# 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, 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).

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.

• Does the same argument hold if I have a holomorphic map $f: \mathbb{P}^n \to \mathbb{C}^{n+1}$? – Kristina Apr 18 '17 at 11:53
• @Kristina: the datum of a holomorphic map $f:\mathbb P^n \to \mathbb C^{n+1}$ is equivalent to the datum of $n+1$ holomorphic functions $f_j=pr_j\circ f:\mathbb P^n \to \mathbb C^{n+1} \to \mathbb C$ . Since the $f_j$ are constant, so is $f$. – Georges Elencwajg Apr 18 '17 at 22:02

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

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.