$\operatorname{Spaces}/BG$ $\sim$ $\operatorname{Spaces}^G$ $\sim$ $??(\Omega G)$

This is a crosspost (with minor alterations).

For a topological group $$G$$, assigning to a $$G$$-space $$X$$ the (canonical) map $$EG\times_GX\to BG$$ establishes an equivalence between the homotopy category of $$G$$-spaces and the homotopy category of spaces over $$BG$$.

The loop space $$\Omega G$$ is in turn equivalent to some topological group $$A$$. Since $$G$$ is a group, $$A$$ can be made homotopy commutative (even braided, if I am not mistaken).

Is there some category with the notion of homotopy $$\mathscr X(A)$$ one may assign to a homotopy commutative topological group $$A$$ such that when $$A$$ is homotopy equivalent to the loop space $$\Omega G$$, then the homotopy category of $$\mathscr X(A)$$ would be equivalent to the homotopy category of $$G$$-spaces?

If needed, assume $$G$$ simply connected or even 2-connected: note that this certainly cannot work for non-connected $$G$$ since switching to $$\Omega G$$ loses all information about everything except the connected component of the unity of $$G$$. For connected non-simply connected $$G$$, I don't know.

First, view a $$G$$-space $$X$$ as a continuous homomorphism $$G\to\operatorname{Aut}(X)$$ to the appropriately topologized group of all self-homeomorphisms of $$X$$. This gives a map $$\Omega G\to\operatorname{aut}_{\operatorname{Aut}(X)}(\operatorname{id}_X)$$, the latter being the (homotopy commutative topological) group of self-homotopies of the identity map of $$X$$. However I do not see a way to recover the action of $$G$$ on $$X$$ from this map.

Second, more straightforward and seemingly more promising approach, but somehow I like it less. Note that any $$G$$-space $$X$$ gives rise to an action of $$\Omega G$$ on $$\Omega X$$. So, one could take for $$\mathscr X(A)$$ the category of $$A$$-groups, i. e. topological groups with a continuous action of $$A$$ by homomorphisms. What confuses me here is that seemingly homotopy commutativity of $$A$$ drops out. In principle such category exists for any topological group $$A$$, so why should this work?

• If $A$ is a topological group, you could still ask what is the category of topological groups with an action of $A$ is equivalent to. Perhaps you can show that it is equivalent to the category of group bundles over $BA$. But $B^2\!A$ does not exist, so there is no guess to make about that. I guess you are hoping that if $A$ has enough commutativity for $B^2\!A$ to exist, then you can show an equivalence to the category of space bundles over $B^2\!A$. Mar 22 at 8:37
• @GregoryArone Interesting suggestion. For any space $X$ one might ask about either $E\to X$ giving rise to an internal group in the homotopy category of $\operatorname{Spaces}/X$ (but then one must suitably interpret pullbacks over $X$ - presumably one must go for homotopy pullbacks), or just principal bundles over $X$, wrt arbitrary topological groups? Mar 22 at 9:05
• If A is braided (and not merely homotopy commutative), we can simply construct the delooping BA as an ∞-group and then take the category of spaces with an action of BA. Since BΩG≃G, this recovers the category of G-spaces. If A is merely homotopy commutative, I have an impression there is not enough data left to reconstruct G-spaces, since G can be reconstructed from the ∞-category of G-spaces, but G cannot be reconstructed from the ∞-group ΩG, unless we equip ΩG with a braiding. (Indeed, the ∞-group ΩG has exactly the data needed to reconstruct G as a space, not ∞-group.) Mar 22 at 16:02
• @DmitriPavlov Yes I agree. Let $A$ be braided. Is it possible to construct some $\mathscr X(A)$ out of $A$ itself, without passing to $BA$ first? Mar 22 at 17:02
• @მამუკაჯიბლაძე: Yes, for example you can take the category of spaces S equipped with a homomorphism of ∞-groups A→Ω(Aut(S)). Mar 22 at 17:05

If $$A$$ is a braided ∞-group, the delooping $$\def\B{{\sf B}}\B A$$ is an ∞-group.

Consider the ∞-category of spaces equipped with an action of the ∞-group $$\B A$$. Since $$\B Ω G≃G$$, this ∞-category is equivalent to the ∞-category of $$G$$-spaces.

The ∞-category of spaces with an action of $$\B A$$ can be formulated without the functor $$\B$$. Indeed, there is an adjunction of ∞-categories $$\B \dashv Ω$$ between braided ∞-groups and ∞-groups. (If we restrict to connected ∞-groups, this adjunction becomes an equivalence.)

An action of $$\B A$$ on a space $$S$$ is a homomorphism of ∞-groups $$\def\Aut{\mathop{\sf Aut}} \B A → \Aut(S)$$, or, equivalently, a homomorphism of braided ∞-groups $$A→Ω\Aut(S)$$.

Thus, the ∞-category of $$G$$-spaces can be recovered as the ∞-category of spaces $$S$$ equipped with a homomorphism of braided ∞-groups $$A→Ω\Aut(S)$$.

If $$A$$ is merely a homotopy commutative ∞-group, there is not enough data left to reconstruct the ∞-category of $$G$$-spaces. Indeed, otherwise we would be able to reconstruct the ∞-group $$G$$ from the ∞-category of $$G$$-spaces. But the ∞-group $$Ω G$$ does not have enough information to reconstruct the ∞-group $$G$$, unless we equip $$Ω G$$ with a braiding. Indeed, thanks to the equivalence between ∞-groups and pointed connected spaces, the ∞-group $$Ω G$$ has exactly the same information as the underlying pointed space of $$G$$.

• A side question - is not every homotopy commutative group equivalent to a braided one, unique up to homotopy? Mar 22 at 17:41
• @მამუკაჯიბლაძე: I think the answer is negative already for 1-categories: there are many nonisomorphic braidings on a monoidal category that induce a commutative monoid structure on isomorphism classes of objects. Mar 22 at 22:48