When is the Jacobian of a hyperelliptic curve $$y^2=x(x-1)(x-a)(x-b)(x-c)$$ a product of two elliptic curves? (This is a sort of reverse to When is a product of elliptic curves isogenous to the Jacobian of a hyperelliptic curve?). Obviously, it implies some algebraic relations between $a,b,c$; the question is: which ones?

P.S. This must be classical, but I am having some trouble figuring it out or finding it in the literature. There is an old example due to Jacobi, but is it all?

EDIT. As pointed out by abx, the set of parameters $(a,b,c)$ for which the Jacobian is isogenous to a product is actually dense (e.g. in $\mathbb{C}^3$ with complex topology). For this reason, the question in this form does not have a reasonable answer. (In hindsight, it is obvious, but I missed it.)

  • $\begingroup$ The locus of p.p.a.v. of dimension 2 isomorphic to a product of elliptic curves (or even to the self-product of an elliptic curve with complex multiplication) is dense in the moduli space, so I am not sure what you can expect apart from (countably many...) examples. $\endgroup$ – abx Oct 2 '16 at 7:03
  • $\begingroup$ I can see two ways of interpreting your question: (a) the Jacobian of $C$ is isogenous to a product of two elliptic curves; (b) the Jacobian of $C$ is isomorphic to a product of two elliptic curves [but the polarization on $J_C$ is not the product polarization on the product of elliptic curves]. Which do you mean? $\endgroup$ – Dan Petersen Oct 2 '16 at 7:09
  • $\begingroup$ Actually, I meant isogenous. $\endgroup$ – Alex Gavrilov Oct 2 '16 at 7:24
  • $\begingroup$ @abx: Is it really dense? I did not expect it. $\endgroup$ – Alex Gavrilov Oct 2 '16 at 7:28
  • $\begingroup$ On second thought, you are right! $\endgroup$ – Alex Gavrilov Oct 2 '16 at 7:32

A genus two curve $C$ has its jacobian isogenous to a product of elliptic curves if and only if there is a nonconstant map from $C$ to an elliptic curve $E$. For each degree of map $C \to E$ there is a hypersurface in the moduli space of genus two curves of such jacobians. So you have a countable union of hypersurfaces, each of which gives you an algebraic relation. Probably one can make these conditions on $a,b,c$ explicit for small degrees. The easiest ones to characterize are the bielliptic ones: in this case, the genus two curve must have a Weierstrass equation of the form $y^2 = f(x^2)$, and the bielliptic involution is $x \mapsto -x$.

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  • $\begingroup$ Thank you. Is there a reason to expect this hypersurface to be irreducible? $\endgroup$ – Alex Gavrilov Oct 2 '16 at 7:53
  • $\begingroup$ Yes, it's irreducible, as long as you consider only curves $C$ which admit a minimal map $C \to E$ of a given degree, where minimal means that the map doesn't factor through an isogeny of elliptic curves. These hypersurfaces are called Humbert surfaces of square discriminant. $\endgroup$ – Dan Petersen Oct 2 '16 at 8:27

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