Is it always possible to find an isogeny from a hyperelliptic curve of genus 4, to a 'normal' elliptic curve (genus 1), or a product of elliptic curves?
Are such isogenies easy to compute?
This question is motivated by a particular instance of a discrete logarithm problem. On hyperelliptic curves, computing group law in its jacobian is a slow operation. I'd like to convert the DLP to a group, where it's easier to solve. Any suggestions on different approaches are welcome too.
This is only a partial answer.
Let $C$ be a hyperelliptic curve and $E$ an elliptic curve. Let $i_1:C\to \mathbb{P}^1$ and $i_2: E\to\mathbb{P}^1$ be the double cover maps. Let $Q_1,\dots,Q_{2g+2}$ be the critical values of $i_1$ (i.e., the images of the Weierstrass points) and $P_1,\dots,P_4$ be the critical values of $i_2$.
Then there is an isogeny $C\to E$ if and only if there is a morphism $\varphi:\mathbb{P}^1\to\mathbb{P}^1$, such that the ramification indices at all the $Q_i$ are odd, at each other point of $\mathbb{P}^1$ the ramification indices are even and for all $i$ we have $\varphi(Q_i)\in \{P_1,\dots,P_2\}$. (This not so hard to prove, the only reference I know is a paper by Chad Schoen in Journal fuer Reine und Angewante Mathematik, you can use this to construct a lot of examples.)
For fixed $E$ and $g$ you can compute the dimension of the locus of hyperelliptic curves that admit an isogeny to $E$, at least over the complex numbers. Over the complex numbers this locus has dimension $g-1$ and therefore the locus of hyperelliptic curves admitting a morphism to an elliptic curve has dimension $g$, whereas the hyperelliptic locus has dimension $2g-1$. Hence a general complex hyperelliptic curve does not admit a morphism to an elliptic curve. I am quite sure a similar results holds true over finite fields, i.e., you need to calculate the dimension of a certain Hurwitz space of coverings $\mathbb{P}^1\to\mathbb{P}^1$. (Details of this calculation are in my paper on Noether-Lefschetz loci of elliptic surfaces, but I would not be suprised if someone had done this before I did this calculation.)
