There are a few equivalent definitions of Stein manifolds. As far as I know they were initially defined as holomorphically convex complex manifolds, and then the other definitions (e.g. complex manifolds which embed holomorphically in a complex Affine space) were proved to be equivalent to the first afterwards. What I am missing is the motivation behind the first definition I mentioned. I know that for a domain in $\mathbb{C}^n$, holomorphic convexity means it is a domain of holomorphy. But why was this definition generalized to a complex manifold, in other words, does holomorphic convexity perhaps imply anything about the complex functions on the manifold? or is a Stein manifold a domain of holomorphy in an ambient complex manifold?

  • $\begingroup$ The tags stein-manifolds and/or cv.complex-variables should be added. $\endgroup$ – Sönke Hansen May 13 '13 at 13:43

Compact complex manifolds are holomorphically convex. Stein's original definition had in addition to holomorphic convexity separation of points by holomorphic functions and local coordinates given by holomorphic functions. Stein himself called these holomorphically complete manifolds. He introduced it to study the solution of Cousin problems on complex manifolds. Remmert showed that a noncompact connected complex manifold is holomorphically convex iff it admits a proper holomorphic mapping to a Stein space. You can find some of these discussed in the book of Grauert and Remmert Theory of Stein spaces.

From Stein's definition one can see that Stein manifolds admit strictly plurisubharmonic exhaustion functions. The converse is a deep theorem of Grauert.

  • $\begingroup$ I think you mean local coordinates given by globally defined holomorphic functions? Moreover, this condition follows from holomorphic convexity and holomorphic separability since these suffice to prove Theorem B. $\endgroup$ – AmorFati Apr 17 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.