Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It's well-known that $SL_2(\mathbb{Z}) \backslash \mathfrak{h}$ is a coarse moduli space for complex elliptic curves. Thus, I would expect this to be related to the pullback of $\mathcal{M}_{ell} \rightarrow \mbox{Spec}(\mathbb{Z})$ along $\mbox{Spec}(\mathbb{C}) \rightarrow \mbox{Spec}(\mathbb{Z})$; for instance, I might expect this pullback to look something like $SL_2(\mathbb{Z}) \backslash \! \! \backslash \mathfrak{h}$. However, I'm pretty sure that $\mathfrak{h}$ isn't actually a complex variety (basically by the Riemann mapping theorem), so at best this would admit a map from the analytification of the actual algebro-geometric pullback.

The answer might be bound up in the $j$-invariant; over $\mathbb{C}$, this is a "biholomorphism" $SL_2(\mathbb{Z}) \backslash \mathfrak{h} \rightarrow \mathbb{C}$, i.e. it is a holomorphic bijection of complex orbifolds. (Around the cone points $i \in \mathfrak{h}$ and $\omega=e^{2 \pi i /3}\in \mathfrak{h}$, the map is locally modeled by $z \mapsto z^2$ and by $z \mapsto z^3$, respectively.) This has always been sort of mysterious to me, but I think the point is just that "biholomorphism" is the wrong notion of equivalence for complex orbifolds; it seems somehow besides the point to me that we happen to have such an equivalence.

share|cite|improve this question
Hmmm, what is your question exactly? – Álvaro Lozano-Robledo Apr 1 '13 at 19:57
The entire question is contained in the title. I don't believe that $\mathfrak{h}$ is a complex-algebraic object, nor does it seem to me to be in the (essential) image of the GAGA analytification functor. The $j$-invariant gives some non-stacky equivalence with $\mathbb{A}^1$ or $\mathbb{P}^1$, but I'm interested in the stack itself. – Aaron Mazel-Gee Apr 2 '13 at 1:24
Of course, please let me know if anything I'm saying is wrong! I'm by no means an algebraic geometer, so this is all somewhat foreign to me. (@ÁlvaroLozano-Robledo) – Aaron Mazel-Gee Apr 2 '13 at 4:05
For $N \geq 3$, the modular curve $X(N)$ is a fine moduli space for elliptic curves with level $N$ structure. And then the stack $X(N)/PSL_2(\mathbb{Z}/N \mathbb{Z})$ is the moduli space of elliptic curves where you forget the level structure. This construction works over $\mathbb{Z}[1/N]$ so, gluing this construction for two relatively prime values of $N$, you get an answer over $\mathbb{Z}$. Is this what you were looking for, or did I misunderstand the question? (Standard references are Katz-Mazur and Deligne-Rapoport.) – David Speyer Aug 3 '15 at 2:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.