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Nov 29, 2017 at 9:52 comment added gdb Once you have an explicit description of Serre duality, you can, in principle, make your map to be explicit as well.
Nov 29, 2017 at 9:50 comment added gdb Did you try to look at 'analytic proof' of Serre duality? It is quite explicit and basically is given by $*$-operator. Though from this point of view it is not clear at all that this isomorphism is canonical.
Nov 25, 2017 at 9:50 answer added Sasha timeline score: 6
Nov 21, 2017 at 14:45 comment added David Loeffler @JasonStarr Thanks for the reference! Hartshorne only seems to consider the case where $X = \mathbf{P}^N$ for some $N$; but I guess the general case probably reduces to this.
Nov 21, 2017 at 11:36 history edited David Loeffler CC BY-SA 3.0
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Nov 21, 2017 at 10:54 comment added Jason Starr The suggestion by @Pooter is correct. The result is, roughly, Lemma III.7.4, p. 242, and Theorem III.7.11, p. 245, of Hartshorne's "Algebraic Geometry." For the pushforward sheaf $\iota_*\omega_Z$, there is a Yoneda-Ext class $a_\iota\in \text{Ext}^c_{\mathcal{O}_X}(\iota_*\omega_Z,\omega_X).$ Tensoring $\mathcal{G}$ with the associated $(c+1)$-term acyclic complex on $X$ again gives an acyclic complex (since $\mathcal{G}$ is flat). Chasing through connecting maps gives $\iota_*$ (up to a sign).
Nov 21, 2017 at 10:41 comment added Pooter Nice question. One more typo: $\omega_Y$ should be $\omega_Z$. One comment: writing $H$ for $G \otimes \omega_X$, the sheaf on $Z$ then becomes (if I calculated correctly) $i^*H \otimes \wedge^{top} N$ where $N$ is the normal bundle of $Z$ in $X$. I don't know if that helps.
Nov 21, 2017 at 10:40 comment added David Loeffler Yes, that's exactly what I'm hoping for.
Nov 21, 2017 at 10:26 comment added Martin Bright Looks a bit like a coherent version of the Gysin map.
Nov 21, 2017 at 10:24 history edited David Loeffler CC BY-SA 3.0
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Nov 21, 2017 at 10:01 history asked David Loeffler CC BY-SA 3.0