Let $X$ be a Riemann surface. Let $M_1$ and $M_2$ be two holomorphic bundles on $X$. Does injectivity of $h^0(M_1)\to h^0(M_2)$ imply $h^0(M_1\otimes L)\to h^0(M_2\otimes L)$ is injective? Where $L$ is a holomorphic line bundle of negative degree on $X$ and $h^0$ is the holomorphic sections of bundles.


Not necessarily. Consider $M_1=\mathcal{O}_X(p)$, $M_2=\mathcal{O}_X(q)$ and $L=M_1^{-1}$ where $p,q\in X$. Then $H^0(M_1)=\mathbb{C}=H^0(M_2)$, but $H^0(M_1\otimes L)=H^0(\mathcal{O}_X)=\mathbb{C}$ and $H^0(M_2\otimes L)=H^0(\mathcal{O}_X(q-p))=0$.

Note: Here I didn't assume you have a morphism between the line bundles, only between their global sections. In the case a morphism exists, the answer is affirmative.

If you have a morphism $M_1\to M_2$, let $K$ be its kernel. So we have an exact sequence $$0\to K\to M_1\to M_2$$ Tensoring with $L$ keeps the sequence exact, since $L$ is locally free. So we get $$0\to K\otimes L\to M_1\otimes L\to M_2\otimes L$$ Taking global sections, since $H^0$ is left-exact, we get an exact sequence $$0\to H^0(K\otimes L)\to H^0(M_1\otimes L)\to H^0(M_2\otimes L).$$ Since $L$ is negative, $H^0(K\otimes L)\subseteq H^0(K)=0$ (from the first exact sequence). Therefore $H^0(M_1\otimes L)$ does inject into $H^0(M_2\otimes L)$.

  • $\begingroup$ I really appreciate your useful answer. But if we have a morphism between them I mean we have a morphism $M_1\to M_2$ ? $\endgroup$ – user105570 Aug 25 '14 at 18:21
  • $\begingroup$ In that case it is affirmative, see my edits. $\endgroup$ – rfauffar Aug 25 '14 at 18:29
  • $\begingroup$ Thank you very much for your detailed explanation. $\endgroup$ – user105570 Aug 25 '14 at 18:36
  • $\begingroup$ You are welcome. $\endgroup$ – rfauffar Aug 25 '14 at 18:57

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