# Why are Stein manifolds/spaces the analog of affine varieties/schemes in algebraic geometry?

I presume this is a GAGA-style result, but I cannot find a reference.

Also like affine varieties, we have:

Theorem. A complex manifold is Stein if and only if it embeds into some $$\mathbb{C}^N$$ as a closed complex submanifold.

For the "only if" direction, see Hörmander, An Introduction to Complex Analysis in Several variables, Theorem 5.3.9. For the converse, an argument is contained on pp 109-110 of Hörmander, immediately after the definition of Stein manifold.

• In particular, the analytification of a complex algebraic variety is a Stein space. May 23, 2019 at 10:31
• @Earthliŋ: You probably mean affine algebraic variety. May 25, 2019 at 20:45
• @JérômePoineau Yes, of course. Somehow "affine" got lost when I added "complex"... May 25, 2019 at 21:50

I believe the reason for this is Cartan's 'Theorem B': for a Stein manifold $$\mathrm{X}$$ sheaf cohomology vanishes, $$\mathrm{H}^n(\mathrm{X},-)=0$$ for $$n\geqslant 1$$, and this property characterises Stein manifolds among complex manifolds. In the same way affine schemes are characterised among (nice) schemes by cohomology vanishing (this is a theorem of Serre). This comes up in the proof of the cohomological comparison result which is part of GAGA, see for example SGA 1, Exposé XII.

• There is no way to be completely tactful here; the vanishing condition is for coherent analytic coefficients only. It is certainly not true for all coefficients, e.g. $X= \mathbb{C}^*$ and cohomology with coefficients in $\mathbb{Z}$ is not zero, but $X$ is Stein. May 25, 2019 at 13:29
• @DonuArapura this was understood... on the scheme side you are also not going to consider sheaves which are not even $\mathcal{O}_X$-modules or which have no finiteness properties.
– ssx
May 25, 2019 at 13:38
• OK, but I'll leave the comment, since not everyone would understand the hidden assumptions. May 25, 2019 at 13:45