Mixing
Question about total branching number
$begingroup$
Given a map $f:mathbb{CP^1} leftarrow mathbb{CP^1}$ by $f(z)=frac{4z^2(z-1)^2}{(2z-1)^2}$ Find all branching points and their degrees.
If my calculation is correct I got 4 branching points and in each point degree of $f$ is 2. Also I have concluded that inifnity is not a branching point. From Riemann Hurwitz formula I get that total branching number is 6 where degree of $f$ is 4. But if I directly compute total branching number I get 4.
Where did I make a mistake?
complex-analysis riemann-surfaces
asked Dec 23 '18 at 15:49
XYZXYZ
1028
$endgroup$
add a comment |
$begingroup$
Given a map $f:mathbb{CP^1} leftarrow mathbb{CP^1}$ by $f(z)=frac{4z^2(z-1)^2}{(2z-1)^2}$ Find all branching points and their degrees.
If my calculation is correct I got 4 branching points and in each point degree of $f$ is 2. Also I have concluded that inifnity is not a branching point. From Riemann Hurwitz formula I get that total branching number is 6 where degree of $f$ is 4. But if I directly compute total branching number I get 4.
Where did I make a mistake?
complex-analysis riemann-surfaces
asked Dec 23 '18 at 15:49
XYZXYZ
1028
$endgroup$
add a comment |
$begingroup$
Given a map $f:mathbb{CP^1} leftarrow mathbb{CP^1}$ by $f(z)=frac{4z^2(z-1)^2}{(2z-1)^2}$ Find all branching points and their degrees.
If my calculation is correct I got 4 branching points and in each point degree of $f$ is 2. Also I have concluded that inifnity is not a branching point. From Riemann Hurwitz formula I get that total branching number is 6 where degree of $f$ is 4. But if I directly compute total branching number I get 4.
Where did I make a mistake?
complex-analysis riemann-surfaces
asked Dec 23 '18 at 15:49
XYZXYZ
1028
$endgroup$
Given a map $f:mathbb{CP^1} leftarrow mathbb{CP^1}$ by $f(z)=frac{4z^2(z-1)^2}{(2z-1)^2}$ Find all branching points and their degrees.
If my calculation is correct I got 4 branching points and in each point degree of $f$ is 2. Also I have concluded that inifnity is not a branching point. From Riemann Hurwitz formula I get that total branching number is 6 where degree of $f$ is 4. But if I directly compute total branching number I get 4.
Where did I make a mistake?
complex-analysis riemann-surfaces
complex-analysis riemann-surfaces
asked Dec 23 '18 at 15:49
XYZXYZ
1028
asked Dec 23 '18 at 15:49
XYZXYZ
1028
asked Dec 23 '18 at 15:49
XYZXYZ
1028
asked Dec 23 '18 at 15:49
XYZXYZ
1028
asked Dec 23 '18 at 15:49
XYZXYZ
1028
1028
add a comment |
add a comment |
1 Answer
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$begingroup$
$f(z_0) = infty$ is in fact a branch point. There will be six ramification points $z_0$:
$$begin{matrix}
z_0 & f(z_0) \
hline
0, 1 & 0 \
frac {1 pm i} 2 & -1 \
frac 1 2, infty & infty
end{matrix},$$
each of index $2$. To see why this is the case, expand $f(z)$ around each $z_0$; the degree of the first non-constant term will be $pm 2$.
answered Feb 1 at 16:09
MaximMaxim
6,2731221
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
$f(z_0) = infty$ is in fact a branch point. There will be six ramification points $z_0$:
$$begin{matrix}
z_0 & f(z_0) \
hline
0, 1 & 0 \
frac {1 pm i} 2 & -1 \
frac 1 2, infty & infty
end{matrix},$$
each of index $2$. To see why this is the case, expand $f(z)$ around each $z_0$; the degree of the first non-constant term will be $pm 2$.
answered Feb 1 at 16:09
MaximMaxim
6,2731221
$endgroup$
add a comment |
$begingroup$
$f(z_0) = infty$ is in fact a branch point. There will be six ramification points $z_0$:
$$begin{matrix}
z_0 & f(z_0) \
hline
0, 1 & 0 \
frac {1 pm i} 2 & -1 \
frac 1 2, infty & infty
end{matrix},$$
each of index $2$. To see why this is the case, expand $f(z)$ around each $z_0$; the degree of the first non-constant term will be $pm 2$.
answered Feb 1 at 16:09
MaximMaxim
6,2731221
$endgroup$
add a comment |
$begingroup$
$f(z_0) = infty$ is in fact a branch point. There will be six ramification points $z_0$:
$$begin{matrix}
z_0 & f(z_0) \
hline
0, 1 & 0 \
frac {1 pm i} 2 & -1 \
frac 1 2, infty & infty
end{matrix},$$
each of index $2$. To see why this is the case, expand $f(z)$ around each $z_0$; the degree of the first non-constant term will be $pm 2$.
answered Feb 1 at 16:09
MaximMaxim
6,2731221
$endgroup$
$f(z_0) = infty$ is in fact a branch point. There will be six ramification points $z_0$:
$$begin{matrix}
z_0 & f(z_0) \
hline
0, 1 & 0 \
frac {1 pm i} 2 & -1 \
frac 1 2, infty & infty
end{matrix},$$
each of index $2$. To see why this is the case, expand $f(z)$ around each $z_0$; the degree of the first non-constant term will be $pm 2$.
answered Feb 1 at 16:09
MaximMaxim
6,2731221
answered Feb 1 at 16:09
MaximMaxim
6,2731221
answered Feb 1 at 16:09
MaximMaxim
6,2731221
answered Feb 1 at 16:09
MaximMaxim
6,2731221
6,2731221
add a comment |
add a comment |
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