# Is the geometry of a variety determined by the counts of rational points?

In Diophantine Geometry: An introduction, Hindry and Silverman write "Geometry Determines Arithmetic" (pg. 2) and "Geometry Governs Arithmetic" (pg. 474).

On pg. 211 of the same book, the authors state the following theorem:

Let $k$ be a number field, let $C/k$ be a smooth curve of genus $g$, and assume that $C(k)$ is not empty. Then there are constants $a$ and $b$, which depend on $C/k$ and on the height used in the counting functions, such that

$N(C(k, T)) \sim aT^b$ if $g = 0$, here $a,b >0$

$N(C(k, T)) \sim a(log(T))^b$ if $g = > 1$, here ($a > 0$ and $b \geq 0$)

$N(C(k, T)) \sim a$ if $g \geq 2$

Here $N(C(k, T))$ counts $k$ rational point of height $\leq T$.

Because one can find a rational point on a genus $0$ curve by passing to a finite extension of the base field and one can find a point of infinite order on a genus $1$ curve by passing to a finite extension of the base field, one sees that the topology of the curve is determined by the asymptotic numbers of rational points in number fields.

In the section with the heading on pg. 474, the authors state the Bombieri-Lang conjecture:

Let $X$ be a variety of general type defined over a number field $k$. Then there is a Zariski open subset $U$ of $X$ such that for all number fields $k'/k$, the set $U(k')$ is finite.

For curves $X$ this is just the Mordell conjecture (since a curve is of general type if and only if its genus is 2 or greater, and Zariski open subsets of curves are just complements of finite sets of points.)

The fact that the converse to the Mordell conjecture is true (in the sense that finiteness of rational points in all finite extensions implies that the curve is of genus 2) suggests that the same might be true for varieties of higher dimensions.

Is the converse to the Bombieri-Lang conjecture stated above true?

More generally,

Let $S$ be the set of varieties of dimension $d$ defined over number fields. To what extent is the geometry of an element of $S$ determined by the asymptotics of the functions that counts $k$ rational points for number fields $k$? (Possibly after passing to Zariski open subsets, etc.)

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For general conjectures in this direction, see the work of F. Campana, e.g., Bull. Belg. Math. Soc. Simon Stevin 13 (2006), no. 5, 827–842. – Felipe Voloch Oct 4 '12 at 18:43
For surfaces, one could just brute force this using the Enriques-Kodaira classification. It is clear that rational surfaces have Zariski dense sets of $K$-rational points in some fields $K$. For a ruled surface, one can find a number field such that $K$-rational points are dense in arbitrarily many fibers, which is sufficient. For elliptic surfaces one can do the same. One can get abelian surfaces, and thus hyperelliptic surfaces, by picking a non-torsion closed point outside of any codimension 1 subset and looking at the variety over its residue field. I don't know how to do $K3$ surfaces. – Will Sawin Oct 5 '12 at 2:09
I'm just going to second Felipe's comment that Campana is the guy who has thought about this very carefully; specifically, he has thought about closely about what the geometric criteria for potential Zariski density of rational points might be. – JSE Oct 5 '12 at 2:27