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I'm studying an introduction to line bundles and I'm struggling with a particular proof. I'm following some notes that present this theorem:

Let $ L \rightarrow X $ be a holomorphic line bundle on a compact manifold $X$. Suppose $ s \in Γ(X, L) $ is not the zero section. Then either one of the following holds:

  1. $L \cong X × C $ is the trivial bundle and $s$ has no zeros.
  2. $Γ(X, L^{−1}) = 0$ and the section $s$ admits at least one zero.

(Note: here $L^{-1}$ denotes the dual bundle of $L$.)

To prove this the author considers a global section $t$ of the dual bundle and, considering local descriptions $L ↔ \{U_α, g_{α\beta}\}$, $L^{-1} ↔ \{U_α, g_{α\beta}^{-1}\}$, $s ↔ \{U_α, s_α\}$ and $t ↔ \{U_α, t_α\}$ he observes that on every $U_\alpha \cap U_\beta$ we have $s_\alpha t_\alpha = s_\beta t_\beta$, so we can glue these $s_\alpha t_\alpha$ in a "global" function $F: X \rightarrow \mathbb{C}$. By the maximum principle, since $X$ is compact, $F \equiv c \in \mathbb{C}$.

Now if $c\neq0$ we have the first result; otherwise if $c=0$ we have that $t$ must be the zero section, thus implying $\Gamma(X,L^{-1}) = 0$. I'm not sure about this last implication: since $F$ depends on the choice of $t$ also $c$ does; what if we have another section $t'\in \Gamma(X,L^{-1})$ s.t. $c_{t'} \neq 0$? Would this lead to a contradiction I'm not able to see?

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Suppose that $L$ is not the trivial bundle, for every section $t$, you can construct $F_t=c_t$, then $c_t$ has to be zero for every chosen $t$, (since $t$ vanishes in an open subset (and $M$ is connected)) we deduce that $t=0$.

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  • $\begingroup$ I'm sorry, I just edited my question (in the last sentences I wanted to know what happens if I find $t'$ s.t. $c_{t'}\neq 0$. By the way, why can I assume that $L$ is not the trivial bundle? I only know that a trivial bundle has always a nowhere vanishing global section: if $s$ has this property and $t$ is s.t. $c_t \neq 0$ I can see no contradiction. $\endgroup$ – M. Rinetti Oct 19 '18 at 13:25
  • $\begingroup$ You want to show that there are two cases: 1. $L$ is trivial, 2. $\Gamma(X,L^{-1})=0$. To achieve this, you can show that if $L$ is not trivial, the second case holds. $\endgroup$ – Tsemo Aristide Oct 19 '18 at 13:28
  • $\begingroup$ If you assume that $L$ is not trivial, you cannot have $c_{t}\neq 0$ for any $t$ since it implies that $L$ is trivial, so $c_t=0$ for every $t$. $\endgroup$ – Tsemo Aristide Oct 19 '18 at 13:31
  • $\begingroup$ What if I have $c = 0$ and $L$ trivial? I also have to show that when $L$ is trivial $s$ has not zeros, but supposing $L$ is not trivial whenever I find $c = 0$ I'm discarding a possible case, am I not? $\endgroup$ – M. Rinetti Oct 19 '18 at 13:43

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