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The equivalence of the category of smooth projective curves over $\mathbb C$ and the category of compact Riemann surfaces is, I believe, well documented. For example, it is mentioned on the wiki page: http://en.wikipedia.org/wiki/Algebraic_curve#Compact_Riemann_surfaces and I believe that Rick Miranda's book Algebraic Curves and Riemann Surfaces goes in to some detail on this.

So I just wondered if when we look at (compact) real manifolds, can we make any relation between them and schemes over $\mathbb R$. I am not aware of any such relation, but on the other hand I do not expect to be enough of an expert on schemes any time soon to spot it myself, and my google searches were fruitless. Apologies if it is too vague/soft, and thank you for any replies :)

EDIT - To give one specific question, are there interesting subcategories of the categories of real manifolds and schemes over $\mathbb R$ that are equivalent as categories.

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Not all real manifolds are algebraic (even over $\mathbb C$ you have to restrict to dimension 1 complex manifolds), so you should give some restrictions. Moreover, the real manifold associated to a connected smooth algebraic scheme over $\mathbb R$ can be disconnected (e.g. elliptic curves) or even empty in which case you certainly can't reverse the process. –  user18119 Feb 26 '13 at 13:00
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And then even when the manifolds in question are as nice as they could possibly be (say, $\mathbb{P}^1(\mathbb{R})$), we would probably have to put restrictions on the mappings we allow, on both sides of the correspondence. –  Zhen Lin Feb 26 '13 at 13:17
    
@QiL'8 As you say, there are some restrictions that would be expected to turn up, but I didn't want to suggest any in case I chose the wrong restriction. When you say "real manifold associated...over $\mathbb R$", what would this association be? If you meant just taking the underlying space then I was assuming that something deeper than this would be required (if it exists). –  Joe Tait Feb 26 '13 at 13:33
    
@JoeTait: yes this what I meant. I think everything coming from the algebraic side (including morphism) is OK for the real side (except for connectedness), but the converse is problematic. Even over $\mathbb C$, how do you charaterize those complex manifolds which are algebraic ? When you restrict to projective complex manifolds, GAGA insures that all morphisms are algebraic and all closed analytic submanifold are algebraic. Over $\mathbb R$ this is certainly false, even for something as nice as elliptic curves. –  user18119 Feb 26 '13 at 14:38
    
@QiL'8 It is my understanding that in the example of smooth projective curves and Riemann surfaces, the equivalence is not one of just underlying spaces - for example, in the case of the torus, as described here en.wikipedia.org/wiki/… it does not seem that it is just about changing the topology - the whole method of getting from one to the other will change the fundamental picture. Am I wrong in thinking this? I mean, I don't see how a curve can have the same underlying space as a surface - one can be embedded where the other can't –  Joe Tait Feb 26 '13 at 14:57

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