Talk:Stein manifold

message: :(copied from my talk page by Oleg Alexandrov (talk) 16:22, 13 December 2005 (UTC)
replies: text: Hi Orthografer. All I did in that article is some spell-checking and linking. :) But from what I read the article it looks that you are right, one must say there that the submanifold is a complex submanifold. I will fix that now. By the way, a good way of noting your remark is also on the article discussion page, assuming that any pople are watching over it. Oleg Alexandrov (talk) 16:22, 13 December 2005 (UTC)
replies: text: I was thinking of the definition of Stein manifold as one that admits a biholomorphic embedding into complex space, without reading the article's definition. I still believe the thing was wrong, but now I wonder how it can be something other than a triviality if we fix it. Thanks for the editing tip, and I agree adding "complex" is better than leaving it alone. --Orthografer 16:26, 13 December 2005 (UTC)
replies: text: One has to clarify once what definition of the several definitions one can find in the literature will be kept here. Eg. is a submanifold by definition closed? Otherwise this has to be expressed explizitly in the entry. Or: Is an embedding a closed mapping by definition? Finally: Is a submanifold of a complex manifold a "complex" submanifold unless otherwise stated? Hottiger 14:54, 6 March 2006 (UTC)
replies: text:  submanifold need not be closed.  For example take the unit ball B in C^n (n>1 of course) and subtract the origin.  That is a wonderful embedded submanifold of C^n that is not Stein.  It is a wonderful abstract complex manifold and B-{0} is the image.  Leaving out words such as closed or complex and assuming they are implied is horrible, especially for an encyclopedia.  You are thinking of the following definition:  M is a Stein manifold if it admits a proper biholomorphic embedding into some finite dimensional complex euclidean space.  I was the one who did yesterdays edit making it a closed complex submanifold --Jiri Lebl 21:17, 19 October 2007 (UTC)
thread_title: Mistake in the article on Stein manifold?

message: I added "closed" at a couple of places; to me a submanifold can be open, and then the statement seems very false. Spakoj 22:42, 23 March 2006 (UTC)
replies: text: There is a theorem due to Eliashberg stating that for n>2 a smooth almost complex 2n-dimensional manifold M admits a Stein structure if and only if it has a proper morse function f:M→[0,∞) which has critical points of index no greater than n. This theorem applies in the case n=2 with restictions on how 2-handles are attached in a handle decomposition of the manifold. This in particular implies that there can be no m-handles in a handle decomposition of the manifold for m>n so that any stein manifold is open or (in the compact case) has boundary. Orthografer 19:52, 18 June 2006 (UTC)

text: You know, I think the main reason people are having trouble understanding this article because the second (main) definition is complicated. After reading closely, I also think there are numerous errors in this article (mainly coming from someone forgetting that the image of a holomorphic function is never compact). I'm going to overhaul this because I think it's wrong and this article has always bothered me. Please feel free to edit without ego! Orthografer 22:17, 26 March 2006 (UTC)
replies: text: Thanks Orthographer, the article has improved decisively. The first Theorem is in fact correct in the way it's written there, but the fact that a complex manifold is holomorphically separable and holomorphically convex (first two items) implies the third condition. Usually this condition is therefore suppressed in the Theorem and stated separately. Hottiger 22:57, 26 March 2006 (UTC)
thread_title: Closed or not

message: From the article:
<I>A consequence of the embedding theorem is the following fact: a connected Riemann surface (i.e. a complex manifold of dimension 1) is a Stein manifold if and only if it is not compact.</I>
Why is this a consequence of the embedding theorem? I do not see
the relation between the embedding theorem and the fact that non-compact Riemann surfaces are Stein.
Joerg Winkelmann 13:37, 13 May 2006 (UTC)
replies: text: I'm not sure either. I wonder if there should be a different embedding theorem used, along with the fact that every real surface admits a canonical complex structure. We would ask JacobSc but it looks like he edited exactly this article and left a year ago. Orthografer 03:42, 1 September 2006 (UTC)
thread_title: Riemann surfaces