Impossible Construction of Riemann Surfaces à la Hurwitz

Question

I'm currently trying to construct simple Riemann Surfaces in the way of Hurwitz (see e.g. here):

Given the complex plane $E$, which originates at $O$ and $w$ non-identical points $a_k$. Cut the plane from $O$ to each $a_k$, which shall be denoted as the cutted complex plane $E^*$. Cutlines are called $l_k$ and have a left and right side called $l^\pm_k$. Assume we have $n$ copies of $E^*$ and they are labelled, call'em $E^*_m$. At each $l_k$ there shall be a mapping $S_k$, i.e. a permutation, for which $$S_k: \{E^*_1,E^*_2,\dots,E^*_n\} \to \{E^*_{S_k(1)},E^*_{S_k(2)},\dots,E^*_{S_k(n)}\} $$ holds. For my personal reason of interest, all my examples would further obey $w=n$. Hurwitz' construction of Riemann surfaces, asks for all permutations $S_k$ of the copies of the cutted complex planes $E^*$ at the branching points $a_k$, to fulfill: $$ S_1S_2S_3\cdots S_w=1, \tag{1} $$ which ensures that there is no ramification at the origin.

So I start out to construct a Riemann surface with $n=w=3$. Additionally, at every branching point $a_k$, there shall be a ramification profile, which permutes two of the $E^*_k$, such that all $S_k$ are unique. Further these transposition shall be "cyclic" [I mean of the form $(k,k+1 \bmod n)$], e.g.: $$S_1=(23),S_2=(13),S_3=(12).$$

It is not possible to me to meet criterion $(1)$ in this case, so I continued with $n=w=4$, but failed again (although there are more possibilities to compose the product in $(1)$, none equaled $1$ as required).

I assume it is not possible to construct Riemann surfaces for any $n$ that way at all. Is that true?

Answer 1

You are correct. With your desired branch data, you will inevitably create branching at the origin $O$ for $n=3$ and $n=4$.

I am not sure what you mean for these permutations to be cyclic, so I will ignore that unless you can clarify. In this interpretation, for general $n$ this is a purely combinatorial problem about the permutation group $S_n$: Can the identity be represented as a product of $n$ distinct transpositions? It turns out that answer is contained in the paper Keeler's theorem and products of distinct transpositions by Evans, Huang, and Nguyen. Theorem 4 says that the identity in $S_n$ is represented as a product of $m$ different transpositions if and only if $m$ is even and $6 \le m \le \binom{n}{2}$. In your case you want $n=m$, so you need $n \ge 6$, in which case this is always possible. Explicitly, you have that $I=(1\;2)(2\;3)(1\;4)(1\;3)(2\;4)(3\;4)$, and you can interpret these as elements of $S_6$. Obviously, in this example two of the sheets are not permuted at all, so I am not sure this is what you want. (The corresponding Riemann surface would be disconnected.)

EDIT: With the added stipulation that the permutations are $(1\;2), (2\;3), \ldots, (n\;1)$, this will never be possible. E.g., the only product of those that fixes $1$ is $\sigma = (1\;2)(2\;3)\cdots(n \;1) = (n \; n-1 \ldots 2)$, which does not fix any other $k$.