# Why do holomorphic functions on Stein manifolds "essentially behave like" holomorphic functions open subsets of $\Bbb C^n$?

I gathered from various books and conversations that the philosophy is that holomorphic functions on open subsets of Stein manifolds "essentially behave like" holomorphic functions on open subsets of $\mathbb{C}^n$ and that this is due to the embedding theorem by Remmert & others (which states that Stein manifolds can be proberly embedded by a holomorphic function into some $\mathbb{C}^n$).

However, I'm not able to grasp the concept. At first I thought it'd be easy, since if $K$ is a Stein manifold and $\Phi: K \rightarrow \mathbb{C}^n$ is a prober holomorphic embedding, $\Omega \subseteq K$ open and $f: \Omega \rightarrow X$ ($X$ a Banach space) is a holomorphic function, you can consider $f \circ \Phi^{-1}: \Phi(\Omega) \rightarrow X$ which is holomorpic in a more classical sense. Then, I noticed that $\Phi(\Omega) \subseteq \mathbb{C}^n$ is (at least in general) not open and only $\Phi(\Omega) \subseteq \Phi(K)$ is open. So the function $f \circ \Phi^{-1}$ is also only holomorphic as a function on the open subset $\Phi(\Omega)$ of the submanifold $\Phi(K) \subseteq \mathbb{C}^n$.

And thus, if I want to understand why $f$ behaves essentially like a holomorphic function on an open subset of $\mathbb{C}^n$ by looking at $f \circ \Phi^{-1}$, I have to understand why holomorphic functions on open subsets of submanifolds of $\mathbb{C}^n$ "behave essentially like" functions on open subsets of $\mathbb{C}^n$. Is that true and if yes why?

Or else, is my whole approach wrong? Or did I misinterpet the philosophy I sketched above?

• By its nature as a closed subset of $\Bbb C^n$, there are many holomorphic functions on $K$; and because it's a properly embedded submanifold, you have eg a Liouville's theorem.
– user98602
Aug 26, 2016 at 17:23
• It is not true at all that holomorphic functions on a Stein manifold behave like holomorphic functions on an open subset $U\subset \mathbb C^n$ (unless of course in the very special case that $U$ is itself a Stein manifold !). I do not believe that any book claims such an idiotic statement and if it does, burn it :-) Feb 3, 2022 at 21:05
• @user98602: I have no idea what you are talking about. There is no Liouville theorem for a general open subset of $\mathbb C^n$ nor for a general Stein manifold. Feb 3, 2022 at 21:08

Suppose $$X$$ is a complex manifold of complex dimension $$n$$ and let $$\mathcal{O}_X$$ denote the ring of holomorphic functions on $$X$$. Recall that we say $$X$$ is a Stein manifold if the following conditions hold:
1. $$X$$ is is holomorphically convex, i.e. for every compact subset $$K\subseteq X$$, the so-called holomorphically convex hull, $${\bar {K}}=\left\{z\in X\,\left|\,|f(z)|\leq \sup _{w\in K}|f(w)|\ \forall f\in {\mathcal {O}}(X)\right.\right\}$$ is also a compact subset of $$X$$.
2. $$X$$ is holomorphically separable, i.e. if $$x\neq y$$ are two points in X, then there exists $$f(x)\in\mathcal{O}(X)$$ $$f(x)\neq f(y)$$.
From the definition we can see that the Stein manifold is the natural generalization of holomorphically convex domain in $$\mathbb{C}^n$$.
So the philosophy you say is not right since not every open set of $$\mathbb{C}^n$$ is holomorphically convex domain.