Sections of pullback bundle

Let $$X$$ be a genus 3 curve canonically embedded in $$\mathbb{CP}^2$$.

Why is it that the line bundle $$L$$ obtained by pulling back the hyperplane bundle $$\mathcal{O}_{\mathbb{P}^2}(1)$$ has 3 independent holomorphic sections, i.e. dim $$H^0(X, L) = 3$$.

Any help would be much appreciated.

• Do you know what "canonically embedded" means? Feb 18 at 22:20
• It should mean the image of $\phi_K$. Now the fact that $X$ has genus 3 means that $K$ has 3 sections, what I don't understand is how to relate them to sections in the line bundle $\phi^*\mathcal{O}(1): A \times_{\mathbb{P}^2} X \to X$ where $\mathcal{O}(1): A \to \mathbb{P}^2$. I clearly should be thinking of this in a different way but I don't know how.
– nope
Feb 18 at 22:39
• And the canonical embedding is constructed how? We take SOMETHING from SOMEWHERE and use them as the coordinate functions to map in to $\Bbb P^{?}$, right? Can you fill in the blanks? Feb 18 at 22:41
• Yeah you should take the three sections $w_1, w_2, w_3 \in H^0(X, K)$ and use them to create the map $\phi_K: p \mapsto [w_1(p), w_2(p), w_3(p)]$
– nope
Feb 18 at 22:44
• Now what I think should happen is that somehow I would be able to conclude that the pullback of the line bundle will be the line bundle generated by the divisor obtained but taking the intersection of $\phi_K(X)$ with some hyperplane.
– nope
Feb 18 at 22:49

Suppose $$X$$ is a variety where every line bundle is the line bundle associated to a divisor. (In particular, one way this happens is if $$X$$ is noetherian and factorial, for instance $$X$$ smooth). I claim that if $$s_0,\cdots,s_d$$ are global sections of $$\mathcal{O}_X(D)$$ which generate $$\mathcal{O}_X(D)$$ at every point, then the map $$\varphi:X\to\Bbb P^l$$ by $$x\mapsto [s_0(x):\cdots:s_d(x)]$$ satisfies $$\varphi^*\mathcal{O}_{\Bbb P^l}(1)\cong\mathcal{O}_X(D)$$.
As $$i^*\mathcal{O}_{\Bbb P^l}(1)$$ is a line bundle on $$X$$, it must be the line bundle associated to some divisor $$D'$$. So what's $$D'$$? Well, if we take some global section of $$\mathcal{O}_{\Bbb P^l}(1)$$ not vanishing identically on $$\varphi(X)$$, then we get a global section of the pullback which cuts out $$D'$$. But this global section is also a global section of $$\mathcal{O}(D)$$, and since any nonzero global section of the line bundle associated to a divisor cuts out a divisor linearly equivalent to that divisor, we must have that $$D\sim D'$$ and therefore $$\mathcal{O}_X(D)\cong\mathcal{O}_X(D')$$.
How does this apply to your situation? Well, $$D=K$$, so $$\varphi_K^*\mathcal{O}(1)\cong \mathcal{O}_X(K)$$, which has $$\dim H^0(\mathcal{O}_X(K))=3$$ by Riemann-Roch.