# Riemann-Roch and quartic

I know very little in algebraic geometry, but I want to learn!! So I know the Riemann-Roch theorem as follow: let $$L(D)=\{\text{ meromorphic functions, s.t. }\operatorname{div}(f)\geq D \}$$ and $$O(D)=\{\text{ meromorphic differentials, s.t. }\operatorname{div}(f)\geq D\}.$$

Then $$\dim L(D^{-1})= \deg(D)-g+1+\dim(O(D)).$$ Then I would like to prove the following: if $Q$ is a quartic (4,0) on $S^2$ with one poles of most order $2$ then $Q$ is zero.

I would like to say $Q=g\,dz^4$ with $g$ meromorphic with a pole of order at most $2$. Then compute $\dim(L(D^{-1}))$ with $D=k\cdot\infty$ with $k\leq 2$ and then conclude that $\dim(L(D^{-1}))=0$ hence $g=0$.

Is the rough idea is correct? how to compute $O(D)$? is there a big difference between quartic and one form? I am ready to read any reasonable approach for somebody with a good background in differential geometry and one variable complex analysis.

• Isn't the defining condition of $L(D)$: $\;\operatorname{div} f \ge -D$? – Bernard Jun 8 '16 at 21:44
• it is corrected. – Paul Jun 9 '16 at 4:52
• not quite: it should be ${}\ge\color{red}- D$. Furthermore the group of divisors is noted additively. – Bernard Jun 9 '16 at 8:48
• Here, math.ucdavis.edu/~kapovich/RS/RiemannRoch.pdf, where my source, but it is not impossible I miss something. – Paul Jun 9 '16 at 9:10