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$\DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\hom}{\mathcal{Hom}} \DeclareMathOperator{\ox}{\mathcal{O}_X}$Let $f:X \to Y$ be a proper morphism. In section 6.4. of Liu's book he introduces the $r$-dualizing sheaf $\omega_f$ for $f$ which satisfies $$f_*\hom_{\ox}(\mathcal{F},\omega_f) \cong \hom_{\mathcal{O}_Y}(R^rf_* \mathcal{F},\mathcal{O}_Y )$$ for all quasi-coherent sheaves $\mathcal{F}$ on $X$.

In the special case of $f$ being finite he proves that the 0-dualizing sheaf is given by $f^!\mathcal{O}_Y = \hom_{\mathcal{O}_Y}(f_*\mathcal{O}_X,\mathcal{O}_Y)$ where this is considered an $\mathcal{O}_X$-module via multiplication into the argument.

Let $X$ be a proper, one-dimensional scheme over the field $k$ and let $Y = \mathbb{P}_k^1$. Let $\omega_f$ denote the 0-dualizing sheaf for $f$. Let $Z$ denote an irreducible component of $X$. Let $j: Z \to X$ denote the corresponding closed immersion. Then $f \circ j$ is again finite and we denote its dualizing sheaf by $\omega_{f \circ j}$.

My question is: How is the restriction of $\omega_f$ to $Z$ (via the pullback $j^*$) related to the 0-dualizing sheaf $\omega_{f \circ j}$ for the morphism $f \circ j$?

Are they isomorphic? In general, I don't think so. But maybe in specific cases they are. Do there exist canonical maps between them?

I am grateful for any kind of help.

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    $\begingroup$ You already wrote the answer in your post: $j_*\omega_{f\circ j}$ equals $\textit{Hom}_{\mathcal{O}_X}(j_*\mathcal{O}_Z,\omega_f)$ considered as a $j_*\mathcal{O}_Z$-module. The homomorphism $\mathcal{O}_X\to j_*\mathcal{O}_Z$ induces a homomorphism $j_*\omega_{f\circ j} \to \omega_f$. Typically these are not isomorphic. For $Z$ equal to a union of irreducible components of a strict normal crossings scheme, $j^*\omega_f$ is the "twist up" of $\omega_{f\circ j}$ by the Cartier divisor $D = Z\cap \overline{X\setminus Z}$. $\endgroup$ – Jason Starr Nov 27 '18 at 12:22
  • $\begingroup$ @JasonStarr Do you know any geometric criteria on which the relation depends? For instance, is the canonical morphism an isomorphism if the irreducible components only intersect transversally? I would really like to get a 'feeling' for the situation. $\endgroup$ – windsheaf Nov 27 '18 at 15:48
  • $\begingroup$ "For instance, is the canonical morphism an isomorphism if ..." No, it is usually not an isomorphism, even if there are just two irreducible components that intersect transversally in a divisor. You still need to "twist up" by the divisor before you get an isomorphism of $j^*\omega_f$ and $\omega_{f\circ j}$, i.e., $j^*\omega_f$ is isomorphic to $\omega_{f\circ j}(\underline{D})$. $\endgroup$ – Jason Starr Nov 27 '18 at 16:44

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