Exact sequence of sheaves in Beauville's "Complex Algebraic Surfaces" Fact I.17

Question

Here $S$ is a complex surface and $C$ is an irreducible curve on $S$. I can't see why "considering the exterior powers" for the sequence $0\to{{\Omega^1}_{S}(-C)}|_C\to {\Omega^1}_{S|C}\to \Omega_C\to 0$ would give $\mathcal O|_S(K+C))|_C=\omega_C$. I think we need to show the sequence is split, but why is that true by considering the exterior powers?

Also, I am a little confused about what the sections of ${\mathcal O}_S(-C)|_C$ look like. I know they are the sections of the line bundle associated with the divisor $-C$ though.

At last, aren't ${\Omega^1}_{S|C}$ and $\omega_C$ just the sheaf of 1-forms on $C$? What's the difference?

Answer 1

This basically follows from Hartshorne Exercise II.5.16(d). This is a more general fact about locally free sheaves: namely, given a short exact sequence of locally free sheaves on a scheme $X$, $$ 0\to \mathcal{F}'\to \mathcal{F}\to \mathcal{F}''\to 0 $$ of ranks $r',r,r''$, respectively, there is an associated isomorphism of sheaves: $$ \bigwedge^r\mathcal{F}\cong \bigwedge^{r'}\mathcal{F}'\otimes_{\mathcal{O}_X}\bigwedge^{r''}\mathcal{F}''. $$ We apply this principle here. $\Omega^1_{S}|_C$ here means the sheaf of differential forms on $S$ restricted to $C$. Note that this is not a line bundle and has rank $2$ (because the ambient space is a surface). Also note that $\bigwedge^2\Omega^1_S|_C\cong K_S|_C$, where $K_S$ is the canonical sheaf. This can also be written as $\mathcal{O}_S(K)|_C$. Now, reading the exact sequence in the book and applying this result about top wedge powers: $$ \mathcal{O}_S(K)|_C \cong \omega_C\otimes \mathcal{O}_S(-C)|_C $$ and so $\omega_C \cong \mathcal{O}_S(K+C)|_C$ as claimed.

Answer 2

To complement the other answer with a remark, and an example:

A tongue-in-cheek answer is that this exact sequence is telling you what the difference is, $\Omega^1_S|_C$ surjects to $\omega_C$ with kernel $\mathcal{O}_S(-C)_C$. Think about the case where $S = \mathbb{CP}^2$ and $C$ is a plane curve. The first sheaf has no global sections, since $\mathbb{CP}^2$ has no global differential forms in the first place, but $\omega_C$ has a $g$-dimensional space of global forms. We can find a concrete example to show this map has a non-trivial kernel: Let's suppose $C$ is of the form $y^2 z = P(x,z)$, where $P$ is a homogenous polynomial of degree $3$. In the affine chart where $z = 1$, we have the equation $P'(x) \ dx - 2y \ dy = 0$ when restricted to $C$, by differentiating the defining equation, but is otherwise a perfectly good differential form on this chart in $\mathbb{CP}^2$.