The Rational Circle.

*Huh*...I like the sound of that.

It vaguely conjures up the image of a clandestine cult, perhaps akin to the Pythagorean Brotherhood (*), encircling an altar of smooth grey slate; each crimson robed member up-raising a dragon's

*fang*of

*chalk*in fearful exultation, only lowering these totems for the transcription of mystical symbols of power.

In short: a typical group of mathematicians sitting around a blackboard.

For the past year, a couple of friends and I have been going through Deligne's proof of the Weil conjectures - a

*finite fields*version of the Riemann hypothesis. We've done Weil I, and we've gone through parts of Weil II and now (most of us) have Weil-fatigue (okay, sure, it should probably be Deligne-fatigue).

So we're currently looking at four potential new topics, one of which will be selected for this coming year: geometric Langlands, Cobordismy stuff (in particular, Madsen Weiss), (Homological) Mirror Symmetry and Topological Recursion (i.e.: Eynard-Orantin invariants and stuff). Oh, and I guess there was a fifth suggestion - we thought about sitting down and proving the Volume Conjecture, but decided that it wasn't exciting enough. =P

And in amidst all the discussion about the merits of each of these topics we started (as you do, assuming that your life is like American television) to reminisce all the good times that we had in the Weil conjecture series. Like that very first session when Norman Do, for whatever reason, gave us a pretty sweet description of the rational points of the unit circle in a passing remark. A description so sweet that we immediately distracted ourselves for a minute or so proving it.

Claim: the rational points on the unit circle \(\mathbb{S}^1\), i.e. the coordinates are fractions (of integers), may all be written in the form:\begin{align}(\frac{u^2-v^2}{u^2+v^2},\frac{2uv}{u^2+v^2}),\,u,v\in\mathbb{Z}.\end{align}

*Huzzah*.

In fact, we can easily see that any point of such a form is a rational circle point, unless if both \(u\) and \(v\) are zero. But I didn't really wanna type all that in the claim because MathJax makes words that you type in mathmode look funny.

Whatevs man, let's just prove this jazz.

To begin with, by the unit circle \(\mathbb{S}^1\), I mean the set:\begin{align}\mathbb{S}^1:=\{(x,y)\in\mathbb{R}^2\,:\,x^2+y^2=1\}.\end{align}

Now, consider the following map from \(\mathbb{S}^1\) to the real projective line \(\mathbb{R}\mathrm{P}^1\) given by:\begin{align}\psi:\mathbb{S}^1\rightarrow\mathbb{R}\mathrm{P}^1,\,(x,y)\mapsto[y:1-x].\end{align}

Heeeey, whaddya know - this is precisely the stereographic projection map of \(\mathbb{S}^1\) from \((1,0)\) onto the \(y\)-axis (plus an extra point at \(\infty\) for the projection of \((1,0)\) itself). So \(\psi\) is totally a bijection.

Now, it's clear that \(\psi(x,y)=[y:1-x]\) can be written in the form \([u:v]\in\mathbb{Z}\mathrm{P}^1\subset\mathbb{R}\mathrm{P}^1\) if \((x,y)\) is a rational circle point. In fact, for any \([u:v]\in\mathbb{Z}\mathrm{P}^1\), the inverse point \(\psi^{-1}[u:v]\) is a rational circle point!

Let's check this by bashing stuff.

So we're looking for some \(x,y)\) such that \([u:v]=[y:1-x]\). Therefore, we have that \(u(1-x)=vy\) and shoving this into \(x^2+y^2=1\) yields that:\begin{align}(u^2+v^2)x^2-2u^2x+u^2-v^2=0.\end{align}This quadratic yields the solutions that:\begin{align}x=\frac{u^2\pm v^2}{u^2+v^2}=1,\frac{u^2-v^2}{u^2+v^2}.\end{align} The case that \(x=1\) is covered by the latter (via \([u:v]=[1:0]\) ) and hence may be excluded.

Repeating the same calculations for \(y\), we get that\begin{align}y=0,\frac{2uv}{u^2+v^2}.\end{align}As before, the case that \(y=0\) is covered by the latter case (via \([u:v]=[\pm1:0]\) ) and may be excluded.

Oh look, not only did we show that the stereographic projection map \(\psi\) gives us a bijection between the rational circle points and \(\mathbb{Z}\mathrm{P}^1\), but we also proved our initial claim.

*-**Insert obligatory proof-ending blank square*

Going back to my story: I think that the reason that Norm brought up this example was to demonstrate to us the following guiding principle: rational solutions to polynomials are all about number theory.

As an example: we've

*just*characterised the rational solutions to the polynomial \(x^2+y^2=1\) as being of the form:\begin{align}(\frac{u^2-v^2}{u^2+v^2},\frac{2uv}{u^2+v^2}).\end{align}And anyone's who's played around with number theory will immediately recognise these as being related to the general form for any Pythagorean triad:\begin{align}(u^2-v^2,2uv,u^2+v^2).\end{align}Which, by the way, is hardly surprising once you think about rescaling Pythagorean triangles so that its hypotenuse is length \(1\).

What might be surprising, is that we started off with a rant about the Pythagorean Brotherhood, wandered about and wound up back at Pythagoras - effectively tracing out a big circle of thought, a

*rational circle*, if you will. =P

Since it's pretty hard to finish on a circle, let's shoot off on a tangent. This characterisation of the rational points of the unit circle also comes as a consequence of Hilbert's Theorem 90 which is a pretty early example of how number theory may be phrased in terms of (and resolved via) cohomology theories.

Sometimes I worry about this.

Sometimes I worry one day we'll show that pretty much everything in maths may be phrased in terms of cohomologies...I mean, cohomology is one of the most straight forward ways that we can extract invariants from stuff...and to be honest, we usually can't do very much in maths apart from the most obvious things. So, who knows...may be my worries are...semi-quasi-pseudo-meta-justifiable?

Who knows, who cares: we thought in a big loop and went off on a perpendicular, thereby forming a Q - that is: Q for quitting.

Let's quit here.

I

*am*become tired and (more) incoherent (than usual).

*: which, despite its name, did include women.

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