Talk:Stein manifold

Mistake in the article on Stein manifold?

 * (copied from my talk page by Oleg Alexandrov (talk) 16:22, 13 December 2005 (UTC))

Hi Oleg - I see you all over and appreciate your work here. You may not be the originator of the statement, but you have edited the same article and seem capable of answering my question: in Stein manifold there is a "property" of Stein manifolds that "any submanifold" of a Stein manifold is Stein. This seems wrong; for example what about odd-dimensional submanifolds, or a regular neighborhood of a 2-sphere in a Stein 4-manifold ($$S^2\times D^2$$ is known to not be Stein)? I don't just want to add the word "complex" before "submanifold" because it seems so trivial. Do you know the proper form for this? --Orthografer 16:13, 13 December 2005 (UTC)
 * 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)
 * 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)
 * 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)
 * 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)

Closed or not
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)
 * There is a theorem due to Eliashberg stating that for n&gt;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&rarr;[0,&#8734;) 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&gt;n so that any stein manifold is open or (in the compact case) has boundary. Orthografer 19:52, 18 June 2006 (UTC)

I think in the sentence "Gathering these facts, one sees, that Stein manifold is a synonym for a closed submanifold of complex space." "complex space" is meant in the sense of finite dimensional complex vector space, right? Hottiger 17:54, 25 March 2006 (UTC)
 * 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)
 * 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)

Serre Examples
Hi 143.167.4.199, I saw your edit and wonder if you could give a reference on the non-affine varieties Serre found. I have "each Stein manifold is a smooth affine algebraic variety" from Robert Gompf's book 4-manifolds and Kirby Calculus. Perhaps we can restore the old GAGA statement unchanged; I don't see the mathematical error there. Further, if Serre's examples actually exist and imply that not every Stein manifold is a smooth complex affine variety, would it be true if we changed affine to analytic (see analytic variety)? Orthografer 21:01, 27 April 2006 (UTC)
 * In a paper I recently read the author refered to the following for "Serres classical construction": Umemura, H.: La dimension cohomologique des surfaces alg´ebriques. Nagoya Math. J. 47 (1972), 155-160.
 * Haven't read it myself though. Spakoj 09:14, 3 May 2006 (UTC)
 * There are Stein manifolds which are not even biholomorphic to any algebraic variety (whether affine or not), e.g. the unit ball in $$\mathbb C^n$$.

Joerg Winkelmann 13:37, 13 May 2006 (UTC)

Riemann surfaces
From the article: 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. 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)
 * 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)

The statement

Now Cartan's theorem B shows that $$H^1(X, \mathcal O_X)= H^2(X, \mathcal O_X)=0 $$, therefore $$H^2(X, \mathbb Z)=0$$.

is false in general. In general one just has

$$H^2(X, \mathbb Z)=H^1(X,\mathcal O_X^*) $$.

More generally, the result is that topologically trivial vector bundles are holomorphically trivial. — Preceding unsigned comment added by 2001:DA8:D800:107:153B:A732:3459:C002 (talk) 13:44, 19 May 2013 (UTC)