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An algebraic variety $V$ is said to be of general type if it is of maximal Kodaira dimension. If $V$ is defined over a number field $K$, then one has the following conjecture due to Lang (Bombieri had made a similar conjecture in the case of surfaces; thus this conjecture is also known as the Bombieri-Lang conjecture):

If $V$ is a variety of general type defined over a number field $K$, then the set of $K$-rational points $V(K)$ of $V$ is not Zariski dense.

When $\dim V = 1$, this is Mordell's conjecture and was famously solved by Faltings in the 1980s.

Has there been any progress towards Lang's conjecture, in any situation beyond the curve case, since then?

"Progress" does not necessarily mean a proof, but even an outline of what a possible proof might look like or a program towards a proof would suffice. For example, Faltings, in his proof of Mordell's conjecture, had carried out a program that was already known at the time: given Parshin's trick, one shows that Mordell's conjecture is implied by Shafarevich's conjecture, and this is what Faltings actually proves. Is there anything like this beyond the curve case?

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    $\begingroup$ Faltings, in his paper "Diophantine approximation on abelian varieties", has proved another conjecture of Lang which implies that the conjecture you state holds for general type subvarieties of abelian varieties (essentially all subvarieties that are not translates of abelian subvarieties). $\endgroup$
    – naf
    Commented Dec 24, 2020 at 1:01
  • $\begingroup$ My understanding is that there is no single smooth projective variety of general type with finite fundamental group (e.g. large degree hypersurfaces) where the conjecture is known to hold or fail. I have also heard the opinion that 'general type' is too weak an assumption and should be replaced by 'hyperbolic' (of some sort), in which case it becomes more natural to look at quasi-projective varieties (and maybe integral points instead of rational?). I hope someone who knows more about this will comment. $\endgroup$ Commented Dec 24, 2020 at 1:10
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    $\begingroup$ The conjectural equivalence of general type to other notions of hyperbolicity is spelled out in this nice survey of Javanpeykar (arxiv.org/pdf/2002.11981.pdf). If one looks at integral points, then there are results of Corvaja, Levin, and Zannier which tell us that once we remove enough ample divisors then there is no set of integral points which is Zariski dense (see e.g., their paper ``Integral points on threefolds and other varieties"). This work builds on previous work of Corvaja--Zannier where they give a new proof of Siegel's theorem using the Schmidt subspace theorem. $\endgroup$ Commented Dec 24, 2020 at 14:14
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    $\begingroup$ @JacksonMorrow The results of Corvaja-Zannier and Levin were subsequently improved by Autissier (as one should probably mention). $\endgroup$ Commented Dec 28, 2020 at 7:44

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Faltings' second proof extends to subvarieties of abelian varieties $A$. (The exceptional locus consists of the translates of abelian subvarieties of $A$.) That's a very special case, but it includes varieties birational with symmetric powers of curves: as long as $d$ is less than the genus of $C$, the $d$-th symmetric product ${\rm Sym}^d C$ is birational with its image in ${\rm Pic}^d(C)$. Rational points of ${\rm Sym}^d C$ come from points of $C$ defined over a varying field extension $K'/K$ of degree at most $d$.

Besides that, if $V$ has an unramified cover $V'$ then one can do a "descent" to find a finite extension $K'/K$ such that $V(K)$ lifts to $V'(K')$. It may be that $V'$ birationally embeds into an abelian variety even though $V$ did not; for example, if $V$ is the quotient of $C \times C'$ by a finite group acting freely on the product of curves $C,C'$ so that the fixed subspace of $H^1$ is trivial. In particular such varieties satisfy the Bombieri-Lang conjecture. But that, too, is a very special case. There might not be any others known yet.

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It may be instructive to think about what we really "know" in the case of projective surfaces. Let $k$ be an algebraically closed field of characteristic zero.

Let me first recall some basic definitions.

A projective (integral) variety over $k$ is of general type if some (hence any) desingularization of $X$ has big canonical bundle. Also, a projective variety is pseudo-Mordellic over $k$ if there is a proper closed subset $\Delta\subsetneq X$ such that, for every finitely generated subfield $K\subset k$ and every model $\mathcal{X}$ for $X$ over $K$, the set $\mathcal{X}(K)\setminus \Delta$ is finite.

If you can take $\Delta$ to be the empty set, then we say that $X$ is Mordellic over $k$. A projective variety $X$ is Mordellic over $k$ if and only if every subvariety of $X$ is pseudo-Mordellic over $k$.

Lang's conjecture (in its most optimistic form) is that a projective variety over $k$ is of general type if and only if it is pseudo-Mordellic over $k$

Due to the birational invariance of both the notions of general type and pseudo-Mordellicity, you could as well stick to smooth projective varieties in this conjecture.

In the case of curves, Lang's conjecture boils down to showing that a projective curve is of general type if and only if it is Mordellic which is what Faltings achieved in 1983.

You ask whether Faltings's proof gives maybe a bit more than just the case of curves. Well, it certainly does. Namely, if $X$ is a projective variety over $k$ which admits a (quasi-)finite morphism to the moduli stack of principally polarized abelian varieties (ppav's), then every subvariety of $X$ is of general type (explanation omitted) and $X$ is Mordellic (by Faltings's theorem). Thus, any projective variety admitting a finite morphism to some moduli stack of ppav's will satisfy Lang's conjecture. Examples of such projective varieties are, for example, compact Shimura varieties of abelian type.

What about Faltings's 1992 result on closed subvarieties of abelian varieties? Let us assume that $X$ is a normal projective surface over $k$, and let $q=q(X)$ be the dimension of its Albanese variety. If $q>2$, then you can show that Lang's conjecture holds. That is:

Theorem. (Faltings) Assume that $X$ is a normal projective surface with $q>2$. Then $X$ is of general type if and only if $X$ is pseudo-Mordellic.

Sketch of proof. The fact that a pseudo-Mordellic surface is of general type follows from properties of K3 surfaces and the classification of surfaces; see Remark 7.8 in https://arxiv.org/abs/1909.12187 . Now, let $Y$ be the image of the Albanese map $X\to \mathrm{Alb}(X)$. Note that $Y$ is not an abelian variety, as this would contradict $q>2$. Note that $\dim Y = 1$ or $\dim Y =2$. If $\dim Y = 1$, then $Y$ is a hyperbolic curve (because it is not an abelian variety). The morphism $X\to Y$ is then a fibration of hyperbolic curves over a hyperbolic curve which readily implies that $X$ is pseudo-Mordellic. If $\dim Y=2$, use Faltings's 1991 theorem and Ueno's fibration theorem. QED

Now, this means that in the case of surfaces, it remains to prove Lang's conjecture whenever $q=0$, $q=1$, or $q=2$. However, if $X$ has a finite etale cover $X'\to X$ with $q(X') >2$, then $X'$ satisfies Lang's conjecture by the above. Since Lang's conjecture is "invariant under finite etale covers", it follows that $X$ also satisfies Lang's conjecture. (This is also what Noam Elkies is saying in his answer.) Thus, if $\dim X=2$, Lang's conjecture is really open if the augmented irregularity $\widehat{q}(X) $ is at most two. (The augmented irregularity is the supremum over $q(Y)$ as $Y$ runs over all finite etale covers of $X$.)

This suggests that the next "reasonable" case to study is that of surfaces with augmented irregularity equal to two. (The smaller the augmented irregularity, the further you are from any auxiliary abelian varieties. This makes life considerably harder.)

Using Stein factorization, in the case of surfaces with augmented irregularity two, one is reduced to studying finite ramified covers of abelian varieties. The next step we want to achieve is the following special case of Lang's conjecture:

Conjecture. Let $A$ be an abelian surface over a number field $K$ and let $X\to A$ be a finite ramified covering with $X$ a normal integral surface of general type over $K$. Then $X(K)$ is not dense. (Actually: we expect that $X_{\overline{K}}$ is pseudo-Mordellic, but peu importe.)

This non-density statement is not known.

Usually, Lang's conjecture is interpreted as suggesting that there are not so many points on a variety of general type. However, since $A(K)$ is dense (for some large enough number field $K$), one may also view Lang's conjecture in this specific situation as predicting that there are actually many rational points on an abelian variety. (That is, Lang predicts that there are so many rational points on an abelian variety that even after removing those coming from a ramified covering, the points remain dense.)

This brings me to my final point that some progress was made on this aspect of Lang's conjecture recently; see Theorem 1.3 in https://arxiv.org/abs/2011.12840 . A special case of that result reads as follows:

Theorem. Let $A$ be an abelian variety over a number field $K$ with $A(K)$ dense. Let $\pi:X\to A$ be a finite ramified morphism with $X$ a normal integral variety over $K$. Then $A(K)\setminus \pi(X(K))$ is dense in $A$.

This result says that there are "more" points in $A(K)$ than in $X(K$), as Lang predicted.

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