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This question is about a link between an open question in low-dimensional topology and a conjecture of Grothendieck, proved by Mochizuki. Let's start by stating them.

Recall that a subgroup $K$ of a group $G$ is characteristic if it is preserved by the action of $\operatorname{Aut}(G)$. In this case, we get an induced homomorphism of outer automorphism groups, $\operatorname{Out}(G)\to \operatorname{Out}(G/K)$. When $G/K$ is finite, the kernel of such a homomorphism is called a congruence subgroup.

Here's the topological question. It makes use of the fact that $\operatorname{Mod}(S)$ is naturally a subgroup of $\operatorname{Out}(\pi_1(S))$, so congruence subgroups make sense there too.

CSP conjecture: Let $S$ be a hyperbolic surface of finite type. Every finite-index subgroup of the mapping class group $\operatorname{Mod}(S)$ contains a congruence subgroup.

Here's the algebro-geometric result. This is far from my area of expertise, so apologies in advance for any errors.

Theorem (Mochizuki, conjectured by Grothendieck): Every smooth algebraic curve over $\mathbb{Q}$ is determined up to isomorphism by its étale fundamental group, equipped with the natural action of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.

In his famous problem list about mapping class groups (posted to the arXiv in 2006), Ivanov writes that he was told by Voevodsky that the above conjecture implies the above theorem. He goes on:

I am not aware of any publication where this conjecture of Grothendieck is deduced from the solution of the congruence subgroup problem for the mapping class groups.

Hence my question.

Has anyone written down a proof of this implication since 2006? If not, where would be a good place to get started learning about this? Best of all would be if someone knows the proof and could explain it!

UPDATE 1 October 2024:

I think, with the aid of @coLaideronette's answer, the question can be sharpened. Here, Voevodsky states a certain "combinatorial conjecture", and from it deduces Grothendieck's conjecture in the genus-zero case. I haven't found time to completely understand Voevodsky's paper, but his "combinatorial conjecture" is plausibly equivalent to, or implied by, the CSP.

[His combinatorial conjecture states that a certain homomorphism is injective. The CSP can be rephrased as: the natural map $\widehat{\operatorname{Mod}(S)}\to \operatorname{Out}(\widehat{\pi_1(S)})$ is injective, where $\widehat{-}$ denotes profinite completion.]

Interestingly, the CSP for genus-zero curves was proved by Diaz--Donager--Harbater in 1989, before Voevodsky's paper, so if this interpretation is correct then perhaps Voevodsky's result can be upgraded to an unconditional proof in that case.

Hence, my question can be rephrased as:

  1. Is my interpretation of Voevodsky's theorem correct?
  2. If so, has anyone generalised Voevodsky's result, that CSP implies Grothendieck's conjecture for genus-zero curves, to curves of higher genus?

As stated above, I am aware that Grothendieck's conjectures have been proved by Mochizuki and others, but my interest is in applications of the CSP. It may be that the answer to 2 is buried in one of the proofs of Grothendieck's conjecture -- if so, I'd appreciate a reference to a specific result, rather than just to the whole paper.

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    $\begingroup$ This seems to rely on the claim that for every two different $\mathbb Q$-points of $\mathcal M_g$, there is a finite \'{e}tale cover of $\mathcal M_g$ that separates them (for example, to which one of them lifts but the other does not). What confuses me here is that I can't think of a method to prove this without using congruence subgroups (i.e. without proving the Grothendieck conjecture directly). $\endgroup$
    – Will Sawin
    Commented Sep 19 at 0:36
  • $\begingroup$ @WillSawin: thanks, this sounds like a start. Excuse my ignorance: when you say one rational point lifts to a cover but another one doesn't, what does this mean? Should I picture an orbifold cover of $\mathcal{M}_g$ (over $\mathbb{C}$), for which one point has rational preimages and the other doesn't? I share your puzzlement: there are very few methods of generating covers of $\mathcal{M}_g$ that aren't a priori congruence (which is why the CSP is hard). $\endgroup$
    – HJRW
    Commented Sep 19 at 7:56
  • $\begingroup$ Yeah, the coverings I'm considering are stacks over $\mathbb Q$ whose $\mathbb C$-points are orbifold covers of $\mathcal M_g$. Depending on the covering the $\mathbb Q$-structure may or may not be unique. $\endgroup$
    – Will Sawin
    Commented Sep 19 at 12:49
  • $\begingroup$ CSP is exactly what you would want to use to prove the Grothendieck conjecture from the section conjecture for $M_g$, to lift an action on the fundamental group to an action through the MCG. This seems silly because the section conjecture is generally harder, but maybe you can use the section conjecture for affine curves to prove it for $M_g$ to prove the Grothendieck conjecture for closed curves? Also, can you run this backwards to prove CSP? $\endgroup$ Commented Sep 19 at 19:42
  • $\begingroup$ @BenWieland: Thanks, this is also interesting. Is the connection between the Grothendieck conjecture, the section conjecture and CSP written down anywhere? $\endgroup$
    – HJRW
    Commented Sep 20 at 9:54

1 Answer 1

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Let $K$ be a field of characteristic zero, $X/K$ a hyperbolic $(g,n)$-curve. There is a famous short exact sequence

$$1\longrightarrow\pi_1(X\otimes \overline{K})\longrightarrow\pi_1(X)\longrightarrow \operatorname{Gal}_K\longrightarrow 1.$$

Let $\Pi_{g,n}$ be the profinite completion of the fundamental group $\pi_1(g,n)$ of a genus $g$ compact Riemann surface of with $n$ points punctured. After fixing an isomorphism $\pi_1(X\otimes \overline{K})\overset{\sim}{\to}\Pi_{g,n}$, one gets two exact sequences

$$1\longrightarrow\Pi_{g,n}\longrightarrow\pi_1(X)\longrightarrow \operatorname{Gal}_K\longrightarrow 1$$

$$1\longrightarrow \operatorname{Inn}(\Pi_{g,n }) \longrightarrow \operatorname{Aut}(\Pi_{g,n}) \longrightarrow \operatorname{Out}(\Pi_{g,n}) \longrightarrow 1$$

and the outer Galois representation

$$\rho_X:\operatorname{Gal}_K\longrightarrow \operatorname{Out}(\Pi_{g,n}).$$

Note that the kernel of $\rho_X$ is independent of the choice of the isomorphism. In his letter to Faltings, Grothendieck conjectured that any hyperbolic curve (actually for any algebraic curve) over a number field should be anabelian i.e. the geometry of any hyperbolic curve $X$ over a number field is determined by $\rho_X$. In order to prove some special cases of Grothendieck’a conjecture, Voevodsky made a conjecture that $\rho_X$ is injective when $X$ is an affine hyperbolic curve and $K$ is a number field. Voevodsky’s conjecture was proved by Matsumoto then extended to the proper case by Hoshi–Mochizuki.

Following the earlier work of Nakamura and Tamagawa, the anabelian conjecture was proved by Mochizuki. Meanwhile, Grothendieck conjectured that the moduli space of hyperbolic curves would be also anabelian. So it’s natural to ask whether Voevodsky’s conjecture holds after replacing $\Pi_{g,n}$ with $\mathscr{M}_{g,n}$, the moduli stack over $K$ of smooth geometrically connected proper curves of genus $g$ with $n$ marked points.

Just like $\Pi_{g,n}$, one has the profinite completion $\mathscr G_{g,n}$ of the oriented mapping class group $\operatorname{MCG}_{g,n} $ of a genus $g$ topological surface with $n$ marked points, which is isomorphic to $\pi_1(\mathscr M_{g,n}\otimes \overline{K})$. Then one gets a Galois representation

$$\rho_{g,n}:\operatorname{Gal}_K\longrightarrow \operatorname{Out}(\mathscr G_{g,n})$$

again. By the work of Matsumoto–Tamagawa, one can study $\rho_{g,n}$ with the associated universal monodromy representation

$$\rho_{g,n}^{\text{um}}:\pi_1(\mathscr{M}_{g,n})\longrightarrow \operatorname{Out}(\Pi_{g,n})$$

given by the exact sequence

$$1\longrightarrow\Pi_{g,n}\longrightarrow \pi_1(\mathscr{M}_{g,n+1})\longrightarrow \pi_1(\mathscr{M}_{g,n})\longrightarrow 1.$$

Using their combinatorial anabelian geometry, Hoshi–Mochizuki showed that $\rho_{g,n}^{\text{um}}$ is injective if and only if $\rho_{g,n}^{\text{um}}|_{\mathscr{G}_{g,n}}$ is injective. Then finally, one just needs to note that the injectivity of $\rho_{g,n}^{\text{um}}|_{\mathscr{G}_{g,n}}$ is equivalent to the congruence subgroup problem for $\operatorname{MCG}_{g,n}$.

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  • $\begingroup$ Thanks for this. Please forgive me -- anabelian geometry is far from my area of expertise, so it will take me some work to understand your answer. At the moment, I don't see how this answers my question. As I understand it, you are combining the CSP with a result of Hoshi--Mochizuki to show that (A) the natural representation of the étale fundamental group of moduli space is injective. However, my question was how to use CSP to deduce that (B) hyperbolic curves are anabelian (proved by Mochizuki as you say). Does (A) imply (B)? If so, could you explain how? $\endgroup$
    – HJRW
    Commented Sep 21 at 12:24
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    $\begingroup$ @HJRW This has been discussed in Voevodsky’s paper, I added the link. $\endgroup$ Commented Sep 21 at 12:57
  • $\begingroup$ Ah, many thanks for the reference. I’ll take a look at Voevodsky’s paper and asked if it answers my question. $\endgroup$
    – HJRW
    Commented Sep 21 at 14:21
  • $\begingroup$ So, in §4 of the linked paper, Voevodsky states a certain "combinatorial conjecture", that a certain morphism is injective. This is plausibly equivalent to the CSP, though I didn't check yet. (CSP is the statement that the natural map $\widehat{\operatorname{Mod}(S)}\to\operatorname{Out}(\widehat{\pi_1(S)})$ is injective.) He then reduces Grothendieck's conjecture to this statement in genus 0. But the CSP was already known in genus 0 when Voevodsky wrote his paper (due to Diaz--Donagi--Harbater)! What about higher genera? $\endgroup$
    – HJRW
    Commented Sep 21 at 15:00
  • $\begingroup$ @HJRW Boggi has proved certain results for higher $g$ and $n$ in his paper The congruence subgroup property for the hyperelliptic modular group: the open surface case and On the procongruence completion of the Teichmüller modular group, but he used the theory of complexes of profinite curves rather than the anabelian geometry. The anabelian proof of the same results is due to Hoshi–Tamagawa, but as far as I know there is no published paper on that. $\endgroup$ Commented Sep 21 at 15:35

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