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It is a folklore result that if $G$ is a discrete group and $BG$ its classifying space, then the free loop space $L(BG)$ is homotopy equivalent to $EG\times_{Ad} G$ where $EG$ is the universal $G$-space and $Ad$ is the conjugation action of G on itself.

A proof is given e.g. in (Klein, John R., Claude L. Schochet, and Samuel B. Smith. "Continuous trace C*-algebras, gauge groups and rationalization." Journal of Topology and Analysis 1.03 (2009): 261-288), see https://doi.org/10.1142/S179352530900014X, resp. https://arxiv.org/abs/0811.0771

  1. Is there a natural $\mathbb{S}^1$-action on $EG\times_{Ad} G$?
  2. If so, is there an equivariant homotopy equivalence $L(BG)\simeq EG\times_{Ad} G$?
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    $\begingroup$ If you look at the proof of Lemma 2.12.1 in my Representations and Cohomology II, you can see that the map $\rho\colon Y/G \to EG\times_G G$ there is a homotopy equivalence and $\mathbb{S}^1$ acts freely on $Y/G$ but not directly on $EG\times_G G$. $\endgroup$ Commented May 9 at 12:32
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    $\begingroup$ The homotopy type of the free loop space of $BG$ can be identified with the cyclic bar construction of $G$, which is a special case of [Nikolaus–Scholze: On Topological Cyclic Homology, Prop IV.3.2], but I do not believe that the formula you wrote has a good way to exhibit the circle action. It is similar to the Hochschild homology case: when we write $\operatorname{HH}(R)$ as $R\otimes_{R^{\mathrm{op}}\otimes^LR}^LR$, the circle action becomes unclear. $\endgroup$ Commented May 9 at 12:39

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The space $EG \times_{Ad} G$ is the geometric realization of the nerve of a category, the action category of $G$ acting on itself via conjugation. Denote this $G //G$. That is the objects are elements of $G$ and a morphism from $g_1$ to $g_2$ is an element $h \in G$ such that $h^{-1}g_1 h = g_2$.

This doesn't have an obvious action by the circle, but it does have an action by the category $pt // \mathbb{Z}$ . This is a category which has one object whose automorphisms is the group $\mathbb{Z}$. Since $\mathbb{Z}$ is abelian this is a symmetric monoidal category and it acts on $G//G$ in a natural way, which I will now describe explicitly.

An action of $pt // \mathbb{Z}$ on a category $\mathcal{C}$ ammounts to a natural isomorphism of the identity functor. The components of such a natural isomorphism are isomorphisms $\eta_c: c \to c$ for all $c \in \mathcal{C}$.

In the case $\mathcal{C} = G//G$, this amounts to a choice for each $g \in G$ of an $h_g \in G$, such that $ h_g^{-1} g h_g = g$. The desired action is given by taking $h_g = g$.

This induces an action on the geometric realizations. The geometric realization of $pt/\mathbb{Z}$ is not the circle but it is a $K(\mathbb{Z},1)$.

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    $\begingroup$ Furthermore, if one composes the map $BZ \times EG \times_G G_[conj} \rightarrow EG \times_GG_{conj}$ described here with the projection to $BG$, and adjoints, one gets a map that can be used to prove the `folk theorem'. $\endgroup$ Commented May 9 at 14:49

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