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Hartshorne makes this sound like a coincidence: we us Cech cohomology on the usual open affine cover $\mathcal{U}$ to get the chain complex

$$\check{C}^\bullet\left(\mathcal{U}, \bigoplus_{n \in \mathbb{Z}} \mathcal{O}(n)\right): 0 \to \bigoplus_i S_{x_i} \to \cdots \to \bigoplus_i S_{x_0\cdots\hat{x_i}\cdots x_r} \to S_{x_0 \cdots x_r}\to 0,$$

calculate the cokernel of the last map to obtain the cohomology groups $\check{H}^r(\mathcal{U}, \mathcal{O}(n))$, and then observe that it's one dimensional when $n = -r - 1$. Voilà! What a coincidence!

Surely there's an actual reason here, though. What is it?

Daniel McLaury
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    Dear Daniel, wee ze French write Voilà, not Viola – Georges Elencwajg Sep 23 '13 at 19:37
  • @Daniel: What do you mean by "actual reason"? The calculation via cech cover gives a perfectly good reason. And of course this was known long before in the setting of de Rham cohomology for the corresponding complex analytic $\mathbb{P}^n$. – Martin Brandenburg Sep 23 '13 at 21:46
  • @GeorgesElencwajg, the sad part is that I not only know that but get really upset when I see other people writing "viola" instead of "voila," yet apparently I've somehow managed to do it myself when distracted... – Daniel McLaury Sep 23 '13 at 22:14
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    Dear Daniel: yes, fate loves to be mischievous. C'est la vie... – Georges Elencwajg Sep 23 '13 at 22:17

1 Answers1

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This is a very long and interesting story.
In a nutshell: it is Serre duality, which asserts that for a smooth projective variety $X$ of dimension $n$ and a locally free sheaf $V$ on $X$ we have a canonical isomorphism $$ H^i(X, V)\cong (H^{n-i}(X, \check{V}\otimes \omega))^* $$ [The Čech sign over $V$ denotes dual vector bundle and the star denotes dual $k$-vector space ]

Your case follows by taking $X=\mathbb P^n, V=\omega, i=n$ and noticing that $\omega=\mathcal O_{\mathbb P^n}(-n-1)$.

Edit
Serre introduced his duality theorem in "Un théorème de dualité" , Comm.Math.Helv. 29 (1955) Théorème 4. He worked in the context of holomorphic manifolds and the cohomology he uses is Dolbeault cohomology.
The algebraic analogue was developed by Grothendieck and his school.
The cohomology they use is derived functor cohomology, as introduced by Grothendieck in his Tohoku paper.

Georges Elencwajg
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  • I don't think that this answers the question. Serre duality is actually the statement that $H^i(X,V) \otimes H^{n-i}(X,\check{V} \otimes \omega) \to H^n(X,\omega)$ is a perfect pairing, and $H^n(X,\omega) \cong k$ is an additional isomorphism, or can be seen as part of Serre duality. But you really have to prove this independently. – Martin Brandenburg Sep 23 '13 at 21:45
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    @Martin: I don't know what allows you to write that "Serre duality is actually ..." something else than what I wrote. In "Un théorème de dualité" , Comm.Math.Helv. 29 (1955) Théorème 4, Serre states his duality theorem in the form that I give (and even slightly more generally) and does not mention perfect pairings.There are many forms of his duality and I said that this was a long and interesting story. Moreover you write "you really have to prove this independently". I have to prove nothing, I'm just stating a result at which there are several ways to arrive. (to be continued) – Georges Elencwajg Sep 23 '13 at 22:05
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    (continued) You should realize that sophisticated theorems have several formulations instead of falsely and not very politely claiming that my answer doesn't answer the question: it does. – Georges Elencwajg Sep 23 '13 at 22:07
  • The issue is that Hartshorne (and I believe Serre) proves Serre duality as a consequence of the result I've stated above. – Daniel McLaury Sep 23 '13 at 22:20
  • I've talked to my professor about it now, though, and he gave some other proofs of Serre duality, at least in less general settings, which helped to motivate it. – Daniel McLaury Sep 23 '13 at 22:29
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    Dear Daniel, this is a long and complicated story: Serre actually proved his duality theorem over $\mathbb C$ in an analytic context (cf. the paper I mention in my response to Martin) and he used Dolbeault cohomology without even mentioning Čech cohomology. So I still believe that the "actual reason" for the calculation you mention is Serre duality, but of course the order of presentation between your calculation versus Serre duality depends on the choices made for developing the theory. – Georges Elencwajg Sep 23 '13 at 22:37
  • Dear Daniel, will you excuse my curiosity if I ask you who your professor is? – Georges Elencwajg Sep 23 '13 at 22:39
  • Dear Georges, instead of explaining $H^n(X,\omega) \cong k$, you mention Serre duality $H^i(X,V) \cong H^{n-i}(X,V^* \otimes \omega)^$, which is much more general. This is also proven in Hartshorne via a direct computation with Cech cohomology. So for me this doesn't explain* anything. Of course I know that you have put the statement into the correct framework etc., but how would you actually explain $H^n(X,\omega) \cong k$? This was the question ... sorry, again I didn't want to be impolite. Maybe it's my english :(. – Martin Brandenburg Sep 23 '13 at 22:56
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    Also, in your version of Serre duality (as compared to with mine) you say that there is a canonical(!) isomorphism $H^i(X,V) \cong H^{n-i}(X,V^* \otimes \omega)^*$. How do you define this, without knowing in advance that $H^n(X,\omega) \cong k$ (or at least that there is a trace map $H^n(X,\omega) \to k$)? See also Daniel's comments. For me, your answer is circular. – Martin Brandenburg Sep 23 '13 at 22:58
  • @GeorgesElencwajg, I propose that you add your comment about Serre's original paper using Dolbeaut cohomology into your answer and then I accept it. Agreeable? – Daniel McLaury Sep 26 '13 at 18:59
  • Dear Daniel, you are right: it is better to incorporate that comment into the answer, which I have just done. – Georges Elencwajg Sep 27 '13 at 21:25