Structure of the module of derivations on the space of Holomorphic functions

Question

Maybe this is well-known, maybe not. Let $\Omega\subset \mathbb{C}$ be connected open and non-empty. It can be shown that if $d\in\mathfrak{Der}(\mathcal{H}(\Omega))$ (i.e. $d$ is a derivation of the algebra $\mathcal{H}(\Omega)$) and is continuous for the topology of compact convergence then $d$ is of the form $d=\varphi(z)\frac{d}{dz}$. My questions are the following

1. Can the continuity be withdrawn ?
2. If yes, for which $\Omega$ ?
3. Can $\Omega$ be replaced by a one dimensional complex manifold ? (in particular, for these manifolds, is there a principal derivation like $\frac{d}{dz}$ ?)

In particular the third question is trivial in the case of a compact connected manifold (because $\mathcal{H}(\Omega)=\mathbb{C}1_{\Omega}$ by maximum principle.).

Answer 1

If you have a smooth manifold $M$, then you have a linear map $\mathfrak{X}(M) \to \text{Der}(C^{\infty}(M))$ from the space of smooth global vector fields to the space of derivations of the algebra of smooth global functions. It sends a vector field $X$ to the operator of differentiation of functions along $X$. It is a standard fact from the smooth manifold theory that this map is an isomorphism.

Now, let $M$ be a complex manifold. You still have a similar map $\mathfrak{X}^{hol}(M) \to \text{Der}(\mathcal{O}(M))$ from global holomorphic vector fields to derivations of the algebra of global holomorphic functions. As I understand, your question is about the surjectivity of this map. First of all, as you mentioned, if $M$ is compact, then there are no nonzero derivations of the algebra of holomorphic functions, but there maybe plenty nonzero holomorphic vector fields, so the map is surjective but not injective in this case. Now, we need the following fact:

If $M$ is Stein, then this map is an isomorphism.

Here is a proof of a similar result for real-analytic manifolds (real-analytic vector fields are exactly derivations of the algebra of real-analytic functions) by David Speyer. He uses the fact that every real-analytic manifold can be properly embedded into a Euclidean space, but the same is true for Stein manifolds (see the Wikipedia article on Stein manifolds; in fact Stein manifolds are exactly those complex manifolds that admit a proper holomorphic embedding into $\mathbb{C}^n$ for $n$ sufficiently large). After that, as far as I see, the proof goes without any changes.

Now, it is known that every non-compact connected Riemannian surface is Stein (see the same Wikipedia article for references). In particular, every domain in $\mathbb{C}$ is Stein. So the answers to your questions are:

1. Yes.
2. For any $\Omega$.
3. For any Riemannian surface $X$, every derivation of $\mathcal{O}(X)$ comes from some holomorphic vector field. It comes from a unique vector field if and only if $X$ has no compact components.

Remark. It is unknown to me if there are any examples of complex manifolds where our map is not surjective. If there is one (connected, WLOG), it must be of dimension greater than 1, non-Stein, and noncompact. Besides, I am interested if there are non-Stein complex manifolds with this map still bijective.

By the way, one can prove that every open subset of $\mathbb{C}$ is Stein without using the deep theorem that noncompact connected Riemannian surfaces are Stein. If you want, you may look for a proof in this wonderful survey into the Levi problem by Harry J. Slatyer.