I am a complex geometer trying to parse Vojta's conjecture on rational points, and I have a very basic misunderstanding (I apologize if this is too easy for MO).


Let $X$ be a variety over a number field. The conjecture states that for all $P$ outside of some Zariski closed subset,


$\sum\_{v \in S} \lambda\_{D,v}(P) + h\_{K\_X}(P) \leq \epsilon h\_H(P) + C$,


where $D$ is an effective divisor, the $\lambda\_{D,v}$ are a finite set of local heights, the $h$'s are Weil heights, $H$ is an ample class, and $\epsilon$ and $C$ are constants.


Here is my confusion. Let's say $X$ is a K3 surface, and take $D = H$. The conjecture is $\sum\_{v \in S} \lambda\_{H,v}(P) \leq \epsilon h\_H(P) + C$, which means that
$\frac{\sum\_{v \in S} \lambda\_{H,v}(P)}{h\_H(P)} \leq \epsilon + \frac{C}{h\_H(P)}$.


Suppose that $C \subset X$ is a rational curve, also defined over our number field. We can pull back the above inequality to $\mathbb P^1$ by functoriality of the various height functions. Now, I don't see why the left side is supposed to be small at all for every $P$, never mind less than epsilon. Aren't there points on $\mathbb P^1$ for which most of the height is accounted for by local heights in $S$? (For example, say $S$ contains only the $2$-adic absolute value, and consider points whose coordinates are powers of $2$.) I would expect that this inequality fails for a dense set of points on $C$ (for any $\epsilon < 1/2$, say).


This itself isn't obviously a problem: $C$ would just need to be in Zariski closed set that we exclude. But $X$ might have infinitely many rational curves defined over the number field, and the argument applies to all of them. And this would contradict the conjecture. So I suspect I am misunderstanding what local heights are on $\mathbb P^1$. Can someone straighten me out?



Local heights in Vojta's conjecture
If your rational curve $C$ intersects $H$ in three or more points, then you in fact won't be able to find infinitely many points in $C(\mathbb Q)$ whose height is entirely (or even mostly) coming from the finitely many places in $S$. For example, if the intersection is 3 points, you'd more or less need lots of solutions to $Au+Bv=C$ with $u$ and $v$ being $S$-units. Also, Vojta says you have to discard a Zariski closed set. So you'd need to discard all rational curves on your surface that intersect $H$ in only one or two points, but there are only finitely many of those. 


