Mixing
X function of Riemann surface $y^3=(x-x_1)*(x-x_2)$
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Im reading a book where they study the X function of the Riemann surface $y^3=(x-x_1)*(x-x_2)$ ($x, y in mathbbS^2$) and $dX$.
They say that $dX$ has a zero of degree 2 in $x_1$ and $x_2$ and "clearly no other zero". However they don't give any explanation.
My reasoning: a theorem says that X is meromorphic of degree 3 (locally in $x_1$, we have $y'^3=x'$ and a meromorphic function is surjective and takes every value n times). So $dX$ has a zero of degree 2 on $x_1$ and $x_2$.
Now if I look at $fracpartial Ppartial x$ (with $P=y^3-(x-x_1)*(x-x_2)$) I see that it is not null except where $x = fracx_1+x_22 = alpha$ so we can write $x=f(y)$ where $x neq alpha$. This gives $y^3=(f(y)-x_1)*(f(y)-x_2)$ and $3y^2 = f'(y)*Q(y)$. Hence $dX = 0$ if $f'(y) = 0$ which implies $y=0$ and $x = x_1$ or $x = x_2$, where $x neq alpha$.
But what about the case $x = alpha$? Is there something clear that I am missing?
riemann-surfaces
asked Aug 2 at 19:04
Thomas
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reputation from Thomas ending ending at 2018-08-12 17:24:47Z">in 7 days.
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Im reading a book where they study the X function of the Riemann surface $y^3=(x-x_1)*(x-x_2)$ ($x, y in mathbbS^2$) and $dX$.
They say that $dX$ has a zero of degree 2 in $x_1$ and $x_2$ and "clearly no other zero". However they don't give any explanation.
My reasoning: a theorem says that X is meromorphic of degree 3 (locally in $x_1$, we have $y'^3=x'$ and a meromorphic function is surjective and takes every value n times). So $dX$ has a zero of degree 2 on $x_1$ and $x_2$.
Now if I look at $fracpartial Ppartial x$ (with $P=y^3-(x-x_1)*(x-x_2)$) I see that it is not null except where $x = fracx_1+x_22 = alpha$ so we can write $x=f(y)$ where $x neq alpha$. This gives $y^3=(f(y)-x_1)*(f(y)-x_2)$ and $3y^2 = f'(y)*Q(y)$. Hence $dX = 0$ if $f'(y) = 0$ which implies $y=0$ and $x = x_1$ or $x = x_2$, where $x neq alpha$.
But what about the case $x = alpha$? Is there something clear that I am missing?
riemann-surfaces
asked Aug 2 at 19:04
Thomas
762611
This question has an open bounty worth +50
reputation from Thomas ending ending at 2018-08-12 17:24:47Z">in 7 days.
This question has not received enough attention.
Friendly "ping" on this, any help really appreciated, thanks!
– Thomas
2 days ago
Are you considering the affine curve or its compactification?
– barto
7 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Im reading a book where they study the X function of the Riemann surface $y^3=(x-x_1)*(x-x_2)$ ($x, y in mathbbS^2$) and $dX$.
They say that $dX$ has a zero of degree 2 in $x_1$ and $x_2$ and "clearly no other zero". However they don't give any explanation.
My reasoning: a theorem says that X is meromorphic of degree 3 (locally in $x_1$, we have $y'^3=x'$ and a meromorphic function is surjective and takes every value n times). So $dX$ has a zero of degree 2 on $x_1$ and $x_2$.
Now if I look at $fracpartial Ppartial x$ (with $P=y^3-(x-x_1)*(x-x_2)$) I see that it is not null except where $x = fracx_1+x_22 = alpha$ so we can write $x=f(y)$ where $x neq alpha$. This gives $y^3=(f(y)-x_1)*(f(y)-x_2)$ and $3y^2 = f'(y)*Q(y)$. Hence $dX = 0$ if $f'(y) = 0$ which implies $y=0$ and $x = x_1$ or $x = x_2$, where $x neq alpha$.
But what about the case $x = alpha$? Is there something clear that I am missing?
riemann-surfaces
asked Aug 2 at 19:04
Thomas
762611
Im reading a book where they study the X function of the Riemann surface $y^3=(x-x_1)*(x-x_2)$ ($x, y in mathbbS^2$) and $dX$.
They say that $dX$ has a zero of degree 2 in $x_1$ and $x_2$ and "clearly no other zero". However they don't give any explanation.
My reasoning: a theorem says that X is meromorphic of degree 3 (locally in $x_1$, we have $y'^3=x'$ and a meromorphic function is surjective and takes every value n times). So $dX$ has a zero of degree 2 on $x_1$ and $x_2$.
Now if I look at $fracpartial Ppartial x$ (with $P=y^3-(x-x_1)*(x-x_2)$) I see that it is not null except where $x = fracx_1+x_22 = alpha$ so we can write $x=f(y)$ where $x neq alpha$. This gives $y^3=(f(y)-x_1)*(f(y)-x_2)$ and $3y^2 = f'(y)*Q(y)$. Hence $dX = 0$ if $f'(y) = 0$ which implies $y=0$ and $x = x_1$ or $x = x_2$, where $x neq alpha$.
But what about the case $x = alpha$? Is there something clear that I am missing?
riemann-surfaces
asked Aug 2 at 19:04
Thomas
762611
asked Aug 2 at 19:04
Thomas
762611
asked Aug 2 at 19:04
Thomas
762611
asked Aug 2 at 19:04
Thomas
762611
762611
This question has an open bounty worth +50
reputation from Thomas ending ending at 2018-08-12 17:24:47Z">in 7 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from Thomas ending ending at 2018-08-12 17:24:47Z">in 7 days.
This question has not received enough attention.
Friendly "ping" on this, any help really appreciated, thanks!
– Thomas
2 days ago
Are you considering the affine curve or its compactification?
– barto
7 hours ago
add a comment |Â
Friendly "ping" on this, any help really appreciated, thanks!
– Thomas
2 days ago
Are you considering the affine curve or its compactification?
– barto
7 hours ago
Friendly "ping" on this, any help really appreciated, thanks!
– Thomas
2 days ago
Friendly "ping" on this, any help really appreciated, thanks!
– Thomas
2 days ago
Are you considering the affine curve or its compactification?
– barto
7 hours ago
Are you considering the affine curve or its compactification?
– barto
7 hours ago
add a comment |Â
1 Answer
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0
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At the points with $x = alpha$, $fracpartial Ppartial x = 0$ so $y$ is no longer a chart around such points. You have to calculate $dX$ in the chart $X$ instead. But $dX$ in the chart $X$ is just $dX$, seen as a differential on an open set of $mathbb C$. By definition, it has no zeroes: its coefficient (=$1$) does not vanish.
Without abuse of notation, if we write $pi_x, pi_y$ for the projections (their restrictions to the curve), what I'm saying is that, if $pi_x : U to V$ is a chart, then $(pi_x)_* (dpi_x|_U) = d(operatornameid_V)$.
answered 7 hours ago
barto
12.5k32378
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
At the points with $x = alpha$, $fracpartial Ppartial x = 0$ so $y$ is no longer a chart around such points. You have to calculate $dX$ in the chart $X$ instead. But $dX$ in the chart $X$ is just $dX$, seen as a differential on an open set of $mathbb C$. By definition, it has no zeroes: its coefficient (=$1$) does not vanish.
Without abuse of notation, if we write $pi_x, pi_y$ for the projections (their restrictions to the curve), what I'm saying is that, if $pi_x : U to V$ is a chart, then $(pi_x)_* (dpi_x|_U) = d(operatornameid_V)$.
answered 7 hours ago
barto
12.5k32378
add a comment |Â
up vote
0
down vote
At the points with $x = alpha$, $fracpartial Ppartial x = 0$ so $y$ is no longer a chart around such points. You have to calculate $dX$ in the chart $X$ instead. But $dX$ in the chart $X$ is just $dX$, seen as a differential on an open set of $mathbb C$. By definition, it has no zeroes: its coefficient (=$1$) does not vanish.
Without abuse of notation, if we write $pi_x, pi_y$ for the projections (their restrictions to the curve), what I'm saying is that, if $pi_x : U to V$ is a chart, then $(pi_x)_* (dpi_x|_U) = d(operatornameid_V)$.
answered 7 hours ago
barto
12.5k32378
add a comment |Â
up vote
0
down vote
up vote
0
down vote
At the points with $x = alpha$, $fracpartial Ppartial x = 0$ so $y$ is no longer a chart around such points. You have to calculate $dX$ in the chart $X$ instead. But $dX$ in the chart $X$ is just $dX$, seen as a differential on an open set of $mathbb C$. By definition, it has no zeroes: its coefficient (=$1$) does not vanish.
Without abuse of notation, if we write $pi_x, pi_y$ for the projections (their restrictions to the curve), what I'm saying is that, if $pi_x : U to V$ is a chart, then $(pi_x)_* (dpi_x|_U) = d(operatornameid_V)$.
answered 7 hours ago
barto
12.5k32378
At the points with $x = alpha$, $fracpartial Ppartial x = 0$ so $y$ is no longer a chart around such points. You have to calculate $dX$ in the chart $X$ instead. But $dX$ in the chart $X$ is just $dX$, seen as a differential on an open set of $mathbb C$. By definition, it has no zeroes: its coefficient (=$1$) does not vanish.
Without abuse of notation, if we write $pi_x, pi_y$ for the projections (their restrictions to the curve), what I'm saying is that, if $pi_x : U to V$ is a chart, then $(pi_x)_* (dpi_x|_U) = d(operatornameid_V)$.
answered 7 hours ago
barto
12.5k32378
edited 7 hours ago
answered 7 hours ago
barto
12.5k32378
answered 7 hours ago
barto
12.5k32378
answered 7 hours ago
barto
12.5k32378
12.5k32378
add a comment |Â
add a comment |Â
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Friendly "ping" on this, any help really appreciated, thanks!
– Thomas
2 days ago
Are you considering the affine curve or its compactification?
– barto
7 hours ago