Mixing
Why is this an open condition of vector bundles?
$begingroup$
Let $S$ be a smooth projective K3 surface. Let $E$ be a vector bundle on $S$ of rank $r+1$ with determinant $L$ such that
a) $H^i(S,E) =0$ for $i=1,2$ and
b) there exists a $Vin G(r+1, H^0(S,E))$ such that the degeneracy locus of $Votimes O_Srightarrow E$ is a smooth curve $C$ in $|L|$.
The claim is that the above is an open condition that is such bundles $E$ form an open subset in a family of vector bundles. How is that? Why are (a) and (b) open conditions.
I have other simpler doubts. Is the morphism $Votimes O_Srightarrow E$ always injective? Also how do we know that $G(r+1,H^0(S,E))$ is non-empty?
algebraic-geometry
asked Jan 6 at 11:01
user52991user52991
336310
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add a comment |
$begingroup$
Let $S$ be a smooth projective K3 surface. Let $E$ be a vector bundle on $S$ of rank $r+1$ with determinant $L$ such that
a) $H^i(S,E) =0$ for $i=1,2$ and
b) there exists a $Vin G(r+1, H^0(S,E))$ such that the degeneracy locus of $Votimes O_Srightarrow E$ is a smooth curve $C$ in $|L|$.
The claim is that the above is an open condition that is such bundles $E$ form an open subset in a family of vector bundles. How is that? Why are (a) and (b) open conditions.
I have other simpler doubts. Is the morphism $Votimes O_Srightarrow E$ always injective? Also how do we know that $G(r+1,H^0(S,E))$ is non-empty?
algebraic-geometry
asked Jan 6 at 11:01
user52991user52991
336310
$endgroup$
add a comment |
$begingroup$
Let $S$ be a smooth projective K3 surface. Let $E$ be a vector bundle on $S$ of rank $r+1$ with determinant $L$ such that
a) $H^i(S,E) =0$ for $i=1,2$ and
b) there exists a $Vin G(r+1, H^0(S,E))$ such that the degeneracy locus of $Votimes O_Srightarrow E$ is a smooth curve $C$ in $|L|$.
The claim is that the above is an open condition that is such bundles $E$ form an open subset in a family of vector bundles. How is that? Why are (a) and (b) open conditions.
I have other simpler doubts. Is the morphism $Votimes O_Srightarrow E$ always injective? Also how do we know that $G(r+1,H^0(S,E))$ is non-empty?
algebraic-geometry
asked Jan 6 at 11:01
user52991user52991
336310
$endgroup$
Let $S$ be a smooth projective K3 surface. Let $E$ be a vector bundle on $S$ of rank $r+1$ with determinant $L$ such that
a) $H^i(S,E) =0$ for $i=1,2$ and
b) there exists a $Vin G(r+1, H^0(S,E))$ such that the degeneracy locus of $Votimes O_Srightarrow E$ is a smooth curve $C$ in $|L|$.
The claim is that the above is an open condition that is such bundles $E$ form an open subset in a family of vector bundles. How is that? Why are (a) and (b) open conditions.
I have other simpler doubts. Is the morphism $Votimes O_Srightarrow E$ always injective? Also how do we know that $G(r+1,H^0(S,E))$ is non-empty?
algebraic-geometry
algebraic-geometry
asked Jan 6 at 11:01
user52991user52991
336310
asked Jan 6 at 11:01
user52991user52991
336310
asked Jan 6 at 11:01
user52991user52991
336310
asked Jan 6 at 11:01
user52991user52991
336310
asked Jan 6 at 11:01
user52991user52991
336310
336310
add a comment |
add a comment |
1 Answer
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$begingroup$
Condition (a) is open by semi-continuity theorem. To check openness of (b) consider a vector bundle $E$ on $S times B$ (where $B$ is an arbitrary base scheme) and rank $r+1$ vector subbundle $V$ in $p_*E$ (where $p colon S times B to B$ is a projection). By adjunction there is a natural morphism $p^*V to E$. Let $C$ be its cokernel and $Z subset S times B$ its support. Then the subscheme of $B$ over which the fibers of $Z$ are $ge 2$-dimensional is closed, and the subscheme where the fibers are 0-dimensional is empty. Therefore, the subscheme $B_1 subset B$ over which the fibers are 1-dimensional is open. Finally, the subscheme of $B_1$ over which the fibers are smooth is open as well.
Surely we don't know that $Gr(r+1,H^0(S,E))$ is non-empty. But if it is empty the corresponding subscheme in $B$ is empty as well, but it is still open.
answered Jan 6 at 12:48
SashaSasha
4,618139
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1 Answer
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1 Answer
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active
oldest
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active
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active
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$begingroup$
Condition (a) is open by semi-continuity theorem. To check openness of (b) consider a vector bundle $E$ on $S times B$ (where $B$ is an arbitrary base scheme) and rank $r+1$ vector subbundle $V$ in $p_*E$ (where $p colon S times B to B$ is a projection). By adjunction there is a natural morphism $p^*V to E$. Let $C$ be its cokernel and $Z subset S times B$ its support. Then the subscheme of $B$ over which the fibers of $Z$ are $ge 2$-dimensional is closed, and the subscheme where the fibers are 0-dimensional is empty. Therefore, the subscheme $B_1 subset B$ over which the fibers are 1-dimensional is open. Finally, the subscheme of $B_1$ over which the fibers are smooth is open as well.
Surely we don't know that $Gr(r+1,H^0(S,E))$ is non-empty. But if it is empty the corresponding subscheme in $B$ is empty as well, but it is still open.
answered Jan 6 at 12:48
SashaSasha
4,618139
$endgroup$
add a comment |
$begingroup$
Condition (a) is open by semi-continuity theorem. To check openness of (b) consider a vector bundle $E$ on $S times B$ (where $B$ is an arbitrary base scheme) and rank $r+1$ vector subbundle $V$ in $p_*E$ (where $p colon S times B to B$ is a projection). By adjunction there is a natural morphism $p^*V to E$. Let $C$ be its cokernel and $Z subset S times B$ its support. Then the subscheme of $B$ over which the fibers of $Z$ are $ge 2$-dimensional is closed, and the subscheme where the fibers are 0-dimensional is empty. Therefore, the subscheme $B_1 subset B$ over which the fibers are 1-dimensional is open. Finally, the subscheme of $B_1$ over which the fibers are smooth is open as well.
Surely we don't know that $Gr(r+1,H^0(S,E))$ is non-empty. But if it is empty the corresponding subscheme in $B$ is empty as well, but it is still open.
answered Jan 6 at 12:48
SashaSasha
4,618139
$endgroup$
add a comment |
$begingroup$
Condition (a) is open by semi-continuity theorem. To check openness of (b) consider a vector bundle $E$ on $S times B$ (where $B$ is an arbitrary base scheme) and rank $r+1$ vector subbundle $V$ in $p_*E$ (where $p colon S times B to B$ is a projection). By adjunction there is a natural morphism $p^*V to E$. Let $C$ be its cokernel and $Z subset S times B$ its support. Then the subscheme of $B$ over which the fibers of $Z$ are $ge 2$-dimensional is closed, and the subscheme where the fibers are 0-dimensional is empty. Therefore, the subscheme $B_1 subset B$ over which the fibers are 1-dimensional is open. Finally, the subscheme of $B_1$ over which the fibers are smooth is open as well.
Surely we don't know that $Gr(r+1,H^0(S,E))$ is non-empty. But if it is empty the corresponding subscheme in $B$ is empty as well, but it is still open.
answered Jan 6 at 12:48
SashaSasha
4,618139
$endgroup$
Condition (a) is open by semi-continuity theorem. To check openness of (b) consider a vector bundle $E$ on $S times B$ (where $B$ is an arbitrary base scheme) and rank $r+1$ vector subbundle $V$ in $p_*E$ (where $p colon S times B to B$ is a projection). By adjunction there is a natural morphism $p^*V to E$. Let $C$ be its cokernel and $Z subset S times B$ its support. Then the subscheme of $B$ over which the fibers of $Z$ are $ge 2$-dimensional is closed, and the subscheme where the fibers are 0-dimensional is empty. Therefore, the subscheme $B_1 subset B$ over which the fibers are 1-dimensional is open. Finally, the subscheme of $B_1$ over which the fibers are smooth is open as well.
Surely we don't know that $Gr(r+1,H^0(S,E))$ is non-empty. But if it is empty the corresponding subscheme in $B$ is empty as well, but it is still open.
answered Jan 6 at 12:48
SashaSasha
4,618139
answered Jan 6 at 12:48
SashaSasha
4,618139
answered Jan 6 at 12:48
SashaSasha
4,618139
answered Jan 6 at 12:48
SashaSasha
4,618139
4,618139
add a comment |
add a comment |
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