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.