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So I've heard in passing that for any congruence modular curve $X$ (over $\mathbb{C}$), there is a $g\in\text{GL}_2(\mathbb{Q})^+$ such that $X$ is birational to a plane curve in $\mathbb{C}^2$ given by the subset $\{(j(\tau),j(g(\tau))) : \tau\in\mathbb{H}\}$

In other words, for every $g\in\text{GL}_2(\mathbb{Q})^+$, the two functions $j,j\circ g: \mathbb{H}\rightarrow\mathbb{C}$ satisfy some polynomial relation $\Phi_g(j,j\circ g)$ (with $\mathbb{C}$-coefficients) such that $\text{Spec }\mathbb{C}[x,y]/\Phi_g(x,y)$ is isomorphic to some affine congruence modular curve, and every congruence modular curve can be obtained this way via some $g\in\text{GL}_2(\mathbb{Q})^+$.

A calculation shows that if $g\in\text{GL}_2(\frac{1}{n}\mathbb{Z})^+$, then $g$ conjugates the principal congruence subgroup $\Gamma(n^2)$ into $\text{SL}_2(\mathbb{Z})$, so $j\circ g$ will be invariant under $\Gamma(n^2)$, so $j,j\circ g$ will generate the function field of some congruence modular curve of level dividing $n^2$. It seems that the corresponding congruence subgroup is then just the largest subgroup of $\text{SL}_2(\mathbb{Z})$ which $g$ conjugates into $\text{SL}_2(\mathbb{Z})$.

  1. Is it true that every congruence modular curve can be obtained in this way? (Is this described anywhere?)

  2. If $K$ is the number field obtained by adjoining the coefficients of $\Phi_g(x,y)$ to $\mathbb{Q}$, then is $\text{Spec }K[x,y]/\Phi_g(x,y)$ the the usual arithmetic model of the modular curve? (e.g., for some $g$ so that $\Phi_g$ gives a principal congruence modular curve $Y(n)$, is $K = \mathbb{Q}$ or $\mathbb{Q}(\zeta_n)$?)

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I think this is true only for the classical modular curves. The reason is that the curve depends only on the double coset $SL_2(\mathbb Z) g SL_2(\mathbb Z)$, as $j$ is $SL_2(\mathbb Z)$-invariant. But every such double coset contains a diagonal matrix (e.g. by Smith normal form). And diagonal matrices give the classical modular curves, for exactly the reason you give about $SL_2(\mathbb Z) \cap g SL_2(\mathbb Z) g^{-1}$.

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  • $\begingroup$ Ah, okay so I can see we get subgroups of the form $\Gamma_{r,s} := \{[[a,b],[c,d]] : a,b,c,d\in\mathbb{Z}, b\equiv 0\text{ mod } r, c \equiv 0\text{ mod } s\}$. Are these what you mean by "classical modular curves"? $\endgroup$
    – Will Chen
    Jul 23 '18 at 18:09
  • $\begingroup$ @WillChen That subgroup is equivalent to the one where the only condition is $b \equiv 0 \mod rs$, by conjugating by a diagonal matrix $[[ r/s,0],[0,1]]$, and that subgroup defines the classical modular curves. $\endgroup$
    – Will Sawin
    Jul 23 '18 at 20:15

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