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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!

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@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. –  Damian Rössler Apr 3 '12 at 8:43
    
@Damian: Thanks for the reference. I will remove my comment. –  ulrich Apr 3 '12 at 9:24
    
@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. –  Damian Rössler Apr 3 '12 at 10:31
    
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. –  Steven Gubkin Apr 3 '12 at 11:25
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1 Answer

up vote 16 down vote accepted

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.

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This answer really clarifies the picture for me. Thank you ! –  Damian Rössler Apr 3 '12 at 16:40
    
Thanks, this is exactly the kind of answer I was hoping for! –  KristianJS Apr 3 '12 at 19:25
    
@Damian and @KristianJS: You're both welcome. I enjoyed thinking about the question. –  ulrich Apr 4 '12 at 5:55
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