# What is nice in projective manifolds?

I am current studying the Calabi-Yau theorem and some of its consequences. One of the most significant results is the one that states that every Calabi-Yau manifold is projective.

Another interesting result, if I am not mistaken, is the one proved by Nash that states that any smooth manifold admits an embedding as an algebraic variety.

My question is: is possible to determine information on the geometry and topology of manifolds by analysing their algebraic descriptions? The fact that Calabi-Yau manifolds are projective suggests that perhaps the correct context for studying such manifolds is the algebraic one.

What do I gain by knowing that a complex manifold is projective?

I am so sorry if this question is vague at all, but this is a truly real question for someone who doesn't know anything about algebraic geometry.

• 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. – 54321user Sep 5 '17 at 23:28
• 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. – roy smith Sep 5 '17 at 23:50
• 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 – roy smith Sep 5 '17 at 23:57
• 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 – roy smith Sep 6 '17 at 0:19
• 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. – Ben McKay Sep 6 '17 at 5:59