@MoBehzadKang I’m not certain, but I don’t think much is known. Some adjacent results pertinent to the real case include Carrick’s analysis of smashing locations in equivariant homotopy theory arxiv.org/abs/1909.08771 & Achim Krause’s analysis of self-maps of compact motivic spectra in his PhD thesis, but even those are a step or two removed

I still don't understand equivariant classifying spaces.

@EricPeterson Thank you, Eric! I suspected the current state of results is what you've stated. Andrew Salch mentioned it to me today as an open problem.

@TylerLawson Nice to meet you! I heard that you once maintained a web page of open problems in algebraic topology. Is this still available somewhere? Thanks in advance.

Let $G$ be a Hopf algebra object in a stable $\infty$-category $(C,\wedge, S)$ which is dualizable as an object of $C$. Then $G = S^{ ad G} \wedge G^\vee$ for some $\wedge$-invertible $L$. Rognes shows this in a certain context with $S^{ad G} = G^{hG}$. Moreover, there's a certain canonical map $S^{ad G} \to S$; if this map is an isomorphism (or more generally if $S^{ad G} = S$, then $G$ is self-dual.

In particular, when $C$ is the $K(n)$-local category, and $G = L_{K(n)} \Sigma^\infty_+ K(\mathbb Z/p, m)$, Rognes shows that the map $S^{ad G} \to S$ is an equivalence when $m=n$ (so that $K(\mathbb Z/p, n)$ is $K(n)$-locally self-dual) but not for $m<n$. However, we know that $K(\mathbb Z / p, m)$ is $K(n)$-locally self-dual for all $m$ (by the yoga of ambidexterity, but probably also in other ways).

This raises the question: is it the case that $S^{ad K(\mathbb Z/p, m)} = S$ for $m < n$ via some other map? Or is $S^{ad K(\mathbb Z/p,m)}$ rather a nontrivial Picard element, and $K(\mathbb Z/p,m)$ is just a nontrivial fixed point of smashing with it?

If it is the case that $S^{ad K(\mathbb Z / p, m)} = S$, then the canonical map to $S$ is an endomorphism of $S$ which is not an automorphism. Which endomorphism is it?

Sorry, "$L$" above was meant to be "$S^{ad G}$"

And $C$ should of course be symmetric monoidal

What's the definition of "$G$ is self-dual"?

Is that Spanier-Whitehead duality?

I just mean that there's an isomorphism between $G$ and $G^\vee$. Yeah, spanier-whitehead

Since there is a canonical map which is an isomorphism, this is saying that it is canonically self-dual?

In the case $m=n$, there's a canonical map which is an isomorphism.

In the case $m<n$, I suppose that ambidexterity theory also gives a canonical map

but I don't know how it relates to this picture

So I'm not asking for any kind of canonicity in the self-duality, but it does seem to be there in interesting cases when it happens.

Are ambidexterity and Rognes producing the same self-duality map when m=n?

I wish I knew!

But when $m\neq n$, $S^{adG}$ is still some kind of invertible object?

Yeah, its inverse is given by the dual construction $(G^\vee)_{hG}$

It's invertible for any dualizable $G$

So for a given $G$, there is some subgroup of the Picard group consisting of $T$ such that there exists an equivalence $G=T\otimes G^\vee$.

Or rather a coset

Which is not trivial in general by what you are telling me?

A coset which contains the identity element if $G$ is self-dual?

Oh -- right -- by the ambidexterity stuff

yeah, by the ambidexterity stuff, $G$ is self-dual in these cases

So Rognes is producing a potentially nontrivial canonical element of this subgroup of the picard group

where "these cases" means $G = L_{K(n)} \Sigma^\infty_+ K(\mathbb Z/p, m)$ in the $K(n)$-local category.

Right.

When $n=m=1$ what is the subgroup?

That's probably a good question to start with.

If $G$ is the trivial group, it seems like the coset only contains one element. Is that right?

Sure -- in that case $G$ is stably $S$

At the prime 2 and height 1, $\Sigma^\infty_+K(Z/2,1)$ is just $S\oplus S$.

Ok -- so in that case the coset must be trivial.

So it's coset also just has one element, although it can be self-dual it a lot of different ways.

Yeah -- and this is consistent with Rognes' result for the $m=n$ case -- we must have $S^{ad G} = S$ here.

I need to read Hopkins-Mahowald-Sadofsky since it occurs to me I don't know how to show a $K(n)$-local Picard element is not torsion, given that the category of $K(n)_\ast$-modules is periodic

Compute the action of $\mathbb{G}_n$ on $E_*T$, where $E$ is Morava $E$_theory.

hm

Basically, when you do this, you tend to get $E_*T\approx \omega^a\otimes \mathrm{det}^b$, where $\omega=E_*S^{-2}$ and $\mathrm{det}$ is the determinant character. If $n\geq2$ and either $a$ or $b$ are non-zero, then it's a non-torsion element.

ok

$a$ and $b$ can be elements of certain completions of $\mathbb{Z}$.

The $K(n)$-local picard group seems to have a natural topology. I wonder if this coset / subgroup stabilizing $G$ is closed.

The topology is induced by tensoring up to various $E^H$, where $H$ ranges over open subgroups of $\mathbb{G}_n$, I think.

I.e., $Pic(S)$ is approximately the inverse limit of the $Pic(E^H)$, and the map $Pic(S) \to \mathrm{lim} Pic(E^H)$ gives the topology.

Hang on. The formula Rognes gives is $S^{ad G} = G^{hG}$. But $K(n)$-locally, this is the same as $G_{hG}$. Which should just be $S$.

Using ambidexterity with respect to $G$

So I think the answer is that indeed $S^{ad K(\mathbb Z/p,m)} = S$ for all $m$, $K(n)$-locally. But it must be via a different map than Rognes' map, at least for $m \neq n$.

@CharlesRezk what's wrong with equivariant classifying spaces?

@DylanWilson Ordinary classifying spaces are simply a consequence of the straightening/unstraightening principle, i.e., that $Fun(X, \mathcal{S})=\mathcal{S}_{/X}$.

Explicitly: the diagonal map $BG\to BG\times BG$, viewed as a morphism in $\mathcal{S}_{/BG}$, corresponds to $EG\to BG$, viewed as a morphism in $Fun(BG,\mathcal{S})$.

I don't know how to describe equivariant classifying spaces by an analogous principle, even accepting that it would need to be more complicated.

Part of the problem is that the total space $P$ of an equivariant principle bundle $P\to X$ is not really in the same category as $X$.

@MoBehzadKang It is still here: topology-octopus.herokuapp.com

Beyond the initial couple of months, however, I haven't heard much from anyone about it. I do know that some of the problems involved have been solved.

@CharlesRezk so I want to describe equivariant principle bundles as doing this: they are Sigma_n-equivariant colimit-preserving functors (G-spaces)^n -> (spaces).

Principal $\Sigma_n$-bundles?

If you write (G-spaces) = Spaces ⊗ O_G, then I think you unravel that this functor category is the category of functors (O_G)^n_{hSigma_n} -> Spaces, and the former is a full subcategory of the orbit category of Sigma_n wreath G (spanned by the subgroups of the form (H_1 x ... x H_n) of G^n).

I'm getting confused about whether there are supposed to be some "op"s in here though.

Yeah, some kind of principal Sigma_n-G-bundles.

That kinda makes sense, though it's not exactly in the direction I was thinking.

I'd like to understand this better also.

But I'm terrible at thinking unstraightened-ly.

So $B(\Sigma_n \wr G)$ is the category of "principal $G$-bundles over spaces which are equivalent to $\{1,\dots,n\}$".

So maybe we are looking at $\Sigma_n$-equivariant principal $G$-bundles?

Yeah. But the isotropy is being constrained so that Sigma_n doesn't really have any

@CharlesRezk @TylerLawson right, so there is an "indexed" version of all this (a la Barwick-Dotto-Glasman-Nardin-Shah). Let's say a G-category is a presheaf of infinity-categories on O_G. This infty-category of G-categories is cartesian closed, and has an internal Hom which we can write as $\underline{\mathsf{Fun}}_G$. Now let $\underline{\mathsf{Spaces}}_G$ be the G-category given by G/H--> $\mathsf{Psh}(\mathrm{Orbit}(G)_{/G/H})$. Then (un)straightening still works: there is an equivalence

$\underline{\mathsf{Fun}}_G(X, \underline{\mathsf{Spaces}}_G) \simeq (\underline{\mathsf{Spaces}}_G)_{/X}$ where

$X$ is a $G$-$\infty$-groupoid (viewed simultaneously as a special type of G-category and as a global section of the cocartesian fibration classified by $\underline{\mathsf{Spaces}}_G$; the overcategory is then formed pointwise)

in particular, evaluating on the orbit G/G, we get $\mathsf{Fun}_G(X, \underline{\mathsf{Spaces}}_G) \simeq (\mathsf{Spaces}_G)_{/X}$

But now it's probably helpful to recognize $\underline{\mathsf{Spaces}}_G$ as `cofree' so that $G$-functors from $X$ to it are the same as $\mathsf{Fun}(\mathrm{colim}^{\mathrm{l.lax}}(X), \mathsf{Spaces})$

(i.e. functors from the total category of the left fibration classified by $X$ to the category of spaces)

this is, I think, what Tyler was trying to unravel?

anyway, you can now apply this to your favorite examples of $G$-$\infty$-groupoids. e.g. the $G$-$\infty$-groupoid given by the functor $G/H \mapsto$ {groupoid of $G$-equivariant covers of $G/H$ with fibers of cardinality $n$}

and other such stuff

to get at $B_G\Gamma$, you can use the functor $G/H \mapsto$ {groupoid of $G$-equivariant principle $\Gamma$-bundles on $G/H$}, i.e. $G\times\Gamma$-equivariant maps to $G/H$ which present the target as the quotient and where $\Gamma$ acts freely on the source.

I'm not sure. I don't think I wanted G fixed. I think I wanted G and H separate groups, and to look at G-(principal H-bundles), which I think are usually described with some extra isotropy condition (they have GxH action and H acts freely). So maybe I'm after a classification of colim-preserving functors from naive H-spaces to genuine G-spaces?

Shoot, I think you just beat me to expressing that.

I don't think that'll do because the 'constant' G-groupoid at BH is gonna be different than B_GH

All right, I think I don't understand equivariant principle bundle theory well enough to say anything intelligent

Maybe it's helpful to point out that the total category of the left fibration classified by $B_G\Gamma$ is just the subcategory of $\mathrm{Orbit}(G\times \Gamma)^{\mathrm{op}}$ spanned by objects on which $\Gamma$ acts freely

so giving a $G$-space over $B_G\Gamma$ is the same as giving a functor out of this category to $\mathsf{Spaces}$. That ought to show how we've mixed the "genuine" equivariance of $G$ with the "Borel" equivariance of $\Gamma$

More generally, for families inside $G \times \Gamma$, you can quotient out by $\Gamma$ and see what family of orbits you get for $G$, and then interpret functors out of the subcategory of $\mathrm{Orbit}(G\times \Gamma)$ as speaking about `some flavor of $\Gamma$-equivariant bundles over some flavor of $G$-spaces' with varying amounts of genuineness tacked on

(turning everything down to free actions of both $\Gamma$ and $G$ just becomes the familiar statement $\mathsf{Fun}(B(G\times \Gamma), \mathsf{Spaces}) \simeq \mathsf{Fun}(B\Gamma, \mathsf{Spaces}_{/BG})$)

anyway, sorry for so much text...

@DylanWilson What is "the cocartesian fibration classified by $\underline{\mathrm{Spaces}}_G$"? Aren't cocartesian fibrations classified by a map from something to $\mathrm{Cat}_\infty$?

Anyway, $B_G\Gamma$ is certainly an object of $\mathrm{Spaces}_G$. But the universal space $E_G\Gamma$ lives in a larger world, which you don't quite name but is a full subcategory of $\mathrm{Spaces}_{G\times \Gamma}$. I want a formalism that explains how this larger world enters the game.

Ideally you would want to generalize this, so that instead of a group $\Gamma$ you have a groupoid. Or perhaps where $\Gamma$ is itself a group object internal to $\mathrm{Spaces}_G$, e.g., $U(n)$ as group object in $C_2$-spaces via complex conjugation.