I've read the section "Serre duality" in Hartshorne's book and have several questions.

1) In Remark 7.1.1 it is claimed that on $X = \mathbb{P}^n$

$\alpha = \frac{x_0^n}{x_1 \cdot ... \cdot x_n} d(\frac{x_1}{x_0}) \wedge ... \wedge d(\frac{x_n}{x_0})$

is a Cech cocycle which generates $H^n(X,\omega_X)$ (I have checked this) and that this is independent from the choice of the basis of $\mathbb{P}^n$. Well it's easy to check that it's independent from the order of $x_0,...,x_n$. But it does not seem to be invariant under other automorphisms of $\mathbb{P}^n$, which are typically given by

$x_0 \mapsto x_0 + \lambda x_1 , x_1 \mapsto x_1, ..., x_n \mapsto x_n$.

Or am I wrong? What is the fix group in $\text{Aut}(\mathbb{P}^n) = PGL(n)$ of $\omega_X$?

2) In Remark 7.12.1 it is shown that Serre duality implies that for a smooth projective variety $X$ of dimension $n$ we have that $H^n(X,\omega_X) \cong k$. It's quite fascinating to get such a concrete result with this abstract machinery of Ext-functors. What can be said about this isomorphism? Is there a "canonical" one? Is it possible to distinguish a generator? What examples do you know?

  • $\begingroup$ It is obvious that your $\alpha$ is not invariant under permutation of the $x_i$s, for it changes sign! $\endgroup$ – Mariano Suárez-Álvarez Feb 15 '11 at 16:47
  • 2
    $\begingroup$ @Mariano Correctly understood, it is. More precisely: permuting the variables preserves $\alpha$ up to sign (this is not meant to be obvious). In a Cech cohomology computation, the cohomology class depends not only on what section we take on $U_0 \cap U_1 \cap U_2 \cap \cdots U_n$, but also on the individual sets $U_i$ and how they are ordered. If we permute the open sets as well, we get precisely the same cohomology class. $\endgroup$ – David E Speyer Feb 15 '11 at 16:53
  • $\begingroup$ @David, I know, I know. I wanted Martin to notice it, though. $\endgroup$ – Mariano Suárez-Álvarez Feb 15 '11 at 17:07
  • $\begingroup$ Ah, sorry. I misread you. $\endgroup$ – David E Speyer Feb 15 '11 at 18:42

The point of this answer is to give a short proof of the following result.

Theorem 1: Let $X$ and $Y$ be smooth $n$-dimensional projective varieties over a field $k$ of characteristic zero; let $B$ be a smooth connected variety; and let $\phi: X \times B \to Y$ be a morphism. Then the induced map $H^q(Y, \Omega_Y^p) \to H^q(X \times \{ b \}, \Omega_X^p)$ is a constant function of $b$. (If the base field is not algebraically closed, this requires a little more language to say correctly; I'm going to gloss over this.)

We need the following, very deep theorem:

Theorem 2: If $X$ is a smooth projective variety over a field of characteristic zero, then the map $H^q(X, \Omega^p) \to H^q(X, \Omega^{p+1})$ induced by $d$ is $0$.

This is usually proved by Hodge theory.

As a corollary of Theorem 1, if $\mathrm{Aut}(X)$ is connected, then $\mathrm{Aut}(X)$ acts trivially on $H^q(X, \Omega^p_X)$. As a corollary of the corollary, $\mathrm{PGL}_{n+1}$ acts trivially on $H^n(\mathbb{P}^n, \omega)$.

Proof (Sketch): We may immediately assume that $B$ is one dimensional, as any two points in a connected variety may be joined by a chain of smooth affine curves. Also, the statement is local on $B$, so we may pass to a smaller base curve whenever it is convenient.

Let $\theta$ be a class in $H^q(Y, \Omega^p_Y)$. Let $\eta = \phi^* \theta \in H^q(X \times B, \Omega^p_{X \times B})$. For any $b \in B$, we can restrict $\eta$ to $X \times \{ b \}$, and get a class in $H^q(X, \Omega^p_X)$. Call this $\eta(b)$. We want to show that $\eta(b)$ is a constant function of $b$. In other words, $\eta(b)$ is a section of $H^q(X, \Omega_X^p) \otimes \mathcal{O}_B$ and we want to show that it is in $H^q(X, \Omega_X^p)$. Since $H^q(X, \Omega_X^p) \otimes \mathcal{O}_B$ is a trivial vector bundle, it makes sense to talk about $d \eta(b)$ (just take $d$ of each component), and we will show that $d \eta(b)=0$.

In case you would like to see this done in more technical language, here it is: We can think of $H^q(Y, \Omega_Y^p)$ as $R^q \pi_* \Omega_Y^p$ where $\pi$ is the map $Y \to \mathrm{Spec} \ k$. So we can pull back to $\theta$ to a class $\eta$ in $R^q \pi_2^* \Omega^p_X$ where $\pi_1$ and $\pi_2$ are the projections of $X \times B$ onto its factors. We have a canonical map $\Omega^p_{X \times B} \to \Omega^p_{X \times B/B}$ (restricting to the vertical fibers). So we get a class in $R^q (\pi_2)_{\ast} \Omega^p_{X \times B/B}$. Also, $\Omega^p_{X \times B/B} \cong \pi_1^* \Omega_X^p$. Finally, if you think about how pushforwards and pullbacks work in products, then $R^q (\pi_2)_{\ast} \pi_1^{\ast} \Omega^p_X \cong H^q(X, \Omega^p_X) \otimes \mathcal{O}_B$. So that's how we get a class in $H^q(X, \Omega^p_X) \otimes \mathcal{O}_B$.

So, we want to compute $d \eta(b)$. Passing to a local neighborhood on $B$, we may assume that $\Omega^1_B$ is free, with generator $dz$ for $z \in \mathcal{O}(B)$. On $X \times B$, we have $$\Omega^p_{X \times B} \cong \pi_1^* \Omega_X^p \oplus \ \pi_1^* \Omega_X^{p-1} \otimes \pi_2^* \Omega_B^1 \quad (*)$$ ($B$ is one dimensional; otherwise there would be more terms.) Locally, write $\eta = \alpha + \beta dz$ where $\alpha$ and $\beta$ are sections of $\pi_1^* \Omega^p_X$ and $\pi_1^* \Omega^{p-1}_X$. So we get $\alpha$ and $\beta$ in $R^q (\pi_2)_{\ast} \Omega^{p}_{X \times B/B}$ and $R^q (\pi_2)_{\ast} \Omega^{p-1}_{X \times B/B}$. (More precisely, $\eta$ is represented by a Cech cocycle $U_{i_0} \cap U_{i_1} \cap \cdots \cap U_{i_p} \mapsto \eta_{i_0 i_1 \cdots i_p}$. Write each $\eta_{I}$ as $\alpha_I + \beta_I dz$. Then $\alpha_I$ and $\beta_I$ are Cech cocycles for the corresponding cohomology groups.)

Write $d'$ for the derivation $\Omega^p_{X} \to \Omega^{p+1}_{X}$ and $d''$ for the analogous derivation on $B$. Using the isomorphism $(*)$, we have $d=d'+d''$. In these terms, we have $$d \eta = d'\alpha + d'' \alpha + d'(\beta) dz. \quad (**)$$ By Theorem 2, $d'(alpha)$ and $d'(\beta)$ are $0$. So $d \eta$, which is a class in $R^q (\pi_2)_{\ast} \Omega^{p+1}_{X}$, is actually a class in $R^q (\pi_2)_{\ast} \pi_1^* \Omega^{p}_{X} \otimes \Omega^1_B$ and, as such, is equal to $d'' \alpha$.

Now, $\eta(b)$ is the restriction of $\eta$ to $X \times \{ b \}$. As $dz$ is $0$ on the vertical fibers, $\eta|_{X \times \{ b \}} = \alpha|_{X \times \{ b \}}$. We wanted to compute $d \eta(b)$, where the $d$ is with respect to the $B$-variables. This makes it plausible, and I leave it to you to check, that $d \eta(b)$ is $d'' \alpha$.

So far, this argument has been valid for any $\eta \in R^q (\pi_2)_{\ast} \Omega^p_{X \times B}$. Now we use that $\eta = \phi^* \theta$. By Theorem 2, $\theta$ is closed. So $d \eta = d \phi^* \theta = \phi^* d \theta = 0$. Looking at $(**)$, and at the compatibilities we have just checked, we see that $d \eta(b)=0$, as desired.


(1) The analogous result in differential geometry is that, if $B$ is connected, and $\phi : X \times B \to Y$ is a smooth map, then the induced maps $H^{k}(Y) \to H^k(X)$ are the same for every point of $B$. The reader might enjoy writing down a deRham proof of this fact, and seeing how it relates to the above argument.

(2) More generally, let $\pi: \mathcal{X} \to B$ be a smooth projective map with $B$ connected; let $\phi: \mathcal{X} \to Y$ be any morphism with $Y$ projective; let $\theta \in H^q(Y, \Omega_Y^p)$; let $\eta = \phi^* \theta \in R^q \pi_{\ast} \Omega^p_{\mathcal{X}}$. Then we want to say that $\eta(b)$ is a constant function of $b$. The right way to say this is that there is a connection on $R^q \pi_{\ast} \Omega^p_{\mathcal{X}}$ which annihilates $\eta$. This is called the Gauss-Manin connection, and I have basically walked you through how to compute it for a trivial family.

(3) Theorem 2 is trivial for $p=n$. However, even if you only care about Theorem 1 for $p=n$, then our argument uses Theorem 2 applied to $\beta$, which is an $n-1$ form. So we still need to use a nontrivial case of Theorem 2, even in that case. (This is what I missed the first time I wrote this up.)

  • $\begingroup$ Is there a typo in the third paragraph of the proof? The notation $R^q\pi_2^*\Omega_X^p$ feels really mysterious. $\endgroup$ – Jiangwei Xue May 19 '11 at 14:55
  • $\begingroup$ awesome answer! $\endgroup$ – Bernie Jan 7 '18 at 14:57

Here are the answers to your questions:

(1) $PGL(n)$ acts trivially on $H^n(\mathbb{P}^n, \omega)$. This will follow from my answer to (2), which is:

For any $n$ dimensional projective variety $X$, the isomorphism $\mathrm{Tr}: H^n(X, \omega_X) \to k$ is completely canonical. More precisely, if $g$ is an automorphism of $X$, and $\eta$ a cohomology class, then $\mathrm{Tr}(g^* \eta) = \mathrm{Tr}(\eta)$. More generally, if $g : X \to Y$ is a degree $d$ map, then $\mathrm{Tr}_Y(g^* \eta) = d \cdot \mathrm{Tr}_X(\eta)$.

I tried to write up a short proof, and failed miserably. I hope someone else will succeed.

  • $\begingroup$ Doesn't Hartshorne's III.7.2 do it? $\endgroup$ – Mariano Suárez-Álvarez Feb 15 '11 at 19:18
  • $\begingroup$ I don't think so? Hartshorne defines a dualizing sheaf as a sheaf equipped with a trace map $t$. (Right above prop III.7.2.) Then Exercise 7.2 doesn't state what the trace map is. The fact that we need to verify is that $f^{!} \omega$ is some relation between the trace maps of $f^{!} \omega$ and $\omega$. I don't see any statement in Hartshorne relating them. Feel free to write it up if you do! $\endgroup$ – David E Speyer Feb 15 '11 at 20:12
  • $\begingroup$ The proposition, I mean. Not the problem :) $\endgroup$ – Mariano Suárez-Álvarez Feb 15 '11 at 20:46
  • $\begingroup$ OK, but I again don't think so. Let $g$ be an automorphism of $X$. Let $\omega = \omega'$ and consider the two trace maps $t$ and $t \circ g^*$. Prop 7.2 states that there is a unique automorphism $\phi$ of $\omega$ such that $t \circ g^* = t \circ H^n(\phi)$. What we want to show is that $\phi = \mathrm{Id}$. How does this follow? $\endgroup$ – David E Speyer Feb 15 '11 at 21:15
  • $\begingroup$ That said, you do make me realize that there is a simple proof in the case of PGL(n). For $g \in PGL(n)$, let $g$ act on $H^n(\omega)$ by $\chi(g)$. Then $\chi$ is a character. The group $PGL(n)$ has no nontrivial characters, so $\chi$ is trivial. $\endgroup$ – David E Speyer Feb 15 '11 at 21:19

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