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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 ( 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)
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)
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)
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)
Joerg Winkelmann 13:37, 13 May 2006 (UTC)
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)
The statement
Now Cartan's theorem B shows that , therefore .
is false in general. In general one just has
.
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)
This article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
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 ( 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)
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)
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)
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)
Joerg Winkelmann 13:37, 13 May 2006 (UTC)
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)
The statement
Now Cartan's theorem B shows that , therefore .
is false in general. In general one just has
.
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)