This might be an easy one. Motivated by this answer. I would like rational pointts on the unit sphere of bounded height: $$ \{ (x,y,z)\in \mathbb{Q}^3: x^2 + y^2 + z^2 = 1 \}$$ This seems straightforward since I could take any three integers $(a,b,c) \in \mathbb{Z}^3$ and re-scale to put them in the unit sphere. However, $$ \frac{(a,b,c)}{\sqrt{a^2 + b^2 + c^2}} \notin \mathbb{Q}^3 $$ Instead if I have a rational point on the sphere I could clear denominators and get an integer solution to: $$ a^2 + b^2 + c^2 = N^2 $$ just the same way rational points on the circle are related to Pythagorean triples.

The standard solution is to start with the point $(0,0,1)\in \mathbb{Q}^3$ since: $$ 0^2 + 0^2 + 1^2 = 1^2 $$ and to introduce a line of rational slope cutting through the sphere. And get more rational points. : $$ L = t\,(0,0,1) + (1-t) (a,b,c)$$

I forget the exact formula. And how do we control the height this way?

In modern language this could be in terms of hyperplane divisors, but the question I am asking here is quite ancient and basic.

  • 4
    $\begingroup$ Can you clarify whether you are now happy? If so, it would be best to cut out the last section of the question and turn it into an answer. (You are allowed to answer your own questions.) If you are not happy, please explain what more you want to know. $\endgroup$ – Neil Strickland Nov 25 '16 at 13:40

This is not the answer, but this elementary construction is related to the question and rather nice to be mentioned.

In the paper "Cubes in an Integer Lattice" (Mathematics and Informatics Quarterly, 1993, 3, 85-89) Horozov gave a parametrization of all mutually perpendicuar vectors $A_j=(x_j,y_j,z_j)$ $(j=1,2,3)$ of equal length (which is $k(a^2+b^2+c^2+d^2)$) \begin{align} x_1=&2k(ac-bd), & y_1=&k(a^2-c^2+d^2-b^2),& z_1=&2k(ab+cd),\\ x_2=&2k(ad+bc), & y_2=&2k(ab-cd),& z_2=&k(b^2-a^2+d^2-c^2),\\ x_3=&k(c^2-a^2+d^2-b^2), & y_3=&2k(ac+bd),& z_3=&2k(bc-ad),\\ \end{align} where $a,b,c,d$ are integers, not all $0$ and $k$ is a positive ineger.

This problem is equivalent to the problem of parametrization of the equation $$x^2+y^2+z^2=0,$$ over Gaussian integers $x=x_1+ix_2$, $y=y_1+iy_2$, $z=z_1+iz_2$.


(unusual move to answer my own question)

In fact this is duplicate:

and the rational points are of course easy to find

$$ x=\frac{2u}{u^2+v^2+1};y=\frac{2v}{u^2+v^2+1};z=\frac{u^2+v^2-1}{u^2+v^2+1} $$

There is even a proof of equidistribution as a result of Duke He shows that:

$$ \frac{\sum_{h(x)\leq T} \psi(x)}{\sum_{h(x)\leq T} \;\;1\;\;} \to \int_{S^2} \psi \, d\mu $$ for any function $\psi: S^2 \to \mathbb{C}$. There is also general discussion of Bjorn Poonen in genral on rational points on varieties

In light of the comments it's worth noting the Manin-Peyre conjecture which discusses rational points on Fano varieties. I don't see how anyone can study rational points on a manifold with no explicity formula. At least, observe that: $$ X^3 + Y^3 + Z^3 = 3 \, T^3$$ does not satisfy weak approximation yet it's a cubic surface in $\mathbb{P}^3$ and therefore Fano.

The full definition of Fano is really fancy involving the anti-canonical divisor and ample line bundles, which I doubt is how Fano or del Pezzo discussed them.

And here is more references:

  • 2
    $\begingroup$ In fact, there is a framework (Manin-Peyre's conjecture) that proposes a conjecture and theorems in a large generality. Your question is a particular case of counting the points of P^3 lying on a quadric x^2+y^2+z^2-t^2=0 (and probably P^2 in this case). This falls under the scope of Franke-Manin-Tschinkel's paper (1989) completed by Peyre (1995). The proof of FMT uses Eisenstein series. Too bad the paper of Duke does not mention this. $\endgroup$ – ACL Nov 25 '16 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.