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.

I thought about two possibilities.

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?

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    $\begingroup$ 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 $B^2\!A$. $\endgroup$ Mar 22 at 8:37
  • $\begingroup$ @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? $\endgroup$ Mar 22 at 9:05
  • $\begingroup$ 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.) $\endgroup$ Mar 22 at 16:02
  • $\begingroup$ @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? $\endgroup$ Mar 22 at 17:02
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    $\begingroup$ @მამუკაჯიბლაძე: Yes, for example you can take the category of spaces S equipped with a homomorphism of ∞-groups A→Ω(Aut(S)). $\endgroup$ Mar 22 at 17:05

1 Answer 1


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 $G$.

  • $\begingroup$ A side question - is not every homotopy commutative group equivalent to a braided one, unique up to homotopy? $\endgroup$ Mar 22 at 17:41
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    $\begingroup$ @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. $\endgroup$ Mar 22 at 22:48

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