Let $X$ Riemann Surface of genus $g$, and $D$ is a divisor on $X$. I try to prove $l(D)=dim H^0 (X,O_D) ≤ 1+degD$ when $−1≤degD≤g−1$ Using the Riemann-Roch theorem and Serre duality. The notations here : $H^0(X,O_D)$ is obviously the $0$-th cohomology group of $X$ with respect to the sheaf $O_D$ where $O_D(U)=\{f∈\mathcal{M}(U)|ordx(f)≥−D(x)∀x∈U\}$ and $\mathcal{M}$ is the sheaf of meromorphic functions.
The statement of Riemann-Roch in O.Forster Book is : for a compact Riemann surface $X$ of genus $g$ and a divisor $D$ on it, $$dimH^0(X,O_D)−dimH^1(X,O^D)=1−g+degD.$$
I tried to use the Serre duality which asserts that $dimH^1(X,O_D)=dimH^0(X,Ω_{−D})$ where $Ω_{−D}$ is the sheaf of meromorphic $1$-forms that are multiples of $−D$. Or, use the the exact sequenece
$$0\to H^0(X,D−p)→H^0(X,D)→\mathbb{C}→0.$$ I find that, if $D=P-N$, where $P,N$ two effective divisors, then $l(D)≤ 1+deg(P)$. But i cannot find the finale result.
Could I please get a hint?
Best Answer
If $D$ is non-negative ok, if not just change $D$ by $D'=div(f)+D$ where $f\in O_D$, then $l(D)=l(D')\leq 1+deg(D')=1+deg(D)$.
rmk$_1$: If $D$ and $D'$ are equivalent divisors on Riemann Surfaces then $O_D(X)\simeq O_{D'}(X)$ given by multiplication of a meromorphic function on $X$.
rmk$_2$: $div(f)$ is the divisor associated a meromorphic function, then $deg(div(f))=0$ (The degree of a holomorphic map is constant in the set of regular values).