Genuine equivariant ambidexterity

Question

A particular case of Lurie and Hopkins' ambidexterity theory is that if $G$ is a finite group acting on a $K(n)$-local spectrum $X$ then the norm map $$ X_{hG} \to X^{hG} $$ is a $K(n)$-local equivalence. If we now replace $G$ with a compact lie group and $X$ with a genuine equivariant free $G$-spectrum (not necessarily $K(n)$-local), then the Adams equivalence asserts that the twisted norm map $$ X_{G} \to (\Sigma^{-ad(G)} X)^G $$ is an equivalence, where $ad(G)$ is the adjoint representation of $G$. One might then be led to ponder the possibility that these two phenomena are somehow part of the same thing. For example, one might consider the following (possibly overly optimistic) conjecture:

Conjecture: Let $G$ be a compact lie group and $X$ a genuine equivariant $K(n)$-local $G$-spectrum (not necessarily free). Then the twisted norm map $$ X_{G} \to (\Sigma^{-ad(G)} X)^G $$ is a $K(n)$-local equivalence.

Is this conjecture known to be true or false? I've had a hard time finding references which talk about genuine equivariance and ambidexterity at the same time, so any literature pointers which might be relevant to the above will be most appreciated.

Answer 1

This is to address Yonatan's question in the comments. Let $G$ be a finite group. To every (genuine) $G$-spectrum $E$, you can associate its (genuine) fixed point spectrum $E^{G}$. This earns its name by virtue of the following compatibility between stable and unstable homotopy theory: $\Omega^{\infty}( E^G ) = (\Omega^{\infty} E)^{G}$.

Now you could ask, could we have some categorically dual gadget $E \mapsto E_{G}$, which satisfied the dual condition $\Sigma^{\infty}_{+} (X/G) = (\Sigma^{\infty}_{+} X)_{G}$ for $X$ a $G$-space? The answer is no, because the construction on the left hand side $X \mapsto \Sigma^{\infty}_{+} (X/G)$ is functorial with respect to maps of $G$-spaces, but not with respect to stable maps of $G$-spaces.

For example, any correspondence of $G$-spaces $X \leftarrow M \rightarrow Y$ where $M$ is a finite covering space of $X$ determines a map of $G$-spectra from $\Sigma^{\infty}_{+}(X)$ to $\Sigma^{\infty}_{+}(Y)$. But there's no functorial procedure to extract from $M$ a map of spectra $\Sigma^{\infty}_{+} (X/G) \rightarrow \Sigma^{\infty}_{+}(Y/G)$. You might say: why not use the correspondence of nonequivariant spaces given by $X/G \leftarrow M/G \rightarrow Y/G$? This does not respect composition of correspondences (observe what happens when you compose the correspondence $\ast \leftarrow G \rightarrow \ast$ with itself).

Answer 2

There's something that goes wrong even when one works strictly in the Borel-equivariant case but for compact Lie groups.

One equivalent form of the vanishing of the Tate construction is the following. Consider the $\infty$-category $\mathrm{Fun}(BG, \mathrm{Sp})$ of spectra equipped with a $G$-action (equivalently, the $\infty$-category of $G$-spectra which are "Borel-complete" in the sense that their fixed points for every choice of subgroup is equivalent to the homotopy fixed points).

Suppose first that $G$ is finite. Fix an object $X \in \mathrm{Fun}(BG, \mathrm{Sp})$. There is a natural norm map $X_{hG} \to X^{hG}$, whose cofiber is the Tate construction $X^{tG}$. There are certain choices of $X$ which will force this to be an equivalence: for example, if $X$ is induced from a spectrum $Y$, so that $X = G_+ \wedge Y$ (with the $G$-action on the first factor). More generally, any object of $X \in \mathrm{Fun}(BG, \mathrm{Sp})$ which belongs to the thick subcategory (i.e., smallest stable subcategory closed under retracts) will have this property, so that the Tate construction vanishes.

Let us call this the category of nilpotent objects in $\mathrm{Fun}(BG, \mathrm{Sp})$. Any nilpotent object has vanishing Tate construction. (Conversely, a ring object with vanishing Tate construction is also nilpotent.) Over the rational numbers, everything is nilpotent as any group action is a retract of an induced one via a transfer construction. There are also nilpotent objects that are more subtle: for example, the $C_2$-action on complex $K$-theory is nilpotent. (The theory of nilpotent objects in the category of genuine $G$-spectra with respect to a family of subgroups is treated in some detail in two joint papers with Naumann and Noel, see here and here, and is related to phenomena such as Quillen stratification in mod $p$ cohomology.)

The $K(n)$-local vanishing of Tate spectra (for finite groups) is equivalent to the following statement: take $K(n)$ with the trivial $G$-action. Then it is nilpotent. (This is due to Greenlees and Sadofsky.)

One can ask the same question when $G$ is now assumed to be a compact Lie group, even a circle: given a ring spectrum with trivial $S^1$-action, is it nilpotent? (That is, does it belong to the thick subcategory of the $\infty$-category $\mathrm{Fun}(BS^1, \mathrm{Sp})$ generated by the induced objects?) Any such object would have the analogous transfer maps be equivalences. However, the answer is never in this case. One way to see this is (e.g., for $R = \mathbb{Q}$ or $K(n)$) is that any $S^1$-action on a spectrum $X$ leads to a homotopy fixed point spectral sequence for computing $\pi_* X^{hS^1}$, and a consequence of nilpotence is that this must degenerate with a horizontal vanishing line at some finite stage. This doesn't happen simply because this spectral sequence (the Atiyah-Hirzebruch spectral sequences) collapses at $E_2$ but with a polynomial class in filtration two. (Note that while the Morava $K$-theory of $BG$ is a finite-dimensional graded vector space when $G$ is finite, but not if $G$ is compact Lie.)