Extension of the trivial bundle by the canonical bundle on a curve

Question

Let $X$ be a smooth projective curve over a field $k$ and $K_X$ be its canonical line bundle. By the Serre duality, $\text{H}^1(X,K_X)$ is a one-dimensional $k$-vector space. On the other hand, $\text{H}^1(X,K_X)=\text{Ext}^1(O_X,K_X)$ so this vector space corresponds to the set of extensions of $O_X$ by $K_X$. Let $V$ be such an extension that corresponds to a nonzero vector in $\text{H}^1(X,K_X)$. This is a $2$-dimensional vector bundle on $X$.

Question. Is there exists a simple/concrete description of $V$?

Answer 1

If $X$ is a smooth curve, the canonical bundle $\omega_X$ is nothing but the bundle of differentials $\Omega_X^1$, and the corresponding extension defines the jet bundle $$ 0 \to \Omega_X^1 \to J_X \to \mathcal{O}_X \to 0. $$

Answer 2

Assume $k=\mathbb C.$ Consider the uniformization of $X,$ which is given by an equivariant developing map of the universal covering of $X$ to the hyperbolic disc sitting inside the projective line $\mathbb CP^1$. Lift the corresponding $PSL(2,\mathbb R)$-representation to a $SL(2,\mathbb R)$-representation. Consider the corresponding flat $\mathrm{SL}(2,\mathbb C)$-bundle $(E,\nabla).$ The tautological line bundle over $\mathbb CP^1$ corresponds to a well-defined holomorphic subbundle $L\subset E$ over $X.$ By its geometric construction, $L^2=K_X$, i.e. $L$ is a spin bundle. Then, $L\otimes E$ is independent of the choice of the monodromy lift, and is a non-trivial extension of $\mathcal O_X$ by $K_X$. $E$ is often called a Gunning bundle on $X$.

Answer 3

This is more of a comment. Let $g$ be the genus of $X$. One can see that the connecting map $H^0(X,O_X)\to H^1(X, K_X)$ is an isomorphism, and therefore that $h^0(V)= h^1(V)=g$. When $g=0$, Grothendieck's theorem implies that $V= O(a)\oplus O(b)$. The previous computation shows that $a,b<0$. Since $\deg V= \deg K_X=-2$, we have $V= O(-1)\oplus O(-1)$.