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I want to show that not every complex manifold is of the form $X_{an}$ where $X$ is some algebraic variety, providing a counterexample.

The candidate for this counterexample seems to be the open unit disk $D$ in $\mathbb{C}$.

But how can I proof that $D$ is not isomorphic to a complex analytic manifold $X_{an}$ for any nonsingular curve $X$?

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    $\begingroup$ The converse of Chow's lemma would say that every projective variety is an analytic subvariety of $\mathbf P^n$, which is clearly true. What statement are you trying to give a counterexample to? $\endgroup$
    – user64687
    Commented Jan 29, 2015 at 15:46
  • $\begingroup$ I've edited my question, because I wrote a stupid thing. I'm sorry!! $\endgroup$ Commented Jan 29, 2015 at 15:54
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    $\begingroup$ Another counter-example would be the complex plane minus a discrete countably infinite subset. If you want compact examples, there aren't any in complex dimension 1, but they exist in dimension 2, say, the Hopf surface. (Any smooth algebraic variety is necessarily Kahler.) $\endgroup$ Commented Jan 31, 2015 at 12:35

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First, $X$ must be smooth of dimension one: since the local ring of a closed point $p\in X$ and the analytic local ring of $p\in X_{an}$ have the same completion, they have the same dimension and the same quotients by powers of their maximal ideals, so we see that the tangent space is one-dimensional for all closed points in $X$. Next, we may observe that $X$ is connected: if $X$ has multiple connected components, then $X_{an}$ must as well. So if there is an $X$ such that $X_{an}\cong D$, then $X$ must be a connected smooth curve.

Now we'll introduce some outside results. We begin with a result about smooth curves: any connected smooth curve can be expressed as a smooth projective curve of genus $g$ minus $r$ points. We'll also need the comparison theorem on $\pi_1$:

Theorem (Grothendieck). For a pointed algebraic variety $(X,x)$ over $\mathbb{C}$ there is a canonical isomorphism between the étale fundamental group $\pi_1^{\text{ét}}(X,x)$ and the profinite completion of the topological fundamental group $\pi_1^{\rm top}(X(\mathbb{C}),x)$.

(This is SGA1, Expose XII, section 5, for instance.) Now we do a little topology: since $D$ is not compact, we must have $r>0$, else $X_{an}$ is compact. Then the fundamental group of a genus $g$ surface with $r>0$ punctures is a free group on $2g+r-1$ generators (you may verify this by drawing a $2g$-gon with edges identified appropriately and then deleting $r$ points in the interior and looking at what happens, for instance). The profinite completion of this group is nontrivial if $2g+r-1>0$, so we must have $g=0$ and $r=1$. But $g=0$ and $r=1$ cannot happen because then we would have $\Bbb C^1$ biholomorphic to $D$, a contradiction to the uniformization theorem.

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There should be an answer (the "right" one) by considering the field of meromorphic functions on $D$, but there was a gap in my earlier logic.

So here's a much cruder answer: if $X$ is an algebraic curve, then it has the form $\overline{X} - \{p_1,\ldots,p_n\}$ where $\overline{X}$ is a compact Riemann surface. We need $X_{an}$ to be simply connected: this gives only one possibility for $\overline{X}$ and $n$, but this can be ruled out by Liouville's theorem.

Let me know if you want more details.

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  • $\begingroup$ So you're telling me this fact: since every algebraic curve can be compactified adding a finite numbers of points, there's a unique way to make this surface simply connected (using uniformization of compact riemann surfaces (Girondo, Gonzalez-Diez pag. 82 Thm 2.1)). But, actually, how can I use this in the example of the unit disk? $\endgroup$ Commented Jan 31, 2015 at 17:09
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    $\begingroup$ @Federico: no, that is not what I meant. I meant that if we compactify $X$ and get a curve of genus $g>0$, then $X$ cannot have been simply connected to begin with, hence not isomorphic to $D$. On the other hand if we compactify and get $\mathbf P^1$, then $X$ is topologically $S^2$ minus $n$ points; if $n>1$ then again $X$ was not simply connected, so again not iso to $D$. Finally, if $n=1$, then $X$ must be $\mathbf C$, which is not iso to $D$ by Liouville. I hope that is clearer. $\endgroup$
    – user64687
    Commented Jan 31, 2015 at 22:16

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