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Is there an example of a complex analytic space $X$ that doesn't have any (not necessarily open or closed) positive dimensional subspace $Y$ which is analytically isomorphic to (the complex analytic space associated to) a scheme?

Edit: after D.Arapura's comment, we require $X$ to have dimension $>1$.

If I remember correctly, there are non "Abelian" complex tori $X=\mathbb{C}^n/\mathrm{Lattice}\;\;$ that do not have any positive dimensional analytic subvariety. Can a counterexample be derived from this?

Also, any complex algebraic space has an open subspace which is a scheme. So the counterexample (if it exists) must be searched outside algebraic spaces.

What if the question is modified by requiring that $X$ has no $Y$ that is locally closed in the analytic Zariski topology (where opens are complements of analytic subspaces)?

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A disk will work, but this is probably not what you want. Did you want $X$ to be compact? –  Donu Arapura May 8 '11 at 13:00
    
+1 to DonuArapura's comment! Perhaps I should've required $X$ to be of dimension > 1 ? –  Qfwfq May 8 '11 at 13:07
    
Does a polidisk work? If so, I'll add the requirement "$X$ compact" as suggested by D.Arapura. –  Qfwfq May 8 '11 at 13:26
    
Yes, I suspect that a polydisk does work, although this would be much harder. Off the top of my head, I might argue that polydisks have many bounded holomorphic functions, while the algebraic examples probably don't (reduce to the quasiprojective case, and apply suitable Riemann/Hartogs extension theorems to extend to the projective closure). –  Donu Arapura May 8 '11 at 13:40
    
I deleted my answer, because the first half of it was wrong: Torus -point is not complex analytic equivalent to a scheme if the tours is non-algebraic. For the second example, I can not prove that non-algebraic torus can not contains something 2-dimensional complex analytic subset X isomorphic to a scheme, so that torus-X is a one dimensional Riemann surface of infinite topological type. –  Dmitri May 8 '11 at 18:17

1 Answer 1

Take a complex 2-torus $X$ without curves, and hence, without non-constant meromorphic functions (see e.g. Shafarevich, Basic algebraic geometry, chapter VIII, \S 1, example 2). The only locally closed subsets of $X$ will be $X$ itself, $\varnothing$, finite subsets and the complements of finite subsets. $X$ minus a finite subset can't be algebraic: if it were, it would have a non-constant meromorphic function, and then so would $X$.

[upd: locally closed here means locally closed in the analytic Zariski topology, so this answers the second question and not the first.]

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Why can not torus have a locally analytic curve of infinite genus, say a leaf of a foliation? –  Dmitri May 21 '11 at 1:24
    
Dmitri -- I'm not sure I understand what a locally analytic curve of infinite genus is. But an analytic Zariski closed subset of a compact complex manifold is homeomorphic to a finite CW-complex –  algori May 21 '11 at 1:44
    
@dmitri. The holomorphic foliations on $X$ are defined by vector fields. To see it, take a vector field and look at the tangency locus with a given foliation. Since $X$ has no curves in it, the foliation and the vector field are everywhere tangent or transverse. But of course we can start with a vector field tangent to the foliation at a given point. –  jvp May 21 '11 at 2:00
    
algori sorry for the confusion. I agree with your remark 100%, but it does not remove my concerns. I had in mind the following. We know that on a hyperbolic genus $g$ surface we can have a collection of geodesics whose complement is a collection of open ideal hyperbolic triangles. This collection is called lamination. Each geodesic is not compact in the surface. So why can not you have a lamination in the torus whose complement is analytically isomorphic to something algebraic? –  Dmitri May 21 '11 at 8:53
    
jvp, you are right about foliations, when I was writing "foliation" I was thinking also about laminations... Is it clear that they don't exist? (by lamination I mean something like an infinite collection of vertical lines on C^2 given by equations $x_n=\frac{1}{n}$) –  Dmitri May 21 '11 at 8:59

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