Let $X$ be an integral proper normal curve over a (perfect) field $F$, of genus $\geq 2$. One variant of Grothendieck's "section conjecture" states that the sections $G_F \rightarrow \pi_1(X)$ of the exact sequence \begin{equation} 1 \rightarrow \pi_1(X_{\bar{F}}) \rightarrow \pi_1(X) \rightarrow G_F \rightarrow 1 \end{equation} are, up to conjugation, in bijection with the $F$-rational points of $X$, where $G_F$ is the absolute Galois group of $F$ and $\pi_1$ is the algebraic fundamental group.

Question: what is the reason for excluding genus 1 curves?

I understand why genus 0 curves must be excluded: if $F$ has characteristic zero, it is a general fact that the 'geometric' fundamental group $\pi_1(X_{\bar{F}})$ is just the profinite completion of the regular topological fundamental group of $X$, seen as a curve over $\mathbb{C}$. For genus 0, the topological fundamental group is trivial, and thus the above exact sequence induces an isomorphism $\pi_1(X) \rightarrow G_F$. Hence there is always at least one section even if $X$ has no rational points whatsoever.

However, I don't know of a good reason why genus 1 curves should be excluded here. The above argument obviously won't do since the topological fundamental group is no longer trivial for genus 1. Are there even so known counter-examples for genus 1 curves? What goes wrong?

I know the philosophy is that one should expect 'anabelian behaviour' only when the fundamental group is 'far from being abelian', which excludes the genus 1 case. But I would be more satisfied with a more concrete, less philosophical, reason!

  • $\begingroup$ @ulrich. Why should that be the case ? Even if that is so, one can show the following: let $X$ be a torsor under an elliptic curve $E$, everything over number field $k$, such that the corresponding element of $H^1(k,E)$ is not divisible in $H^1(k,E)$. Then the exact sequence of fundamental groups written down in the question does not split. See Harari-Szamuely, "Galois sections\dots", Math. Ann. (2009) 344:779-800, Th. 1.2. $\endgroup$ Apr 3, 2012 at 8:43
  • $\begingroup$ @Damian: Thanks for the reference. I will remove my comment. $\endgroup$
    – naf
    Apr 3, 2012 at 9:24
  • $\begingroup$ @KristianJS. I think that this is a good question. The only place where I know the hypothesis on the genus to appear is in Mochizuki's theorem 19.1 in his article "The local pro-p anabelian geometry of curves." Invent. Math. 138 (1999), no. 2, 319–423. In all the other references that I know (like Esnault-Wittenberg, Harari-Szamuely, Koenigsmann), the genus plays no role. I am no expert though and I might not be aware of a classical counterexample. $\endgroup$ Apr 3, 2012 at 10:31
  • $\begingroup$ In complex geometry, curves of genus 0 have the sphere as universal cover, curves of genus 1 have the plane, and all other curves have the upper half plane as universal cover. So a lot of theorems about Riemann surfaces start off "For curves of genus >1". Not sure why in this particular case, but it is something to keep in mind. $\endgroup$ Apr 3, 2012 at 11:25
  • $\begingroup$ By the way, the section conjecture holds for Abelian varieties over finite fields, see math.uni-frankfurt.de/~stix/research/preprints/… Remark 2 (4). $\endgroup$
    – user19475
    Sep 26, 2018 at 12:12

1 Answer 1


I think that Grothendieck had already observed that the map from rational points to sections is injective (for curves of genus at least 2 over a number field) and I believe that his proof works even for curves of genus $1$, so the thing that fails for curves of genus $1$ is surjectivity.

Consider any exact sequence of groups $1 \to A \to G \to H \to 1$ with $A$ abelian and assume that there is a section $\sigma:H \to G$. A simple calculation shows that if $\sigma$ is a section and $f: H \to A$ is any map, then the function $\tau: H \to G$ given by $\tau(h) = f(h)\sigma(h)$ is a section (i.e. also a homomorphism) iff $f$ is a $1$-cocycle. If the cocyle is not a coboundary then $\sigma$ and $\tau$ are not conjugate, so what we really care about is $H^1(H,A)$.

We now apply the foregoing in the situation of the question, so we are led to consider the group $C = H^1(Gal(\bar{F}/F), \pi_1(X_{\bar{F}}))$. Since $\pi_1(X_{\bar{F}})$ is a $\hat{\mathbb{Z}}$ module, so is $C$. Now suppose the curve $X$ has infinitely many rational points, so the group $C$ is also infinite (but finitely generated by the Mordell-Weil theorem). However, there are no fintely generated but infinite $\hat{\mathbb{Z}}$ modules. It follows that the map from rational points to sections modulo conjugacy cannot be surjective.

  • $\begingroup$ This answer really clarifies the picture for me. Thank you ! $\endgroup$ Apr 3, 2012 at 16:40
  • $\begingroup$ Thanks, this is exactly the kind of answer I was hoping for! $\endgroup$
    – KristianJS
    Apr 3, 2012 at 19:25
  • $\begingroup$ @Damian and @KristianJS: You're both welcome. I enjoyed thinking about the question. $\endgroup$
    – naf
    Apr 4, 2012 at 5:55
  • $\begingroup$ @ulrich may you clarify why there are no finitely generated and infinite $\mathbb{Z}$^ modules? $\endgroup$ Jan 3, 2018 at 20:03
  • $\begingroup$ @proofromthebook: The precise statement is that any finitely generated abelian group $C$ which is a $\hat{\mathbb{Z}}$ module is finite. Using that $\hat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$, this follows from the corresponding fact for $\mathbb{Z}_p$ (use structure theorem for modules over a PID). $\endgroup$
    – naf
    Jan 6, 2018 at 6:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.