Let $C$ be a smooth (geometrically irreducible) projective curve of genus $g>1$ over a number field $K$. The Mordell conjecture (first proved by Faltings) says that for any finite field extension $L/K$ (say, inside a fixed algebraic closure $\overline{K}$) there is a function $\{$(isomorphism classes of) finite extensions $L/K$ inside $\overline{K}\} \rightarrow\ \{$nonnegative integers$\}$ defined by $L \mapsto \#C(L).$ Does this function determine (the isomorphism class of) $C/\overline{K}$, or otherwise any geometric properties of $C$ such as the genus?

3$\begingroup$ The answer is no for a silly reason, the curves $y^2 = x^5 \pm ix +1$ have the same number of points over any extension of $\mathbb{Q}(i)$. But once one identifies Galois conjugate curves, the question comes back. $\endgroup$ – Felipe Voloch Feb 20 '20 at 18:55

1$\begingroup$ Right, thanks @FelipeVoloch, I had $K=\mathbb{Q}$ in mind when I wrote the question. $\endgroup$ – David Lampert Feb 20 '20 at 19:11

$\begingroup$ Just a remark: the answer is no for affine curves: consider $C\{P\}$ for a projective curve $C$ without automorphisms and 2 different $P \in C(K)$. $\endgroup$ – David Lampert Feb 20 '20 at 19:39

2$\begingroup$ You can try to recover the gonality of the curve by looking at how the number of points behaves over extensions of $K$ of a fixed degree. For instance, lots of quadratic points should imply that the curve is hyperelliptic. $\endgroup$ – damiano Feb 20 '20 at 21:31

3$\begingroup$ @FelipeVoloch I don't think this is true  given a field extension of $\mathbb Q(i)$, applying the nontrivial automorphism of $\mathbb Q(i)$ may produce a different extension. $\endgroup$ – Will Sawin Feb 25 '20 at 18:47
Heuristically, we can compute the genus. This follows essentially the srategy of damiano and David Lampert in the comments.
First note that it is possible to compute from this function the number $\#'C(L)$ of points over $L$ with field of definition exactly $L$. (Here the field of definition is the smallest field containing $K$ that the point is defined over.) This follows from inclusion exclusion.
Now observe that if we let $$ r_d = \lim \sup_{X\to \infty} \frac{\log \sum_{L/K, \Delta_{L/K}<X} \#'C(L) }{\log X} $$
and let $d$ be minimal with $r_d>0$, I think it is reasonable to expect that $d$ is the gonality of $C$ and that $r_d = \frac{1}{ g+ d1}$, so that the genus of $C$ is $\frac{1}{r_d} +1d$.
Why do I think this? Consider first a degree $d$ map $\pi: C \to \mathbb P^1$. Associated to a point of $x\in \mathbb P^1(K)$, of which there are $\approx Y^2$ of height $<Y$, we obtain a degree $d$ divisor in $C$. By Hilbert's irreducibility theorem, for almost all such points, the Galois action on this divisor is transitive, so this divisor defines a point of $C$ with field of definition a degree $d$ field extension $L$ of $K$.
Associated to $\pi : C \to \mathbb P^1$ is a branch divisor counting the ramification of $\pi$ over each point. By RiemannHurwitz, it has degree $2g2+ 2d$. The ramification points of $L$ are at most the intersection points of $x$ with the branch divisor in the scheme $\mathbb P^1_{\mathcal O_K}$, and the ramification index is at most the intersection number, with equality if the intersection is transverse.
Thus the discriminant of $L$ over $K$ is at most the arithmetic intersection number, which is simply a degree $2g2+2d$ polynomial evaluated at $x$, and thus is $O( Y^{2g2+2d})$. It follows that the number of points arising this way with discriminant less than $X$ is at least $\approx X^{ \frac{1}{ g1+d}}$.
Furthermore, for points which are generic in the sense that their intersection with the branch divisor is transverse, and which are not too close to the branch divisor in the archimedean sense, this is sharp, and so it's reasonable to expect that the number of points arising this way with discriminant less than $X$ is $\approx X^{ \frac{1}{ g1+d}}$. This is the first heuristic.
Furthermore, for points arising from covers of an elliptic curve, because the number of rational points on an elliptic curve with height $<Y$ is $O( Y^{\epsilon})$, it's reasonable to expect the number of points with discriminant $<X$ is $O(X^{\epsilon})$. This is the second heuristic.
Combining these, and using Falting's theorem, we see that all the points of degree less than the gonality are explained by finitely many points and elliptic curves and thus $r_d=0$ for $d$ less than the gonality, and for $d$ equal to the gonality we have finitely many covering maps to $\mathbb P^1$, all of which have the same number $X^{ \frac{1}{g+d1}}$ of points, and so $r_d = \frac{1}{g+d1}
It might be possible to extend this idea further to calculate the isogeny class of the Jacobian of $C$ by using the average rank of simple abelian varieties when pulled back to fields $L$ of degree $d$, weighted by $\#'C(L)$, but this would be even more heuristic.

$\begingroup$ Thanks @WillSawin for this interesting answer which I'll try to digest. $\endgroup$ – David Lampert Feb 26 '20 at 0:44