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There are many theorems in algebraic geometry which were proved using cohomology. I would like to know examples of such theorems which have been proved only by using cohomology. In other words, those theorems which have not been so far proven without using cohomology.

I am asking a big list of such examples.

Makoto Kato
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  • How about the fact that $\Gamma(X,\mathscr{F})$ is a coherent $A$-module, if $\mathscr{F}$ is a coherent sheaf on the projective $A$-scheme $X$. Can you explain what your motivation for this question is? Are you trying to justify to yourself why cohomology is a useful tool? I would suggest reading chapter 18 of Vakil, if so. He does a good job of motivating why a cohomology theory of sheaves is a useful thing. – Alex Youcis Mar 30 '14 at 06:53
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    @AlexYoucis [Are you trying to justify to yourself why cohomology is a useful tool?] Not necessarily. Cohomology is obviously a powerful tool. I'm interested in whether those theorems can be proven without using it. Here's an analogy. The prime number theorem was first proved using complex analysis. Many number theorists(including Hardy) thought it was impossible to prove it by an elementary method. Later it was found that it was not the case. – Makoto Kato Mar 30 '14 at 07:21
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    Yes, but the "elementary proof" by Erdos is incomprehensible, and doesn't contain the intuitive, and more widely applicable techniques presented in the complex analytic proof. The question is not just whether or not cohomology can prove things which are 'unprovable' otherwise, but also if cohomology makes clearer proofs which were previously muddled, and unintuitive, and which lacked any clear generalization of techniques. The $p^aq^b$-theorem of group theory has a non-rep theory proof, but no one in their right minds thinks it's superior to the character theoretic one. The method is ad hoc – Alex Youcis Mar 30 '14 at 07:55
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    @AlexYoucis It's just an analogy. Please don't take it too seriously. The question is clearly stated. I don't think I need to explain my motivation in detail and persuade everybody that he should agree with me. – Makoto Kato Mar 30 '14 at 08:02
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    That's fine, I was just giving a perspective, a naive one perhaps, as to why that's the wrong question. Good luck. – Alex Youcis Mar 30 '14 at 08:49
  • @AlexYoucis [I was just giving a perspective, a naive one perhaps, as to why that's the wrong question.] I don't understand why you think that is the wrong question. Are you saying that a non-cohomological proof of a theorem which can be proved using cohomology is useless? – Makoto Kato Mar 30 '14 at 09:41
  • @MakotoKato: Perhaps there is a weaker statement here, that a non-cohomological proof of a theorem which can be proved using cohomology is not automatically useful. – RghtHndSd Mar 30 '14 at 15:56
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    @rghthndsd Of course. However, I don't think it is not an enough reason to reject the question. Please don't misunderstand me. I'm not saying that everybody should be interested in this question. – Makoto Kato Mar 30 '14 at 20:59

1 Answers1

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Here are two results usually proved by using cohomology and for which I am not aware of any other proof.

1) A noetherian scheme $X$ is affine if and only if its reduction $X_{red}$ is affine
The proof uses Serre's characterization of noetherian affine schemes as those noetherian schemes for which $H^i(X,\mathcal F)=0$ for all coherent sheaves $\mathcal F$ on $X$ and all $i\gt 0$ .

2) A compact complex manifold $M$ is projective algebraic if and only if it is a Hodge manifold
A Hodge manifold is a compact complex manifold admitting of a Kähler metric whose associated fundamental form $\phi$ has an integral De Rham cohomology class: $[\phi]\in H^2_{DR}(M,\mathbb Z)\subset H^2_{DR}(M,\mathbb C)$.
The heart of the proof is that on aHodge manifold there exists a suitable positive line bundle $L$ and for such a positive line bundle we have $H^q(M,\Omega_M^n\otimes L)=0$ for $q\gt 0$ ("Kodaira's vanishing theorem") .

This had been conjectured by Hodge.
Kodaira's proof of that conjecture certainly played a large role in his being awarded a Fields medal in 1954.

Georges Elencwajg
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  • is a good example. When one first learns the cohomological characterization of ampleness on proper things, it initially seems much more complicated than the other definitions. But, after seeing how much easier it is to manipulate cohomology, as in the standard proof of 2), you start to appreciate it.
    • – Alex Youcis Mar 30 '14 at 16:04
  • Dear Georges, judging from this question, it seems that Görtz-Wedhorn's book knows how to prove (1) without using cohomology. – Makoto Kato Mar 30 '14 at 22:00
  • I'm not 100% sure but this paper seems to prove (2) without using cohomology. An analytical proof of Kodaira's embedding theorem for Hodge manifolds by Noboru Tanaka, Hokkaido Mathematical Journal Vol. 13 (1984) p. 232-240 – Makoto Kato Mar 30 '14 at 22:22
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    Dear Makoto, I have no reason to believe that Görtz-Wedhorn know how to avoid cohomology in (1). They say nothing of the sort and in any case they definitely use cohomology to prove (1) : look at their Lemma 12.38, page 337. In the comment you link to, all I can see is that "YBL", some anonymous commenter, thinks it can be done by hand. This does not constitute a proof ! – Georges Elencwajg Mar 30 '14 at 22:35
  • Yes, Tanaka seems to prove Kodaira's embedding theorem without using cohomology. I can only say that: 1) this proof came 30 years after the cohomological proof and, from superficially browsing it, seems to use very hard analysis techniques which I imagine are not especially familiar to algebraic geometers. What I am certain of is that these techniques go way over my head... 2) I don't think that there exists a theorem in algebraic geometry of which one could prove that it cannot be proved without using cohomology, nor even that this is a meanigful statement. (to be continued) – Georges Elencwajg Mar 30 '14 at 22:54
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    (continued) I am taking your question rather as a challenge to find results for which the easiest or best known or most natural proof is cohomological. This is a very interesting challenge and it was an excellent idea to ask about it: +1 ! And I am certainly looking forward to other users giving new examples of such results. – Georges Elencwajg Mar 30 '14 at 23:02
  • Dear Georges, I checked the book. (1) is in Exercise 3.31. p.92. I tried to prove it without using cohomology, but failed. – Makoto Kato Apr 01 '14 at 10:17