rank $2$ bundle being an extension of line bundles
$begingroup$
Let $X$ be a smooth projective variety of dimension $geq 2$. Fix an ample divisor $H$ and let $E$ be a $mu$-semi-stable vector bundle of rank $2$ with $c_1(E)=0$. Assume $H^0(E)neq 0$. I am looking for a proof of the following statement:
If $0neq sin H^0(E)$ vanishes no where, then $E$ is an extension of line bundles.
My attempt: Suppose such a section exists, then we have a surjection from $E^*to mathcal{O}_X$ (this follows from the definition of section of vector bundle), but I do not know how to proceed from here.
Thanks!
algebraic-geometry
asked Jan 24 at 3:45
Rust QRust Q
1398
$endgroup$
add a comment |
$begingroup$
Let $X$ be a smooth projective variety of dimension $geq 2$. Fix an ample divisor $H$ and let $E$ be a $mu$-semi-stable vector bundle of rank $2$ with $c_1(E)=0$. Assume $H^0(E)neq 0$. I am looking for a proof of the following statement:
If $0neq sin H^0(E)$ vanishes no where, then $E$ is an extension of line bundles.
My attempt: Suppose such a section exists, then we have a surjection from $E^*to mathcal{O}_X$ (this follows from the definition of section of vector bundle), but I do not know how to proceed from here.
Thanks!
algebraic-geometry
asked Jan 24 at 3:45
Rust QRust Q
1398
$endgroup$
$begingroup$
If such a section exists (with no further assumptions), we have $mathcal{O}to E$ given by this section. The cokernel is a line bundle $L$, since the section is nowhere vanishing. So, the vector bundle is an extension of $mathcal{O}$ and $L=E/mathcal{O}$.
$endgroup$
– Mohan
Jan 24 at 3:59
$begingroup$
@Mohan Could you explain a little more on why the section is no where vanishing implies the cokernel is a line bundle?
$endgroup$
– Rust Q
Jan 24 at 4:56
2
$begingroup$
If $sin H^0(E)$ is nowhere vanishing for a rank $r$ vector bundle $E$, then the cokernel is a (rank $r-1$) vector bundle. This is a local checking. Locally, $E$ is a free module and nowhere vanishing section means $s$ is part of a basis of $E$ by Nakayama's lemma. Thus, the cokernel is a free module.
$endgroup$
– Mohan
Jan 24 at 14:26
add a comment |
$begingroup$
Let $X$ be a smooth projective variety of dimension $geq 2$. Fix an ample divisor $H$ and let $E$ be a $mu$-semi-stable vector bundle of rank $2$ with $c_1(E)=0$. Assume $H^0(E)neq 0$. I am looking for a proof of the following statement:
If $0neq sin H^0(E)$ vanishes no where, then $E$ is an extension of line bundles.
My attempt: Suppose such a section exists, then we have a surjection from $E^*to mathcal{O}_X$ (this follows from the definition of section of vector bundle), but I do not know how to proceed from here.
Thanks!
algebraic-geometry
asked Jan 24 at 3:45
Rust QRust Q
1398
$endgroup$
Let $X$ be a smooth projective variety of dimension $geq 2$. Fix an ample divisor $H$ and let $E$ be a $mu$-semi-stable vector bundle of rank $2$ with $c_1(E)=0$. Assume $H^0(E)neq 0$. I am looking for a proof of the following statement:
If $0neq sin H^0(E)$ vanishes no where, then $E$ is an extension of line bundles.
My attempt: Suppose such a section exists, then we have a surjection from $E^*to mathcal{O}_X$ (this follows from the definition of section of vector bundle), but I do not know how to proceed from here.
Thanks!
algebraic-geometry
algebraic-geometry
asked Jan 24 at 3:45
Rust QRust Q
1398
asked Jan 24 at 3:45
Rust QRust Q
1398
edited Jan 24 at 4:55
Rust Q
asked Jan 24 at 3:45
Rust QRust Q
1398
asked Jan 24 at 3:45
Rust QRust Q
1398
asked Jan 24 at 3:45
Rust QRust Q
1398
1398
$begingroup$
If such a section exists (with no further assumptions), we have $mathcal{O}to E$ given by this section. The cokernel is a line bundle $L$, since the section is nowhere vanishing. So, the vector bundle is an extension of $mathcal{O}$ and $L=E/mathcal{O}$.
$endgroup$
– Mohan
Jan 24 at 3:59
$begingroup$
@Mohan Could you explain a little more on why the section is no where vanishing implies the cokernel is a line bundle?
$endgroup$
– Rust Q
Jan 24 at 4:56
2
$begingroup$
If $sin H^0(E)$ is nowhere vanishing for a rank $r$ vector bundle $E$, then the cokernel is a (rank $r-1$) vector bundle. This is a local checking. Locally, $E$ is a free module and nowhere vanishing section means $s$ is part of a basis of $E$ by Nakayama's lemma. Thus, the cokernel is a free module.
$endgroup$
– Mohan
Jan 24 at 14:26
add a comment |
$begingroup$
If such a section exists (with no further assumptions), we have $mathcal{O}to E$ given by this section. The cokernel is a line bundle $L$, since the section is nowhere vanishing. So, the vector bundle is an extension of $mathcal{O}$ and $L=E/mathcal{O}$.
$endgroup$
– Mohan
Jan 24 at 3:59
$begingroup$
@Mohan Could you explain a little more on why the section is no where vanishing implies the cokernel is a line bundle?
$endgroup$
– Rust Q
Jan 24 at 4:56
2
$begingroup$
If $sin H^0(E)$ is nowhere vanishing for a rank $r$ vector bundle $E$, then the cokernel is a (rank $r-1$) vector bundle. This is a local checking. Locally, $E$ is a free module and nowhere vanishing section means $s$ is part of a basis of $E$ by Nakayama's lemma. Thus, the cokernel is a free module.
$endgroup$
– Mohan
Jan 24 at 14:26
$begingroup$
If such a section exists (with no further assumptions), we have $mathcal{O}to E$ given by this section. The cokernel is a line bundle $L$, since the section is nowhere vanishing. So, the vector bundle is an extension of $mathcal{O}$ and $L=E/mathcal{O}$.
$endgroup$
– Mohan
Jan 24 at 3:59
$begingroup$
If such a section exists (with no further assumptions), we have $mathcal{O}to E$ given by this section. The cokernel is a line bundle $L$, since the section is nowhere vanishing. So, the vector bundle is an extension of $mathcal{O}$ and $L=E/mathcal{O}$.
$endgroup$
– Mohan
Jan 24 at 3:59
$begingroup$
@Mohan Could you explain a little more on why the section is no where vanishing implies the cokernel is a line bundle?
$endgroup$
– Rust Q
Jan 24 at 4:56
$begingroup$
@Mohan Could you explain a little more on why the section is no where vanishing implies the cokernel is a line bundle?
$endgroup$
– Rust Q
Jan 24 at 4:56
2
2
$begingroup$
If $sin H^0(E)$ is nowhere vanishing for a rank $r$ vector bundle $E$, then the cokernel is a (rank $r-1$) vector bundle. This is a local checking. Locally, $E$ is a free module and nowhere vanishing section means $s$ is part of a basis of $E$ by Nakayama's lemma. Thus, the cokernel is a free module.
$endgroup$
– Mohan
Jan 24 at 14:26
$begingroup$
If $sin H^0(E)$ is nowhere vanishing for a rank $r$ vector bundle $E$, then the cokernel is a (rank $r-1$) vector bundle. This is a local checking. Locally, $E$ is a free module and nowhere vanishing section means $s$ is part of a basis of $E$ by Nakayama's lemma. Thus, the cokernel is a free module.
$endgroup$
– Mohan
Jan 24 at 14:26
add a comment |
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$begingroup$
If such a section exists (with no further assumptions), we have $mathcal{O}to E$ given by this section. The cokernel is a line bundle $L$, since the section is nowhere vanishing. So, the vector bundle is an extension of $mathcal{O}$ and $L=E/mathcal{O}$.
$endgroup$
– Mohan
Jan 24 at 3:59
$begingroup$
@Mohan Could you explain a little more on why the section is no where vanishing implies the cokernel is a line bundle?
$endgroup$
– Rust Q
Jan 24 at 4:56
2
$begingroup$
If $sin H^0(E)$ is nowhere vanishing for a rank $r$ vector bundle $E$, then the cokernel is a (rank $r-1$) vector bundle. This is a local checking. Locally, $E$ is a free module and nowhere vanishing section means $s$ is part of a basis of $E$ by Nakayama's lemma. Thus, the cokernel is a free module.
$endgroup$
– Mohan
Jan 24 at 14:26