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The Grothendieck-Verdier duality: $$ Rf_*\big(R\mathcal{H}\textit{om}_X^\bullet(\mathcal{E}^\bullet,f^!\mathcal{F}^\bullet)\big) \cong R\mathcal{H}\textit{om}^\bullet_Y(Rf_*\mathcal{E}^\bullet,\mathcal{G}^\bullet) $$ is known to hold for $f:X\to Y$ being a proper map of noetherian schemes.

Is there a way to get rid of this requirement on schemes to be noetherian, possibly by putting some extra conditions on the morphism $f$?

For example: let $X$ be smooth (projective in necessary) and consider the projection to the second factor $f:X\times U\to U$, where $U$ is some affine scheme. Does one have the duality in this setting?

Edit: After reading about the topic in Neeman's and Lipman's work, I have not managed to find an explicit construction of the right adjoint $f^\times$. In what cases is the explicit construction of $f^\times$ known? What would it be in the above example?

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    $\begingroup$ The name Grothendieck-Verdier is wrong. Grothendieck studied duality for complexes of coherent sheaves on schemes while Verdier developed duality for constructible sheaves on locally compact spaces. I understand they are different theorems and should be named appropriately. $\endgroup$
    – Leo Alonso
    Oct 14, 2019 at 8:44
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    $\begingroup$ I just used the name that Huybrecht has given the standard formulation of the theorem. $\endgroup$
    – Arkadij
    Oct 14, 2019 at 10:17

1 Answer 1

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Does one have the duality in this setting?

Yes, we do have duality in a very general setting. Your question is equivalent to asking for the existence of a right adjoint to the derived pushforward functor $\mathbf{R}f_*\colon \mathbf{D}_{qc}(X) \to \mathbf{D}_{qc}(Y)$. It turns out that it exists for any morphism $f\colon X \to Y$ of qcqs schemes. Look at tag 0A9D for more details.

P. S. It is not a good idea to call this adjoint functor by $f^!$ unless $f$ is proper. Usually, another functor is denoted by $f^!$ that actually differs from the right adjoint $f^\times$ (SP denotes this functor by $a(f)$).

UPD: If $Y$ is qcqs and $f\colon X \to Y$ is proper and smooth of (pure) relative dimension $d$, then $f^!(-)\cong f^{\times}(-)\cong \Omega^d_{X/Y}[d]\otimes_{\mathcal O_X} f^*(-)$. Look at tag 0BRT and tag 0A9U for a proof in the noetherian case. The proof in the non-noetherian setup is essentially the same. Just note that noetherianness of $Y$ is used only to show that $\mathbf{R}f_*$ preserves perfect objects. However, it is true without this assumption as it is briefly explained in Example 2.2 of the paper Quasi-perfect scheme-maps and boundedness of the twisted inverse image functor.

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  • $\begingroup$ In the example I mention, is the functor $a(f)$ given in the usual way that one would expect? That is, by derived pullback and tensoring with the canonical bundle on $X$ shifted by $\text{dim}(X)$? $\endgroup$
    – Arkadij
    Oct 13, 2019 at 8:06
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    $\begingroup$ @Arkadij Bojko That's correct if we assume that $f$ is proper. I've updated my answer. $\endgroup$
    – gdb
    Oct 14, 2019 at 4:45

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