# Why is Oka's coherence theorem a deep result?

This is a very naive question.

Let $X$ be a complex manifold. Let $\mathcal{O}_X$ be the structure sheaf of $X$, a sheaf of rings whose sections over opens $U\subset X$ are just the holomorphic functions $U\rightarrow\mathbb{C}$.

A sheaf of $\mathcal{O}_X$-modules $F$ is coherent if:

1. It is locally finitely generated: Ie, there is an open cover $\{U_i\}$ of $X$ such that $F|_{U_i}$ admits surjections $\mathcal{O}_{U_i}\rightarrow F|_{U_i}$ for each $i$.
2. For any open $V\subset X$, and any morphism $f : \mathcal{O}_V^s\rightarrow F|_V$, $\text{Ker}(f)$ is a locally finitely generated sheaf on $V$ (ie, satisfies condition 1).

Then, Oka's coherence theorem states that the structure sheaf $\mathcal{O}_X$ is coherent.

When $X$ is a scheme, this statement is almost tautological. What is it about the setting of complex manifolds that makes this theorem deep?

• On a scheme, you need that the structure sheaf be locally Noetherian or something for the sheaf to be coherent, no? – Mariano Suárez-Álvarez Oct 31 '17 at 7:07
• On a scheme, the structure sheaf is coherent when it is, well, coherent. Coherence is strictly weaker than local noetherianness. – Fred Rohrer Oct 31 '17 at 7:17
• Well, sure. But apart from that tautological condition, you need, in real life, something. – Mariano Suárez-Álvarez Oct 31 '17 at 7:25
• Everything is deeper and harder for complex analytic spaces. What makes you think this should be any different? – Denis Nardin Oct 31 '17 at 7:55
• Coherence of $\mathcal{O}_X$ is a nontrivial finiteness condition, even for schemes. Does Georges Elencwajg's answer mathoverflow.net/a/129390/121 help at all? – S. Carnahan Oct 31 '17 at 8:09

## 1 Answer

In scheme theory applied to complex geometry one usually does not encounter coherent rings which are not noetherian as well. However if $X$ is (for example) a Stein manifold then the ring $R = \mathcal{O}(X)$ of global holomorphic functions is typically non-noetherian, which makes remarkable the fact that $R$ is nonetheless coherent.

• Since the question posed comes at the start of the complex-analytic theory, whereas Stein spaces lie much further into the theory, perhaps it is better to just try to prove coherence of $\mathscr{O}_{\mathbf{C}^2}$ by hand and then read the proof of Oka's theorem (such as in the book Coherent Analytic Sheaves) to sense the depth. It is somehow because complex manifolds are not built out of rings as schemes are that coherence properties of their sheaf theory are much more difficult; e.g., the analogous coherence theorem in rigid-analytic geometry isn't quite as deep. – nfdc23 Oct 31 '17 at 14:35
• Ah, I was recalling Hartshorne's definition of coherent as just a finitely generated quasi-coherent (so structure sheaves would be tautologically coherent), but forgot that he assumes all his schemes to be noetherian. – Will Chen Oct 31 '17 at 17:01
• @oxeimon: there isn't even a very robust notion of "quasi-coherence" in the analytic setting (one can make a definition and prove some reasonable properties, but it lacks others one would want for a fully satisfactory notion). You gave the correct general definition of coherence in the question posed, due to Serre (motivated by experience in the complex-analytic case with Oka's work!); it does not require any notion of "quasi-coherence". – nfdc23 Oct 31 '17 at 22:17