Hurwitz' Construction of Riemann Surfaces, but $S_1S_2cdots S_wneq 1$












3












$begingroup$


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:
$$
S_1S_2S_3cdots S_w=1, tag{1}
$$

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?



What happens if we relax $(1)$ to
$$
(S_1S_2S_3cdots S_w)^N=1? tag{1*}
$$





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 and neither for the given nor for any other reordering of $A,B,C$ and $D$
$$
S_DS_CS_BS_A=1
$$

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 Hurwitz!










share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    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:
    $$
    S_1S_2S_3cdots S_w=1, tag{1}
    $$

    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?



    What happens if we relax $(1)$ to
    $$
    (S_1S_2S_3cdots S_w)^N=1? tag{1*}
    $$





    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 and neither for the given nor for any other reordering of $A,B,C$ and $D$
    $$
    S_DS_CS_BS_A=1
    $$

    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 Hurwitz!










    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      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:
      $$
      S_1S_2S_3cdots S_w=1, tag{1}
      $$

      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?



      What happens if we relax $(1)$ to
      $$
      (S_1S_2S_3cdots S_w)^N=1? tag{1*}
      $$





      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 and neither for the given nor for any other reordering of $A,B,C$ and $D$
      $$
      S_DS_CS_BS_A=1
      $$

      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 Hurwitz!










      share|cite|improve this question











      $endgroup$




      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:
      $$
      S_1S_2S_3cdots S_w=1, tag{1}
      $$

      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?



      What happens if we relax $(1)$ to
      $$
      (S_1S_2S_3cdots S_w)^N=1? tag{1*}
      $$





      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 and neither for the given nor for any other reordering of $A,B,C$ and $D$
      $$
      S_DS_CS_BS_A=1
      $$

      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 Hurwitz!







      riemann-surfaces






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      edited Jan 14 at 11:56







      draks ...

















      asked Dec 10 '18 at 15:14









      draks ...draks ...

      11.4k644130




      11.4k644130






















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