Let $\mathscr{M}_*(N)$ be the Deligne Mumford stack of elliptic curves with the level $\Gamma_*(N)$.
($* = \varnothing, 0, 1$ or bal.$1$, see Katz-Mazur.)
Then this has the coarse moduli scheme $Y$ over $\mathbb{Z}$.
(not complex manifold)
I want to show that as Riemann surfaces, $Y(\mathbb{C}) \cong \mathbb{H}/\Gamma$, for suitable group $\Gamma$.
(e.g., $\Gamma(N) = \operatorname{ker}(\operatorname{SL}_2\mathbb{Z} \to \operatorname{SL}_2\mathbb{Z}/N), \Gamma_0(N) = \{ \gamma \in \operatorname{SL}_2\mathbb{Z} | \gamma \equiv
$$\begin{bmatrix}* & *\\0 & *\end{bmatrix} \mod N \}$ etc.)
(Where by "the Riemann surface defined by an affine smooth algebraic variety $X$ over $\mathbb{C}$", I mean the Riemann surface induced by a closed immersion $X \subseteq \mathbb{A}_\mathbb{C}^N$ and the identification $\mathbb{A}^N(\mathbb{C}) = \mathbb{C}^N$. See appendix B in Hartshorne's AG.)
Here is what I have tried:
First, by the definition of coarse moduli and by some fundamental properties about classical modular forms (e.g., see theorem 1.5.1 of Diamond-Shurman's text), we have
$\require{AMScd}$
\begin{CD}
|\mathscr{M}_*(N)(\mathbb{C})| @>{\text{bijective}}>> Y(\mathbb{C}) \\
@V{\text{bijective}}VV \\
\mathbb{H}/\Gamma.
\end{CD}
So if $\mathbb{H}/\Gamma$ is algebraic over $\mathbb{C}$ (write it by $Z$) and if we have a map $\mathscr{M}_*(N) \to Z$, then $Y \cong Z$, and so as Riemann surfaces $Y(\mathbb{C}) \cong \mathbb{H}/\Gamma$.
(By the universal property of a coarse moduli, we have $Y \to Z$.
By above argument this is bijective on $\mathbb{C}$ rational points.
Thus is an isomoprhism, by Zariski Main theorem.)
Because $\mathbb{H}^*/\Gamma$ ($\mathbb{H}^*$ is the upper half plane with the cusps) is algebraic, it seems that $\mathbb{H}/\Gamma$ is also algebraic.
However I have no idea how to get a map $\mathscr{M}_*(N) \to Z$.
(To define this map, by noetherian reduction and by the sheaf condition of representable functors on fpqc sites, it sufficies to define maps $\mathscr{M}_*(N)(S) \to Z(S)$ functorially for all affine scheme $S$ over $\mathbb{C}$ of finite type, I think.)
Thank you very much!
You have to specify what you mean by $Y({\mathbb C})\cong {\mathbb H}/\Gamma$.
I assume that you mean that $\Gamma$ is a discrete torsion-free subgroup of $PSL(2, {\mathbb R})$.
Then for $N=1$ the modular curve is biholomorphic to ${\mathbb C}$ and, hence, is not biholomorphic to the quotient of the hyperbolic plane by a discrete torsion-free subgroup of $PSL(2, {\mathbb R})$. (This is a consequence of the Liouville's theorem plus the covering theory.)
In contrast, for $N\ge 2$, $\Gamma(N)$ projects to a discrete torsion-free subgroup $\bar{\Gamma}(N)< PSL(2, {\mathbb R})$ and, hence, $Y({\mathbb C})\cong {\mathbb H}/\bar{\Gamma}(N)$. (The group $\Gamma(2)$ itself does have torsion, namely its center.)
More generally:
Definition. A Riemann surface $S$ is said to be of hyperbolic type (this terminology is not universally accepted) if $S$ is biholomorphic to the quotient ${\mathbb H}/\Gamma$ where $\Gamma$ is a discrete torsion-free subgroup of $PSL(2, {\mathbb R})$.
The uniformization theorem implies that a connected Riemann surface $S$ is of hyperbolic type if and only if $\chi_{top}(S)<0$ (possibly $-\infty$).