Let $C$ be smooth curve, and let $F$ be a stable rank 2 vector bundle of degree equal to $2c+1$, $c\in\mathbb{N}$, and fix a point $p\in C$:

Can one choose an epi-morphism $u:F\rightarrow \mathbb C_p$ such that $u$ does not vanish on all the sub-line bundles of $F$ of degree $c$? (where $\mathbb C_p$ is the skyscraper sheaf)

Thank you.


1 Answer 1


Yes, because $F$ has at most 2 sub-line bundles of degree $c$, so you have plenty of choices. The reason is the following. Twisting by a line bundle of degree $-c$ you reduce to the case $c=0$. You can assume that $F$ contains at least one sub-line bundle of degree $0$, and twisting again that this is $\mathcal{O}_C$, so that you have an exact sequence $$0\rightarrow \mathcal{O}_C\rightarrow F\rightarrow L\rightarrow 0$$with $\deg(L)=1$. This extension is given by a nonzero extension class $e\in H^1(C,L^{-1})$.

Now suppose that $F$ contains another line bundle of degree $0$. It must map non-trivially to $L$, hence it is of the form $L(-q)$ for some point $q\in C$. This means that your extension splits when pulled back to $L(-q)$, or equivalently that the extension class $e$ goes to $0$ in $H^1(C,L^{-1}(q))$. Dually, $e$ defines a hyperplane $e^*$ in $H^0(C,K_C\otimes L)$, and this hyperplane must be equal to the image of $H^0(C,K_C\otimes L(-q))$; in other words, the image of $q$ by the map $\varphi :C\rightarrow |K\otimes L|^*$ defined by the linear system $|K\otimes L|$ must be equal to $e^*$. Since $\deg(K\otimes L)=2g-1$ it is an easy exercise to show that there are at most 2 such points $q$.

Edit: as explained in the comment below, there can actually be 3 such points if $C$ is hyperelliptic.

  • $\begingroup$ If $C$ is elliptic there is a map from any line bundle of degree $0$ to $F$, just by Riemann-Roch. $\endgroup$
    – Sasha
    Feb 5, 2015 at 21:15
  • $\begingroup$ Right, I assume $g\geq 2$ -- my argument fails for $g=1$ because the linear system $|K\otimes L|$ is just a point. $\endgroup$
    – abx
    Feb 6, 2015 at 5:08
  • $\begingroup$ thanks @abx; if there are two such point, so $F$ has 3 sub-line bundles $\mathcal O_C$, $L(-p)$ and $L(-q)$ ?? Could you please explain why there are at most 2 such points? $\endgroup$
    – Z.A.Z.Z
    Feb 6, 2015 at 13:45
  • $\begingroup$ Actually I forgot one case in which there are 3 points $q$ (hence 4 sub-line bundles). With the above notation, the possible line bundles are $\mathcal{O}_C$ and $L(-q)$ with $\varphi (q)=e^*$, with $\varphi $ defined by the line bundle $L':=K_C\otimes L$ of degree $2g-1$. If there are 3 points $q_1,q_2,q_3$ with this property, one has $\dim H^0(C,L'(-q_1-q_2-q_3))=\dim H^0(C,L')-1= g-1$. By Riemann-Roch this is possible only if $C$ is hyperelliptic and $L=H^{-1}(q_1+q_2+q_3)$. By Riemann-Roch again 4 points in $\varphi ^{-1}(e^*)$ is impossible. $\endgroup$
    – abx
    Feb 6, 2015 at 15:54

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