Resolution of a torsion sheaf
$begingroup$
Let $J$ be the hyperplane divisor in $mathbb{C}P^2$, and let $i:C hookrightarrow mathbb{C}P^2$ be the closed immersion of a smooth generic curve of degree 2. We know that $Csimeq mathbb{C}P^1$, and let $H$ be the hyperplane divisor in $C$. So loosely $H sim frac{1}{2}J|_C$. The question is, what is the projective (locally free) resolution of $i_* mathcal{O}_C(H)$? in other words, we have the following resolution in general,
begin{equation}
0longrightarrow mathcal{L}_2longrightarrowmathcal{L_1}longrightarrowmathcal{L}_0longrightarrow i_* mathcal{O}_C(H) longrightarrow 0,
end{equation}
what are the locally free sheaves $mathcal{L}_i$?
ag.algebraic-geometry projective-resolution
asked Dec 3 '18 at 5:21
Mohsen KarkheiranMohsen Karkheiran
21610
$endgroup$
add a comment |
$begingroup$
Let $J$ be the hyperplane divisor in $mathbb{C}P^2$, and let $i:C hookrightarrow mathbb{C}P^2$ be the closed immersion of a smooth generic curve of degree 2. We know that $Csimeq mathbb{C}P^1$, and let $H$ be the hyperplane divisor in $C$. So loosely $H sim frac{1}{2}J|_C$. The question is, what is the projective (locally free) resolution of $i_* mathcal{O}_C(H)$? in other words, we have the following resolution in general,
begin{equation}
0longrightarrow mathcal{L}_2longrightarrowmathcal{L_1}longrightarrowmathcal{L}_0longrightarrow i_* mathcal{O}_C(H) longrightarrow 0,
end{equation}
what are the locally free sheaves $mathcal{L}_i$?
ag.algebraic-geometry projective-resolution
asked Dec 3 '18 at 5:21
Mohsen KarkheiranMohsen Karkheiran
21610
$endgroup$
$begingroup$
It is possible to put some topological constraints on $mathcal{L}_i$'s, but I'm curious to see whether it is possible to say more....
$endgroup$
– Mohsen Karkheiran
Dec 3 '18 at 5:24
add a comment |
$begingroup$
Let $J$ be the hyperplane divisor in $mathbb{C}P^2$, and let $i:C hookrightarrow mathbb{C}P^2$ be the closed immersion of a smooth generic curve of degree 2. We know that $Csimeq mathbb{C}P^1$, and let $H$ be the hyperplane divisor in $C$. So loosely $H sim frac{1}{2}J|_C$. The question is, what is the projective (locally free) resolution of $i_* mathcal{O}_C(H)$? in other words, we have the following resolution in general,
begin{equation}
0longrightarrow mathcal{L}_2longrightarrowmathcal{L_1}longrightarrowmathcal{L}_0longrightarrow i_* mathcal{O}_C(H) longrightarrow 0,
end{equation}
what are the locally free sheaves $mathcal{L}_i$?
ag.algebraic-geometry projective-resolution
asked Dec 3 '18 at 5:21
Mohsen KarkheiranMohsen Karkheiran
21610
$endgroup$
Let $J$ be the hyperplane divisor in $mathbb{C}P^2$, and let $i:C hookrightarrow mathbb{C}P^2$ be the closed immersion of a smooth generic curve of degree 2. We know that $Csimeq mathbb{C}P^1$, and let $H$ be the hyperplane divisor in $C$. So loosely $H sim frac{1}{2}J|_C$. The question is, what is the projective (locally free) resolution of $i_* mathcal{O}_C(H)$? in other words, we have the following resolution in general,
begin{equation}
0longrightarrow mathcal{L}_2longrightarrowmathcal{L_1}longrightarrowmathcal{L}_0longrightarrow i_* mathcal{O}_C(H) longrightarrow 0,
end{equation}
what are the locally free sheaves $mathcal{L}_i$?
ag.algebraic-geometry projective-resolution
ag.algebraic-geometry projective-resolution
asked Dec 3 '18 at 5:21
Mohsen KarkheiranMohsen Karkheiran
21610
asked Dec 3 '18 at 5:21
Mohsen KarkheiranMohsen Karkheiran
21610
asked Dec 3 '18 at 5:21
Mohsen KarkheiranMohsen Karkheiran
21610
asked Dec 3 '18 at 5:21
Mohsen KarkheiranMohsen Karkheiran
21610
asked Dec 3 '18 at 5:21
Mohsen KarkheiranMohsen Karkheiran
21610
21610
$begingroup$
It is possible to put some topological constraints on $mathcal{L}_i$'s, but I'm curious to see whether it is possible to say more....
$endgroup$
– Mohsen Karkheiran
Dec 3 '18 at 5:24
add a comment |
$begingroup$
It is possible to put some topological constraints on $mathcal{L}_i$'s, but I'm curious to see whether it is possible to say more....
$endgroup$
– Mohsen Karkheiran
Dec 3 '18 at 5:24
$begingroup$
It is possible to put some topological constraints on $mathcal{L}_i$'s, but I'm curious to see whether it is possible to say more....
$endgroup$
– Mohsen Karkheiran
Dec 3 '18 at 5:24
$begingroup$
It is possible to put some topological constraints on $mathcal{L}_i$'s, but I'm curious to see whether it is possible to say more....
$endgroup$
– Mohsen Karkheiran
Dec 3 '18 at 5:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write the equation of $C$ as $ XY-Z^2=0$, and consider the homomorphism $u: mathcal{O}_{mathbb{P}^2}(-1)^2rightarrow mathcal{O}_{mathbb{P}^2}^2$ given by the matrix $begin{pmatrix}
X & Z\Z& Y
end{pmatrix}$. It is invertible outside $C$, and has rank 1 at every point of $C$. Thus $u$ is injective, and its cokernel is $i_*L$, where $L$ is a line bundle on $C$. The exact sequence $0rightarrow mathcal{O}_{mathbb{P}^2}(-1)^2xrightarrow{ u } mathcal{O}_{mathbb{P}^2}^2rightarrow i_*Lrightarrow 0 $ gives $h^0(L)= 2$, thus $L=mathcal{O}_C(H)$, and this exact sequence gives the resolution you are asking for.
answered Dec 3 '18 at 6:24
abxabx
23.3k34884
$endgroup$
add a comment |
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
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active
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votes
$begingroup$
Write the equation of $C$ as $ XY-Z^2=0$, and consider the homomorphism $u: mathcal{O}_{mathbb{P}^2}(-1)^2rightarrow mathcal{O}_{mathbb{P}^2}^2$ given by the matrix $begin{pmatrix}
X & Z\Z& Y
end{pmatrix}$. It is invertible outside $C$, and has rank 1 at every point of $C$. Thus $u$ is injective, and its cokernel is $i_*L$, where $L$ is a line bundle on $C$. The exact sequence $0rightarrow mathcal{O}_{mathbb{P}^2}(-1)^2xrightarrow{ u } mathcal{O}_{mathbb{P}^2}^2rightarrow i_*Lrightarrow 0 $ gives $h^0(L)= 2$, thus $L=mathcal{O}_C(H)$, and this exact sequence gives the resolution you are asking for.
answered Dec 3 '18 at 6:24
abxabx
23.3k34884
$endgroup$
add a comment |
$begingroup$
Write the equation of $C$ as $ XY-Z^2=0$, and consider the homomorphism $u: mathcal{O}_{mathbb{P}^2}(-1)^2rightarrow mathcal{O}_{mathbb{P}^2}^2$ given by the matrix $begin{pmatrix}
X & Z\Z& Y
end{pmatrix}$. It is invertible outside $C$, and has rank 1 at every point of $C$. Thus $u$ is injective, and its cokernel is $i_*L$, where $L$ is a line bundle on $C$. The exact sequence $0rightarrow mathcal{O}_{mathbb{P}^2}(-1)^2xrightarrow{ u } mathcal{O}_{mathbb{P}^2}^2rightarrow i_*Lrightarrow 0 $ gives $h^0(L)= 2$, thus $L=mathcal{O}_C(H)$, and this exact sequence gives the resolution you are asking for.
answered Dec 3 '18 at 6:24
abxabx
23.3k34884
$endgroup$
add a comment |
$begingroup$
Write the equation of $C$ as $ XY-Z^2=0$, and consider the homomorphism $u: mathcal{O}_{mathbb{P}^2}(-1)^2rightarrow mathcal{O}_{mathbb{P}^2}^2$ given by the matrix $begin{pmatrix}
X & Z\Z& Y
end{pmatrix}$. It is invertible outside $C$, and has rank 1 at every point of $C$. Thus $u$ is injective, and its cokernel is $i_*L$, where $L$ is a line bundle on $C$. The exact sequence $0rightarrow mathcal{O}_{mathbb{P}^2}(-1)^2xrightarrow{ u } mathcal{O}_{mathbb{P}^2}^2rightarrow i_*Lrightarrow 0 $ gives $h^0(L)= 2$, thus $L=mathcal{O}_C(H)$, and this exact sequence gives the resolution you are asking for.
answered Dec 3 '18 at 6:24
abxabx
23.3k34884
$endgroup$
Write the equation of $C$ as $ XY-Z^2=0$, and consider the homomorphism $u: mathcal{O}_{mathbb{P}^2}(-1)^2rightarrow mathcal{O}_{mathbb{P}^2}^2$ given by the matrix $begin{pmatrix}
X & Z\Z& Y
end{pmatrix}$. It is invertible outside $C$, and has rank 1 at every point of $C$. Thus $u$ is injective, and its cokernel is $i_*L$, where $L$ is a line bundle on $C$. The exact sequence $0rightarrow mathcal{O}_{mathbb{P}^2}(-1)^2xrightarrow{ u } mathcal{O}_{mathbb{P}^2}^2rightarrow i_*Lrightarrow 0 $ gives $h^0(L)= 2$, thus $L=mathcal{O}_C(H)$, and this exact sequence gives the resolution you are asking for.
answered Dec 3 '18 at 6:24
abxabx
23.3k34884
answered Dec 3 '18 at 6:24
abxabx
23.3k34884
answered Dec 3 '18 at 6:24
abxabx
23.3k34884
answered Dec 3 '18 at 6:24
abxabx
23.3k34884
23.3k34884
add a comment |
add a comment |
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$begingroup$
It is possible to put some topological constraints on $mathcal{L}_i$'s, but I'm curious to see whether it is possible to say more....
$endgroup$
– Mohsen Karkheiran
Dec 3 '18 at 5:24