# Zeros of a global section on a holomorphic line bundle

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?

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 $$t=0$$.
• 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. – M. Rinetti Oct 19 '18 at 13:25
• 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. – Tsemo Aristide Oct 19 '18 at 13:28
• 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 $t$. – Tsemo Aristide Oct 19 '18 at 13:31
• 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? – M. Rinetti Oct 19 '18 at 13:43