Comparison: Formal Wirthmüller isomorphism of Fausk-Hu-May vs. Balmer et. al

Question

$\newcommand{\Cc}{\mathcal{C}}$ $\newcommand{\Dd}{\mathcal{D}}$ $\newcommand{\tensor}{\otimes}$ $\DeclareMathOperator{\Sp}{Sp}$ This question is about comparing the approaches for a formal Wirthmüller isomorphism by Fausk-Hu-May [FHM] in isomorphisms between left and right adjoints and by Balmer-Dell'Ambrogio-Sanders [BDS] in Grothendieck-Neeman duality and the Wirthmüller isomorphism.

In both articles, a formal Wirthmüller isomorphism is studied: we are given a tensor-exact functor $f^*: \Dd \to \Cc$ between tensor-triangulated categories that has both a left adjoint $f_!: \Cc \to \Dd$ and a right adjoint $f_*: \Cc \to \Dd$ and we are looking for a `twisted isomorphism' between $f_!$ and $f_*$. In [FHM] this takes the form of an isomorphism $f_*(-) \overset{\cong}{\implies} f_!(- \otimes C)$ for some object $C \in \Cc$, and in [BDS] this takes the form of an isomorphism $f_*(- \tensor \omega_f) \overset{\cong}{\implies} f_!(-)$ for another object $\omega_f \in \Cc$.

In [BDS] a comparison is given with the approach of [FHM]: it is stated in proposition 4.4 that if the situations of [BDS] and [FHM] both apply simultaneously, then the object $\omega_f$ is the dual in $\Cc$ of the object $C$. However, I don't understand the rather short proof: it is claimed that the formal assumptions already give an isomorphism $f_* \cong f_!(- \tensor C)$ by [FHM, Thm. 8.1]. But it seems to me that [FHM, Thm. 8.1] requires some non-formal input, namely one needs to explicitly check that $f_*(G) \to f_!(G \tensor C)$ is an isomorphism for some set of generating compact dualizable objects $\{G\}$ of $\Cc$. Indeed, it takes May in the follow-up article The Wirthmüller isomorphism revisited quite some time to prove that the latter condition is satisfied in the setting of equivariant homotopy theory. It seems that Proposition 4.4 of [BDS] is cutting this short.

Question 1: Why does it follow in [BDS,Prop. 4.4] that $f_*(-) \implies f_!(- \otimes C)$ is a natural isomorphism?

Question 2: If it doesn't follow, is it otherwise possible to compare $C$ and $\omega$ without checking that $f_*(-) \implies f_!(- \otimes C)$ is a natural isomorphism on all compact generators?

Let's assume that we know that both isomorphisms hold and additionally that $\omega_f$ is invertible (with inverse $C$). Then both approaches give a natural isomorphism $f_*(-) \implies f_!(- \otimes C)$.

Question 3: (Why) do these two isomorphisms agree?

Answer 1

Thanks for pointing this out and sorry for the confusion. Proposition 4.4 of our paper does appear to be misstated. The meaning of "Wirthmüller context in the sense of [FHM]" needs to be strengthened to include the isomorphism that [FHM, Theorem 8.1] doesn't provide for free. The Proposition could be (correctly) stated as: "Suppose the basic adjunction $f^* \dashv f_*$ satisfies a Wirthmüller isomorphism in the sense of [FHM]; i.e., suppose that $f^*$ has a left adjoint $f_!$ and that there exists an object $C \in \mathcal C$ together with an isomorphism $f_!(C\otimes -)\simeq f_*(-)$. Then its dual $\mathrm{hom}(C,1)$ is isomorphic to $\omega_f$."

The point is that we can recognize the dualizing object from existing isomorphisms that have been established in specific examples by specific non-formal means. (See e.g., Remark 2.16 of my other paper, "The compactness locus of a geometric functor and the formal construction of the Adams isomorphism".)

In any case, you are correct that there appears to be no reason why the choice of isomorphism $f_!(C)\simeq f_*(1)$ would imply an isomorphism $f_!(C\otimes -)\simeq f_*(-)$. Indeed, this is one of the points about these works --- the $C$ that appears in [FHM] is non-canonical and the isomorphism of Theorem 8.1 doesn't come for free; in contrast, the relative dualizing object is canonical and the isomorphism of [BDS] comes essentially for free (in our rigidly-compactly generated setting).

Also, the real point of our Remark 4.3 and (misstated) Proposition 4.4 is to show that the Wirthmüller isomorphism (à la [FHM]) implies the Grothendieck-Neeman isomorphism, contrary to the discussion in [FHM], about these being distinct settings. I think this is one of the things that was clarified by our paper.

Finally, as you mention (in your answer) the example from May's second paper doesn't exactly fit into our setup since the equivariant stable homotopy category over the $N$-fixed universe is not rigidly-compactly generated --- not all of the generators are dualizable. Nevertheless, one can give an example of an $f^*:\mathcal D \rightarrow \mathcal C$ between rigidly-compactly generated categories which admits an invertible object $C\in \mathcal C$ such that $f_!(C) \simeq f_*(1)$ and yet $\mathrm{hom}(C,1) \not\simeq \omega_f$. I'll give this example as an answer to your other MathOverflow question. This will thus provide an example (fitting into our setup) where the associated [FHM] map $f_* \rightarrow f_!(C\otimes -)$ is not an isomorphism.

Answer 2

$\newcommand{\Cc}{\mathcal{C}}$ $\newcommand{\Dd}{\mathcal{D}}$ $\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\tensor}{\otimes}$

Let me write down what I (think I) know so far.

Answer to 1: I think that this does not follow, i.e. that the additional assumptions that $f^*$ has a left adjoint $f_{(1)}$ (denoted $f_!$ by [FHM]) and that there is an object $C \in \mathcal{C}$ with $f_{(1)}(C) \simeq f_*(\mathbb{1}_{\mathcal{C}})$ are not enough to guarantee that the map $f_* \to f_{(1)}(- \otimes C)$ constructed in [FHM] is an equivalence. In fact, it seems that May has explicitly written down a counterexample to this statement in section 2 of `The Wirthmüller isomorphism revisited'.

(I should mention though that I think not all compact generators are strongly dualizable in May's example, so not all conditions of [BDS] are satisfied. Still, [BDS] don't give an indication why their extra conditions would make a difference.)

Edit on 25.03.22: For a counterexample in which the compact generators are strongly dualizable, see Beren Sander's answer below (in which he refers to my other MO-post mentioned in the next paragraph).

Answer to 2: Because of section 2 of May's article mentioned above, I expect that the condition $f_*(\mathbb{1}_{\mathcal{C}}) \simeq f_!(C)$ does not uniquely specify $C$, but I would like to see a counterexample. I have asked this as a separate question.

Answer to 3: These isomorphisms do indeed agree after the identification $D(C) \simeq \omega_f$. Explicitly, letting $\eta: 1_{\Dd} \to f_!C$ denote the structure map of $C$ as in [FHM], this identification is given by $$D(C) \to f^{(1)}f_*(DC) \simeq f^{(1)}D(f_!C) \xrightarrow{\eta^*} f^{(1)}D(1_{\Dd}) = \omega_f,$$ and in particular the counit $f_*\omega_f \to 1_{\Dd}$ becomes under this equivalence simply $$f_*D(C) \simeq D(f_!C) \xrightarrow{\eta^*} D(1_{\Dd}) = 1_{\Dd}.$$ We should thus compare the following two maps: $$f_*(X \tensor DC) \xrightarrow{\eta} f_*(X \tensor DC) \tensor f_!C \simeq f_!(f^*f_*(X \tensor DC) \tensor C) \to f_!(X \tensor DC \tensor C) \to f_!(X)$$ and $$f_*(X \tensor DC) \to f_*(f^*f_!X \tensor DC) \simeq f_!X \tensor f_*DC \simeq f_!X \tensor D(f_!C) \xrightarrow{\eta^*} f_!X,$$ where are the ommited maps are either adjunction units/counits or left/right projection formulas. One can now write down a huge commuting diagram that shows the equivalence of these two maps.