6
$\begingroup$

Let $V/\mathbb{Q}$ be a subvariety of $\mathbb{P}^n$. There are many plausible choices of height function, some differing only by constant factors: $\max |x_i|$ (for $(x_0,x_1,\dotsc,x_n)$, $\gcd(x_1,\dotsc,x_n)=1$, representing a given point $P$), or also $\sqrt{\sum_i |x_i|^2}$, etc.

Given a height function $H$, we can define $N_H(V;x)$ to be the number of points $P$ in $V(\mathbf{Q})$ with $H(P)\leq x$. For $H_1\ll H_2\ll H_1$, we clearly have $N_{H_1}(V;x)\ll O(1)^n N_{H_2}(V;x)$ and $N_{H_2}(V;x)\ll O(1)^n N_{H_1}(V;x)$. Could the behavior of $N_{H_1}(V;x)$ and $N_{H_2}(V;x)$ nevertheless be qualitatiely different? For example -- could it be that $N_{H_1}(V;x)$ has an asymptotic, whereas $N_{H_2}(V;x)$ does not (i.e. it oscillates)?

$\endgroup$
0

1 Answer 1

5
$\begingroup$

Surely this behaviour can never happen, but it will be near impossible to prove this. Conjectures of Manin and others predict that there is an asymptotic formula for these functions in many cases, and the shape of the asymptotic only depends on the choice of embedding and not the choice of norm. (Changing the norm just changes the leading constant in the asymptotic formula).

A nice trick however is that if you can prove an asymptotic for a "dense" set of norms, then you get an asymptotic for all norms via standard approximation arguments. In particular, e.g. it suffices to prove an asymptotic for all norms which are given by infinitely differentiable functions.

$\endgroup$
2
  • $\begingroup$ What about norms that do not differ from each other by a constant factor? Surely one can have a situation where $\max_{i\ne i_0} |x_i|$, say, is a sensible norm (e.g., a finite number of points will be of bounded norm), yet the norm of a point may vary by more than a constant factor depending on the choice of $i_0$. This is not a problem in the case of elliptic curves, say, but it still requires a bit of work even then. $\endgroup$ Commented Nov 28, 2020 at 15:24
  • $\begingroup$ I don't understand your comment. It is a general fact that for heights $H_1,H_2$ associated to the same very ample line bundle (= same embedding) satisfy $H_1 \asymp H_2$. So if $H_1$ satisfies the Northcott property then so does $H_2$. $\endgroup$ Commented Nov 28, 2020 at 15:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .