Mixing
Hurwitz' Construction of Riemann Surfaces, but $S_1S_2cdots S_wneq 1$
$begingroup$
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!
riemann-surfaces
asked Dec 10 '18 at 15:14
draks ...draks ...
11.4k644129
$endgroup$
add a comment |
$begingroup$
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!
riemann-surfaces
asked Dec 10 '18 at 15:14
draks ...draks ...
11.4k644129
$endgroup$
add a comment |
$begingroup$
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!
riemann-surfaces
asked Dec 10 '18 at 15:14
draks ...draks ...
11.4k644129
$endgroup$
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
riemann-surfaces
asked Dec 10 '18 at 15:14
draks ...draks ...
11.4k644129
asked Dec 10 '18 at 15:14
draks ...draks ...
11.4k644129
edited Jan 14 at 11:56
draks ...
asked Dec 10 '18 at 15:14
draks ...draks ...
11.4k644129
asked Dec 10 '18 at 15:14
draks ...draks ...
11.4k644129
asked Dec 10 '18 at 15:14
draks ...draks ...
11.4k644129
11.4k644129
add a comment |
add a comment |
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