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Let $C$ be a curve over $\mathbb Q$ with a point $P$ on $Pic^1$. For each $\mathbb Q$-rational point $Q$, $Q-P$ is a point on the Jacobian $J$. We can use the map $H^0(\mathbb Q, J) \to H^1(\mathbb Q,J[n])$ to obtain a class in $H^1(\mathbb Q,J[n])$ associated to each rational point.

Rather than trying to find a local obstruction to existence of rational points on $C$, we could try to find a local obstruction to existence of rational points in each $H^1(\mathbb Q, J[n])$ class. A local obstruction at $\mathbb Q_p$ would occur when the class, projected down to $H^1(\mathbb Q_p, J[n])$ does not arise from any $\mathbb Q_p$-rational point of $C$.

If $C$ does not have rational points, must there be some $n$ such that each class in $H^1(\mathbb Q, J[n])$ has a local obstruction at some $p$ or at $\infty$?

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  • $\begingroup$ Is this inspired at all by the descent obstruction to rational points? $\endgroup$ – stankewicz Jun 14 '13 at 19:17
  • $\begingroup$ This is inspired by work of Manjul Bhargava on hyperelliptic curves. I guess this is the universal descent obstruction for finite abelian groups? $\endgroup$ – Will Sawin Jun 14 '13 at 20:09
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    $\begingroup$ Yes. This is the finite abelian descent obstruction. It's been conjectured that it's the only obstruction to the existence of rational points. $\endgroup$ – Felipe Voloch Jun 14 '13 at 20:25
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As was pointed out by Felipe Voloch in the comments, this obstruction is equivalent to the finite abelian descent obstruction [Stoll, Thm. 6.4 and the discussion before Cor. 6.2]. It is also shown that for curves, this agrees with the Brauer–Manin obstruction [Stoll, Cor. 7.3]. Finally, Stoll conjectures the following:

Conjecture [Stoll, Conj. 9.1]. Let $C$ be a smooth projective geometrically connected curve over a number field $K$. Then $\overline{C(K)} = C(\mathbf A_K)_\bullet^{\text{f-ab}}$ (see [Stoll, Def. 5.4 and §2] for notation).

That is, the finite abelian descent obstruction is the only obstruction to the Hasse principle and weak approximation.

What is known.

  • If $g = 0$, then $C$ satisfies the Hasse principle $C(K) = \varnothing \Leftrightarrow C(\mathbf A_K)_\bullet = \varnothing$. If $C(\mathbf A_K)_\bullet = \varnothing$ then there is nothing to prove, and if $C \cong \mathbf P^1_K$ then $C(K)$ is dense in $C(\mathbf A_K)_\bullet$.
  • If $g = 1$, then the conjecture is true if and only if $Ш(K,\operatorname{Pic}^0_C)_{\text{div}} = 0$ [Stoll, Cor. 6.2 and the discussion after Cor. 6.3]. In particular, this is a weak form of the Tate–Shafarevich conjecture.
  • If $g \geq 2$, then $C(K)$ is finite by Faltings, so $\overline{C(K)} = C(K)$. Thus, the conjecture says that $C(\mathbf A_K)_\bullet^{\text{f-ab}} = C(K)$. I'm not even sure if we know whether $C(\mathbf A_K)_\bullet^{\text{f-ab}}$ is finite.

Finally, Stoll proves the conjecture in the following case:

Theorem [Stoll, Thm. 8.6]. Assume $C$ admits a nonconstant map $C \to A$ into an abelian variety $A$ such that $A(K)$ is finite and $Ш(K,A)_{\text{div}} = 0$. Then $C(K) = C(\mathbf A_K)_\bullet^{\text{f-ab}}$.

This is for example the case when $K = \mathbf Q$ and $A$ is modular of analytic rank $0$. As mentioned before, the condition $Ш(K,A)_{\text{div}}$ is expected to always be true, but the condition that $A(K)$ is finite is a serious restriction on the generality of the theorem.

Recent progress?

I haven't worked on rational points for a while, so I don't know what the current status is. I would be surprised if there were a great breakthrough on the conjecture in its general form, though.


References.

[Stoll] Stoll, Michael, Finite descent obstructions and rational points on curves, Algebra Number Theory 1, No. 4, 349-391 (2007). ZBL1167.11024.

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  • $\begingroup$ Shouldn't this answer have $C(\mathbb A_K)_\bullet^{\text{f-ab}}$ in every single place where you wrote $C(\mathbb A_K)_\bullet$? $\endgroup$ – M.D. Mar 30 at 21:39
  • $\begingroup$ @M.D. you're absolutely right! Fixed now. $\endgroup$ – R. van Dobben de Bruyn Mar 31 at 2:16
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This is a known difficult problem as Felipe Voloch mentions in the comments and is as far as I know still open today.

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  • $\begingroup$ Reference? What is known? .... $\endgroup$ – Chris Wuthrich Jul 31 '18 at 8:08

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