Let $\mathfrak{M}_g$ denote the moduli stack of Riemann surfaces of genus $g$, it is a smooth complex analytic stack, and is the analytic stack underlying $\mathsf{M}_g$, the moduli stack of complex algebraic curves of genus $g$. Alternatively, it is the quotient stack $\mathcal{T}_g // \Gamma_g$ of the mapping class group acting on Teichmuller space (by biholomorphisms).

There are two things we might mean by the Picard group of $\mathfrak{M}_g$, using either holomorphic line bundles or algebraic line bundles. The first is $Pic_{hol} := H^1(\mathfrak{M}_g;\mathcal{O}^\times_{\mathfrak{M}_g})$ and the second is $Pic_{alg} := H^1(\mathsf{M}_g;\mathcal{O}^\times_{\mathsf{M}_g})$. As $\mathfrak{M}_g$ has a finite cover by the quasiprojective variety $\mathfrak{M}_g[3]$ of curves with a level 3 structure, I believe the latter group can also be defined to be the subgroup of $Pic_{hol}$ of those holomorphic line bundles which become algebraic on $\mathfrak{M}_g[3]$. If not, this defines yet another notion of Picard group.

It seems to me that it is $Pic_{alg}$ which is usually discussed, for example in Mumford's paper on Picard groups of moduli problems. Is $Pic_{hol}$ discussed explicitly anywhere? On the other hand, presumably these groups agree (certainly $Pic_{alg} = \mathbb{Z}$ is a summand of $Pic_{hol}$). Is this explicitly stated and proved anywhere?

Any other observations about this distinction are welcome too.

EDIT: As a related question, suppose that one only knew about the analytic object $\mathfrak{M}_g$. Can one directly show that $c_1 : Pic_{hol} \to H^2(\mathfrak{M}_g;\mathbb{Z})$ is injective? (Granted $H^1(\mathfrak{M}_g;\mathbb{Q})=0$, say.)

  • $\begingroup$ Shall we assume $g \geq 2$? $\endgroup$ – S. Carnahan Nov 15 '10 at 10:30
  • $\begingroup$ @Scott: I suppose so. $\endgroup$ – Oscar Randal-Williams Nov 15 '10 at 10:32
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    $\begingroup$ "I believe [that $\Pic_{alg}$] can also be defined to be the subgroup of $Pic_{hol}$ of those holomorphic line bundles which become algebraic on $\mathfrak{M}_g[3]$": What is the rationale behind this belief? Why should two holomorphically equivalent algebraic line bundles be algebraically equivalent? The moduli space of level 3 curves is not proper over $Spec (\mathbb{C})$... $\endgroup$ – Johannes Ebert Nov 15 '10 at 11:37
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    $\begingroup$ @Johannes: I don't know an a priori reason, but the composition $Pic_{alg} \to Pic_{hol} \to H^2(\mathfrak{M}_g;\bZ)$ is known to be an isomorphism, so $Pic_{alg} \to Pic_{hol}$ is indeed injective. $\endgroup$ – Oscar Randal-Williams Nov 15 '10 at 12:20

This will be a bit sketchy, but hopefully you can fill in the necessary steps. If you see a problem, let me know.

To avoid getting distracted with stacks, let me use a level $n\ge 3$ structure, and write $M=M_{g}[n]$. Let $j:M\hookrightarrow \bar M$ be the Satake (not Deligne-Mumford) compactification. This is the normalization of the closure of the image of $M$ in the Satake compactification of $A_{g}[n]$. The point is that $codim (\bar M-M)\ge 2$. So given a holomorphic line bundle $L$ on $M$, $j_*L$ would be a coherent analytic sheaf. Therefore by GAGA $j_*L$ is coherent algebraic, so that $L$ is an algebraic line bundle. This implies surjectivity of map $Pic_{alg}\to Pic_{hol}$ in your notation.

This should finish it, since as you said, you already know injectivity of the above map.

  • $\begingroup$ At what point have you used that the codimension is at least 2? $\endgroup$ – Oscar Randal-Williams Nov 15 '10 at 18:44
  • $\begingroup$ Coherence of $j_*L$. $\endgroup$ – Donu Arapura Nov 15 '10 at 18:51
  • $\begingroup$ The coherence of $j_*L$ follows from Hartogs extension theorem, correct? $\endgroup$ – Johannes Ebert Nov 15 '10 at 21:28
  • $\begingroup$ Yes, exactly. I'll try to expand the answer when I have a bit more time. $\endgroup$ – Donu Arapura Nov 15 '10 at 21:56

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