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Let $\Gamma$ be a congruence subgroup of $\operatorname{SL}_2(\mathbb Z)$, and let $\mathfrak H$ be the upper half-plane. Let $X(\Gamma)$ be the compactification of $\Gamma\backslash\mathfrak H$. Then $X(\Gamma)$ is a compact Riemann surface.

Suppose that $X(\Gamma)$ has genus $0$. Then also according to a general theorem there exists an analytic isomorphism $X(\Gamma)\longrightarrow \mathbb P_\mathbb C^1$. In the case of $\Gamma=\operatorname{SL}_2(\mathbb Z)$ this isomorphism is given by the $j$-invariant.

  1. What is the simplest way to prove that there exists an analytic isomorphism $X(\Gamma)\longrightarrow \mathbb P_\mathbb C^1$ in this special case? That is, how much advantage can our knowledge that $X(\Gamma)$ is a modular curve give us?
  2. Is there an algorithm to determine such isomorphism in terms of known functions (such as Dedekind $\eta$-function or $j$-invariant)?

For example, consider $$\Gamma=\bigg\lbrace\gamma\in \operatorname{SL}_2(\mathbb Z)\colon \gamma\equiv \begin{pmatrix}1 & 0 \\ 0 &1\end{pmatrix},\begin{pmatrix}0 & 1 \\ 1 &0\end{pmatrix}\text{mod }2\bigg\rbrace.$$ Then an isomorphism $X(\Gamma)\longrightarrow \mathbb P_\mathbb C^1$ is determined by the function $$j_\Gamma(\tau)=-\frac{\eta\left(\frac{\tau+1}{2}\right)^{24}}{\eta(\tau)^{24}}=q^{-1/2}\prod_{n=1}^{\infty}(1+q^{n-1/2}).$$

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    $\begingroup$ Can't you just take the $j$ invariant to answer both questions? $\endgroup$ – Will Sawin Jun 7 '20 at 20:52
  • $\begingroup$ @WillSawin Thank you, I will modify my question. $\endgroup$ – Shimrod Jun 7 '20 at 20:55
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    $\begingroup$ probably what you are looking for is the notion of Hauptmodul. $\endgroup$ – Henri Cohen Jun 7 '20 at 21:25

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