Base point free for $g^1_2$ for hyperelliptic curve
1
$begingroup$
Let $C$ be a curve of genus lager than 1. $C$ is called hyperelliptic if it contains a $g_2^1$ linear system, meaning that $D$ is of degree $2$ with $dim|D|=2$ if $D$ is such a divisor in this linear system. Then the book said that it has to be base point free. I did not get this result. I used Riemann-Roch showed that $l(K-D)>0$. But I did not see why it is base point free.
algebraic-geometry algebraic-curves line-bundles
asked Dec 22 '18 at 0:04
Peter LiuPeter Liu
305114
$endgroup$
add a comment |
1
$begingroup$
Let $C$ be a curve of genus lager than 1. $C$ is called hyperelliptic if it contains a $g_2^1$ linear system, meaning that $D$ is of degree $2$ with $dim|D|=2$ if $D$ is such a divisor in this linear system. Then the book said that it has to be base point free. I did not get this result. I used Riemann-Roch showed that $l(K-D)>0$. But I did not see why it is base point free.
algebraic-geometry algebraic-curves line-bundles
asked Dec 22 '18 at 0:04
Peter LiuPeter Liu
305114
$endgroup$
$begingroup$
You are starting from a function $f in mathcal{L}(Q_1+Q_2)subset k(C)$ such that $div(f) = P_1+P_2 -Q_1-Q_2 $ ?
$endgroup$
– reuns
Dec 22 '18 at 1:19
1
$begingroup$
This question is really badly written. Please read it and try to edit so that it makes sense. And tell us what $D$ you're working with.
$endgroup$
– Ted Shifrin
Dec 22 '18 at 1:33
$begingroup$
@TedShifrin Sorry for my bad writing. I try to improve it a bit. Here $D$ is a divisor in such a linear system. As I know the degree of the divisor, and I know $l(D)$ as it is same as dimension of the linear system +1, so I apply the Riemann-Roch to it.
$endgroup$
– Peter Liu
Dec 22 '18 at 9:52
$begingroup$
Hint: suppose there is a nonempty base locus. Now remove it. What happens to your $mathfrak{g}^{1}_2$?
$endgroup$
– Samir Canning
Dec 22 '18 at 17:33
$begingroup$
@SamirCanning I try to argue like this. Let $B$ be a base locus, nonempty. First the degree will not be 2, as $deg(D-B)=0$ which will have dimension of global sections is one, contradiction to $r=1$. if $deg(B)=1$, then there is a meromorphic function with only one pole and no others, then the curve would be of genus 0, which is a contradiction. Is the argument correct? And is there a proof without argument depending on degree?
$endgroup$
– Peter Liu
Dec 22 '18 at 21:40
add a comment |
1
1
1
$begingroup$
Let $C$ be a curve of genus lager than 1. $C$ is called hyperelliptic if it contains a $g_2^1$ linear system, meaning that $D$ is of degree $2$ with $dim|D|=2$ if $D$ is such a divisor in this linear system. Then the book said that it has to be base point free. I did not get this result. I used Riemann-Roch showed that $l(K-D)>0$. But I did not see why it is base point free.
algebraic-geometry algebraic-curves line-bundles
asked Dec 22 '18 at 0:04
Peter LiuPeter Liu
305114
$endgroup$
Let $C$ be a curve of genus lager than 1. $C$ is called hyperelliptic if it contains a $g_2^1$ linear system, meaning that $D$ is of degree $2$ with $dim|D|=2$ if $D$ is such a divisor in this linear system. Then the book said that it has to be base point free. I did not get this result. I used Riemann-Roch showed that $l(K-D)>0$. But I did not see why it is base point free.
algebraic-geometry algebraic-curves line-bundles
algebraic-geometry algebraic-curves line-bundles
asked Dec 22 '18 at 0:04
Peter LiuPeter Liu
305114
asked Dec 22 '18 at 0:04
Peter LiuPeter Liu
305114
edited Dec 22 '18 at 9:57
Peter Liu
asked Dec 22 '18 at 0:04
Peter LiuPeter Liu
305114
asked Dec 22 '18 at 0:04
Peter LiuPeter Liu
305114
asked Dec 22 '18 at 0:04
Peter LiuPeter Liu
305114
305114
$begingroup$
You are starting from a function $f in mathcal{L}(Q_1+Q_2)subset k(C)$ such that $div(f) = P_1+P_2 -Q_1-Q_2 $ ?
$endgroup$
– reuns
Dec 22 '18 at 1:19
1
$begingroup$
This question is really badly written. Please read it and try to edit so that it makes sense. And tell us what $D$ you're working with.
$endgroup$
– Ted Shifrin
Dec 22 '18 at 1:33
$begingroup$
@TedShifrin Sorry for my bad writing. I try to improve it a bit. Here $D$ is a divisor in such a linear system. As I know the degree of the divisor, and I know $l(D)$ as it is same as dimension of the linear system +1, so I apply the Riemann-Roch to it.
$endgroup$
– Peter Liu
Dec 22 '18 at 9:52
$begingroup$
Hint: suppose there is a nonempty base locus. Now remove it. What happens to your $mathfrak{g}^{1}_2$?
$endgroup$
– Samir Canning
Dec 22 '18 at 17:33
$begingroup$
@SamirCanning I try to argue like this. Let $B$ be a base locus, nonempty. First the degree will not be 2, as $deg(D-B)=0$ which will have dimension of global sections is one, contradiction to $r=1$. if $deg(B)=1$, then there is a meromorphic function with only one pole and no others, then the curve would be of genus 0, which is a contradiction. Is the argument correct? And is there a proof without argument depending on degree?
$endgroup$
– Peter Liu
Dec 22 '18 at 21:40
add a comment |
$begingroup$
You are starting from a function $f in mathcal{L}(Q_1+Q_2)subset k(C)$ such that $div(f) = P_1+P_2 -Q_1-Q_2 $ ?
$endgroup$
– reuns
Dec 22 '18 at 1:19
1
$begingroup$
This question is really badly written. Please read it and try to edit so that it makes sense. And tell us what $D$ you're working with.
$endgroup$
– Ted Shifrin
Dec 22 '18 at 1:33
$begingroup$
@TedShifrin Sorry for my bad writing. I try to improve it a bit. Here $D$ is a divisor in such a linear system. As I know the degree of the divisor, and I know $l(D)$ as it is same as dimension of the linear system +1, so I apply the Riemann-Roch to it.
$endgroup$
– Peter Liu
Dec 22 '18 at 9:52
$begingroup$
Hint: suppose there is a nonempty base locus. Now remove it. What happens to your $mathfrak{g}^{1}_2$?
$endgroup$
– Samir Canning
Dec 22 '18 at 17:33
$begingroup$
@SamirCanning I try to argue like this. Let $B$ be a base locus, nonempty. First the degree will not be 2, as $deg(D-B)=0$ which will have dimension of global sections is one, contradiction to $r=1$. if $deg(B)=1$, then there is a meromorphic function with only one pole and no others, then the curve would be of genus 0, which is a contradiction. Is the argument correct? And is there a proof without argument depending on degree?
$endgroup$
– Peter Liu
Dec 22 '18 at 21:40
$begingroup$
You are starting from a function $f in mathcal{L}(Q_1+Q_2)subset k(C)$ such that $div(f) = P_1+P_2 -Q_1-Q_2 $ ?
$endgroup$
– reuns
Dec 22 '18 at 1:19
$begingroup$
You are starting from a function $f in mathcal{L}(Q_1+Q_2)subset k(C)$ such that $div(f) = P_1+P_2 -Q_1-Q_2 $ ?
$endgroup$
– reuns
Dec 22 '18 at 1:19
1
1
$begingroup$
This question is really badly written. Please read it and try to edit so that it makes sense. And tell us what $D$ you're working with.
$endgroup$
– Ted Shifrin
Dec 22 '18 at 1:33
$begingroup$
This question is really badly written. Please read it and try to edit so that it makes sense. And tell us what $D$ you're working with.
$endgroup$
– Ted Shifrin
Dec 22 '18 at 1:33
$begingroup$
@TedShifrin Sorry for my bad writing. I try to improve it a bit. Here $D$ is a divisor in such a linear system. As I know the degree of the divisor, and I know $l(D)$ as it is same as dimension of the linear system +1, so I apply the Riemann-Roch to it.
$endgroup$
– Peter Liu
Dec 22 '18 at 9:52
$begingroup$
@TedShifrin Sorry for my bad writing. I try to improve it a bit. Here $D$ is a divisor in such a linear system. As I know the degree of the divisor, and I know $l(D)$ as it is same as dimension of the linear system +1, so I apply the Riemann-Roch to it.
$endgroup$
– Peter Liu
Dec 22 '18 at 9:52
$begingroup$
Hint: suppose there is a nonempty base locus. Now remove it. What happens to your $mathfrak{g}^{1}_2$?
$endgroup$
– Samir Canning
Dec 22 '18 at 17:33
$begingroup$
Hint: suppose there is a nonempty base locus. Now remove it. What happens to your $mathfrak{g}^{1}_2$?
$endgroup$
– Samir Canning
Dec 22 '18 at 17:33
$begingroup$
@SamirCanning I try to argue like this. Let $B$ be a base locus, nonempty. First the degree will not be 2, as $deg(D-B)=0$ which will have dimension of global sections is one, contradiction to $r=1$. if $deg(B)=1$, then there is a meromorphic function with only one pole and no others, then the curve would be of genus 0, which is a contradiction. Is the argument correct? And is there a proof without argument depending on degree?
$endgroup$
– Peter Liu
Dec 22 '18 at 21:40
$begingroup$
@SamirCanning I try to argue like this. Let $B$ be a base locus, nonempty. First the degree will not be 2, as $deg(D-B)=0$ which will have dimension of global sections is one, contradiction to $r=1$. if $deg(B)=1$, then there is a meromorphic function with only one pole and no others, then the curve would be of genus 0, which is a contradiction. Is the argument correct? And is there a proof without argument depending on degree?
$endgroup$
– Peter Liu
Dec 22 '18 at 21:40
add a comment |
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$begingroup$
You are starting from a function $f in mathcal{L}(Q_1+Q_2)subset k(C)$ such that $div(f) = P_1+P_2 -Q_1-Q_2 $ ?
$endgroup$
– reuns
Dec 22 '18 at 1:19
1
$begingroup$
This question is really badly written. Please read it and try to edit so that it makes sense. And tell us what $D$ you're working with.
$endgroup$
– Ted Shifrin
Dec 22 '18 at 1:33
$begingroup$
@TedShifrin Sorry for my bad writing. I try to improve it a bit. Here $D$ is a divisor in such a linear system. As I know the degree of the divisor, and I know $l(D)$ as it is same as dimension of the linear system +1, so I apply the Riemann-Roch to it.
$endgroup$
– Peter Liu
Dec 22 '18 at 9:52
$begingroup$
Hint: suppose there is a nonempty base locus. Now remove it. What happens to your $mathfrak{g}^{1}_2$?
$endgroup$
– Samir Canning
Dec 22 '18 at 17:33
$begingroup$
@SamirCanning I try to argue like this. Let $B$ be a base locus, nonempty. First the degree will not be 2, as $deg(D-B)=0$ which will have dimension of global sections is one, contradiction to $r=1$. if $deg(B)=1$, then there is a meromorphic function with only one pole and no others, then the curve would be of genus 0, which is a contradiction. Is the argument correct? And is there a proof without argument depending on degree?
$endgroup$
– Peter Liu
Dec 22 '18 at 21:40