Recover the $C_k$-action of a cyclic object as from the $S^1$-action on Hochschild chain

Question

$\newcommand{\Fun}{\operatorname{Fun}}$Let $X\in\Fun(\Lambda^\mathrm{op}, \mathcal{C})$ be a cyclic object in the ($\infty$-)symmetric monoidal category $\mathcal{C}$, where $\Lambda$ is the cyclic category of Connes.

Then, there exists a functor $\Fun(\Lambda^\mathrm{op}, \mathcal{C}) \rightarrow \Fun(BS^1, \mathcal{C})$, whose underline object is given by $colim_{\Delta^\mathrm{op}} X$.

On one hand, there is a natural functor $\Fun(BS^1, \mathcal{C})\rightarrow \Fun(BC_k, \mathcal{C})$ given by restriction to cyclic subgroup $C_k\subset S^1$.

On the other hand, I might construction a functor by $$\Fun(\Lambda^\mathrm{op}, \mathcal{C})\xrightarrow{sd_k} \Fun(\Lambda_k^\mathrm{op}, \mathcal{C})\rightarrow \Fun^{BC_k}(\Lambda_k^\mathrm{op}, \mathcal{C}^{BC_k}) \rightarrow \mathcal{C}^{BC_k}=\Fun(BC_k, \mathcal{C}),$$ where the first functor is edge subdivision, and the last is the evaluation at $0$ (or truncation whatever).

My first question I am not very sure is: Is the section functor well-defined?

Next, we assume further that all arrows of the diagram of $X$ are isomorphisms in $\mathcal{C}$, i.e. $X\in \Fun(\Lambda^\mathrm{op}, \mathcal{C})_\simeq \simeq \Fun(BS^1, \mathcal{C})$. Then are these two functor naturally equivalent?

My motivation is: if we take an algebraic object in $C$, such that $m: A\otimes A \rightarrow A$ is an isomorphism, for example $A$ is a field or something more advance. We take $X$ as the cyclic bar complex, then the result of the first functor is $HH(A)$; and the result of the second functor gives us the cyclic permutation on $A^{\otimes k}$.

In general, I should not expect they are isomorphic cyclic equivariantly, but we assume $m: A\otimes A \rightarrow A$ is an isomorphism, we should have $A^{\otimes k}\simeq A$ for all $k$. Then we can expect the cyclic permutation on $A^{\otimes k}$ comes from the restriction on $S^1$-action of $HH(A)$. I think this is true if we really working on chain complexes. But I am not sure how to write a proof for general categories. Is this possible true, or any possible references?

Or, maybe go back to the question in title, how can we obtain the cyclic permutation on $A^{\otimes k}$ from the $S^1$-action of $HH(A)$.

Answer 1

A recent paper by Zihong Chen https://arxiv.org/abs/2402.06183 answers the question in the title.

To be precise, let us consider another category $_{k}\Lambda$ which is defined as $(\Delta^{op})^k$ jointed with the $C_k$-action (see p14 of the paper.) Notice that in the paper, $_{k}\Lambda$ is different from the $\Lambda_k$ I mentioned above for edge subdivision.

The author shows in Lemma 5.1 that there exists a functor $_{k}\Lambda \rightarrow \Lambda$, whose groupoid completion is equivalent to the functor $BC_k\rightarrow BS^1$ (induced by the natural inclusion of groups).

Therefore, the pullback of $X\in\text{Fun}(\Lambda^\mathrm{op}, \mathcal{C})$ alone $_{k}\Lambda \rightarrow \Lambda$ gives the functor that restricts to the $C_k$-action for a cyclic object. Moreover, if we take the cyclic object to be the Hochschild chain, the author also explains that the resulting object is the $C_k$-equivariant Hochschild chain. In particular, if we have $A^{\otimes k}\simeq A $, then we can recover the cyclic permutation on $A^{\otimes k} $.

So far, the paper answers what I actually concerned about. But maybe still a little more needs to be said if the functor I explained above is the same functor as the pullback alone $_{k}\Lambda \rightarrow \Lambda$? I think the answer might be no since there are different topological natures. But I feel like there are still some relations here.