# Confusion in the definition of base point free line bundle

Let $C$ be an non-hyperelliptic algebraic curve (complete, non-singular, one dimensional scheme over an algebraically closed field) of genus $g\geq 3$ and let $P\in C$.

We have that $$\mathrm{deg}(\omega^2_C(-P))=4g-5\geq 2g+1.$$ It follows that $\omega^2_C(-P)$ is very ample, and therefore base point free.

Nevertheless, we may understand $$H^0(\omega^2_C(-P))=\{s\in H^0(\omega^2_C):s(P)=0\}.$$ It follows that $\omega^2_C(-P)$ is not base point free, because all of its sections have a common zero.

What am I misunderstanding?

• Notation $\Omega^2_C$ is very confusing --- it looks like the sheaf of 2-forms. It is better to write $\omega^2_C$, or $O_C(2K_C)$. – Sasha Sep 18 '17 at 13:10

## 1 Answer

The problem is that in this way you identify your sheaf as a subsheaf of $\omega_C^2$, i.e., the image of the embedding $\omega^2_C(-P) \to \omega_C^2$. This image, indeed, is not base point free.

You can get a similar picture in a much simpler situation. Let $C = {\mathbb{P}}^1$ with homogeneous coordinates $(x:y)$ and consider the sheaf $O_C(1)$. Definitely, it is base point free (in fact, it is very ample). On the other hand, consider its embedding into $O_C(2)$, given by multiplication with $x$. Then $H^0(C,O_C(1))$ identifies with the subspace of $H^0(C,O_C(2))$ spanned by $x^2$ and $xy$, which generate a linear system which is not base point free.