# Surjectivity of the Abel-Prym map

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, $$d=g-1$$?

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.

• 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? May 8, 2020 at 19:43
• Because this is a question about cohomology classes, hence invariant by deformation. But you can also compute $\theta^{*h}=h!$ directly.
– abx
May 8, 2020 at 19:48
• 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? Aug 15, 2021 at 13:36
• @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.
– abx
Aug 15, 2021 at 19:39
• @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 $P$. Aug 16, 2021 at 20:02