does the finite dimensionlity of the first cohomology group ($ H^1 $) of the sheaf of meromorphic sections of a holomorphic line bundle on a compact riemann surface follow easily from the finite dimensionality of the cohomologies of the sheaf of holomorphic functions on the surface?

The derivation is indeed "easy", in the sense that no additional results from analysis are needed. Consider the sheaf map $\mathcal{O}(L) \to \mathcal{M}(L)$, it is injective and the cokernel is the sheaf $\mathcal{H}(L)$ of principal parts of meromorphic sections of $L$. The sheaf $\mathcal{H}$ is fine. Therefore you get from the long exact sequence associated with $$0\to \mathcal{O}(L) \to \mathcal{M}(L) \to \mathcal{H}(L) \to 0$$ a short exact sequence $$ H^1 (X;\mathcal{O}(L)) \to H^1 (X;\mathcal{M}(L)) \to H^1 (X;\mathcal{H}(L))=0 $$ and the finitedimensionality of $H^1 (X;\mathcal{O}(L))$ implies the result. Addendum: it is known that any line bundle admits a nonzero meromorphic section $f$. Multiplication by $f$ induces an isomorphism $\mathcal{M}\cong \mathcal{M}(L)$, so $dim (H^1 (X;\mathcal{M}(L)))$ does not depend on $L$. Since there is a line bundle $L$ with $H^1 (X;\mathcal{O}(L))=0$ (this happens if the degree of $L$ is a sufficiently large positive number). This show that $H^1 (X;\mathcal{M}(L))=0$ for ANY line bundle. The proof that $H^1 (X;\mathcal{O})$ is finitedimensional (which requires quite a bit of analysis) generalizes to general line bundles. Or one can use the result for trivial line bundles plus the existence of meromorphic sections, see the answer by Francesco. The existence of meromorphic sections is not easy to establish; it follows from RiemannRoch. There is also a more direct argument, using the analysis involved in the proof of RiemannRoch. 


The answer is "more or less yes", depending on your definition of "easily". The following is a possible approach. We use the language of divisors, and we assume the standard facts that the $H^0$ of a skyscreaper sheaf is finitedimensional and that its $H^1$ is zero. Assume first that $\mathcal{O}(D)$ is effective. From the short exact sequence $0 \to \mathcal{O} \to \mathcal{O}(D) \to \mathcal{O}_D \to 0$ and from the vanishing of $H^1(\mathcal{O}_D)$ one obtains a surjective homomorphism $H^1(\mathcal{O}) \twoheadrightarrow H^1(\mathcal{O}(D))$, hence the finitedimensionality of the first group implies the finitedimensionality of the second. Now assume that $D$ is any divisor, and write $D=E F$, with $E$, $F$ effective. Then by $0 \to \mathcal{O}(D) \to \mathcal{O}(E) \to \mathcal{O}_F \to 0$ we deduce as before $H^0(\mathcal{O}_F) \to H^1(\mathcal{O}(D)) \to H^1(\mathcal{O}(E)) \to 0$. Since $H^0(\mathcal{O}_F)$ and $H^1(\mathcal{O}(E))$ are both finitedimensional, the claim follows. EDIT. This argument proves the finitedimensionality of the sheaf of holomorphic sections. The finitedimensionality of the sheaf of meromorphic sections $\mathcal{M}(D)$ follows from the fact that the cokernel of the natural injection $\mathcal{O}(D) \to \mathcal{M}(D)$ has trivial $H^1$, see Johannes Ebert's answer. 

