It is well known that the Abel-Jacobi map restricted to $\text{Eff}_g(C)$ surjects onto the Jacobian $\text{Jac}(C)$, since every divisor of degree $g$ is effective.

Is there an analogous statement for Prym varieties? That is, given an unramified double cover $\widetilde C\to C$ with involution $\tau$, consider the map $f:\text{Eff}_d(\widetilde{C})\to\text{Prym}(\widetilde{C}/C)$ given by $f(D)=D-\tau(D)$. Is $f$ surjective if, for instance, $d=g-1$?


1 Answer 1


First of all, note that your definition is not correct: when $d$ is odd, the image of your map does not land in the Prym variety -- you have to add a constant term. When this is done, the answer is yes, for the following reason. Let $X$ be the image of $\tilde{C} $ in $P:=\operatorname{Prym}(\tilde{C}/C ) $. Let me put $h:=g-1=\dim P$. What you want to prove is that the addition map $X^{h}\rightarrow P$ is surjective, that is, of degree $>0$. Now this degree is computed by the Pontryagin product $[X]^{*h}$, where $[X]$ is the class of $X$ in $H^{2h-2}(P,\mathbb{Z})$. We know that this class is $2\dfrac{\theta ^{h-1}}{(h-1)!} $, where $\theta $ is the class of the principal polarization.

So we just have to prove that $\theta ^{*h}\in H^{2h}(P,\mathbb{Z})$ is nonzero. This is true for any principally polarized abelian variety $(P,\theta )$ of dimension $h$: it suffices to prove it for a Jacobian $J\Gamma $, and this amounts to say that the Abel-Jacobi map $\Gamma ^h\rightarrow J\Gamma $ is surjective, as you recall in your post.

  • $\begingroup$ Thanks for your answer! You're right, I meant to say that either there is a parity condition on the degree, or that the image falls in a translation of the Prym. Why does it suffice to prove the claim for a Jacobian? $\endgroup$
    – Yoav Len
    May 8, 2020 at 19:43
  • $\begingroup$ Because this is a question about cohomology classes, hence invariant by deformation. But you can also compute $\theta^{*h}=h!$ directly. $\endgroup$
    – abx
    May 8, 2020 at 19:48
  • $\begingroup$ I thought the Prym is the kernel of the norm map, and since a divisor of the form $D - \tau(D)$ always maps to $0$ via the norm map it represents a point in the Prym by definition? $\endgroup$
    – zudumazics
    Aug 15, 2021 at 13:36
  • 1
    $\begingroup$ @zudumazics: no, the kernel of the norm map has two components, the Prym being the component of zero. So $D-\tau(D)$ is in the Prym variety iff $\deg(D)$ is even. $\endgroup$
    – abx
    Aug 15, 2021 at 19:39
  • $\begingroup$ @abx thanks for the correction. Why in the above did you assume $\dim P = g-1?$, i.e. the genus of $C$ is 1? Also, can you elaborate a bit more about why $[X]^{\ast h}$ has that expression in terms of $\theta$? To me a priori $X$ seems quite arbitrary a subset of $P$. $\endgroup$
    – zudumazics
    Aug 16, 2021 at 20:02

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