## Stein Manifolds and Affine Varieties

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When is a Stein manifold a complex affine variety? I had thought that there was a theorem saying that a variety which is Stein and has finitely generated ring of regular functions implies affine, but in the comments to my answer here, Serre's counterexample was brought up. I'm guessing that the answer is that the ring of regular functions must be nontrivial somehow, like it must separate points, but I'm curious about what the exact condition is.

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I have to ask: what is "gaga"? Google did not turn up anything useful.  – Darsh Ranjan Oct 26 2009 at 5:41
GAGA refers to Serre's paper "Géométrie algébrique et géométrie analytique" and more generally, the philosophy that it embodies that there is a correspondence between complex analytic geometry and complex algebraic geometry.  Serre's paper, I believe, focuses on proving that for any variety and sheaf, there is an analytic space and sheaf defined naturally to be the analytifications of what you started with, and that cohomology doesn't change.  That is, you can compute cohomology of coherent sheaves in the complex topology. – Charles Siegel Oct 26 2009 at 11:52

## 1 Answer

Charlie, it is funny answering this way but here it is.

The criterion you are thinking about is a criterion that is relative to an embedding. It says that if $X$ is a quasi-affine complex normal variety, whose associated analytic space $X^{an}$ is Stein, then $X$ is affine if (and only if) the algebra $\Gamma(X,\mathcal{O}_{X})$ is finitely generated. This is a theorem of Neeman.

You can reformulate the requirement of $X$ being quasi-affine as a separation of points property: for any point $x \in X$ consider the subset $S_{x} \subset X$ defined as the set of all points $y \in X$ such that all regular functions on $X$ have equal values at $x$ and $y$. Then by an old theorem of Goodman and Hartshorne $X$ is quasi-affine if $S_{x}$ is finite for all $x$. So you can say that $X$ is affine if it satisfies: 1) $X^{an}$ is Stein; 2) $S_{x}$ is finite for all $x \in X$; 3) $\Gamma(X,\mathcal{O}_{X})$ is finitely generated.

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Answering this way (rather than in person) means more people get to see it! – Allen Knutson Feb 17 2010 at 19:15
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