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Let $C$ be a smooth projective curve of genus $g$, we know that for a general line bundle $\mathcal{L}$ of degree $g-1$, $\mathcal{L}$ has no global sections, i.e. $\text{H}^0(C, \mathcal{L})=0$.

My question is that, in some particular case, for example, $C$ is a plane algebraic curve and I know the defining equation of $C$, How can I explicitly find such a line bundle $\mathcal{L}$? By explicitly I mean to find points on $C$ and write down the divisor form of $\mathcal{L}$.

Or if I pick a divisor of $C$ randomly, for example, divisors of the form $D=p_1+\cdots + p_g-q$, do I have a criterion to determine whether $\text{H}^0(C, \mathcal{O}_C(D))=0$?

Xueqing Wen
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  • Well by Riemann-Roch, the trick is to find a divisor $D$ of that degree with no $h^1$, or by Serre Duality one such that $h^0(K - D) = 0$. I think if you choose $D$ at random, i.e. so that it consists of $g-1$ reduced points that impose independent conditions on $K$, you are good to go. I would suggest thinking through the case $g=3$ in detail since you can model the whole situation with smooth plane quartics, and in this case the canonical divisors are cut out by lines in the plane. – Tabes Bridges Oct 11 '20 at 14:53
  • Thanks for comments, but I don't understand what does "impose independent conditions on $K$" mean. You mean for a divisor it is somehow difficult to calculate the dimension of global sections, but for a divisor of the form $K-D$, we have methods to determine the space of global sections, right? – Xueqing Wen Oct 11 '20 at 15:39
  • A point $p$ imposes independent conditions on a divisor $L$ if $h^0(L-p) =h^0(L) - 1$ (the two $h^0$'s could be equal if $p$ is a base point). A set of $d$ points imposes independent conditions if imposing them as zeroes causes $h^0$ to drop by $d$. – Tabes Bridges Oct 11 '20 at 15:42
  • This is probably easiest to understand precisely in terms of the subspaces $H^0(L-p_i) \subset H^0(L)$. If these are 1) all codimension one, and 2) all intersect generically, then $p_1,...,p_d$ is said to impose independent conditions on $L$, or sometimes people say "on sections of $L$." – Tabes Bridges Oct 11 '20 at 15:50
  • I think I know what you mean now, thanks a lot! – Xueqing Wen Oct 14 '20 at 08:41

1 Answers1

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I will assume that $C$ is a smooth plane curve of degree $d$. Let $C'$ be a general curve of degree $d-3$ so that the intersection $$ C \cap C' = \{P_1,\dots,P_{d(d-3)}\} $$ is transverse. Note that $d(d-3) = 2g - 2$ by adjunction. Choose $g$ points out of $P_i$ such that $C'$ is the only curve of degree $d-3$ passing through them. Assume these are points $P_1,\dots,P_g$. Then for any point $Q \not\in\{P_1, \dots, P_g\}$ the divisor class $P_1 + \dots + P_g - Q$ has no cohomology.

Sasha
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  • Thanks for answering! I don't if my comprehending is right: the pull back of $\mathcal{O}{\mathbb{P}^2}(C^{\prime})$ on $C$ which is the line bundle corresponding the divisor $P_1+\cdots+P{d(d-3)}$, is isomorphic to $\omega_C$. So $\mathcal{O}{C}(P_1+\cdots+P_g)\cong \omega_C(-P{g+1}-\cdots-P_{d()d-3})$, by the choice of points $P_1, \cdots, P_g$, we see that $\text{H}^0(C,\mathcal{O}_{C}(P_1+\cdots+P_g))=1$(this is where I'm not so sure). Another question is that how to determine that whether $C^{\prime}$ is the only curve passing there points. It seems very difficult to me. – Xueqing Wen Oct 14 '20 at 08:34
  • @XueqingWen: This is correct. Every point on a plane provides a hyperplane in the space of degree $d-3$ curves, i.e., it gives a linear equation on the coefficients of the equation of the curve. The number of curves passing through the set of points corresponds to the dimension of the space of solutions of the corresponding system of linear equations. It might be hard to find this by hand, of course. – Sasha Oct 14 '20 at 11:00