# Parametrization of extensions of vector bundles

Suppose $$M$$ is a line bundle on a smooth projective curve $$C$$ and $$N$$ is a vector bundle on $$C$$. Suppose the degree of $$M$$ is sufficiently big so that $${\rm Hom}(M, N)=0$$ and $${\rm Ext}^1(N, M)=0$$. Suppose $$x$$ is a point of $$C$$. Question: what are extensions of the form $$0 \to M \oplus N \to X \to {\mathcal O}_x \to 0$$ parametrized by, where $$X$$ has the form $$M_1 \oplus N$$ such that $$M_1$$ is a line bundle of degree larger by $$1$$ than the degree of $$M$$? Are such extensions parametrized by $${\rm Ext}^1({\mathcal O}_x, M)$$ or by something else?

Since in the map $$M\oplus N\to M_1\oplus N$$, one has no non-zero maps from $$M\to N$$, one must have an exact sequence $$0\to M\to M_1\to O_x\to 0$$. The rest should be easy.
This is just an elaboration of the above asked by the OP. Given an exact sequence $$0\to M\oplus N\to M_1\oplus N\to O_x\to 0$$ as in the question, we see that the left map $$M\to M_1\oplus N$$ is given by $$(i,0)$$ and $$N\to M_1\oplus N$$ by $$(\alpha,\beta)$$. Then it is clear that $$\beta$$ must be an isomorphism. Then, take the isomorphism $$M_1\oplus N\to M_1\oplus N$$, by $$(Id_{M_1}, \beta^{-1})$$ and then one sees that the original sequence is equivalent to one given by $$M\to M_1\oplus N$$ by $$(i,0)$$ and $$N\to M_1\oplus N$$ by $$(\alpha, Id_N)$$, (possibly for a different $$\alpha)$$. Now, one can take the isomorphism $$M_1\oplus N$$ to itself by $$M_1\to M_1\oplus N$$, given by $$(Id_{M_1}, 0)$$ and $$N\to M_1\oplus N$$ by $$(-\alpha, Id_N)$$ and then you get an equivalent extension where $$M\to M_1\oplus N$$ is $$(i,0)$$ and $$N\to M_1\oplus N$$ is $$(0,Id_N)$$ and the rest should be clear.
• Thank you very much for offering to help! I am sorry, I understand that there is an exact sequence $0 \to M \to M_1 \to O_x \to 0$ but I don't think this means that my extensions are parametrized by $Ext^1(O_x, M)$. Indeed, what if the induced map $N \to M_1$ from the map $M \oplus N \to X$ in my extension is nonzero? Maybe in that case one could still have a splitting $X=M_1 \oplus N$? Or one couldn't? Could you please elaborate on the rest of the answer? Thank you again! Jul 4, 2019 at 20:34
• First of all, my answer is correct. Have you tried to work out that from what I said, you can show what you want? Can you show that the extension you have is equivalent to the extension $0\to M\oplus N\to M_1\oplus N\to O_x\to 0$ where the first map is $M\to M_1\oplus N$, given by $(i,0)$ and the second map $N\to M_1\oplus N$ is given by $(0, Id_N)$? Jul 5, 2019 at 0:35