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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