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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.

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  • $\begingroup$ Do you know what "canonically embedded" means? $\endgroup$
    – KReiser
    Feb 18 at 22:20
  • $\begingroup$ 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. $\endgroup$
    – nope
    Feb 18 at 22:39
  • $\begingroup$ 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? $\endgroup$
    – KReiser
    Feb 18 at 22:41
  • $\begingroup$ 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 $\phi_K: p \mapsto [w_1(p), w_2(p), w_3(p)]$ $\endgroup$
    – nope
    Feb 18 at 22:44
  • $\begingroup$ 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. $\endgroup$
    – nope
    Feb 18 at 22:49

1 Answer 1

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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.

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