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I'd like to know examples of non-trivial families of abelian varieties over rational bases (e.g. open subschemes of the projective line P^1).

One can generate many examples as Jacobians of rational families of curves (e.g. the hyperellitpic family, plane curves, complete intersections). Prym varieties are another example.

Are there any examples which are not obviously Jacobians of a family of curves? I would like to know both principally polarized and non principally polarized examples.

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  • $\begingroup$ David, can you explain the question a bit? Thanks. $\endgroup$
    – shenghao
    Commented Oct 16, 2009 at 20:56
  • $\begingroup$ By "rational base" you mean a rational variety like affine spaces, not Spec of the rational numbers, right? So you want a non-constant k-morphism from some rational k-variety to like the moduli space A_{g,d} of dimension g degree d polarized abelian varieties? $\endgroup$
    – shenghao
    Commented Oct 17, 2009 at 0:49

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David, I don't know if you are still interested in this, it's been over a year. I just stumbled upon your question in the depths of MO. I often found that Weil restrictions of elliptic curves give nice families of examples on which you can test things.

E.g. take a family of elliptic curves $y^2=x^3+t$ and Weil restrict it from ${\mathbb Q}(i)$ to ${\mathbb Q}$. Writing $x=x_1+x_2i$ and similarly for $y$ and $t$, expanding the equation and breaking it into real and imaginary parts, you get a family of 2-dimensional abelian varieties over ${\mathbb Q}(t_1,t_2)$ given by two equations in a 4-dimensional space, $$ y_1^2-y_2^2 = x_1^3 - 3x_1 x_2^2 + t_1, \qquad 2y_1y_2 = x_1^2x_2 - 3x_2^3 + t_2. $$ Alternatively, you fix the elliptic curve, but you let the extension vary with $t$ (e.g. ${\mathbb Q}(t^{1/3})$), or both, and you also get interesting families.

The really nice thing is that as opposed to Jacobians, Weil restrictions are trivial to write down in terms of equations. Over the algebraic closure they are isogenous isomorphic to products of elliptic curves (making them boring), but for arithmetic applications they are interesting. There is a small extension of this construction, when you do not base change the elliptic curve but you "tensor it with a ${\mathbb Z}-$module with a Galois action", which is not necessarily a permutation module. This is explained in Milne's paper "On the arithmetic of abelian varieties" (Invent. Math. 1972) section 2, and it is useful if you want to write down non-principally polarised examples.

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    $\begingroup$ Actually, over the algebraic closure, a Weil restriction of an elliptic curve becomes isomorphic (not just isogenous) to a product of elliptic curves (that are Galois-conjugates of the original one). $\endgroup$
    – Yuri Zarhin
    Commented Feb 5, 2011 at 22:33
  • $\begingroup$ You are absolutely right. Thank you, fixed! $\endgroup$
    – Tim Dokchitser
    Commented Feb 5, 2011 at 23:17
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It is known that $A_g$ is unirational for $g\leq 5$, so at least for these g, we have some nontrivial families over open subsets of projective spaces. For $g> 3$, $\dim M_g < \dim A_g$, so these families are not Jacobians.

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  • $\begingroup$ Do these all come from Prym's? $\endgroup$
    – David Zureick-Brown
    Commented Oct 17, 2009 at 20:43
  • $\begingroup$ I don't know what are Prym varieties...but you can read papers proving A_g are unirational, and see the construction there. $\endgroup$
    – shenghao
    Commented Oct 18, 2009 at 0:26
  • $\begingroup$ They don't all come from Pryms -- Pryms arise from Jacobians of lower-genus curves, so the locus of Pryms in A_g has dimension strictly smaller than 3g-3. Once g > 3, the Jacobians and Pryms form a finite union of proper subvarieties, so there are certainly rational curves not contained in this locus if A_g is unirational. (If you think of families of abelian varieties ISOGENOUS to Jacobians as also "coming from Jacobians," then you have a countable union of proper subvarieties you have to miss; it's still clear you can do it with a rational curve over C, not so easy over Qbar..) $\endgroup$
    – JSE
    Commented Oct 19, 2009 at 16:14
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    $\begingroup$ In fact they do all come from Pryms. The point is that to get a g dimensional abelian variety as a Prym one has to consider etale double covers of curves of genus g+1; one gets all ppavs of dimension g as Pryms exactly when g <= 5. $\endgroup$
    – naf
    Commented Oct 20, 2009 at 6:32
  • $\begingroup$ unknown above is right, sorry for the mistake! $\endgroup$
    – JSE
    Commented Oct 24, 2009 at 1:53
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There are nice examples of Gross and Popescu of Calabi-Yau threefolds fibered over $\mathbb P^1$ with generic fibers being abelian surfaces. See arXiv:math/000108 and arXiv:0904.3354.

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  • $\begingroup$ In a similar vein, Horrocks--Mumford quintics: see e.g. arxiv.org/abs/math/0102055 $\endgroup$
    – user5117
    Commented Mar 11, 2014 at 12:27
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Section 5 of Faltings: Arakelov's theorem for abelian varieties is dedicated to the construction of an elaborate example of a family of abelian varieties that do not satisfy a condition Faltings calls $(*)$. I believe the condtion implies that the abelian varieties in the family, at least the general fiber, cannot be a Jacobian. The base of the family is $\mathbb H$ the upper-half plane, so it won't give you a family over a rational curve, but the construction itself is pretty involved so undertsanding it might help with understanding families of abelian varieties in general and might also give you some ideas for this problem.

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    $\begingroup$ There is an explicit example of a certain family of abelian eightfolds at the very end of my paper (Proc. London Math. Soc. 96 (2008), no. 2, 312--334) where the general fiber is the square of a Jacobian while the condition (*) does not hold. $\endgroup$
    – Yuri Zarhin
    Commented Feb 5, 2011 at 21:52
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    $\begingroup$ Using a fact that every abelian variety is isogenous to a quotient of a Jacobian, one may construct families of Jacobians that do not satisfy the condition (*). $\endgroup$
    – Yuri Zarhin
    Commented Feb 6, 2011 at 4:18
  • $\begingroup$ Just for ease, the paper mentioned in a comment above is titled "Isogeny classes of abelian varieties over function fields" and is accessible at doi:10.1112/plms/pdm049, arXiv:math/0504523 [math.AG], CiteSeerX (Zbl 1145.14034). $\endgroup$
    – The Amplitwist
    Commented Sep 19, 2022 at 3:45
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How about intermediate Jacobians of cubic threefolds? It's easy to write down rational families of cubic hypersurfaces in P^4; it might be much harder to say anything about the corresponding family of abelian varieties, depending on what features you're looking for.

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    $\begingroup$ There are different ways to make intermediate Jacobians. But as far as I know, either they produce abelian varieties, that is they come with a polarization (the Weil ones) or deform holomorphically in families (the Griffiths ones), but not both at the same time. $\endgroup$
    – Abdó Roig-Maranges
    Commented Oct 19, 2009 at 16:23
  • $\begingroup$ Hmm. So for this family of intermediate Jacobians of cubic hypersurfaces, can one calculate the monodromy? $\endgroup$
    – David Zureick-Brown
    Commented Oct 19, 2009 at 17:24
  • $\begingroup$ These intermediate Jacobians are 5 dimensional ppavs. There are general results about the monodromy action on the middle cohomology of families of hypersurfaces of any degree (for example, by Beauville). The monodromy is usually as big as possible; in the cubic threefold case I think it might be all of Sp(10,Z). $\endgroup$
    – naf
    Commented Oct 20, 2009 at 6:40
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    $\begingroup$ @Abdó: Cubic threefolds have $h^{0,3}=h^{3,0}=0$ so the Hodge structure of $H^3$ is a Tate twist of an abelian variety and there are no problems with getting both a polarisation and holomorphic variation. I think (but am not totally sure) that in general when there are only two non-zero Hodge numbers, the intermediate Jacobians are both polarised and deform holomorphically. It is only when there are at least four non-zero Hodge numbers that one has to involve complex conjugation in the Weil construction. $\endgroup$
    – Torsten Ekedahl
    Commented Nov 26, 2010 at 6:14
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    $\begingroup$ @DavidZureick-Brown There are two papers by Jeffrey Achter which determine all the five-dimensional abelian varieties associated to a cubic surface (via the intermediate Jacobian of the cubic threefold associated to the cubic surface); these are available on arXiv at arxiv.org/abs/1208.2974 and arxiv.org/abs/1005.2131 . $\endgroup$
    – Ariyan Javanpeykar
    Commented Mar 11, 2014 at 12:12
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Just for the curiosity, into which category falls the fibration which has an elliptic curve with given j-invariant over a point $j \in \mathbb{P}^1$?

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  • $\begingroup$ An elliptic curve is its own Jacobian, if that helps. $\endgroup$
    – Anweshi
    Commented Feb 5, 2010 at 13:19
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One can construct some families over rational bases which are not Jacobians by taking quotients:

For example, let $A$ be a fixed abelian variety of dimension $> 1$ and let $S$ be the space of all smooth complete interesection curves for some very ample line bundle on $A$. For any $s \in S$, let $C_s$ denote the corresponding curve in $A$. The inclusion of $C_s$ in $A$ induces a surjective morphism from $J(C_s)$ (the Jacobian of $C_s$) to $A$ and so by duality a morphism $A^{t}$ to $J(C_s)$ where $A^{t}$ is the dual abelian variety of $A$. The quotient of $J(C_s)$ by the image of $A^t$ gives a family of abelian varieties over $S$. It can be shown using monodromy that this family is non-trivial.

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  • $\begingroup$ "..let S be the space of all smooth complete interesection curves for some very ample line bundle on A." What does this mean? $\endgroup$
    – David Zureick-Brown
    Commented Oct 19, 2009 at 16:34
  • $\begingroup$ Let L be a very ample line bundle on A, let T = P(H^0(A,L)) and S be the open subset of T^{d-1}, d = dim(A), corresponding to those intersections of d-1 hypersurfaces which are smooth. $\endgroup$
    – naf
    Commented Oct 20, 2009 at 6:27

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