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Let $N \ge 1$ be an integer and $\mathscr{M}_*(N)$ the stack of elliptic curves with the level $\Gamma(N), \Gamma_0(N), \Gamma_1(N)$ or $\Gamma_{\text{bal.}1}(N)$. (For its definition see Katz-Mazur.)
Then there is the coarse moduli scheme of $\mathscr{M}_*(N)$, denoted by $ Y = Y_*(N)$, over $\mathbb{Z}$, and it is a connected regular affine scheme of pure dimension 2, and is flat of finite type of relative dimension 1.
Moreover, $Y \times \mathbb{Z}[1/N]$ is smooth over $\mathbb{Z}[1/N]$.

I showed this proposition except the connectedness. I want to show it.

Here is what I've tried:
Since $Y \times \mathbb{Q} \to Y$ has the dense image, it sufficies to show that $Y \times \mathbb{C}$ is connected. And so since the Euclid toplogy is finer than the Zariski topology, it sufficies to show that the complex manifold $Z$ induced by $Y \times \mathbb{C}$ (i.e., the closed submanifold induced by $Y(\mathbb{C}) \subseteq \mathbb{A}^N(\mathbb{C}) = \mathbb{C}^N$.) is connected.

Next,it seems that, as Riemann surfaces, $Z \cong \mathbb{H}/ \Gamma_*(N)$, at least for $* = 0, 1, \varnothing$. (Where $\mathbb{H}$ is the upper half plane.) If so, trivially $Z$ is connected, and so $Y$ is connected.

So it sufficies to show $Z \cong \mathbb{H}/ \Gamma_*(N).$
The "coarse moduli map" $\mathscr{M}_*(N) \to Y$ induces $|\mathscr{M}_*(N)(\mathbb{C})| $ $\cong Y(\mathbb{C}) \cong Z$. On the other hand, by a fundamental proposition of modular forms, we have $|\mathscr{M}_*(N)(\mathbb{C})| \cong \mathbb{H}/\Gamma_*(N)$.
So as sets, we have $Z \cong \mathbb{H}/ \Gamma_*(N)$.

It's hard for me to show this map is holomorphic (or even it is continuous, which is enough in order to show $Z$ is connected), because I cannot understand the topology on $Z$ "moduli theoritically". (i.e., the interpretation such as "$U \subseteq Z$ is open iff elliptic curves in $U$ are such-and-such...")


P.S. I showed other properties as follows:
First, the separated Deligne-Mumford stack $\mathscr{M}_*(N)$ is regular of dimension $2$ at every closed point, and is affine flat of relative dimension 1 over $\mathbb{Z}$ of finite type. (By (5.1.1) of Katz-Mazur.)

Next, for $r \ge 1$, the diagram $\require{AMScd}$ \begin{CD} \mathscr{M}_{*, \varnothing} (N, r)[1/r] @>>> \mathscr{M}(r)[1/r] \\ @VVV @V{ \text{etale finite}}VV\\ \mathscr{M}_*(N)[1/r] @>{ \text{finite flat}}>> \mathscr{M}[1/r] \end{CD} is 2-Cartesian, and $\mathscr{M}(r)[1/r]$ is a regular scheme. So $\mathscr{M}_{*, \varnothing} (N, r)[1/r]$ is also a regular scheme, $T$.

It's easy to show that $(|\mathscr{M}_{*, \varnothing} (N, r)[1/r]| / \operatorname{GL}_2(\mathbb{Z}/r) )^\# \to |\mathscr{M}_*(N)[1/r]|^\#$ is isomorphism. (Where $|\mathscr{X}|$ is the functor taking $S$ to the isomoprhism classes of $\mathscr{X}(S)$, and $\mathscr{F}^\#$ is the sheafification of $\mathscr{F}$.)

So $T/ \operatorname{GL}_2(\mathbb{Z}/r)$ is the coarse moduli scheme over $\mathbb{Z}[1/r]$. And since this is the quotient by etale finite group scheme, it inherits many properties of $T$.
Therefore, patching these schemes for relatively prime $r, r'$, we have the result.

I'm not familiar with the stack theory, so if my "proof" is wrong, please correct it.

And can I show the connectedness stack-theoritically? For example, by 4.14 of Deligne-Mumford, if $\mathscr{M}_*(N) \times \mathbb{C}$ is connected as a stack, then it seems that the coarse moduli scheme is connected as a scheme. (But I can't show the stack is connected.)

Thank you very much!

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    $\begingroup$ To check the map is holomorphic, you want to use that there exists a universal family of elliptic curves over the moduli stack. You can check the map is holomorphic locally n an analytic neighborhood of each point. in such a neighborhood you can trivialize the homology of the family of elliptic curves, so the map actually lifts to the upper half plane, and you want to show that this lift is holomorphic. This can be done by writing the coordinate as a ratio of two integrals over the elliptic curve. $\endgroup$ – Will Sawin Feb 13 at 17:46
  • $\begingroup$ @WillSawin Would you explain it in a little bit more detail as an answer? $\endgroup$ – k.j. Feb 13 at 21:40
  • $\begingroup$ @WillSawin What is "trivialize the homology"? I think that it is "take isomorphism $\mathbb{Z}^2 \to H_1(E^\text{an}, \mathbb{Z})$ for all $E$ in a neighborhood $U \subseteq Z$ which is isomorphic to $\mathbb{C}$, in some sense continuously". And I think that this "continuous" is precisely the condition which induces the holomorphy of the map $Z \to \mathbb{H}/\Gamma$. $\endgroup$ – k.j. Feb 14 at 15:06
  • $\begingroup$ I don't really have the energy to write a full answer. For continuously, there are many ways to say it. One is that if you have a loop representation a homology class in one fiber, and deform it to a nearby fiber, it represents the same homology class (regardless of how you deform it). Another is that you have a trivialization of the pushforward sheaf of $\mathbb Z$. This continuity is really what's needed to give a holomorphic map $Z \to \mathbb H$. Maybe you want to think about this in terms of the Gauss-Manin connection. $\endgroup$ – Will Sawin Feb 14 at 18:48
  • $\begingroup$ @WillSawin Thank you very much. I'll try it. Finally, do you know some references which explain it? $\endgroup$ – k.j. Feb 14 at 22:25

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