Fresh out of the arXiv press is the remarkable result of Manjul Bhargava saying that most hyperelliptic curves over $\mathbf{Q}$ have no rational points. Don Zagier suggests the paraphrase : Most hyperelliptic curves are pointless.

Crucial to the precise mathematical formulation of the statement is a kind of canonical equation for hyperelliptic curves (of a fixed genus) permitting one to define the density of those which have no rational points.

What is the corresponding statement for all curves over $\mathbf{Q}$ ?

Addendum (2013/09/28) A very nice introduction to the work of Bhargava can be found in How many rational points does a random curve have? by Wei Ho.

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    $\begingroup$ Presumably you'll want to embed the curves into $\mathbb{P}^{g-1}$ via the canonical embedding. Whether one can prove anything in the non-hyperelliptic case is another matter. $\endgroup$ – Michael Zieve Aug 5 '13 at 12:56
  • $\begingroup$ I had used the word "canonical" advisedly, and I'm looking for a statement in terms of the canonical embedding. I'm not asking for a proof, ... $\endgroup$ – Chandan Singh Dalawat Aug 5 '13 at 15:09
  • $\begingroup$ I just saw this. I had also asked this question at mathoverflow.net/questions/121104/… . To state the obvious, in any flat family $X \to B$ of geometrically irreducible affine curves over $\mathbb{Q}$ (over an integral base $B$), one of the following alternatives should hold: (1) there is a section to $X \to B$; or (2) the pointless curves in the family have a full density (as in JSE's answer below); or (3) the geometric genus of the generic member is $\leq 1$. One could also ask about the possible densities for the genera $0$ and $1$. $\endgroup$ – Vesselin Dimitrov Sep 28 '13 at 14:18
  • $\begingroup$ About the issue of letting $g \to +\infty$: For each individual $g > 1$, surely the density of pointless hyperelliptic curves of genus $g$ (denoted $\rho_g = 1-o(2^{-g})$ in Bhargava's paper) should equal $1$. Wouldn't this imply, a fortiori, the statement "most hyperelliptic curves are pointless," taken in any reasonable sense? This refers to your comment following JSE's answer. $\endgroup$ – Vesselin Dimitrov Sep 28 '13 at 14:25
  • $\begingroup$ I'm taking "most curves" in the sense "density 1". $\endgroup$ – Chandan Singh Dalawat Sep 29 '13 at 1:45

My paper with Bjorn Poonen (which is referenced and discussed in Bjorn's answer to this MO question: Are most cubic plane curves over the rationals elliptic?) has a precise statement for plane curves. You can follow Mike's suggestion in his comment to make a statement for all curves, but this has a problem. Namely, the moduli space of curves of genus $g$ is of general type for $g>22$ (or something like that) so, if you believe Lang's conjecture (or some weakening of it) then there no (or very few) "general" curves of genus $g$ defined over $\mathbb{Q}$, so one expects that most curves of genus $g$ defined over $\mathbb{Q}$ are restricted to rational subvarieties of the moduli space and the biggest one is the hyperelliptic locus, so maybe in some weird sense "most" curves over $\mathbb{Q}$ are hyperelliptic.

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    $\begingroup$ Regarding "most curves over $\mathbb{Q}$ are hyperelliptic": don't we expect the moduli space of genus-$g$ curves to contain infinitely many rational subvarieties (even if we restrict to rational subvarieties not properly contained in another such)? If that's the case, then maybe it makes more sense to say that, while there are infinitely many "types" of genus-$g$ curves over $\mathbb{Q}$, the type containing the "most" curves is the hyperelliptic locus. (I say "most" because all these sets are countable, hence have the same cardinality.) $\endgroup$ – Michael Zieve Aug 5 '13 at 16:47
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    $\begingroup$ @MichaelZieve Lang's conjecture predicts that there are a finite set of maximal rational and abelian subvarieties of a variety of general type containing all others. Of course, one may not believe this conjecture. I am not sure what's been proved for $M_g$, although a lot is known in this case. $\endgroup$ – Felipe Voloch Aug 5 '13 at 17:00
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    $\begingroup$ Although I agree in spirit with Felipe's comment, actually the locus of trigonal curves is a unirational subvariety of $\mathcal{M}_g$ that has even larger dimension than the locus of hyperelliptic curves. So, in this sense, "more" curves over $\mathbb{Q}$ are trigonal than are hyperelliptic. $\endgroup$ – Jason Starr Aug 5 '13 at 19:32
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    $\begingroup$ And if you were to count points on M_g(Q) of bounded height, with respect to a suitable ample divisor, it's not totally clear to me that the trigonal locus would "win" -- this depends on more subtle things than dimension, right? A line in P^3 has more points than the rest of the cubic surface it sits on, though both are rational and the cubic surface has more dimension to it. $\endgroup$ – JSE Aug 6 '13 at 2:17
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    $\begingroup$ ALL HAIL THE LINES $\endgroup$ – JSE Aug 6 '13 at 19:43

I suppose I would say this. Let h: M_g(Q) -> R be the height function corresponding to the Hodge class. Let S(N) be the set of points of M_g(Q) of height at most N. Finally, let P be the image of the projection from M_{g,1}(Q) to M_g(Q). (i.e. "the set of pointy curves.") Then one version of the assertion you're looking for would be:

$\lim_{N \rightarrow \infty} \frac{|S(N) \cap P|}{|S(N)|} = 0$.

But as Felipe says, it's possible that asymptotically 100% of the points in $S(N)$ are hyperelliptic, which makes this not very general at all. So you could go more hardcore and try this.

For all subvarieties X of M_g with infinitely many rational points, either:

  • $X(\mathbf{Q}) \subset P$ (i.e. every curve parametrized by X has a point)


  • $\lim_{N \rightarrow \infty} \frac{|S(N) \cap P \cap X|}{|S(N) \cap X|} = 0$

That is, either every curve in the family X has a point, or almost none of them do.

Of course, I have no idea whether this is true -- just trying to write something down in the vein of your question which is not "secretly just about hyperelliptic curves."

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  • $\begingroup$ You are fixing the genus $g$ whereas in Bhargava's theorem $g\to+\infty$ is required. $\endgroup$ – Chandan Singh Dalawat Aug 6 '13 at 3:18
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    $\begingroup$ Hey, I'm just writing what might be true, not what I have any idea how to prove! If you want, you can weaken the above to say the limsup is less than 1, even uniformly so in X, and then that the value of that limsup goes to 0 as g grows, which would be a little more like what Manjul gets. $\endgroup$ – JSE Aug 6 '13 at 3:37
  • $\begingroup$ I just saw this. I had asked the exact same question ("either every curve in the family X has a point, or almost none of them do") in mathoverflow.net/questions/121104/… . $\endgroup$ – Vesselin Dimitrov Sep 28 '13 at 14:03

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