Stable extensions by line bundles on Riemann surfaces












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$begingroup$


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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    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Nov 17 '18 at 8:00









      swalkerswalker

      33718




      33718






















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          $begingroup$

          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









          $endgroup$













          • $begingroup$
            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 $g(X) > 1$.
            $endgroup$
            – swalker
            Nov 18 '18 at 2:44










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










          • $begingroup$
            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.
            $endgroup$
            – abx
            Nov 21 '18 at 6:32











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          1 Answer
          1






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          active

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          active

          oldest

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          10












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            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 $g(X) > 1$.
            $endgroup$
            – swalker
            Nov 18 '18 at 2:44










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










          • $begingroup$
            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.
            $endgroup$
            – abx
            Nov 21 '18 at 6:32
















          10












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            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 $g(X) > 1$.
            $endgroup$
            – swalker
            Nov 18 '18 at 2:44










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










          • $begingroup$
            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.
            $endgroup$
            – abx
            Nov 21 '18 at 6:32














          10












          10








          10





          $begingroup$

          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









          $endgroup$



          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 '18 at 9:58









          abxabx

          23.5k34885




          23.5k34885












          • $begingroup$
            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 $g(X) > 1$.
            $endgroup$
            – swalker
            Nov 18 '18 at 2:44










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










          • $begingroup$
            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.
            $endgroup$
            – abx
            Nov 21 '18 at 6:32


















          • $begingroup$
            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 $g(X) > 1$.
            $endgroup$
            – swalker
            Nov 18 '18 at 2:44










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










          • $begingroup$
            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.
            $endgroup$
            – abx
            Nov 21 '18 at 6:32
















          $begingroup$
          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 $g(X) > 1$.
          $endgroup$
          – swalker
          Nov 18 '18 at 2:44




          $begingroup$
          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 $g(X) > 1$.
          $endgroup$
          – swalker
          Nov 18 '18 at 2:44












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




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












          $begingroup$
          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.
          $endgroup$
          – abx
          Nov 21 '18 at 6:32




          $begingroup$
          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.
          $endgroup$
          – abx
          Nov 21 '18 at 6:32


















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