# Simple proofs of $H^1(X, \mathcal{O}^*)=0$ when $X$ is an open Riemann Surface

I am trying to get proof that $$H^1(X, \mathcal{O}^*)=0$$ when $$X$$ is an Open Riemann surface.

Looking at some books I have seen the following two approaches:

• Use Mittag-Leffer distributions, Runge theorem and some functional analysis to prove the Mittag-Leffer and Weierstrass theorems (this is, that every divisor is the divisor of some meromorphic function).

• Use Runge theorem to prove that $$H^1(X, \mathcal{O})=0$$, and assume that $$H^2(X, \mathbb{Z})=0$$ and then the long cohomology sequence associated to the exponential exact sequence proves the claim.

I was wondering if there is a proof that is more or less self contained, where Runge theorem on open Riemann Surfaces and the fact that $$H^1(X, \mathcal{O}^*)=0$$ iff every line bundle is trivial are assumed.

For example, I don't find the second one satisfactory because $$H^2(X, \mathbb{Z})=0$$ depends on somewhat hard results: That sheaf cohomology agrees with singular cohomology and Poincaré duality with coefficients in $$\mathbb{Z}$$, whereas the first one depends too heavily on finding weak solutions to PDEs.

Let $$L \to X$$ be a holomorphic line bundle and $$\mathcal{U}$$ an open cover of $$X$$ such that:
• On each $$U_i$$ there is defined a nowhere section of $$L$$ $$s_i$$
• Each $$U_i$$ is contained in a holomorphic chart
• Each $$U_i$$ and each $$U_i \cap U_j$$ is simply connected
Let $$h_{ij} = s_i/s_j$$. Because $$h_{ij}$$ is nowhere 0 on a simply connected domain, $$h_{ij} = e^{g_{ij}}$$. Let $$\xi_{ij}$$ be a partition of unity subordinate to the covering $$U_{ij}$$ for each $$i$$, and define $$g_i = \sum_j \xi_{ij}g_{ij}$$. Then $$g_i - g_j = g_{ij}$$, so $$\bar{\partial}g_i = \bar{\partial} g_j$$ because each $$g_{ij}$$ is holomorphic. Therefore we can define some $$(0,1)$$ form $$\alpha$$ such that $$\alpha_{\mid U_i} = \bar{\partial} g_i$$. Let $$f$$ be a solution to $$\bar{\partial} f = \alpha$$ (which exists by a direct application of Runge theorem) and let $$h_i = g_i - f$$. Then $$h_i$$ is holomorphic and $$h_i - h_j = g_{ij}$$. If $$e_i = e^{-h_i}s_i$$ then $$\frac{e_i}{e_j} = e^{-g_{ij}}\frac{s_i}{s_j} = h_{ij}^{-1} \frac{s_i}{s_j} =1$$So the $$e_i$$ can be glued to obtain a global holomorphic section