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Let $X \subset \mathbb{P}^n_k$ be a normal projective subscheme over $k$ of dimension $n$. The dualizing sheaf is in context of Serre duality a pair $(\omega_X,t)$ (which exists in that case) consisting of a coherent sheaf $\omega_X$ and a linear functional $t: H^n(X,\omega_X) \to k$ giving rise to natural isomorphism $\operatorname{Hom}_X(\mathcal{F}, \omega_X) \simeq \operatorname{H}^n(X, \mathcal{F})^*$ for each coherent sheaf $\mathcal{F} $ via naturality & composition. Obviously the pair is unique up to isomorphism by usual argument as characterized by universal property.

This map $t$ is also called trace, and is sometimes denoted by an integral symbol $\int$. This naturally leads me to blackboxophobic

Question: What is the original reason - presumably having it's roots in differential geometry - to call $t$ a "trace" and label it by an integral symbol?

Asking plainly: If we go back to classical theory of vector bundles over projective submanifolds $X \subset \mathbb{P}^n_k$, which integral is the "analogon or precursor" of this map $t$ in algebraic context of Serre duality "legitlimating" to call $t$ trace and labeling it by an integral symbol?
Ie, is $t$ adapted literally from certain integral resp. trace map (term "trace" there should be synonymously to a contraction in diff geometric sense).

Would be completely happy to see it elaborated on "baby example" $X=\mathbb{P}^n$ with $\omega_X= \wedge^n \Omega_X$.
Motivation: Intuition

The confusing part is the following: in classical setting there exist a (up to scalar multiples canonical) volume form $vol$, a (for nice enough manifolds) which can be considered as a nowhere vanishing section of bundle $\wedge^n \Omega_X$ generating the $k$-space of global sections of the latter bundle. The algebro-geometric pendant to this space is surely $H^0(X, \wedge^n \Omega_X)$.

At first glance it look rather natural as one might think that the motivation for labeling the algebraic trace map with integral symbol comes from that the classical pendant is given by integrating the volume form of the mfd (resp it's multiples).

But there is a problem so far I can see (and don't know how to satisfactorily overcome, more precisely a degree shifting issue:
The volume form $vol$ mentioned above lives in $H^0(X, \wedge^n \Omega_X)$, in contrast the trace map "integrates" the classes living in $H^n(X, \wedge^n \Omega_X)$.

Can this discrepancy be somehow naturally resolved in order to take integration of volume form (resp it's multiples) as legitime precursor of the algebraic trace map? (if it really the case. Maybe it has a different origin; the one with $vol$ was just the only one came into my mind pondering about this question).

Besides, is there any (historic) background to call it "trace"?

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You are implicitly asking two different questions: why trace, and why $\int$?

For the first, I only have a guess. One can extend Grothendieck duality to the relative setting. For a finite flat map $f:X\to Y$ of schemes with dualizing sheaves, one has a trace map $tr: f_*\omega_X\to \omega_Y$. In the simplest case $Spec K\to Spec L$, where $L\subset K$ is a finite field extension, this is literally the trace $tr:K\to L$.

Now suppose, we work with smooth projective varieties over $\mathbb{C}$. By GAGA and the Dolbeault isomorphism $$H^n(X,\Omega_X^n) \cong \mathcal{E}^{n,n}(X)/\bar\partial \mathcal{E}^{n,n-1}(X)$$ where $\mathcal E(X)^{p,q}(X)$ is the space of $C^\infty$ forms of type $(p,q)$. So classes on the left are represented by forms $\alpha$ of type $(n,n)$. Given such a form $\int\alpha$ is a number. This is the trace.

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  • $\begingroup$ The original reference for this is iSerre's paper "Un théorème de dualité." in Commentarii Mathematici Helvetici 29 (1954) pp.9-26. In it, the dimension of $H^n(X, \Omega^n_X)$ is shown to have dimension 1 for a smooth compact complex manifold. $\endgroup$
    – Leo Alonso
    Commented Apr 26 at 17:05
  • $\begingroup$ a question about the example with finite field extension $L \subset K$ as function fields over some base field $k$ (...assume for sake of simplicity $K/L$ to be separable, otherwise the relative $\Omega_{K/L}$ might be nonzero). Then $\Omega_{L/k}$ and $\Omega_{K/k}$ are $L$- resp $K$-vspaces of same dimension $n \ge \text{trans}_k(L)$. Wouldn't the the induced map $f_* \Omega_{K/k} \to \Omega_{L/k}$ be given by usual trace map for finite field extensions, but componentwise? And in turn $f_*\omega_{K/k} \to \omega_{L/k}$ as $n$-product/wedge of trace maps? Or do I confusing something there? $\endgroup$
    – user267839
    Commented Apr 26 at 23:41

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