Does $deg_k F(-D)/F = deg_k D$ hold for effective divisors $D$ and coherent, torsion-free...
$DeclareMathOperator{F}{mathcal{F}}DeclareMathOperator{o}{mathcal{O}}$Let $X$ be a reduced, pure dimensional, projective curve over some field $k$.
Let $F$ be a coherent and torsion-free $o_X$-module.
Let $D$ be a non-zero effective Cartier divisor on $X$ with regular global section $s$ which defines the embedding $F stackrel{cdot s}{to} F otimes o_X(-D)$. We write $F(-D) := F otimes_{o_X} o_X(-D)$.
Do we have $$dim_k H^0(X,F(-D)/F) = deg_k Dquad ?$$
This is obviously true for invertible $F$. But what about the more general case of coherent and torsion-free sheaves?
algebraic-geometry sheaf-theory algebraic-curves divisors-algebraic-geometry
add a comment |
$DeclareMathOperator{F}{mathcal{F}}DeclareMathOperator{o}{mathcal{O}}$Let $X$ be a reduced, pure dimensional, projective curve over some field $k$.
Let $F$ be a coherent and torsion-free $o_X$-module.
Let $D$ be a non-zero effective Cartier divisor on $X$ with regular global section $s$ which defines the embedding $F stackrel{cdot s}{to} F otimes o_X(-D)$. We write $F(-D) := F otimes_{o_X} o_X(-D)$.
Do we have $$dim_k H^0(X,F(-D)/F) = deg_k Dquad ?$$
This is obviously true for invertible $F$. But what about the more general case of coherent and torsion-free sheaves?
algebraic-geometry sheaf-theory algebraic-curves divisors-algebraic-geometry
2
In the case $mathcal F = mathcal O_X^r$, do we not get $(mathrm{deg}_k D)^r$ instead?
– Marc Paul
Nov 29 '18 at 15:41
2
In addition to @MarcPaul comment, I think what you write $mathcal{O}(-D)$ is usually written as $mathcal{O}(D)$.
– Mohan
Nov 29 '18 at 16:19
@MarcPaul Could you elaborate that? So is the formula rather $deg F (D)/F = (deg D)^{text{rank} F} $?
– windsheaf
Nov 30 '18 at 7:49
@Mohan You write $mathcal{O}(D)$ for the integral ideal sheaf corresponding to $D$? In the books I've read they (for instance Görtz/Wdhorn and Liu) it is denoted by $mathcal{O}(-D)$. Or am I mistaken something? Do you mean $mathcal{O}(D) supset mathcal{O}$ or $mathcal{O}(D) subset mathcal{O}$?
– windsheaf
Nov 30 '18 at 11:05
The former. $mathcal{O}(-D)$ is the ideal sheaf of $D$, so you have this a subsheaf of $mathcal{O}$ and dually, you have $mathcal{O}subsetmathcal{O}(D)$.
– Mohan
Nov 30 '18 at 14:00
add a comment |
$DeclareMathOperator{F}{mathcal{F}}DeclareMathOperator{o}{mathcal{O}}$Let $X$ be a reduced, pure dimensional, projective curve over some field $k$.
Let $F$ be a coherent and torsion-free $o_X$-module.
Let $D$ be a non-zero effective Cartier divisor on $X$ with regular global section $s$ which defines the embedding $F stackrel{cdot s}{to} F otimes o_X(-D)$. We write $F(-D) := F otimes_{o_X} o_X(-D)$.
Do we have $$dim_k H^0(X,F(-D)/F) = deg_k Dquad ?$$
This is obviously true for invertible $F$. But what about the more general case of coherent and torsion-free sheaves?
algebraic-geometry sheaf-theory algebraic-curves divisors-algebraic-geometry
$DeclareMathOperator{F}{mathcal{F}}DeclareMathOperator{o}{mathcal{O}}$Let $X$ be a reduced, pure dimensional, projective curve over some field $k$.
Let $F$ be a coherent and torsion-free $o_X$-module.
Let $D$ be a non-zero effective Cartier divisor on $X$ with regular global section $s$ which defines the embedding $F stackrel{cdot s}{to} F otimes o_X(-D)$. We write $F(-D) := F otimes_{o_X} o_X(-D)$.
Do we have $$dim_k H^0(X,F(-D)/F) = deg_k Dquad ?$$
This is obviously true for invertible $F$. But what about the more general case of coherent and torsion-free sheaves?
algebraic-geometry sheaf-theory algebraic-curves divisors-algebraic-geometry
algebraic-geometry sheaf-theory algebraic-curves divisors-algebraic-geometry
asked Nov 29 '18 at 15:10
windsheafwindsheaf
602312
602312
2
In the case $mathcal F = mathcal O_X^r$, do we not get $(mathrm{deg}_k D)^r$ instead?
– Marc Paul
Nov 29 '18 at 15:41
2
In addition to @MarcPaul comment, I think what you write $mathcal{O}(-D)$ is usually written as $mathcal{O}(D)$.
– Mohan
Nov 29 '18 at 16:19
@MarcPaul Could you elaborate that? So is the formula rather $deg F (D)/F = (deg D)^{text{rank} F} $?
– windsheaf
Nov 30 '18 at 7:49
@Mohan You write $mathcal{O}(D)$ for the integral ideal sheaf corresponding to $D$? In the books I've read they (for instance Görtz/Wdhorn and Liu) it is denoted by $mathcal{O}(-D)$. Or am I mistaken something? Do you mean $mathcal{O}(D) supset mathcal{O}$ or $mathcal{O}(D) subset mathcal{O}$?
– windsheaf
Nov 30 '18 at 11:05
The former. $mathcal{O}(-D)$ is the ideal sheaf of $D$, so you have this a subsheaf of $mathcal{O}$ and dually, you have $mathcal{O}subsetmathcal{O}(D)$.
– Mohan
Nov 30 '18 at 14:00
add a comment |
2
In the case $mathcal F = mathcal O_X^r$, do we not get $(mathrm{deg}_k D)^r$ instead?
– Marc Paul
Nov 29 '18 at 15:41
2
In addition to @MarcPaul comment, I think what you write $mathcal{O}(-D)$ is usually written as $mathcal{O}(D)$.
– Mohan
Nov 29 '18 at 16:19
@MarcPaul Could you elaborate that? So is the formula rather $deg F (D)/F = (deg D)^{text{rank} F} $?
– windsheaf
Nov 30 '18 at 7:49
@Mohan You write $mathcal{O}(D)$ for the integral ideal sheaf corresponding to $D$? In the books I've read they (for instance Görtz/Wdhorn and Liu) it is denoted by $mathcal{O}(-D)$. Or am I mistaken something? Do you mean $mathcal{O}(D) supset mathcal{O}$ or $mathcal{O}(D) subset mathcal{O}$?
– windsheaf
Nov 30 '18 at 11:05
The former. $mathcal{O}(-D)$ is the ideal sheaf of $D$, so you have this a subsheaf of $mathcal{O}$ and dually, you have $mathcal{O}subsetmathcal{O}(D)$.
– Mohan
Nov 30 '18 at 14:00
2
2
In the case $mathcal F = mathcal O_X^r$, do we not get $(mathrm{deg}_k D)^r$ instead?
– Marc Paul
Nov 29 '18 at 15:41
In the case $mathcal F = mathcal O_X^r$, do we not get $(mathrm{deg}_k D)^r$ instead?
– Marc Paul
Nov 29 '18 at 15:41
2
2
In addition to @MarcPaul comment, I think what you write $mathcal{O}(-D)$ is usually written as $mathcal{O}(D)$.
– Mohan
Nov 29 '18 at 16:19
In addition to @MarcPaul comment, I think what you write $mathcal{O}(-D)$ is usually written as $mathcal{O}(D)$.
– Mohan
Nov 29 '18 at 16:19
@MarcPaul Could you elaborate that? So is the formula rather $deg F (D)/F = (deg D)^{text{rank} F} $?
– windsheaf
Nov 30 '18 at 7:49
@MarcPaul Could you elaborate that? So is the formula rather $deg F (D)/F = (deg D)^{text{rank} F} $?
– windsheaf
Nov 30 '18 at 7:49
@Mohan You write $mathcal{O}(D)$ for the integral ideal sheaf corresponding to $D$? In the books I've read they (for instance Görtz/Wdhorn and Liu) it is denoted by $mathcal{O}(-D)$. Or am I mistaken something? Do you mean $mathcal{O}(D) supset mathcal{O}$ or $mathcal{O}(D) subset mathcal{O}$?
– windsheaf
Nov 30 '18 at 11:05
@Mohan You write $mathcal{O}(D)$ for the integral ideal sheaf corresponding to $D$? In the books I've read they (for instance Görtz/Wdhorn and Liu) it is denoted by $mathcal{O}(-D)$. Or am I mistaken something? Do you mean $mathcal{O}(D) supset mathcal{O}$ or $mathcal{O}(D) subset mathcal{O}$?
– windsheaf
Nov 30 '18 at 11:05
The former. $mathcal{O}(-D)$ is the ideal sheaf of $D$, so you have this a subsheaf of $mathcal{O}$ and dually, you have $mathcal{O}subsetmathcal{O}(D)$.
– Mohan
Nov 30 '18 at 14:00
The former. $mathcal{O}(-D)$ is the ideal sheaf of $D$, so you have this a subsheaf of $mathcal{O}$ and dually, you have $mathcal{O}subsetmathcal{O}(D)$.
– Mohan
Nov 30 '18 at 14:00
add a comment |
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2
In the case $mathcal F = mathcal O_X^r$, do we not get $(mathrm{deg}_k D)^r$ instead?
– Marc Paul
Nov 29 '18 at 15:41
2
In addition to @MarcPaul comment, I think what you write $mathcal{O}(-D)$ is usually written as $mathcal{O}(D)$.
– Mohan
Nov 29 '18 at 16:19
@MarcPaul Could you elaborate that? So is the formula rather $deg F (D)/F = (deg D)^{text{rank} F} $?
– windsheaf
Nov 30 '18 at 7:49
@Mohan You write $mathcal{O}(D)$ for the integral ideal sheaf corresponding to $D$? In the books I've read they (for instance Görtz/Wdhorn and Liu) it is denoted by $mathcal{O}(-D)$. Or am I mistaken something? Do you mean $mathcal{O}(D) supset mathcal{O}$ or $mathcal{O}(D) subset mathcal{O}$?
– windsheaf
Nov 30 '18 at 11:05
The former. $mathcal{O}(-D)$ is the ideal sheaf of $D$, so you have this a subsheaf of $mathcal{O}$ and dually, you have $mathcal{O}subsetmathcal{O}(D)$.
– Mohan
Nov 30 '18 at 14:00