Hurwitz' construction of Riemann surfaces (see e.g. here), 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.
How would ramification of the origin disturb the way Hurwitz constructs his Riemann surfaces? What if I face a situation where this doesn't hold and how to adapt Hurwitz' way to still get a nicely defined Riemann surface?
When my permutations $S_k$ are given and fixed, the result of the product depends on the ordering of the product. Let the ordering be fixed by other means as well, so we end up in a situation, where the product is most problably $\neq 1$.
What happens if we relax $(1)$ to $$ (S_1S_2S_3\cdots S_w)^N=1? \tag{1*} $$
Any help or hints are welcome...
Example:
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 $w=n$.
Example 1 $n=w=4$ and $$ \begin{eqnarray} S_A= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix}\\ S_B= \begin{pmatrix} 1 & 2 & 3 & 4 \\ \color{blue}3 & \color{blue}1 & \color{blue}2 & 4 \end{pmatrix}\\ S_C= \begin{pmatrix} 1 & 2 & 3 & 4 \\ \color{blue}4 & 2 & \color{blue}1 & \color{blue}3 \end{pmatrix}\\ S_D= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 4 & 2 \end{pmatrix}\\ \end{eqnarray} $$ I checked all and for all orderings of $A,B,C$ and $D$ $$ S_DS_CS_BS_A\neq 1 $$ holds (so this is not a Riemann surface in the sense of Hurwitz), where as e.g. $(S_AS_BS_CS_D)^3=1$ would.
Example 2: $S_B$ and $S_C$ are altered to
$$ \begin{eqnarray} S_A= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix}\\ S_B= \begin{pmatrix} 1 & 2 & 3 & 4 \\ \color{red}2 & \color{red}3 & \color{red}1 & 4 \end{pmatrix}\\ S_C= \begin{pmatrix} 1 & 2 & 3 & 4 \\ \color{red}3 & 2 & \color{red}4 & \color{red}1 \end{pmatrix}\\ S_D= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 4 & 2 \end{pmatrix}\\ \end{eqnarray} $$
Now $S_CS_DS_BS_A=1$, is a Riemann Surface in the sense of Hurwitz!
For those who are interested, the different permutations originally come from to $3$-edge colorings of the cube graph...
