A Stein manifold is a manifold which is holomorphically separable and convex. It is well known that a product of two holomorphically convex (resp. Stein) manifolds is again holomorphically convex (resp. Stein). Also, compact complex manifolds automatically have the property of being holomorphically convex but they are not holomorphically separable, of course.

This has led me to the following question:

Given a Stein manifold $X$ and a compact complex manifold $Y$ (say of dimension $\geq 1$), when is $X \times Y$ Stein?

Due to the remark made above, one only has to find a condition for $X \times Y$ to be holomorphically separable. Let $(x,y) \neq (x',y')$ be points of $X \times Y$. If $x \neq x'$, then everything is easy, as one could simply project onto $X$ and choose a separating function there. But what if $x = x'$ and $y \neq y'$? One cannot do the same for $Y$, as every holomorphic function on $Y$ is constant. So, when is it possible to separate points in this case? Never? Always? Does one have to add more conditions on $X$ and $Y$?

Remark: I asked myself this question in the much more special situation where $X \subset \mathbb{C}$ is a disc and $Y = \mathbb{P}^1$, but I decided to pose the question in this more general fashion.

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    $\begingroup$ I think that the answer should be no, as a closed submanifold of a Stein manifold is Stein, and $Y$ is a closed submanifold of $ X \times Y$. $\endgroup$
    – Daniele A
    Feb 16, 2015 at 20:28
  • $\begingroup$ I guess that makes sense. Thanks! $\endgroup$
    – Slash_
    Feb 16, 2015 at 20:34

1 Answer 1


Besides from @DanieleA's anwer in the comments:

If $f$ is holomorphic on $X \times Y$, then in particular $y \mapsto f(x,y)$ is holomorphic on $Y$ for each $x \in X$, and hence constant. In other words if $x' = x$ then $f(x,y) = f(x',y')$ for all $y, y' \in Y$, so there is no hope of separating points.


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