picard group of moduli of elliptic r-prym curves

Let $\overline{\mathcal{M}}_{1,1}$ be the DM compactification of the moduli stack of elliptic curves. Its Picard group is $\mathbb{Z}$. Let us now consider stack of $r$-prym curves $\overline{\mathcal{M}}_{1,1}^r$, parametrizing elliptic curves plus an $r$-root of $\mathcal{O}$. It is compactified naturally as an Hurwitz stack. Is its Picard group known?

• For spin curves, the result seems to be in section 4 of Jarvis, "The Picard Group of the Moduli of Higher Spin Curves", NY Journ. Math 2001. Do you expect the result to be very different? Oct 27, 2013 at 22:40
• so, as I stated below, Matthieu was right. they should be not different at all! Oct 31, 2013 at 16:28

The stack $\overline{\mathcal M}_{1,1}^r$ is usually denoted by $X_1(r)$. It is a modular curve. In fact one can write down by hand an isomorphism between the moduli functors for the Hurwitz stack interpretation of $\overline{\mathcal M}_{1,1}^r$ that you mention, and the modular interpretation of $X_1(r)$ described in Deligne-Rapoport: this isomorphism is described in Section 3 of http://www.ams.org/mathscinet-getitem?mr=2968638 . I should also say that there is a minor issue here -- there are two common definitions of $\overline{\mathcal M}_{1,1}^r$, and one is a $\mu_r$-gerbe over the other. For the isomorphism $\overline{\mathcal M}_{1,1}^r \cong X_1(r)$ you should use the "rigidified" version.
Anyway, the Picard group of a smooth twisted curve $\mathscr C$ is an extension $$0 \to \mathrm{Pic}(C)\to \mathrm{Pic}(\mathscr C) \to \prod_{i=1}^k \mu_{r_i} \to 1,$$ if there are $k$ stacky points with stabilizers of order $r_1, \ldots, r_k$, and $C$ is the coarse space. Most textbooks on modular forms will describe the genus of $X_1(r)$, its number of cusps and the respective stabilizers. For instance this is described very explicitly in Diamond-Shurman.
• thanks, I will read. I guess that if $r\geq 3$ then the modular curve is already rigid right? I only see the $\pm Id$ automorphism, when $r=2$. Oct 28, 2013 at 7:25
• Actually, $r\ge 4$ Oct 28, 2013 at 10:31