Do elements of the absolute Galois group $G=\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ of $\mathbf{Q}$ induce automorphisms of $\mathbf{C}$ if we extend the morphisms trivially to the rest of $\mathbf{C}$? Here we consider an element of $G$ as an automorphism of $\overline{\mathbf{Q}}\subset \mathbf{C}$ (and we fix this embedding).

Does this define an action on the moduli space of compact connected Riemann surfaces of genus $g$ by pulling back a Riemann surface via this action?

That is, let $\sigma$ be in $G$. Let $X$ be compact connected Riemann surface. Then we can let $\sigma$ act on $X$ algebraically by pulling back $X/\mathrm{Spec} \ \mathbf{C}$ along $\sigma$.

It is clear that $G$ acts on the $\overline{\mathbf{Q}}$-points of the moduli space $M_g$.

  • 2
    $\begingroup$ What do you mean by "extend the morphisms trivially to the rest of $\mathbb{C}$"? If $\alpha$ is an algebraic number and an element of $G$ acts trivially on $\pi$ and $\pi - \alpha$, then necessarily it also acts trivially on $\alpha$... $\endgroup$ – Qiaochu Yuan Mar 8 '12 at 21:36
  • $\begingroup$ Anyway, there is an interesting relationship between the absolute Galois group of $\mathbb{Q}$ and moduli of curves, but this isn't how it goes. The relevant keywords are "Belyi's theorem" and "Grothendieck-Teichmuller." $\endgroup$ – Qiaochu Yuan Mar 8 '12 at 21:37
  • $\begingroup$ Ow, so there is actually no "canonical" way to extend a non-trivial element of $G$ to an endomorphism of $\mathbf{C}$... $\endgroup$ – seporhau Mar 8 '12 at 21:38
  • $\begingroup$ I know Belyi's theorem: the set of dessins d'enfants associated to a $\overline{\mathbf{Q}}$-point of $M_g$ is non-empty. What do you mean by Grothendieck-Teichmuller? $\endgroup$ – seporhau Mar 8 '12 at 21:40
  • $\begingroup$ I don't feel particularly qualified to explain, but a Google search will turn up lots of resources. $\endgroup$ – Qiaochu Yuan Mar 8 '12 at 21:43

Every automorphism of $\overline{\mathbb{Q}}$ extends to an automorphism of $\mathbb{C}$, but in many ways. In fact, one has a short exact sequence

$$1 \rightarrow \operatorname{Aut}(\mathbb{C}/\overline{\mathbb{Q}}) \rightarrow \operatorname{Aut} \mathbb{C} \rightarrow \operatorname{Aut}(\overline{\mathbb{Q}}) \rightarrow 1.$$

The group $K = \operatorname{Aut}(\mathbb{C}/\overline{\mathbb{Q}})$ has cardinality $2^c = 2^{2^{\aleph_0}}$, so is truly enormous. By a theorem of D. Lascar, it is a simple group. I do not know whether the exact sequence splits; if it doesn't, then this would give a very strong sense in which there is no "natural" extension of an automorphism of $\overline{\mathbb{Q}}$ to an automorphism of $\mathbb{C}$.

As you say, the absolute Galois group $\mathfrak{g}_{\mathbb{Q}} = \operatorname{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ acts on the $\overline{\mathbb{Q}}$-points of $M_g$,..as is the case for any scheme over $\mathbb{Q}$. It also acts on the $\ell$-adic cohomology groups of $M_g$, which is, via a comparison theorem, reasonably close to saying that it acts on the cohomology of $M_g$ in the usual topological sense.


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