5

Let $\mathscr{C}$ and $\mathscr{D}$ be closed symmetric monoidal categories. We fix a strong symmetric monoidal functor $f^*:\mathscr{D}\to\mathscr{C}$ with a right adjoint $f_*:\mathscr{C}\to\mathscr{D}$. In this context, there is a formal isomorphism $\hom(Y,f_*X)\cong f_*\hom(f^*Y,X)$.

Now, suppose that we're given a second adjoint pair $(f_!,f^!)$ relating $\mathscr{C}$ and $\mathscr{D}$. It would be nice if we had a internal adjunction $\hom(f_!X,Y)\cong f_*\hom(X,f^!Y)$.

While this holds in most 6 functor formalisms, this doesn't follow formally from our suppositions.

In Isomorphisms between left and right adjoints, H. Fausk, P. Hu, and J.P. May affirm that it suffices to give the existence of one of the three arrows $$f_*\hom(X,f^!Y)\to \hom(f_!X,Y),\quad \hom(f^*Y,f^!Z)\to f^!\hom(Y,Z),\quad\pi:Y\otimes f_!X\to f_!(f^*Y\otimes X)$$ for all three to exist. Moreover, if one of them is an isomorphism, so are them all.

I wonder what condition holds in practice for us to have at least the existence of these morphisms.

Perhaps the fact that we usually have a morphism $f_!\to f_*$ suffices for us to construct these morphisms? (That's how we sometimes construct the projection formula. But we usually have that this morphism is injective and this doesn't hold for $D$-modules, for example.)

Perhaps a base change theorem suffices as in Ryan Reich's answer in Ubiquity of the push-pull formula? (This answer doesn't completely solve my problem since the formula $(g\times h)_!(X\boxtimes Y)\cong g_! X\boxtimes h_! Y$ is not clear to me as well.)

Gabriel
  • 943
  • What a priori relationship is there between $f^\dashv f_$ and $f^!\dashv f_!$? What are the "6 functors"? Is this question a matter of skill with diagram-chasing or does it depend on knowing algebraic geometry or something like that? – Paul Taylor Sep 20 '21 at 12:42
  • We suppose the existence of 3 pairs of adjunct functors (our 6 functors): $f^\dashv f_$, $f_!\dashv f^!$ (observe that $f^$ is the left adjoint but $f^!$ is the right adjoint), and $\otimes\dashv\hom$. A priori, there's no relation between $f^\dashv f_$ and $f_!\dashv f^!$, so there's no reason for the existence of the wanted morphism. What I'm asking if the existence of a morphism $f_!\to f_$ is enough. If not, what else can we do? (The latter question depends on the knowledge of the contexts which we're modelling, but not the former.) – Gabriel Sep 20 '21 at 12:50
  • I would also like to remark that $f^\dashv f_$ interacts with $\otimes\dashv\hom$ since $f^$ is supposed to be monoidal. This allows us to construct a morphism $Y\otimes f_X\to f_(f^ Y\otimes X)$, for example. – Gabriel Sep 20 '21 at 12:57
  • Do you know an example where $f^!$ is not just right adjoint to $f_*$? – Drew Heard Sep 21 '21 at 19:55
  • @DrewHeard The case where $f_=f_!$ (that is, $f^!$ is right adjoint to $f_$) is usually called the "Grothendieck context", since that happens in Grothendieck duality. But in Verdier duality, in the context of $D$-modules, in étale cohomology, and in many other cases we don't have $f_*=f_!$ for non-proper $f$. – Gabriel Sep 22 '21 at 08:16
  • 1
    @Gabriel In any case, if you are not aware of it (although you probably are) the paper by Balmer--Dell'Ambrogio--Sanders should be of interest (https://arxiv.org/pdf/1501.01999.pdf) - Equation 3.7 is what you want, although they are working under stronger conditions than what you want. But maybe you can find something useful in there anyway. – Drew Heard Sep 22 '21 at 08:26
  • (I got my adjunctions the wrong way around in the previous comment, I meant a case where $f_!$ is not right adjoint to $f_*$) – Drew Heard Sep 22 '21 at 08:29
  • @DrewHeard While indeed very interesting, I think that this paper doesn't solve my problem. Indeed, it's only in the "Verdier-Grothendieck context" (using the language of Fausk-Hu-May) that my desired morphism is not formal (from the hypothesis of the same article). And the paper by Balmer-Dell'Ambrogio-Sanders does not consider this case. – Gabriel Sep 22 '21 at 09:26
  • Even in coherent duality, it is not morally correct that $f_!=f_*$ unless the map $f$ is proper. Of course, the classical $f_!$ might not exist when $f$ is not proper, but this is corrected in Clausen-Scholze. I tend to believe that shrieks have their geometrical meaning and not just something as formal as stars. – Z. M Sep 25 '21 at 11:51
  • @Z.M I'm not an expert in this, but in Grothendieck duality things are weird for non-proper $f$. For example, the functor $f^!$ which interests us is only the right adjoint of $Rf_*$ when $f$ is proper. – Gabriel Sep 25 '21 at 12:04

1 Answers1

-1

Consider $f\colon X \to Y$ to be a smooth map between smooth complex projective varieties and $f_{*}, f_{!} \colon D(X) \rightleftarrows D(Y) \colon f^{!}$, $f^{*}$ be the associated derived functors on the level of bounded derived categories of coherent sheaves. Assuming also that $f$ is perfect, the following adjuctions hold: $f_{!}\dashv f^{*}\dashv f_{*}\dashv f^{!}$. The first (from left to right) is explained in https://arxiv.org/pdf/1004.3052.pdf, the remaining three in Huybrecht's book about Fourier Mukai kernels.

In this setup, holds $f_{*}(A\otimes B) \xrightarrow{\simeq} f_{*}(A \otimes f^{*}B)$, called projection formula. For a proof see Hartshorne's Residues and duality.

Christos
  • 55
  • Dear @Christos, I'm not supposing to be in the Grothendieck duality context. My question is purely categorical; I wonder which structure I must add in the formalism I've described in order to obtain the projection formula (or any other morphism described by Fausk-Hu-May). – Gabriel Sep 24 '21 at 19:56
  • Moreover, while I know that this is somewhat usual, I would advise against denoting the left adjoint of $fˆ$ by $f_!$. In the context of Grothendieck duality, the most natural functor to be called $f_!$ is $f_$. – Gabriel Sep 24 '21 at 19:57
  • @Gabriel I think that the paper of Fausk-Hu-May abstracted the various cases, and when I was reading it, was trying to find concrete examples in which this formalism can be applied. So, on the one hand, you need to specify suitable geometry (in order the projection formula map would be an iso.)

    As far as it concerns the left adjoint of $f^{}$, then this is defined as $f_{!}:=f_{}(f^{!}(\mathcal{O}_{Y}) \otimes -)$.

     – Christos Sep 24 '21 at 22:53
  • the "best" concrete examples of this formalism is Verdier duality, the six functors in étale cohomology, in D-modules (and, I think, in some motivic stuff). In the context of Grothendieck duality, we have that $f_!=f_*$ (in the notation of Fausk-Hu-May, which disagrees with yours), so this is somewhat a particular case. And in this particular case we have a formal projection formula. – Gabriel Sep 25 '21 at 08:38
  • Surely we need something more in order for the projection formula. I'm just asking what this may be. For example, perhaps it suffices to consider an axiom of proper base change? Or perhaps it suffices to give a morphism $f_!\to f_*$ (which is compatible with composition)? (The latter, as simple as it may seem, is already a very non-trivial result in the context of D-modules.) – Gabriel Sep 25 '21 at 08:39