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Often number theorists can bound the number of solutions to a diophantine equation based on the size of the points and the size of the coefficients. But this, as I understand it, can be a bit of a red herring - the genus of a curve is what is really controlling the number of points.

What I would like is some intuition as to why. I must admit that I don't understand genus very well, but perhaps someone can illustrate (in a somewhat elementary fashion) why genus is the right tool for bounding solutions.

anon
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  • The answer probably has something to do with Falting's Theorem - https://en.wikipedia.org/wiki/Faltings%27s_theorem. This is a pretty hard theorem, so it might be tough to give an elementary answer to this question. – Paul Siegel Oct 29 '15 at 18:14
  • Relevant: http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/ – dhy Oct 29 '15 at 18:22
  • Why is it always considers only 2 variables? – individ Oct 29 '15 at 18:31

1 Answers1

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Even for curves, the situation becomes clearer if you also allow affine curves and look at integral points, or more generally, $S$-integral points in number field. If the curve is projective, that's the same as looking at rational points. Also, it's better to look at Euler characteristic $\chi(C)$ (so as to get integer values), so

$\chi(\text{projective curve of genus $g$ with $n$ points removed}) = 2-2g-n.$

Then we have

$\chi(C)=2$ for $\mathbb P^1$, which is projective and has lots of points.

$\chi(C)=1$ for $\mathbb A^1=\mathbb P^1\setminus\{pt\}$ is the additive group, so also has lots of points.

$\chi(C)=0$ means that $C$ has the structure of a group, either an elliptic curve (which is projective) or the multiplicative group $\mathbb G_m=\mathbb P^1\setminus\{pt_1,pt_2\}$. In both cases, the set of integral points forms a finitely generated group, so is often infinite, but much sparser than the $\chi(C)>0$ cases.

$\chi(C)<0$ means $C$ is either a projective curve of genus at least 2, or an elliptic curve minus one or more points, or $\mathbb P^1$ minus at least 3 points. In all three cases, there are finitely many integral points. These three major theorems are due, respectively, to Faltings, Siegel, and Siegel, although the last one (which is solutions to the unit equation) follows for units in number fields from Thue's theorem.

So for quasi-projective curves, there fairly coarse Euler characteristic determines the qualitative behavior of the integral points.

In higher dimensions, one has similar conjectures, where the Euler characteristic is replaced, roughly, by the extent to which the canonical bundle is ample, or anti-ample, or 0, or something "in between". This is all made quite precise in Vojta's conjectures. However, in higher dimensions one only gets statements that apply to all points outside of a proper Zariski closed subset.

Summary: the intuition is that "as the geometry of the variety gets more complicated, the integral/rational points get more sparse."

Joe Silverman
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  • So here is what I am taking away, maybe you can correct me if I am wrong. Height is a pretty simple measure of the complexity of a number. If an equation has large coefficients, it is pretty complicated and so if it is unlikely to have simple solutions (solutions of low height). A better measure of the complexity of an equation is something geometric, like genus. Integers, and by extension rationals, are the least complicated numbers, so in the same vein as above, they are unlikely to live on curves of higher genus? – anon Oct 29 '15 at 19:38
  • One thing that puzzles me: if I think of integers as being simple, that is a more algebraic or arithmetic fact. Why is it that because a curve is geometrically complex, it should also be arithmetically complex? – anon Oct 29 '15 at 19:40