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.


This never happens. Pick a point $p\in X$; the exact sequence $0\rightarrow L^{2}\rightarrow L^{2}(p)\rightarrow \mathbb{C}_p\rightarrow 0$ gives rise to an exact sequence $0\rightarrow \mathbb{C}\xrightarrow{\ \partial \ } H^1(L^2)\longrightarrow H^1(L^2(p))\rightarrow 0$. 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.