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Given an irreducible smooth complex-projective curve $X$, I will say that a subgroup $\Gamma< SL(2, {\mathbb R})$ weakly uniformizes $X$ if [corrected] there exists a nonconstant holomorphic map from ${\mathbb H}^2/\Gamma$ to $X$.

One formulation of the Shumura-Taniyama conjecture is that if $X$ is (the set of complex points of) an elliptic curve defined over ${\mathbb Q}$, then there exists a congruence-subgroup $\Gamma< SL(2, {\mathbb Z})$ which weakly uniformizes $X$.

(I learned this from the paper by Barry Mazur "Number theory as a gadfly, but misremembered the statement.)

Question: What if we remove the assumption that $X$ is elliptic: Does this result still hold?

Remark. 1. One reformulation of Belyi's theorem is that if $X$ is a complex-projective curve over $\bar{\mathbb Q}$ then $X$ is uniformized by a finite index subgroup $\Gamma$ of $SL(2, {\mathbb Z})$, in the sense that ${\mathbb H}^2/\Gamma$ is biholomorphic to an open (and dense) subset of $X$.

  1. I am a topologist, not a number-theorist, so my interest in this question is purely casual.

Update. (Thanks to a comment by François Brunault). The paper

Baker, Matthew H.; González-Jiménez, Enrique; Gonzáles, Josep; Poonen, Bjorn, Finiteness results for modular curves of genus at least 2, Am. J. Math. 127, No. 6, 1325-1387 (2005). ZBL1127.11041.

contains some results suggesting that the answer is strongly negative, i.e. that for each genus $\ge 2$ only finitely many curves over ${\mathbb Q}$ are weakly uniformized by congruence subgroups of $SL(2, {\mathbb Z})$. They conjecture this finiteness property under the extra assumption that the holomorphic map is a morphism defined over ${\mathbb Q}$. I have no feel for how strong this restriction is. They prove this conjecture under certain extra assumptions.

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    $\begingroup$ I don't think your statement of the Shimura-Taniyama conj is correct. Rademacher's conjecture (proven by Dennin) says that there are only finitely many congruence subgps of a given genus. Since there are inf. many elliptic curves over $\mathbb{Q}$, they can't all be isom to a congruence modular curve. This also of course implies that the statement for general $X$ is also false... $\endgroup$
    – Will Chen
    Mar 6, 2021 at 17:24
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    $\begingroup$ ... The modularity theorem says that every elliptic curve over $\mathbb{Q}$ admits a congruence modular curve as a branched cover (so the elliptic curve is isogenous to a factor of the jacobian of the modular curve). In all but finitely many cases the modular curve has higher genus. $\endgroup$
    – Will Chen
    Mar 6, 2021 at 17:32
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    $\begingroup$ The curves of genus $g \geq 2$ which can be uniformised by (congruence) modular curves have been studied by Baker, Gonzalez-Jimenez, Gonzalez, Poonen: www-math.mit.edu/~poonen/papers/finiteness.pdf Essentially, for a given genus $\geq 2$, there are only finitely many such curves, in contrast with elliptic curves. $\endgroup$ Mar 6, 2021 at 17:42
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    $\begingroup$ Will Chen is completely right - the correct claim replaces "biholomorphic" with "admits a nonconstant/surjective holomorphic map". $\endgroup$
    – Will Sawin
    Mar 6, 2021 at 18:46
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    $\begingroup$ I'm sorry to hijack the conversation but I was curious about a related problem: in this comment JS Milne mentions that a version of modularity does not hold even if we consider Shimura varieties in place of modular curves. However looking at the referenced paper of Blasius I was not able to decipher an argument for that. Can anyone clarify that or point to a reference which discusses it in more detail? $\endgroup$
    – Wojowu
    Mar 6, 2021 at 21:40

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