Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to know where are essential applications of the theory of Stein spaces in algebraic geometry. I heard Cartan's theorem A & B were used in Serre's GAGA, but are there any other applications?

share|cite|improve this question
up vote 5 down vote accepted

I would say that technically speaking Stein spaces are essentially not used in algebraic geometry but that they had, and still have, a strong motivational role. Let me elaborate on these polemical assertions:

First of all algebraic geometry, and especially scheme theory, does not presuppose that the base field is $\mathbb C$ and even less that it has characteristic zero. So Stein manifolds can't even be defined.

Stein manifolds are replaced in algebraic geometry by affine varieties (or schemes), and theorems for them are infinitely easier than the corresponding theorems for Stein spaces: the vanishing of cohomology for coherent sheaves (and even for quasi-coherent sheaves !) is proved in two pages over general affine noetherian schemes (not only varieties) in Hartshorne, whereas the corresponding theorem for Stein spaces was developed over two decades by Oka, Cartan, Serre, ... and takes half a book to be proved (cf. the book by Grauert-Remmert on Stein spaces or that by L.Kaup-B.Kaup on holomorphic functions of several variables) .

And finally the empirical evidence is overwhelming: of the numerous algebraic geometry books which handle cohomology of coherent sheaves over an algebraic variety exactly zero (to my lnowledge) use Stein manifolds to prove their results.

That said, historically the first use of the cohomology of coherent sheaves was indeede in complex manifold theory : Cartan realized that Oka's results on pseudo-convex domains could be interpreted in the language of Leray's recently discovered sheaves and Cartan's student Serre adapted these techniques to algebraic geometry in his ground-breaking article Faisceaux Algébriques Cohérents.
But let me emphasize again that Serre never uses any result from Stein theory in that article.

share|cite|improve this answer
Georges, unless I am mistaken, Joseph Taylor, in his book Several Complex Variables with Connections to Algebraic Geometry and Lie Groups, shows the essential surjectivity of $\text{Coh}(X)\to\text{Coh}(X^\text{an})$ (as in GAGA) by appealing to the Cartan-Serre Theorem (as I mentioned below). Just a remark :) – Alex Youcis May 4 '14 at 12:50
Dear Alex, yes, Taylor's Chapter 13 is devoted (as he states at the head of the chapter) to proving Serre's theorem. The presentation is more leisurely and self-contained than the article. You are right to mention this excellent (and maybe somewhat underrated) textbook. – Georges Elencwajg May 4 '14 at 13:14
Thanks for your answers. I roughly saw where the position of Stein spaces is in the AG. That Taylor's book seems to be very interesting, so I'll read it. – Tei Huang May 4 '14 at 16:10

How about the Cartan-Serre theorem:

Theorem(Cartan-Serre): Let $X$ be a compact holomorphic variety, and $\mathscr{F}$ a coherent sheaf on $X$. Then, $H^p(X,\mathscr{F})$ is a finite dimensional space for all $p\geqslant 0$.

This is, undeniably, one of the most important theorems in any introduction to analytic geometry. The proof of this theorem makes huge use of Stein spaces.

The guiding principle behind the use of Stein spaces in the proof is simple. Whenever we want to compute cohomology, we like to have a more concrete way of getting a handle on it.

Usually this presents itself in the form of a Leray cover $\mathfrak{U}$ for the sheaf $\mathscr{F}$ (a cover where the restriction of the sheaf to every finite intersection is acyclic) on a space $X$. This is because Leray's theorem then tells us that the Čech cohomology $\check{H}^i(\mathfrak{U},\mathscr{F})$ coincides with the usual definition of cohomology as $H^i(X,\mathscr{F})=(R^i\Gamma)(\mathscr{F})$.

In algebraic geometry, this principle manifests itself fairly aggressively. I am sure you have appealed to the fact that if $X$ is a separated quasicompact scheme, and $\mathscr{F}$ a coherent sheaf on $X$, then $H^i(X,\mathscr{F})$ can be computed as $\check{H}^i(\mathfrak{U},\mathscr{F})$, where $\mathfrak{U}$ is any finite open cover of $X$. This is merely Leray's theorem, along with the observation that Serre's criterion for affiness implies that $\mathscr{U}$ is a Leray cover for $\mathscr{F}$ on $X$.

Stein spaces and Cartan's theorems play the exact same role in complex manifolds that affine varieties and Serre's criterion play in algebraic geometry. Namely, Cartan's theorems tell us that if $X$ is an analytic space and $\mathscr{F}$ a coherent sheaf, then any cover $\mathscr{U}$ by Stein manifolds will be a Leray cover. Thus, we may appeal to Leray's theorem to conclude that we can compute the sheaf cohomology as the Čech cohomology of this cover.

share|cite|improve this answer
Dear Alex, Stein manifolds are indeed used to considerable success in complex holomorphic geometry especially, as you explain, in cohomology theory. But Stein spaces are not used in algebraic geometry, for reasons I explain in my answer, where they can't even be defined. Their role is taken up by affine varieties, as you mention yourself. – Georges Elencwajg May 4 '14 at 12:39
@GeorgesElencwajg I, and I believe the OP, are using 'algebraic geometry' loosely :) In particular, whenever I said algebraic geometry with respect to Stein manifolds, I meant complex analytic geometry. – Alex Youcis May 4 '14 at 12:41
@GeorgesElencwajg I guess to be explicit, by algebraic geometry, I am referring to the connection being, explicitly, algebraic geometry over $\mathbb{C}$, and its relation to transcendental techniques via the GAGA principle. – Alex Youcis May 4 '14 at 12:44
Dear Alex, yes your use of "algebraicaic geometry" is absolutely legitimate and I only wanted you to explicitly clarify your usage of the term in your excellent answer. – Georges Elencwajg May 4 '14 at 12:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.