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Suppose you have a curve $C$ such that deg$K_C =0$ and $\Gamma(C,\Omega_C^1) \neq 0$. Does this automatically imply that $\vartheta_C \equiv \Omega_C^1$? My thought is yes, I've seen a proposition (Stanford AG course notes) that $\vartheta_C \equiv \Omega_C^1$ for a nonsingular plane cubic, but the proof is done in a particular case, and I must think there is easier way to show this. In particular, what is the morphism?

Thank you, just trying to make sense of this concept.

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I suggest a more descriptive title for this question. – Andrew Critch Dec 14 '09 at 1:46

2 Answers 2

up vote 2 down vote accepted

On a complete nonsingular curve over an algebraically closed field, a line bundle of degree zero with a global section is necessarily the trivial bundle. This is lemma IV.1.2 in Hartshorne's Algebraic Geometry: if $\mathrm{deg} D = 0$ and $\mathcal{L} = O_C(D)$ has a global section, then $D$ is linearly equivalent to an effective divisor $E$ of the same degree; but the only effective divisor of degree zero is the zero divisor. Hence $\mathcal{L} \cong \mathcal{O}_C(E) = \mathcal{O}_C$.

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@Akhil: pay attention, you are thinking about a nowhere vanishing global section. – Andrea Ferretti Dec 14 '09 at 1:15
@Andrea: Ack, sorry. I've deleted the previous comment. – Akhil Mathew Dec 14 '09 at 1:42

I am sure that this is true. That $\Gamma(C,\Omega^1_C)\neq 0$ implies $K_C\in \Gamma(C,\Omega^1_C)$ is a effective divisor. So $\mathcal{O}_C\subset\mathcal{O}_C(K_C)$. Since $\deg K_C =0$, so the quotient must be trivial, which means $\mathcal{O}_C=\mathcal{O}_C(K_C)$.

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