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

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3 Answers 3

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

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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.

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However, this works only for $\dim_{\mathbb{C}} M=1$, not? –  Peter Franek Jul 29 at 17:17
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You can use the maximum modulus principle linked in Jonas Meyer's answer with the same proof in the genral case. –  jef808 Jul 29 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 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.

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