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Suppose $V$ is a projective algebraic variety over $\mathbb{Q}$ which does not contain any rational curves. Is it true/provable that the number of rational points on $V$ of Weil height at most $T$ grows slower than $T^{\delta}$ for any $\delta>0$?

I'm not sure what is implied by all the variations of Lang's conjectures out there, so clarifications on this would also be much appreciated.

Thanks!!

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  • $\begingroup$ I take it your height is not logarithmic, i.e. the height of the point $n$ on $\mathbb P^1$ is $n$ and not $\log n$? $\endgroup$
    – Will Sawin
    Commented Jul 30, 2017 at 13:34
  • $\begingroup$ One case to consider for this problem is that of Calabi-Yau varieties without rational curves. These are at least conjecturally the furthest you can get from general type without containing a rational curve. Their existence is discussed here mathoverflow.net/questions/69716/… The examples mentioned are quotients of abelian varieties, so maybe there is hope of verifying this for these, although the statement for the quotient does not follow from the known statement for the variety. $\endgroup$
    – Will Sawin
    Commented Jul 30, 2017 at 19:36
  • $\begingroup$ "Does not contain any rational curves" over $\bf Q$ or over $\overline{\bf Q}$? If you mean the former, there may already be many examples among cubic surfaces, though it's probably intractable to prove that there is even one such surface with a rational point and no rational curves. $\endgroup$ Commented Jul 30, 2017 at 22:40
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    $\begingroup$ @Noam D.Elkies: Any smooth cubic surface over $\mathbb{Q}$ with a rational point contains many rational curves: just take the tangent hyperplane to a general rational point. $\endgroup$ Commented Jul 31, 2017 at 7:22
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    $\begingroup$ Surely not a cubic threefold because once you have a rational point you have a hyperplane section that's a cubic surface with a rational point and then @DanielLoughran's observation applies. $\endgroup$ Commented Aug 1, 2017 at 0:36

2 Answers 2

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My guess is that this is true, however there is basically no hope in proving this at all in general. The most relevant conjecture here is Manin's conjecture [1].

As an example, take $S$ to be a smooth surface of general type. Then Lang's conjecture predicts that the rational points on $S$ are not Zariski dense. In particular, the closure $Z \subset X$ of the set of rational points consists of a finite union of curves on $S$. The number of rational points on $S$ of height at most $T$ which do not lie on a rational curve is therefore $O( (\log T)^{r/2})$, where $r$ is the maximum of the Mordell-Weil ranks of the elliptic curves contained in $Z$. In particular what you want should be true in this case.

However Lang's conjecture is only known in some very special cases. E.g. it is unknown for smooth surfaces of degree $d \geq 5$ in $\mathbb{P}^3$. For such surfaces one has no idea how to prove sub-polynomial growth away from the rational curves. The best known method for getting upper bounds here is the determinant method of Bombieri-Pila/Heath-Brown/Salberger, however this gives good results for the "worst case", so it does not know the difference between a smooth surface and a singular one. (It can be used to throw away rational curves, but will still give polynomial bounds for the remainder).

The case of K3 surfaces (e.g. smooth quartic surfaces in $\mathbb{P}^3$), is very interesting as here the rational points are expected to be Zariski dense as soon as there is a single rational point. Here Manin's conjecture [1,Thm. 3.5] predicts that for all $\delta > 0$, there exists an open dense subset $U_\delta \subset S$ such that $U_\delta$ contains $O(T^\delta)$ rational points of height at most $T$. Note that in the conjecture as $\delta$ decreases it says that you need to throw away more and more curves (as $U_\delta$ depends on $\delta$). However if $S$ contains no rational curves, then you see that at each step you are only removing finitely many curves of positive genus. From this one easily sees that the result you want follows from Manin's conjecture in this case. However again no one can prove this conjecture for any K3 surface!

[1] - Batyrev, Manin - Sur le nombre des points rationnels de hauteur borné des variétés algébriques

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I found it in an a paper of Mckinnon: https://arxiv.org/pdf/1011.5825.pdf

Apparently this is known as the "rational curve conjecture" and Mckinon almost proves that it follows from Vojta's conjectures. Specifically, assuming Vojta's conjecture he proves that any variety $X$ of non-negative Kodaira-dimension has an open subset $U\subset X$ containing sub-polynomially many points. If one then assumes that all varieties with negatvie Kodaira dimensio are ruled( apparently a conjecture of the minimal model program) one almost gets the result (you get rational curves, but perhaps not defined over $\mathbb{Q}$.

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