Exact sequences of $mathcal{O}_X$-modules, sections over X minus a point, and splitting
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Let $X$ be a (let's say irreducible) scheme, let $x$ be a closed point, put $U = X - {x}$. Let $0tomathcal{F}tomathcal{G}tomathcal{H}to 0$ be an exact sequence of $mathcal{O}_X$-modules. If the exact sequence of sections over $U$ is split, is the original sequence of $mathcal{O}_X$-modules split? The case I am most interested in is when $X$ is projective.
sheaf-theory schemes
asked Dec 6 '18 at 19:12
vgty6h7uijvgty6h7uij
583515
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$begingroup$
Let $X$ be a (let's say irreducible) scheme, let $x$ be a closed point, put $U = X - {x}$. Let $0tomathcal{F}tomathcal{G}tomathcal{H}to 0$ be an exact sequence of $mathcal{O}_X$-modules. If the exact sequence of sections over $U$ is split, is the original sequence of $mathcal{O}_X$-modules split? The case I am most interested in is when $X$ is projective.
sheaf-theory schemes
asked Dec 6 '18 at 19:12
vgty6h7uijvgty6h7uij
583515
$endgroup$
1
$begingroup$
No, why would a section of $mathcal{Gto H}$ extends to $X$ ? Take $X=mathbb{P}^1, x$ the point at infinity, then $0tomathcal{O}_{mathbb{P}^1}(-1)tomathcal{O}_{mathbb{P}^1}to mathcal{O}_xto 0$ splits if we restrict to $U$ because it is $0to mathcal{O}_{mathbb{A}^1}tomathcal{O}_{mathbb{A}^1}to 0$. There are also counter-example a short exact sequence of vector bundles (take the Euler sequence over $mathbb{P}^1$).
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– Roland
Dec 6 '18 at 20:36
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Thanks, I was looking for a counter-example but couldn't think of one.
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– vgty6h7uij
Dec 6 '18 at 20:45
add a comment |
$begingroup$
Let $X$ be a (let's say irreducible) scheme, let $x$ be a closed point, put $U = X - {x}$. Let $0tomathcal{F}tomathcal{G}tomathcal{H}to 0$ be an exact sequence of $mathcal{O}_X$-modules. If the exact sequence of sections over $U$ is split, is the original sequence of $mathcal{O}_X$-modules split? The case I am most interested in is when $X$ is projective.
sheaf-theory schemes
asked Dec 6 '18 at 19:12
vgty6h7uijvgty6h7uij
583515
$endgroup$
Let $X$ be a (let's say irreducible) scheme, let $x$ be a closed point, put $U = X - {x}$. Let $0tomathcal{F}tomathcal{G}tomathcal{H}to 0$ be an exact sequence of $mathcal{O}_X$-modules. If the exact sequence of sections over $U$ is split, is the original sequence of $mathcal{O}_X$-modules split? The case I am most interested in is when $X$ is projective.
sheaf-theory schemes
sheaf-theory schemes
asked Dec 6 '18 at 19:12
vgty6h7uijvgty6h7uij
583515
asked Dec 6 '18 at 19:12
vgty6h7uijvgty6h7uij
583515
asked Dec 6 '18 at 19:12
vgty6h7uijvgty6h7uij
583515
asked Dec 6 '18 at 19:12
vgty6h7uijvgty6h7uij
583515
asked Dec 6 '18 at 19:12
vgty6h7uijvgty6h7uij
583515
583515
1
$begingroup$
No, why would a section of $mathcal{Gto H}$ extends to $X$ ? Take $X=mathbb{P}^1, x$ the point at infinity, then $0tomathcal{O}_{mathbb{P}^1}(-1)tomathcal{O}_{mathbb{P}^1}to mathcal{O}_xto 0$ splits if we restrict to $U$ because it is $0to mathcal{O}_{mathbb{A}^1}tomathcal{O}_{mathbb{A}^1}to 0$. There are also counter-example a short exact sequence of vector bundles (take the Euler sequence over $mathbb{P}^1$).
$endgroup$
– Roland
Dec 6 '18 at 20:36
$begingroup$
Thanks, I was looking for a counter-example but couldn't think of one.
$endgroup$
– vgty6h7uij
Dec 6 '18 at 20:45
add a comment |
1
$begingroup$
No, why would a section of $mathcal{Gto H}$ extends to $X$ ? Take $X=mathbb{P}^1, x$ the point at infinity, then $0tomathcal{O}_{mathbb{P}^1}(-1)tomathcal{O}_{mathbb{P}^1}to mathcal{O}_xto 0$ splits if we restrict to $U$ because it is $0to mathcal{O}_{mathbb{A}^1}tomathcal{O}_{mathbb{A}^1}to 0$. There are also counter-example a short exact sequence of vector bundles (take the Euler sequence over $mathbb{P}^1$).
$endgroup$
– Roland
Dec 6 '18 at 20:36
$begingroup$
Thanks, I was looking for a counter-example but couldn't think of one.
$endgroup$
– vgty6h7uij
Dec 6 '18 at 20:45
1
1
$begingroup$
No, why would a section of $mathcal{Gto H}$ extends to $X$ ? Take $X=mathbb{P}^1, x$ the point at infinity, then $0tomathcal{O}_{mathbb{P}^1}(-1)tomathcal{O}_{mathbb{P}^1}to mathcal{O}_xto 0$ splits if we restrict to $U$ because it is $0to mathcal{O}_{mathbb{A}^1}tomathcal{O}_{mathbb{A}^1}to 0$. There are also counter-example a short exact sequence of vector bundles (take the Euler sequence over $mathbb{P}^1$).
$endgroup$
– Roland
Dec 6 '18 at 20:36
$begingroup$
No, why would a section of $mathcal{Gto H}$ extends to $X$ ? Take $X=mathbb{P}^1, x$ the point at infinity, then $0tomathcal{O}_{mathbb{P}^1}(-1)tomathcal{O}_{mathbb{P}^1}to mathcal{O}_xto 0$ splits if we restrict to $U$ because it is $0to mathcal{O}_{mathbb{A}^1}tomathcal{O}_{mathbb{A}^1}to 0$. There are also counter-example a short exact sequence of vector bundles (take the Euler sequence over $mathbb{P}^1$).
$endgroup$
– Roland
Dec 6 '18 at 20:36
$begingroup$
Thanks, I was looking for a counter-example but couldn't think of one.
$endgroup$
– vgty6h7uij
Dec 6 '18 at 20:45
$begingroup$
Thanks, I was looking for a counter-example but couldn't think of one.
$endgroup$
– vgty6h7uij
Dec 6 '18 at 20:45
add a comment |
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No, why would a section of $mathcal{Gto H}$ extends to $X$ ? Take $X=mathbb{P}^1, x$ the point at infinity, then $0tomathcal{O}_{mathbb{P}^1}(-1)tomathcal{O}_{mathbb{P}^1}to mathcal{O}_xto 0$ splits if we restrict to $U$ because it is $0to mathcal{O}_{mathbb{A}^1}tomathcal{O}_{mathbb{A}^1}to 0$. There are also counter-example a short exact sequence of vector bundles (take the Euler sequence over $mathbb{P}^1$).
$endgroup$
– Roland
Dec 6 '18 at 20:36
$begingroup$
Thanks, I was looking for a counter-example but couldn't think of one.
$endgroup$
– vgty6h7uij
Dec 6 '18 at 20:45