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Let $k$ be a field, $X$, $Y$, $Z$ smooth geometrically connected curves, and $f: Z \to X$, $g : Z \to Y$ finite morphisms. Suppose that $f$ is separable.
Then we have $f_* \circ g^* : \Gamma(Y, \Omega_Y) \to \Gamma(Z, \Omega_Z) \to \Gamma(X, \Omega_X)$.

By definition, the Hecke operator is this map for certain morphisms.

The definition of $f_*$ is as follows:
Now $\Omega_{Z, \zeta} = \Omega_{X, \xi} \otimes_{k(X)} k(Z)$. Using this, $f_*$ is defined as $\Gamma(Z, \Omega_Z) \to \Omega_{Z, \zeta} = \Omega_{X, \xi} \otimes_{k(X)} k(Z) \to \Omega_{X, \xi}$, where $\Omega_{X, \xi} \otimes_{k(X)} k(Z) \to \Omega_{X, \xi}$ is $1 \otimes \operatorname{tr}$.

But why this map factors through $\Gamma(X, \Omega_X)$? I've found a reference (Zannier - A note on traces of differential forms (MSN)), but its proof is too long. I wonder whether we can give a more elementary proof (using $\dim = 1$).

And I've heard that this $f_*$ is the Serre dual of $H^1(X, \mathscr{O}_X) \to H^1(Z, \mathscr{O}_Z)$. Is this true? And how can I show these two maps are the same?

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  • $\begingroup$ You say "The definition of $f_*$ is: Now $\Omega_{Z, \zeta} = \Omega_{X, \xi} \otimes_{k(X)} k(Z)$." Is that intentional? $\endgroup$
    – LSpice
    Commented Apr 22, 2020 at 18:37
  • $\begingroup$ @LSpice It's because of lack of my English skills. I edited it. And thank you for editing. $\endgroup$
    – k.j.
    Commented Apr 22, 2020 at 19:12
  • $\begingroup$ The article of Zannier is incorrect. $\endgroup$
    – Jason Starr
    Commented Apr 22, 2020 at 19:59
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    $\begingroup$ It's a pity that neither the journal web-page for Zannier's paper, nor the MathSciNet review, has any kind of note attached acknowledging the error. $\endgroup$
    – David Loeffler
    Commented Apr 23, 2020 at 7:24
  • $\begingroup$ @JasonStarr So does not the map $f_*$ factor through $\Gamma(X, \Omega_X)?$ $\endgroup$
    – k.j.
    Commented Apr 23, 2020 at 10:53

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