# When is a product of elliptic curves isogenous to the Jacobian of a hyperelliptic curve?

David's question Families of genus 2 curves with positive rank jacobians reminded me of a question that once very much interested me: when is a product of elliptic curves isogenous to the jacobian of a hyperelliptic curve?

How is this related to David's question? Well, if we can multiply two elliptic curves over $\mathbb{Q}(t)$ with large rank, and the result is isogenous to the jacobian of a hyperelliptic curve, then this will probably produce record families answering David's question, i.e. genus two curves with very large rank. It is also interesting for all genera, so don't restrict answers to 2. On the other hand, answers containing arithmetic information, for example on elliptic curves over the rationals, are more than welcome.

• A nice example due to Jacobi: the Jacobian of $y^2=x(1-x)(1+ax)(1+bx)(1-abx)$ is a product of elliptic curves. Aug 10, 2010 at 13:50
• By the way, the strategy of Gouvea, Mazur, Rubin, Silverberg for generating lots of quadratic twists of high rank is to find a hyperelliptic curve C on E^r (whose Jacobian thus has E^r as an isogeny quotient.) Then the presence of lots of points on C over quadratic fields (plus some care about linear independence under specialization) shows that E acquires rank at least r over lots of quadratic fields.
– JSE
Aug 11, 2010 at 0:42

Over the complex numbers every product of two elliptic curves is isogenous to the jacobian of a genus 2 (hyperelliptic) curve. Indeed, the corresponding Siegel upper half-space $H_2$ is an orbit of the real symplectic group $Sp(4,R)$. Since the subgroup $Sp(4,Q)$ of its rational points is everywhere dense in $Sp(4,R)$, every $Sp(4,Q)$-orbit is everywhere dense in $H_2$ and therefore meets the (Torelli) open subset $T_2$ of $H_2$ that parametrized" the Jacobians. Now one has only to notice that if points $x,y \in H_2$ correspond to principally polarized abelian surfaces $A_x$ and $A_y$ respectively then $A_y$ is isogenous to $A_x$ if $y$ lies in the $Sp(4,Q)$-orbit of $x$. (And, of course, one should take $x$ with $A_x$ being a product of two elliptic curves.)

Similar arguments prove that every product of three elliptic curves is isogenous to the jacobian of a genus 3 curve. (These arguments were used in Sect. 2, Remark 3 on pp. 60--61 of arXiv:0912.4325v1 [math.NT] in order to prove certain properties of the modular height.)

• Thanks for being the first to not restrict to the product of only 2 elliptic curves. As you say, 2 and 3 are not hard (at least over $\mathbb{C}$). Do you know anything non-trivial when the product is of more curves? (I realize this is harder since the Torelli subset now has positive codimension) Feb 8, 2012 at 22:46
• @Dror Speiser You are welcome. Alas, I don't know anything non-trivial when the product is of more curves except some negative results that tell that jacobians of certain superelliptic curves are not isogenous to products of elliptic curves. Feb 8, 2012 at 22:57

As damiano makes implicit in his answer, a curve of genus two admits a map of degree N to an elliptic curve if and only if there is an isogeny of degree N^2 of its Jacobian and a product of two elliptic curves. The case N=2 also admits the following simple description: a genus two curve (over an alg. closed field) is a double cover of an elliptic iff it can be written as $y^2 = f(x^2)$, where $f$ is a squarefree cubic with $f(0) \neq 0$.

One can say slightly more. Given a map $f : C \to E_1$ of degree N, we get also a map $f_\ast : \mathrm{Jac} \; C \to E_1^\vee$. If we assume that f does not factor through an isogeny then the kernel is connected so we get a second elliptic curve E2 lying as a subgroup on the Jacobian. So the Jacobian has two elliptic subgroups and it turns out that they intersect exactly in their N-torsion points. We get an induced isomorphism between the N-torsion subgroups which turns out to invert the Weil pairing, and this is the data needed to reconstruct C. Namely, given two elliptic curves and an isomorphism of their N-torsion subgroups inverting the Weil pairing, one may consider the graph of this isomorphism in the product of the elliptic curves. The condition on the Weil pairing ensures that the graph is a maximally isotropic subgroup, so the quotient will have a principal polarization, and (when the quotient is not again a product of two elliptic curves) this is exactly the Jacobian of C with its principal polarization.

This is pretty classical stuff which is well explained in several articles by Ernst Kani and Gerhard Frey. Start with "Curves of genus 2 covering elliptic curves and an arithmetical application." It can also be fruitful to think about this in a general context of Prym varieties -- when we have a degree N map from a genus two curve to an elliptic curve, the second elliptic curve which appears is of course exactly the Prym variety, which in this case is principally polarized.

I know in particular that Frey and Kani worked out some rather precise conditions on when a principally polarized abelian surface arising as a quotient of a product of two elliptic curves as above actually is the Jacobian of a curve. I am not really sure in what article it can be found though. It was something like: pairs of elliptic curves with an isomorphism of their N-torsion are parametrised by $Y(N) \times Y(N) / SL(2,{\mathbb Z}/N)$, and the locus of such pairs which do not give rise to the Jacobian of a curve is the union of certain Hecke correspondences on Y(N). The description of exactly which Hecke correspondences was pretty complicated but when N is prime it was at least workable. The case of N=2 is easy though: then the "bad" locus is just the image of the diagonal in $Y(2) \times Y(2)$.

Here is one easy case of a curve of genus two whose Jacobian is isogenous to a product of two elliptic curves.

Let $p,q,r$ be homogeneous separable polynomials of degree two in two variables $s,t$ that are pairwise relatively prime and let $C$ be the smooth projective model of the curve with equations $$\begin{array}{rcl} x^2 & = & pq , \cr y^2 & = & pr . \end{array}$$ Clearly, each of the equations defining $C$, taken individually, defines a curve of genus one, and the curve $C$ maps to both these curves with degree two. Also, it is easy to check that such morphisms ramify at two points (above the roots of the polynomial $q$ or $r$). Thus $C$ is a curve of genus two with two morphisms to curves of genus one. It follows that the Jacobian of $C$ is isogenous to the product of the Jacobians of the two curves of genus one described above.

You can play similar tricks with more equations, to obtain curves with lots of morphisms to curves of genus one. Unfortunately, I do not know what conditions guarantee that the resulting curve be hyperelliptic, nor that the Jacobian of the whole curve be isogenous to the product of all the Jacobians of the genus one curves.

• If we let $D \to C$ be an unramified morphism of degree two, where $D$ is connected and projective, and $C$ is one of the curves given in the answer above, then $D$ has genus three and its Jacobian is isogenous to a product. Indeed, it is isogenous to the product of the Jacobian of $C$ and the cokernel of the pull-back map $Jac(C) \to Jac(D)$, which is a abelian variety of dimension one. It is unclear to me, though, what would be the conditions implying that $D$ is hyperelliptic (and it might even be that it is never hyperelliptic...). Aug 10, 2010 at 15:50
• Re: your last sentence, in fact the very opposite is true!! Whenever a curve of genus three is a double cover of a genus two curve, it is automatically hyperelliptic. This is shown on p. 147 of Maurizio Cornalba's "On the locus of curves with automorphisms". Aug 11, 2010 at 7:41
• ...and on the other hand, by playing around with the Hurwitz formula and using the fact that the h.e. involution is central, one can show that a hyperelliptic curve of genus at least four can never be a double cover of a genus one curve. So after genus three the strategy of using Prym varieties to find elliptic subgroups of the Jacobian will no longer work. Aug 11, 2010 at 7:47
• Very nice, thanks for the pointer! I see that Cornalba dedicated a special case to this situation! Also, it seems that the problem with degree two has been pushed to the extreme: very satisfying! Aug 11, 2010 at 8:02

Many examples can be found in the recent paper of Paulhus, "Elliptic Factors in Jacobians of Hyperelliptic Curves with Certain Automorphism Groups." She gives examples in genus 3,4,5,6, and 9. The genus 9 curve actually has Jacobian isogenous to the four copies of one elliptic curve plus five copies of another!

The 1993 paper of Ekedahl and Serre remains the most thorough general investigation of curves (not just hyperelliptic curves) whose Jacobians are isogenous to products of elliptic curves.

With 2^2 isogeny it never works:

• F. Richelot, De transformatione integralium Abelianorum primi ordinis comentatio. J. reine angew. Math. 16 (1837) 221-341
• G. Humbert, Sur la transformation ordinaire des fonctions abeliennes. J. de Math. (5) 7 (1901), 359 - 417 )
• J.-B. Bost and J.-F. Mestre, Moyenne arithmetico-geometrique et periodes des courbes de genre 1 et 2, Gaz. Math. 38 (1988), 36-64

With 3^2 it always does (unless the kernel of the isogeny splits on the two components) - see 3.3 in http://arxiv.org/abs/0710.1298 ( = AMS contemp. math. series, volume 465 (2008): Curves and abelian varieties, p 51-69)

The paper of Hayashida and Nishi "Existence of curves of genus two in the product of smooth elliptic curves", J. Math. Soc. Jap 17, partially answers the question:

"When is the product of two elliptic curves $isomorphic$ to the Jacobian of a genus $2$ curve?"

The result is the following:

THEOREM Consider a product of two elliptic curves $A:=E \times F$, whose ring of endomorphism is isomorphic to the principal order of an imaginary quadratic field $Q(\surd-m)$. Then $A$ can be a Jacobian for all values of $m$ except $1, 3, 7$ and $15$. Moreover, there are only finitely many curves of genus $2$ on $A$ up to isomorphism.

• To be clear, this (nice) paper answers the question only in a subbcase of the case in which the two elliptic curves have complex multiplication (CM). This makes the situation much more arithmetic in nature (since CM abelian varieties do not vary in moduli). There has been a lot of work on this problem in general, but I think the answer is probably out of reach. Aug 10, 2010 at 21:51

In two papers dated 2014 and 2016 (i.e., after 6 years the question was posted), this question was completely solved in genus 2 case by E. Kani. As Francesco Polizzi mentioned, this question was partially solved in 1965 by Hayashida and Nishi. They only worked on the abelian varieties $$A$$ over $$\mathbb{C}$$ with the product of two $$CM$$ elliptic curves by the maximal order. So there are two things to remove the assumptions: 1) $$A$$ may be $$E_1\times E_2$$ with their endomorphism rings are not necessarily isomorphic. 2) Their endomorphism ring may be any order, i.e., not necessarily the maximal order.

By using the refined Humbert invariant which is an integral quadratic form, Kani translated this geometric problem into an arithmetic problem and removed these two assumptions.

Here is the statement for the $$CM$$ case from [JT]:=Jacobians isomorphic to a product of two elliptic curves and ternary quadratic forms, 2014.:

Let $$E_1\sim E_2$$ be two isogenous $$CM$$ elliptic curves over an algebraically closed field $$K$$ with its characteristic is zero. Then there is no genus 2 curve on $$E_1\times E_2$$ if and only if the degree map on $$Hom(E_1,E_2)$$ is equivalent to one of the 15 forms $$f(x,y)=ax^2+bxy+cy ^2$$ whose coefficients $$(a,b,c)$$ are in the following list: $$\mathcal{S}=\{k(1,1,1):k=1,2,4,6,10\}\cup\{k(1,0,1):k\in 1,2,6\}\cup\{(1,1,2),(1,1,4)\}\cup\{2(1,1,c):c=3,9\}\cup\{2(1,0,c):c=2,5\}\cup\{2(2,0,3)\}.$$

The statement for the nonCM case can be found in Theorem 1 of [JT]. Its proof can be found in the article of The moduli spaces of Jacobians isomorphic to a product of two elliptic curves, 2016.

Some notes: The assumption of the characteristic 0 can be removed. This was also done by Ibukiyama/Katsura/Oort.

The exact list is only known under the GRH in the nonCM case. But Kani's solution doesn't depend on the GRH.