Stable extensions by line bundles on Riemann surfaces
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Is there a compact Riemann surface $X$ and a line bundle $L$ of negative degree on $X$, such that for any nontrivial extension
$$ 0 rightarrow L rightarrow E rightarrow L^{-1} rightarrow 0, $$
$E$ is a stable vector bundle on $X$? Any comment and reference is welcome, thank you.
ag.algebraic-geometry vector-bundles riemann-surfaces
asked 2 days ago
swalker
33018
add a comment |
up vote
4
down vote
favorite
Is there a compact Riemann surface $X$ and a line bundle $L$ of negative degree on $X$, such that for any nontrivial extension
$$ 0 rightarrow L rightarrow E rightarrow L^{-1} rightarrow 0, $$
$E$ is a stable vector bundle on $X$? Any comment and reference is welcome, thank you.
ag.algebraic-geometry vector-bundles riemann-surfaces
asked 2 days ago
swalker
33018
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Is there a compact Riemann surface $X$ and a line bundle $L$ of negative degree on $X$, such that for any nontrivial extension
$$ 0 rightarrow L rightarrow E rightarrow L^{-1} rightarrow 0, $$
$E$ is a stable vector bundle on $X$? Any comment and reference is welcome, thank you.
ag.algebraic-geometry vector-bundles riemann-surfaces
asked 2 days ago
swalker
33018
Is there a compact Riemann surface $X$ and a line bundle $L$ of negative degree on $X$, such that for any nontrivial extension
$$ 0 rightarrow L rightarrow E rightarrow L^{-1} rightarrow 0, $$
$E$ is a stable vector bundle on $X$? Any comment and reference is welcome, thank you.
ag.algebraic-geometry vector-bundles riemann-surfaces
ag.algebraic-geometry vector-bundles riemann-surfaces
asked 2 days ago
swalker
33018
asked 2 days ago
swalker
33018
asked 2 days ago
swalker
33018
asked 2 days ago
swalker
33018
asked 2 days ago
swalker
33018
33018
add a comment |
add a comment |
1 Answer
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10
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This never happens. Pick a point $pin X$; the exact sequence $0rightarrow L^{2}rightarrow L^{2}(p)rightarrow mathbb{C}_prightarrow 0$ gives rise to an exact sequence $0rightarrow mathbb{C}xrightarrow{ partial } H^1(L^2)longrightarrow H^1(L^2(p))rightarrow 0$. The class
$e:=partial (1)$ in $H^1(L^2)cong operatorname{Ext}^1(L^{-1},L) $ maps to $0$ in $operatorname{Ext}^1(L^{-1}(-p),L) $, hence it defines a nontrivial extension of $L^{-1}$ by $L$ which becomes trivial when pulled back to $L^{-1}(-p)$. This means that the extension bundle $E$ contains $L^{-1}(-p)$, hence is not stable.
answered 2 days ago
abx
22.6k34581
Thank you very much! By the openness of stability, we know that the unstable extensions form a subvariety of Ext$^1(L^{-1},L)$ of codimension $ge 1$. Can we expect that the subvariety has codimension $>1$ for some general Riemann surfaces? For example $g(X) > 1$.
– swalker
yesterday
If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
– swalker
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
This never happens. Pick a point $pin X$; the exact sequence $0rightarrow L^{2}rightarrow L^{2}(p)rightarrow mathbb{C}_prightarrow 0$ gives rise to an exact sequence $0rightarrow mathbb{C}xrightarrow{ partial } H^1(L^2)longrightarrow H^1(L^2(p))rightarrow 0$. The class
$e:=partial (1)$ in $H^1(L^2)cong operatorname{Ext}^1(L^{-1},L) $ maps to $0$ in $operatorname{Ext}^1(L^{-1}(-p),L) $, hence it defines a nontrivial extension of $L^{-1}$ by $L$ which becomes trivial when pulled back to $L^{-1}(-p)$. This means that the extension bundle $E$ contains $L^{-1}(-p)$, hence is not stable.
answered 2 days ago
abx
22.6k34581
Thank you very much! By the openness of stability, we know that the unstable extensions form a subvariety of Ext$^1(L^{-1},L)$ of codimension $ge 1$. Can we expect that the subvariety has codimension $>1$ for some general Riemann surfaces? For example $g(X) > 1$.
– swalker
yesterday
If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
– swalker
yesterday
add a comment |
up vote
10
down vote
This never happens. Pick a point $pin X$; the exact sequence $0rightarrow L^{2}rightarrow L^{2}(p)rightarrow mathbb{C}_prightarrow 0$ gives rise to an exact sequence $0rightarrow mathbb{C}xrightarrow{ partial } H^1(L^2)longrightarrow H^1(L^2(p))rightarrow 0$. The class
$e:=partial (1)$ in $H^1(L^2)cong operatorname{Ext}^1(L^{-1},L) $ maps to $0$ in $operatorname{Ext}^1(L^{-1}(-p),L) $, hence it defines a nontrivial extension of $L^{-1}$ by $L$ which becomes trivial when pulled back to $L^{-1}(-p)$. This means that the extension bundle $E$ contains $L^{-1}(-p)$, hence is not stable.
answered 2 days ago
abx
22.6k34581
Thank you very much! By the openness of stability, we know that the unstable extensions form a subvariety of Ext$^1(L^{-1},L)$ of codimension $ge 1$. Can we expect that the subvariety has codimension $>1$ for some general Riemann surfaces? For example $g(X) > 1$.
– swalker
yesterday
If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
– swalker
yesterday
add a comment |
up vote
10
down vote
up vote
10
down vote
This never happens. Pick a point $pin X$; the exact sequence $0rightarrow L^{2}rightarrow L^{2}(p)rightarrow mathbb{C}_prightarrow 0$ gives rise to an exact sequence $0rightarrow mathbb{C}xrightarrow{ partial } H^1(L^2)longrightarrow H^1(L^2(p))rightarrow 0$. The class
$e:=partial (1)$ in $H^1(L^2)cong operatorname{Ext}^1(L^{-1},L) $ maps to $0$ in $operatorname{Ext}^1(L^{-1}(-p),L) $, hence it defines a nontrivial extension of $L^{-1}$ by $L$ which becomes trivial when pulled back to $L^{-1}(-p)$. This means that the extension bundle $E$ contains $L^{-1}(-p)$, hence is not stable.
answered 2 days ago
abx
22.6k34581
This never happens. Pick a point $pin X$; the exact sequence $0rightarrow L^{2}rightarrow L^{2}(p)rightarrow mathbb{C}_prightarrow 0$ gives rise to an exact sequence $0rightarrow mathbb{C}xrightarrow{ partial } H^1(L^2)longrightarrow H^1(L^2(p))rightarrow 0$. The class
$e:=partial (1)$ in $H^1(L^2)cong operatorname{Ext}^1(L^{-1},L) $ maps to $0$ in $operatorname{Ext}^1(L^{-1}(-p),L) $, hence it defines a nontrivial extension of $L^{-1}$ by $L$ which becomes trivial when pulled back to $L^{-1}(-p)$. This means that the extension bundle $E$ contains $L^{-1}(-p)$, hence is not stable.
answered 2 days ago
abx
22.6k34581
answered 2 days ago
abx
22.6k34581
answered 2 days ago
abx
22.6k34581
answered 2 days ago
abx
22.6k34581
22.6k34581
Thank you very much! By the openness of stability, we know that the unstable extensions form a subvariety of Ext$^1(L^{-1},L)$ of codimension $ge 1$. Can we expect that the subvariety has codimension $>1$ for some general Riemann surfaces? For example $g(X) > 1$.
– swalker
yesterday
If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
– swalker
yesterday
add a comment |
Thank you very much! By the openness of stability, we know that the unstable extensions form a subvariety of Ext$^1(L^{-1},L)$ of codimension $ge 1$. Can we expect that the subvariety has codimension $>1$ for some general Riemann surfaces? For example $g(X) > 1$.
– swalker
yesterday
If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
– swalker
yesterday
Thank you very much! By the openness of stability, we know that the unstable extensions form a subvariety of Ext$^1(L^{-1},L)$ of codimension $ge 1$. Can we expect that the subvariety has codimension $>1$ for some general Riemann surfaces? For example $g(X) > 1$.
– swalker
yesterday
Thank you very much! By the openness of stability, we know that the unstable extensions form a subvariety of Ext$^1(L^{-1},L)$ of codimension $ge 1$. Can we expect that the subvariety has codimension $>1$ for some general Riemann surfaces? For example $g(X) > 1$.
– swalker
yesterday
If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
– swalker
yesterday
If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
– swalker
yesterday
add a comment |
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