Zeros of a global section on a holomorphic line bundle

Question

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?

Answer 1

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 $t=0$.