# Holomorphic functions on a complex compact manifold are only constants

Is there a simple proof that every holomorphic function $M\to\mathbb{C}$ on a compact complex manifold $M$ is constant?

-

You mean assuming that $M$ is also connected. Yes, a simple proof exists if you can use the maximum modulus principle.

-

It depends on what you mean by simple but if you take the maximum principle for holomorphic functions on $\mathbb{C}$ for granted (a non-constant holomorphic function doesn't admit local maxima), the statement follows easily: Suppose $f : M \to \mathbb{C}$ is a holomorphic function from a compact connected Riemann surface $M$. Then by compactness of $M$, the function $|f| : M \to \mathbb{R}$ attains a maximum at some point $p \in M$. If $(U,\varphi)$ is a holomorphic coordinate patch around $p$, say $\varphi : \mathbb{C} \supset V \overset{\sim}\to U \subset M$, then this gives a holomorphic map $f \circ \varphi : V \to \mathbb{C}$ with a maximum at $0$, hence $f \circ \varphi \equiv C$ is constant on $V$, so $f \equiv C$ is constant on $U$. Since $U$ is open and $M$ is connected, analytic continuation implies that $f \equiv C$ on all of $M$.

Edit: I just realised you asked for complex manifolds in general but the proof is the same.

-
However, this works only for $\dim_{\mathbb{C}} M=1$, not? – Peter Franek Jul 29 '14 at 17:17
You can use the maximum modulus principle linked in Jonas Meyer's answer with the same proof in the genral case. – jef808 Jul 29 '14 at 17:19
Yes, I got it. Thanks a lot (it was hard to decide which answer to accept, both are ok)! – Peter Franek Jul 29 '14 at 17:20

Non-constant holomorphic functions on connected complex manifolds are open maps.
So, if $M$ were compact and $f:M\to \mathbb C$ were non-constant, its image would be an open, compact non-empty subset $f(M)\subset\mathbb C$. Such a beast does not exist.

-

jef808's proof works, but if you don't like using analytic continuation at the last step, one can refer to Theorem 1.11 in Well's Differential Analysis on Complex Manifolds. One shows the set $$\{ x \in M \mid f(x) = f(x_0) \}$$ is all of $M$, using Maximum Principle (where $x_0$ where the maximum modulus is achieved.

-