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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 $BG$.

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?

I tried this path: 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.

Another possibility could be to consider the category of $A$-groups, given that any $G$-space $X$ gives rise to an action of $\Omega G$ on $\Omega X$...

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.

მამუკა ჯიბლაძე
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  • If you say that $A$ is a homotopy commutative topological group, do you mean that all the group axioms hold strictly for $A$, and it is on top of that commutative up to homotopy? And do you assume that the homotopy equivalence $A\simeq \Omega G$ preserves group structures in a homotopical way? – Daniël Apol Mar 11 '24 at 22:01
  • @DaniëlApol Yes. In fact, if I am not mistaken, every $A$ of the form $\Omega G$ is also braided, i. e. satisfies a coherence condition on triples of elements: one can choose a homotopy $\tau$ of maps $A\times A\to A$ between $(x,y)\mapsto xy$ and $(x,y)\mapsto yx$ in such a way that $\tau(x,y_1y_2)$ is the composite of $\tau(x,y_1)y_2$ with $y_1\tau(x,y_2)$, and $\tau(x_1x_2,y)$ is the composite of $x_1\tau(x_2,y)$ with $\tau(x_1,y)x_2$. But I am not sure, and also I am not sure if this is relevant here, so I abstained from mentioning this – მამუკა ჯიბლაძე Mar 12 '24 at 07:20
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    The corresponding post on MO: https://mathoverflow.net/q/467492 – Martin Sleziak Mar 22 '24 at 07:32
  • It is recommended to avoid titles consisting only of math formulas, see Title and $\LaTeX$ on meta. – Martin Sleziak Mar 22 '24 at 07:32
  • @MartinSleziak Changed. Hope there is enough non-formula stuff in the new title... – მამუკა ჯიბლაძე Mar 22 '24 at 08:29

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