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What are the possible automorphism groups of a principally polarized abelian variety $(A,\lambda)$ of dimension $g,$ say an abelian surface ($g=2$) over the complex numbers or algebraic closure of a finite field? The fact that the moduli stack $A_g$ is of finite diagonal (over the integers) implies that the automorphism groups are all finite, but do we know more? Like the size.

When $g=1$ this is given in Silverman I, p.103.

Edit: Let me make the question more specific. Let $(A,\lambda)$ be an $\mathbb F_q$-point of $A_g$ (i.e. an abelian variety $A$ over $\mathbb F_q$ of dimension $g$ and a principal polarization $\lambda$). We want to consider its automorphism group (over $\mathbb F_q$).

Let $\pi:A_{g,N}\to A_g$ be the natural projection, where $A_{g,N}$ is the moduli stack of p.p.a.v. of dimension $g$ with a level $N$ structure (a symplectic isomorphism $H^1(A,Z/N)\to(Z/N)^{2g}$). We always assume $q$ is prime to $N.$ Note that $\pi$ is a $G$-torsor, for $G=GSp(2g,Z/N),$ so it gives a surjective homomorphism $\pi_1(A_g)\to G.$ The sheaf $\pi_*\mathbb Q_l$ on $A_g$ is lisse (even locally constant), corresponding to the representation of $\pi_1(A_g)$ obtained from the regular representation $\mathbb Q_l[G]$ of $G$ and the projection $\pi_1(A_g)\to G.$ For any $\mathbb F_q$-point $x$ of $A_g,$ the local trace $\text{Tr}(Frob_x,(\pi_*\mathbb Q_l)_{\overline{x}})$ is either $|G|$ or 0, depending on $Frob_x\in\pi_1(A_g)$ is mapped to 1 in $G$ or not.

We have isomorphisms $H^i_c(A_{g,N},\mathbb Q_l)=H^i_c(A_g,\pi_*\mathbb Q_l).$ By Lefschetz trace formula, applied to both $\mathbb Q_l$ on $A_{g,N}$ and $\pi_*\mathbb Q_l$ on $A_g,$ we have

$$|A_{g,N}(\mathbb F_q)|=|G|\sum_{x\in S} 1/\#Aut(A_x,\lambda_x),$$

where $S$ is the subset of $[A_g(\mathbb F_q)]$ consisting of points $x$ such that all $N$-torsion points of the abelian variety $A_x$ are rational over $\mathbb F_q$ (i.e. $|A_x[N](\mathbb F_q)|=N^{2g}$), and $(A_x,\lambda_x)$ is the pair corresponding to $x.$ This equation gives some constraints (one for each $N$) that $|Aut(A,\lambda)|$ must satisfy. In particular, when $g=N=2$ and $q=3,$ we have $|A_{2,2}(\mathbb F_3)|=10$ and $|G|=720$ (in this case $G$ is the symmetric group $S_6$), and this becomes a puzzle of solving $$ 1/72 = \sum 1/n_i, $$ and the $n_i$'s satisfy some additional conditions. Any idea on how to solve it? I'm considering the contributions of the two parts in $A_2,$ one for Jacobians of smooth genus 2 curves and one for Jacobians of stable singular ones $E_1\times E_2$. Any suggestion is appreciated.

Edit: Maybe it's easier to solve it over $\mathbb F_5,$ since the (orders of the) automorphism groups of smooth genus 2 curves over finite fields of characteristic 5 is known.

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One coarse bound is that, for p not the characteristic, the automorphism group of the polarized abelian variety embeds into the symplectic group of 2g x 2g matrices over Z_p via its action on the torsion. You know the automorphism group is finite, and some logarithm-esque argument implies any torsion element that reduces to the identity mod p (or 4 if p=2) is trivial. You also know that a p-complete cyclotomic polynomial for the order of the torsion point has to have order dividing 2g. But these are pretty darn coarse as they only apply to orders of individual elements. – Tyler Lawson Nov 2 '09 at 4:27
@shenghao: If you used LaTeX and the sign "$" around your math formulas, it would be much easier to read... – Sándor Kovács Nov 8 '10 at 16:40
@shenghao: For automorphisms over $\mathbb{C}$ you can also have a look at the book by Birkenhake and Lange, Complex Abelian Varieties. Their Chapter 13 is about this topic; e.g. they give rather precise information in the case of abelian surfaces in Section 13.4. – Lennart Meier Apr 23 '13 at 20:10
up vote 12 down vote accepted

The standard proof of finiteness goes as follows: the polarization defines a positive involution * on the endomorphism algebra, and so the automorphisms are the elements of End(A) tensor the reals that are in End(A) and satisfy a*a=1. Thus the set of automorphisms is the intersection of a discrete set and a compact set. Let phi(n) be the degree of the field you get by adjoining an nth root of 1 to Q. Then certainly there exist abelian varieties of dimension g on which the nth roots of 1 act if phi(n) divides 2g (for any CM-field E there is an abelian variety with complex multiplication by E). However, unlike the elliptic curve case, there are other possibilities.

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It's great to see you here, James. And thank you for your reply. I edited to make the question more specific (maybe you don't like this). It's about "small g and N".... – shenghao Nov 2 '09 at 20:01

If we start with the same problem over the complex numbers, then the automorphism groups of principally polarised abelian varieties of dimension $g$ are exactly the stabilisers of $\mathrm{Sp}_{2g}(\mathbb Z)$ in its action on the Siegel upper half space $\mathbb H_g$. These are finite groups and it is a general fact that a finite subgroup of $\mathrm{Sp}_{2g}(\mathbb Z)$ fixes a point of $\mathbb H_g$. Hence the answer in that case are that the automorphism groups are exactly the maximal finite subgroups of $\mathrm{Sp}_{2g}(\mathbb Z)$. Note that just as for $\mathrm{SL}_g(\mathbb Z)$ (which is a subgroup of $\mathrm{Sp}_{2g}(\mathbb Z)$) the classification of such finite subgroups quickly becomes untractable (for increasing $g$).

If we try to use this to get lower bounds for the case when the base field is the closure of $\mathbb Z/p$ we can start with a pair $(A,G)$ of a p.p.a.v. and a finite group $G$ of automorphisms over a field of characteristic. We can then (after appropriately changing fields) assume that the base field is the fraction field of a DVR $R$ of mixed characteristic with finite residue field of characteristic $p$ and that $A$ has semi-stable reduction over $R$. The action of $G$ extends and we then get $G$ as a subgroup of of the automorphism group of a semi-abelian variety with a principal polarisation of the abelian part of characteristic $p$. Note that if $G$ is large (for some definition of "large") then the reduction is necessarily an abelian variety as the automorphism of a properly semi-abelian variety is "smaller". This should give a lot of characteristic $p$ examples.

This however will not get everything as there are (already in the case of $g=1$) pairs $(A,G)$ that don't lift to characteristic $0$. It does however give all such pairs for which the order of $G$ is prime to $p$. In principle it should be possible to resolve the problem with the approach suggested in Milne's answer. However, I think there is a problem (apart from the fact that the algebraic problem quickly becomes intractable) in that it is a somewhat tricky problem to figure out which pairs consisting of a Rosati involution and a stable order in the isogeny algebra are realisable by principal polarisations. Still there are some constructions that can be used: One can take the product of two principally polarised automorphism group. One can also tensor a principally polarised AV with a positive definite unimodular hermitian form over the endomorphism ring with the Rosati involution. The latter includes starting with a positive definite integral unimodular form of rank $g$ and realising its automorphism group as the automorphism group of the product of $g$ copies of any elliptic curve. Starting instead with a supersingular elliptic curve probably give larger examples.

The problem for finite fields instead of the algebraic closure of one is even more unpredictable.

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If you fix the 2-torsion points, then all you have is the x->-x involution; so Z/2 times SP_2g(2) is a bound from above. On the other hand, if you take A to be g-fold "power" of some elliptic curve, I think you get rather close to this bound.

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Grushevsky's paper GEOMETRY OF Ag AND ITS COMPACTIFICATIONS, remark 2.9 sketches why the diagonal of A_g is finite; maybe you can make that proof effective?

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