Timeline for When is a product of elliptic curves isogenous to the Jacobian of a hyperelliptic curve?
Current License: CC BY-SA 2.5
5 events
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| Aug 11, 2010 at 8:02 | comment | added | damiano | 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 7:47 | comment | added | Dan Petersen | ...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 7:41 | comment | added | Dan Petersen | 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 10, 2010 at 15:50 | comment | added | damiano | 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 7:12 | history | answered | damiano | CC BY-SA 2.5 |