Jacobians of twisted coverings

Question

Given a compact Riemann surface $M$ and two double coverings $\hat\pi\colon \hat M\to M$ and $\tilde\pi\colon \tilde M\to M$ which are branched over the same points $p_1,..,p_n\in M.$ As is well-known, the two coverings are related by an element in $H_1(M,\mathbb Z_2).$ Is there any (explicitely) known relationship between the Jacobians of $\hat M$ annd $\tilde M$? Is there an isogeny between them?

Answer 1

They are not isogeneous in general. For an explicit example, take the case $n=0$ and $C$ hyperelliptic, so that we have a double covering $C\rightarrow \mathbb{P}^1$ branched along a subset $B$ of $\mathbb{P}^1$. Étale double coverings $\tilde{C}\rightarrow C $ correspond to partitions $B=B'\cup B''$, with $\#B'$ and $\#B''$ even; the Jacobian of $\tilde{C} $ is isogeneous to $JC\times JC'\times JC''$, where $C'$ and $C''$ are the double cover of $\mathbb{P}^1$ branched along $B'$ and $B''$ (this is well explained in Mumford's Prym varieties I, §7). Take for instance $g(C)=2$, $B=\{0,1,\infty,\lambda_1 ,\lambda _2,\lambda _3\} $; you get some double étale coverings $\tilde{C}_i $ of $C$ with $J\tilde{C}_i $ isogeneous to $JC\times E_i$, where $E_i$ is the double cover of $\mathbb{P}^1$ branched along $\{0,1,\infty,\lambda_i\}$. It is pretty clear that you can choose the $\lambda _i$ so that the elliptic curves $E_i$ are not isogeneous, and therefore the only common factor of $J\tilde{C}_1 $ and $J\tilde{C}_2 $ is $JC$.

Answer 2

Here is an example of a hyperelliptic $M$ that admits two Prym varieties that are not isogenous (actually, there are no nonzero homomorphisms between them). Namely, let $g \ge 8$ be an integer, $f(x)$ an irreducible degree $(2g+2)$ polynomial over the rationals, whose Galois group is the full symmetric group $\mathbf{S}_{2g+2}$ of permutations on $(2g+2)$ letters. (E.g., one may take $f(x)=x^{2g+2}-x-1$, thanks to results of Selmer and Osada.) Take as $M$ the (compact) genus $g$ Riemann surface of $\sqrt{f(x)}$, i.e., the smooth projective model of the plane algebraic curve $y^2=f(x)$. Then (thanks to Mumford and Dalalyan) the unramified double coverings of $M$ are indexed by partions of the set $R_f$ of of roots of $f(x)$ into a disjoint union of nonempty even cardinalty sets $T$ and $S$ such that the corresponding Prym variety is a product $J_S\times J_T$ of the jacobians of hyperelliptic curves $C_S:y^2=\prod_{\alpha\in S}(x-\alpha)$ and $C_T:y^2=\prod_{\alpha\in T}(x-\alpha)$ respectively. The dimensions of $J_S$ and $J_T$ are $(\#(S)-2)/2$ and $(\#(T)-2)/2$ respectively.

Assume that each of $S$ and $T$ has, at least, 4 elements, i.e., both $J_S$ and $J_T$ are positive-dimensional. By results of MR2961409 (arXiv:1012.3731 [math.AG])(Sect.5), the Galois property of $f(x)$ imply that both $J_S$ and $J_T$ are absolutely simple complex abelian varieties and there are no nonzero homomorphism between them. Now let us choose another partition of $R_f$ into a disjoint union of $(2g)$-element set $S^{\prime}$ and two-element $T^{\prime}$. Then the corresponding Prym variety is $J_{S^{\prime}}$, which (by results of MR1748293, arXiv:math/9909052 [math.AG]) is an absolutely simple $(g-1)$-dimensional complex abelian variety. Since both $\dim(J_S)$ and $\dim(J_T)$ are strictly less than $g-1$ (actually, they are both positive and their sum is $g-1$), there are no nonzero homomorphisms between $J_S$ and $J_{S^{\prime}}$, between $J_T$ and $J_{S^{\prime}}$, and therefore between Prym varieties $J_S\times J_T$ and $J_{S^{\prime}}$. More generally, one may choose another partition of $R_f$ into a disjoint union of $S_1$ and $T_1$ in such a way that $\#(S)\ne \#(S_1), \#(S) \ne \#(T_1)$ and then the dimension arguments imply that there are no nonzero homomorphisms between absolutely simple $J_S$ and $J_{S_1}$, $J_S$ and $J_{T_1}$, $J_T$ and $J_{T_1}$, $J_T$ and $J_{S_1}$. This implies that there are no nonzero homomorphisms between Prym varieties $J_S\times J_T$ and $J_{S_1}\times J_{T_1}$.

One should add that the jacobian of the corresponding covering space is isogenous to a product of the jacobian $J(M)$ of $M$ and the corresponding Prym variety. Notice that in our case $J(M)$ is an absolutely simple $g$-dimensional complex abelian variety.