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?



Jacobians of twisted coverings
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 $JC$.


