Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm looking for an accessible reference for the fact that the moduli stack of principally polarized abelian varieties is in fact an algebraic stack. Faltings/Chai sketch two possible proofs in their book on degenerations of abelian varieties but I don't think I will be able to get the details from this source. Maybe anyone more experienced worked this out and published his thoughts?

share|improve this question
Not really an answer, because I haven't checked it, but Olsson has a LNM fairly recently that I think talks about this. amazon.com/Compactifying-Abelian-Varieties-Lecture-Mathematics/… –  Charles Siegel Nov 22 '09 at 19:48
Olsson's monograph math.berkeley.edu/~molsson/mono020807.pdf –  David Roberts May 22 '12 at 6:05
add comment

4 Answers

The method is really the same as what Deligne-Mumford do to handle the moduli space of curves (creating a smooth cover from a part of a Hilbert scheme), except facts about curves (e.g., Riemann-Roch and cohomological vanishing results) are replaced with analogues for abelian varieties. Below is a guide to relevant literature for why $\mathcal{A}_{g,d}$ is a separated DM stack of finite type over $\mathbf{Z}$. (Question had $d = 1$.)

Everything needed about abelian varieties is in Mumford's book on abelian varieties and the (self-contained!) Chapter 6 on abelian schemes in his GIT book. The key ingredients are (i) the "Riemann-Roch and Vanishing" theorems from section 16 of Mumford's book on abelian varieties, (ii) Proposition 6.13 in Mumford's GIT book (a relativization of results proved over an algebraically closed field in his book on abelian varieties), (iii) Proposition 6.11 in GIT, and (iv) Proposition 6.14 in GIT.

The references (i) and (ii) ensure that if you consider an abelian scheme $f:A \rightarrow S$ of relative dimension $g > 0$ (with dual $A^{\rm{t}}$ and Poincar\'e bundle $\mathcal{P}$) and a polarization $\phi:A \rightarrow A^{\rm{t}}$ on $A$ of degree $d^2$ then for the resulting $S$-ample line bundle $\mathcal{L} = (1, \phi)^{\ast}(\mathcal{P})$ on $A$ the pushforward $\mathcal{E} = f_{\ast}(\mathcal{L}^{\otimes 3})$ is a vector bundle on $S$ whose formation commutes with any base change, has rank determined entirely by $g$ and $d$, and defines a closed immersion $A \hookrightarrow \mathbf{P}(\mathcal{E})$. (In these arguments with relative ampleness, EGA IV$_3$, 9.6.4 is very useful for bootstrapping from fields to a general base.) Thus, upon making a universal (smooth) base change to equip $\mathcal{E}$ with a global frame, one gets a map from the new base to a certain Hilbert scheme for a projective space (over $\mathbf{Z}$) and Hilbert polynomial determined by $g$ and $d$. Since an abelian scheme has a section, this map naturally factors through the universal object $X$ over that Hilbert scheme (which represents the functor of also specifying a section).

Now (iv) comes in: it says that over $X$, the condition on the universal marked object that it admit a (necessarily unique) structure of abelian scheme with the marked section as the identity is represented by an open subscheme. Restricting to this locus (and again using the Hilbert polynomial to track the rank of the pushforward of its $\mathcal{O}(1)$) gives a kind of first approximation to a "universal" abelian scheme $\mathcal{A}$, but we need to universally equip it with a polarization whose cube recovers the $\mathcal{O}(1)$ from the projective embedding as its "associated" ample line bundle (via the construction with the Poincar\'e bundle). That's where (iii) comes in: it says that such a polarization is unique if it exists and that its existence is represented by a closed subscheme of the base. Over that subscheme we get a degree-$d^2$ polarization on an abelian scheme of relative dimension $g$, and the way we arrived at it showed that the base of the family (of finite type over $\mathbf{Z}$) is a smooth cover of the moduli stack (the "smooth cover" coming about because of the arrangement to get a global frame for $\mathcal{E}$).

Since one checks before any of the above that the moduli stack does satisfy effective descent (thanks to descent theory with the ample line bundle associated to the polarization; see 6.1/7 in the book "Neron Models"), we indeed have an Artin stack of finite type over $\mathbf{Z}$. The finite \'etale property of automorphism groups of geometric points implies it is actually a DM-stack (using the "unramified diagonal" criterion for an Artin stack to be a DM stack: Theorem 8.1 in the Laumon/Moret-Bailly book). Finally, to check separatedness one uses the valuative criterion, which is the N\'eronian property for abelian schemes over a discrete valuation ring (1.2/8 in "Neron Models"). QED

share|improve this answer
add comment

The standard source is Alexeev's annals paper (based on the previous paper by Alexeev and Nakamura). It's more readable then Faltings and Chai, but definitely not a light reading.

share|improve this answer
add comment

Mumford has a series of three Inventiones papers from the late '60s called "On the equations defining abelian varieties I/II/II" that may be what you're looking for. It uses sufficient-level-structure language rather than stack language, but that's just a detail.

If you're talking about a compactified moduli, you'll need someone who's more of an expert to refer you.

share|improve this answer
add comment

@Tyler: Mumford (with an extra paper of Kempf On Linear Systems on Abelian vareities) proves the existence of a fine moduli space for Abelian varieties equipped with a totally symmetric relatively ample line bundle and a symmetric "theta structure" of type 2d with d an integer. Unfortunately this moduli problem only makes sense away from characteristic 2 and so it does not suffice in general.

share|improve this answer
My answer above is a relic of a time, long ago, when I would occasionally answer questions, outside my areas of expertise, based on what knowledge I had. The intervening time has made it very clear that such efforts are unnecessary, based on how many excellent contributors the site has. –  Tyler Lawson Oct 29 '13 at 3:54
add comment

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.