# Complex analytic space with no (positive dim.) subscheme ?

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)?

• A disk will work, but this is probably not what you want. Did you want $X$ to be compact? May 8, 2011 at 13:00
• +1 to DonuArapura's comment! Perhaps I should've required $X$ to be of dimension > 1 ? May 8, 2011 at 13:07
• Does a polidisk work? If so, I'll add the requirement "$X$ compact" as suggested by D.Arapura. May 8, 2011 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). May 8, 2011 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. May 8, 2011 at 18:17

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 $X$.
• @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. May 21, 2011 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? May 21, 2011 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 $x_n=\frac{1}{n}$) May 21, 2011 at 8:59