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.










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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.










    share|cite|improve this question
























      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.










      share|cite|improve this question













      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






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      asked Nov 17 at 8:00









      swalker

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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.






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          • 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
            Nov 18 at 2:44










          • If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
            – swalker
            Nov 18 at 2:48










          • Yes, I think that quite generally the subvariety of unstable extensions has high codimension. You might have a look at a paper by A. Bertram, Moduli of rank-2 vector bundles, theta divisors, and the geometry of curves in projective space (J. Differential Geom. 35 (1992), no. 2, 429-469). He does a detailed analysis (in a slightly different situation) of the stability of extensions.
            – abx
            yesterday











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          1 Answer
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          active

          oldest

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          up vote
          10
          down vote



          accepted










          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.






          share|cite|improve this answer





















          • 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
            Nov 18 at 2:44










          • If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
            – swalker
            Nov 18 at 2:48










          • Yes, I think that quite generally the subvariety of unstable extensions has high codimension. You might have a look at a paper by A. Bertram, Moduli of rank-2 vector bundles, theta divisors, and the geometry of curves in projective space (J. Differential Geom. 35 (1992), no. 2, 429-469). He does a detailed analysis (in a slightly different situation) of the stability of extensions.
            – abx
            yesterday















          up vote
          10
          down vote



          accepted










          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.






          share|cite|improve this answer





















          • 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
            Nov 18 at 2:44










          • If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
            – swalker
            Nov 18 at 2:48










          • Yes, I think that quite generally the subvariety of unstable extensions has high codimension. You might have a look at a paper by A. Bertram, Moduli of rank-2 vector bundles, theta divisors, and the geometry of curves in projective space (J. Differential Geom. 35 (1992), no. 2, 429-469). He does a detailed analysis (in a slightly different situation) of the stability of extensions.
            – abx
            yesterday













          up vote
          10
          down vote



          accepted







          up vote
          10
          down vote



          accepted






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 17 at 9:58









          abx

          22.7k34681




          22.7k34681












          • 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
            Nov 18 at 2:44










          • If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
            – swalker
            Nov 18 at 2:48










          • Yes, I think that quite generally the subvariety of unstable extensions has high codimension. You might have a look at a paper by A. Bertram, Moduli of rank-2 vector bundles, theta divisors, and the geometry of curves in projective space (J. Differential Geom. 35 (1992), no. 2, 429-469). He does a detailed analysis (in a slightly different situation) of the stability of extensions.
            – abx
            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
            Nov 18 at 2:44










          • If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
            – swalker
            Nov 18 at 2:48










          • Yes, I think that quite generally the subvariety of unstable extensions has high codimension. You might have a look at a paper by A. Bertram, Moduli of rank-2 vector bundles, theta divisors, and the geometry of curves in projective space (J. Differential Geom. 35 (1992), no. 2, 429-469). He does a detailed analysis (in a slightly different situation) of the stability of extensions.
            – abx
            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
          Nov 18 at 2:44




          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
          Nov 18 at 2:44












          If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
          – swalker
          Nov 18 at 2:48




          If $X$ is a Riemann surface of genus $g>1$, I proved that there always exist stable extension bundles.@abx
          – swalker
          Nov 18 at 2:48












          Yes, I think that quite generally the subvariety of unstable extensions has high codimension. You might have a look at a paper by A. Bertram, Moduli of rank-2 vector bundles, theta divisors, and the geometry of curves in projective space (J. Differential Geom. 35 (1992), no. 2, 429-469). He does a detailed analysis (in a slightly different situation) of the stability of extensions.
          – abx
          yesterday




          Yes, I think that quite generally the subvariety of unstable extensions has high codimension. You might have a look at a paper by A. Bertram, Moduli of rank-2 vector bundles, theta divisors, and the geometry of curves in projective space (J. Differential Geom. 35 (1992), no. 2, 429-469). He does a detailed analysis (in a slightly different situation) of the stability of extensions.
          – abx
          yesterday


















           

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