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Even if in dimension 2, complex structure is equivalent to algebraic structure for surfacs, but when studying deformation theory or moduli theory for surface, they are different, for example, the concept of the family of Riemann surface and the family of complex smooth algebraic curve are different, but it seems that we can study the deformation theory (or moduli question) for Riemann surface in the language of algebraic geometry (or just consider it is an algebraic geometry object), I don't know if it is true or it is my misunderstanding. So my question is if the deformation theory for Riemann surface is the same as the deformation theory for smooth complex algebraic curve? Thank you for any answer for this question which confuses me a long time!

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mathabc -- what is the question? –  algori Jul 4 '10 at 22:04
    
The question is if the study of deformation theory of Riemann surface is the same as the deformation theory of complex smooth algebraic curve. –  HYYY Jul 4 '10 at 22:09
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@mathabc: please write the question in the question! –  Mariano Suárez-Alvarez Jul 4 '10 at 22:54
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Every 1-dimensional compact Hausdroff complex-analytic space is algebraic, and more specifically the fully faithful (by general GAGA) analytification functor from proper $\mathbf{C}$-schemes to compact Hausdorff complex-analytic spaces hits all objects of dimension at most 1. (This requires some work when there are no reducedness hypotheses.) Since completion identifies algebraic and analytic local rings, and completion suffices to test flatness, it follows that the infinitesimal flat deformation theories coincide under analytification of such objects, so the formal deformation rings agree. –  BCnrd Jul 4 '10 at 23:07
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1 Answer

For compact Riemann surfaces and projective algebraic curves, deformation theory will be the same.

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Is it really? I just see the concept of family of Riemann surfaces is different from the concept of family of projective smooth curves. So.....? –  HYYY Jul 8 '10 at 3:42
    
See BCnrd's comment. "What he said." The categories are equivalent. –  Ketil Tveiten Jul 8 '10 at 13:19
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