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$\mathstrut$Hi, guys!

Let $C$, $C^\prime$ be projective smooth irreducible algebraic curves over an algebraically closed field $k$ ($\mathrm{char}(k) \neq 2$), $\phi : C$ $\to$ $C^\prime$ a two-sheeted covering, that is a surjective morphism of degree $2$, and $\phi_* : J(C) \to J(C^\prime)$ induced to Jacobians direct image morphism.

We know that there is the covering of $J(C)$ by symmetric power $\mathrm{sym}^{g(C)}(C)$. Consider Prym variety $\mathrm{Prym}(C/C^\prime) := \mathrm{ker}^0(\phi_*) \subseteq J(C)$ of the covering $\phi$. Is there a covering of $\mathrm{Prym}(C/C^\prime)$ by $\mathrm{sym}^{d}(C)$, where $d = \dim(\mathrm{Prym}(C/C^\prime))$?

Thank you!

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  • $\begingroup$ I just want to clarify the question. There exists a finite, closed subgroup scheme $\Gamma \subset \text{Prym}(C/C')$ and a flat morphism of group schemes $\pi:J(C)\to \text{Prym}(C/C')/\Gamma$ that restricts to the quotient morphism $q:\text{Prym}(C/C')\to\text{Prym}(C/C')/\Gamma$. Certainly the composition of $\pi$ with the Abel map $\text{sym}^d(C)\to J(C)$ is surjective. Are you asking whether this surjection factors through $q$? $\endgroup$
    – Jason Starr
    Commented Jan 16, 2016 at 14:19

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The answer is positive: there is a surjective, generically finite morphism $\text{sym}^d(C)\to \text{Prym}(C/C')$, at least away from small characteristics. Fix a $k$-point $x$ of $C$, and use that to define an Abel map, $\alpha_x:C \to J(X)$. The induced composition morphism $$C^g \xrightarrow{\alpha^g} J(C)^g \xrightarrow{\Sigma} J(C),$$ is surjective and generically étale. Also the projection morphism, $$\pi:J(C) \to \text{Prym}(C/C')/\Gamma,$$ is a smooth morphism. Since the Zariski tangent space of $C^g$ is generated by the Zariski tangent spaces of the $g$ fibers or the $g$ projections $C^g\to C^{g-1}$, up to a permutation of the factors, for sufficiently general $(x_{d+1},\dots,x_g)\in C^{g-d}$, the induced morphism $$C^d \times\{(x_{d+1},\dots,x_g)\} \to J(C) \to \text{Prym}(C/C')/\Gamma$$ is surjective and generically étale. Of course the morphism $$C^d \times \{(x_{d+1},\dots,x_g)\} \to J(C)$$ factors through the morphism $C^d \to \text{sym}^d(C)$ since the group law on $J(C)$ is Abelian. Thus, in all, there is a surjective, generically étale morphism $$\psi:\text{sym}^d(C) \to \text{Prym}(C/C')/\Gamma.$$ Observe that, up to composing this morphism with a translation of $\text{Prym}(C/C')$, the morphism is independent of the choice of permutation or general point $(x_{d+1},\dots,x_g) \in C^{g-d}$.

Finally, the finite subgroup scheme $\Gamma$ is contained in the $N$-torsion subgroup scheme for some integer $N$ (that can be bounded just in terms of the topological data of $\phi$). Thus, assuming the characteristic is larger than $N$, the multiplication by $N$ morphism, $$ \text{Prym}(C/C') \to \text{Prym}(C/C'),$$ is surjective and étale, and it factors through a surjective, étale morphism $$\chi:\text{Prym}(C/C')/\Gamma \to \text{Prym}(C/C').$$ Therefore, there is a surjective, étale morphism, $$\chi\circ \psi:\text{sym}^d(C) \to \text{Prym}(C/C').$$

Of course that may not be the answer you want. You still need to work out $\Gamma$ and $N$. Also, "canonically" you only obtain a morphism from $\text{sym}^d(C)$ to a torsor for $\text{Prym}(C/C')$; trivializing the torsor depends on choosing a $k$-point of $C$ (at least as specified above). So if you want to do this in families or over a non-algebraically closed field, you will need to do some work. Finally, there is the question, in small characteristics, of what to do when the group scheme $\Gamma$ is not étale.

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    $\begingroup$ nice answer. Over the complex numbers at least, there is a topological approach as well generalizing the topological proof of jacobi inversion. I.e. the g fold pontrjagin product of the homology class of the abel curve C in J(C) is, as I recall, g! times the fundamental class of J(C), hence the abel map from C^(g)-->J(C) is surjective of degree one. Since the abel prym curve C in the prym variety has twice this minimal class, the same computation apparently shows that d fold sums of the abel prym curve give the prym variety via a map of degre 2^(d) from the d fold symmetric procduct. $\endgroup$
    – roy smith
    Commented Jan 16, 2016 at 21:31
  • $\begingroup$ Jason, thank you for answering my question. How can I explicitely construct the scheme $\Gamma$ and the morphism $\pi$? $\endgroup$
    – Dimitri Koshelev
    Commented Jan 17, 2016 at 15:43
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    $\begingroup$ If j is the involution of J(C) induced by the sheet interchange involution on the curve C, then (1-j):J(C)-->P is a surjection whose composition with the abel map gives an abel-prym map C-->P. Adding the images of any d points of C, gives a map C^(d)-->P which, if my computation is right, maps the fundamental class of C^(d) to 2^d times the fundamental class of P, hence the map is surjective and finite. This may avoid the need to understand the finite subgroups. $\endgroup$
    – roy smith
    Commented Jan 19, 2016 at 2:13
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    $\begingroup$ Explicitly, if D' = j(D) denotes the involution applied to a divisor D, and q is a chosen point of C, the map C^(d)-->P takes a divisor D to (the class of ) D-D'-d(q-q'). If you are willing to map to the other component of the kernel, just send D to the class of D-D'. $\endgroup$
    – roy smith
    Commented Jan 19, 2016 at 17:09

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