5
$\begingroup$

Let $X$ be a smooth projective algebraic variety over a field $k$, of dimension $n$, and let $Z$ be a smooth closed subvariety of dimension $m$, with $i: Z \hookrightarrow X$ the inclusion map.

For any locally free coherent sheaf $\mathcal{F}$ on $X$, there is a pullback map $$\imath^*: H^i(X, \mathcal{F}) \to H^i(Z, \iota^* \mathcal{F});$$ and via Serre duality we have isomorphisms $H^i(X, \mathcal{F})^\vee = H^{n-i}(X, \mathcal{F}^\vee \otimes \omega_X)$ and $H^i(Z, \iota^* \mathcal{F})^\vee = H^{m-i}(Z, \iota^* \mathcal{F}^\vee \otimes \omega_Z)$, where $\omega_X$ and $\omega_Z$ are the dualising sheaves. Setting $j=m-i$ and $\mathcal{G} = \mathcal{F}^\vee$, we conclude that there is a pushforward map $$\imath_*: H^j(Z, \iota^* \mathcal{G} \otimes \omega_Z) \to H^{j + c}(X, \mathcal{G} \otimes \omega_X),$$ for any $j$ and any locally free coherent sheaf $\mathcal{G}$ on $X$, where $c = n-m$ is the codimension of $Z$ in $X$.

Does this map have an intrinsic description (not using Serre duality)? Can it be defined without assuming that $X$ be projective, or that $\mathcal{G}$ be locally free?

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Looks a bit like a coherent version of the Gysin map. $\endgroup$
    – Martin Bright
    Nov 21, 2017 at 10:26
  • 2
    $\begingroup$ Yes, that's exactly what I'm hoping for. $\endgroup$
    – David Loeffler
    Nov 21, 2017 at 10:40
  • 3
    $\begingroup$ 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. $\endgroup$
    – Pooter
    Nov 21, 2017 at 10:41
  • 1
    $\begingroup$ 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). $\endgroup$
    – Jason Starr
    Nov 21, 2017 at 10:54
  • $\begingroup$ @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. $\endgroup$
    – David Loeffler
    Nov 21, 2017 at 14:45

1 Answer 1

Reset to default
5
$\begingroup$

The map is induced by the right adjoint $i^!$ of the pushforward functor and the adjunction morphism $i_*i^! \to \mathrm{id}$, in view of the formula $i^!(F) \cong i^*(F) \otimes \omega_{Z/X}[\dim Z/X]$. This works for any locally complete intersection closed embedding.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ In other words, the map is induced by the residue along $Z$. $\endgroup$
    – Leo Alonso
    Nov 25, 2017 at 10:30
  • $\begingroup$ @Sasha Defining the map as being "induced by the right adjoint of the pushforward functor" is essentially re-stating Serre duality. What I was hoping for was a direct description of $\iota_*$ (one not going via duality). $\endgroup$
    – David Loeffler
    Nov 25, 2017 at 13:35
  • $\begingroup$ @DavidLoeffler Not at all! Serre duality, as you mentioned, works only in projective setting, while the adjoint functor for a closed embedding always exists. Of course, when Serre functors exist you can express the right adjoint functor in terms of the left adjoint and duality isomorphisms, but I would say that right adjoint functor is more fundamental than Serre duality. $\endgroup$
    – Sasha
    Nov 26, 2017 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.