Why should the anabelian geometry conjectures be true?

I had probed friends of mine about Grothendieck's motivation for making the anabelian geometry conjectures, and they gave me the following explanation:

If $X$ is a hyperbolic curve over some field $K$ (think projective and of genus $\geq 2$), then, intuitively, its universal cover is the upper half plane. This means that to distinguish between any two hyperbolic curves, it suffices to distinguish between the actions on the upper-half plane that induce those two hyperbolic curves. In some vague way, this should be the same as distinguishing between their fundamental groups.

This seems a little tenuous to me. Is there a modification of the above argument that gives a moral reason for why anabelian geometry should be correct? Is there a completely different moral reason for anabelian geometry? If so, what is it? What intuitive reason should I have to believe anabelian geometry (beside the mounting evidence that it is indeed true)?

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I agree that that seems a tenuous argument. Any convincing argument has to be able to make a distinction between the genus $1$ and higher genus cases. – Torsten Ekedahl Nov 12 '11 at 5:18
I remember reading somewhere (perhaps in Faltings' Bourbaki report on Mochizuki's Theorem?) that Mostow's Rigidity theorem (en.wikipedia.org/wiki/Mostow_rigidity_theorem) was one of the motivations. – Lars Nov 12 '11 at 11:03

I can only offer a "strengthening" of your friends' explanation. Let me first remark that I am not an expert in this field and I am sure that there are some grave mistakes in my argument. However, it is much too long for a comment, so I post it as an answer.

Let us first consider the simpler case of (co)homology instead of fundamental groups. When talking about étale (say, $\ell$-adic) cohomology together with its Galois action, the transcendental analogue is generally taken to be not just the singular cohomology groups, but these groups together with their Hodge structure.

Similarly, consider a hyperbolic curve $X$ over a number field $K$. For simplicity, assume we are given a $K$-rational base point $x\in X(K)$. The fundamental group one considers is either the group $\pi_1^{et}(X,x)$ as an abstract profinite group, or the group $\pi_1^{et}(X\otimes\overline{K},x)$ together with its action of $\operatorname{Gal}(\overline{K}|K)$. The former can be reconstructed from the latter. The weakest version of Grothendieck's anabelian conjecture for curves says (roughly) that we can reconstruct $X$ from $\pi_1^{et}(X,x)$.

Let me explain why we can reconstruct $X$ from $\pi_1^{et}(X\otimes\overline{K},x)$ with its Galois action. The abelianization of this group with Galois action is just the product over all $\ell$ of the $\ell$-adic Tate modules $T_{\ell }(\operatorname{Jac}X)$. These are the $\ell$-adic analogues of the Hodge structures on first homomology, which bear the same information as the Jacobian itself. Thus it is not surprising (although very difficult!) that we can reconstruct $\operatorname{Jac}X$ from these data, and the Jacobian determines the isomorphism class of the curve by Torelli's theorem. [Edit: As Torsten Ekedahl has pointed out in the comments, it is not true that you can recover an abelian variety from its Tate module.]

Now there are certainly some points where the above argument does not work as simply as presented, but the morals is that the analogue of the arithmetic fundamental group over $\mathbb{C}$ should be the topological fundamental group with a "Hodge structure on groups". I do not know if this has ever been worked out, but there is a good understanding of the "Hodge structure on the nilpotent completion of the fundamental group", introduced by Hain and Zucker.

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That's interesting, thanks. I do wonder though, harking back to Torsten's comment, what in this argument uses the fact that $X$ is hyperbolic. – Makhalan Duff Nov 12 '11 at 16:11
Nothing, and that is why I think my "answer" misses the point somehow. What I was thinking about is what Deligne does in Le groupe fondamental ..., and there he writes in the introduction that this theory is far away from Grothendieck's anabelian dream, but instead his philosophy of motives can be applied ... I'm still eager to see someone with more knowledge about these things give an answer to your question. – Robert Kucharczyk Nov 13 '11 at 13:25
Note that in the genus $1$ case you do not recover the elliptic curve from its Tate modules. If you twist a (CM-)curve by a rank $1$ projective module over its endomorphism ring then the result will have the same Tate modules but only be isomorphic to the curve if the module is free. The same in some sense holds true in higher genus only that the twists may not be Jacobians. Hence it is crucial to look at the fundamental group and not its abelianisation. – Torsten Ekedahl Nov 14 '11 at 10:50

I think you should take a look at: http://www.renyi.hu/~szamuely/heid.pdf

The section there about anabelian geometry gives several reasons why one might believe it is true.

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