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According to the Wikipedia article on complex varieties, it is not possible to a compact connected complex manifold $M$ have an holomorphic embedding in $\mathbb{C}^{N}$.

The first step of the proof presented on the article establishes that any holomorphic function on $M$ is locally constant. Then it says that

"if we had a holomorphic embedding of $M$ into $\mathbb{C}^{N}$, then the coordinate functions of $\mathbb{C}^{N}$ would restrict to nonconstant holomorphic functions on $M$ (...)"

Could more details can be given why it is so?

shamisen
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2 Answers2

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By compactness, any component of the embedding $f$ achieves a maximum modulus, contradicting the open mapping property of holomorphic functions. (Restrict to a little complex disk centered at the maximum point in a chart.)

Ted Shifrin
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Let $M$ be a compact connected complex manifold of positive dimension. Suppose $f : M \to \mathbb{C}^N$ is a holomorphic embedding and let $z_i : \mathbb{C}^N \to \mathbb{C}$ be a coordinate function, $z_i(a_1, \dots, a_N) = a_i$. Then the composition $z_i\circ f : M \to \mathbb{C}$ is a holomorphic function. If $z_i\circ f$ is a constant map with value $b_i$, then

$$f(M) \subseteq z_i^{-1}(b_i) = \{(a_1, \dots, a_{i-1}, b_i, a_{i+1}, \dots, a_N) \mid a_1, \dots, a_{i-1}, a_{i+1}, \dots, a_N \in \mathbb{C}\}.$$

If $z_i\circ f$ is constant for all $i = 1, \dots, N$, then $f(M)$ is a single point which is a contradiction as $f$ is an embedding (in particular, injective). Therefore, $z_i\circ f$ is nonconstant for some $i \in \{1, \dots, N\}$.

Once you know that you have a nonconstant holomorphic function $M \to \mathbb{C}$, you arrive at a contradiction because $M$ is compact so no such things exist.

Michael Albanese
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