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It is a well known fact that proper scheme $X$ over $k$ has a up to isomorphism unique dualizing sheaf (EGA I, Hartshorne).

This dualizing sheaf $\omega_X$ comes with two striking properties:

(i) There is a homomorphism $t : H^n(X, \omega_X ) \to k$ (also called the trace) such that for every coherent $\mathcal{O}_X$-Module $\mathcal{F}$ the following holds: There exists a canonical bilinear map

$$ \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) \times H^n(X, \mathcal{F}) \to H^n(X, \omega_X) $$

which gives an isomorphism:

$$ \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) \cong H^n(X, \mathcal{F})^*$$

by composing with t. Here, * means the dual vector space over $k$.

(ii) In addition for every integer $i ≥ 0$ and coherent $\mathcal{F}$ , there exists a canonical isomorphism: $\operatorname{Ext}^i(\mathcal{F}, \omega_X) \cong H^{n-i}(X, \mathcal{F})^*$ if and only if $X$ is Cohen– Macaulay.

In other words (i) means $H^n(X, \mathcal{F})^*$ is representable. Now $H^n(X, \mathcal{F})^*$ carries structure of a $k$-vector space. If we assume that $\dim_k H^n(X, \mathcal{F}) < \infty$, then $H^n(X, \mathcal{F}) \cong H^n(X, \mathcal{F}) ^*$ and in following we will not differ between $H^n(X, \mathcal{F})$ and it's dual.

A one dimensional $k$- subspace $V_1 \subset H^n(X, \mathcal{F})$ correspond in high tec language to an orbit of a non zero vector $v \in H^n(X, \mathcal{F})$ by action of $k$ on $H^n(X, \mathcal{F})$ via multiplication, ie $V_1= k \cdot v$.

Now my first question is what are the special morphisms in $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) $ which correspond to $V_1$ aka to the orbit of $k$-action on $v$. How are they related to each other as objects in $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) $?

In other words if $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) $ inherits the $k$-action from $H^n(X, \mathcal{F})$, are the elements from the same orbit related to each other in certain "deep" way? Any intuition how one can think about these orbits (except of the boring answer "lines in $H^n(X, \mathcal{F})$")?

The second question is if we take $\mathcal{F}= \omega_X$, then $id_{\omega_X} \in \operatorname{Hom}_(\omega_X,\omega_X)$. Is it's image in $H^n(X, \mathcal{F})$ "special" in certain way? What can we say about this element considered as vector?

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First of all, dualizing sheaves are unfortunately not treated in EGA. The treatment in Hartshorne has some limitations. Perhaps some of them are related to your questions.

For pointers to more recent and complete treatments of duality, I suggest you to look at the MO question "Serre duality in families".

Let me start by your second question. The defining property of $\omega_X$, namely $$ H^n(X, \mathcal{F})^* \cong \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) $$ expresses the fact that the functor $H^n(X, -)^*$ is representable. This means that there is a representing pair $(\omega_X, \int_X)$ with $\int_X \colon H^n(X, \omega_X) \to k$ a canonical isomorphism that, as you explain, indices the previously displayed isomorphism. This is what you denote "$t$" in your post. Notice that $\int_{X}$ is what corresponds to $\operatorname{id_{\omega_X}}$ in the isomorphism $$ H^n(X, \omega_X)^* \cong \operatorname{Hom}_{\mathcal{O}_X}(\omega_X,\omega_X) $$

The fact that the representing object of a functor is unique up to unique isomorphism means that there is no choice for it, once you have a concrete description of $\omega_X$ it forces a unique description of $\int_X$.

How to get such a description? On the projective space $\mathbb{P}^n_k$ one gets a characterization of $\Omega_{\mathbb{P}^n_k|k}$ as $\mathcal{O}_{\mathbb{P}^n_k}(-n-1)$ and from this characterization a canonical isomorphism: $$ \int_{\mathbb{P}^n_k} \colon H^n(\mathbb{P}^n_k, \Omega^n_{\mathbb{P}^n_k|k}) \longrightarrow k $$ Therefore $\omega_{\mathbb{P}^n_k|k} = \Omega^n_{\mathbb{P}^n_k|k}$. Once you get this description you extend it to other projective varieties and with ba little more work to proper varieties over $k$. This is explained under the assumption that $k$ is perfect in J. Lipman's blue book, Dualizing sheaves, differentials and residues on algebraic varieties, Astérisque No. 117 (1984).

Now for your fist question. The pairing $$ \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) \times H^n(X, \mathcal{F}) \to k $$ that assigns to a linear map $\varphi \colon \mathcal{F} \to \omega_X$ and a cohomology class $\alpha \in H^n(X, \mathcal{F})$ the element $\int_{X} (\alpha \circ \varphi)$ where "$\circ$" denotes Yoneda composition. So the fact that $H^n(X, \omega_X)$ is 1-dimensional means, essentially, that the integral is unique up to "rescaling", so any time you take a multiple of $\alpha$ you are basically recalling it. In a perhaps more abstract point of view one may interpret $\alpha \colon \mathcal{O}_X \to \mathcal{F}[n]$ (in the derived category), therefore $\alpha \circ \varphi \colon \mathcal{O}_X \to \omega_X[n]$ is just a scalar multiple of the "volume form": the element in $H^n(X, \omega_X)$ whose image by $\int_{X}$ is $1 \in K$.

The fact that you are dealing with canonical maps suggests that one should avoid identifying a space with its dual, unless there is canonical choice of the isomorphism. This is crucial in this theory.

What is fascinating to me is that this story makes sense in any characteristic, and that there is an interesting counterpoint between the abstract aspects (dualizing sheaves, representable functors) and the more concrete ones in the sense of computations with cohomology classes, traces and differentials.

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  • $\begingroup$ Thank you very much for your great answer! Now I see the reason why $t$ is called the trace by considering $\alpha \circ \phi$ as a kind of "diagonal". $\endgroup$ – MortyPB Apr 23 at 14:24
  • $\begingroup$ @MortyPB In the relative case, for a finite separable morphism "$t$" is the usual trace. This, together with global Noether normalization is exploited in the blue book to give an explicit description of the integral for projective varieties. $\endgroup$ – Leo Alonso Apr 23 at 14:28

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