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I'm trying to learn a little about Grothendieck duality. One version of the theorem states that if $f: X \to Y$ is a proper morphism of schemes, then the induced functor on derived categories $f_*: D^+(\mathrm{QCoh}(X)) \to D^+(\mathrm{QCoh}(Y))$ has a right adjoint $f^!$ (and under nice hypotheses, these will preserve the subcategories with coherent cohomology). The existence of an adjoint can be proved via adjoint functor arguments, even without assuming $f$ proper; this was done, e.g., by Neeman. (The point is that a triangulated functor between nice triangulated categories (or, stable $\infty$-categories) which preserves coproducts is a left adjoint.)

However, in trying to identify $f^{!}$, we might want to be able to localize on $X$ and $Y$, and thus deal with the non-proper case. My understanding is that the upper-shriek functor $f^{!}$ there is not supposed to be the right adjoint to $f_{\ast}$: for example, for an open immersion it should be the upper-star $f^*$.

In the topological version, one can define a $f_{!}$ functor for sheaves (push-forward with compact support) and $f^!$ is the right adjoint to $f_{!}$. Is there any "functorial" way to interpret $f^{!}$ when $f$ is not proper?

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  • $\begingroup$ I'm afraid there's not really more to say than you've said. You have descriptions for open embeddings and proper morphisms and that's enough to describe it in general (say, for schemes finite type over a field). But $j^*=j^!$ doesn't commute with products (for $j$ an open embedding) and therefore doesn't have a left adjoint. Do you have a more precise question in mind? $\endgroup$
    – Moosbrugger
    Commented Aug 3, 2012 at 1:58
  • $\begingroup$ Not really. I was just curious if there was any way of making the formalism clearer (e.g., a way of defining this functor which would generalize nicely to a derived setting). $\endgroup$
    – Akhil Mathew
    Commented Aug 3, 2012 at 2:26
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    $\begingroup$ Doing it right is a non-trivial issue, discussed in some detail e.g. in arxiv.org/abs/1105.4857. But note that just defining it naively is easy: Nagata compactification exists for derived schemes as well. The analysis that this (homotopy) doesn't depend on the compactification essentially follows from the analysis in Section 6.2.2 of loc. cit. $\endgroup$
    – Moosbrugger
    Commented Aug 3, 2012 at 3:07

2 Answers 2

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Classically, the functor $f^!$ is indeed not a right adjoint in general. Clausen and I have recently found a way to make it a right adjoint in general, by enlarging the category of modules to that of solid modules, and constructing a general $f_!$ functor on solid modules directly. Solid modules are a version of "completed topological modules", but with excellent categorical properties. As an example, for $f: \mathbb A^1 =\mathrm{Spec} \mathbb Z[T]\to \mathrm{Spec}\mathbb Z$, one has $$ f_! \mathcal O_{\mathbb A^1} = \left[\mathbb Z[T]\to \mathbb Z((T^{-1}))\right], $$ the idea being that one wants to look at those functions $f\in \mathbb Z[T]$ that vanish to all orders at $\infty$, i.e. lie in the kernel of $\mathbb Z[T]\to \mathbb Z((T^{-1}))$. This kernel is of course trivial, but there's an interesting cokernel, so it gives an interesting functor on the derived level. And it's really necessary to consider $\mathbb Z((T^{-1}))$ not as an abstract $\mathbb Z$-module, but with its topological (or rather condensed) structure. See here for an account of this approach to coherent duality.

(I should say that a version of this is due to Deligne in the appendix to Hartshorne "Residues and Duality", working with pro-coherent sheaves.)

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  • $\begingroup$ Given a map of (finitely generated) rings, you have the universal compactification given by the discrete Huber pair. Given a map of discrete Huber pairs, seemingly Huber's argument will also give a discrete Huber pair as the universal compactification, but sorry for being a beggar: do you have a very simple way to write this? More generally, do you have the "universal compactification" for all (reasonable) maps of analytic (animated) rings? $\endgroup$
    – Z. M
    Commented Apr 12, 2021 at 20:27
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    $\begingroup$ For a map $(\mathcal A,\mathcal M)\to (\mathcal B,\mathcal N)$ of analytic animated rings, you can always factor it over the induced analytic ring structure $(\mathcal A,\mathcal M)\otimes_{\mathcal A} \mathcal B$, which plays the role of the "universal compactification". For Huber pairs $(A,A^+)\to (B,B^+)$, it just amounts to taking the pair of $B$ and the integral closure of $A^+$ in $B$. $\endgroup$
    – Peter Scholze
    Commented Apr 13, 2021 at 7:44
  • $\begingroup$ Thanks. In this general context $(\mathcal A,\mathcal M)\to(\mathcal B,\mathcal N)$, let $(B,\mathcal M_{\mathcal B})$ denote the object that you get. Then do we have the following analogue: a compact idempotent commutative algebraic object $\mathcal B_{\infty/A}$ of which the category of modules coincides with the kernel of $D(\mathcal B,\mathcal M_{\mathcal B})\to D(\mathcal B,\mathcal N)$, and the similar 6-functor formalism for all analytic animated rings? $\endgroup$
    – Z. M
    Commented Apr 13, 2021 at 18:09
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    $\begingroup$ This does not happen in general; for example not for $\mathbb Z\to \mathbb Z_\blacksquare$, but also in some more geometric situations; asking for the existence of $\mathcal B_\infty$ is like asking for the map on analytic spectra to be an open immersion, roughly; this need not be the case, for example the map might only be pro-open. But it's a reasonable condition, satisfied in many cases of interest, and some statements ought to extend to this general context. $\endgroup$
    – Peter Scholze
    Commented Apr 13, 2021 at 20:53
  • $\begingroup$ Thanks @PeterScholze for this answer! (I had forgotten having asked this question...) I also wondered if there might be an analog of your and Clausen's approach for Grothendieck duality via six functors on solid modules in prismatic (or de Rham) cohomology? $\endgroup$
    – Akhil Mathew
    Commented Jun 20, 2021 at 11:12
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I don't have an answer, but maybe these notes by Lipman help.

From what I understand the upper pling functor $f^!$ is a sort of Frankenstein, definitely for etale maps it's given by ordinary pullback. More generally for smooth maps you have to tensor with the relative canonical bundle.

The only description I know of is via dualising complexes, but that's perhaps not categorical enough for what you want.

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  • $\begingroup$ The ! is probably the gadget with more pronounciations in all of current notation :-) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Aug 2, 2012 at 23:36
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    $\begingroup$ Isn't that symbol used to represent the "click" sound in click languages? $\endgroup$
    – Lee Mosher
    Commented Aug 2, 2012 at 23:55
  • $\begingroup$ Tyler Lawson suggested the pronunciation "surprise", because you should be (pleasantly) surprised that the functor exists. $\endgroup$
    – S. Carnahan ♦
    Commented Aug 3, 2012 at 8:54
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    $\begingroup$ When I first heard someone say "shriek" for "exclamation point" I broke out laughing (certainly interrupting an explanation they were giving). It's such an evocative word, which one easily forgets as a mathematician repeating it frequently (e.g., the person explaining it to me had never quite thought about the word choice). I also appreciated that in our science we have a much more poetical term for this punctuation mark than in poetry. $\endgroup$
    – Moosbrugger
    Commented Aug 3, 2012 at 12:50
  • $\begingroup$ Jacob, maybe it would be worth writing down the definition by $f^*$ and dualizing complexes for others to see? $\endgroup$
    – Karl Schwede
    Commented Aug 3, 2012 at 17:49

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