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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