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Let $\mathscr{L}$ be a line bundle on a curve $C$ such that $h^0(C, \mathscr{L}) = 1$ and deg $\mathscr{L} = 0$. Why does it imply that $\mathscr{L}$ is the trivial line bundle?

I found some explanation in Vakil's notes here, which I don't quite understand. We know that degree of $\mathscr{L}$ can be computed by counting zeros and poles of any rational section. Say $s$ is a section of $\mathscr{L}$, then $s$ must have no pole, and since $\mathscr{L}$ has degree 0, $s$ must also have no zero. That means $s$ is invertible. But why does this then imply that $\mathscr{L}$ is the trivial line bundle?

Aaron Johnson
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3 Answers3

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Isn't $\phi:C\times\mathbb{C}\rightarrow{\cal L}$ defined as $\phi(x,\zeta)=\zeta s(x)$ an isomorphism?

Andrea Mori
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  • I’m not sure what you mean... Can you explain more? – Aaron Johnson Jan 14 '18 at 09:25
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    A (complex) line bundle over $C$ is a space $\cal L$ together with a map $\pi:{\cal L}\rightarrow C$ such that every $x\in C$ is contained in an open set $U$ with $\pi^{-1}(U)\simeq U\times\mathbb{C}$ and $\pi$ is the first projection. Given that ${\cal L}$ admits a non-vanishing section $s:C\rightarrow{\cal L}$, I wrote an isomorphism between $\cal L$ and the trivial bundle on $C$. – Andrea Mori Jan 14 '18 at 09:40
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In the language of sheaves, you can define an isomorphism of sheaves $\mathscr{O}_C \rightarrow \mathscr{L}$ where on each open $U$ sends $a \mapsto as|_U$.

loch
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Since $s$ is nowhere vanishing, you can essentially declare it to be the unit constant section $1$, and the assignment $s \mapsto 1$ is enough to define a global sheaf map which is clearly nowhere $0$. To me, this perspective actually illuminates the motivating notion of a free $R$-module of rank $1$.

Having written the preceding paragraph, I recall now that I answered essentially the same question in a slightly more rigorous and algebraic manner here: Line bundle with a nowhere vanishing global section is trivial.

Tabes Bridges
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