subfields of the real numbers, the top half of the circle corresponds to elements $> 0$ and the bottom half corresponds to elements $< 0$ (adding two numbers on the top half gives something on the top half, and adding two numbers on the bottom yield again something on top).

Fundamentally you are building something out of the equations $x^2+y^2 = 1$ and $y=0$, so you really do get a nice algebro-geometric object (once you remove the relevant points). But as I said, this is not projective, so you can't apply Mordell-Weil (which is good, since $\mathbb{Q}^*$ is not a finitely generated group!). Recall the only one dimensional abelian varieties are elliptic curves.

In terms of generalizing these ideas, it is worth noting that the actual circle you use matters if you want to give a group structure to the rational points though. It so happens that with the circle $x^2+y^2=1$ and the line $y=0$, adding rational points will yield rational points. It is not clear this works for $x^2+y^2 = 5$ (it is also often false that an arbitrary circle will have any rational points at all).

#### mathbalarka

##### Well-known member
MHB Math Helper
Nice answer, both this and the other. +1.

Turgul said:
Recall the only one dimensional abelian varieties are elliptic curves.
Ah, yes, thanks. But (Q, *) is not a very nice object really, so I think Opalg was partly correct when he said so.

Balarka
.