# Computation of $L(p+q+r)$ on a smooth projective curve

Let $$X$$ be a smooth projective curve in $$\mathbb{P}^2(\mathbb{C})$$ of degree $$4$$ and $$p,q,r \in X$$. What's $$L(p+q+r)$$?

With a standard computation, the genus of $$X$$ is $$3$$, so applying Riemann-Roch theorem, we obtain $$\dim L(p+q+r)=1+L(K-p-q-r)$$, with $$K$$ a canonical divisor. As $$\deg(K)=4$$, we have $$\deg(K-p-q-r)=1$$ and so $$\dim L(K-p-q-r) \leq 1$$. Indeed, we have two possibilities: $$\lvert K-p-q-r\rvert=\lvert s\rvert$$ for some $$s \in X$$, so that $$L(K-p-q-r)=L(s)=\mathbb{C}$$, because the genus of $$X$$ is strictly bigger than $$0$$; or $$L(K-p-q-r)=0$$.

I don't know how to conclude from this or if this the right way to proceed.

If the points $$p$$, $$q$$, and $$r$$ in $$\mathbb{P}^2$$ are collinear, then $$\dim H^0(X,K_X(-p-q-r)) = 1$$, and hence by Riemann-Roch $$\dim H^0(X,O_X(p+q+r)) = 2$$. Otherwise, $$\dim H^0(X,K_X(-p-q-r)) = 0$$, and $$\dim H^0(X,O_X(p+q+r)) = 1$$.
• These are synonyms: $H^0 = L$. – Sasha Dec 17 '18 at 9:18