Resolution of a torsion sheaf











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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 $mathcal{L}_i$?










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  • 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....
    – Mohsen Karkheiran
    2 hours ago

















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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 $mathcal{L}_i$?










share|cite|improve this question






















  • 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....
    – Mohsen Karkheiran
    2 hours ago















up vote
3
down vote

favorite









up vote
3
down vote

favorite











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 $mathcal{L}_i$?










share|cite|improve this question













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






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asked 2 hours ago









Mohsen Karkheiran

17710




17710












  • 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....
    – Mohsen Karkheiran
    2 hours ago




















  • 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....
    – Mohsen Karkheiran
    2 hours ago


















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....
– Mohsen Karkheiran
2 hours ago






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....
– Mohsen Karkheiran
2 hours ago












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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.






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    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.






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      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.






      share|cite|improve this answer























        up vote
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        up vote
        3
        down vote









        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        abx

        22.7k34681




        22.7k34681






























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