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I have been going through basic Algebraic Geometry, and although I have been using Serre duality for quite a lot of time, I still do not understand how one came up with the dualizing sheaf for a Projective scheme $X \hookrightarrow \mathbb{P}^N.$ The way we prove Serre duality in Hartshorne is like this: (I) First prove it for Projective Space

(II) Define dualizing sheaf, and then prove that for a projective scheme $\mathcal{E}xt(i_*\mathcal{O}_X,\omega_{\mathbb{P}^N})$ is a dualizing sheaf.

(III) Then for the duality we need ample $\mathcal{O}(1)$ and Cohen-Macaulayness.

The proof went smooth, but I did not understand how Hartshorne came up with that dualizing sheaf. I know how in derived categories we look for the right adjoint of $f_*$ to prove Serre duality, but how does that influence the definition of the dualizing sheaf in Hartshorne is still not clear to me. Any comments are welcome.

I am aware of almost all the questions on dualizing sheaf and Serre duality on StackExchange, but none of them seemed to clear my confusion. Thank you for your time.

Rio
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  • The thing you are not understanding... Is it the general definition of a "dualizing sheaf"? Or is it the reason why that Ext group is a dualizing sheaf in the projective case? – Compacto Feb 16 '22 at 11:53
  • look at lemma 7.4 and its proof, as well as Theorem 7.11 and corollary 7.12. The starting point is that for a smooth scheme you want the dualizing sheaf to be the canonical sheaf, which is motivated from complex geometry and Serre duality there – Aitor Iribar Lopez Feb 17 '22 at 11:13

2 Answers2

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Actually, we've already seen a sheaf on projective space which satisfies some of the properties we're after! That is, we've seen a sheaf $\mathcal{E}$ on $\Bbb P^n_k$ and a morphism $t:H^n(\Bbb P^n_k,\mathcal{E})\to k$ which for some class of coherent sheaves $\mathcal{F}$ on $\Bbb P^n$ satisfies the properties of a dualizing sheaf, namely that $$\operatorname{Hom}(\mathcal{F},\mathcal{E})\times H^n(\Bbb P^n_k,\mathcal{F})\to H^n(\Bbb P^n_k,\mathcal{E})$$ followed by $t:H^n(\Bbb P^n_k,\mathcal{E})\to k$ gives an isomorphism $\operatorname{Hom}(\mathcal{F},\mathcal{E})\to H^n(\Bbb P^n_k,\mathcal{F})'$.

In the computation of the cohomology of sheaves of the form $\mathcal{O}(d)$ on projective space $\Bbb P^n_A$ (theorem III.5.1), we showed that the natural map $$H^0(\Bbb P^n_A,\mathcal{O}(d))\times H^n(\Bbb P^n_A,\mathcal{O}(-n-d-1))\to H^n(\Bbb P^n_A,\mathcal{O}(-n-1))\cong A$$ is a perfect pairing of finitely generated free $A$-modules. Reindexing slightly, we can rewrite this as $$H^0(\Bbb P^n_A,\mathcal{O}(-n-d-1))\times H^n(\Bbb P^n_k,\mathcal{O}(d))$$ and recognizing that $\operatorname{Hom}(\mathcal{O}(d),\mathcal{O}(-n-1))\cong \operatorname{Hom}(\mathcal{O},\mathcal{O}(-n-d-1))\cong H^0(\Bbb P^n_A,\mathcal{O}(-n-d-1))$, we've exactly found that $\mathcal{E}=\mathcal{O}(-n-1)$ and the isomorphism $H^n(\Bbb P^n_k,\mathcal{O}(-n-1))\cong k$ satisfy what we ask of a dualizing sheaf when $\mathcal{F}$ is among the sheaves of the form $\mathcal{O}(d)$. As $\mathcal{O}(-n-1)\cong\bigwedge^n\Omega_{\Bbb P^n_k/k}$ is the canonical sheaf of projective space, this would be a sign that we're on the right track if we're trying to invent this theory.

There's also good intuition coming from the manifold case: remember, Serre duality is supposed to be kind of like an analogue of Poincare duality. How does Poincare duality work? Cap with the fundamental class! What is the fundamental class? It's an element of the top-dimensional homology which tracks orientations. Orientations are described by an ordered basis of the tangent space at each point up to equivalence, which can be measured by top-dimensional differential forms. So poking around top-dimensional differential forms to try and invent a duality theory is a natural avenue for experimentation.

Finally, let me point out that I've never actually asked Serre, Hartshorne, or any of the other mathematicians who developed this theory about how they got to the idea or read accounts of their thought processes, so I can't be sure that what I've related here was actually a factor in how they got to these concepts. But I think this material is certainly at least suggestive of a direction to explore - once you have the motivation of "can I prove some sort of duality for varieties?" and some of this evidence in the form of calculations showing some preliminary success, it would be natural to investigate further.

KReiser
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  • I think I have not clearly stated my question. My apologies. I understand the motivation behind the dualizing sheaf, and I certainly get how the theory has been developed from Projective spaces to Projective schemes. For projective spaces, we got the dualizing sheaf via the calculations of the cohomology of the projective. From there I do not understand how someone comes up with the dualizing sheaf $\mathcal{E}xt^r(\mathcal{O}X,\omega{\mathbb{P}^N}).$ I was thinking maybe there is a more in-depth explanation behind this, but now I think it just came up in plain algebraic manipulation. – Rio Feb 18 '22 at 16:34
  • @RioDutta Sorry for not addressing what you meant to ask. In my understanding, basically what happens is once you come up with the three requirements for duality mentioned in theorem III.7.1, the choice of a dualizing sheaf is forced for all closed subschemes of $\Bbb P^n$: the rest of the results in the section, like III.7.4, show this. That is, if you want to be able to do duality on a projective scheme $X$ in the same way you do it on $\Bbb P^n$, there's only one possible choice of $\omega_X$ you could ever pick, and you can figure out what it is from the conditions you've imposed. – KReiser Feb 23 '22 at 02:56
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As mentioned in the question, we are seeking the right adjoint $f^!$ of $f_!$; connection to the classical form is well explained here. I want to write down my understanding of why the dualizing sheaf or complex $f^!(\mathcal{O}_Y)$ should take the above form, or a shift of the canonical line bundle when $f$ is smooth. Of course I'm missing many details, and much of the analogy and intuition is subjective.

We start from the simplest case. Let $f:X\to Y$ be a finite separable morphism. In fact, there is no need to go to the derived category in this case (to some extent $-$ without the flatness, derived functors are not so simple). That is, we have the adjunction $\text{Hom}_X(\mathcal{F},f^!\mathcal{G})\simeq\text{Hom}_Y(f_*\mathcal{F}, \mathcal{G})$ for sheaves if we define $f^!(-)=\mathcal{Hom}_Y(f_*\mathcal{O}_X, \mathcal{G})$, Hartshorne III. ex 6.10. This is more or less the Hom-tensor adjunction $\text{Hom}_B(N, \text{Hom}_A(B, M))\simeq\text{Hom}_A(N\otimes_B B, M)$ where $A\to B$ is a ring map and $M\in (A-\text{mod})$, $N\in (B-\text{mod})$. We need the fact that $f$ is affine to reduce like that. I believe this adjunction only requires the affine hypothesis; but $f$ should be proper to have $f_!=f_*$, and affine+proper implies finite.

Here the counit, or the trace morphism is very close to the usual trace (when $f$ is flat), as explaned here. Since $f$ is of relative dimension zero, (e.g. in better cases a finite etale cover), there is no shift of degree (as we already know the answer). This trace map matches well with our intuition. For an analogy, let $\pi: E\to M$ be a (oriented) vector bundle over a smooth manifold. Then $R\pi_!$ can be calculated by integration along fiber, which is the Thom isomorphism. It gives the trace map for the Poincare duality of $\mathbb{R}^n$, which can be glued by Mayer-Vietoris arguments. If we imagine $E$ being something like rank 0 bundle (e.g. orientation sheaf), integration should be the trace.

Now we move to the more general case, where $f$ is proper flat. There are two homological lemmas about $f^!$. First, we can reduce to the calculation of $f^!(\mathcal{O}_Y)$, the relative dualizing complex: SP 0A9T. Next, we can relate $f^!$ to upper shriek of the diagonal morphism $\Delta:X\to X\times_Y X$, SP 0E2P.

So when $f$ is smooth, we somehow reduce to the regular embedding case. Hence our second paragraph applies, we have to calculate $R\mathcal{Hom}_Y(\mathcal{O}_X,\mathcal{O}_Y)$. Now here is where the differential sheaf comes into play. By some properties of exterior power and yoneda product of $\text{Ext}$, we reduce to the case of embedding of an effective Cartier divisor $i: D\to X$. Since $R\mathcal{Hom}(\mathcal{O}_X,\mathcal{O}_X)=\mathcal{O}_X$ and $R\mathcal{Hom}(\mathcal{I}_D,\mathcal{O}_X)=\mathcal{O}_X(D)$, by $0\to \mathcal{I}_D\to \mathcal{O}_X\to\mathcal{O}_D\to 0$, $R\mathcal{Hom}(\mathcal{O}_D,\mathcal{O}_X)=(\mathcal{O}_X\to\mathcal{O}_X(D))$. The twist gives $0\to \mathcal{O}_X\to \mathcal{O}_X(D)\to i_*\mathcal{N}\to 0$ which proves $R\mathcal{Hom}(\mathcal{O}_D,\mathcal{O}_X)\simeq \mathcal{N}[-1]$.

For me, this procedure looks like some kind of formal dual of integration. Regular embedding is dual to smooth morphism, in the same sense that cofiber sequence is dual to fiber sequence. Instead of integrating the forms, the trace map uses the normal bundle to describe $\mathcal{O}_X$ (in some sense it does one-dimension less, since it has no information along $D$). Maybe this explanation is nonsense, but I feel the situation is analogous to the invention of negative numbers. Finite flat case lies in the middle of them.

Finally, the cotangent sheaf is defined as the relative conormal sheaf of the diagonal, hence it is not so surprising that after above lemmas, for smooth $f:X\to Y$ we get $f^!(-)=f^*(-)\otimes\Omega_{X/Y}[d]$. Specializing $Y=\text{Spec }k$ gives the classical result.

okabe rintarou
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