Timeline for What is nice in projective manifolds?
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Sep 6, 2017 at 5:59 | comment | added | Ben McKay | The Nash theorem makes any manifold into the real points of an algebraic variety. But there are no compact complex submanifolds of Euclidean space, so there is no complex variant of Nash's theorem. | |
Sep 6, 2017 at 5:57 | history | edited | Ben McKay | CC BY-SA 3.0 |
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Sep 6, 2017 at 0:19 | comment | added | roy smith | here are some notes by an expert, discussing special topological properties of projective manifolds. see e.g. p. 29, math.columbia.edu/~thaddeus/seattle/voisin.pdf | |
Sep 6, 2017 at 0:11 | comment | added | roy smith | the file I linked states that no algebraic proof of KVT exists, but one has apparently existed some 30 years now, but I am definitely not expert. (the only proof i have worked through is Kodaira's.) uni-due.de/~mat903/books/esvibuch.pdf | |
Sep 5, 2017 at 23:57 | comment | added | roy smith | oops, 54321 is correct, one does indeed need a theorem to deduce that an analytic projective manifold is algebraic, namely Chow's theorem. w3.impa.br/~massaren/files/Kodaira.pdf | |
Sep 5, 2017 at 23:50 | comment | added | roy smith | complex projective varieties are algebraic by definition, (the Kodaira thorem gives a criterion for manifolds to be projective in terms of existence of certain cohomology classes). since projective space has a flag of subspaces, every projective manifold inherits similar sub varieties, and thus being projective does imply some information about the geometry and topology of a manifold. | |
Sep 5, 2017 at 23:28 | comment | added | 54321user | Complex projective manifolds are always algebraic by the Kodaira embedding theorem. Also, you should not get excited about a smooth manifold embedding into a variety. By the whitney embed into some $\mathbb{R}^n$, and so if your manifold is compact, you can embed it into a neighborhood of a smooth point of some variety. | |
Sep 5, 2017 at 22:54 | history | asked | L.F. Cavenaghi | CC BY-SA 3.0 |