Conjecturally, there exists an integer $n$ such that the number of rational points of a genus $2$ curve over $\mathbf{Q}$ is at most $n$. (This follows from the Bombieri-Lang conjecture.)

We are very far from proving the existence of such an integer, let alone find an explicit value which works.

My question is:

What is the best known lower bound for $n$?

One way to obtain a lower bound $m$ for $n$ is to prove the existence of a curve of genus $2$ over $\mathbf{Q}$ with at least $m$ rational points.

  • $\begingroup$ I thought the Bombieri-Lang conjecture implied the statement that the number of rational points lying on a genus $2$ curves over $\bf Q$ is bounded by a number depending on the rank of the Mordell-Weil group of the jacobian of the curve (by the work of Caporaso-Harris-Mazur). How would you prove that there is an absolute constant ? Even the conjectural "effective Mordell" does not seem to imply that. $\endgroup$ – Damian Rössler Jul 27 '12 at 21:26
  • $\begingroup$ Theorem 1.1. in "UNIFORMITY OF RATIONAL POINTS" by CAPORASO HARRIS and MAZUR states that, assuming the weak Lang conjecture, there exists a real number $c(K,g)$ such that for all curves $X$ over $K$ of genus $g\geq 2$, the number of $K$-rational points of $X$ is bounded by $c(K,g)$. So conjecturally, there really is a uniform bound on the number of rational points. $\endgroup$ – Harry Jul 29 '12 at 8:35
  • $\begingroup$ @Harry. Thank you for the reference. I wasn't aware of that result. $\endgroup$ – Damian Rössler Jul 29 '12 at 10:07
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    $\begingroup$ "One way to obtain a lower bound m for n is to prove the existence of a curve of genus 2 over Q with at least m rational points." Is there another way? $\endgroup$ – JSE Aug 27 '12 at 3:16
  • $\begingroup$ @DamianRössler You may be thinking of an earlier conjecture of Lang's, which says that the number of integer points on a minimal equation for an elliptic curve is bounded in terms of the rank. This was generalized to the statement you made, which is weaker than the C-H-M result (contingent on Bombieri-Lang). Marc Hindry and I proved an analogue of Lang's elliptic curve conjecture for characteristic 0 function fields. $\endgroup$ – Joe Silverman Nov 18 '15 at 14:07

I believe it is 642. See http://www.mathe2.uni-bayreuth.de/stoll/recordcurve.html

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    $\begingroup$ This is correct (for smooth projective curves). $\endgroup$ – Michael Stoll Nov 18 '15 at 16:44

Here is some more information.

  1. The curve that establishes the current record is obtained from a K3 surface $S$ that was found by Noam Elkies. $S$ is a double cover of ${\mathbb P}^2$ ramified above a smooth sextic $B$ that has lots of tritangent lines and also higher degree curves meeting it with even intersection multiplicity in all points. Pulling back any rational line $L$ in ${\mathbb P}^2$ to $S$ that is not tangent to $B$ gives a genus 2 curve $C$ on which all the tritangents induce pairs of rational points, and the same is true for the higher degree curves when they intersect $L$ in rational points. Of course, there can be additional rational points on $C$ that do not arise in this way. I found the record curve by a systematic search through rational lines of relatively small height. The previous record was 588 points, due to Keller and Kulesz. Their curve is of a special form and has 12 automorphisms defined over $\mathbb Q$; the 588 points come in 49 orbits. By contrast, the new record curve has minimal automorphism group (only the hyperelliptic involution).

  2. The result of Caporaso, Harris and Mazur needs the weak Lang conjecture for varieties of arbitrarily large dimension. (I have seen Bombieri protest against the name `Bombieri-Lang Conjecture', saying that he only ever made the conjecture for surfaces.) We know essentially nothing in this direction beyond what is covered by Faltings' result on subvarieties of abelian varieties (or can be deduced from that).

  3. What is perhaps more convincing is the conjecture that there should be a bound in terms of the genus $g$ and the rank $r$ for the number of (geometric) points on a genus $g$ curve $C$ mapping into a rank $r$ subgroup of its Jacobian (under some embedding given by a base-point on $C$). This follows from the Zilber-Pink conjecture for families of abelian varieties. This of course implies a bound for the number of rational points on a genus 2 curve in terms of the Mordell-Weil rank of its Jacobian. If these ranks are bounded, then one would expect a bound on the number of rational points. But this is another open question. Perhaps the data in this paper might be interesting.

  4. Such bounds do exist (and are even explicit) when the rank $r$ is sufficiently small compared to $g$; concretely for $r \le g-3$ (which unfortunately does not tell us anything about the case $g = 2$). See this and this paper. For hyperellptic curves, the bound can be taken to be $33(g-1) + 8rg \pm 1$ ($+$ when $r = 0$, $-$ otherwise).

  • $\begingroup$ According to sagemath your projective curve with many points is not smooth. Is this bug in sage? It claims the curve is not smooth at $(0:1:0)$. $\endgroup$ – joro Oct 15 '17 at 9:37
  • $\begingroup$ @joro The closure of the affine curve in the standard projective plane is not smooth, indeed. If you take the closure in the weighted projective plane with weights 1 for $x$, 3 for $y$ and 1 for the "new" variable, you obtain a smooth projective model of it. If the equation is $y^2 = f(x)$, then the other affine chart is given by $u^2 = v^6 f(1/v)$. You can find more about this in mathe2.uni-bayreuth.de/stoll/teaching/ArithHypKurven-SS2014/… (these are some lecture notes of mine). $\endgroup$ – Michael Stoll Oct 15 '17 at 10:36
  • $\begingroup$ Thank you. Can some CAS do the computations for me for similar curves? $\endgroup$ – joro Oct 15 '17 at 10:51
  • $\begingroup$ @joro You can use my "ratpoints" (which is also available in SAGE and Magma) to find the rational points up to height (= maximum of absolute value of numerator and denominator of the $x$-coordinate) $10^5$ or perhaps $10^6$. To find the larger points on the record curve, I used a supplementary program that uses known points to try to find new ones. $\endgroup$ – Michael Stoll Oct 15 '17 at 12:02
  • $\begingroup$ I meant CAS to tell me if the weighted projective curve is smooth. $\endgroup$ – joro Oct 15 '17 at 12:05

The way the question is formulated, the answer is unbounded, since rational singular points can be unbounded.

For natural $n$, define $j(n)=\prod_{i=1}^n(x-i)$.

Consider the curve $C_n : j(n)^2(x^5+13)=y^2$.

It is birationally equivalent to $x'^5+13=y'^2$ which is genus $2$, so $C_n$ is genus $2$.

$C_n$ has the rational (singular) points $(1,0),(2,0),\ldots(n,0)$ which are unbounded, since $n$ is unbounded.


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