# Rational points on surfaces of general type

The weak Lang conjecture asserts that rational points on a variety of general type defined over $\mathbb{Q}$ are not Zariski dense (same replacing $\mathbb{Q}$ with a number field). This one is proved in dimension $1$, and in dimension $2$ it would in particular mean that the rational points of surface of general type are contained in the union of a finite number of rational and elliptic curves, plus a finite set of "isolated points".

It seems natural for me that a surface of general type has "less points" because the polynomials that define it are of "higher" degree. However, is there some more precise evidence for this conjecture ?

And in which particular cases is the conjecture known to be true?

I am mostly interested in the case of surfaces, but higher dimension case is also interesting.

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A Theorem of Faltings states that any proper subvariety of an abelian variety has finitely many rational points provided this subvariety does not contain a translate of any non-trivial proper abelian subvariety:

G. Faltings, The general case of S. Lang’s conjecture, pages 175–182 of Barsotti Symposium in Algebraic Geometry (Abano Terme, 1991), Perspect. Math. 15, Academic Press, San Diego, CA, 1994.

Harris and Silverman use this to show that if C is curve of genus $\ge 2$ that is neither hyperelliptic nor bielliptic, then the set of rational points on $C^{(2)}$ is finite. Here $C^{(2)}$ is the symmetric square of $C$.

J. Harris and J. H. Silverman, Bielliptic curves and symmetric products, Proceedings of the American Mathematical Society 112 (1991), no. 2, 347–356.

My understanding is that if $C$ has genus $3$ (say a plane quartic) then $C^{(2)}$ is a surface of general type.

I admit however that symmetric powers of curves are really special and probably not what the OP is hoping for.

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In other words, Faltings's Theorem treats the case of varieties of general type which can be embedded in an Abelian variety. –  ACL Mar 7 '13 at 13:50
Even if the variety of general type just admits a non-constant morphism to an Abelian variety, that can be enough. Thus a variety that fibers a hyperbolic curve cannot have a Zariski dense set of rational points, since the rational points must lie in the finitely many fibers over the finitely many rational points of the base curve. –  Jason Starr Mar 7 '13 at 16:13
Thanks for the answer, I like it. However, I would like to have other results. For example if you take hypersurfaces of $\mathbb{P}^3$ of degree $d>4$, with singularities which are not too bad (so that it is of general type), can we say something? And apart from the proofs in some particular cases (which is the second question), is there a reason a priori why general type should imply non-Zariski dense rational points? (first question) –  Jérémy Blanc Mar 7 '13 at 17:55
Dear Jeremy: I think this was the insight of Serge Lang to relate positivity properties of the canonical bundle, hyperbolicity and rational points in higher dimensions. He wrote a paper on that subject, projecteuclid.org/euclid.bams/1183553166 –  ACL Mar 7 '13 at 23:47
Thanks Antoine for the comment and the reference, I downloaded the article and will read it. –  Jérémy Blanc Mar 8 '13 at 5:31