2
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

I found a proof only involving the existence of a good cover.

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .