Holomorphic functions on a complex compact manifold are only constants

Question

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

Answer 1

You mean assuming that $M$ is also connected. Yes, a simple proof exists if you can use the maximum modulus principle, by showing the set $$ \{ x \in M \mid f(x)=f(x_0) \} $$ is all of $M$ (where $x_0$ is a point where the maximum modulus is achieved).

Answer 2

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.

Answer 3

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.

Answer 4

proof. ‎$‎\left| f \right|:X \to R‎$‎ is a continuous function, ‎$X‎‎$ ‎is ‎compact, ‎‎$‎‎\Rightarrow‎ \left\{ {\left| {f\left( x \right)} \right|:x \in X} \right\}‎$‎ is compact, hence bounded. Thus, there is an ‎$‎x_0 \in X$‎ such that ‎$‎‎\left| {f(x_0 )} \right| = M$ ‎is ‎maximal. ‎Let ‎‎$‎‎a = f(x_0 ) \in \mathbb{C}‎$ .‎ ‎‎Obviously, ‎$‎‎f^{ - 1} (a)$‎ is closed in ‎$X$‎ (it is a pre-image of a point); if we are able to show that $‎‎f^{ - 1} (a)$ is open, too, then $‎‎f^{ - 1} (a) = X$, that implies ‎$‎‎f(x) = a~~~~\forall x \in X$‎. Let $x \in ‎‎f^{ - 1} (a)$ and ‎$ (U,z) $ be a chart with ‎$ z(x)=0 $‎. Then ‎$ F:foz^{ - 1} :z(U) \to ‎\mathbb{C}‎ $‎ is holomorphic on the open subset ‎$ z(U) \subseteq C^n ;F(0)=f(x) = a $ ‎and‎ ‎$ \left| F \right| $‎ has a maximum in ‎$ z=0 $.‎ Let‎ ‎$ \varepsilon > 0 $ ‎such ‎that‎ ‎$ B_\varepsilon : = \left\{ {y \in C^n :\left\| y \right\| < \varepsilon } \right\} \subseteq z(U) $. ‎For‎ ‎$ y \in B_\varepsilon $‎, the function ‎$ g(t): = F(ty) $‎ is holomorphic on ‎$ \left\{ {t \in C:\left\| {ty} \right\| < \varepsilon } \right\} $ ‎and‎ ‎$ \left| g \right| $‎ takes its maximum in ‎$ t=0 $.‎ By the "maximum principle", ‎$ g $‎ is constant ‎$ \Rightarrow a = g(0) = g(1) = F(y) $‎, that means ‎$ F(y) = a~~~ \forall y \in B_‎\varepsilon‎ $‎. Hence ‎$ f \equiv a $ ‎on‎ ‎$ z^{ - 1} (B_\varepsilon ) $‎, an open subset of ‎$ X $‎ containing ‎$ x $‎. Hence ‎$ f^{ - 1} (a) $‎ is open.