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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.


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...

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There is also a visual proof of this fact: Topological (the Riemann surface structure of $\Sigma_g$ is determined by the cross-ratio of the four branch images, in the case of the Lawson surface it is 2), the map is the same as the quotient by the $g+1$ symmetry of the Lawson minimal surface $\xi_{g,1}.$ But this symmetry is given by rotation by $2\pi/(g+1)$ around a great circle, and one sees that $f$ is regular. – Sebastian Mar 5 '13 at 7:45
@Sebastian That's an interesting point. My motivation is actually coming from minimal surfaces so it might make sense to think about things this way. – Rbega Mar 5 '13 at 8:38
@Sebastian Can you expand on your comment? I'm having a hard time convincing myself that $\Sigma_g$ and $\xi_{g,1}$ are conformally equivalent when the cross-ratio of the branch points is $2$. Maybe I am misunderstanding... – Rbega Mar 5 '13 at 9:34
This is just a consequence of Riemann-Hurwitz and the construction of the Lawson surface $\xi_{g,1}:$ the $(g+1)-$symmetry is isometric and orientation preserving (both on the surface and in 3 space), hence it gives a holomorphic symmetry of the Riemann surface. By construction, it has 4 fix points, and all are of order (g+1). The quotient (as a Riemann surface) $\xi_{g,1}/\mathbb Z_{g+1}$ is the projective line. – Sebastian Mar 5 '13 at 10:11
@Sebastian I understand that $\xi_{g,1}$ has the claimed symmetry. I'm just not convinced that showing $\Sigma_{g}$ and $\xi_{g,1}$ are conformally equivalent does not depend on knowing that the cover is regular. As I said maybe I'm missing something... – Rbega Mar 5 '13 at 10:20
up vote 6 down vote accepted

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.)

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This seems pretty compelling to me, but I'm pretty confused about this stuff. – Rbega Mar 5 '13 at 15:04
I agree that my answer was incorrect. I withdrew it. Sorry. – Alexandre Eremenko Mar 5 '13 at 16:26

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