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Let $\phi_g : \mathcal{M}_g \rightarrow \mathcal{A}_g$ be the period mapping from the open moduli space of genus $g$ Riemann surfaces to the moduli space of $g$-dimensional principally polarized abelian varieties over $\mathbb{C}$. Thus for a Riemann surface $S$ the image $\phi_g(S)$ is the Jacobian of $S$. The Schottky problem consists in determining the image of $\phi_g$.

It is classical that $\text{Im}(\phi_2)$ is exactly the set of abelian varieties that are not isomorphic to a product of elliptic curves. This is asserted in many places, but I have not been able to find a nice discussion of it in the literature. Does anyone know one? The more down-to-earth, the better.

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5 Answers

up vote 6 down vote accepted

This will need expansion by a more knowledgable person, but as memory serves, it was proved by Mayer and Mumford that the closure in Ag of the locus of traditional Jacobians is the set of products of Jacobians. This is probably exposed first in a talk in the 1964 Woods Hole talks on James Milne's site. (I see Mumford credits it there, on page 4 of his talk, in part three of the Woods Hole notes, to Matsusaka and Hoyt. Apparently Mayer and Mumford computed the closure in the Satake compactification.) But let us try to explain this more in dim two.

A two diml ppav is a compact 2 torus A containing a curve C carrying the homology class a1xb1 + a2xb2, where the aj,bj are a basic symplectic homology basis of H1(A).

It follows from the topological Pontrjagin product that the induced map from the Albanese variety of C to A, has topological degree one, hence is an isomorphism. (I.e. the map from the Cartesian product of C with itself g times to A, has image whose class is the g fold Pontrjagin product of [C], which equals g! times the fundamental class of A. Hence the induced map from the g fold symmetric product of C, has image with exactly the fundamental class of A. Hence this map has degree one as does that induced from the Jacobian.)

Since it also induces the identity map on C, it also preserves the polarization.

Let me speculate on the special cases. If C is reducible it is known (Complex abelian varieties and theta functions, George Kempf, p. 89, Cor. 10.4) that A is a product of elliptic curves. If C is irreducible and singular then I guess the normalization map extends to a map of the Albanese of C to A. But that seems to imply the image of C in A does not span, a contradiction.

So it seems that any irreducible curve C contained in a two diml ppav A and carrying the class of a principal polarization, is smooth and induces an isomorphism from the Albanese (i.e. Jacobian) of the curve to the ppav.

I hope there is some useful information in this.

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This was very helpful. Thanks! –  G Fiori Jan 25 '12 at 20:21
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By (a possible) definition, a principal polarization on an abelian surface is a curve with self-intersection 2. So, if smooth, it is a genus two curve and the abelian surface is a jacobian. You have to rule out the case of a singular irreducible curve and the remaining possibility is a union of two elliptic curves meeting at a point and then you show that in this case, the surface is the product of the two curves.

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According to MR0364265 (51 #520) Oort, Frans; Ueno, Kenji "Principally polarized abelian varieties of dimension two or three are Jacobian varieties" (J. Fac. Sci. Univ. Tokyo Sect. IA Math. 20 (1973), 377–381) covers this (I don't have access to the paper and am going solely by the review).

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This was very helpful to me, and I had trouble deciding whether to accept it or roy smith's answer. Thanks! –  G Fiori Jan 25 '12 at 20:22
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Geoffrey Mess' paper "The Torelli group of genus 2 or 3 surfaces" provides two proofs of this fact---one cohomological and the other topological. But I understand neither. Maybe somebody could provide some commentary on his demonstrations?

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I just looked at his paper, and I also understand neither of his proofs. But they look like exactly what I want! Hopefully an expert will come along and unpack them. –  G Fiori Jan 24 '12 at 21:50
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In genus 2 there is NO Schottky problem.

g(g+1)/2 = 3g-3.

Dim(Abelian) = dim(moduli curves)

PS Also in g=2 any curve is hyper-elliptic - again dimension count

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Yes, this argument shows that the image of $\phi_2$ is dense in $\mathcal{A}_2$. But why is its complement as I described above (ie the Jacobians of curves of compact type)? –  G Fiori Jan 24 '12 at 17:39
    
As far as I understand the the product of elliptic curves will have decomposable period matrix "B". I.e. B = diag(t1,t2) While it is not decomposable for any curve or any other abelian variety. So we see, that image is contained in the desired set. But you want more... To show that it coincides we need somehow to restore the curve from the abelian variety... Hmmm, how to do this ? –  Alexander Chervov Jan 24 '12 at 20:07
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