Question

This is probably trivial but has been bothering me all day.

Suppose $f:\Sigma_g\to \mathbb{S}^2$ is a $g+1$ fold branched conformal map with $\Sigma_g$ a connected genus $g$ surface and $f$ having $4$ branch points (all of order $g+1$).

Is it true that $f:\Sigma_g^* \to (\mathbb{S}^2)^*$ (the cover obtained by removing the branch points) is a regular cover (i.e. so the deck group acts transitively on the fibers of $f$)?

I thought this was obvious but then realized I didn't know how to prove it (admittedly this is a bit outside my area of expertise). Part of the problem is that I don't have a good example of a branched map from a connected surface which is not a regular cover away from the branch points.

Edit

Based on some comments of Sebastian. It would be enough for my purposes to prove that the conformal structure of $\Sigma_g$ was uniquely determined by the cross-ratio of the branch points. However, its not clear to me whether this can be shown without knowing that the cover is regular...

Answer 1

The answer is no:

The monodromy group of the cover, as a permutation group on the fiber $f^{-1}(b)$ of a non-branch point $b$, is generated by four elements $s_1,s_2,s_3,s_4$ with $s_1s_2s_3s_4=1$. By Riemann's existence theorem, the question is equivalent to the following, with $n=g+1$:

Let $s_1,s_2,s_3,s_4$ be $n$-cycles in the symmetric group on $n$ letters with $s_1s_2s_3s_4=1$. Does this imply that the group generated by $s_1, s_2,s_3, s_4$ has order $n$?

This does not hold for $n=4$: Take for example $s_1=(1, 2, 3, 4)$, $s_2=(1, 3, 4, 2)$, $s_3=s_2^{-1}$, $s_4=s_1^{-1}$. Then $s_1s_2s_3s_4=1$, but $s_1s_2=(2, 4, 3)$.

This works for any $n\ge4$: Simply take $s_1$ an $n$-cycle and $s_2$ another $n$-cycle which is not a power of $s_1$, and $s_3$, $s_4$ the inverses of $s_2$ and $s_1$ as above.

(Note: I use the right action of permutations, so in $s_1s_2$, one first applies $s_1$, and then $s_2$ afterwards.)