The question in the title doesn't quite make sense. I was always wondering if the scheme is the generalization of manifold. The precise statement should be like following:

If $X$ is a complex (smooth)manifold, is there a scheme $Y$ over $\mathbb{C}$ such that the corresponding analytic space $Y^{an}$ isomorphic to $X$?

I guess if $X$ is projective, then this is Chow's theorem. And I doubt the question has an affirmative answer if $X$ is non-compact (I guess that $X: y-e^x=0$ in $\mathbb{C}^2$ might be an example, but I do not know how to show this).

Edit 1: There are also example (as @hunter pointed out) of a complex torus but not an abelian variety shows that a compact manifold might not be a variety, however, is it possible to be a scheme ?(according to Hartshorne, the category of varieties is a subcategory of quasi-projective, integral, separated scheme. But could there be some non-quasiprojective scheme($Y$ has to be integral, separated in our case according to GAGA, I guess) )?

Edit 2: Another question I think of is: is (Artin) Stack a generalization of manifold? I really want to know if manifold could be think of some subcategory of some category of with "algebraic objects".

  • $\begingroup$ A part of the problem here might be that you're trying to compare general complex manifolds to algebraic schemes. This tries to do two things at once: (1) express a manifold as a scheme and (2) show that the scheme is algebraic. The second part fails quite often, as evidenced by non-projective complex tori or (some) non-Kahler manifolds. However, a complex manifold is an analytic scheme, where the defining equations are holomorphic functions. (See en.wikipedia.org/wiki/Analytic_space for some details; Grauert-Remmert and that Romanian book for more.) $\endgroup$ Nov 22, 2013 at 17:50
  • $\begingroup$ (Or, well, a complex manifold + a generic point is a scheme. There's a subtlety there somewhere (that's the same one as for algebraic variety + generic point = scheme) that's nicely explained in Shafarevich's "Basic algebraic geometry".) $\endgroup$ Nov 22, 2013 at 17:53
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    $\begingroup$ The unit disc in $\mathbb{C}$ is surely a Riemann surface but not a complex variety. (Note, it is not isomorphic to $\mathbb{C}$ as a Riemann surface!) $\endgroup$
    – Zhen Lin
    Nov 22, 2013 at 18:01
  • $\begingroup$ Zhen Lin: But isn't it a complex analytic scheme? It is a complex analytic space, and even a local model of one (by taking the defining equations to be $f = 0$ on the unit disk). Is there a difference between complex analytic spaces and schemes? $\endgroup$ Nov 22, 2013 at 18:09
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    $\begingroup$ Dear @Gunnar: I don't think anybody (and definitely not Shafarevich) has ever used your terminology "complex analytic scheme". There exist complex analytic spaces and Grauert has introduced non-reduced ones. The Wikipedia article you refer to only says that their "definition is similar to the definition of a scheme". So, to answer your question: yes, there is a huge difference between complex analytic spaces and schemes. For example I don't believe that there is any scheme to which you can reasonably associate a disc, even if you add to the disc a generic point (whatever that means). $\endgroup$ Nov 22, 2013 at 18:55

1 Answer 1


This is a very interesting question and it doesn't have an easy answer. I will just give an overview over some facts.

A necessary condition for $X$ to be algebraic (i.e. it is the analytification of a scheme) is that the transcendence degree of its field of meromorphic functions $K(X)$ is equal to the dimension of $X$ (because in algebraic geometry those two numbers are always equal for nice schemes).

It is not enough for $X$ to be compact, for instance the general complex torus of dimension $\geq 2$ (all of which are compact) has no non-constant meromorphic functions at all.

On the other hand, a non-compact analytic manifold may be algebraic. This is the case e.g. for the complex line $\mathbb{C}$.

The condition $\mbox{trdeg}_{\mathbb{C}} K(X) = \dim(X)$ is also not sufficient. Compact manifolds having this property are called Moishezon manifolds but not all of them are algebraic (a famous example is called Hironaka's example). The nice thing is that all of them come from something called 'algebraic spaces' which is a generalization of a scheme.

There are also various other properties a complex manifold can have. It can be a Kähler manifold (i.e. being equipped with a special metric) and together with being Moishezon it implies being projective which is equal to being projective algebraic (this is Chow's theorem). Also there is the notion of Hodge manifold (a special kind of Kähler manifold) which is actually equivalent to being projective. This is called the Kodaira embedding theorem.

If you want to know more about all this, I recommend you have a look at the book 'Complex geometry' by Daniel Huybrechts. All the terms I used (except for algebraic spaces) are explained there.

  • $\begingroup$ Thank you for your answer, you explain the varieties is not a generalization of manifold. However, what about schemes? $\endgroup$
    – Li Yutong
    Nov 22, 2013 at 17:20
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    $\begingroup$ Actually, if you want $X^{\text{an}}$ to be a smooth manifold, $X$ needs to be a smooth variety. I have never seen the analytification functor for something else than schemes of finite type over $\mathbb{C}$. $\endgroup$ Nov 22, 2013 at 17:34
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    $\begingroup$ @Gregor: neither have I. $\endgroup$ Nov 22, 2013 at 18:56
  • $\begingroup$ This question is GAGA ncatlab.org/nlab/show/GAGA Check out ncatlab.org/nlab/show/analytification#Existence . $\endgroup$ May 22, 2014 at 19:45

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