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I have seen a quote saying that

Every smooth projective curve over a number field is a modular curve, i.e. (compactification of) $\Gamma\backslash\mathcal{H}$ for some finite index subgroup $\Gamma<\operatorname{SL}_2(\mathbb{Z})$ and $\mathcal{H}$ is the upper half plane.

Here, $\Gamma$ is not necessarily a congruence subgroup. The author says that this is because of Belyi's Theorem. Could anyone explain how one can get such implication from Belyi's Theorem? Any reference would be appreciated.

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Belyi's theorem says that every smooth projective curve $X$ over $\overline{\mathbb{Q}}$ is a cover of $\mathbb{P}^1$ ramified above 3 points. The key is to find a modular curve that is a thrice-punctured $\mathbb{P}^1$ without elliptic points. The classic example is the modular curve $\mathcal{H}/\Gamma(2)$. Let $X^*$ be the preimage of the branch points in $\mathbb{P}^1$, then $X^*\rightarrow\mathbb{P}^1 - \{3 \text{ points}\}\cong\mathcal{H}/\Gamma(2)$ is an unramified cover. Since the image of $\Gamma(2)$ in $\text{Aut}(\mathcal{H})\cong\text{PSL}_2(\mathbb{C})$ acts freely on $\mathcal{H}$, $\mathcal{H}$ is the universal cover of $\mathcal{H}/\Gamma(2)$, and every intermediate cover takes the form $\mathcal{H}/\Gamma$ for some $\Gamma\le\Gamma(2)$. In particular, applying this to $X^*$, we find that $X^*\rightarrow\mathcal{H}/\Gamma(2)$ is of the form $\mathcal{H}/\Gamma\rightarrow\mathcal{H}/\Gamma(2)$. Taking smooth compactifications, we find that $X$ is the smooth compactification of $\mathcal{H}/\Gamma$.

Note: Since $-I\in\Gamma(2)$, it might be cleaner to use an index 2 subgroup of $\Gamma(2)$ which does not contain $-I$. There are four such subgroups. One of them is the Sanov subgroup, freely generated by $[[1,2],[0,1]]$ and $[[1,0],[2,1]]$. It acts freely on $\mathcal{H}$ and has the same image in $\text{Aut}(\mathcal{H})$ as $\Gamma(2)$.

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  • $\begingroup$ Do you know that if the full list of $\Gamma$s' such that the associated modular curve is isomorphic to $\mathbb{P}^1-\{3\text{ points}\}$ is known? $\endgroup$
    – User0829
    Commented Oct 31, 2023 at 15:10
  • $\begingroup$ @User0829 For the congruence ones, you can find a complete list of the genus 0 congruence modular curves here. In general, you can think of modular curves as given by permutation representations of $SL(2,\mathbb{Z})$. The ramification above 0,1728,infinity can be calculated from the images of the standard generators $(E,V,EV)$, where $E$ has order 4, $V$ has order 6, and $EV$ is conjugate to $[[1,1],[0,1]]$. From this you can calculate the genus, number of cusps,...etc $\endgroup$
    – Will Chen
    Commented Oct 31, 2023 at 16:47

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