Example 1: A smooth scheme

To try to understand this, I want to also 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 pushout of two copies of $D^1$ glued along $S^0$ $$S^1 \simeq * \coprod_{* \coprod *} *$$ and then using some general principle that pushouts in the source of a mapping stack become products, find that $$\mathcal{L}(X) = \text{Map}(S^1, X) \simeq X \times_{X \times X} X$$ Taking the bar resolution then gives the usual Hochschild complex. One drawback to this approach seems to be that the $S^1$ action seems somewhat obscured by this asymmetric presentation. One might hope for a more direct way to relate, for example, the cyclic structure on the Hochschild complex with the derived $S^1$-action above. Question: how does one understand the cyclic set structure Hochschild chain complex from DAG first principles?

If $G = \mathbb{G}_a$, then there is some nontrivial group cohomology: $\mathcal{O}(\mathbb{G}_a/\mathbb{G}_a) \simeq k[x, \eta]$, the free dg-commutative ring generated by $|x| = 0$ and $|\eta| = 1$. Here there is room for a $S^1$-action and my guess would be that it sends $\eta \mapsto x$, using HKR since $B\mathbb{G}_a$ is an affine stack (though it's not clear to me that "HKR" makes sense in this context). Question: How would one see this purely from first principles, or otherwise?

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 $Y \times Y$, but 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?

(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.

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?

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?