Let $X$ be a derived stack. There is a $S^1$-action on the derived loop space $\mathcal{L}(X) = \text{Maps}(S^1, X)$. In particular, $\mathcal{O}(\mathcal{L} X)$ should be quasi-isomorphic to a mixed complex.
I want to consider three specific examples, all over a field of characteristic zero. Let $G$ be an affine algebraic group. (1) A warm-up, where $X$ is a classical scheme (let's assume it is smooth and affine to make things easier), (2) $X = BG$, and in particular $X = B\mathbb{G}_a$ and (3) $X = Y/G$ where say, $G$ is a reductive and $Y$ is affine (to make things easier).
(Edit: I made many edits below, after thinking about this some more)
Example 1: A smooth scheme
I want to understand how one obtains the Connes $B$-operator (via HKR?) on functions on the derived loop space of a smooth (say, affine, to make things simpler) scheme $X$ from DAG first principles (rather than say, cyclic sets). To compute the loop space, we can write $S^1$ as the homotopy colimit of the simplicial set corresponding to the presentation of $S^1$ by two simplices (of dimension zero and one). I'll denote this simplicial set by $C$, so that we have $$\mathcal{L}(X) = \text{Map}(S^1, X) \simeq \text{colim}_C \text{Map}(\text{Spec}(k), X) \simeq \lim_{C^{op}} X$$ In particular, if $X$ is affine, one gets the usual Hochschild complex. Edit: In light of the comment below by Marc, it seems likely that some combination of his notes and Loday's book should tell how we obtain Connes' mixed complex.
Example 2: $BG$
We have that $\mathcal{L}(BG) = G/G$, and that $\mathcal{O}(G/G) = k[G]^G$, where the invariants are derived. In particular, $\mathcal{O}(G/G)$ is the rational cohomology of the $G$-representation $k[G]$. If $G$ is reductive, then it has no higher cohomology, so the $S^1$-action on this vector space is necessarily trivial.
Another example: if $B$ is a Borel subgroup of a reductive algebraic group, then $B/B \simeq \tilde{G}/G$. It's known that the higher cohomology of $\tilde{G}$ vanishes, and since $G$ is reductive taking invariants is exact, so again the $S^1$-action is trivial.
In search of a less trivial example, let us take $G = \mathbb{G}_a$; then there is some nontrivial group cohomology: $\mathcal{O}(\mathbb{G}_a/\mathbb{G}_a) \simeq C^\bullet(\mathbb{G}_a k[\mathbb{G}_a]) \simeq k[x, \eta]$, the free dg-commutative ring generated by $|x| = 0$ and $|\eta| = 1$. Here, there is a presentation of $\mathcal{O}(G/G)$ as a cocyclic set (as discussed in Jantzen's book) and the "dual" Connes degree $-1$ operator $B^*$ (I don't know if this is written down anywhere) sends $\eta \mapsto x$.
However, I don't understand why this cocyclic set should have anything to do with the circle action on loops. In particular, it seems somewhat orthogonal to the first example: there, the cyclic structure came from taking a left-derived functor, and here the cocyclic arises from taking a right-derived functor. Question: How does this (co)cyclic structure arise from first DAG principles?
Example 3: $Y/G$
The loop space of $Y/G$ is again a derived scheme stacky-modulo a group action of $G$. We can identify this scheme by the derived fiber product $Y/G \times_{BG} \text{Spec}(k) = Y \times_{Y \times Y} (G \times Y)$. If $Y$ is affine, we can take a bar resolution of $Y$ over $Y \times Y$.
However, the resulting complex after tensoring with the right factor is not a cyclic set in any obvious way to me. For example, taking $Y = \mathbb{A}^1$ and $G = \mathbb{G}_m$, the first differential in the resulting complex is a map $k[x] \otimes k[x] \otimes k[z^{\pm 1}] \rightarrow k[x] \otimes k[z^{\pm 1}]$, sending (assume $f, g$ homogeneous) $$f \otimes g \otimes 1 \mapsto fg \otimes (1 - z^{|g|})$$ which does not intertwine with any rotation homomorphism I can think to define (even "twisted" ones). Question: what is the $S^1$ action on this (sort of) derived loop space?
