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 rank2 vector bundles, theta divisors, and the geometry of curves in projective space (J. Differential Geom. 35 (1992), no. 2, 429469). He does a detailed analysis (in a slightly different situation) of the stability of extensions. $\endgroup$ – abx Nov 21 '18 at 6:32