# Local heights in Vojta's conjecture

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?

• At least for arithmetic surfaces, the contribution of the Neron functions is not in anyway obviously functorial. I have never seen anywhere in literature where this is proved. – Bombyx mori Jun 18 at 17:36
• I see. I had in mind Theorem B.8.1 in Hindry-Silverman, which says $\lambda_{\phi^\ast D,v} = \lambda_{D,v} \circ \phi + O_v(1)$. I want to apply this to $\phi : C \to X$ the embedding of $C$. Of course if $C$ has high degree, then $\phi^\ast H$ is $\mathcal O_{\mathbb P^1}(d)$ for some very large $d$, but you get the same $d$ term on the $h_H(P)$. And the constant term $O_v(1)$ should wash out since $h_H(P)$ can be large. – user142054 Jun 18 at 17:43
• I did not know about this. But I suspect their result is for finite places, not for places at infinity. You can still talk about other heights like Neron-Tate height, Faltings' height, etc, which does behave nicely as they are in some sense global invariants. The issue with the Neron functions is that they are defined with respect to the canonical metric (or Peterson metric if you prefer), and estimating Green functions effectively is a big issue. Papers to look up are the ones by Jorgenson and Kramer in Composito. But that paper showed up around 2004 or later. – Bombyx mori Jun 18 at 17:56
• I'll take a look, thanks. I don't think they mention that it must be a finite place, but in any case I am happy to only include finite places in my $S$, so I don't think that's the source of my problems. – user142054 Jun 18 at 18:11
• I actually looked up Hindry-Silverman, and I realized they defined it slightly differently. I will read your post again to see what your issue is. – Bombyx mori Jun 18 at 18:12

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.
• Ah, thank you! In case someone else stumbles upon this, my issue appears to be that the Weil height machine works at the level of Pic(X), but the local ones really do need Div(X). So my computations of the local heights on $\mathbb P^1$ were mistaken, as I assumed I could move $\phi^\ast H$ to be supported at a point. – user142054 Jun 18 at 18:28
• That's right. The local ht $\lambda_{H,v}(P)$ is essentially $$-\log(\text{v-adic distance from P to H}).$$ So it indeed depends on $H$, and not on the linear equivalence class of $H$. But when you add over all $v$, that dependence more-or-less goes away because of the product formula. – Joe Silverman Jun 18 at 18:33