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Let $f\in\mathbb{Z}[X_1,\ldots,X_n]$ be a Diophantine equation which, for the purposes of this question, I will assume is homogeneous and nonsingular on $\mathbb{R}^n\setminus\{0\}$ (so that $\nabla f\not=0$). Supposing that it has infinitely many primitive integer zeros, we can posit that they are smoothly distributed in an asymptotic sense. Writing $V(R)\subseteq R^n$ for the set of primitive solutions to $f(x)=0$ in a ring $R$, the integer solutions $V(\mathbb{Z})$ clearly lie on the manifold $V(\mathbb{R})$. So, I am looking for a density $\rho\colon V(\mathbb{R})\to\mathbb{R}$ with $$ \vert V(\mathbb{Z})\cap U\vert\sim\int_{V(\mathbb{R})\cap U}\rho(x)\,d\sigma(x),\qquad\qquad{\rm(1)} $$ for subsets $U\subseteq\mathbb{R}^n$, where $d\sigma$ is the standard surface integral on $V(\mathbb{R})$. This should hold asymptotically as $U$ is scaled up, and for reasonably regular regions $U$.

My question is regarding a simple (but incorrect -- see below) heuristic argument for calculating $\rho$. Choosing positive integer $N$ and real $a\gg N$ then, for large regions $U$, the set of $x\in U$ with $\vert f(x)\vert < 2a$ has volume about $2a\int_{V(\mathbb{R})\cap U}\Vert\nabla f\Vert^{-1}\,d\sigma$, so should contain about that number of integer points. The probability of a random $x\in\mathbb{Z}^n$ being relatively prime to $N$ and satisfying $f(x)=0$ (mod $N$) is $N^{-n}\vert V(\mathbb{Z}/N\mathbb{Z})\vert$. Conditional on $\vert f(x)\vert < 2a$ and $f(x)=0$ (mod $N$), it seems reasonable to suppose that $f(x)=0$ with probability $N/(2a)$. Multiplying these terms together and taking the limit as $N$ increases to include all prime-powers as factors, we get the following expression for $\rho$. $$ \begin{align} &\rho(x)=\Vert\nabla f(x)\Vert^{-1}\prod_p c_p,\qquad\qquad{\rm(2)}\\ &c_p=\lim_{r\to\infty}p^{-r(n-1)}\left\lvert V(\mathbb{Z}/p^r\mathbb{Z})\right\rvert. \end{align} $$ The product is taken over all primes $p$. This seems like a very neat expression, and can be seen that it gives the correct result for linear equations. However, it is not correct in general. Just looking at quadratic forms for $f$, the expression given by (2) is wrong. I do not have any good feeling as to where exactly this heuristic goes astray, and if it is possible to fix it. Maybe this approach and the reason that it does not quite work is well known. This is not an area in which I am any kind of expert, so maybe others on MathOverflow would be able to help?

For example, consider $f=x^2+y^2-z^2$, so that we are looking for primitive Pythagorean triples. Euclid's parameterization $(x,y,z)=(a^2-b^2,2ab,a^2+b^2)$ can be used to show that $\rho=\sqrt{2}\pi^{-2}\vert z\vert^{-1}$. However, on $V(\mathbb{R})$ we have $\Vert\nabla f\Vert = 2\sqrt{2}\vert z\vert$ and you can calculate $c_2=1$ and $c_p=1-p^{-2}$ for odd prime $p$. Using (2) would lead to $\rho=2\sqrt{2}\pi^{-2}\vert z\vert^{-1}$, which is out by exactly a factor of 2. If we look at Pythagorean quadruples $f=w^2+x^2+y^2-z^2$ instead, then we can calculate $c_p=(1-p^{-1})(1+2p^{-1}1_{\{p\equiv1{\rm\ mod\ }4\}}+p^{-2})$ for odd primes $p$, so the product in (2) is not unconditionally convergent.

Is there a known or, even, just conjectural expression for the asymptotic density $\rho$? And, is it possible to explain precisely how the heuristic used to derive (2) fails?

It would be great if my expression (2) above could be fixed. Heuristics like the one used here are often very useful to understand what the integer solutions to Diophantine equations look like, and it is a bit worrying that it gives the wrong answer in this case. It is also consistent with the idea that find rational solutions to an equation, you should first check for solutions in the completions of $\mathbb{Q}$, according to the Hasse principle. Also, it so nearly works (only being a factor of 2 out for Pythagorean triples) and gives perfectly sensible looking results in many cases, that I am loath to give up and just accept that it doesn't work without a good reason as to why. For example, it does seem perfectly consistent with Falting's theorem (as given in my answer to a previous MO question) and with the Birch and Swinnerton-Dyer conjecture. In the case where $f$ is a a cubic describing an elliptic curve, then $c_p=(1-p^{-1})N_p/p$ for all but finitely many primes $p$, where $N_p$ is the number of $\mathbb{F}_p$-points on the elliptic curve reduced modulo $p$. Then, up to finitely many terms, $\prod_pc_p$ coincides with the Euler product at $s=1$ of $(L(s)\zeta(s))^{-1}$, where $L$ is the L-function of the curve. According to the Birch and Swinnerton-Dyer conjecture, I would expect this to be zero, finite, or infinite when the curve has rank $r=0$, $r=1$ and $r>1$ respectively. Putting this back into (2) is consistent with $\vert V(\mathbb{Z})\cap B_R\vert$ growing at rate $(\log R)^r$, which you would expect for an elliptic curve of rank $r$.

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Here's a possible failure point of the heuristic: if the sets $\{ x : |f(x)| < 2a \}$ are sufficiently irregular there's no reason to expect their volume to be a good estimate for the number of integer points they contain. As JSE says it seems like this might be less of a problem if the number of variables is large relative to the degree. –  Qiaochu Yuan Jun 21 '11 at 2:39
    
Sorry for this very naive comment, but did you take into account that Euclid's parametrization only gives you half of all primitive Pythagorean triples (those with $x$ odd and $y$ even)? –  Dan Petersen Jun 21 '11 at 10:53
    
@Qiaochu: It does seem that it is more likely to work when the number of variables is large relative to the degree. I wasn't sure how large but, for quadratic forms, it seems that you need at least 5 variables. I'm still not sure why it fails for fewer variables. Maybe because, as you suggest, the number of integer points in $\{x\colon\vert f(x)\vert < 2a\}$ differs from the estimate used in the heuristic, whether it is because they are not evenly distributed mod N, or whether conditional on $\vert f(x)\vert < 2a$ and $f(x)=0$ (mod N), the probability that $f(x)=0$ is different from $2a/N$. –  George Lowther Jun 21 '11 at 20:17
    
@Dan: Yes, I was careful to keep track of the various factors of 2, and the answer agrees with the formula given by Lehmer in 1900 for the number of triples with hypotenuse less than N (mathworld.wolfram.com/PythagoreanTriple.html). Also, after a bit of a search using the keywords mentioned by JSE, I now see that it is correct that I was out by a factor of 2. –  George Lowther Jun 21 '11 at 20:52
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3 Answers

up vote 13 down vote accepted

You are on the way to redeveloping the singular series, which does indeed give the correct asymptotic for integral solutions to many flavors of Diophantine equation -- they key words here are "Hardy-Littlewood method" or "circle method," which you can read about in any text on analytic number theory, such as the book of Iwaniec and Kowalski.

Loosely speaking -- when the number of variables is very large relative to the degree of the equation, the singular series is known to give you the right asymptotic. When the number of variables is somewhat large relative to the degree of the equation, it is expected to give the right asymptotic but there are no proofs outside very special cases.

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Thanks! That's very helpful. A bit of searching with those keywords has been very helpful, and my product $\prod_pc_p$ is indeed the singular series. I have found one reference in particular which proves the result for quadratics, where for 3 variables you do need to multiply by 1/2 (so, it is correct that I was out by a factor of 2), and for 4 variables you have to multiply $c_p$ by an extra term to get unconditional convergence. For 5 variables, the heuristic is correct. (dx.doi.org/10.1515/crll.1996.481.149) –  George Lowther Jun 21 '11 at 20:24
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In 1989, Manin and his collaborators formed a series of conjectures on the asymptotic behaviour of the number of solutions to diophantine equations. Let $X\subset \mathbb{P}^n$ be a fano variety (that is $-K_X$ is ample) under its anticanonical embedding, and let $H$ be the associated height function. Then it is expected that there exists a zariski open subset $U \subset X$ such that the number of rational points of height less than $B$ (e.g. the number of solutions in an expanding ball or box) is asymptotic to $c_X B(\log B)^{r_X},$ as $B \to \infty$ for some constants $c_X$ and $r_X$.

This result is true in some cases, for example complete intersections with many variables and small degree via the circle method, quadratic forms, toric varieties, flag varieties and also for some del Pezzo surfaces. But there are counter-examples showing that it is not true for all fano varieties (namely one expects $r_X=\textrm{rank } \textrm{Pic}(X)-1$, but this is not true in general).

At any rate, Peyre formed a conjecture on the leading constant $c_X$ which occurs in the asymptotic formula. It is very close to what you describe. One defines a measure on the set of adelic points on $X$, and then the leading constant is essentially the volume of the closure of $X(\mathbb{Q})$ inside the adeles. This is really an adelic integral and not a real integral, but for suitable varieties (namely those which satisfy weak approximation), the local factors at the primes come out as the $c_p$ in the way you describe. In general though one needs to introduce convergence factors to insure that the product over the $c_p$ converges. These come from an Artin L-function associated to the Picard group.

There are however some extra factors $\alpha$ and $\beta$ present in the constant, related to the position of the anticanonical divisor in the effective cone and the Brauer group of $X$. For conics with a rational point, we have $\alpha=1/2$ and $\beta =1$. This might explain your missing factor of two.

Papers:

J. Franke, Y. I. Manin and Y. Tschinkel, Rational Points of Bounded Height on Fano Varieties. Invent. Math. 95, 421--435 (1989).

E. Peyre, Hauteurs et measures de Tamagawa sur les variétiés de Fano. Duke Math. J., 79(1), 101--218 (1995).

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I should also say that if you want a good overview of what Peyre's constant should look like for hypersurfaces, I would recommend reading Section VI.5 of Joerg Jahnel's habilitation thesis, which can be found on his webpage. –  Daniel Loughran Jun 21 '11 at 11:27
    
Many thanks, that's a very useful answer. I'll have a look through those references. I've already accepted JSE's answer -- unfortunately I can only choose one, and it can be tricky to tell which to accept before properly understanding everything that's being refered to. It sounds like there's some interesting stuff to read about here. –  George Lowther Jun 21 '11 at 20:35
    
However, this concerns rational points of bounded height, and not integral points. General asymptotics for integral points on quasi-projective varieties are not yet conjectured. However, in a recent paper, Yuri Tschinkel and myself developed part of what is needed to state such conjectures in general. In particular, we defined convergence factors and established various volume asymptotics. The reference is Igusa integrals and volume asymptotics in analytic and adelic geometry, Confluentes Mathematici 2 (2010), p. 351–429. arXiv:0909.1568. –  ACL Jun 21 '11 at 22:17
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Searching including the key phrases "Hardy-Littlewood circle method" and "singular series", as suggested by the other answers, turned up some interesting references which shed light on the question and why I obtained the results I did for the cases mentioned. As the question is already quite long and would become rather unmanageable to add this as a large update, I'm adding it as an answer here.

The product $\prod_pc_p$ is indeed called the singular series and $\int\Vert\nabla f\Vert^{-1}\\,d\sigma$ is the singular integral. The product does asymptotically give the number of solutions subject to hypotheses and/or conditions on $f$, generally seeming to work better when the number of variables is large relative to the degree. For quadratic forms, the paper A new form of the circle method, and its application to quadratic forms by D.R. Heath Brown (Journal für die reine und angewandte Mathematik, 1996. Preprint available here) gives expressions for the asymptotic density which show why I obtained the results I did for Pythagorean triples and quadruples mentioned in the question. For a quadratic form $f$ in $n$ variables, they show the following.

  • For $n\ge5$, expression (2) for the asymptotic density given in the question is correct, so the heuristic works! (Theorem 5 of the D.R. Heath Brown paper).
  • For $n=4$ and the determinant of $f$ not a perfect square, you should multiply the terms $c_p$ by $1-\chi(p)p^{-1}$ and the overall expression by $L(1,\chi)$ (Theorem 6 of the paper). Here, the character $\chi$ is the Jacobi symbol $\left({\rm det}(f)\over\ast\right)$. Leaving out these terms will give a product which is not unconditionally convergent, as happened for Pythagorean quadruples.
  • For $n=4$ and the determinant of $f$ a perfect square, the product $\prod_pc_p$ will diverge to infinity (assuming that there are solutions in every $p$-adic field, so $c_p\not=0$). Instead, $c_p$ should be multiplied by $1-p^{-1}$ and the overall density by $\log\Vert x\Vert$ (Theorem 7 of the paper).
  • For $n=3$ then expression (2) given in the question works after multiplying by a factor of $\frac12$ (Theorem 8 and Corollary 2 of the paper)! This is why I was out by a factor of 2 for Pythagorean triples. The D.R. Heath Brown paper has the following to say on this.

    ...It therefore remains to understand the appearance of the factor $\frac12$ in the case $n=3$, which can be thought of as corresponding to a Tamagawa number of 2. In the proof of Theorem 8 this factor arises from the residue at s = 0 of $$\zeta(2s+1)\frac{P^s}{s}.$$

I'm not very familiar with Tamagawa numbers and am not yet sure whether this is the same as the factor $\alpha=\frac12$ mentioned in Daniel's answer or how it comes into the heuristic derivation.

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Similar results can be obtained for certain homogenous spaces of semisimple groups, see my paper with Rudnick math.tau.ac.il/~borovoi/papers/hardy.pdf . Note also that a similar asymptotic formula holds when you count integral points subject to congruence conditions. –  Mikhail Borovoi Jun 22 '11 at 0:45
    
@George. Out of interest there are geometrical reasons why the case $n=4$ is special. Let $X$ be a non-singular projective variety given by a quadratic form in $n\geq3$ variables, over an algebraically closed field. Then $Pic(X) \cong \mathbb{Z}$ if $n \neq 4$ and $Pic(X) \cong \mathbb{Z}^2$ otherwise. The condition on the determinant of $f$ appearing is determining whether or not $Pic(X)\cong \mathbb{Z}$ or $Pic(X) \cong \mathbb{Z}^2$ over the ground field. The appearence of the extra logarithm in the latter case is what I alluded to in my answer. –  Daniel Loughran Jun 22 '11 at 11:01
    
More generally we have $Pic(X) \cong \mathbb{Z}$ if the number of variables is large compared to the degree. It is in these cases where the circle method generally works (although it can sometimes also be made to work when $Pic(X)$ is larger). –  Daniel Loughran Jun 22 '11 at 11:02
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