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!



Why does the Section Conjecture exclude curves of genus 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.


