Configuration of the branch locus of a branched covering of an elliptic curve

Question

Let $C$ be a curve of genus 3 and suppose that it admits a branched cover $\varphi:C\rightarrow E$ with $E$ elliptic and such that $\varphi$ does not factor through any \' etale cover. Then the degree of the ramification divisor $R$ is $4$ and the branch divisor $B\subseteq E$ is supported in a set of cardinality between 1 and 4. I wonder if all 4 hypothesis are possible and given a branch locus, which is the minimun degree for which it can be achieved. Obviously if $B$ is supported on 4 point I can take a double cover ramified on 4 point. Let us suppose that $B$ is supported on 3 point. Then if I understood well the theory, the existence of a branched covering of degree $n$ such as mine is equivalent to the existence of a primitive, transitive subgroup of $S_n$ generated by 5 element $a,b,g_1,g_2,g_3$ such that $[a,b]g_1g_2g_3=\mathrm{id}$. Is this correct? Does this groupo exist? for which $n$? If the above is correct then I have also to find a primitive transitive subgroup of $S_n$ generated by $a,b,g_1, g_2$ satisfying $[a,b]g_1g_2=\mathrm{id}$. And finally I would need a primitive transitive subgroup of $S_n$ generated by three element with relation $[ab]g=\mathrm{id}$. Thank you very much for your help!

Answer 1

According to Sofia's and my comments to the original question, we look for instance for elements $a,b\in S_n$ which generate a primitive subgroup of $S_n$ and where the commutator $[a,b]=a^{-1}b^{-1}ab$ is a $5$-cycle.

This is possible for each $n\ge5$: Set $a=(1,2\,3,\cdots,n)$, $b=(1,2,\cdots,n-4,n-3,n-1,n-2)$. Then $[a,b]=(1,2,n-2,n-1,n)$. Note that in this computation, I use the right action of the symmetric group.

Note that $a$ is an $n$--cycle, so $G=\langle a,b\rangle$ is transitive. Furthermore, $b$ is an $(n-1)$-cycle, so $G$ is doubly transitive and hence primitive. Actually, it is not difficult to show that even $G=S_n$.

So this is an example for a branch locus of size $1$. The size $2$ can be obtained by replacing the $5$-cycle $g_1$ by two $3$-cycles $h_1$, $h_2$ with $g_1=h_1h_2$. The branch locus of size $3$ or $4$ will be obtained by further splitting up $h_1$ or $h_1$ and $h_2$ into a product of two transpositions.

The cases of $2$, $3$ and $4$ branch points however can be realized with smaller degrees, namely $n=3$ for $2$ and $3$ branch points, and $n=2$ for $4$ branch points. These are the minimal degrees.

Answer 2

It is often easier to construct Galois covers of degree $n$.

In the case where $E$ is an elliptic curve and we have $t$ branch points, by Riemann existence theorem a Galois cover with Galois group $G$ is defined by the following data:

$\bullet$ a finite group $G$ of order $n$;

$\bullet$ elements $a, b, g_1, \ldots, g_t$ such that $G= \langle a, b, g_1, \ldots, g_t \rangle$ and $[a,b]g_1g_2 \ldots g_t=e$.

An important remark is that for a Galois cover the local monodromies around points in the same fibre are conjugate transpositions, i.e. points over the same branch points have monodromies with the same cyclic structure (this is in general false for non-Galois covers). Moreover, over the branch point $b_i$ such a structure is a cycle of order equal to the order of the element $g_i$.

Let us see now how to construct Galois coverings in the cases you are interested in.

Three branch points, that is $t=3$. For each $n$ we can construct a cyclic cover of degree $n$: simply choose $$G=\mathbb{Z}/n \mathbb{Z}=\langle x | x^n=1 \rangle $$ (I use the multiplicative notation) and take $$a=x, \quad b=x, \quad g_1=x, \quad g_2=x, \quad g_3=x^{-2}.$$

Two branch points, that is $t=2.$ Again, we have a cyclic cover for any $n$. Choose $G$ as above and take $$a=x, \quad b=x, \quad g_1=x, \quad g_2=x^{-1}$$.

One branch point, that is $t=1.$ A moment of reflection shows that we cannot take $G$ abelian in this case, otherwise $[a,b]g_1=e$ implies $g_1=e$, that is the cover would be unramified, contradiction. We can however construct a dihedral cover of degree $n=2m$ for any $m$. In fact, set $$D_{2m}= \langle x, y | x^2=y^m=1, \quad xy=y^{-1}x \rangle$$ and take $$a=x, \quad b=y, \quad g_1=y^2.$$

regarding your last comment, observe that when $m=5$ you have a Galois cover of degree $10$ with a unique branch point, whose branching order is $5$. In other words, over the unique branch point $b$ one has precisely two points $q_1, q_2$ and the local monodromy around each of the $q_i$ is by construction a $5$-cycle.

One can also construct Galois covers of odd degree with a unique branch point, for instance taking more general metacyclic groups (i.e. non-trivial semidirect products of cyclic groups). The dihedral covers are in fact a particular case of this construction.

Remark. I misread the question and I did not see the assumption that the covers must be of genus $3$. Only few of my constructions give genus $3$ curves (actually, it is not difficult to find all of them by using Hurwitz formula).

I will leave the answer anyway, since maybe it can be useful for other pourposes.